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Theorem fourierdlem115 39114
Description: Fourier serier convergence, for piecewise smooth functions. (Contributed by Glauco Siliprandi, 11-Dec-2019.)
Hypotheses
Ref Expression
fourierdlem115.f (𝜑𝐹:ℝ⟶ℝ)
fourierdlem115.t 𝑇 = (2 · π)
fourierdlem115.per ((𝜑𝑥 ∈ ℝ) → (𝐹‘(𝑥 + 𝑇)) = (𝐹𝑥))
fourierdlem115.g 𝐺 = ((ℝ D 𝐹) ↾ (-π(,)π))
fourierdlem115.dmdv (𝜑 → ((-π(,)π) ∖ dom 𝐺) ∈ Fin)
fourierdlem115.dvcn (𝜑𝐺 ∈ (dom 𝐺cn→ℂ))
fourierdlem115.rlim ((𝜑𝑥 ∈ ((-π[,)π) ∖ dom 𝐺)) → ((𝐺 ↾ (𝑥(,)+∞)) lim 𝑥) ≠ ∅)
fourierdlem115.llim ((𝜑𝑥 ∈ ((-π(,]π) ∖ dom 𝐺)) → ((𝐺 ↾ (-∞(,)𝑥)) lim 𝑥) ≠ ∅)
fourierdlem115.x (𝜑𝑋 ∈ ℝ)
fourierdlem115.l (𝜑𝐿 ∈ ((𝐹 ↾ (-∞(,)𝑋)) lim 𝑋))
fourierdlem115.r (𝜑𝑅 ∈ ((𝐹 ↾ (𝑋(,)+∞)) lim 𝑋))
fourierdlem115.a 𝐴 = (𝑛 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 / π))
fourierdlem115.b 𝐵 = (𝑛 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 / π))
fourierdlem115.s 𝑆 = (𝑘 ∈ ℕ ↦ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋)))))
Assertion
Ref Expression
fourierdlem115 (𝜑 → (seq1( + , 𝑆) ⇝ (((𝐿 + 𝑅) / 2) − ((𝐴‘0) / 2)) ∧ (((𝐴‘0) / 2) + Σ𝑛 ∈ ℕ (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))))) = ((𝐿 + 𝑅) / 2)))
Distinct variable groups: 𝐴,𝑘 𝐵,𝑘 𝑘,𝐹,𝑛,𝑥 𝑘,𝐺,𝑥 𝑘,𝐿 𝑅,𝑘 𝑇,𝑘,𝑥 𝑘,𝑋,𝑛,𝑥 𝜑,𝑘,𝑥
Allowed substitution hints: 𝜑(𝑛) 𝐴(𝑥,𝑛) 𝐵(𝑥,𝑛) 𝑅(𝑥,𝑛) 𝑆(𝑥,𝑘,𝑛) 𝑇(𝑛) 𝐺(𝑛) 𝐿(𝑥,𝑛)
Proof of Theorem fourierdlem115
Dummy variables 𝑧 𝑓 𝑔 𝑤 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fourierdlem115.f . . . 4 (𝜑𝐹:ℝ⟶ℝ)
2 fourierdlem115.t . . . 4 𝑇 = (2 · π)
3 fourierdlem115.per . . . 4 ((𝜑𝑥 ∈ ℝ) → (𝐹‘(𝑥 + 𝑇)) = (𝐹𝑥))
4 fourierdlem115.g . . . 4 𝐺 = ((ℝ D 𝐹) ↾ (-π(,)π))
5 fourierdlem115.dmdv . . . 4 (𝜑 → ((-π(,)π) ∖ dom 𝐺) ∈ Fin)
6 fourierdlem115.dvcn . . . 4 (𝜑𝐺 ∈ (dom 𝐺cn→ℂ))
7 fourierdlem115.rlim . . . 4 ((𝜑𝑥 ∈ ((-π[,)π) ∖ dom 𝐺)) → ((𝐺 ↾ (𝑥(,)+∞)) lim 𝑥) ≠ ∅)
8 fourierdlem115.llim . . . 4 ((𝜑𝑥 ∈ ((-π(,]π) ∖ dom 𝐺)) → ((𝐺 ↾ (-∞(,)𝑥)) lim 𝑥) ≠ ∅)
9 fourierdlem115.x . . . 4 (𝜑𝑋 ∈ ℝ)
10 fourierdlem115.l . . . 4 (𝜑𝐿 ∈ ((𝐹 ↾ (-∞(,)𝑋)) lim 𝑋))
11 fourierdlem115.r . . . 4 (𝜑𝑅 ∈ ((𝐹 ↾ (𝑋(,)+∞)) lim 𝑋))
12 fourierdlem115.a . . . . 5 𝐴 = (𝑛 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 / π))
13 oveq1 6556 . . . . . . . . . . 11 (𝑛 = 𝑘 → (𝑛 · 𝑥) = (𝑘 · 𝑥))
1413fveq2d 6107 . . . . . . . . . 10 (𝑛 = 𝑘 → (cos‘(𝑛 · 𝑥)) = (cos‘(𝑘 · 𝑥)))
1514oveq2d 6565 . . . . . . . . 9 (𝑛 = 𝑘 → ((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) = ((𝐹𝑥) · (cos‘(𝑘 · 𝑥))))
1615adantr 480 . . . . . . . 8 ((𝑛 = 𝑘𝑥 ∈ (-π(,)π)) → ((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) = ((𝐹𝑥) · (cos‘(𝑘 · 𝑥))))
1716itgeq2dv 23354 . . . . . . 7 (𝑛 = 𝑘 → ∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 = ∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑘 · 𝑥))) d𝑥)
1817oveq1d 6564 . . . . . 6 (𝑛 = 𝑘 → (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 / π) = (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑘 · 𝑥))) d𝑥 / π))
1918cbvmptv 4678 . . . . 5 (𝑛 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 / π)) = (𝑘 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑘 · 𝑥))) d𝑥 / π))
2012, 19eqtri 2632 . . . 4 𝐴 = (𝑘 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑘 · 𝑥))) d𝑥 / π))
21 fourierdlem115.b . . . . 5 𝐵 = (𝑛 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 / π))
2213fveq2d 6107 . . . . . . . . . 10 (𝑛 = 𝑘 → (sin‘(𝑛 · 𝑥)) = (sin‘(𝑘 · 𝑥)))
2322oveq2d 6565 . . . . . . . . 9 (𝑛 = 𝑘 → ((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) = ((𝐹𝑥) · (sin‘(𝑘 · 𝑥))))
2423adantr 480 . . . . . . . 8 ((𝑛 = 𝑘𝑥 ∈ (-π(,)π)) → ((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) = ((𝐹𝑥) · (sin‘(𝑘 · 𝑥))))
2524itgeq2dv 23354 . . . . . . 7 (𝑛 = 𝑘 → ∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 = ∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑘 · 𝑥))) d𝑥)
2625oveq1d 6564 . . . . . 6 (𝑛 = 𝑘 → (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 / π) = (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑘 · 𝑥))) d𝑥 / π))
2726cbvmptv 4678 . . . . 5 (𝑛 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 / π)) = (𝑘 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑘 · 𝑥))) d𝑥 / π))
2821, 27eqtri 2632 . . . 4 𝐵 = (𝑘 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑘 · 𝑥))) d𝑥 / π))
29 fourierdlem115.s . . . 4 𝑆 = (𝑘 ∈ ℕ ↦ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋)))))
30 eqid 2610 . . . 4 (𝑘 ∈ ℕ ↦ {𝑤 ∈ (ℝ ↑𝑚 (0...𝑘)) ∣ (((𝑤‘0) = -π ∧ (𝑤𝑘) = π) ∧ ∀𝑧 ∈ (0..^𝑘)(𝑤𝑧) < (𝑤‘(𝑧 + 1)))}) = (𝑘 ∈ ℕ ↦ {𝑤 ∈ (ℝ ↑𝑚 (0...𝑘)) ∣ (((𝑤‘0) = -π ∧ (𝑤𝑘) = π) ∧ ∀𝑧 ∈ (0..^𝑘)(𝑤𝑧) < (𝑤‘(𝑧 + 1)))})
31 id 22 . . . . . 6 (𝑦 = 𝑥𝑦 = 𝑥)
32 oveq2 6557 . . . . . . . . 9 (𝑦 = 𝑥 → (π − 𝑦) = (π − 𝑥))
3332oveq1d 6564 . . . . . . . 8 (𝑦 = 𝑥 → ((π − 𝑦) / 𝑇) = ((π − 𝑥) / 𝑇))
3433fveq2d 6107 . . . . . . 7 (𝑦 = 𝑥 → (⌊‘((π − 𝑦) / 𝑇)) = (⌊‘((π − 𝑥) / 𝑇)))
3534oveq1d 6564 . . . . . 6 (𝑦 = 𝑥 → ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇) = ((⌊‘((π − 𝑥) / 𝑇)) · 𝑇))
3631, 35oveq12d 6567 . . . . 5 (𝑦 = 𝑥 → (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)) = (𝑥 + ((⌊‘((π − 𝑥) / 𝑇)) · 𝑇)))
3736cbvmptv 4678 . . . 4 (𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇))) = (𝑥 ∈ ℝ ↦ (𝑥 + ((⌊‘((π − 𝑥) / 𝑇)) · 𝑇)))
38 eqid 2610 . . . 4 ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺)) = ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))
39 eqid 2610 . . . 4 ((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1) = ((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1)
40 isoeq1 6467 . . . . 5 (𝑔 = 𝑓 → (𝑔 Isom < , < ((0...((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1)), ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) ↔ 𝑓 Isom < , < ((0...((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1)), ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺)))))
4140cbviotav 5774 . . . 4 (℩𝑔𝑔 Isom < , < ((0...((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1)), ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺)))) = (℩𝑓𝑓 Isom < , < ((0...((#‘({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))) − 1)), ({-π, π, ((𝑦 ∈ ℝ ↦ (𝑦 + ((⌊‘((π − 𝑦) / 𝑇)) · 𝑇)))‘𝑋)} ∪ ((-π[,]π) ∖ dom 𝐺))))
421, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 28, 29, 30, 37, 38, 39, 41fourierdlem114 39113 . . 3 (𝜑 → (seq1( + , 𝑆) ⇝ (((𝐿 + 𝑅) / 2) − ((𝐴‘0) / 2)) ∧ (((𝐴‘0) / 2) + Σ𝑘 ∈ ℕ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋))))) = ((𝐿 + 𝑅) / 2)))
4342simpld 474 . 2 (𝜑 → seq1( + , 𝑆) ⇝ (((𝐿 + 𝑅) / 2) − ((𝐴‘0) / 2)))
44 nfcv 2751 . . . . 5 𝑘(((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))))
45 nfmpt1 4675 . . . . . . . . 9 𝑛(𝑛 ∈ ℕ0 ↦ (∫(-π(,)π)((𝐹𝑥) · (cos‘(𝑛 · 𝑥))) d𝑥 / π))
4612, 45nfcxfr 2749 . . . . . . . 8 𝑛𝐴
47 nfcv 2751 . . . . . . . 8 𝑛𝑘
4846, 47nffv 6110 . . . . . . 7 𝑛(𝐴𝑘)
49 nfcv 2751 . . . . . . 7 𝑛 ·
50 nfcv 2751 . . . . . . 7 𝑛(cos‘(𝑘 · 𝑋))
5148, 49, 50nfov 6575 . . . . . 6 𝑛((𝐴𝑘) · (cos‘(𝑘 · 𝑋)))
52 nfcv 2751 . . . . . 6 𝑛 +
53 nfmpt1 4675 . . . . . . . . 9 𝑛(𝑛 ∈ ℕ ↦ (∫(-π(,)π)((𝐹𝑥) · (sin‘(𝑛 · 𝑥))) d𝑥 / π))
5421, 53nfcxfr 2749 . . . . . . . 8 𝑛𝐵
5554, 47nffv 6110 . . . . . . 7 𝑛(𝐵𝑘)
56 nfcv 2751 . . . . . . 7 𝑛(sin‘(𝑘 · 𝑋))
5755, 49, 56nfov 6575 . . . . . 6 𝑛((𝐵𝑘) · (sin‘(𝑘 · 𝑋)))
5851, 52, 57nfov 6575 . . . . 5 𝑛(((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋))))
59 fveq2 6103 . . . . . . 7 (𝑛 = 𝑘 → (𝐴𝑛) = (𝐴𝑘))
60 oveq1 6556 . . . . . . . 8 (𝑛 = 𝑘 → (𝑛 · 𝑋) = (𝑘 · 𝑋))
6160fveq2d 6107 . . . . . . 7 (𝑛 = 𝑘 → (cos‘(𝑛 · 𝑋)) = (cos‘(𝑘 · 𝑋)))
6259, 61oveq12d 6567 . . . . . 6 (𝑛 = 𝑘 → ((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) = ((𝐴𝑘) · (cos‘(𝑘 · 𝑋))))
63 fveq2 6103 . . . . . . 7 (𝑛 = 𝑘 → (𝐵𝑛) = (𝐵𝑘))
6460fveq2d 6107 . . . . . . 7 (𝑛 = 𝑘 → (sin‘(𝑛 · 𝑋)) = (sin‘(𝑘 · 𝑋)))
6563, 64oveq12d 6567 . . . . . 6 (𝑛 = 𝑘 → ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))) = ((𝐵𝑘) · (sin‘(𝑘 · 𝑋))))
6662, 65oveq12d 6567 . . . . 5 (𝑛 = 𝑘 → (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋)))) = (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋)))))
6744, 58, 66cbvsumi 14275 . . . 4 Σ𝑛 ∈ ℕ (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋)))) = Σ𝑘 ∈ ℕ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋))))
6867oveq2i 6560 . . 3 (((𝐴‘0) / 2) + Σ𝑛 ∈ ℕ (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))))) = (((𝐴‘0) / 2) + Σ𝑘 ∈ ℕ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋)))))
6942simprd 478 . . 3 (𝜑 → (((𝐴‘0) / 2) + Σ𝑘 ∈ ℕ (((𝐴𝑘) · (cos‘(𝑘 · 𝑋))) + ((𝐵𝑘) · (sin‘(𝑘 · 𝑋))))) = ((𝐿 + 𝑅) / 2))
7068, 69syl5eq 2656 . 2 (𝜑 → (((𝐴‘0) / 2) + Σ𝑛 ∈ ℕ (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))))) = ((𝐿 + 𝑅) / 2))
7143, 70jca 553 1 (𝜑 → (seq1( + , 𝑆) ⇝ (((𝐿 + 𝑅) / 2) − ((𝐴‘0) / 2)) ∧ (((𝐴‘0) / 2) + Σ𝑛 ∈ ℕ (((𝐴𝑛) · (cos‘(𝑛 · 𝑋))) + ((𝐵𝑛) · (sin‘(𝑛 · 𝑋))))) = ((𝐿 + 𝑅) / 2)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ≠ wne 2780 ∀wral 2896 {crab 2900 ∖ cdif 3537 ∪ cun 3538 ∅c0 3874 {ctp 4129 class class class wbr 4583 ↦ cmpt 4643 dom cdm 5038 ↾ cres 5040 ℩cio 5766 ⟶wf 5800 ‘cfv 5804 Isom wiso 5805 (class class class)co 6549 ↑𝑚 cmap 7744 Fincfn 7841 ℂcc 9813 ℝcr 9814 0cc0 9815 1c1 9816 + caddc 9818 · cmul 9820 +∞cpnf 9950 -∞cmnf 9951 < clt 9953 − cmin 10145 -cneg 10146 / cdiv 10563 ℕcn 10897 2c2 10947 ℕ0cn0 11169 (,)cioo 12046 (,]cioc 12047 [,)cico 12048 [,]cicc 12049 ...cfz 12197 ..^cfzo 12334 ⌊cfl 12453 seqcseq 12663 #chash 12979 ⇝ cli 14063 Σcsu 14264 sincsin 14633 cosccos 14634 πcpi 14636 –cn→ccncf 22487 ∫citg 23193 limℂ climc 23432 D cdv 23433 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-inf2 8421 ax-cc 9140 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 ax-addf 9894 ax-mulf 9895 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-fal 1481 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-iin 4458 df-disj 4554 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-se 4998 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-isom 5813 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-ofr 6796 df-om 6958 df-1st 7059 df-2nd 7060 df-supp 7183 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-2o 7448 df-oadd 7451 df-omul 7452 df-er 7629 df-map 7746 df-pm 7747 df-ixp 7795 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-fsupp 8159 df-fi 8200 df-sup 8231 df-inf 8232 df-oi 8298 df-card 8648 df-acn 8651 df-cda 8873 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-4 10958 df-5 10959 df-6 10960 df-7 10961 df-8 10962 df-9 10963 df-n0 11170 df-xnn0 11241 df-z 11255 df-dec 11370 df-uz 11564 df-q 11665 df-rp 11709 df-xneg 11822 df-xadd 11823 df-xmul 11824 df-ioo 12050 df-ioc 12051 df-ico 12052 df-icc 12053 df-fz 12198 df-fzo 12335 df-fl 12455 df-mod 12531 df-seq 12664 df-exp 12723 df-fac 12923 df-bc 12952 df-hash 12980 df-shft 13655 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-limsup 14050 df-clim 14067 df-rlim 14068 df-sum 14265 df-ef 14637 df-sin 14639 df-cos 14640 df-pi 14642 df-struct 15697 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-mulr 15782 df-starv 15783 df-sca 15784 df-vsca 15785 df-ip 15786 df-tset 15787 df-ple 15788 df-ds 15791 df-unif 15792 df-hom 15793 df-cco 15794 df-rest 15906 df-topn 15907 df-0g 15925 df-gsum 15926 df-topgen 15927 df-pt 15928 df-prds 15931 df-xrs 15985 df-qtop 15990 df-imas 15991 df-xps 15993 df-mre 16069 df-mrc 16070 df-acs 16072 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-submnd 17159 df-mulg 17364 df-cntz 17573 df-cmn 18018 df-psmet 19559 df-xmet 19560 df-met 19561 df-bl 19562 df-mopn 19563 df-fbas 19564 df-fg 19565 df-cnfld 19568 df-top 20521 df-bases 20522 df-topon 20523 df-topsp 20524 df-cld 20633 df-ntr 20634 df-cls 20635 df-nei 20712 df-lp 20750 df-perf 20751 df-cn 20841 df-cnp 20842 df-t1 20928 df-haus 20929 df-cmp 21000 df-tx 21175 df-hmeo 21368 df-fil 21460 df-fm 21552 df-flim 21553 df-flf 21554 df-xms 21935 df-ms 21936 df-tms 21937 df-cncf 22489 df-ovol 23040 df-vol 23041 df-mbf 23194 df-itg1 23195 df-itg2 23196 df-ibl 23197 df-itg 23198 df-0p 23243 df-ditg 23417 df-limc 23436 df-dv 23437 This theorem is referenced by: fourierd 39115 fourierclimd 39116
Copyright terms: Public domain W3C validator | 9,037 | 13,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.137255 |
https://stacks.math.columbia.edu/tag/05H8 | 1,718,528,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00144.warc.gz | 489,415,681 | 6,190 | Lemma 38.4.5. Let $S$, $X$, $\mathcal{F}$, $x$, $s$ be as in Definition 38.4.2. Let $(Z, Y, i, \pi , \mathcal{G}, z, y)$ be a one step dévissage of $\mathcal{F}/X/S$ at $x$. Let $(S', s') \to (S, s)$ be a morphism of pointed schemes which induces an isomorphism $\kappa (s) = \kappa (s')$. Let $(Z', Y', i', \pi ', \mathcal{G}')$ be as constructed in Lemma 38.4.4 and let $x' \in X'$ (resp. $z' \in Z'$, $y' \in Y'$) be the unique point mapping to both $x \in X$ (resp. $z \in Z$, $y \in Y$) and $s' \in S'$. If $S'$ is affine, then $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ at $x'$.
Proof. By Lemma 38.4.4 $(Z', Y', i', \pi ', \mathcal{G}')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ over $s'$. Properties (1) – (4) of Definition 38.4.2 hold for $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ as the assumption that $\kappa (s) = \kappa (s')$ insures that the fibres $X'_{s'}$, $Z'_{s'}$, and $Y'_{s'}$ are isomorphic to $X_ s$, $Z_ s$, and $Y_ s$. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 476 | 1,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.439099 |
https://www.physicsforums.com/threads/show-that-the-kronecker-delta-retains-its-form-under-any-transformation.976139/ | 1,726,267,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00702.warc.gz | 867,496,015 | 28,124 | # Show that the Kronecker delta retains its form under any transformation
• blue24
In summary: I dunno what you're trying to do.In summary, the textbook says that the Kronecker delta has the property that "it can easily be verified (see Exercise 1.8) that the Kronecker delta ##d_{ij}## has such a property."
blue24
Homework Statement
Show that 2nd order tensor, a*d_ij, where a is an arbitrary constant, retains its form under any transformation, Q_ij. This form is then an isotropic 2nd order tensor.
Relevant Equations
a'_ij = Q_ip * Q_jq * a_pq (General transformation relation for 2nd order tensor)
Backstory - I have not been in school for 5ish years, and am returning to take some grad classes in the field of Solid Mechanics. I am freaking out a bit about the math (am rusty). I have not started class yet, but figured I would get my books and start working through problems. This problem is from my Theory of Elasticity Class.
The book provides the general transformation relation for a 2nd order tensor. Applying a rotation to the given tensor, a*d_ij:
a*d'_ij = Q_ip * Q_jq * a*d_pq
I am not sure where to go from here. My understanding is that the repeated indices "p" and "q" imply summation, so then the first term in the matrix a*d'_ij would be:
a*d'_11 = Q_1p*Q_1q*a*d_pq = a*[Q_11*Q_11*d_11 + Q_12*Q_12*d_22 + Q_13*Q_13*d_33]
Since d_11 = d_22 = d_33 = 1, then a*d'_11 = a*[Q_11*Q_11 + Q_12*Q_12 + Q_13*Q_13]
But my guess is I've gone wrong somewhere, because I don't know what I do with this. My apologies if this is a dumb question. Still trying to get my head wrapped around this. Thank you.
What does your book give as the definition of ##Q_{ij}## ?
It is best to absorb notions of raised indices (contravariant) and lowered indices (covariant) first, before answering questions about kronecker-delta (and levi-civta).
strangerep said:
What does your book give as the definition of ##Q_{ij}## ?
Thank you for the response. The book defines ##Q_{ij}## as the angle between the ##x^{'}_i## and ##x_j## axes.
Michael Price said:
It is best to absorb notions of raised indices (contravariant) and lowered indices (covariant) first, before answering questions about kronecker-delta (and levi-civta).
Thank you for the reply. Unfortunately those terms, "covariant" and "contravariant", are not in my textbook. I have a pdf (legally purchased) and I searched for them. I will do some searching online.
What is throwing me is that the textbook literally says, "[referring to isotropic tensors] It can easily be verified (see Exercise 1.8) that the Kronecker delta ##d_{ij}## has such a property." So I figured, hey I better do Exercise 1.8, it'll be easy. And I'm stuck.
Can anyone give me any hints as to where I am going wrong, what steps I should follow, etc? I want to repeat - I am not even in the class yet, first day of class is September 6. Most likely this will never actually be an assigned homework problem. I am just trying to work ahead because I know I will need to do so in order to get help. So I am not posting on here trying to cut corners and get you guys to do my homework for me or anything like that.
Thanks for any help you can provide!
Andrew
blue24 said:
Thank you for the response. The book defines ##Q_{ij}## as the angle between the ##x^{'}_i## and ##x_j## axes.
Well, either that's not exactly what the book says, or it's being extremely sloppy.
You should probably post a link so I can look at what the book actually says.
I suspect what's meant is that ##Q_{ij}## denotes the transformation coefficients corresponding to a certain 3D rotation. I.e., probably something like
$$Q_{ij} ~=~ \frac{\partial x'_i}{\partial x_j} ~~.$$ You also need to know that for any matrix with components ##A_{ij}##, we have $$A_{ij} \delta_{jk} ~=~ A_{ik} ~,$$ where there is an implicit summation over repeated indices, in this case ##\sum_j##. The above is true because is just matrix multiplication of ##A## by the identity matrix ##1## (all 1's on the diagonal).
blue24 and Michael Price
blue24 said:
Thank you for the reply. Unfortunately those terms, "covariant" and "contravariant", are not in my textbook. I have a pdf (legally purchased) and I searched for them. I will do some searching online.
What is throwing me is that the textbook literally says, "[referring to isotropic tensors] It can easily be verified (see Exercise 1.8) that the Kronecker delta ##d_{ij}## has such a property." So I figured, hey I better do Exercise 1.8, it'll be easy. And I'm stuck.
Can anyone give me any hints as to where I am going wrong, what steps I should follow, etc? I want to repeat - I am not even in the class yet, first day of class is September 6. Most likely this will never actually be an assigned homework problem. I am just trying to work ahead because I know I will need to do so in order to get help. So I am not posting on here trying to cut corners and get you guys to do my homework for me or anything like that.
Thanks for any help you can provide!
Andrew
Upon reflection the contravariant/covariant thing might not be relevant, since you are doing 3D tensors. They come into their own for 4D tensor transformations.
strangerep said:
Well, either that's not exactly what the book says, or it's being extremely sloppy.
My apologies! The book defines ##Q_{ij}## as the cosine of the angle between the ##x^{'}_{i}## and ##x_j## axes. See attachment.
#### Attachments
• Qij.png
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Ok, so I'd like to zoom out here and get a better understanding of ##Q_{ij}##. I'm trying to do a basic example problem where the solution is given in the book (see attached image).
I can calculate the transformation matrix, ##Q_{ij}## (did it without looking at the solution, woot!), and do the transformation of the vector, ##a_{i}##, no problem. However I am stumped by the transformation of the 2nd order tensor, ##a_{ij}##. The solution given points you to the equation ##a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}##. I don't understand how this turns into ##Q_{ij}a_{ij}Q^{T}_{ij}##.
I guess one of the things I am fundamentally not understanding is what is the difference between ##Q_{ip}## and ##Q_{jq}## in the equation you are supposed to use? And how are they different than ##Q_{ij}##?
#### Attachments
• Transformation example.png
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blue24 said:
I can calculate the transformation matrix, ##Q_{ij}## (did it without looking at the solution, woot!), and do the transformation of the vector, ##a_{i}##, no problem. However I am stumped by the transformation of the 2nd order tensor, ##a_{ij}##. The solution given points you to the equation ##a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}##. I don't understand how this turns into ##Q_{ij}a_{ij}Q^{T}_{ij}##.
It doesn't. It turns into
$$a^{'}_{ij} ~=~ Q_{ip} a_{pq} Q_{jq} ~=~ Q_{ip} a_{pq} Q^T_{qj} ~=~ \Big(Q a Q^T)_{ij} ~.$$
The ##i,j,## are ordinary indices -- they must match up on both sides of the equality. The ##p,q## indices are dummy summation indices (implicit sum over repeated indices). I.e., there's an implicit ##\sum_p \sum_q##. See Einstein summation convention. (That reference is for a more general case where there are both upstairs and downstairs indices, whereas in your case all indices are downstairs. Same principle though.)
I guess one of the things I am fundamentally not understanding is what is the difference between ##Q_{ip}## and ##Q_{jq}## in the equation you are supposed to use? And how are they different than ##Q_{ij}##?
Trying sticking those explicit "##\sum##" symbols back into the equation, and then relate it to how one multiplies matrices.
Btw, I sense that you really need to study some books on linear algebra and vector/tensor calculus. The Schaum Outline series are usually good for covering a lot of material quickly, with example and exercises. Otherwise, ask for suggestions in the Academic Guidance forum.
blue24 and Michael Price
blue24 said:
Ok, so I'd like to zoom out here and get a better understanding of ##Q_{ij}##. I'm trying to do a basic example problem where the solution is given in the book (see attached image).
I can calculate the transformation matrix, ##Q_{ij}## (did it without looking at the solution, woot!), and do the transformation of the vector, ##a_{i}##, no problem. However I am stumped by the transformation of the 2nd order tensor, ##a_{ij}##. The solution given points you to the equation ##a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}##. I don't understand how this turns into ##Q_{ij}a_{ij}Q^{T}_{ij}##.
I guess one of the things I am fundamentally not understanding is what is the difference between ##Q_{ip}## and ##Q_{jq}## in the equation you are supposed to use? And how are they different than ##Q_{ij}##?
Have a thorough read of
https://en.m.wikipedia.org/wiki/Matrix_multiplication
blue24
strangerep said:
Btw, I sense that you really need to study some books on linear algebra and vector/tensor calculus. The Schaum Outline series are usually good for covering a lot of material quickly, with example and exercises. Otherwise, ask for suggestions in the Academic Guidance forum.
Thank you for the feedback. This is exactly what I am trying to figure out. This class is going to be a heavy lift for me, and so I need to figure out what the gaps are in my understanding and start working on addressing those. I have not taken a linear algebra class (was optional in my undergrad), and so I think that is a definite weakness. I just ordered Schaum's Outline of Linear Algebra.
strangerep said:
It doesn't. It turns into
$$a^{'}_{ij} ~=~ Q_{ip} a_{pq} Q_{jq} ~=~ Q_{ip} a_{pq} Q^T_{qj} ~=~ \Big(Q a Q^T)_{ij} ~.$$
I don't understand how you got from ##Q_{ip} Q_{jq} a_{pq}## to ##Q_{ip} a_{pq} Q_{jq}##. Why did the order of the terms change?
blue24 said:
I don't understand how you got from ##Q_{ip} Q_{jq} a_{pq}## to ##Q_{ip} a_{pq} Q_{jq}##. Why did the order of the terms change?
It's because of the transpose (you dropped it in your question). What happens here is that ##Q_{jq} = (Q^T)_{qj} ## which is just saying that the jq entry of a matrix is equal to the qj entry of the transpose of the matrix.
nrqed said:
It's because of the transpose (you dropped it in your question). What happens here is that ##Q_{jq} = (Q^T)_{qj} ## which is just saying that the jq entry of a matrix is equal to the qj entry of the transpose of the matrix.
The given equation from the book is: ##a^{'}_{ij}=Q_{ip}Q_{jq}a_{pq}## (Eq 1)
Per strangerrep's post from Yesterday at 1:55PM, this "turns into ##a^{'}_{ij}=Q_{ip}a_{pq}Q_{jq}##..." (Eq 2)
He does not introduce the transpose until after shifting the ##a_{pq}## term before the ##Q_{jq}## term. I don't understand what allows for this shift. In other words, how are equations 2 and 3 equal?
I do understand that ##Q_{jq} = (Q^T)_{qj} ##
blue24 said:
The given equation from the book is: ##a^{'}_{ij}=Q_{ip}Q_{jq}a_{pq}## (Eq 1)
Per strangerrep's post from Yesterday at 1:55PM, this "turns into ##a^{'}_{ij}=Q_{ip}a_{pq}Q_{jq}##..." (Eq 2)
He does not introduce the transpose until after shifting the ##a_{pq}## term before the ##Q_{jq}## term. I don't understand what allows for this shift. In other words, how are equations 2 and 3 equal?
I do understand that ##Q_{jq} = (Q^T)_{qj} ##
I am not sure what Eqs 2 and 3 are. I Guess you mean Eqs 1 and 2.
These are just numbers so you can move them around and put them in any order you want ##Q_{jq} a_{pq}=a_{pq}Q_{jq}## for example.
@blue24 : When asking questions out of a textbook like you're doing in this thread, it's wise (even mandatory) to say exactly which textbook it is, because sometimes questions can be answered more effectively by pointing you to another section of the book.
(I haven't responded to your most recent questions above, because nrqed already did.)
nrqed said:
I am not sure what Eqs 2 and 3 are. I Guess you mean Eqs 1 and 2.
These are just numbers so you can move them around and put them in any order you want ##Q_{jq} a_{pq}=a_{pq}Q_{jq}## for example.
Sorry, yes, Eqs 1 and 2.
Thanks. I am definitely confused on this point. I thought the terms were matrices, not numbers. If you look at the book solution (my post from yesterday at 12:46), it appears that they are matrices. And I thought that order matters for matrix multiplication.
strangerep said:
@blue24 : When asking questions out of a textbook like you're doing in this thread, it's wise (even mandatory) to say exactly which textbook it is, because sometimes questions can be answered more effectively by pointing you to another section of the book.
(I haven't responded to your most recent questions above, because nrqed already did.)
Good idea. The book I am using is Elasticity: Theory, Applications, and Numerics; Third Edition; by Martin Sadd.
blue24 said:
Sorry, yes, Eqs 1 and 2.
Thanks. I am definitely confused on this point. I thought the terms were matrices, not numbers. If you look at the book solution (my post from yesterday at 12:46), it appears that they are matrices. And I thought that order matters for matrix multiplication.
As soon as we have indices, like ##Q_{pq}##, we are talking about the entries of the matrix, which are just numbers. If we write ##Q## , we mean the matrix and the order matters.
nrqed said:
As soon as we have indices, like ##Q_{pq}##, we are talking about the entries of the matrix, which are just numbers. If we write ##Q## , we mean the matrix and the order matters.
Thanks for the reply. That was my understanding before I got the textbook for this class and started reading through it. However, this textbook seems to use a notation where ##a_{ij}## represents a matrix. See the image below.
Also, in order to calculate the solution, we get to ##Q_{ip}a_{pq}Q^{T}_{qj}##. We are transposing the third symbol in the term, ##Q_{qj}##. It only makes sense to transpose ##Q_{qj}## if it is a matrix, and not a number, right?
I have no doubt that you are right and I am wrong. Just trying to highlight the things I don't understand, and I'm hoping to see the light.
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• matrix.png
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Last edited:
You have good questions. The problem is the sloppiness of notation that is unfortunately widespread, which makes things confusing indeed.
1) When we write ##Q_{qj}^T##, we mean ##(Q^T)_{qj}##. The parentheses are important, but people usually drop them. So we mean the qj entry of the matrix ##Q^T##.
2) As for the notation used by the book, you are right that it is confusing. I am a physicist often working with mathematicians and I have been scolded by them more than once for writing things this way. Now I realize they they are entirely right, one should never write something like that! When people write a_{ij} = a matrix, they are badly abusing notation. What they mean is that if we let the indices range over all their values, we span the entire matrix. But it is incorrect to write that a_{ij} is equal to a matrix. Unfortunately, this is an abuse often done by physicists.
To be precise, a_{ij} is always an entry in the matrix, when we work with equations and manipulate terms.
blue24
Thank you. That is a really helpful answer. So when I am looking at equations which are written in index notation, I should treat the elements of the equation as numbers, not matrices.
For me, that begs a new question. In order to arrive at the final solution (see first attached image), matrices are used. I don't understand the logic that determines what order the matrices are multiplied in or why the one matrix is transposed.
I think I understand the basic concepts of Einstein notation (is explained in the book, and strangerep explained it above as well). In order to convince myself that I understand Einstein notation, I calculated the first term of the transformed matrix, ##a^{'}_{11}## using summation, and then also calculated it using ##(QaQ^{T})_{ij}## and found them to be the same (see second attached image).
So I understand that ##(QaQ^{T})_{ij}## gets you the final answer. I just don't understand the thought process for getting to ##(QaQ^{T})_{ij}##.
My guess is that I simply need to shore up my fundamentals, which I am planning on doing. If there is no clear answer to my question, no worries. I will give up on this problem for now and come back when I have the background to understand. Thanks everyone for the help.
#### Attachments
• example.png
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• a11.png
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blue24 said:
I don't understand the logic that determines what order the matrices are multiplied in or why the one matrix is transposed.
OK. Consider the column vector in your example, whose components are denoted ##a_i##. Do you understand why $$a'_i ~=~ Q_{ij} a_j ~~~?$$ And what is this in matrix language? (Once you're clear on this, we'll move on to 2nd rank tensors.)
## What is the Kronecker delta?
The Kronecker delta is a mathematical symbol denoted by Δ or δ that represents a function of two variables. It is used to represent the number 1 when the two variables are equal, and 0 when they are not equal. It is commonly used in linear algebra and calculus.
## What does it mean for the Kronecker delta to "retain its form"?
When we say that the Kronecker delta retains its form, we mean that it remains unchanged or unaltered under various transformations. These transformations can be changes in coordinates, basis vectors, or any other mathematical operations.
## Why is it important to show that the Kronecker delta retains its form under any transformation?
It is important to show this because the Kronecker delta is a fundamental tool in many mathematical equations and proofs. If it did not retain its form under transformations, it would complicate many calculations and make it difficult to use in various applications.
## How can we prove that the Kronecker delta retains its form under any transformation?
The proof involves using the definition of the Kronecker delta and applying it to the various transformations. It also involves using properties of the transformations, such as the commutative and distributive properties, to show that the Kronecker delta remains unchanged.
## Can the Kronecker delta be used in any type of transformation?
Yes, the Kronecker delta can be used in any type of transformation as long as the underlying properties of the transformation are preserved. This includes linear transformations, coordinate transformations, and basis vector transformations.
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3K | 4,859 | 18,640 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.917167 |
https://www.jiskha.com/questions/1643991/hey-bob-heres-what-i-got-so-far-3agno3-na3po4-ag3po4-3nano3-silver-nitrate-and | 1,653,561,372,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00248.warc.gz | 957,001,067 | 7,184 | # Chemistry
Hey bob- Here's what I got so far:
3AgNO3 + Na3PO4 ---> Ag3PO4 + 3NaNO3
Silver nitrate and sodium phosphate are reacted in equal amounts of 200. g each. How many grams of silver phosphate are produced?
1. What is the limiting?
2. How much silver phosphate is produced?
3. How much is in excess?
200gAGNO3/170= 1.18 AGNO3
200gNA3PO4/164= 1.22 Ag3PO4
Not sure of the next step....here's what I got:
419 AG3PO4 x 3 = 1257 AG3PO4
Is AG3PO4 the limiting reagent??
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1. 200gAGNO3/170= 1.18 AGNO3
200gNA3PO4/164= 1.22 Ag3PO4
3AgNO3 + Na3PO4 ---> Ag3PO4 + 3NaNO3
One needs three time the silver nitrate, you dont have near that, so silver nitrate is the limiting reageant. So you will consume all you have 1.18 moles, and you will get 1.18/3 moles of silver phosphate.
On the sodium phosphate, you will consume 1.18/3 moles of it, you had 1.22 to start, so the left over is the difference, and you cna convert that to grams
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bobpursley
1. Ok I think I understand. Still a little unsure tho where we're getting the 3 times the silver nitrate from for silver nitrate to be limiting reagent.
2. 1.18/3=.39 x 419 Ag3PO4 = 163g Ag3PO4 produced??
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3. 3. 200g Na3PO4 - 163 = 36g Ag3PO4??
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4. The three times is because your balanced equation says you need three silver nitrate molecules for every sodium phosphate molecule.
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5. You do not have that much silver nitrate so it limits the reaction.
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6. Gotcha, thanks. Can you one of you guys see if I got 2 & 3 correct from the tips bob was giving me.
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7. so you use 1.18 mols of silver nitrate
that needs 1.18/3 = .393 mols of sodium phosphate. For every mol of sodium phosphate you get a mol of silver phosphate so you get ,393 mols of Silver Phosphate
you have 1.22 mols of sodium phosphate
so you will have 1.22 - .39 = .83 mols of sodium phosphate left over.
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8. I will leave you to convert mols to grams :)
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9. I only do mols.
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10. 3.) Cool that makes sense. So I believe this is how you convert mole to gram:
.83 x 164g/mol = 136g Na3PO4 left over
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11. I assume your 164 g/mol is correct. I know O is 16 and Na is 23 but do not remember P . Not about to Google.
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12. Phosphorus is 31g.
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13. Ya and that equals 164. Thank you a ton for the help I've stomped on this whole problem for like a half an hour and I got lots more to do. This was my 2nd limiting reagent problem I've ever done so I'm still learning how to handle these.
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14. Would you have another 5 minutes to help me one more problem I'm stumped on???
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15. I'll post it as a new problem. Its a Stoichemetric Problem that I'm completely stumped on. I think I have part of correct which i'd be happy to see if its correct so far or not.
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https://www.powershow.com/view4/845ef4-ZTI4N/Elementary_Linear_Algebra_powerpoint_ppt_presentation | 1,660,474,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00675.warc.gz | 832,623,079 | 24,270 | # Elementary Linear Algebra - PowerPoint PPT Presentation
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## Elementary Linear Algebra
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Title: Elementary Linear Algebra
1
Elementary Linear Algebra
Howard Anton Chris Rorres
2
Chapter Contents
• 1.1 Introduction to System of Linear
• Equations
• 1.2 Gaussian Elimination
• 1.3 Matrices and Matrix Operations
• 1.4 Inverses Rules of Matrix Arithmetic
• 1.5 Elementary Matrices and a Method for
• Finding
• 1.6 Further Results on Systems of Equations
• and Invertibility
• 1.7 Diagonal, Triangular, and Symmetric
• Matrices
3
1.1 Introduction to
• Systems of Equations
4
Linear Equations
• Any straight line in xy-plane can be represented
algebraically by an equation of the form
• General form define a linear equation in the n
variables
• Where and b are real
constants.
• The variables in a linear equation are sometimes
• called unknowns.
5
Example 1Linear Equations
• The equations
and
• are linear.
• Observe that a linear equation does not involve
any products or roots of variables. All variables
occur only to the first power and do not appear
as arguments for trigonometric, logarithmic, or
exponential functions.
• The equations
• are not linear.
• A solution of a linear equation is a sequence of
n numbers
• such that the equation is
satisfied. The set of all solutions of the
equation is called its solution set or general
solution of the equation
6
Example 2Finding a Solution Set (1/2)
• Find the solution of
• Solution(a)
• we can assign an arbitrary value to x and
solve for y , or choose an arbitrary value for y
and solve for x. If we follow the first approach
and assign x an arbitrary value, we obtain
.
• Or, if we follow the second approach, we
obtain
• arbitrary numbers are called parameter.
• for example
7
Example 2Finding a Solution Set (2/2)
• Find the solution of
• Solution(b)
• we can assign arbitrary values to any two
variables and solve for the third variable.
• for example
• where s, t are arbitrary values
8
Linear Systems (1/2)
• A finite set of linear equations in the variables
• is called a system of linear equations or a
linear system .
• A sequence of numbers
• is called a solution of
the system.
• A system has no solution is said to be
inconsistent if there is at least one solution
of the system, it is called consistent.
An arbitrary system of m linear equations
in n unknowns
9
Linear Systems (2/2)
• Every system of linear equations has either no
solutions, exactly one solution, or infinitely
many solutions.
• A general system of two linear equations
(Figure1.1.1)
• Two lines may be parallel -gt no solution
• Two lines may intersect at only one point
• -gt one solution
• Two lines may coincide
• -gt infinitely many solution
10
Augmented Matrices
• The location of the s, the xs, and the s can
be abbreviated by writing only the rectangular
array of numbers.
• This is called the augmented matrix for the
system.
• Note must be written in the same order in each
equation as the unknowns and the constants must
be on the right.
1th column
1th row
11
Elementary Row Operations
• The basic method for solving a system of linear
equations is to replace the given system by a new
system that has the same solution set but which
is easier to solve.
• Since the rows of an augmented matrix correspond
to the equations in the associated system. new
systems is generally obtained in a series of
steps by applying the following three types of
operations to eliminate unknowns systematically.
These are called elementary row operations.
• 1. Multiply an equation through by an nonzero
constant.
• 2. Interchange two equation.
• 3. Add a multiple of one equation to another.
12
Example 3Using Elementary row Operations(1/4)
13
Example 3Using Elementary row Operations(2/4)
14
Example 3Using Elementary row Operations(3/4)
15
Example 3Using Elementary row Operations(4/4)
• The solution x1,y2,z3 is now evident.
16
1.2 Gaussian Elimination
17
Echelon Forms
• This matrix which have following properties is in
reduced row-echelon form (Example 1, 2).
• 1. If a row does not consist entirely of zeros,
then the first nonzero number in the row is a 1.
We call this a leader 1.
• 2. If there are any rows that consist entirely
of zeros, then they are grouped together at the
bottom of the matrix.
• 3. In any two successive rows that do not
consist entirely of zeros, the leader 1 in the
lower row occurs farther to the right than the
leader 1 in the higher row.
• 4. Each column that contains a leader 1 has
zeros everywhere else.
• A matrix that has the first three properties is
said to be in row-echelon form (Example 1, 2).
• A matrix in reduced row-echelon form is of
necessity in row-echelon form, but not conversely.
18
Example 1Row-Echelon Reduced Row-Echelon form
• reduced row-echelon form
• row-echelon form
19
Example 2More on Row-Echelon and Reduced
Row-Echelon form
• All matrices of the following types are in
row-echelon form ( any real numbers substituted
for the s. )
• All matrices of the following types are in
reduced row-echelon form ( any real numbers
substituted for the s. )
20
Example 3aSolutions of Four Linear Systems (a)
Suppose that the augmented matrix for a system of
linear equations have been reduced by row
operations to the given reduced row-echelon form.
Solve the system.
Solution (a) the corresponding system of
equations is
21
Example 3bSolutions of Four Linear Systems (b1)
Solution (b) 1. The corresponding system of
equations is
free variables
22
Cont Example 3bSolutions of Four Linear Systems
(b2)
2. We see that the free variable can be assigned
an arbitrary value, say t, which then determines
3. There are infinitely many solutions, and the
general solution is given by the formulas
23
Example 3cSolutions of Four Linear Systems (c1)
• Solution (c)
• The 4th row of zeros leads to the equation places
no restrictions on the solutions (why?). Thus, we
can omit this equation.
24
Cont Example 3cSolutions of Four Linear Systems
(c2)
• Solution (c)
• Solving for the leading variables in terms of the
free variables
• 3. The free variable can be assigned an
arbitrary value,there are infinitely many
solutions, and the general solution is given by
the formulas.
25
Example 3dSolutions of Four Linear Systems (d)
Solution (d) the last equation in the
corresponding system of equation is Since this
equation cannot be satisfied, there is no
solution to the system.
26
Elimination Methods (1/7)
• We shall give a step-by-step elimination
procedure that can be used to reduce any matrix
to reduced row-echelon form.
27
Elimination Methods (2/7)
• Step1. Locate the leftmost column that does not
consist entirely of zeros.
• Step2. Interchange the top row with another row,
to bring a nonzero entry to top of the column
found in Step1.
Leftmost nonzero column
The 1th and 2th rows in the preceding matrix were
interchanged.
28
Elimination Methods (3/7)
• Step3. If the entry that is now at the top of the
column found in Step1 is a, multiply the first
row by 1/a in order to introduce a leading 1.
• Step4. Add suitable multiples of the top row to
the rows below so that all entires below the
The 1st row of the preceding matrix was
multiplied by 1/2.
-2 times the 1st row of the preceding matrix was
29
Elimination Methods (4/7)
• Step5. Now cover the top row in the matrix and
begin again with Step1 applied to the submatrix
that remains. Continue in this way until the
entire matrix is in row-echelon form.
Leftmost nonzero column in the submatrix
The 1st row in the submatrix was multiplied by
-1/2 to introduce a leading 1.
30
Elimination Methods (5/7)
• Step5 (cont.)
-5 times the 1st row of the submatrix was added
to the 2nd row of the submatrix to introduce a
The top row in the submatrix was covered, and we
returned again Step1.
Leftmost nonzero column in the new submatrix
The first (and only) row in the new submetrix was
multiplied by 2 to introduce a leading 1.
• The entire matrix is now in row-echelon form.
31
Elimination Methods (6/7)
• Step6. Beginning with las nonzero row and working
upward, add suitable multiples of each row to the
rows above to introduce zeros above the leading
1s.
7/2 times the 3rd row of the preceding matrix was
-6 times the 3rd row was added to the 1st row.
5 times the 2nd row was added to the 1st row.
• The last matrix is in reduced row-echelon form.
32
Elimination Methods (7/7)
• Step1Step5 the above procedure produces a
row-echelon form and is called Gaussian
elimination.
• Step1Step6 the above procedure produces a
reduced row-echelon form and is called
Gaussian-Jordan elimination.
• Every matrix has a unique reduced row-echelon
form but a row-echelon form of a given matrix is
not unique.
33
Example 4Gauss-Jordan Elimination(1/4)
• Solve by Gauss-Jordan Elimination
• Solution
• The augmented matrix for the system is
34
Example 4Gauss-Jordan Elimination(2/4)
• Adding -2 times the 1st row to the 2nd and 4th
rows gives
• Multiplying the 2nd row by -1 and then adding -5
times the new 2nd row to the 3rd row and -4 times
the new 2nd row to the 4th row gives
35
Example 4Gauss-Jordan Elimination(3/4)
• Interchanging the 3rd and 4th rows and then
multiplying the 3rd row of the resulting matrix
by 1/6 gives the row-echelon form.
• Adding -3 times the 3rd row to the 2nd row and
then adding 2 times the 2nd row of the resulting
matrix to the 1st row yields the reduced
row-echelon form.
36
Example 4Gauss-Jordan Elimination(4/4)
• The corresponding system of equations is
• Solution
• The augmented matrix for the system is
• We assign the free variables, and the general
solution is given by the formulas
37
Back-Substitution
• It is sometimes preferable to solve a system of
linear equations by using Gaussian elimination to
bring the augmented matrix into row-echelon form
without continuing all the way to the reduced
row-echelon form.
• When this is done, the corresponding system of
equations can be solved by solved by a technique
called back-substitution.
• Example 5
38
Example 5 ex4 solved by Back-substitution(1/2)
• From the computations in Example 4, a row-echelon
form from the augmented matrix is
• To solve the corresponding system of equations
• Step1. Solve the equations for the leading
variables.
39
Example5ex4 solved by Back-substitution(2/2)
• Step2. Beginning with the bottom equation and
working upward, successively substitute each
equation into all the equations above it.
• Substituting x61/3 into the 2nd equation
• Substituting x3-2 x4 into the 1st equation
• Step3. Assign free variables, the general
solution is given by the formulas.
40
Example 6Gaussian elimination(1/2)
• Solve by Gaussian
elimination and
• back-substitution. (ex3 of Section1.1)
• Solution
• We convert the augmented matrix
• to the ow-echelon form
• The system corresponding to this matrix is
41
Example 6Gaussian elimination(2/2)
• Solution
• Solving for the leading variables
• Substituting the bottom equation into those above
• Substituting the 2nd equation into the top
42
Homogeneous Linear Systems(1/2)
• A system of linear equations is said
• to be homogeneous if the constant
• terms are all zero that is , the
• system has the form
• Every homogeneous system of linear equation is
consistent, since all such system have
• as a solution. This solution is called the
trivial solution if there are another solutions,
they are called nontrivial solutions.
• There are only two possibilities for its
solutions
• The system has only the trivial solution.
• The system has infinitely many solutions in
43
Homogeneous Linear Systems(2/2)
• In a special case of a homogeneous linear system
of two linear equations in two unknowns
(fig1.2.1)
44
Example 7Gauss-Jordan Elimination(1/3)
• Solve the following homogeneous system of linear
equations by using Gauss-Jordan elimination.
• Solution
• The augmented matrix
• Reducing this matrix to reduced row-echelon form
45
Example 7Gauss-Jordan Elimination(2/3)
• Solution (cont)
• The corresponding system of equation
• Solving for the leading variables is
• Thus the general solution is
• Note the trivial solution is obtained when st0.
46
Example7Gauss-Jordan Elimination(3/3)
• Two important points
• Non of the three row operations alters the final
column of zeros, so the system of equations
corresponding to the reduced row-echelon form of
the augmented matrix must also be a homogeneous
system.
• If the given homogeneous system has m equations
in n unknowns with mltn, and there are r nonzero
rows in reduced row-echelon form of the augmented
matrix, we will have rltn. It will have the form
47
Theorem 1.2.1
• A homogeneous system of linear equations with
more unknowns than equations has infinitely many
solutions.
• Note theorem 1.2.1 applies only to homogeneous
system
• Example 7 (3/3)
48
Computer Solution of Linear System
• Most computer algorithms for solving large linear
systems are based on Gaussian elimination or
Gauss-Jordan elimination.
• Issues
• Reducing roundoff errors
• Minimizing the use of computer memory space
• Solving the system with maximum speed
49
1.3 Matrices and
• Matrix Operations
50
Definition
• A matrix is a rectangular array of numbers.
The numbers in the array are called the entries
in the matrix.
51
Example 1Examples of matrices
• Some examples of matrices
• Size
• 3 x 2, 1 x 4, 3 x 3, 2 x 1,
1 x 1
entries
row matrix or row vector
column matrix or column vector
columns
rows
52
Matrices Notation and Terminology(1/2)
• A general m x n matrix A as
• The entry that occurs in row i and column j of
matrix A will be denoted . If
is real number, it is common to be referred
as scalars.
53
Matrices Notation and Terminology(2/2)
• The preceding matrix can be written as
• A matrix A with n rows and n columns is called a
square matrix of order n, and the shaded entries
• are said to be on the main diagonal of A.
54
Definition
• Two matrices are defined to be equal if they
have the same size and their corresponding
entries are equal.
55
Example 2Equality of Matrices
• Consider the matrices
• If x5, then AB.
• For all other values of x, the matrices A and B
are not equal.
• There is no value of x for which AC since A and
C have different sizes.
56
Operations on Matrices
• If A and B are matrices of the same size, then
the sum AB is the matrix obtained by adding the
entries of B to the corresponding entries of A.
• Vice versa, the difference A-B is the matrix
obtained by subtracting the entries of B from the
corresponding entries of A.
• Note Matrices of different sizes cannot be added
or subtracted.
57
• Consider the matrices
• Then
• The expressions AC, BC, A-C, and B-C are
undefined.
58
Definition
• If A is any matrix and c is any scalar, then
the product cA is the matrix obtained by
multiplying each entry of the matrix A by c. The
matrix cA is said to be the scalar multiple of A.
59
Example 4Scalar Multiples (1/2)
• For the matrices
• We have
• It common practice to denote (-1)B by B.
60
Example 4Scalar Multiples (2/2)
61
Definition
• If A is an mr matrix and B is an rn matrix,
then the product AB is the mn matrix whose
entries are determined as follows.
• To find the entry in row i and column j of AB,
single out row i from the matrix A and column j
from the matrix B .Multiply the corresponding
entries from the row and column together and then
62
Example 5Multiplying Matrices (1/2)
• Consider the matrices
• Solution
• Since A is a 2 3 matrix and B is a 3 4 matrix,
the product AB is a 2 4 matrix. And
63
Example 5Multiplying Matrices (2/2)
64
Examples 6Determining Whether a Product Is
Defined
• Suppose that A ,B ,and C are matrices with the
following sizes
• A B C
• 3 4 4 7 7 3
• Solution
• Then by (3), AB is defined and is a 3 7 matrix
BC is defined and is a 4 3 matrix and CA is
defined and is a 7 4 matrix. The products AC ,CB
,and BA are all undefined.
65
Partitioned Matrices
• A matrix can be subdivided or partitioned into
smaller matrices by inserting horizontal and
vertical rules between selected rows and columns.
• For example, below are three possible partitions
of a general 3 4 matrix A .
• The first is a partition of A into
• four submatrices A 11 ,A 12,
• A 21 ,and A 22 .
• The second is a partition of A
• into its row matrices r 1 ,r 2,
• and r 3 .
• The third is a partition of A
• into its column matrices c 1,
• c 2 ,c 3 ,and c 4 .
66
Matrix Multiplication by columns and by Rows
• Sometimes it may b desirable to find a particular
row or column of a matrix product AB without
computing the entire product.
• If a 1 ,a 2 ,...,a m denote the row matrices of A
and b 1 ,b 2, ...,b n denote the column matrices
of B ,then it follows from Formulas (6)and (7)that
67
Example 7Example5 Revisited
• This is the special case of a more general
procedure for multiplying partitioned matrices.
• If A and B are the matrices in Example 5,then
from (6)the second column matrix of AB can be
obtained by the computation
• From (7) the first row matrix of AB can be
obtained by the computation
68
Matrix Products as Linear Combinations (1/2)
69
Matrix Products as Linear Combinations (2/2)
• In words, (10)tells us that the product A x of
a matrix
• A with a column matrix x is a linear
combination of
• the column matrices of A with the
coefficients coming
• from the matrix x .
the
• product y A of a 1m matrix y with an mn
matrix A
• is a linear combination of the row matrices
of A with
• scalar coefficients coming from y .
70
Example 8Linear Combination
71
Example 9Columns of a Product AB as Linear
Combinations
72
Matrix form of a Linear System(1/2)
• Consider any system of m
• linear equations in n unknowns.
• Since two matrices are equal if
• and only if their corresponding
• entries are equal.
• The m1 matrix on the left side
• of this equation can be written
• as a product to give
73
Matrix form of a Linear System(1/2)
• If w designate these matrices by A ,x ,and b
,respectively, the original system of m equations
in n unknowns has been replaced by the single
matrix equation
• The matrix A in this equation is called the
coefficient matrix of the system. The augmented
matrix for the system is obtained by adjoining b
to A as the last column thus the augmented
matrix is
74
Definition
• If A is any mn matrix, then the transpose of A
,denoted by ,is defined to be the nm matrix
that results from interchanging the rows and
columns of A that is, the first column of
is the first row of A ,the second column of
is the second row of A ,and so forth.
75
Example 10Some Transposes (1/2)
76
Example 10Some Transposes (2/2)
• Observe that
• In the special case where A is a square matrix,
the transpose of A can be obtained by
interchanging entries that are symmetrically | 5,027 | 18,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-33 | latest | en | 0.835058 |
https://sites.google.com/site/jgilberteconomics/Home/excel/step-by-step | 1,405,095,127,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776428273.44/warc/CC-MAIN-20140707234028-00060-ip-10-180-212-248.ec2.internal.warc.gz | 647,652,001 | 7,662 | Search this site
Step-by-Step Example Using Solver This page gives a step-by-step tutorial on using the Solver based models. We will consider simulating the Stolper-Samuelson theorem using the single economy version of the HOS model. First open the sheet in Excel. You should see the following screen (we are using Excel 2007): Click on the image to see it in full size. You should have the Solver add-in installed following theTo simulate Stolper-Samuelson we use the spinner next to cell C14 to increase the price of good X by (say) 20 percent: As we change the price, note that the graphs move in response. The unit value isoquant for X moves inward, while the isoprice for X moves outward. The prevailing factor prices are not a solution. Now, from the ribbon select the Data tab and click on the Solver button under Analysis: If you are using Excel 2003, choose Solver from the Tools menu. The following dialog appears: This is the model, written in Excel. Click the Solve button in the top right, and the following dialog appears: This indicates that a solution was found. Clicking OK returns us to the sheet: We observe that the price of capital has risen by more than 20 percent, while the price of labor has fallen. Examination of the factor use matrix confirms that X is capital-intensive, so the result is consistent with what the Stolper-Samuelson theorem predicts. The graphs depict the new equilibrium. We also observe that output of X has risen in response to the price rise, while output of Y has fallen, and so on. Any other simulation can be conducted in the same way. | 353 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2014-23 | longest | en | 0.905681 |
http://conversion.org/time/month-sidereal/nanoseconds | 1,642,543,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00078.warc.gz | 17,109,319 | 7,001 | # month (sidereal) to nanoseconds conversion
Conversion number between month (sidereal) and nanoseconds [ns] is 2.3605915104 × 10+15. This means, that month (sidereal) is bigger unit than nanoseconds.
### Contents [show][hide]
Switch to reverse conversion:
from nanoseconds to month (sidereal) conversion
### Enter the number in month (sidereal):
Decimal Fraction Exponential Expression
month (sidereal)
eg.: 10.12345 or 1.123e5
Result in nanoseconds
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 month (sidereal) = (2360591.5104) / (0.000000001) = 2.3605915104 × 10+15 nanoseconds
• 1 nanoseconds = (0.000000001) / (2360591.5104) = 4.2362263678164 × 10-16 month (sidereal)
• ? month (sidereal) × (2360591.5104 ("s"/"month (sidereal)")) / (0.000000001 ("s"/"nanoseconds")) = ? nanoseconds
### High precision conversion
If conversion between month (sidereal) to second and second to nanoseconds is exactly definied, high precision conversion from month (sidereal) to nanoseconds is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### month (sidereal) to nanoseconds conversion chart
Start value: [month (sidereal)] Step size [month (sidereal)] How many lines? (max 100)
visual:
month (sidereal)nanoseconds
00
102.3605915104 × 10+16
204.7211830208 × 10+16
307.0817745312 × 10+16
409.4423660416 × 10+16
501.1802957552 × 10+17
601.41635490624 × 10+17
701.65241405728 × 10+17
801.88847320832 × 10+17
902.12453235936 × 10+17
1002.3605915104 × 10+17
1102.59665066144 × 10+17
Copy to Excel
## Multiple conversion
Enter numbers in month (sidereal) and click convert button.
One number per line.
Converted numbers in nanoseconds:
Click to select all
## Details about month (sidereal) and nanoseconds units:
Convert Month (sidereal) to other unit:
### month (sidereal)
Definition of month (sidereal) unit: ≈ 27.321661 days. "It is the time it takes the Moon to go around Earth measured from a fixed point. Sideric month is not completely constant, an average value has been used for calculating."
Convert Nanoseconds to other unit:
### nanoseconds
Definition of nanoseconds unit: ≡ 10-9 s. billionth part or a second
← Back to Time units | 709 | 2,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.545926 |
https://datasciencewiki.net/lattice-path/ | 1,720,825,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00469.warc.gz | 162,392,682 | 9,933 | # Lattice Path
## Lattice Path :
A lattice path is a path that moves only horizontally or vertically on a grid. The path follows the grid lines, meaning it can only move to adjacent cells that share an edge. This type of path is commonly used in combinatorial problems, where the goal is to find the number of ways to reach a certain point on the grid.
One example of a lattice path is the movement of a knight on a chessboard. The knight can move in an “L” shape, meaning it can move two squares horizontally and one square vertically, or two squares vertically and one square horizontally. This movement pattern is constrained to the grid of the chessboard, making it a lattice path. The number of possible moves the knight can make from a certain position can be determined using combinatorial techniques.
Another example of a lattice path is the movement of a robot on a grid of streets. The robot can only move north, south, east, or west on the grid, following the lines of the streets. This movement is also constrained to the grid, making it a lattice path. The number of possible paths the robot can take from one point on the grid to another can be determined using combinatorial techniques.
In both examples, the movement is constrained to the grid, following only horizontal and vertical paths. This type of movement is useful for problems involving counting the number of possible paths or for problems involving optimization, such as finding the shortest path from one point to another.
Lattice paths can also be extended to three-dimensional grids, such as a cube. In this case, the path can move in the horizontal and vertical directions, as well as up and down along the z-axis. The number of possible paths on a three-dimensional grid can also be determined using combinatorial techniques.
Overall, lattice paths are useful for problems involving movement on a grid, allowing for the counting and optimization of possible paths. | 391 | 1,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.929649 |
http://lambdapage.org/29498/composting-worksheets-high-school/kids-can-compostwen-chia-parker-fellow-master-gardener-within-composting-worksheets-high-school/ | 1,618,594,802,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00064.warc.gz | 58,351,158 | 7,573 | # kids can compostwen chia parker fellow master gardener within composting worksheets high school
##### kids can compostwen chia parker fellow master gardener within composting worksheets high school.
Thus, you should simply the template of picture tracing worksheet and have fun with your kid. After the tracing process is done, you also can engage your kids to color the pictures.
The picture is actually simple. You can make an easy demonstration for the kid. Ensure that you also give a rule for the kid to keep their concentration on you.
The line also has various shape. It usually comes with a simple tracing that can easily followed for the beginner.
If your kids started to confuse, then you can point the correct way to finish the project. Do this only if your kids confuse. Let them finish the project and make a final evaluation.
March 3, 2021
March 11, 2021
March 6, 2021
March 7, 2021
### Free Learning To Write Worksheets
March 4, 2021
It’s another activity that will stimulate their memory and recognize the colors. Use the crayon to finish this project. As a reward, kids are allowed to hang their creation on their wall. It will make they enjoy and want to learn more using the other worksheets.
Learning to write the letter is the first skill that should be done for kids. This worksheet has a simple instruction for tracing the letter. You can make a demonstration to the kid before instruct them to do it by themselves. Remember, this is need a good patience.
It can be done for making your kids understand about the alphabet. As the pre-requisite, you have to ensure that your kids already recognize about the color and alphabet.
As a pre-requisite, you need to ensure that kids already understand about the color. Now give a simple instruction that they need to coloring the object that match with the number given.
March 5, 2021
March 6, 2021
March 9, 2021
March 6, 2021
### Free Plural Nouns Worksheets
March 10, 2021
If your kids understand on how to trace the line, then you can let them work individually. You can also evaluate about the number that they’ve already made.
After you’ve done with the line tracing, now it’s time for more advanced exercise. The spiral tracing can be challenging for your kids. It’s also need a concentration with the good coordination for tracing the lines.
If they done with their drawing, you also can ask them to color their drawing. Remember, you can’t expect the color or drawing in a really good picture. Usually, kids in preschool age draw some random objects.
The instruction is simple. You can ask your kid to put the color by saying the letter. For example, instruct them to color the “A” letter using the blue. Don’t point where is the A, let them pick the letter. If they pick the right letter, then they can continue to color the alphabet.
March 11, 2021
March 6, 2021
March 3, 2021
March 4, 2021
March 5, 2021
March 6, 2021
March 7, 2021
March 5, 2021
March 8, 2021
March 4, 2021
March 8, 2021
March 11, 2021 | 719 | 3,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-17 | latest | en | 0.952997 |
https://hextobinary.com/unit/angularacc/from/turnpms2/to/milpms2/976 | 1,719,261,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00436.warc.gz | 249,612,276 | 17,036 | # 976 Turn/Square Millisecond in Mil/Square Millisecond
Angular Acceleration
Turn/Square Millisecond
Mil/Square Millisecond
976 Turn/Square Millisecond = 6246400 Mil/Square Millisecond
## How many Mil/Square Millisecond are in 976 Turn/Square Millisecond?
The answer is 976 Turn/Square Millisecond is equal to 6246400 Mil/Square Millisecond and that means we can also write it as 976 Turn/Square Millisecond = 6246400 Mil/Square Millisecond. Feel free to use our online unit conversion calculator to convert the unit from Turn/Square Millisecond to Mil/Square Millisecond. Just simply enter value 976 in Turn/Square Millisecond and see the result in Mil/Square Millisecond.
## How to Convert 976 Turn/Square Millisecond to Mil/Square Millisecond (976 turn/ms2 to mil/ms2)
By using our Turn/Square Millisecond to Mil/Square Millisecond conversion tool, you know that one Turn/Square Millisecond is equivalent to 6400 Mil/Square Millisecond. Hence, to convert Turn/Square Millisecond to Mil/Square Millisecond, we just need to multiply the number by 6400. We are going to use very simple Turn/Square Millisecond to Mil/Square Millisecond conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Turn/Square Millisecond} = \text{6400 Mil/Square Millisecond}$$
$$\text{976 Turn/Square Millisecond} = 976 \times 6400 = \text{6246400 Mil/Square Millisecond}$$
## What is Turn/Square Millisecond Unit of Measure?
Turn per square millisecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one turn per square millisecond, its angular velocity is increasing by one turn per millisecond every millisecond.
## What is the symbol of Turn/Square Millisecond?
The symbol of Turn/Square Millisecond is turn/ms2. This means you can also write one Turn/Square Millisecond as 1 turn/ms2.
## What is Mil/Square Millisecond Unit of Measure?
Mil per square millisecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one mil per square millisecond, its angular velocity is increasing by one mil per millisecond every millisecond.
## What is the symbol of Mil/Square Millisecond?
The symbol of Mil/Square Millisecond is mil/ms2. This means you can also write one Mil/Square Millisecond as 1 mil/ms2.
## Turn/Square Millisecond to Mil/Square Millisecond Conversion Table (976-985)
Turn/Square Millisecond [turn/ms2]Mil/Square Millisecond [mil/ms2]
9766246400
9776252800
9786259200
9796265600
9806272000
9816278400
9826284800
9836291200
9846297600
9856304000
## Turn/Square Millisecond to Other Units Conversion Table
Turn/Square Millisecond [turn/ms2]Output
976 turn/square millisecond in degree/square second is equal to351360000000
976 turn/square millisecond in degree/square millisecond is equal to351360
976 turn/square millisecond in degree/square microsecond is equal to0.35136
976 turn/square millisecond in degree/square nanosecond is equal to3.5136e-7
976 turn/square millisecond in degree/square minute is equal to1264896000000000
976 turn/square millisecond in degree/square hour is equal to4553625600000000000
976 turn/square millisecond in degree/square day is equal to2.6228883456e+21
976 turn/square millisecond in degree/square week is equal to1.285215289344e+23
976 turn/square millisecond in degree/square month is equal to2.4299523673344e+24
976 turn/square millisecond in degree/square year is equal to3.4991314089615e+26
976 turn/square millisecond in radian/square second is equal to6132388859.81
976 turn/square millisecond in radian/square millisecond is equal to6132.39
976 turn/square millisecond in radian/square microsecond is equal to0.0061323888598073
976 turn/square millisecond in radian/square nanosecond is equal to6.1323888598073e-9
976 turn/square millisecond in radian/square minute is equal to22076599895306
976 turn/square millisecond in radian/square hour is equal to79475759623102000
976 turn/square millisecond in radian/square day is equal to45778037542907000000
976 turn/square millisecond in radian/square week is equal to2.2431238396024e+21
976 turn/square millisecond in radian/square month is equal to4.2410669476616e+22
976 turn/square millisecond in radian/square year is equal to6.1071364046327e+24
976 turn/square millisecond in gradian/square second is equal to390400000000
976 turn/square millisecond in gradian/square millisecond is equal to390400
976 turn/square millisecond in gradian/square microsecond is equal to0.3904
976 turn/square millisecond in gradian/square nanosecond is equal to3.904e-7
976 turn/square millisecond in gradian/square minute is equal to1405440000000000
976 turn/square millisecond in gradian/square hour is equal to5059584000000000000
976 turn/square millisecond in gradian/square day is equal to2.914320384e+21
976 turn/square millisecond in gradian/square week is equal to1.42801698816e+23
976 turn/square millisecond in gradian/square month is equal to2.699947074816e+24
976 turn/square millisecond in gradian/square year is equal to3.887923787735e+26
976 turn/square millisecond in arcmin/square second is equal to21081600000000
976 turn/square millisecond in arcmin/square millisecond is equal to21081600
976 turn/square millisecond in arcmin/square microsecond is equal to21.08
976 turn/square millisecond in arcmin/square nanosecond is equal to0.0000210816
976 turn/square millisecond in arcmin/square minute is equal to75893760000000000
976 turn/square millisecond in arcmin/square hour is equal to273217536000000000000
976 turn/square millisecond in arcmin/square day is equal to1.57373300736e+23
976 turn/square millisecond in arcmin/square week is equal to7.711291736064e+24
976 turn/square millisecond in arcmin/square month is equal to1.4579714204006e+26
976 turn/square millisecond in arcmin/square year is equal to2.0994788453769e+28
976 turn/square millisecond in arcsec/square second is equal to1264896000000000
976 turn/square millisecond in arcsec/square millisecond is equal to1264896000
976 turn/square millisecond in arcsec/square microsecond is equal to1264.9
976 turn/square millisecond in arcsec/square nanosecond is equal to0.001264896
976 turn/square millisecond in arcsec/square minute is equal to4553625600000000000
976 turn/square millisecond in arcsec/square hour is equal to1.639305216e+22
976 turn/square millisecond in arcsec/square day is equal to9.44239804416e+24
976 turn/square millisecond in arcsec/square week is equal to4.6267750416384e+26
976 turn/square millisecond in arcsec/square month is equal to8.7478285224038e+27
976 turn/square millisecond in arcsec/square year is equal to1.2596873072262e+30
976 turn/square millisecond in sign/square second is equal to11712000000
976 turn/square millisecond in sign/square millisecond is equal to11712
976 turn/square millisecond in sign/square microsecond is equal to0.011712
976 turn/square millisecond in sign/square nanosecond is equal to1.1712e-8
976 turn/square millisecond in sign/square minute is equal to42163200000000
976 turn/square millisecond in sign/square hour is equal to151787520000000000
976 turn/square millisecond in sign/square day is equal to87429611520000000000
976 turn/square millisecond in sign/square week is equal to4.28405096448e+21
976 turn/square millisecond in sign/square month is equal to8.099841224448e+22
976 turn/square millisecond in sign/square year is equal to1.1663771363205e+25
976 turn/square millisecond in turn/square second is equal to976000000
976 turn/square millisecond in turn/square microsecond is equal to0.000976
976 turn/square millisecond in turn/square nanosecond is equal to9.76e-10
976 turn/square millisecond in turn/square minute is equal to3513600000000
976 turn/square millisecond in turn/square hour is equal to12648960000000000
976 turn/square millisecond in turn/square day is equal to7285800960000000000
976 turn/square millisecond in turn/square week is equal to357004247040000000000
976 turn/square millisecond in turn/square month is equal to6.74986768704e+21
976 turn/square millisecond in turn/square year is equal to9.7198094693376e+23
976 turn/square millisecond in circle/square second is equal to976000000
976 turn/square millisecond in circle/square millisecond is equal to976
976 turn/square millisecond in circle/square microsecond is equal to0.000976
976 turn/square millisecond in circle/square nanosecond is equal to9.76e-10
976 turn/square millisecond in circle/square minute is equal to3513600000000
976 turn/square millisecond in circle/square hour is equal to12648960000000000
976 turn/square millisecond in circle/square day is equal to7285800960000000000
976 turn/square millisecond in circle/square week is equal to357004247040000000000
976 turn/square millisecond in circle/square month is equal to6.74986768704e+21
976 turn/square millisecond in circle/square year is equal to9.7198094693376e+23
976 turn/square millisecond in mil/square second is equal to6246400000000
976 turn/square millisecond in mil/square millisecond is equal to6246400
976 turn/square millisecond in mil/square microsecond is equal to6.25
976 turn/square millisecond in mil/square nanosecond is equal to0.0000062464
976 turn/square millisecond in mil/square minute is equal to22487040000000000
976 turn/square millisecond in mil/square hour is equal to80953344000000000000
976 turn/square millisecond in mil/square day is equal to4.6629126144e+22
976 turn/square millisecond in mil/square week is equal to2.284827181056e+24
976 turn/square millisecond in mil/square month is equal to4.3199153197056e+25
976 turn/square millisecond in mil/square year is equal to6.2206780603761e+27
976 turn/square millisecond in revolution/square second is equal to976000000
976 turn/square millisecond in revolution/square millisecond is equal to976
976 turn/square millisecond in revolution/square microsecond is equal to0.000976
976 turn/square millisecond in revolution/square nanosecond is equal to9.76e-10
976 turn/square millisecond in revolution/square minute is equal to3513600000000
976 turn/square millisecond in revolution/square hour is equal to12648960000000000
976 turn/square millisecond in revolution/square day is equal to7285800960000000000
976 turn/square millisecond in revolution/square week is equal to357004247040000000000
976 turn/square millisecond in revolution/square month is equal to6.74986768704e+21
976 turn/square millisecond in revolution/square year is equal to9.7198094693376e+23
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 2,924 | 10,679 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.802923 |
http://www.expertsmind.com/questions/area-under-curve-30110326.aspx | 1,637,999,863,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00252.warc.gz | 108,837,864 | 14,713 | area under curve, C/C++ Programming
Assignment Help:
Write a program to find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.
```#include float start_point, /* GLOBAL VARIABLES */ end_point, total_area; int numtraps; main( ) { void input(void ); float find_area(float a,float b,int n); /* prototype */ print(“AREA UNDER A CURVE”); input( ); total_area = find_area(start_point, end_point, numtraps); printf(“TOTAL AREA = %f”, total_area); } void input(void ) { printf(“\n Enter lower limit:”); scanf(“%f”, &start_point); printf(“Enter upper limit:”); scanf(“%f”, &end_point); printf(“Enter number of trapezoids:”); scanf(“%d”, &numtraps); } float find_area(float a, float b, int n) { float base , lower, h1, h2; /* LOCAL VARIABLES */ float function_x(float x); /* prototype */ float trap_area(float h1,float h2,float base );/*prototype*/ base = (b-1)/n; lower = a; for (lower =a; lower <= b-base ; lower = lower + base ) { h1 = function_x(lower); h1 = function_x(lower + base ); total_area += trap_area(h1, h2, base ); } return (total_area); float trap_area(float height_1,float height_2,float base ) { float area; /* LOCAL VARIABLE */ area = 0.5 * (height_1 + height_2) * base ; return (area); } float function_x(float x) { /* F(X) = X * X + 1 */ return (x*x + 1); } Output AREA UNDER A CURVE Enter lower limit: 0 Enter upper limit: 3 Enter number of trapezoids: 30 TOTAL AREA = 12.005000 AREA UNDER A CURVE Enter lower limit: 0 Enter upper limit: 3 Enter number of trapezoids: 100 TOTAL AREA = 12.000438```
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Write a program to find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b. The area under a curve between two points can b
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Define an asset base class, a) Define an Asset base class that provides the...
a) Define an Asset base class that provides the following method: class Asset { public: virtual double getValue()=0; }; This will be the base class for both stock an | 1,007 | 3,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-49 | latest | en | 0.58266 |
https://www.wiley.com/en-af/The+Art+and+Craft+of+Problem+Solving%2C+3rd+Edition-p-9781119094845 | 1,568,955,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00289.warc.gz | 1,083,423,801 | 26,366 | The Art and Craft of Problem Solving, 3rd Edition
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Appealing to everyone from college-level majors to independent learners, The Art and Craft of Problem Solving, 3rd Edition introduces a problem-solving approach to mathematics, as opposed to the traditional exercises approach. The goal of The Art and Craft of Problem Solving is to develop strong problem solving skills, which it achieves by encouraging students to do math rather than just study it. Paul Zeitz draws upon his experience as a coach for the international mathematics Olympiad to give students an enhanced sense of mathematics and the ability to investigate and solve problems.
Related Resources
Instructor
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Contact your Rep for all inquiries
Student
View Student Companion Site
1 What This Book Is About and How to Read It 1
1.1 “Exercises” vs. “Problems” 1
1.2 The Three Levels of Problem Solving 3
1.3 A Problem Sampler 6
1.4 How to Read This Book 9
2 Strategies for Investigating Problems 12
2.1 Psychological Strategies 12
Mental Toughness: Learn from Pólya’s Mouse 13
Creativity 15
2.2 Strategies for Getting Started 23
The First Step: Orientation 23
I’m Oriented. Now What? 24
2.3 Methods of Argument 37
Common Abbreviations and Stylistic Conventions 37
Deduction and Symbolic Logic 38
Mathematical Induction 42
2.4 Other Important Strategies 49
Draw a Picture! 49
Pictures Don’t Help? Recast the Problem in Other Ways! 51
Change Your Point of View 55
3 Tactics for Solving Problems 58
3.1 Symmetry 59
Geometric Symmetry 60
Algebraic Symmetry 64
3.2 The Extreme Principle 70
3.3 The Pigeonhole Principle 80
Basic Pigeonhole 80
Intermediate Pigeonhole 82
3.4 Invariants 88
Parity 90
Modular Arithmetic and Coloring 95
Monovariants 97
4 Three Important Crossover Tactics 105
4.1 Graph Theory 105
Connectivity and Cycles 107
Eulerian and Hamiltonian Walks 108
The Two Men of Tibet 111
4.2 Complex Numbers 116
Basic Operations 116
Roots of Unity 122
Some Applications 123
4.3 Generating Functions 128
Introductory Examples 129
Recurrence Relations 130
Partitions 132
4.4 Interlude: A few Mathematical Games 138
5 Algebra 143
5.1 Sets, Numbers, and Functions 143
Sets 143
Functions 145
5.2 Algebraic Manipulation Revisited 147
The Factor Tactic 148
Manipulating Squares 149
Substitutions and Simplifications 150
5.3 Sums and Products 157
Notation 157
Arithmetic Series 158
Geometric Series and the Telescope Tool 158
Infinite Series 161
5.4 Polynomials 164
Polynomial Operations 165
The Zeros of a Polynomial 165
5.5 Inequalities 174
Fundamental Ideas 174
The AM-GM Inequality 177
Massage, Cauchy-Schwarz, and Chebyshev 181
6 Combinatorics 189
6.1 Introduction to Counting 189
Permutations and Combinations 189
Combinatorial Arguments 192
Pascal’s Triangle and the Binomial Theorem 193
Strategies and Tactics of Counting 195
6.2 Partitions and Bijections 197
Counting Subsets 197
Information Management 200
Balls in Urns and Other Classic Encodings 203
6.3 The Principle of Inclusion-Exclusion 207
Count the Complement 207
PIE with Sets 208
PIE with Indicator Functions 212
6.4 Recurrence 215
Tiling and the Fibonacci Recurrence 215
The Catalan Recurrence 217
7 Number Theory 224
7.1 Primes and Divisibility 224
The Fundamental Theorem of Arithmetic 224
GCD, LCM, and the Division Algorithm 226
7.2 Congruence 232
What’s So Good About Primes? 233
Fermat’s Little Theorem 234
7.3 Number Theoretic Functions 236
Divisor Sums 237
Phi and Mu 238
7.4 Diophantine Equations 242
General Strategy and Tactics 242
7.5 Miscellaneous Instructive Examples 249
Can a Polynomial Always Output Primes? 249
If You Can Count It, It’s an Integer 250
A Combinatorial Proof of Fermat’s Little Theorem 250
Sums of Two Squares 251
8 Geometry for Americans 258
8.1 Three “Easy” Problems 258
8.2 Survival Geometry I 259
Points, Lines, Angles, and Triangles 260
Parallel Lines 262
Circles and Angles 265
Circles and Triangles 267
8.3 Survival Geometry II 271
Area 271
Similar Triangles 275
Solutions to the Three “Easy” Problems 277
8.4 The Power of Elementary Geometry 283
Concyclic Points 284
Area, Cevians, and Concurrent Lines 287
Similar Triangles and Collinear Points 290
Phantom Points and Concurrent Lines 293
8.5 Transformations 297
Symmetry Revisited 297
Rigid Motions and Vectors 299
Homothety 306
Inversion 308
9 Calculus 316
9.1 The Fundamental Theorem of Calculus 316
9.2 Convergence and Continuity 318
Convergence 319
Continuity 324
Uniform Continuity 325
9.3 Differentiation and Integration 329
Approximation and Curve Sketching 329
The Mean Value Theorem 332
A Useful Tool 335
Integration 336
Symmetry and Transformations 338
9.4 Power Series and Eulerian Mathematics 342
Don’t Worry! 342
Taylor Series with Remainder 344
Eulerian Mathematics 347
Beauty, Simplicity, and Symmetry: The Quest for a Moving Curtain 350
References 355
Index 357
The new material for this edition was inspired by Paul’s experience over the past decade working with teachers and students in math circles. In these math circles, they investigated many different topics ranging from simple magic tricks based on parity to understanding random variables to deep algebraic discoveries relying on the interplay between number theory and complex numbers. With the third edition, Paul has attempted to share hours and hours of fun, frustration, and discovery with these new problems.
• Substantial additions have been made to the problems, with several new themes that allow the reader to explore a wide variety of topics
• New section in Chapter 4, investigating many different topics: Mathematical Games
Resources
Instructor Resources
• Instructor’s Manual
• Hints to Selected Problems
Student Resources
• Hints to Selected Problems | 1,607 | 6,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | latest | en | 0.808205 |
http://www.studymode.com/essays/Department-Of-Electrical-And-Computer-Engineering-1526288.html | 1,524,423,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00028.warc.gz | 536,883,779 | 26,015 | # Department of Electrical and Computer Engineering: Final Examination
Topics: Roman numerals, MOSFET, Trigraph Pages: 22 (1959 words) Published: March 20, 2013
University of Waterloo
Department of Electrical & Computer Engineering
E&CE 231 Final Examination - Spring 2000
Aids: Formula Sheets (attached), Scientific Calculator
Time Allowed: 3 hours
Exam Type: Closed Book
Instructor: C. R. Selvakumar
Date: August 10, 2000
Max Marks: 100
Instructions:
Answer all questions in PART-A and any two questions in full from PART-B.
State your assumptions clearly. Be concise, precise and clear in your answers General assumptions to be made when not specified in a question: (a) Assume that the semiconductor is Silicon.
(b) Assume that the temperature T = 300K
(c) Use the data given in the formula sheets where needed.
(d) Use the following expressions for the Effective Density of States in the Conduction Band (NC) and in the Valence Band (NV) respectively: 3
2
3
3
3
m T 2 −3
N C = 2.5 × 1019
cm
m 0 300
*
n
m* 2 T 2
p
−3
19
N V = 2.5 × 10
m 300 cm
0
PART -A
1a) Consider a Silicon p+-n diode with the following doping densities: NA = 1019 cm-3 and ND is 1016 cm-3. The diode has an area of 100 µm by 20 µm.
(i)
Without doing any calculations, sketch the capacitance versus reverse voltage (VR) starting from VR = 0.
(4 marks)
(ii)
Calculate the voltage at which you will obtain the minimum
capacitance and also determine (calculate) the minimum
capacitance at that voltage.
(10 marks)
(iii)
Derive the mathematical relations you use in calculating the quantities in (ii) above.
(16 marks)
1b) Assuming that the p+ region and the n-region of the diode described in 1a) above are ‘long’ compared to the minority carrier diffusion lengths in those regions, show how you would obtain the complete Current-Voltage (I-V) Characteristic of the diode. You can assume that there is no recombination in the space-charge layer and you need not solve the continuity equation. Sketch the electron and hole current distributions in the entire device.
(10 marks)
Page 1
PART B
2a) Draw a clearly labelled band diagram of an n-p-n transistor under thermal equilibrium and superimpose on it a band diagram of the same transistor when it is under normal forward active mode of operations.
(8 marks)
2b) Derive an expression for the common emitter current gain \$ (\$ = IC/IB), in terms of the doping densities in the different regions, thickness and carrier diffusivities and diffusion lengths. Assume that there is no recombination in the neutral base or in the space-charge layers. Also, assume that the conventional reverse saturation current of the reverse-biased diode, IC0, is negligible. Assume that short-region approximation is valid in the base and that the bandgap narrowing in the emitter is important. No need to solve continuity equations and you can assume the expected carrier distributions. (12 marks)
2c) Obtain the modified Ebers-Moll (EM) equations from the original EM equations given in the formula sheet. Sketch Common-Base output
characteristics based on the modified EM equations and show the Forward Active Region of operation, Saturation Region and Cut-off Region. (10 marks) 3a) A silicon n-p-n transistor has an emitter doping NDE = 1020 cm-3 and a base doping NAB = 1016 cm-3. The emitter is 1 µm thick and assume that the hole diffusion length in the emitter is 0.1 :m. The base is 0.35 :m thick and you can use the values of mobilities and lifetimes given in the tables in the formula sheet to determine the electron diffusion length in the base. Verify that the short-region approximation is applicable to the base. Assume that the carrier recombinations in the neutral base an in the emitter-base depletion layer are zero. When this transistor is operating in the normal forward active mode with 0.6 volts forward bias across the emitter-base junction and a 2 volt reverse bias across the collector-base junction, what is the collector current density (JC) and the base current density (JB) ? You can assume that the... | 1,061 | 4,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-17 | latest | en | 0.841518 |
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# Exam 1 - P-v T-v and P-T planes Label each state in your...
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NAME: AME 20231 Thermodynamics Examination 1 Prof. J. M. Powers 12 February 2010 “The advantageous use of Steam-power is, unquestionably, a modern discovery. And yet, as much as two thousand years ago the power of steam was not only observed, but an ingenius toy was actually made and put in motion by it, at Alexandria in Egypt.” Abraham Lincoln, 6 April 1858 Bloomington, Illinois Happy 201st Birthday! 1. (20) Diatomic nitrogen, N 2 , exists at T = 65 . 9 K , v = 0 . 4 m 3 /kg . Find the pressure. 2. (40) A mass, 10 kg , of H 2 O initially at T 1 = 30 C , v 1 = 0 . 001080 m 3 /kg is heated isochorically to state 2 where T 2 = 140 C . It then undergoes an isobaric process to state 3 where T 3 = 250 C . (a) Find the ±nal speci±c volume. (b) Accurately sketch the total process in the
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Unformatted text preview: P-v , T-v , and P-T planes. Label each state in your sketch giving numerical values for P,T,v . Include the vapor dome in its correct position. (c) Find the work done in the total process. 3. (40) A mass of 0 . 01 kg of helium at P 1 = 100 kPa , T 1 = 300 K exists inside of the piston-cylinder arrangement of Fig. 1. The piston has a cross-sectional helium Figure 1: Piston-cylinder arrangement. area of A = 0 . 2 m 2 . The helium is heated until T 2 = 2000 K . The motion of the piston is resisted by a linear spring. The spring exerts no force at state 1, and has a spring constant of 1000 kN/m . (a) Find the ±nal pressure. (b) Find the total work done....
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Ask a homework question - tutors are online | 534 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-09 | latest | en | 0.886786 |
https://socratic.org/questions/how-do-you-calculate-pka-using-the-henderson-hasselbalch-equation | 1,701,834,949,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00781.warc.gz | 587,060,500 | 6,197 | # How do you calculate PKa using the Henderson-Hasselbalch equation?
## PH=6.98. Mix 25ml of 0.1M $N a {H}_{2} P {O}_{4}$ and 25ml of 0.1M $N {a}_{2} H P {O}_{4}$ and dilute with water in a 250ml Volumetric flask.
May 16, 2016
$p {K}_{a} = 6.98 - \log \left(\frac{0.05 M}{0.05 M}\right) = 6.98$
#### Explanation:
The Henderson-Hasselbalch equation is:
$p H = p {K}_{a} + \log \left(\frac{\left[B a s e\right]}{\left[A c i d\right]}\right)$
Therefore, $p {K}_{a} = p H - \log \left(\frac{\left[B a s e\right]}{\left[A c i d\right]}\right)$
$p {K}_{a} = 6.98 - \log \left(\frac{0.05 M}{0.05 M}\right) = 6.98$
Note that the new concentrations are divided by $2$ because the total volume doubles ${V}_{s o \ln .} = 25 m L + 25 m L = 50 m L$
Notice that n this case the $p H = p {K}_{a}$ because the $\left[N {a}_{2} H P {O}_{4}\right] = \left[N a {H}_{2} P {O}_{4}\right]$ and the $\log$ of the ratio is equal to zero.
Here is a video that explains more this topic (topic starts at minute 9:02):
Acid - Base Equilibria | Buffer Solution. | 397 | 1,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-50 | latest | en | 0.663001 |
https://www.lsbin.com/4572.html | 1,721,810,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00034.warc.gz | 725,380,882 | 17,119 | # Python OpenCV cv2.rectangle()方法绘制矩形示例
2021年4月7日18:54:42 发表评论 999 次浏览
OpenCV的Python是旨在解决计算机视觉问题的Python绑定库。Python OpenCV cv2.rectangle()方法用于在任何图像上绘制矩形。
Python OpenCV绘制矩形语法：
cv2.rectangle(image, start_point, end_point, color, thickness)
image：这是要在其上绘制矩形的图像。
start_point：矩形的起始坐标。坐标表示为两个值的元组, 即(X坐标值, Y坐标值)。
end_point：矩形的终点坐标。坐标表示为两个值的元组, 即(X坐标值, Y坐标值)。
color：是要绘制的矩形的边界线的颜色。对于BGR, 我们传递一个元组。例如：(255, 0, 0)为蓝色。
thickness：是矩形边框线的粗细(以像素为单位)。 -1 px的厚度将用指定的颜色填充矩形。
Python OpenCV cv2.rectangle示例1：
``````# Python program to explain cv2.rectangle() method
# importing cv2
import cv2
# path
path = r 'C:\Users\Rajnish\Desktop\lsbin\geeks.png'
# Reading an image in default mode
# Window name in which image is displayed
window_name = 'Image'
# Start coordinate, here (5, 5)
# represents the top left corner of rectangle
start_point = ( 5 , 5 )
# Ending coordinate, here (220, 220)
# represents the bottom right corner of rectangle
end_point = ( 220 , 220 )
# Blue color in BGR
color = ( 255 , 0 , 0 )
# Line thickness of 2 px
thickness = 2
# Using cv2.rectangle() method
# Draw a rectangle with blue line borders of thickness of 2 px
image = cv2.rectangle(image, start_point, end_point, color, thickness)
# Displaying the image
cv2.imshow(window_name, image)``````
``````# Python program to explain cv2.rectangle() method
# importing cv2
import cv2
# path
path = r 'C:\Users\Rajnish\Desktop\lsbin\geeks.png'
# Reading an image in grayscale mode
image = cv2.imread(path, 0 )
# Window name in which image is displayed
window_name = 'Image'
# Start coordinate, here (100, 50)
# represents the top left corner of rectangle
start_point = ( 100 , 50 )
# Ending coordinate, here (125, 80)
# represents the bottom right corner of rectangle
end_point = ( 125 , 80 )
# Black color in BGR
color = ( 0 , 0 , 0 )
# Line thickness of -1 px
# Thickness of -1 will fill the entire shape
thickness = - 1
# Using cv2.rectangle() method
# Draw a rectangle of black color of thickness -1 px
image = cv2.rectangle(image, start_point, end_point, color, thickness)
# Displaying the image
cv2.imshow(window_name, image)`````` | 744 | 2,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.449913 |
https://quantumcomputing.stackexchange.com/questions/2627/is-running-a-large-random-brute-force-on-quantum-computer-possible-at-the-moment/2657 | 1,563,435,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525524.12/warc/CC-MAIN-20190718063305-20190718085305-00007.warc.gz | 519,921,956 | 38,995 | # Is running a large random brute force on quantum computer possible at the moment?
I want to run a experiment like this:
1. Generate a bunch of random 12-character passwords like $\texttt{<Bb\{Q,r2Qp8}".$
2. Write an algorithm to randomly generate & compare value on quantum computer.
3. If the value was found, return number of time it take to generate, let's say $6102111820800.$
The only available quantum computer I know of is IBM's quantum computing cloud service.
Questions:
1. Is it possible to run this program on existing quantum computers?
2. If so, how fast would it be?
• "12 random string" or "a string composed of 12 random chars"? You want to generate this string with a quantum computer? And then to re-generate random strings using the same procedure until the same one is found? – Nelimee Jul 5 '18 at 15:46
• @Nelimee it's "a string composed of 12 random chars", generate it as fast & many as the system can & see if the value you want show up, it could take billion or trillion time. I just learn about quantum computer recently & wonder if it can do this kind of operation – Huang Lee Jul 5 '18 at 16:18
• Sounds like you hope that quantum computers can run through a list of possibilities very quickly to find a needle in a haystack. Is that about right? – Niel de Beaudrap Jul 5 '18 at 17:00
• I tweaked your question a bit as far as I could understand it, though it still seems a bit unclear to me. For one spot, is that number, $6102111820800$, meant to be a count of attempts or an amount of time? And if time, is it measured in clock cycles or something? – Nat Jul 5 '18 at 17:01
• @NieldeBeaudrap yeah, i just found a quantum slimutalor service which is pretty similar to my question demo.riverlane.io – Huang Lee Jul 5 '18 at 17:04
I want to run a experiment like this:
1. Generate a bunch of random 12-character passwords like ";Bb{Q,r2Qp8" (changed the first character because it interefed with the citation style).
Let's say your characters are encoded in extended ASCII, i.e. they have a value between 0 and 255. You need 8 classical bits to represent one character. One could expect that you could encode this value on 3 qubits ($2^3$), but you need to do a compromise here:
1. If you have access to an external source of randomness, then you can use it to generate a random quantum state (by applying random gates to the initial quantum state for example). In this case, the amplitudes of the obtained quantum state may represent your character (you still need to find how to represent a random character from complex non-integer numbers) and depending on the encoding you use you may need less than 8 qubits.
2. If you don't have access to an external source of randomness or if you want your random numbers to be "perfect", you can use quantum superposition and measurement to generate perfect random integers. This can be done by taking 8 qubits, applying the H gate to the 8 qubits and measuring them in the computational basis. With this algorithm, you will have 8 qubits in the state $\vert0\rangle$ or $\vert1\rangle$ (i.e. a random number on 8 qubits). With this method, 1 character = 8 qubits.
1. Write an algorithm to randomly generate & compare value on quantum computer.
The generation can be done in the same way as above. For the comparison, it depends on the method you used for the generation and how you represent your characters:
1. As the characters are encoded in the amplitudes, you can use the SWAP test to check for closeness.
2. Here, the qubits are just classical bits so you could just measure them and check classically for equality.
1. If the value was found, return number of time it take to generate, let's say 6102111820800.
Again, depending on what you want (but the bullet points here are not related to the methods above):
1. You could count the try-fails in a classical register, just by incrementing a classical counter at each fail. If you used the SWAP test, you can measure the ancillary qubits and update the counter.
2. If you want to encode your counter in a quantum state, you need a circuit that will increment the value of a register. You can find a way to construct such a circuit here for example. Then, if you used the SWAP test you can either read the qubit and apply the increment operation or directly apply the increment operation controlled by the state of the ancillary qubit.
The only available quantum computer I know of is IBM's quantum computing cloud service.
Questions:
1. Is it possible to run this program on existing quantum computers?
It depends on the method you take: the first method may be able to run on an existing chips, but the second one would need at least $96 = 12*8$ qubits to store the 12 characters, which is above the maximum number of qubits currently available.
1. If so, how fast would it be?
It will be sloooooooooooow. The first method may be able to use quantum superposition to speed-up the computations, but the second method uses quantum superposition only to generate random numbers, and then treat them classically.
• i keep hearing that qubits take lots of space to store data compare to classical computer? How does this make quantum computer the future? – Huang Lee Jul 6 '18 at 7:57
• i read that the ibm service has 20qbits available, so it can run 2^20 operation. Is that correct? How different it is compare to classical computer? – Huang Lee Jul 6 '18 at 7:59
• Can we use quantum computer to store & search data? – Huang Lee Jul 6 '18 at 7:59
• Each comment above deserve its own question on the main site because they are not really related to your original question :) – Nelimee Jul 6 '18 at 8:04
• ok, i keep that in mind – Huang Lee Jul 6 '18 at 8:08
From the question title, it sounds like you're interested in brute-force password cracking.
There is a quantum algorithm for this that outperforms brute force, in principle. It's called Grover's algorithm and it was one of the earliest quantum algorithms to be discovered.
However, to crack a password, you need as many qubits in your quantum computer as you'd need bits in a traditional computer to hash the passwords (actually more, since you have to hash them reversibly, and that involves some overhead). This is orders of magnitude more qubits than any present-day quantum computer has, even for simple password hashing techniques. Also, the computation lasts for far longer than any present-day quantum computer can maintain the integrity of its qubits. And even if it worked, it's not very fast: testing $n$ passwords with Grover's algorithm takes about as long as testing $\sqrt{n}$ passwords by brute force.
• no, i'm not even hash, just generate & compare value – Huang Lee Jul 8 '18 at 15:08 | 1,599 | 6,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | longest | en | 0.909084 |
http://mathhelpforum.com/trigonometry/218354-determine-all-te-possible-value-x-where-0-degree-x-360-degree-such.html | 1,511,292,303,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806422.29/warc/CC-MAIN-20171121185236-20171121205236-00740.warc.gz | 187,487,421 | 11,367 | # Thread: determine all te possible value of x where 0 degree <= x <= 360 degree such that
1. ## determine all te possible value of x where 0 degree <= x <= 360 degree such that
tan 2x = 6 cot x.
-is it change the tan 2x or 6 cot x first?
2. ## Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that
Originally Posted by choi105
tan 2x = 6 cot x.
-is it change the tan 2x or 6 cot x first?
Ummm....Use the double angle formula to replace tan(2x). I think that's what you were asking.
-Dan
3. ## Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that
\displaystyle \begin{align*} \tan{(2x)} &= 6\cot{(x)} \\ \frac{\sin{(2x)}}{\cos{(2x)}} &= \frac{6\cos{(x)}}{\sin{(x)}} \\ \frac{2\sin{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= \frac{6\cos{(x)}}{\sin{(x)}} \\ \frac{2\sin^2{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= 6\cos{(x)} \\ \frac{2\sin^2{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)} } - 6\cos{(x)} &= 0 \\ 2\cos{(x)} \left[ \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 \right] &= 0 \\ \cos{(x)} = 0 \textrm{ or } \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 &= 0 \end{align*}
Case 1:
\displaystyle \begin{align*} \cos{(x)} &= 0 \\ x &= \left( 2m + 1 \right) 90^{\circ} \textrm{ where } m \in \mathbf{Z} \end{align*}
Case 2:
\displaystyle \begin{align*} \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 &= 0 \\ \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= 3 \\ \sin^2{(x)} &= 3 \left[ \cos^2{(x)} - \sin^2{(x)} \right] \\ \sin^2{(x)} &= 3 \left[ 1 - 2\sin^2{(x)} \right] \\ \sin^2{(x)} &= 3 - 6\sin^2{(x)} \\ 7\sin^2{(x)} &= 3 \\ \sin^2{(x)} &= \frac{3}{7} \\ \sin{(x)} &= \pm \sqrt{ \frac{3}{7}} \\ \sin{(x)} &= \pm \frac{\sqrt{21}}{7} \\ x &= \left\{ \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} + \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 360^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} \right\} + 360^{\circ} n \textrm{ where } n \in \mathbf{Z} \end{align*}
So in the domain \displaystyle \begin{align*} 0^{\circ} \leq x \leq 360^{\circ} \end{align*}, we have
\displaystyle \begin{align*} x = \left\{ \arcsin{ \left( \frac{\sqrt{21}}{7} \right) }^{\circ} , 90^{\circ}, 180^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} + \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 270^{\circ} , 360^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} \right\} \end{align*}
4. ## Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that
Thank you very much.
5. ## Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that
Hello, choi105!
$\text{Solve for }x\text{ on }[0^o,\,360^o]\!:\;\tan 2x \:=\: 6\cot x$
We have: . $\frac{2\tan x}{1-\tan^2\!x} \;=\;\frac{6}{\tan x} \quad\Rightarrow\quad 2\tan^2\!x \;=\;6 - 6\tan^2\!x$
. . $8\tan^2\!x \:=\:6 \quad\Rightarrow\quad \tan^2\!x \:=\:\frac{3}{4} \quad\Rightarrow\quad \tan x \:=\:\pm\frac{\sqrt{3}}{2}$
Hence: . $x \;=\;\tan^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right) \;=\;40.89339465^o \;\approx\;40.9^o$
Therefore: . $x \;\approx\;\begin{Bmatrix}40.9^o \\ 139.1^o \\ 220.9^o \\ 319.1^o \end{Bmatrix}$ | 1,444 | 3,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-47 | longest | en | 0.416515 |
http://www.chegg.com/homework-help/questions-and-answers/person-pushes-236-kg-shopping-cart-constant-velocity-fora-distance-184-m-flat-horizontal-s-q124383 | 1,472,685,132,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982939756.54/warc/CC-MAIN-20160823200859-00002-ip-10-153-172-175.ec2.internal.warc.gz | 374,035,585 | 13,350 | A person pushes a 23.6-kg shopping cart at a constant velocity fora distance of 18.4 m on a flat horizontal surface. She pushes in adirection 27.7 ° below the horizontal. A 44.1-N frictionalforce opposes the motion of the cart. (a) What isthe magnitude of the force that the shopper exerts? Determine thework done by (b) the pushing force,(c) the frictional force, and (d)the gravitational force. | 104 | 396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-36 | latest | en | 0.825072 |
https://answers.ros.org/question/11350/is-there-any-way-to-calculate-inertial-property-of-a-robot-to-simulate-it-in-gazebo/ | 1,675,603,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00356.warc.gz | 126,635,082 | 16,259 | ROS Resources: Documentation | Support | Discussion Forum | Index | Service Status | Q&A answers.ros.org
# Is there any way to calculate inertial property of a robot to simulate it in Gazebo
I have got urdf/xacro files for pioneer3dx robot from "p2os" stack, modified them by replacing erratic_robot differential drive plugin and adding transmission for swivel and hubcap (which was not there as many reported a problem in visualizing swivel/hubcap in rviz). Then I tried to simulate pioneer3dx model in gazebo by using erratic_robot_teleop_keyboard node and also using navigation stack (by configuring yaml files).
My robot is behaving very strangely as it topples, lift backside up....etc. I am sure this is because of inertial properties provided in the model.
Is there any way to calculate these properties without estimating them by hand...like provide dimensions to any software then it 'll calculate for you??
Also I have model for schunk powercube arm but without any inertial properties. I don't think schunk 'll provide these information, I have to calculate myself to simulate it in gazebo.
Looking forward for a quick solution.
Thanks V.N.
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One general rule of thumb I use for checking my inertia tensors is: If total mass of the rigid body is m, and the dimension of the corresponding body is d, then check to see if ixx, iyy, izz are near m*(d/2)^2.
This is by no means correct, but a sanity check to make sure the moment of inertia are the right order of magnitudes, so the model behaves somewhat physically realistically.
I've seen too many simulation where mass is 1(kg), and ixx,iyy,izz are also 1(kg*m^2), while the robot is only 10cm in diameter. This setup usually result in a robot that does not respond to rotational commands easily (given that izz of 1 means there is a 1kg mass at the end of a 1 meter pole).
more
I know that SolidWorks can calculate this value for you, and I'm assuming that any other CAD program can as well.
As a general rule of thumb, I try to follow a number of guidelines when creating urdf models. These are mostly based on experience, so they should only be taken for what they are.
1. Don't make objects too large
2. Don't make objects too small
3. Avoid modifying the gravity
4. Keep masses for dynamic objects as close as possible. (i.e. the simulator will not like it if you place a 1000kg weight on four 1kg wheels.
5. When in doubt, use the identity mass moment of inertia <1 0 0 1 0 1>
You might also want to try the following to make your links more "solid":
<gazebo reference="back_left_wheel">
<kp>1000000.0</kp>
<kd>1.0</kd>
</gazebo>
more
You can even use the free Meshlab for this: Filters->Quality Measure and Computations->Compute Geometric Measures. However, Meshlab doesn't document which mass does it suppose (I hope it just uses unit mass) and it provides you the inertia tensor with respect to the origin, not with respect to CoM.
( 2014-07-01 08:11:45 -0600 )edit
It assumes unit density. So basically the volume is the mass of the object.
( 2016-07-04 10:20:18 -0600 )edit
I implemented a python script reading an 3D file like stl or dae and outputting the full URDF XML inertial tag based on the gazebosim tutorial. and pymeshlab.
At first the script calculates the center of mass of the object. After that it scales the mesh by a factor of 100 to increase the numerical accuracy. You could manually change that, if you need to scale it more up. Now the convex hull of your object will be calculated. For the hull the geometric properties will be calculated, scaled back down and outputted as URDF XML inertial tag like that (the precision could also be changed in the script)
<inertial>
<origin xyz="0.00000002 -0.00000001 0.00000000"/>
<mass value="5.00000000"/>
<inertia ixx="7.90866056" ixy="0.00000006" ixz="0.00000000" iyy="7.90866060" iyz="0.00000000" izz="2.48398783"/>
</inertial>
Check it out here: https://github.com/vonunwerth/MeshLab...
more | 1,020 | 4,021 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | longest | en | 0.92939 |
https://issp-center-dev.github.io/HPhi/manual/develop/tutorial/en/html/zero_temperature/spin_dimer.html | 1,719,041,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00456.warc.gz | 286,767,206 | 3,887 | # 1.1. Spin 1/2 Dimer¶
Let’s solve the following spin 1/2 dimer model (2-site Heisenberg model).
(1.1)$H = J {\bf S}_{0}\cdot{\bf S}_{1}$
The input file (samples/tutorial_1.1/stan1.in) is as follows:
L=2
model = "Spin"
method = "FullDiag"
lattice = "chain"
J = 0.5
2Sz = 0
2S = 1
It should be noted that the L=2 chain has two sites connected by two bonds with each other, and so we let J=0.5 to simulate the dimer model with $$J=1$$.
You can execute HPhi as follows
HPhi -s stan.in
## 1.1.1. Check the energy¶
Please check whether the energies are given as follows.
$$E_{\rm min}=-3/4$$ (singlet)
$$E_{\rm max}=1/4$$ (triplet)
## 1.1.2. Check S dependence¶
By changing 2S=1 in stan.in, you can treat spin-S dimer (eg. 2S=2 means S=1, see samples/tutorial_1.1/stan2.in). Please check whether the energies are given as follows.
$$E_{\rm min}=-S(S+1)$$
$$E_{\rm max}=S^2$$
## 1.1.3. Add magnetic field H¶
By adding H in stan.in, selecting model as “SpinGC”, and deleting “2Sz=0” you can examine the effects of the magnetic field in the dimer model. An example of the input file (samples/tutorial_1.1/stan3.in) is as follows:
L=2
model = "SpinGC"
method = "FullDiag"
lattice = "chain"
J = 0.5
2S = 1
H = 2
Please check whether the ground state becomes polarized state (Sz=1).
## 1.1.4. Try to use Lanczos method¶
By selecting method as “Lanczos” you can perform the Lanczos calculations. An example of the input file (samples/tutorial_1.1/stan4.in) is as follows:
L=2
model = "SpinGC"
method = "Lanczos"
lattice = "chain"
J = 0.5
2S = 1
H = 2
Please check the Lanczos method reproduces the energy (energy is output in *output/zvo_energy.dat ).
This is just a pedagogical example. By changing H = 20 (very large magnetic field), please examine what will happen. This calculation may fail! Please think why the Lanczos method fails for large magnetic field.
## 1.1.5. Try to use LOBCG method¶
LOBCG is locally optimal block conjugate gradient method. By selecting method as “CG”, you can perform the LOBCG calculations. An example of the input file (samples/tutorial_1.1/stan5a.in) is as follows:
L=2
model = "SpinGC"
method = "CG"
lattice = "chain"
J = 0.5
2S = 1
H = 2
Please check the LOBCG method reproduces the energy (energy is output in *output/zvo_energy.dat ).
This is just a pedagogical example. By changing H = 20 (very large magnetic field), please examine what will happen. In contrast to the Lanczos method, LOBCG will work well ! Please think why the LOBCG method works well for the large magnetic field.
Please also check the excited states can be correctly obtained by using LOBCG method. (Please compare the energies obtained by the full diagonalization.) An example of the input file (samples/tutorial_1.1/stan5b.in) is as follows:
L=2
model = "SpinGC"
method = "CG"
lattice = "chain"
J = 0.5
2S = 1
H = 2
exct = 4
Here, exct represents the number of excited states, which are obtained by the LOBCG method. | 931 | 2,970 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.836619 |
https://www.goconqr.com/quiz/7068569/triangle-congruencies-proofs | 1,610,776,500,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00790.warc.gz | 791,205,570 | 11,905 | # Triangle Congruencies Proofs
Quiz by Cassidy Paine, updated more than 1 year ago
Created by Cassidy Paine about 4 years ago
462
2
### Description
Proofs for triangle congruencies
## Resource summary
### Question 1
Question
Check which congruence postulate you would use to prove that the two triangles are congruent.
• SAS
• ASA
• AAS
• SSA
### Question 2
Question
Check which congruence postulate you would use to prove that the two triangles are congruent given the markings only.
• AAS
• ASA
• AA
### Question 3
Question
Check which congruence postulate you would use to prove that the two triangles are congruent.
• ASA
• AAS
• AA
### Question 4
Question
Check which congruence postulate you would use to prove that the two triangles are congruent.
• SSS
• SSA
• ASA
• SAS
### Question 5
Question
Fill in the missing reasons
• Given
• Alternate Interior
• Vertical angles
• Given
• Reflexive
• Linear Pair
• Given
• Linear Pair
• Symmetric Property
• Reflexive Property
• ASA
• AAS
• SSA
### Question 6
Question
Fill in the missing reasons
• Given
• Def. of Midpoint
• Def. of Bisector
• Given (2)
• Vertical Angles
• Given (3)
• Corresponding Angles
• Same Side Interior
• ASA
• AAS
### Question 7
Question
Fill in the missing reasons
• Given
• Given (2)
• def. of midpoint
• def. of bisector
• Vertical Angles
• Corresponding Angles
• Linear Pair
• def. of perpendicular
• def. of parallel
• All right angles are congruent
• Alternate Interior
• ASA
• HL
• SAS
• AAS
### Question 8
Question
Fill in the missing reasons
• Given
• Reflexive
• Parallel
• Given
• Def. of Midpoint
• Bisector
• def. of midpoint
• def. of bisector
• reflexive
• transitive
• def. of midpoint
• def. of bisector
• reflexive
• subsititution
• SSS
• SAS
• HL
• Side Theorem
• CPCTC
• Prove
### Question 9
Question
Determine the reasons of the following proof
• Given
• Parallel Lines
• Given (2)
• def of bisector
• def of midpoint
• symmetric
• reflexive
• transitive
• SAS
• SSA
• ASA
### Question 10
Question
Select the method which will prove the triangles congruent, if possible.
• ASA
• AAS
• SAS
• HL
• Not enough information
### Question 11
Question
Select the method which will prove the triangles congruent, if possible.
• SSS
• SAS
• ASA
• AAS
• HL
### Question 12
Question
Select the method which will prove the triangles congruent, if possible.
• HL
• SAS
• Not enough info
• SSA
• ASA
### Question 13
Question
Select the method which will prove the triangles congruent, if possible.
• SSS
• SAS
• Not enough info
• SSA
• ASA
### Question 14
Question
Select the method which will prove the triangles congruent, if possible.
• SAS
• SSA
• SSS
• ASA
• HL
• Not enough information
### Question 15
Question
Select the method which will prove the triangles congruent, if possible.
• Not enough info
• SSA
• ASA
• SAS
• AAS
### Question 16
Question
Select the method which will prove the triangles congruent, if possible.
• Triangle PQC
• Triangle PCQ
• Triangle QPC
• Triangle QCP
• SAS
• SSS
• ASA
### Question 17
Question
What additional information is needed to prove triangle congruency by ASA
• EF≅ST
• DE≅RS
• FD≅TR
### Question 18
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• AAS
### Question 19
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• SAS
### Question 20
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• HL
### Question 21
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• SSS
### Question 22
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• NONE
### Question 23
Question
Determine which method you would use to prove the two triangles congruent. If none of the methods apply, write NONE.
• SAS
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Algebra 2 Quality Core | 1,194 | 4,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-04 | latest | en | 0.803325 |
https://studylib.net/doc/7761533/co2008w3t | 1,620,913,397,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989916.34/warc/CC-MAIN-20210513111525-20210513141525-00102.warc.gz | 506,099,219 | 11,620 | # CO2008w3t
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```Computer Organisation 1 Week 3 Tutorial Questions
www.csis.ul.ie/modules/cs4211
Pure Binary:
1. Add 1001 to 0011.
2. add 1000 1101 to 0001 0111
3. add 1001 0110 to 1100 0111.
Translate the numbers above to hexadecimal and repeat the additions. Check you are
correct by translating to decimal.
Pure Binary:
4. Subtract 0011 from 1011
5. Subtract 0010 from 1000
6. Subtract 0001 0111 from 1000 0011
7. Subtract 0110 1111 from 1111 0000
Translate the numbers above to hexadecimal and repeat the subtractions.
Pure Binary:
8. Multiply 1101 by 11
9. Multiply 1000 by 1101
10. Multiply 1001 1100 by 1001
11. Multiply 0011 1110 by 0000 0110
Translate the numbers above to hexadecimal and repeat the calculations.
12. What is the maximum decimal integer that can be held as a pure binary 6 bits?
10 bits?
13. What is the two’s complement byte representation of 23, -5, 127, -1, -128,
132?
14. Tabulate the twos complement values in a 6-bit representation.
``` | 307 | 981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-21 | latest | en | 0.641391 |
https://en.m.wikipedia.org/wiki/Prior_Analytics | 1,571,469,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00458.warc.gz | 477,520,652 | 19,749 | # Prior Analytics
Aristotle Prior Analytics in Latin, 1290 circa, Biblioteca Medicea Laurenziana, Florence
Page from a 13th/14th-century Latin transcript of Aristotle's Opera Logica.
The Prior Analytics (Greek: Ἀναλυτικὰ Πρότερα; Latin: Analytica Priora) is Aristotle's work on deductive reasoning, which is known as his syllogistic. Being one of the six extant Aristotelian writings on logic and scientific method, it is part of what later Peripatetics called the Organon. Modern work on Aristotle's logic builds on the tradition started in 1951 with the establishment by Jan Łukasiewicz of a revolutionary paradigm. His approach was replaced in the early 1970s in a series of papers by John Corcoran and Timothy Smiley[1] —which inform modern translations of Prior Analytics by Robin Smith in 1989 and Gisela Striker in 2009.[2]
The term "analytics" comes from the Greek words ἀναλυτός (analytos "solvable") and ἀναλύω (analyo "to solve", literally "to loose"). However, in Aristotle's corpus, there are distinguishable differences in the meaning of ἀναλύω and its cognates. There is also the possibility that Aristotle may have borrowed his use of the word "analysis" from his teacher Plato. On the other hand, the meaning that best fits the Analytics is one derived from the study of Geometry and this meaning is very close to what Aristotle calls έπιστήμη episteme, knowing the reasoned facts. Therefore, Analysis is the process of finding the reasoned facts.[3]
Aristotle's Prior Analytics represents the first time in history when Logic is scientifically investigated. On those grounds alone, Aristotle could be considered the Father of Logic for as he himself says in Sophistical Refutations, "... When it comes to this subject, it is not the case that part had been worked out before in advance and part had not; instead, nothing existed at all."[4]
A problem in meaning arises in the study of Prior Analytics for the word "syllogism" as used by Aristotle in general does not carry the same narrow connotation as it does at present; Aristotle defines this term in a way that would apply to a wide range of valid arguments. Some scholars prefer to use the word "deduction" instead as the meaning given by Aristotle to the Greek word συλλογισμός syllogismos. At present, "syllogism" is used exclusively as the method used to reach a conclusion which is really the narrow sense in which it is used in the Prior Analytics dealing as it does with a much narrower class of arguments closely resembling the "syllogisms" of traditional logic texts: two premises followed by a conclusion each of which is a categorial sentence containing all together three terms, two extremes which appear in the conclusion and one middle term which appears in both premises but not in the conclusion. In the Analytics then, Prior Analytics is the first theoretical part dealing with the science of deduction and the Posterior Analytics is the second demonstratively practical part. Prior Analytics gives an account of deductions in general narrowed down to three basic syllogisms while Posterior Analytics deals with demonstration.[5]
In the Prior Analytics, Aristotle defines syllogism as "... A deduction in a discourse in which, certain things being supposed, something different from the things supposed results of necessity because these things are so." In modern times, this definition has led to a debate as to how the word "syllogism" should be interpreted. Scholars Jan Lukasiewicz, Józef Maria Bocheński and Günther Patzig have sided with the Protasis-Apodosis dichotomy while John Corcoran prefers to consider a syllogism as simply a deduction.[6]
In the third century AD, Alexander of Aphrodisias's commentary on the Prior Analytics is the oldest extant and one of the best of the ancient tradition and is available in the English language.[7]
In the sixth century, Boethius composed the first known Latin translation of the Prior Analytics. No Westerner between Boethius and Bernard of Utrecht is known to have read the Prior Analytics.[8] The so-called Anonymus Aurelianensis III from the second half of the twelfth century is the first extant Latin commentary, or rather fragment of a commentary.[9]
## The syllogism
The Prior Analytics represents the first formal study of logic, where logic is understood as the study of arguments. An argument is a series of true or false statements which lead to a true or false conclusion.[10] In the Prior Analytics, Aristotle identifies valid and invalid forms of arguments called syllogisms. A syllogism is an argument that consists of at least three sentences: at least two premises and a conclusion. Although Aristotles does not call them "categorical sentences," tradition does; he deals with them briefly in the Analytics and more extensively in On Interpretation.[11] Each proposition (statement that is a thought of the kind expressible by a declarative sentence)[12] of a syllogism is a categorical sentence which has a subject and a predicate connected by a verb. The usual way of connecting the subject and predicate of a categorical sentence as Aristotle does in On Interpretation is by using a linking verb e.g. P is S. However, in the Prior Analytics Aristotle rejects the usual form in favor of three of his inventions: 1) P belongs to S, 2) P is predicated of S and 3) P is said of S. Aristotle does not explain why he introduces these innovative expressions but scholars conjecture that the reason may have been that it facilitates the use of letters instead of terms avoiding the ambiguity that results in Greek when letters are used with the linking verb.[13] In his formulation of syllogistic propositions, instead of the copula ("All/some... are/are not..."), Aristotle uses the expression, "... belongs to/does not belong to all/some..." or "... is said/is not said of all/some..."[14] There are four different types of categorical sentences: universal affirmative (A), particular affirmative (I), universal negative (E) and particular negative (O).
• A - A belongs to every B
• E - A belongs to no B
• I - A belongs to some B
• O - A does not belong to some B
A method of symbolization that originated and was used in the Middle Ages greatly simplifies the study of the Prior Analytics. Following this tradition then, let:
a = belongs to every
e = belongs to no
i = belongs to some
o = does not belong to some
Categorical sentences may then be abbreviated as follows:
AaB = A belongs to every B (Every B is A)
AeB = A belongs to no B (No B is A)
AiB = A belongs to some B (Some B is A)
AoB = A does not belong to some B (Some B is not A)
From the viewpoint of modern logic, only a few types of sentences can be represented in this way.[15]
### The three figures
Depending on the position of the middle term, Aristotle divides the syllogism into three kinds: Syllogism in the first, second and third figure.[16] If the Middle Term is subject of one premise and predicate of the other, the premises are in the First Figure. If the Middle Term is predicate of both premises, the premises are in the Second Figure. If the Middle Term is subject of both premises, the premises are in the Third Figure.[17]
Symbolically, the Three Figures may be represented as follows:
First figure Second figure Third figure
Predicate — Subject Predicate — Subject Predicate — Subject
Major premise A ------------ B B ------------ A A ------------ B
Minor premise B ------------ C B ------------ C C ------------ B
Conclusion A ********** C A ********** C A ********** C
[18]
## Syllogism in the first figure
In the Prior Analytics translated by A. J. Jenkins as it appears in volume 8 of the Great Books of the Western World, Aristotle says of the First Figure: "... If A is predicated of all B, and B of all C, A must be predicated of all C."[19] In the Prior Analytics translated by Robin Smith, Aristotle says of the first figure: "... For if A is predicated of every B and B of every C, it is necessary for A to be predicated of every C."[20]
Taking a = is predicated of all = is predicated of every, and using the symbolical method used in the Middle Ages, then the first figure is simplified to:
If AaB
and BaC
then AaC.
Or what amounts to the same thing:
AaB, BaC; therefore AaC[21]
When the four syllogistic propositions, a, e, i, o are placed in the first figure, Aristotle comes up with the following valid forms of deduction for the first figure:
AaB, BaC; therefore, AaC
AeB, BaC; therefore, AeC
AaB, BiC; therefore, AiC
AeB, BiC; therefore, AoC
In the Middle Ages, for mnemonic reasons they were called respectively "Barbara", "Celarent", "Darii" and "Ferio".[22]
The difference between the first figure and the other two figures is that the syllogism of the first figure is complete while that of the second and fourth is not. This is important in Aristotle's theory of the syllogism for the first figure is axiomatic while the second and third require proof. The proof of the second and third figure always leads back to the first figure.[23]
## Syllogism in the second figure
This is what Robin Smith says in English that Aristotle said in Ancient Greek: "... If M belongs to every N but to no X, then neither will N belong to any X. For if M belongs to no X, neither does X belong to any M; but M belonged to every N; therefore, X will belong to no N (for the first figure has again come about)."[24]
The above statement can be simplified by using the symbolical method used in the Middle Ages:
If MaN
but MeX
then NeX.
For if MeX
then XeM
but MaN
therefore XeN.
When the four syllogistic propositions, a, e, i, o are placed in the second figure, Aristotle comes up with the following valid forms of deduction for the second figure:
MaN, MeX; therefore NeX
MeN, MaX; therefore NeX
MeN, MiX; therefore NoX
MaN, MoX; therefore NoX
In the Middle Ages, for mnemonic reasons they were called respectively "Camestres", "Cesare", "Festino" and "Baroco".[25]
## Syllogism in the third figure
Aristotle says in the Prior Analytics, "... If one term belongs to all and another to none of the same thing, or if they both belong to all or none of it, I call such figure the third." Referring to universal terms, "... then when both P and R belongs to every S, it results of necessity that P will belong to some R."[26]
Simplifying:
If PaS
and RaS
then PiR.
When the four syllogistic propositions, a, e, i, o are placed in the third figure, Aristotle develops six more valid forms of deduction:
PaS, RaS; therefore PiR
PeS, RaS; therefore PoR
PiS, RaS; therefore PiR
PaS, RiS; therefore PiR
PoS, RaS; therefore PoR
PeS, RiS; therefore PoR
In the Middle Ages, for mnemonic reasons, these six forms were called respectively: "Darapti", "Felapton", "Disamis", "Datisi", "Bocardo" and "Ferison".[27]
## Table of syllogisms
Figure Major premise Minor premise Conclusion Mnemonic name
First Figure AaB BaC AaC Barbara
AeB BaC AeC Celarent
AaB BiC AiC Darii
AeB BiC AoC Ferio
Second Figure MaN MeX NeX Camestres
MeN MaX NeX Cesare
MeN MiX NoX Festino
MaN MoX NoX Baroco
Third Figure PaS RaS PiR Darapti
PeS RaS PoR Felapton
PiS RaS PiR Disamis
PaS RiS PiR Datisi
PoS RaS PoR Bocardo
PeS RiS PoR Ferison
[28]
## The fourth figure
"In Aristotelian syllogistic (Prior Analytics, Bk I Caps 4-7), syllogisms are divided into three figures according to the position of the middle term in the two premises. The fourth figure, in which the middle term is the predicate in the major premise and the subject in the minor, was added by Aristotle's pupil Theophrastus and does not occur in Aristotle's work, although there is evidence that Aristotle knew of fourth-figure syllogisms."[29]
## Boole’s acceptance of Aristotle
Commentaria in Analytica priora Aristotelis, 1549
George Boole's unwavering acceptance of Aristotle’s logic is emphasized by the historian of logic John Corcoran in an accessible introduction to Laws of Thought[30] Corcoran also wrote a point-by-point comparison of Prior Analytics and Laws of Thought.[31] According to Corcoran, Boole fully accepted and endorsed Aristotle’s logic. Boole’s goals were “to go under, over, and beyond” Aristotle’s logic by 1) providing it with mathematical foundations involving equations, 2) extending the class of problems it could treat—from assessing validity to solving equations--, and 3) expanding the range of applications it could handle—e.g. from propositions having only two terms to those having arbitrarily many.
More specifically, Boole agreed with what Aristotle said; Boole’s ‘disagreements’, if they might be called that, concern what Aristotle did not say. First, in the realm of foundations, Boole reduced the four propositional forms of Aristotle's logic to formulas in the form of equations—-by itself a revolutionary idea. Second, in the realm of logic’s problems, Boole’s addition of equation solving to logic—-another revolutionary idea—-involved Boole’s doctrine that Aristotle’s rules of inference (the “perfect syllogisms”) must be supplemented by rules for equation solving. Third, in the realm of applications, Boole’s system could handle multi-term propositions and arguments whereas Aristotle could handle only two-termed subject-predicate propositions and arguments. For example, Aristotle’s system could not deduce “No quadrangle that is a square is a rectangle that is a rhombus” from “No square that is a quadrangle is a rhombus that is a rectangle” or from “No rhombus that is a rectangle is a square that is a quadrangle”.
## Notes
1. ^ "We should not let modern standard systems force us to distort our interpretations of the ancient doctrines. A good example is the Corcoran-Smiley interpretation of Aristotelian categorical syllogistic which permits us to translate the actual details of the Aristotelian exposition almost sentencewise into modern notation (Corcoran 1974a; Smiley 1973). Lukasiewicz (1957) once thought that most of Aristotle's more specific methods were inadequate because they could not be formulated in the modern systems then known. He arrived at such a formulation only by distorting Aristotle's thought to a certain degree. In this respect Corcoran's interpretation is far superior in that it is very near to the texts while being fully correct from the point of view of modern logic." Urs Egli, "Stoic Syntax and Semantics." In Jacques Brunschwig (ed.), Les Stoiciens et leur logique, Paris, Vrin, 1986, pp. 135-147 (2nd edition 2006, pp. 131-148).
2. ^ *Review of "Aristotle, Prior Analytics: Book I, Gisela Striker (translation and commentary), Oxford UP, 2009, 268pp., \$39.95 (pbk), ISBN 978-0-19-925041-7." in the Notre Dame Philosophical Reviews, 2010.02.02.
3. ^ Patrick Hugh Byrne (1997). Analysis and Science in Aristotle. SUNY Press. p. 3. ISBN 0-7914-3321-8. ... while "decompose" - the most prevalent connotation of "analyze" in the modern period — is among Aristotle's meanings, it is neither the sole meaning nor the principal meaning nor the meaning which best characterizes the work, Analytics.
4. ^ Jonathan Barnes, ed. (1995). The Cambridge Companion to Aristotle. Cambridge University Press. p. 27. ISBN 0-521-42294-9. History's first logic has also been the most influential...
5. ^ Smith, Robin (1989). Aristotle: Prior Analytics. Hackett Publishing Co. pp. XIII–XVI. ISBN 0-87220-064-7. ... This leads him to what I would regard as the most original and brilliant insight in the entire work.
6. ^ Lagerlund, Henrik (2000). Modal Syllogistics in the Middle Ages. BRILL. pp. 3–4. ISBN 978-90-04-11626-9. In the Prior Analytics Aristotle presents the first logical system, i.e., the theory of the syllogisms.
7. ^ Striker, Gisela (2009). Aristotle: Prior Analytics, Book 1. Oxford University Press. p. xx. ISBN 978-0-19-925041-7.
8. ^ R. B. C. Huygens (1997). Looking for Manuscripts... and Then?. Essays in Medieval Studies: Proceedings of the Illinois Medieval Association. 4. Illinois Medieval Association.
9. ^ Ebbesen, Sten (2008). Greek-Latin philosophical interaction. Ashgate Publishing Ltd. pp. 171–173. ISBN 978-0-7546-5837-5. Authoritative texts beget commentaries. Boethus of Sidon (late first century BC?) may have been one of the first to write one on Prior Analytics.
10. ^ Nolt, John; Rohatyn, Dennis (1988). Logic: Schaum's outline of theory and problems. McGraw Hill. p. 1. ISBN 0-07-053628-7.
11. ^ Robin Smith. Aristotle: Prior Analytics. p. XVII.
12. ^ John Nolt/Dennis Rohatyn. Logic: Schaum's Outline of Theory and Problems. pp. 274–275.
13. ^ Anagnostopoulos, Georgios (2009). A Companion to Aristotle. Wiley-Blackwell. p. 33. ISBN 978-1-4051-2223-8.
14. ^ Patzig, Günther (1969). Aristotle's theory of the syllogism. Springer. p. 49. ISBN 978-90-277-0030-8.
15. ^ The Cambridge Companion to Aristotle. pp. 34–35.
16. ^ The Cambridge Companion to Aristotle. p. 35. At the foundation of Aristotle's syllogistic is a theory of a specific class of arguments: arguments having as premises exactly two categorical sentences with one term in common.
17. ^ Robin Smith. Aristotle: Prior Analytics. p. XVIII.
18. ^ Henrik Legerlund. Modal Syllogistics in the Middle Ages. p. 4.
19. ^ Great Books of the Western World. 8. p. 40.
20. ^ Robin Smith. Aristotle: Prior Analytics. p. 4.
21. ^ The Cambridge Companion to Aristotle. p. 41.
22. ^ The Cambridge Companion to Aristotle. p. 41.
23. ^ Henrik Legerlund. Modal Syllogistics in the Middle Ages. p. 6.
24. ^ Robin Smith. Aristotle: Prior Analytics. p. 7.
25. ^ The Cambridge Companion to Aristotle. p. 41.
26. ^ Robin Smith. Aristotle: Prior Analytics. p. 9.
27. ^ The Cambridge Companion to Aristotle. p. 41.
28. ^ The Cambridge Companion to Aristotle. p. 41.
29. ^ Russell, Bertrand; Blackwell, Kenneth (1983). Cambridge essays, 1888-99. Routledge. p. 411. ISBN 978-0-04-920067-8.
30. ^ George Boole. 1854/2003. The Laws of Thought, facsimile of 1854 edition, with an introduction by J. Corcoran. Buffalo: Prometheus Books (2003). Reviewed by James van Evra in Philosophy in Review.24 (2004) 167–169.
31. ^ JOHN CORCORAN, Aristotle's Prior Analytics and Boole's Laws of Thought, History and Philosophy of Logic, vol. 24 (2003), pp. 261–288.
## Bibliography
Translations
• Aristotle, Prior Analytics, translated by Robin Smith, Indianapolis: Hackett, 1989.
• Aristotle, Prior Analytics Book I, translated by Gisela Striker, Oxford: Clarendon Press 2009.
Studies
• Corcoran, John, (ed.) 1974. Ancient Logic and its Modern Interpretations., Dordrecht: Reidel.
• Corcoran, John, 1974a. "Aristotle's Natural Deduction System". Ancient Logic and its Modern Interpretations, pp. 85-131.
• Lukasiewicz, Jan, 1957. Aristotle s Syllogistic from the Standpoint of Modern Formal Logic. 2nd edition. Oxford: Clarendon Press.
• Smiley, Timothy. 1973. "What is a Syllogism?", Journal of Philosophical Logic, 2, pp.136-154. | 4,743 | 18,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-43 | latest | en | 0.95643 |
https://forum.duet3d.com/topic/10934/5-axis-3d-printing/?page=1 | 1,576,177,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540545146.75/warc/CC-MAIN-20191212181310-20191212205310-00434.warc.gz | 363,822,566 | 21,306 | # 5-axis 3D printing
• Hello community
I am a master student from Germany currently working on my master thesis which is about getting a self-made 5-axis 3D Printer work at my University.
At the moment we struggle with the slicer software for a 5-axis print.
Is there anyone out there with interest to develop something in this direction?
Thanks a lot for your responses.
Best regards,
Harald Schmidt
• what do your 4th and 5th axis do?
• Probably rotation of the nozzle about x and y axis for conformal printing or repair onto non-planar surfaces such as rollers/cylinders.
• Hi Harald,
Some months ago, came the idea to build a 5 axis machine, mainly to print BB-8 panels, which are pure sphere portions.
My idea was to tilt the bed along X and Y. If I could have a tilt radius equal of the panel radius, the nozzle could remain static (so, it would not be a 5 axis machine!). But it is not easy to do, as the bed would collide the Z axis. So, using a shorter tilt radius would still require to move the head.
On the slicer side, my idea was to print planar parts, and have a specific kinematic to map the plane onto the sphere. This is not ideal, beause I loose precision when I go off center (or, to be more precise, the planar part does not have all the informations, as parts far from the center have to be stretched to map onto the sphere).
So, I'm very interested to follow your devs on the slicer, but I have no clue how a slicer works
PS: I didn't start to work on the mechanic yet. I had a quick look at RRF firmware, but the kinematic model for CoreXY machines has changed, recently, so I have to dig again.
PS2: tilting the head would work too, but seems more complicated...
• @veti
Hello veti,
the 4th axis is a rotation axis which rotates the Panel.
The 5th axis is to swivel the Panel.
• I currently program 5 axis vertical milling machines XYZ+A&B axis rotary trunnion A is +or- 999 degrees, B is +or- 120 degrees. I see no reason why you could not use 5 axis CNC software to generate your G code, I use both Surfcam and Edgecam. both have fully integrated customizable Post Processor generators.
• CNC and 3D printing does not uses same tools... For example, CAM softwares don't know how to manage bridges or so.
I don't think driving the 5 axis is a problem; it's more a matter of a clever slicer to take advantage of additionnal axis.
• @haraldschmidt said in 5-axis 3D printing:
getting a self-made 5-axis 3D Printer work at my University.
What's your goal? Being able to print shapes that can't be printed on a regular 3D printer? Any example?
• Look autodesk powermill additive 2019/2020 with netfabb strategies.
It should be able to handle 3 + 2 axis interpolations for print
• Hi,
I started to work on a pseudo 5 axis 3D printing stuff. This is a work in progress, but I share it so you may give me some advices
• As you can see on the wiki page, the idea is to map a portion of sphere onto the plane: this part works fine, and I can have the part sliced the usual wayยน.
Now, I need to implement the kinematic. My first idea was to post-process the G-Code file, and transform the X/Y/Z coordinates to X/Y/Z/A/B coordinates with a Python script. But it is not that simple, as the additionnal bed rotations make the kinematic non linear So, the only way is to implement that in the firmware.
Some months ago, I started to look in the CoreXY code, but I'm wondering if it would not be easier to build a 6 axis Delta or Stewart platform... Delta kinematic is already implemented, with all the segmention stuff, and adding 3 axis is maybe easier than tweeking the CoreXY code to make it segmented. A 6 axis Delta is maybe also easier to build than a 2 axis bed rotation.
David, what do you think? Would it be hard to extend the 3 axis Delta kinematic code to a 6 axis kinematic?
ยน BTW, I made some tests with my sphere_to_plan mapper, and found that slicing the portion of sphere mapped to plan uses much less filament : on my little test part, I found 3m40 vs 4m12 ; this is 17% less, and this is done without support for the sphere! If my Python code is correct, this is an argument to develop 5 axis 3D printers.
• Sorry but I don't understand what you want to do with xyzab.
At most you can exchange a linear axis with a rotary one.
I've been working on 5-axis milling machines for several years and what you say doesn't make any sense
• The goal is to print a sphere as if it was a plan. A/B axis are bed (or hotend) tilt along X/Y, and are used to ensure the hotend is always normal to the sphere at the printing point.
Once the STL file of the sphere is mapped to the plan (which works fine) and sliced, the X/Y coordinates of the G-Code describe the sphere of the original stl file. The firmware should then convert these X/Y coordinates (which the printer still see as a plan) to new X/Y/Z/A/B coordinates to re-create the sphere in the printer space.
• @fma Have you thought this through? You'll need to tilt the head, not the bed. Otherwise the part will fall off. To start the sphere, the print head will need to be tilted and positioned such that it is inside the sphere which could be difficult if the part is solid. If you start with the head outside the sphere, then the head will need to be below the bed. Half way up the sphere, you'll need to switch the head from inside to outside.
• There is no problem to tilt the bed +-35ยฐ, the part won't fall, as it is very thin compare to its width. I've seen people putting their (small) printer upside down while printing!
The bed rotations would be inside the sphere, more or less at the flat bed height (flat bed which won't exist anymore). So, the hotend have to raise while the bed tilts.
BTW, I plan to make a sphercial bed, so there is no need for supports. This bed can be 3D-printing, and by adding a raft for each print, can be re-used several times, giving very good adhesion.
• One concern with rotating the part might be that you loose (or vary) the force of gravity to help squish the layers as well, but if parts fall off then bed adhesion might be worth looking into in any case.
• I don't think gravity plays a big role in layer adhesion; it is more a matter of temperature and speed... And this part will have no strenth on it, so even if layer adhesion is bad, this is not a problem. The goal is 1) save time and filament 2) give a nice finish 3) have fun while designing the printing process
• When printing in a xy plane as you will be doing its at least a constant force, I'm just relaying some of the stuff I've read with regards to "true" 5-axis printing. | 1,612 | 6,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-51 | latest | en | 0.955704 |
https://stats.stackexchange.com/questions/325953/how-to-find-approximation-of-variance-of-ith-order-statistic?noredirect=1 | 1,713,482,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00078.warc.gz | 482,618,254 | 33,097 | # How to find approximation of variance of $i^{th}$ order statistic [duplicate]
Given PDF and CDF of a distribution, how does one find an approximation of $(\operatorname{Var}(X_i))$ using a normal approximation of $(X_i)$?
According to "Mathematica Laboratories for Mathematical Statistics" by Jenny Baglivo (Theorem 9.1 on page 120), it is said that:
$\operatorname{Var}(X_i) \approx p(1-p) / (n+2)f(x)^2$ where $f$ is PDF, $x$ is $p^\text{th}$ quantile of the $X$ distribution, and $p = i/(n+1)$
I have $0$ ideas on how we get to this result. I am absolutely drawing at a blank. Help is very much appreciated!
• Welcome to CV, blamp. The community treats questions about homework and self-study a little differently. I recommend editing your questions (use the "edit" link in the lower left) to include the [self-study] tag. Cool? Cool. Jan 30, 2018 at 22:43 | 246 | 866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.890823 |
http://woodworkingplans2013.com/medieval-furniture-plans-valet-woodworking-plans.html | 1,560,980,112,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00098.warc.gz | 187,811,182 | 6,489 | This time I surprised one of my favorite dice games and took it outside. I made a set of wooden dice in just a few hours, and instead of sitting in the room and doing nothing, we are taking our dice game out into the yard. With this set of wooden dice, dice games are becoming our favorite backyard game. Check out the step by step tutorial below so you can make your own.
These things may be tiny in size, but building one is not that easy. It takes some serious woodworking knowledge and skill to build a nice wooden mobile stand. When I first saw one online, I just couldn’t resist thinking of buying one. But when I saw the price, I was forced to rethink. Also, a woodwork lover like me cannot be contained with just one piece and I was not willing to spend on more than one. So instead I decided to build myself one. Yes, it took some doing but the final result was satisfying. Luckily, I found this awesome tutorial online that helped me build my first ever wooden phone holder.
How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
```Some types of wood filler can be hard to get off your hands after they dry, especially if you use your fingers to push it into small cracks and holes. When that happens, I reach for fine grit sandpaper and sand it off my fingers. It’s great for removing dried-on polyurethane glue and canned foam from your hands, too. — Chris James. We’ve got great solutions for removing super glue, too.
```
Some tools that are required for this project are Miter saw, drilling machine, pencil, tape measure, screws, etc. Those, who prefer a video tutorial instead, can visit the below link to a YouTube video tutorial that illustrates the process of creating a DIY Beer Bottle Crate. The video tutorial explains every step properly so that anyone can make a Beer bottle crate easily.
Not happy with the selection of sanding blocks at the hardware store, I made a few of my own from hardwood scraps left over from a woodworking project. I cut each one to 3/4 in. x 1-1/2 in. x 4-1/2 in.—which is just the right size to wrap a quarter sheet of sandpaper around. And the “kerf” cut helps hold the sandpaper in place until I’m ready to change it. —Tim Olaerts. Here are 41 more genius sanding tips you need to know.
###### With strength, sturdiness, and durability, maple is a common material for furniture for the bedroom and even china cabinets. Maple is moisture-resistant and frequently displays stand-out swirls in the wood grain, an aesthetically pleasing differentiator from other hardwoods. While most commonly a lighter color, maple also can take stains and paint well.[13]
The procedure is very easy to understand and follow for anyone with a little woodworking knowledge. Make sure to collect all the items you need before you start with the project. You may even ask Tracy your queries directly in the comment section of the tutorial post. Or you can ask them here. Either way, I hope that you manage to build this one nicely.
The shelf in the first picture is made of red oak plywood. You can choose the wood type, color and design as you like for your project. In case if you need more help understanding this project, you can refer the source link below. It discusses various items used, steps and tips and personal experience of the author who personally built a Zigzag shelf.
```Nightstand table plans have everything you need to create a bedside table to keep every needy thing at reach at night time. This Nigh Stand plan is quite different in design from the most of the other plans. This stand has not only the three regular drawers but also having a hidden drawer that uses a secret locking mechanism to keep contents securely.
```
Instead of permanently mounting my 6-in. vise to a work-bench, I attached it to scrap plywood so I can clamp it wherever I need it. Stack two pieces of 3/4-in. plywood and screw them together with 1-1/4 in. drywall screws. Mark the vise-mounting holes on the plywood and drill 3/4-in. guide holes through both pieces. Recess the nut by drilling through the bottom sheet with a 1-in. spade bit using the 3/4-in. hole as a guide. Fasten the vise to the plywood with bolts sized to match the vise-mounting holes. If the bolt shafts are too long, cut them off with a hacksaw. — LuAnn Aiu. Plus: Learn how to use vise grips to pull nails.
The procedure is very easy to understand and follow for anyone with a little woodworking knowledge. Make sure to collect all the items you need before you start with the project. You may even ask Tracy your queries directly in the comment section of the tutorial post. Or you can ask them here. Either way, I hope that you manage to build this one nicely.
Along with stone, clay and animal parts, wood was one of the first materials worked by early humans. Microwear analysis of the Mousterian stone tools used by the Neanderthals show that many were used to work wood. The development of civilization was closely tied to the development of increasingly greater degrees of skill in working these materials.
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
```Cut off a 21-in.-long board for the shelves, rip it in the middle to make two shelves, and cut 45-degree bevels on the two long front edges with a router or table saw. Bevel the ends of the other board, cut dadoes, which are grooves cut into the wood with a router or a table saw with a dado blade, cross- wise (cut a dado on scrap and test-fit the shelves first!) and cut it into four narrower boards, two at 1-3/8 in. wide and two at 4 in.
```
Although refrigerators long ago rendered them obsolete, antique oak ice boxes remain popular with collectors, even though they’re expensive and hard to find. This do-it-yourself version is neither: it’s both inexpensive and easy to build. An authentic reproduction of an original, the project is especially popular when used as a bar, but it has many
Making a garden arched footbridge out of some wood boards can be fun, hard working plan and also it’s quite rewarding. We are providing the project tutorial for how to build an arched footbridge without rails or having rails. If you take your hands of work and have some basic woodworking skills you can easily build this type of bridge. While this garden bridge is too small to walk over but it can make a really stunning addition to your lush yard or garden.
Building a Wooden Office Desk Organizer is an easy task for a professional woodworker, but not so much for normal people like you and me. But that doesn’t mean you cannot do this. Two years ago, I had almost no woodworking experience, but now I make most of my household and office wooden items by myself. This saves me a lot of money. And believe me when I say this; you can also manage to make wonderful wood items with a little practice and some woodworking experience.
```The engineering involved in building this garden bench is pretty simple, and we have provided some links to get a full cut list and plans with photos to help you along the way. Additionally, to the stock lumber, you will need wood screws, barrel locks, and hinges to complete the table. A miter saw or hand saw is also extremely helpful for cutting down your stock to the correct angle and length.
```
Do you want to use an oil stain, a gel stain, a water-based stain or a lacquer stain? What about color? Our ebook tells you what you really need to know about the chemistry behind each wood stain, and what to expect when you brush, wipe or spray it on. It’s a lot simpler than you think! This is the comprehensive guide to all the varieties of stain you will find at the store and how to use them. | 1,854 | 8,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-26 | latest | en | 0.950874 |
https://sudonull.com/post/167340-The-idea-is-Liquid-core-we-change-the-approach-to-computing | 1,611,538,973,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00174.warc.gz | 573,319,032 | 5,132 | # The idea is “Liquid core”, we change the approach to computing
The idea of a subject came the other day, I'm not sure that the topic has not been touched on before. Therefore, I apologize if I haven’t googled enough, and the topic is pulled by a bearded button accordion.
In a nutshell: we create a processor with programmable logic, as in FPGAs, but with a dynamic configuration. The program will not be a sequence of commands, but a sequence of configurations.
##### What do we have
Now all processors have a certain set of instructions. Any program, in any programming language, turns precisely into a sequence of these commands (naturally, not counting virtual platforms, for example Java). For example, in C ++ there will be only one operator for calculating the sine, but the compiler implements this calculation from many consecutive elementary instructions of the processor.
There is FPGA , which logic is configured, you can calculate the sine in one! tact. But since the logical structure is set at startup, and sometimes hard-coded into the FPGA, in the end we can do nothing more than calculate the sine.
##### What I offer
To create a processor with a core from a sufficient number (how much I can’t even imagine) of the elementary logical elements - “AND”, “NOT”, “OR” ... But without any predefined relationships between them. Communications will be carried out dynamically, and in any form - i.e. any element can be associated with any other. What do we get. We need to calculate the sine - load the appropriate configuration. Moreover, you can calculate not only mathematical functions, but anything, for example, work with regular expressions. The compiler will simply have to turn the high-level command / instruction set into the optimal configuration. As a result, the program does not turn into a set of processor instructions, but into a set of kernel configurations.
In practice, you can add such a core to the classical ones, and perform “liquid” calculations as necessary.
##### The general essence of "liquid computing"
To create not a solution algorithm on a specific computer, but to create a computer to solve a specific problem.
In general, the idea outlined. I am pleased to hear the opinion of the habr community.
Z.Y. Liquid core , liquid computing - it sounds like that! | 475 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.923258 |
http://math.stackexchange.com/questions/285579/domain-of-a-complex-function | 1,469,599,893,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00081-ip-10-185-27-174.ec2.internal.warc.gz | 165,373,143 | 17,141 | # Domain of a complex function
1) Why do the domain of a complex function has to be a disk (circular neighborhood of Zo)? $$|z-z_0|<p$$ 2) Domain is an open connected set. An open set D is said to be connected if every pair of points $z_1$,$z_2$ in S can be joined by a polygonal path that lies entirely in S.
so why are the following sets are not domains?
a) $-1<Im( z )<= 1$
b) $(Re(z))^2 >1$
Thank you
-
is (a) open? is (b) connected? – Maesumi Jan 24 '13 at 4:32
It's worth noting that in this post, "domain" is used in two different contexts. In one, it is simply the subset of the complex plane over which a given function is defined. In two, it is (as stated) an open connected set. – Cameron Buie Jan 24 '13 at 4:39
The domain of a complex function does not have to be a disk. It can be any subset of $\mathbb C$. | 250 | 828 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-30 | latest | en | 0.919606 |
https://www.roseindia.net/answers/viewqa/Java-Beginners/22113-sum.html | 1,669,478,060,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00736.warc.gz | 1,060,082,165 | 9,036 | # sum
s=135.............17
(mul of 1 to 17 odd numbers)
sum
sum a program to find the sum of the alternative diagit of it ex- no=123456 sum=1+3+5=9
sum
; sum=sum+s; if(i==n){ System.out.print(s...+"+"); } System.out.print("\nSum of the series : "+sum
sum in JTable
sum in JTable how to calculate sum from JTable's one field like total
sum of the series
sum of the series Program to sum the series 1+1/1!+1/2!+1/3!+....n
Sum of seris
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Sum of integers
Sum of integers A Java program that reads an integer value from the user and displays i) the sum of all even integers between 1 and the input value, both inclusive. ii) The sum of all odd integers between 1 and the input
ModuleNotFoundError: No module named 'sum'
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Calculate sum and Average in java
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sum of fibonacci series Write a Java program to print Fibonacci series upto n and find their sum also. 0+1+1+2+3+5+8+13+21����=sum Hi, Please see the thread Fibonacci program
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Sum of positive and negative numbers Hi if someone could help... the sum of all the positive integers and the sum of all the negative integers...("Sum of positive numbers: "+sum2); System.out.println("Sum of negative
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Hibernate criteria sum. Here, we will introduce you to about the sum() function . It returns the sum of total value of column. Syntax : Criteria... message as shown below: Hibernate: select sum(this_.fee
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Sum database colum in java
Sum database colum in java please i have two columns-col1 and col2... the database to sum the Items and display output so that it will be like... have asked. select item, sum(quqantity) from `order` group by item Here item
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Sum - JSP-Servlet
Sum Hi Guys, thank you all for support for us all. Please I want to sum up from a particular row from the database, get all the rows affected... = con.createStatement(); ResultSet res = st.executeQuery("SELECT SUM(amount) FROM
ModuleNotFoundError: No module named 'akerun-sum'
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ModuleNotFoundError: No module named 'djl-sum'
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ModuleNotFoundError: No module named 'hn_sum'
ModuleNotFoundError: No module named 'hn_sum' Hi, My Python..._sum' How to remove the ModuleNotFoundError: No module named 'hn_sum'... to install padas library. You can install hn_sum python with following command
ModuleNotFoundError: No module named 'hn_sum'
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ModuleNotFoundError: No module named 'simple-sum'
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ModuleNotFoundError: No module named 'sum-walker'
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ModuleNotFoundError: No module named 'sum_x'
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ModuleNotFoundError: No module named 'sum-py'
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ModuleNotFoundError: No module named 'sum_toon'
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ModuleNotFoundError: No module named 'sum_x'
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SQL Aggregate Sum
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Python sum dictionary values by key
Python sum dictionary values by key Hi, In python we can easily... these dictionaries into one and sum the dictionary values by key. What is the code for Python sum dictionary values by key? Thanks (adsbygoogle | 2,312 | 9,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.621738 |
https://www.kalakadu.com/2016/02/implicit-cost-is-cost-of-self-owned.html | 1,719,331,929,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00320.warc.gz | 736,380,573 | 13,396 | ## Pages
### Implicit cost is the cost of self owned resources of the production used in production process. Or estimated value of inputs supplied by owner itself.
Implicit cost is the cost of self owned resources of the production used in production process. Or estimated value of inputs supplied by owner itself.
Total cost refers to total amount of money which is incurred by a firm on production of a given amount of a good.
Total cost is the sum of total fixed cost and total variable cost.
TC = TFC + TVC or TC = AC × Q
Total fixed cost remains constant at all levels of output. It is not zero even at zero output level. Therefore, TFC curve is parallel to OX-axis.
TFC = TC – TVC or TFC = AFC × Q
Total variable cost is the cost which vary with the quantity of output produced. It is zero at zero level of output. TVC curve is parallel to TC curve.
TVC = TC – TFC or TVC = AVC × Q
Average cost is per unit of production of a commodity. It is the sum of average fixed cost and average variable cost.
AC = TC or AC = AFC + AVC Q
Average fixed cost is per unit of fixed cost of production of a commodity.
TFC
AFC = Qor AFC = AC – AVC
Per unit of variable of production of a commodity is called average variable cost.
TVC
AVC = Qor AVC = AC – AFC
MC-It refers to change in TC, due to additional unit of a commodity is produced. MC = DTC/DQ or MCn = TCn – TCn–1. But under short run, it is calculated from TVC.
TVCMCn TVCn– TCn–1 or MC
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http://cn.metamath.org/mpeuni/sgrp2rid2ex.html | 1,660,558,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00328.warc.gz | 12,005,805 | 7,772 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sgrp2rid2ex Structured version Visualization version GIF version
Theorem sgrp2rid2ex 17622
Description: A small semigroup (with two elements) with two right identities which are different. (Contributed by AV, 10-Feb-2020.)
Hypotheses
Ref Expression
mgm2nsgrp.s 𝑆 = {𝐴, 𝐵}
mgm2nsgrp.b (Base‘𝑀) = 𝑆
sgrp2nmnd.o (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵))
sgrp2nmnd.p = (+g𝑀)
Assertion
Ref Expression
sgrp2rid2ex ((♯‘𝑆) = 2 → ∃𝑥𝑆𝑧𝑆𝑦𝑆 (𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦))
Distinct variable groups: 𝑥,𝑆,𝑦 𝑥,𝐴,𝑦 𝑥,𝐵,𝑦 𝑥,𝑀 𝑥, ,𝑦 𝑧,𝐴 𝑧,𝐵 𝑧,𝑆 𝑧, ,𝑥,𝑦
Allowed substitution hints: 𝑀(𝑦,𝑧)
Proof of Theorem sgrp2rid2ex
StepHypRef Expression
1 mgm2nsgrp.s . . 3 𝑆 = {𝐴, 𝐵}
21hashprdifel 13388 . 2 ((♯‘𝑆) = 2 → (𝐴𝑆𝐵𝑆𝐴𝐵))
3 simp1 1130 . . 3 ((𝐴𝑆𝐵𝑆𝐴𝐵) → 𝐴𝑆)
4 simp2 1131 . . 3 ((𝐴𝑆𝐵𝑆𝐴𝐵) → 𝐵𝑆)
5 simpl3 1231 . . . . 5 (((𝐴𝑆𝐵𝑆𝐴𝐵) ∧ 𝑦𝑆) → 𝐴𝐵)
65ralrimiva 3115 . . . 4 ((𝐴𝑆𝐵𝑆𝐴𝐵) → ∀𝑦𝑆 𝐴𝐵)
7 mgm2nsgrp.b . . . . . . 7 (Base‘𝑀) = 𝑆
8 sgrp2nmnd.o . . . . . . 7 (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵))
9 sgrp2nmnd.p . . . . . . 7 = (+g𝑀)
101, 7, 8, 9sgrp2rid2 17621 . . . . . 6 ((𝐴𝑆𝐵𝑆) → ∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦)
11 oveq2 6801 . . . . . . . . . 10 (𝑥 = 𝐴 → (𝑦 𝑥) = (𝑦 𝐴))
1211eqeq1d 2773 . . . . . . . . 9 (𝑥 = 𝐴 → ((𝑦 𝑥) = 𝑦 ↔ (𝑦 𝐴) = 𝑦))
1312ralbidv 3135 . . . . . . . 8 (𝑥 = 𝐴 → (∀𝑦𝑆 (𝑦 𝑥) = 𝑦 ↔ ∀𝑦𝑆 (𝑦 𝐴) = 𝑦))
1413rspcv 3456 . . . . . . 7 (𝐴𝑆 → (∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦 → ∀𝑦𝑆 (𝑦 𝐴) = 𝑦))
1514adantr 466 . . . . . 6 ((𝐴𝑆𝐵𝑆) → (∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦 → ∀𝑦𝑆 (𝑦 𝐴) = 𝑦))
1610, 15mpd 15 . . . . 5 ((𝐴𝑆𝐵𝑆) → ∀𝑦𝑆 (𝑦 𝐴) = 𝑦)
17163adant3 1126 . . . 4 ((𝐴𝑆𝐵𝑆𝐴𝐵) → ∀𝑦𝑆 (𝑦 𝐴) = 𝑦)
18 oveq2 6801 . . . . . . . . . 10 (𝑥 = 𝐵 → (𝑦 𝑥) = (𝑦 𝐵))
1918eqeq1d 2773 . . . . . . . . 9 (𝑥 = 𝐵 → ((𝑦 𝑥) = 𝑦 ↔ (𝑦 𝐵) = 𝑦))
2019ralbidv 3135 . . . . . . . 8 (𝑥 = 𝐵 → (∀𝑦𝑆 (𝑦 𝑥) = 𝑦 ↔ ∀𝑦𝑆 (𝑦 𝐵) = 𝑦))
2120rspcv 3456 . . . . . . 7 (𝐵𝑆 → (∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦 → ∀𝑦𝑆 (𝑦 𝐵) = 𝑦))
2221adantl 467 . . . . . 6 ((𝐴𝑆𝐵𝑆) → (∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦 → ∀𝑦𝑆 (𝑦 𝐵) = 𝑦))
2310, 22mpd 15 . . . . 5 ((𝐴𝑆𝐵𝑆) → ∀𝑦𝑆 (𝑦 𝐵) = 𝑦)
24233adant3 1126 . . . 4 ((𝐴𝑆𝐵𝑆𝐴𝐵) → ∀𝑦𝑆 (𝑦 𝐵) = 𝑦)
25 r19.26-3 3214 . . . 4 (∀𝑦𝑆 (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦) ↔ (∀𝑦𝑆 𝐴𝐵 ∧ ∀𝑦𝑆 (𝑦 𝐴) = 𝑦 ∧ ∀𝑦𝑆 (𝑦 𝐵) = 𝑦))
266, 17, 24, 25syl3anbrc 1428 . . 3 ((𝐴𝑆𝐵𝑆𝐴𝐵) → ∀𝑦𝑆 (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦))
273, 4, 263jca 1122 . 2 ((𝐴𝑆𝐵𝑆𝐴𝐵) → (𝐴𝑆𝐵𝑆 ∧ ∀𝑦𝑆 (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦)))
28 neeq1 3005 . . . . 5 (𝑥 = 𝐴 → (𝑥𝑧𝐴𝑧))
29 biidd 252 . . . . 5 (𝑥 = 𝐴 → ((𝑦 𝑧) = 𝑦 ↔ (𝑦 𝑧) = 𝑦))
3028, 12, 293anbi123d 1547 . . . 4 (𝑥 = 𝐴 → ((𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦) ↔ (𝐴𝑧 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝑧) = 𝑦)))
3130ralbidv 3135 . . 3 (𝑥 = 𝐴 → (∀𝑦𝑆 (𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦) ↔ ∀𝑦𝑆 (𝐴𝑧 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝑧) = 𝑦)))
32 neeq2 3006 . . . . 5 (𝑧 = 𝐵 → (𝐴𝑧𝐴𝐵))
33 biidd 252 . . . . 5 (𝑧 = 𝐵 → ((𝑦 𝐴) = 𝑦 ↔ (𝑦 𝐴) = 𝑦))
34 oveq2 6801 . . . . . 6 (𝑧 = 𝐵 → (𝑦 𝑧) = (𝑦 𝐵))
3534eqeq1d 2773 . . . . 5 (𝑧 = 𝐵 → ((𝑦 𝑧) = 𝑦 ↔ (𝑦 𝐵) = 𝑦))
3632, 33, 353anbi123d 1547 . . . 4 (𝑧 = 𝐵 → ((𝐴𝑧 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝑧) = 𝑦) ↔ (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦)))
3736ralbidv 3135 . . 3 (𝑧 = 𝐵 → (∀𝑦𝑆 (𝐴𝑧 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝑧) = 𝑦) ↔ ∀𝑦𝑆 (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦)))
3831, 37rspc2ev 3474 . 2 ((𝐴𝑆𝐵𝑆 ∧ ∀𝑦𝑆 (𝐴𝐵 ∧ (𝑦 𝐴) = 𝑦 ∧ (𝑦 𝐵) = 𝑦)) → ∃𝑥𝑆𝑧𝑆𝑦𝑆 (𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦))
392, 27, 383syl 18 1 ((♯‘𝑆) = 2 → ∃𝑥𝑆𝑧𝑆𝑦𝑆 (𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 382 ∧ w3a 1071 = wceq 1631 ∈ wcel 2145 ≠ wne 2943 ∀wral 3061 ∃wrex 3062 ifcif 4225 {cpr 4318 ‘cfv 6031 (class class class)co 6793 ↦ cmpt2 6795 2c2 11272 ♯chash 13321 Basecbs 16064 +gcplusg 16149 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-8 2147 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-rep 4904 ax-sep 4915 ax-nul 4923 ax-pow 4974 ax-pr 5034 ax-un 7096 ax-cnex 10194 ax-resscn 10195 ax-1cn 10196 ax-icn 10197 ax-addcl 10198 ax-addrcl 10199 ax-mulcl 10200 ax-mulrcl 10201 ax-mulcom 10202 ax-addass 10203 ax-mulass 10204 ax-distr 10205 ax-i2m1 10206 ax-1ne0 10207 ax-1rid 10208 ax-rnegex 10209 ax-rrecex 10210 ax-cnre 10211 ax-pre-lttri 10212 ax-pre-lttrn 10213 ax-pre-ltadd 10214 ax-pre-mulgt0 10215 This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-3or 1072 df-3an 1073 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-mo 2623 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ne 2944 df-nel 3047 df-ral 3066 df-rex 3067 df-reu 3068 df-rmo 3069 df-rab 3070 df-v 3353 df-sbc 3588 df-csb 3683 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-pss 3739 df-nul 4064 df-if 4226 df-pw 4299 df-sn 4317 df-pr 4319 df-tp 4321 df-op 4323 df-uni 4575 df-int 4612 df-iun 4656 df-br 4787 df-opab 4847 df-mpt 4864 df-tr 4887 df-id 5157 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-pred 5823 df-ord 5869 df-on 5870 df-lim 5871 df-suc 5872 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-riota 6754 df-ov 6796 df-oprab 6797 df-mpt2 6798 df-om 7213 df-1st 7315 df-2nd 7316 df-wrecs 7559 df-recs 7621 df-rdg 7659 df-1o 7713 df-oadd 7717 df-er 7896 df-en 8110 df-dom 8111 df-sdom 8112 df-fin 8113 df-card 8965 df-cda 9192 df-pnf 10278 df-mnf 10279 df-xr 10280 df-ltxr 10281 df-le 10282 df-sub 10470 df-neg 10471 df-nn 11223 df-2 11281 df-n0 11495 df-z 11580 df-uz 11889 df-fz 12534 df-hash 13322 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 4,171 | 5,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-33 | latest | en | 0.202072 |
https://ukm.pure.elsevier.com/en/publications/similarity-solutions-for-the-stagnation-point-flow-and-heat-trans | 1,582,658,873,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00459.warc.gz | 606,771,673 | 7,429 | # Similarity solutions for the stagnation-point flow and heat transfer over a nonlinearly stretching/shrinking sheet
Norfifah Bachok, Anuar Mohd Ishak
Research output: Contribution to journalArticle
9 Citations (Scopus)
### Abstract
This paper presents a numerical analysis of a stagnation-point flow towards a nonlinearly stretching/shrinking sheet immersed in a viscous fluid. The stretching/shrinking velocity and the external flow velocity impinges normal to the stretching/shrinking sheet are assumed to be in the form U ∼ x m, where m is a constant and x is the distance from the stagnation point. The governing partial differential equations are converted into ordinary ones by a similarity transformation, before being solved numerically. The variations of the skin friction coefficient and the heat transfer rate at the surface with the governing parameters are graphed and tabulated. Different from a stretching sheet, it is found that the solutions for a shrinking sheet are non-unique for m > 1/3.
Original language English 1297-1300 4 Sains Malaysiana 40 11 Published - Nov 2011
### Fingerprint
Stretching
Heat transfer
Skin friction
Flow velocity
Partial differential equations
Numerical analysis
Fluids
### Keywords
• Boundary layer
• Dual solutions
• Nonlinear stretching/shrinking
• Similarity solution
• General
### Cite this
In: Sains Malaysiana, Vol. 40, No. 11, 11.2011, p. 1297-1300.
Research output: Contribution to journalArticle
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title = "Similarity solutions for the stagnation-point flow and heat transfer over a nonlinearly stretching/shrinking sheet",
abstract = "This paper presents a numerical analysis of a stagnation-point flow towards a nonlinearly stretching/shrinking sheet immersed in a viscous fluid. The stretching/shrinking velocity and the external flow velocity impinges normal to the stretching/shrinking sheet are assumed to be in the form U ∼ x m, where m is a constant and x is the distance from the stagnation point. The governing partial differential equations are converted into ordinary ones by a similarity transformation, before being solved numerically. The variations of the skin friction coefficient and the heat transfer rate at the surface with the governing parameters are graphed and tabulated. Different from a stretching sheet, it is found that the solutions for a shrinking sheet are non-unique for m > 1/3.",
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N2 - This paper presents a numerical analysis of a stagnation-point flow towards a nonlinearly stretching/shrinking sheet immersed in a viscous fluid. The stretching/shrinking velocity and the external flow velocity impinges normal to the stretching/shrinking sheet are assumed to be in the form U ∼ x m, where m is a constant and x is the distance from the stagnation point. The governing partial differential equations are converted into ordinary ones by a similarity transformation, before being solved numerically. The variations of the skin friction coefficient and the heat transfer rate at the surface with the governing parameters are graphed and tabulated. Different from a stretching sheet, it is found that the solutions for a shrinking sheet are non-unique for m > 1/3.
AB - This paper presents a numerical analysis of a stagnation-point flow towards a nonlinearly stretching/shrinking sheet immersed in a viscous fluid. The stretching/shrinking velocity and the external flow velocity impinges normal to the stretching/shrinking sheet are assumed to be in the form U ∼ x m, where m is a constant and x is the distance from the stagnation point. The governing partial differential equations are converted into ordinary ones by a similarity transformation, before being solved numerically. The variations of the skin friction coefficient and the heat transfer rate at the surface with the governing parameters are graphed and tabulated. Different from a stretching sheet, it is found that the solutions for a shrinking sheet are non-unique for m > 1/3.
KW - Boundary layer
KW - Dual solutions
KW - Nonlinear stretching/shrinking
KW - Similarity solution
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https://easyhsc.com.au/easyphys/advanced-mechanics-2/circular-motion/investigate-the-relationship-between-the-rotation-of-mechanical-systems-and-the-applied-torque/ | 1,709,074,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00499.warc.gz | 218,150,909 | 19,510 | # Investigate the relationship between the rotation of mechanical systems and the applied torque
Torque : It is a measure of the force that can cause an object to rotate about an axis.
It is also known as moment or moment of force.
• As you can see the force F can be broken up into two components.
• Fcosθ, which is parallel to the rod and cannot rotate the rod.
• Fsin θ, which is perpendicular to the rod and contributes to the rotation.
• r is the radial distance from the axis of rotation to the point of application of the force
• As the r is increased , the moment or torque produced increased.
Thus torque produced is
Where ‘’ denotes the component of the force which is perpendicular to the radial distance (of rod)
Extract from Physics Stage 6 Syllabus © 2017 NSW Education Standards Authority (NESA) | 187 | 817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | longest | en | 0.903541 |
http://forum.arduino.cc/index.php?topic=133319.0;prev_next=prev | 1,472,317,740,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982924605.44/warc/CC-MAIN-20160823200844-00154-ip-10-153-172-175.ec2.internal.warc.gz | 87,449,825 | 12,162 | Go Down
Topic: Re: Random() (Read 2651 times)previous topic - next topic
Docedison
#15
Nov 18, 2012, 11:43 pm
dhenry doesn't need facts, he has an opinion about everything and he IS NEVER wrong.
Bob
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard
NV1T
#16
Nov 19, 2012, 11:19 pm
If you want true randomness, try biasing a diode near the "knee" of the I-V curve (around 0.7V for a silicon diode). The current will fluctuate randomly
due to thermal noise.
73,
NV1T
dhenry
#17
Nov 19, 2012, 11:22 pm
You already have that diode, in some mcus with a temperature sensor;
But the thermal noise is too small to produce a truly random figure by the onboard adc.
NV1T
#18
Nov 20, 2012, 03:14 am
Yeah, you might need to amplify it with an op-amp circuit. I built one of these many moons ago. Y'all should read this informative article:
http://en.wikipedia.org/wiki/Hardware_random_number_generator
73,
NV1T
dhenry
#19
Nov 20, 2012, 03:28 am
There is an easier way: reverse connect a npn to form an esaki diode. Power that up via a capacitor + a low. It will form a flasher. The period of its flashes is random and can be easily measured by a timer.
NV1T
#20
Nov 20, 2012, 03:36 am
Yeah, you can do it with an Esaki (aka tunnel) diode, or an avalanche diode, or a Zener diode, or...
Always more than one way to skin a cat.
However, for iluvplanes' application, sounds like a simple random permutation generator would be sufficient. I don't think he's really looking for
a cryptographically strong RNG.
73,
NV1T
Docedison
#21
Nov 20, 2012, 07:59 am
No I don't either But *dhenry* Loves to talk. He is one of the most popular people here.
Bob
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard
Nick Gammon
#22
Nov 20, 2012, 08:19 am
Enough insults, thanks.
Please post technical questions on the forum, not by personal message. Thanks!
http://www.gammon.com.au/electronics
dhenry
#23
Nov 20, 2012, 12:55 pm
Quote
Yeah, you can do it with an Esaki (aka tunnel) diode, or an avalanche diode, or a Zener diode, or...
Zenre is tough. Avalanche slightly easier but it oscillates at too high a frequency to be measured by a mcu.
An oscillator with lots of phase noise (a relaxation oscillator for example) would be ideal for this.
Quote
However, for iluvplanes' application, sounds like a simple random permutation generator would be sufficient. I don't think he's really looking for
a cryptographically strong RNG.
True.
dhenry
#24
Nov 21, 2012, 12:21 am
As expected, here is the output from a rc oscillator (100k+1n) on a hc132: 6000 data points.
Looks pretty random to me.
dhenry
#25
Nov 21, 2012, 12:29 am
This one doesn't use a timer. Instead it counts a variable in ram.
5000 data points.
Code: [Select]
`//generate a random seedunsigned char rseed2(void) { unsigned char mask = 0x80; unsigned char tmp = 0; unsigned char cnt=0; //tmr0_init(); //reset the tmr do { cnt = 0; //reset the counter while (digitalRead(IN_PIN)==0) continue; //wait for the pin to go high //TCCR2B = 0x01; //start the timer, 1:1 prescaler while (digitalRead(IN_PIN)==1) cnt+=1; //wait for the pin to go low //TCCR2B = 0x00; //stop the timer if (cnt & 0x01) tmp |= mask; mask = mask >> 1; } while (mask); return tmp;}`
Nick Gammon
#26
Nov 21, 2012, 12:33 am
Quote
Looks pretty random to me.
I think you need to run that through a proper mathematical analysis. "Looking" random isn't really enough. It could just be evenly distributed.
Please post technical questions on the forum, not by personal message. Thanks!
http://www.gammon.com.au/electronics
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Please enter a valid email to subscribe | 1,155 | 3,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-36 | longest | en | 0.886762 |
https://leanprover-community.github.io/mathlib_docs/analysis/normed_space/add_torsor.html | 1,709,030,147,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00824.warc.gz | 359,414,017 | 45,275 | # Torsors of normed space actions. #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
theorem affine_subspace.is_closed_direction_iff {W : Type u_4} {Q : Type u_5} [metric_space Q] [ Q] {𝕜 : Type u_6} [normed_field 𝕜] [ W] (s : Q) :
@[simp]
theorem dist_center_homothety {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( c) p₂) = c * p₂
@[simp]
theorem nndist_center_homothety {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( c) p₂) = c‖₊ * p₂
@[simp]
theorem dist_homothety_center {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_dist.dist ( c) p₂) p₁ = c * p₂
@[simp]
theorem nndist_homothety_center {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_nndist.nndist ( c) p₂) p₁ = c‖₊ * p₂
@[simp]
theorem dist_line_map_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c₁ c₂ : 𝕜) :
has_dist.dist ( p₂) c₁) ( p₂) c₂) = c₂ * p₂
@[simp]
theorem nndist_line_map_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c₁ c₂ : 𝕜) :
has_nndist.nndist ( p₂) c₁) ( p₂) c₂) = c₂ * p₂
theorem lipschitz_with_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) :
lipschitz_with p₂) p₂)
@[simp]
theorem dist_line_map_left {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_dist.dist ( p₂) c) p₁ = c * p₂
@[simp]
theorem nndist_line_map_left {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_nndist.nndist ( p₂) c) p₁ = c‖₊ * p₂
@[simp]
theorem dist_left_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( p₂) c) = c * p₂
@[simp]
theorem nndist_left_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( p₂) c) = c‖₊ * p₂
@[simp]
theorem dist_line_map_right {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_dist.dist ( p₂) c) p₂ = 1 - c * p₂
@[simp]
theorem nndist_line_map_right {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_nndist.nndist ( p₂) c) p₂ = 1 - c‖₊ * p₂
@[simp]
theorem dist_right_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( p₂) c) = 1 - c * p₂
@[simp]
theorem nndist_right_line_map {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( p₂) c) = 1 - c‖₊ * p₂
@[simp]
theorem dist_homothety_self {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_dist.dist ( c) p₂) p₂ = 1 - c * p₂
@[simp]
theorem nndist_homothety_self {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
has_nndist.nndist ( c) p₂) p₂ = 1 - c‖₊ * p₂
@[simp]
theorem dist_self_homothety {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( c) p₂) = 1 - c * p₂
@[simp]
theorem nndist_self_homothety {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] (p₁ p₂ : P) (c : 𝕜) :
( c) p₂) = 1 - c‖₊ * p₂
@[simp]
theorem dist_left_midpoint {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
(midpoint 𝕜 p₁ p₂) = 2⁻¹ * p₂
@[simp]
theorem nndist_left_midpoint {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
(midpoint 𝕜 p₁ p₂) = 2‖₊⁻¹ * p₂
@[simp]
theorem dist_midpoint_left {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
has_dist.dist (midpoint 𝕜 p₁ p₂) p₁ = 2⁻¹ * p₂
@[simp]
theorem nndist_midpoint_left {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
has_nndist.nndist (midpoint 𝕜 p₁ p₂) p₁ = 2‖₊⁻¹ * p₂
@[simp]
theorem dist_midpoint_right {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
has_dist.dist (midpoint 𝕜 p₁ p₂) p₂ = 2⁻¹ * p₂
@[simp]
theorem nndist_midpoint_right {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
has_nndist.nndist (midpoint 𝕜 p₁ p₂) p₂ = 2‖₊⁻¹ * p₂
@[simp]
theorem dist_right_midpoint {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
(midpoint 𝕜 p₁ p₂) = 2⁻¹ * p₂
@[simp]
theorem nndist_right_midpoint {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ : P) :
(midpoint 𝕜 p₁ p₂) = 2‖₊⁻¹ * p₂
theorem dist_midpoint_midpoint_le' {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ p₃ p₄ : P) :
has_dist.dist (midpoint 𝕜 p₁ p₂) (midpoint 𝕜 p₃ p₄) p₃ + p₄) / 2
theorem nndist_midpoint_midpoint_le' {V : Type u_2} {P : Type u_3} [ P] {𝕜 : Type u_6} [normed_field 𝕜] [ V] [invertible 2] (p₁ p₂ p₃ p₄ : P) :
has_nndist.nndist (midpoint 𝕜 p₁ p₂) (midpoint 𝕜 p₃ p₄) p₃ + p₄) / 2‖₊
theorem antilipschitz_with_line_map {W : Type u_4} {Q : Type u_5} [metric_space Q] [ Q] {𝕜 : Type u_6} [normed_field 𝕜] [ W] {p₁ p₂ : Q} (h : p₁ p₂) :
p₂)
theorem eventually_homothety_mem_of_mem_interior {W : Type u_4} {Q : Type u_5} [metric_space Q] [ Q] (𝕜 : Type u_6) [normed_field 𝕜] [ W] (x : Q) {s : set Q} {y : Q} (hy : y ) :
∀ᶠ (δ : 𝕜) in nhds 1, δ) y s
theorem eventually_homothety_image_subset_of_finite_subset_interior {W : Type u_4} {Q : Type u_5} [metric_space Q] [ Q] (𝕜 : Type u_6) [normed_field 𝕜] [ W] (x : Q) {s t : set Q} (ht : t.finite) (h : t ) :
∀ᶠ (δ : 𝕜) in nhds 1, δ) '' t s
theorem dist_midpoint_midpoint_le {V : Type u_2} [ V] (p₁ p₂ p₃ p₄ : V) :
has_dist.dist p₁ p₂) p₃ p₄) p₃ + p₄) / 2
theorem nndist_midpoint_midpoint_le {V : Type u_2} [ V] (p₁ p₂ p₃ p₄ : V) :
has_nndist.nndist p₁ p₂) p₃ p₄) p₃ + p₄) / 2
noncomputable def affine_map.of_map_midpoint {V : Type u_2} {P : Type u_3} {W : Type u_4} {Q : Type u_5} [ P] [metric_space Q] [ Q] [ V] [ W] (f : P Q) (h : (x y : P), f x y) = (f x) (f y)) (hfc : continuous f) :
A continuous map between two normed affine spaces is an affine map provided that it sends midpoints to midpoints.
Equations | 3,204 | 6,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.395445 |
http://mathhelpforum.com/algebra/169109-simplify-expression-involving-fractional-exponents-factoring.html | 1,524,807,130,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00544.warc.gz | 203,732,358 | 11,141 | # Thread: Simplify an expression involving fractional exponents by factoring
1. ## Simplify an expression involving fractional exponents by factoring
This one's driving me up the wall. It's not for homework, but I'm re-entering the math world after a liberal arts stint and am trying to be a perfectionist about it.
The expression: $\displaystyle (x^3 + 1)^2 (x - 1)^(-1/2) + 2x (x - 1)^(1/2) (x^2+1)$
(the -1/2 and 1/2 following the raised left parentheses are both supposed to be fractional exponents)
The book's answer: $\displaystyle [(x^2 + 1) (3x^2 - 2x + 1)] / sqrt(x-1)$
I'd just scan and post an image of the pages I've devoted solely to this problem if they were at all legible. Instead, I'll summarize my approaches to this problem:
I've tried putting (x-1)^1/2 in the denominator, then factoring everything out. I've tried putting that in the denominator, then adding like bases and factoring it out. I've tried just adding like bases to begin with, which totally prevented me from putting (x-1)^1/2 in the denominator. I've also tried squaring the whole problem to get rid of any (1/2) exponent and factoring it out. I even bought the Bagatrix "Algebra Solved!" software and had it try to simplify the problem, yet its answer doesn't match my books; its process matches my factoring process. I've double checked whether I've copied the initial problem correctly and I have.
P.S. Sorry about the formatting. Just blame my n00bhood.
2. Originally Posted by wolfe
This one's driving me up the wall. It's not for homework, but I'm re-entering the math world after a liberal arts stint and am trying to be a perfectionist about it.
The expression: $\displaystyle (x^3 + 1)^2 (x - 1)^(-1/2) + 2x (x - 1)^(1/2) (x^2+1)$
(the -1/2 and 1/2 following the raised left parentheses are both supposed to be fractional exponents)
The book's answer: $\displaystyle [(x^2 + 1) (3x^2 - 2x + 1)] / sqrt(x-1)$
I'd just scan and post an image of the pages I've devoted solely to this problem if they were at all legible. Instead, I'll summarize my approaches to this problem:
I've tried putting (x-1)^1/2 in the denominator, then factoring everything out. I've tried putting that in the denominator, then adding like bases and factoring it out. I've tried just adding like bases to begin with, which totally prevented me from putting (x-1)^1/2 in the denominator. I've also tried squaring the whole problem to get rid of any (1/2) exponent and factoring it out. I even bought the Bagatrix "Algebra Solved!" software and had it try to simplify the problem, yet its answer doesn't match my books; its process matches my factoring process. I've double checked whether I've copied the initial problem correctly and I have.
P.S. Sorry about the formatting. Just blame my n00bhood.
the original expression is ...
$\displaystyle (x^2 + 1)^2 (x - 1)^{-1/2} + 2x (x - 1)^{1/2} (x^2+1)$
... not an $\displaystyle x^3$ in the first factor.
common factors are $\displaystyle (x^2+1)$ and $\displaystyle (x-1)^{-1/2}$
$\displaystyle (x^2+1)(x-1)^{-1/2}[(x^2+1) + 2x(x-1)]$
$\displaystyle (x^2+1)(x-1)^{-1/2}[x^2+1 + 2x^2-2x]$
$\displaystyle (x^2+1)(x-1)^{-1/2}[3x^2-2x+1]$
$\displaystyle \displaystyle \frac{(x^2+1)(3x^2-2x+1)}{\sqrt{x-1}}$
3. Originally Posted by skeeter
the original expression is ...
$\displaystyle (x^2 + 1)^2 (x - 1)^{-1/2} + 2x (x - 1)^{1/2} (x^2+1)$
... not an $\displaystyle x^3$ in the first factor.
common factors are $\displaystyle (x^2+1)$ and $\displaystyle (x-1)^{-1/2}$
$\displaystyle (x^2+1)(x-1)^{-1/2}[(x^2+1) + 2x(x-1)]$
$\displaystyle (x^2+1)(x-1)^{-1/2}[x^2+1 + 2x^2-2x]$
$\displaystyle (x^2+1)(x-1)^{-1/2}[3x^2-2x+1]$
$\displaystyle \displaystyle \frac{(x^2+1)(3x^2-2x+1)}{\sqrt{x-1}}$
Oh sorry for the typo. You obviously know your stuff to have picked up on it!
The math makes beautiful sense to me once you rearranged it that first time. However, I'm having trouble understanding what's happening with all the motion. Particularly, where do the exponents 2 and 1/2 go in $\displaystyle (x^2 + 1)^2$ and $\displaystyle 2x(x-1)^{-1/2}$?
EDIT: I think I can reason that the 1/2 'vanishes' since, working backwards with the multiplication of exponents rule.. $\displaystyle (x-1)^{-1/2} * (x-1) = (x-1)^{-1/2} * (x-1)^{2/2} = (x-1)^{(-1/2) + (2/2)} = (x-1)^{1/2}$ .. however, it seems like in the original expression there are essentially 3 powers of $\displaystyle (x^2 + 1)$ (what with it being squared in the left side and multiplied later on the right).. and working backwards from your first rearrangement, I can only account for two powers (just the squared portion on the left side of the expression). My apologies if I'm being too deeply confusing.
4. Originally Posted by wolfe
Oh sorry for the typo. You obviously know your stuff to have picked up on it!
The math makes beautiful sense to me once you rearranged it that first time. However, I'm having trouble understanding what's happening with all the motion. Particularly, where do the exponents 2 and 1/2 go in $\displaystyle (x^2 + 1)^2$ and $\displaystyle 2x(x-1)^{1/2}$ ... fixed the last exponent
note that $\displaystyle (x^2+1)^2 = (x^2+1)(x^2+1)$
and $\displaystyle (x-1)^{1/2} = (x-1)^{-1/2} (x-1)$
5. OH. Hahaha. It just clicked. I rewrote that second portion of your advice as a radical and saw the light. Thanks for your patience skeeter. I'm officially in love with this site. | 1,694 | 5,405 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-17 | latest | en | 0.951837 |
https://www.tutorpace.com/algebra/absolute-value-function-definition-online-tutoring | 1,550,579,596,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490107.12/warc/CC-MAIN-20190219122312-20190219144312-00432.warc.gz | 1,018,882,713 | 10,777 | # Absolute value Function Definition
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Definition: -By absolute value function of a real number we mean its numerical value.
For example: - Suppose a function is f (a) = | 4 a |
When a = 1 then f (1 ) = | 4 (1) | = 4
When a = - 1 then f (- 1) = | 4 (- 1) | = | - 4 | = 4
Other example: - If f (x) = | x | + 2, find the value of the function at x = 5 and x = - 5.
Solution: - f ( x) = | x | + 2
When x = 5 then f ( 5 ) = | 5 | + 2 = 5 + 2 = 7
When x = - 5 then f ( - 5 ) = | - 5 | + 2 = 5 + 2 = 7
Properties of absolute value function: -
If x and y are two real numbers then
i) | x | = | - x |
For example, if x = 2 then | x | = | 2 | = 2 and | - x | | - 2 | = 2.
Hence | x | = | - x |
ii) | x – y | = | y – x |
iii) | x – y | = x – y when x > y
| x – y | = y – x when x < y | 358 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-09 | longest | en | 0.558319 |
https://www.abebooks.it/9780415526449/Structural-Analysis-Principles-Methods-Modelling-0415526442/plp | 1,477,002,025,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717954.1/warc/CC-MAIN-20161020183837-00077-ip-10-171-6-4.ec2.internal.warc.gz | 903,487,914 | 17,528 | # Structural Analysis: Principles, Methods and Modelling
## Ranzi, Gianluca; Gilbert, Raymond Ian
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( su 0 valutazioni fornite da GoodReads )
Provides Step-by-Step Instruction
Structural Analysis: Principles, Methods and Modelling outlines the fundamentals involved in analyzing engineering structures, and effectively presents the derivations used for analytical and numerical formulations. This text explains practical and relevant concepts, and lays down the foundation for a solid mathematical background that incorporates MATLAB® (no prior knowledge of MATLAB is necessary), and includes numerous worked examples.
Effectively Analyze Engineering Structures
Divided into four parts, the text focuses on the analysis of statically determinate structures. It evaluates basic concepts and procedures, examines the classical methods for the analysis of statically indeterminate structures, and explores the stiffness method of analysis that reinforces most computer applications and commercially available structural analysis software. In addition, it covers advanced topics that include the finite element method, structural stability, and problems involving material nonlinearity.
MATLAB® files for selected worked examples are available from the book’s website. Resources available from CRC Press for lecturers adopting the book include:
• A solutions manual for all the problems posed in the book
• Nearly 2000 PowerPoint presentations suitable for use in lectures for each chapter in the book
• Revision videos of selected lectures with added narration
• Figure slides
Structural Analysis: Principles, Methods and Modelling exposes civil and structural engineering undergraduates to the essentials of structural analysis, and serves as a resource for students and practicing professionals in solving a range of engineering problems.
Le informazioni nella sezione "Riassunto" possono far riferimento a edizioni diverse di questo titolo.
Book Description:
"This book gives a good in-depth explanation of the fundamental principles of structural analysis. Topics are dealt with in considerable detail and illustrated with copious examples."
––Dr Robert Vollum, Department of Civil & Environmental Engineering Imperial College London, United Kingdom
"… explains very well and in simple terms topics which are often perceived by young students to be complicated and confusing, without sacrificing the formal mathematical treatment of the subject. … will also serve as a reference for all those practitioners who would like to revisit or gain deeper insight into the theoretical basis of the main calculation methods nowadays adopted for the design of structures."
?Massimiliano Bocciarelli, Politecnico di Milano
"… presents in a comprehensive way topics of structural analysis that are basic for civil and building engineers. The authors bring students toward a deep understanding of difficult issues in a very "natural" way. Final chapters, which introduce advanced analysis tools as the finite element method and issues like stability and plasticity of structures, give a clear perception of the behaviour complexity of a real structure. MATLAB tools allow facilitating and multiplying the experiences necessary to develop an intuitive approach to the structural design."
?Graziano Leoni, University of Camerino, Italy
Gianluca Ranzi is an associate professor and the director of the Centre for Advanced Structural Engineering at the University of Sydney, specializing in the analysis and design of concrete and composite steel-concrete structures.
Raymond Ian Gilbert is an emeritus professor at the University of New South Wales. He has over 35 years’ experience in teaching structural analysis and design and is a specialist in the analysis and design of reinforced and prestressed concrete structures.
.
Le informazioni nella sezione "Su questo libro" possono far riferimento a edizioni diverse di questo titolo.
## 1.Structural Analysis Principles, Methods and Modelling
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Descrizione libro Taylor Francis Ltd, United Kingdom, 2014. Paperback. Condizione libro: New. 252 x 178 mm. Language: English . Brand New Book. Provides Step-by-Step Instruction Structural Analysis: Principles, Methods and Modelling outlines the fundamentals involved in analyzing engineering structures, and effectively presents the derivations used for analytical and numerical formulations. This text explains practical and relevant concepts, and lays down the foundation for a solid mathematical background that incorporates MATLAB(R) (no prior knowledge of MATLAB is necessary), and includes numerous worked examples. Effectively Analyze Engineering Structures Divided into four parts, the text focuses on the analysis of statically determinate structures. It evaluates basic concepts and procedures, examines the classical methods for the analysis of statically indeterminate structures, and explores the stiffness method of analysis that reinforces most computer applications and commercially available structural analysis software. In addition, it covers advanced topics that include the finite element method, structural stability, and problems involving material nonlinearity. MATLAB(R) files for selected worked examples are available from the book s website. Resources available from CRC Press for lecturers adopting the book include: * A solutions manual for all the problems posed in the book * Nearly 2000 PowerPoint presentations suitable for use in lectures for each chapter in the book * Revision videos of selected lectures with added narration * Figure slides Structural Analysis: Principles, Methods and Modelling exposes civil and structural engineering undergraduates to the essentials of structural analysis, and serves as a resource for students and practicing professionals in solving a range of engineering problems. Codice libro della libreria AA69780415526449
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Descrizione libro Taylor Francis Ltd, United Kingdom, 2014. Paperback. Condizione libro: New. 252 x 178 mm. Language: English . Brand New Book. Provides Step-by-Step Instruction Structural Analysis: Principles, Methods and Modelling outlines the fundamentals involved in analyzing engineering structures, and effectively presents the derivations used for analytical and numerical formulations. This text explains practical and relevant concepts, and lays down the foundation for a solid mathematical background that incorporates MATLAB(R) (no prior knowledge of MATLAB is necessary), and includes numerous worked examples. Effectively Analyze Engineering Structures Divided into four parts, the text focuses on the analysis of statically determinate structures. It evaluates basic concepts and procedures, examines the classical methods for the analysis of statically indeterminate structures, and explores the stiffness method of analysis that reinforces most computer applications and commercially available structural analysis software. In addition, it covers advanced topics that include the finite element method, structural stability, and problems involving material nonlinearity. MATLAB(R) files for selected worked examples are available from the book s website. Resources available from CRC Press for lecturers adopting the book include: * A solutions manual for all the problems posed in the book * Nearly 2000 PowerPoint presentations suitable for use in lectures for each chapter in the book * Revision videos of selected lectures with added narration * Figure slides Structural Analysis: Principles, Methods and Modelling exposes civil and structural engineering undergraduates to the essentials of structural analysis, and serves as a resource for students and practicing professionals in solving a range of engineering problems. Codice libro della libreria AA69780415526449
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Destinazione, tempi e costi | 2,831 | 12,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-44 | longest | en | 0.83673 |
https://www.enotes.com/homework-help/find-volume-solid-first-octant-bounded-by-graphs-z-323782 | 1,669,756,680,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00224.warc.gz | 798,656,747 | 18,201 | # Find the volume of the solid in the first octant bounded by the graphs z=1-(y^2), y=2x and x=3.
You need to use triple integral to evaluate the volume of solid such that:
`dV = dxdydz`
Notice that the solid is in first octant, hence, `xgt0, ygt0,zgt0` . Since `zgt0 and ygt0 =gt z = y^2 - 1 gt 0 `
`V = int_0^3 dx int_0^(2x) dy int_0^(y^2-1)dz`
You need to evaluate the inner integral `int_0^(y^2-1)dz` such that:
`int_0^(y^2-1)dz = z |_0^(y^2-1)`
`int_0^(y^2-1)dz = y^2-1`
You need to evaluate the middle integral such that:
`int_0^(2x) (y^2-1) dy = int_0^(2x) y^2dy - int_0^(2x) dy`
`int_0^(2x) (y^2-1) dy = (y^3/3 - y)|_0^(2x)`
`int_0^(2x) (y^2-1) dy = (8x^3)/3 - 2x`
You need to evaluate the outer integral such that:
`int_0^3 ((8x^3)/3 - 2x)dx =int_0^3 (8x^3)/3 dx - int_0^3 2xdx`
`int_0^3 ((8x^3)/3 - 2x) dx = ((8/3)(x^4/4) - x^2)|_0^3`
`int_0^3 ((8x^3)/3 - 2x) dx = (2x^4/3 - x^2)|_0^3`
`int_0^3 ((8x^3)/3 - 2x) dx = 2*27 - 9 = 45`
Hence, evaluating the volume of solid in the first octant yields V = 45.
Approved by eNotes Editorial Team | 502 | 1,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-49 | latest | en | 0.632416 |
http://tutor1.net/wikibooks/175984 | 1,603,193,002,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00415.warc.gz | 108,898,369 | 4,344 | [<< wikibooks] Aeroacoustics/Wave Equation and Green's function
We will show later that the propagation of acoustic waves (small amplitude fluctuations) in fluids is basically governed by the celebrated Wave equation. Therefore, in this chapter we will review some mathematical aspects of the wave equation which will help us to understand the physics in later chapters. It must be noted that we consider the wave propagation is happening in a three dimensional space, unless otherwise stated.
== Wave Equation ==
Wave equation is the simplest, linear, hyperbolic partial differential equation [1] which governs the linear propagation of waves, with finite speed, in media. Consider p(x,t) to be physical quantity (like pressure disturbance) which is propagating in space as a linear wave. Formally, the wave equation can be written as
∂
2
p
∂
t
2
−
c
0
2
∇
2
p
=
0
,
{\displaystyle {\frac {\partial ^{2}p}{\partial t^{2}}}-c_{0}^{2}\nabla ^{2}p=0\,,}
where
c
0
{\displaystyle c_{0}}
is the speed of wave propagation. This is the homogeneous wave equation which governs the propagation of a wave in a quiescent medium. If there is a source or a distribution of sources in the field which produce the wave, the wave equation will take the nonhomogeneous form
∂
2
p
∂
t
2
−
c
0
2
∇
2
p
=
f
(
x
,
t
)
,
{\displaystyle {\frac {\partial ^{2}p}{\partial t^{2}}}-c_{0}^{2}\nabla ^{2}p=f(x,t)\,,}
where f(x,t) represents the existence of the sources. To solve a nonhomogenous PDE like the one above, we need to utilize a mathematical tool called the Green's function.
== Green's Function ==
Green's function [2] is named after the British mathematician George Green [3], who first developed the concept in the 1830s. The concept of Green's function is one of the most powerful mathematical tools to solve boundary value problems.
Suppose, we have a linear differential equation given by:
L
u
=
f
,
{\displaystyle L\,u=f,}
where L is the differential operator. The main idea is to find a function G, called Green's function, such that the solution of the above differential equation can be determined from
u
(
x
)
=
∫
G
(
x
,
s
)
f
(
s
)
d
s
{\displaystyle u(x)=\int G(x,s)f(s)ds}
To find the appropriate green function for a given differential equation, one should solve
L
G
(
x
,
s
)
=
δ
(
x
−
s
)
{\displaystyle L\,G(x,s)=\delta (x-s)}
with the same boundary condition as the original problem. For example, the free-space Green's function of the wave equation, is the solution of the wave equation with an impulsive point source
δ
(
x
−
y
)
δ
(
t
−
τ
)
{\displaystyle \delta (\mathbf {x} -\mathbf {y} )\delta (t-\tau )}
(which is located at point
y
{\displaystyle \mathbf {y} }
and generates an impulse at time
τ
{\displaystyle \tau }
). Therefore,
◻
2
G
(
x
,
y
,
t
,
τ
)
=
δ
(
x
−
y
)
δ
(
t
−
τ
)
{\displaystyle \Box ^{2}G(\mathbf {x} ,\mathbf {y} ,t,\tau )=\delta (\mathbf {x} -\mathbf {y} )\delta (t-\tau )}
where
◻
2
=
∂
2
∂
t
2
−
c
0
2
∂
2
∂
x
2
{\displaystyle \Box ^{2}={\frac {\partial ^{2}}{\partial \,t^{2}}}-c_{0}^{2}{\frac {\partial ^{2}}{\partial \,x^{2}}}}
.
The solution of the above equation is
G
(
x
,
y
,
t
,
τ
)
=
1
4
π
r
δ
(
t
−
τ
−
r
c
0
)
{\displaystyle G(\mathbf {x} ,\mathbf {y} ,t,\tau )={\frac {1}{4\pi \,r}}\delta \left(t-\tau -{\frac {r}{c_{0}}}\right)}
where
r
=
|
x
−
y
|
{\displaystyle r=|\mathbf {x} -\mathbf {y} |}
The above relation represents an impulsive, spherically symmetric wave which is expanding from source
y
{\displaystyle \mathbf {y} }
. It is also important to mention that the argument of the delta function,
τ
+
r
c
0
{\displaystyle \tau +{\frac {r}{c_{0}}}}
is called the retorted time and is equal to the time required for a wave, generated by source
y
{\displaystyle \mathbf {y} }
at time
τ
{\displaystyle \mathbf {\tau } }
, to arrive at point
x
{\displaystyle \mathbf {x} }
at time
t
{\displaystyle \mathbf {t} }
.
Therefore, if we have a general problem (let's say noise generated by a turbulent jet), which is governed by
◻
2
p
=
Q
(
x
,
t
)
{\displaystyle \Box ^{2}p=Q(x,t)}
the instantaneous solution at any point in space is given by
p
(
x
,
t
)
=
1
4
π
∫
∫
−
∞
∞
Q
(
y
,
τ
)
G
(
x
,
y
,
t
,
τ
)
d
3
y
d
τ
{\displaystyle p(x,t)={\frac {1}{4\pi }}\int \int _{-\infty }^{\infty }Q(\mathbf {y} ,\tau )G(\mathbf {x} ,\mathbf {y} ,t,\tau )d^{3}\,\mathbf {y} d\tau }
In the case of radiation in free-space,
p
(
x
,
t
)
=
1
4
π
∫
−
∞
∞
∫
Q
(
y
,
τ
)
|
x
−
y
|
δ
(
t
−
τ
−
|
x
−
y
|
c
0
)
d
3
y
d
τ
=
1
4
π
∫
−
∞
∞
Q
(
y
,
t
−
|
x
−
y
|
/
c
0
)
|
x
−
y
|
d
3
y
{\displaystyle p(x,t)={\frac {1}{4\pi }}\int _{-\infty }^{\infty }\int {\frac {Q(\mathbf {y} ,\tau )}{|\mathbf {x} -\mathbf {y} |}}\delta \left(t-\tau -{\frac {|\mathbf {x} -\mathbf {y} |}{c_{0}}}\right)d^{3}\,\mathbf {y} d\tau ={\frac {1}{4\pi }}\int _{-\infty }^{\infty }{\frac {Q(\mathbf {y} ,t-|\mathbf {x} -\mathbf {y} |/c_{0})}{|\mathbf {x} -\mathbf {y} |}}d^{3}\,\mathbf {y} }
== References ==
M. S. Howe, "Theory of Vortex Sound," Cambridge Texts in Applied Mathematics 2003. | 1,802 | 5,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-45 | latest | en | 0.916226 |
https://www.btechguru.com/GATE--electronics-and-communication--calculus--multiple-integrals--introduction-to-sequential-circuits-video-lecture--6933--21--109.html | 1,660,096,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00000.warc.gz | 625,114,783 | 11,555 | # Plan, Prepare, Practice and Perform
Engineering Courses, Campus Placement Preparation, Bank exam & GATE Preparation
India's No.1 Platform for Online Learning, Served more than 1.1 lakh Premium Users, Unique platform for students in higher education in India
Chapter 1: Mean value theorems
Mean Value Theorem
Mean Value Theorem
Chapter 2: Evaluation of definite and improper intergrals
10 - Mod-1 Lec-10 Evaluation Of Real Improper Integrals-1 [58:32]
9 - Mod-1 Lec-9 Evaluation Of Real Integrals [45:43]
14 - Mod-1 Lec-14 Evaluation Of Real Integrals-revision [1:01:56]
Pn Junction Diodes
Application Of Shift Registers
Chapter 3: Partial derivatives
S-r, J-k And D Flip Flops
Chapter 4: Maxima and minima
Module - 3 Lecture - 2 Properties Of Surfaces - Ii
Maxima - Minima
Module - 5 Lecture - 1 Motion Of Particles Planar Polar Coordinates
Maxima Minima
Chapter 5: Multiple integrals
Bjt Small Signal Analysis
Introduction To Sequential Circuits
Shift Registers
Multiple Inegrals
Chapter 7: Line
Mathematics
Array Multiplier
Chapter 8: Surface and volume intergrals
Atm Networks
Surface Integrals
Chapter 9: Stokes
Mathematics
Module - 4 Lecture - 1 Method Of Virtual Work
2's Complement Subtractor And Bcd Adder
Lecture 16
Chapter 10: Gauss and green's theorems
Module - 2 Lecture - 3 Friction
Module - 3 Lecture - 1 Properties Of Surfaces - I
Subtractors
Greens Theorem
Guass Divergence Theorem
Chapter 11: Mean value Theorems
Mathematics
Mathematics
Chapter 12: Theorem of intergral calculus
3 - Mod-1 Lec-3 Cauchy`s Integral Theorem [49:43]
Module - 3 Lecture - 3 Properties Of Surfaces - Iii
Chapter 13: Evaluation of definite and improper intergrals
11 - Mod-1 Lec-11 Evaluation Of Real Improper Integrals-2 [50:04]
12 - Mod-1 Lec-12 Evaluation Of Real Improper Integrals-3 [53:08]
13 - Mod-1 Lec-13 Evaluation Of Real Improper Integrals-4 [52:53] | 538 | 1,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-33 | latest | en | 0.567519 |
https://ebrary.net/243477/education/analysing_quantitative_data | 1,685,998,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00045.warc.gz | 256,831,871 | 32,227 | # Analysing quantitative data
## Introduction
This chapter focuses on quantitative data analysis. It gives you an overview of the ways in which you can analyse the data that you have spent so much time and energy collecting. Using real-life data and examples, the chapter provides a guide to essential statistical techniques commonly used in undergraduate dissertation. By providing the essential steps and procedures in analysing your data, the chapter seeks to develop your skills in handling and making sense of your data, whether collected through questionnaires, surveys, structured interviews, observations, existing secondary data or any other methods that you may have used to collect the data. All examples and case studies in this chapter are based on real-life data. All statistical analyses are done using IBM-SPSS programme Version 24. The emphasis here is to develop your knowledge and understanding of when to use different analytical techniques and the practical skills and procedures required to analyse your data and how to interpret or draw inferences from your results.
By the end of the chapter, you will have a better understanding of how to:
• • Prepare for quantitative data analysis by organising and coding your data;
• • Define different types of variables and entering data into IBM-SPSS statistical programme;
• • Explore the distribution of your data and using descriptive statistics to summarise your data;
• • Use IBM-SPSS to construct simple graphs to visually present your data;
• • Explore any statistical relationships between variables; through univariate and bivariate analyses;
• • Draw inferences and conclusions from quantitative statistical analysis of your data.
## Variables
Not all numbers are the same. An understanding of the differences between variables is important when thinking of analysing your data and choosing which calculations to do with your data. There are three main types of variable;
• 1. Nominal. This is when numbers are used like names. In a questionnaire, for example, certain questions might be coded with numbers to represent different categories. In a question on country of birth, Afghanistan might be coded as 1, Albania as 2, Algeria as 3, etc. The numbers 1, 2 and 3 have no numeric value and have been chosen arbitrarily. We could easily have chosen to code Albania as 9 and Algeria as 11. These categories cannot be rank ordered, and it would be meaningless to carry out certain statistical tests (such as calculating the mean) on nominal data. Other examples of nominal data include ethnicity, eye colour and housing tenure.
• 2. Ordinal. For these variables, the numbers represent categories again, but this time they can be rank ordered through the use of Likert-type scales. For example, levels of satisfaction can be numbered from 1 to 5, where 1 = extremely satisfied, 2 = satisfied, 3 = neither satisfied nor not satisfied, 4 = not satisfied, 5 = not satisfied at all. With these kinds of data it is possible to describe people’s level of satisfaction, e.g. '67 per cent of respondents were very satisfied with the service.’ It is important to remember, however, that the distances across the categories might not be equal. The researcher cannot judge whether someone who gives a 5 for satisfaction is five times less satisfied than someone who gives a 1. This means that, as with nominal data, certain calculations, such as mean and standard deviation, cannot be carried out on ordinal data. Other examples of ordinal data include: age categories (21-30, 31-40, etc.) or frequency of doing something (never, rarely, often, frequently).
• 3. IntervallRatio. Here, the differences between the numbers are equal across the range. If someone is 21 and someone else is 18, the difference is three years. These three years are equal to the three-year difference between someone who is 35 and someone else who is 32. The distinction between interval and ratio data is that the zero in interval data is arbitrary. For example, on a thermometer, the zero for Fahrenheit and Celsius scales is different. In social science research, most variables will have a fixed zero — so they are ratio variables. It is possible to carry out more complex calculations and statistical tests on interval and ratio data. Examples of ratio data include age, income, height or weight.
Think about your data and decide which types of variables you are working with before starting your data analysis. The volume of numbers from which you need to create order and meaning can be intimidating at the start of data analysis. You need to find ways to summarise the data so that you can more easily see what the data is telling you. As you describe and summarise your data, you will be making it more readable, comprehensible and clear. Here we will look at how you can describe one variable and then compare two variables. Univariate analysis is looking at one variable and describing tendencies, patterns and trends whereas bivariate analysis looks at relationships between two variables. Multivariate analysis looks at more than two relationships simultaneously.
## Data exploration and descriptive statistics
Quantitative data analysis is generally divided into two categories — descriptive and inferential statistics. Descriptive statistics enable you to explore and understand your data before carrying out any further or detailed statistical analysis that may be required. For some dissertations, it could well be that all is required are descriptive statistics without any further complicated statistical analysis. There are several ways you can explore and make a judgement on the nature of your data in terms of its distribution, underlying patterns and structures. One of these may involve manual calculations of simple statistics that measure averages such as the mean value, weighted mean, the median, the mode, percentile, etc. Other manual calculations may involve the use of statistics that measure variability such as the range, standard deviation and the variance. Each of these descriptive statistics are explained below, using examples of real-life data.
## Averages and measures of central tendency
You can explore the distribution of quantitative data using simple statistics that measure characteristics such as:
• • Average value
• • Variability
• • Skewness
• • Kurtosis.
Average
An average is the one value that best represents an entire group of scores or values in your dataset. Statistics based on averages tend to measure central tendency of the data. There are three forms of averages:
• • The mean
• • The median
• • The mode.
Each of these averages will produce a different type of information about the nature of your data and its distribution.
The mean
The mean (also known as arithmetic mean) is the most common type of average computed in social science undergraduate dissertations. It is the sum of all the values in your data set, divided by the number of values or cases in that group. This is mathematically expressed as:
Where X bar is the mean value of the group of scores or simply, the mean;
y is the summation sign denoted by the Greek letter sigma;
The X is each individual score in the group or dataset;
The n is the size of the sample or number of cases relating to your dataset. In some publications, the mean is sometimes represented or denoted by the letter M. Technically, the arithmetic mean is defined as the point at which the sum of the deviations from the mean is equal to zero.
For example, a researcher interviewed a group of ten people whose ages (in years) are recorded as 18, 18, 37, 40, 47, 54, 62, 70, 74, and 80. The mean age in this sample is 50, and the sum of the deviation of each score from the mean is zero (i.e. adding up: -.32, -32, -1.3, -10, -3, 4, 12, 20, 24, .30).
Weighted mean
Weighted mean is used in situations where you have occurrence of more than one value. For example, a Sociology lecturer interested in calculating the weighted mean score in her Sociology and Change module over a period of three years recorded the data in Table 9-1.
The weighted mean was obtained by multiplying each score (module grade) by the frequency of its occurrence (frequency), adding the total of all the occurrences, and then divided by the total number of occurrences. The weighted mean grade, in this case, is 61.24% (6124 divided by 100).
The median
The median is defined as the midpoint in a set of values or scores. The median divides your data set into two equal halves such that one-half, or 50%, of the scores or values in your data set fall above the median point and the other half or 50% fall below the median point. Although there is no standard formula for computing the median, it can be determined by:
Table 9.1 Weighted mean of students’ grades enrolled on Society and Change module 2017-2019
Module grade (%) Frequency (no. of students with corresponding grade score) Grade x frequency 47 5 235 50 12 600 54 11 594 60 20 1200 63 29 1827 70 11 770 71 8 568 80 2 160 84 1 84 86 1 86 Total 100 6124
The weighted mean grade is 61.24%.
• • Listing all the values in order, either from highest to lowest or lowest to highest
• • Finding the middle-most score
• • Averaging between the two middle values, if the number of values is even.
While the mean measures the middle point of a set of values, the median is the middle point of a set of cases. In relation to the previous example and the ages of the of ten people recorded as 18, 18, 37, 40, 47, 54, 62, 70, 74, and 80, the median will be between 47 and 54, i.e. 51.5.
Percentile
A percentile is a statistical measure of distribution that shows the value below which a given percentage of observations in a set of data or group of observation fall. It is used to define the percentage of cases equal to and below a certain point in a distribution or set of scores. If a score is at the 75th percentile, it means that the score is at or above 75% of the other scores in your data set or sample. The median is the 50th percentile, i.e. the point below which 50% of your data sample fall. The 25th percentile is known as the First Quartile (Qi) which is the middle number between the smallest number and the median of your data set. The 50th percentile is the Second Quartile (Qj) which is also the median of the data. The 75th percentile is referred to as the Third Quartile (Q3) which is the middle value between the median and the highest value of your data sample.
The mode
The mode is the value that occurs most frequently. To compute the mode, you need to:
• • List all the values in your data set or distribution (list each value only once)
• • Tally the number of times that each value occurs
• • Note the value that occurs most often.
If every value in a distribution contains the same number of occurrences, then there is no modal value. If more than one value appears with equal frequency, the distribution is multi-modal. If two modes exist in a set of scores, then the distribution is bimodal. The mode in the interviewed group of ten people recorded as 18, 18, 37, 40, 47, 54, 62, 70, 74 and 80 would be 18.
## How do I know which measure of central tendency to use?
The type of averages or central tendency measures you need for your dissertation will depend on the type of data you've collected. For categorical or nominal data such as hair colour, income bracket, voting preference, racial group, etc., the mode is a more practical measure of central tendency to use. For interval/ratio data such as income levels, age, test score, height, weight, etc., the median and mean are best used. Generally, the mean is a more precise measure than the median, and the median is a more precise measure than the mode.
## Measures of variability (spread or dispersion)
Measures of variability reflect how scores differ from one another. So, for example, if a researcher was interested in the variations in life expectancy at birch across different countries around the world, they could randomly select six countries each from sub-Saharan Africa, Asia and Europe as recorded in Table 9-2.
For purpose of illustration, the life expectancy figures in Table 9-2 show different variability within each of the three groups of countries in sub-Saharan Africa, Asia and Europe. While the data shows a relatively less degree of variation in life expectancy for the selected countries in Europe, countries in sub-Saharan Africa have greater variations. The selected Asian countries have no variability in their life expectancy data at all. Technically, variability is a measure of how much each score in a group of scores differs
Table 9.2 Life expectancy at birth, total (years), for selected countries in sub-Saharan Africa, Asia and Europe (2017)
Sub-Saharan Africa Asia Europe Cameroon 59 Mongolia 69 Belgium 81 Eritrea 66 India 69 Bulgaria 75 Liberia 63 Timor-Leste 69 France 83 Mali 58 Cambodia 69 Ireland 82 Nigeria 54 Indonesia 69 Italy 83 Sierra Leone 52 Philippines 69 United Kingdom 81
Source: The World Bank, 2018. Accessible at: https://data.worldbank.org/indicator/SP.
DYN.LEOO.IN
from the mean. Therefore, both average and variability are used to describe the characteristics of a distribution or data set.
The three most commonly used measures of variability are:
• • The range
• • The standard deviation
• • The variance
The range
The range gives an idea of how far apart scores are from one another.
It is computed by subtracting the lowest score in a distribution from the highest score
where r is the range, h is the highest score and 1 is the lowest score in the data set.
There are two kinds of ranges:
• • exclusive range
• • inclusive range.
Inclusive range is calculated using the formula:
Exclusive range is highest score minus the lowest score plus 1. This is computed using the formula:
The range gives only a general estimate on how wide or different scores are from one another. They should not be used to reach any conclusions regarding how individual scores differ from each other.
The standard deviation (SD)
The standard deviation represents the average amount of variability in a set of scores. In technical terms, the SD is the average distance from the mean. The larger the standard deviation, the larger the average distance each data point is from the mean of the distribution. The SD is computed using the formula:
Where s is the standard deviation, S is sigma — the summation sign, X is each individual score,
X is the mean of all the scores and n is the sample size of your data.
In order to manually compute the SD, for your data set, you will need to:
• • Find the difference between each individual score and the mean (X-X)
• • Square each difference and sums them all together
• • Divide the sum by the size of the sample (minus 1)
• • Then take the square root of the results.
The mean deviation
The mean deviation (also called the mean absolute deviation) is the sum of the absolute value of the deviations from the mean divided by the number of data points. The sum of the deviations from the mean is always equal to 0.
The variance
The variance is the standard deviation squared. This can be computed using the formula:
Distribution curves and skewness
Skewness and kurtosis are statistical terms used to describe the shape of a distribution. Most statistical analysis assumes a normal distribution of data with a symmetric bell-shaped pattern as shown in Figure 9-1. With normal distribution the data tends to be around the mean with no bias
Figure 9. / Normal distribution curve
left or right — so, for example, if you were to take a class of 50 people and measure their heights, if there was a normal distribution, then most people would have heights clustered around the mean.
However, many data may not conform to the normal distribution assumption and it may be necessary to establish the degree to which your data deviates from normal distribution. Skewness is a measure of the lack of symmetry, or the lopsidedness, of a distribution. This occurs when one 'tail’ of the distribution is longer that another.
Figure 9-2 shows two forms of distribution with varying degrees of skewness. While the normal distribution curve in Figure 9-1 has equal lengths of tails and no skewness, curve A in Figure 9-2 has a longer right tail than left. This suggests a smaller number of occurrences at the high end of the distribution. This kind of distribution is referred to as positively skewed. Conversely, the distribution B in Figure 9-2 has a shorter right tail than left. This means a larger number of occurrences at the high end of the distribution. Therefore, curve B denotes a negatively skewed distribution.
The location of the mean value in relation to the median value will indicate the direction of skewness. Generally, if the mean is greater than the median, the distribution is positively skewed. Conversely, if the median is greater than the mean, the distribution is positively skewed.
Figure 9.2 Distribution curves with positive and negative skew
In mathematical terms, skewness is computed by subtracting the value of the median from the mean. For example, if the mean value of a distribution is 95 and the median is 86, the skewness value is 9, i.e. 95—86. This means the distribution is positively skewed. Similarly, if the mean of a distribution is 67 and the median is 74, the skewness value is -7, i.e. 67—74. That will suggest that the distribution is negatively skewed.
To compare the skewness of one distribution to another, in absolute terms, the following formula is often used:
Where SK is Pearson's measure of skewness (correlation); X is the mean value, M is the median and S is the standard deviation.
For example, if the mean value of a distribution is 100, the median 105, and the standard deviation is 10, its skewness will be -5. This means the distribution is negatively skewed. In the same vein, if a distribution has a mean value of 120, the median of 116 and the standard deviation of 10, its skewness will be 4. This means the distribution is positively skewed.
Kurtosis
Kurtosis relates to how flat or peaked a distribution appears. Figure 9-3 shows three different distribution curves with different kurtosis. The term platykurtic is used to refer to a distribution curve that is relatively flat, compared to a normal or bell-shaped distribution. A normal bellshaped distribution is described as mesokurtic, while the term leptokurtic refers to a distribution that is relatively peaked compared to a flatshaped or bell-shaped distribution.
Generally, data sets that are platykurtic are relatively more dispersed than those that are not. Distributions that are leptokurtic are less variable or dispersed relative to others. While skewness and kurtosis are used mostly as descriptive terms, there are mathematical indicators or measures that can be computed to indicate how skewed or kurtotic your data distribution is.
## Using software to analyse your data
Do not despair if you have read through this chapter so far and wondered how on earth you would be able to do all of the mathematical calculations and produce the complicated curves, tables and graphs you need for your dissertation — there is software to help you!
Figure 9.3 Distribution curves and kurtosis
The most commonly used statistical computer program designed originally for social scientists is IBM-SPSS. It is relatively easy to use and there are many good books that will introduce you to the program and provide step-by-step guides on how you can use it to analyse your data. While it is out of the scope of this book to provide you with the training and skills to use IBM-SPSS, we have, where necessary, offered some tips to get you started. IBM-SPSS is a powerful piece of software which has functionality way beyond what you will need for your research. Your institution might well have a licence for IBM-SPSS or other similar general-purpose statistical software such as Stata. There are many online guides, textbooks and chapters in data analysis books that may also help you with learning IBM-SPSS. It is worth taking a look at these texts and working through some of the examples before you start to analyse your own data.
There are other statistical packages available to you, some of which are free access — for example. Openstat and Excel in the Microsoft Office package. The statistics functionality of these programs is good, and the freely available add-ons can give more advanced features. Using IBM-SPSS or any of the other statistical programs will enable you to do different calculations and you can also produce graphs, plots and tables to visually present your data. For the rest of this chapter, we have provided a number of case studies to illustrate how to use IBM-SPSS to analyse your data.
The computer package only works with what you input. So, it is important you understand the principles and techniques for using IBM-SPSS to achieve your data analysis objective or indeed any other statistical package. That is why we have concentrated more on what the different techniques show rather than demonstrating how you carry them out in different packages. It is really important, therefore, that you understand the tests you are asking IBM-SPSS software to do and how to interpret the results and present your own findings. Read, carefully, each of the following case studies and examples of how IBM-SPSS was used to answer specific research questions.
## Preparing data for computer analysis
The first stage in using computers to analyse your data is getting the data into a format that a computer can read. This usually involves creating a spreadsheet to input all your data. This initial data organisation could be done using Microsoft Excel or any specialised software that allows data entry and storage. If your data collection instrument is a questionnaire, you may need to design a coding scheme to enter all your questionnaire data in a format that the computer will be able to read and process the information. This initial questionnaire data management and organisation is referred to as coding.
Coding and coding schemes
Coding is the transformation of the information contained in your questionnaires into a numeric or alpha-numeric format that a computer can understand and use in statistical analysis. It relates to the method of assigning numerical values/symbols to various answer categories in a questionnaire. For each response option to a question, a letter code (a, b, c) or preferably numerical code (1,2, 3) is usually assigned. Coding is an important stage in data processing; hence care should be taken while assigning codes to your questionnaire information to make your analysis meaningful.
For example, in a questionnaire survey of undergraduate students’ educational experience and course choice in a UK University, the coding scheme in Table 9-3 was used to record and enter some of the respondents’ answers to the survey questions.
The essence of a coding scheme is to facilitate computer data entry and to ensure data is correctly entered. The coding scheme allows questionnaire information to be entered in a consistent format that can be read and analysed by computer programmes such as IBM-SPSS in the form of a spreadsheet.
## Introducing IBM-SPSS, defining variables and inputting data
To make the most of IBM-SPSS, you need to first know how the program works and the various interface to use it to analyse your data. It is worth consulting books that introduce you to the program in full and that helps you learn how to use IBM-SPSS software to define your variables, input your data and save your data file. To get you started, here are some useful tips:
• • Log on to IBM-SPSS.
• • A dialogue box will appear to either open an existing SPSS data file or create a new data file.
• • To create a new SPSS data file, you need to define each of your variables in the variable window.
Table 9.3 Sample coding scheme used in a survey of students' educational experience and course choice in a UK university
Survey question Variable name used to define question Possible response options to question Codes used to define options Gender Gender Male 1 Female 2 Your age category Age 18-21 21-25 2 25-35 3 35-45 4 45-55 5 55-60 6 60-65 7 65 plus 8 Studentship status Status Undergraduate Postgraduate 2 Other 3 Mode of study Study mode Full time 1 Part time 2 Distance learning 3 E-learning 4 Other 5 Current stage/year of study Study stage Year 1 1 Year 2 2 Year 3 3 Year 4 4 Other Year 5 How important is the course design a factor in choosing your programme? Course design Very important 1 Important 2 Not important 3 How important is the University reputation and standing in the League table in choosing your course? Uni reputation Very important 1 Important 2 Not important 3
(Continued)
Table 9.3 (Cont.)
Survey question Variable name used to define question Possible response options to question Codes used to define options How important to you is job prospect in choosing your course? Job prospect Very important 1 Important 2 Not important 3
Source: extract from a student experience survey (Jegede, 2018)
• • At the bottom left hand corner, you can swap between variable view window and data view window.
• • In the first row of the variable view window, define your first variable by specifying the variable name (variable name cannot be more than eight characters).
• • Choose the variable type (numeric for numbers or strings for texts or letters).
• • Define the width of your variable — variable width must be equal to or greater than the largest number of digits in the data set for the variable including the decimal point. For example, to enter 267.84 will require a variable width of 6 while 7.9 is 3 in width.
• • Define the number of decimal places in your data set. The programme default is two decimal places, but this can be changed as required. If there is no decimal place, the value of zero should be entered.
• • Label your variable if required. (You have the option to label your variable with a longer name containing more information.)
• • To define other variables, repeat the above procedure on row two, row three, etc.
• • Once you have defined and entered all the essential information for each of your variables, you can click on the data view tab in the bottom left hand corner, to start entering your data.
• • Remember to save your data.
• • If a variable name is not fully displayed, you can increase the width of the field by holding down the left button on the variable name cell and dragging it to the right.
• The main menu options are located at the top of the screen where you will select all the SPSS commands needed for your analysis.
All the tips provided in this book are based on IBM-SPSS Statistics Version 24. For a practical guide to computing descriptive statistics using real-life data, read case study 91. It illustrates how IBM-SPSS can be used to generate descriptive statistics to summarise your data.
Case Study 9.1 Computing descriptive statistics using crime data derived from the Crime Survey for England and Wales and Police Recorded Crime Data
Problem definition
A criminologist interested in analysing the volume of violent crime dealt with by the police from year ending March 2003 to year ending March 2015 in England and Wales extracted the data in Table 9-2 from the Crime Survey data for England and Wales.
The objective is to use IBM-SPSS descriptive statistics to analyse and summarise the data.
See Table 9.4
Tips for IBM-SPSS procedure for computing descriptive statistics:
• • Log on and enter your data into IBM-SPSS.
• • Define your variables, e.g. crime figures (numeric); crime record period (string); crime category (string), e.g. violence with injury = 1; violence without injury = 2; stalking and harassment = 3-
• • Data — split file, compare groups, groups based on crime category.
• • Analyse.
• • Descriptive statistics; crime figures.
• • Options — check mean, sum, std deviation, minimum, maximum, range, kurtosis and skewness boxes.
• • Extract results from IBM-SPSS output window.
The result of the SPSS descriptive statistics computed using this data is summarised in Table 9-3. The researcher also used the same data to construct a histogram and boxplots for each of the crime category as shown in Figure 9-4 and Figure 9-5.
Tips for IBM-SPSS procedure for boxplots:
• • Log on and enter your data into IBM-SPSS.
• • Define your variables, e.g. crime figures (numeric); crime record period (string); crime category (string), e.g. violence with injury = 1; violence without injury = 2; stalking and harassment = 3-
• • Graph > legacy dialogues > boxplot > simple.
• • Select summaries for groups of cases.
• • Define.
• • A new dialogue box opens up.
• • Move crime figures variable into variable box.
• • Move crime category variable into category axis.
• • Move crime record period variable into label cases by box.
• • Click OK.
• • Copy your boxplots and note the size of each of the boxplots, the location of the median line, the length of the whiskers and any outliers.
Table 9.4 Volume of violent crime dealt with by the police from year ending March 2003 to year ending March 2015
Period Violence with injury Violence without injury Stalking and harassment Total violence April02- March03 371,774 302,450 33,002 708,742 April03- March04 457,223 300,090 40,522 799,247 April04- MarchOS 514,638 277,569 52,1 17 845,673 AprilOS- March06 543,044 237,218 57,192 838,674 April06- March07 505,848 249,632 58,150 814,865 April07- March08 451,806 241,226 54,531 748,779 April08- March09 420,184 236,943 50,758 709,008 April09- MarchlO 400,703 241,818 55,329 699,01 1
(Continued)
Table 9.4 (Cont.)
Period Violence with injury Violence without injury Stalking and harassment Total violence April 10- March 11 367,847 243,426 53,144 665,486 April 1 1-Marchl2 337,709 238,276 49,766 626,720 AprilO 12-Marchl3 31 1,740 232,466 56,032 601,141 April 13- Marchl4 322,362 248,616 62,656 634,625 April 14-Marchl5 373,509 317,166 86,368 778,172
Data extracted from the Crime Survey for England and Wales and Police Recorded Crime Data.
Office of National Statistics. Accessed 4 June 2019 from: www.ons.gov.uk/peoplepopula tionandcommunity/crimeandjustice/datasets/crimeinenglandandwalesbulletintables. (Data used for this exercise is based on Police recorded crime, Home Office, licensed under the Open Government Licence and available from Office of National Statistics.
Accessed 4 June 2019 from: www.ons.gov.uk/peoplepopulationandcommunity/crimeand justice/datasets/crimeinenglandandwalesbulletintables)
The result of the SPSS descriptive statistics used in Case study 9-1 is summarised in Table 9-5. Using the same data, construct a histogram for each of the crime categories as shown in Figure 9-4.
## Graphs and graphical display of data
Visual aids are an important part of data exploration and can help you make sense of your data. Graphs generated through IBM-SPSS or Excel can help you summarise your data and highlight key areas that you may focus your attention. Different types of graphs can be used to make your data more visually appealing and accessible.
The most common types of graphs used in dissertations are:
• • Histogram
• • Bar graph
• • Line graph
• • Pie chart
• • Box plots.
Table 9.5 Summary of descriptive statistics for violent crime, England and Wales, from year ending March 2003 to year ending March 2015
Violence with injury Violence without injury Stalking and harassment Total Sample size n 13 13 13 13 Minimum 31 1,740 23,2466 33,002 33,002 Maximum 543,044 317,166 86,368 543,044 Mean (arithmetic) 413,722.08 258,992.00 54,582.08 242,432.05 Standard deviation 75,804.35 29,507.28 12,271.53 156,014.40 Sum 5,378,387 3,366,896 709,567 9,454,850 Range 231,304 84,700 53,366 510,042 Skewness 0.36 III 1.05 117 Kurtosis -1.08 -0.40 3.87 -1.081
Source: Extracts from IBM-SPSS Descriptive Statistics Output.
Frequency distributions
One way of presenting your data is through frequency distributions. This will show the number of people and the percentage for each category in your variable. Frequency distributions can be used for all types of variable (nominal, ordinal, interval/ratio) mentioned above. The way you present your frequency distribution will depend on the variables you are describing. You can present your data in tables or graphs (pie charts, bar charts, histograms). A good rule of thumb is to use a table unless a graph can put across the message more clearly.
Boxplot
The box plot shows all of the following:
• • The smallest observation (the bottom horizontal line)
• • The bottom 25% (the section between the lowest observation and the grey box)
• • The interquartile range (the grey box)
• • The mean (thick black line inside the box)
• • The top 25% (section above the grey box)
• • The highest observation (upper horizontal line).
Figure 9.4 Histogram of violent crime in England and Wales 2003-2015
Figure 9.4 (Continued)
The box plot shows whether your data is a symmetrical or skewed distribution. In this example, the boxplot shows that violent crime data in England and Wales is skewed, suggesting there is more spread for all categories of crime in the upper 25%. The box plot also indicates where there might be outliers. Outliers are cases which are very different from the rest of the cases. They are shown here by circles and stars. In our example, the stalking and harassment boxplot shows an outlier.
Exploring your data through descriptive statistics and graphical presentation as shown in this chapter can enable you to make a judgement on the nature of your data in terms of its distribution, underlying patterns and structures. Therefore, you may consider descriptive statistics as the first step in quantitative analysis. As an exploratory tool, descriptive statistics can uncover hidden patterns in your data and help you decide on further analysis that may be needed.
Figure 9.5 Boxplots of violent crime in England and Wales 2003-2015
Univariate, bivariate and multivariate analysis
As mentioned earlier it is important to make a distinction between different forms of analyses in relation to the number of variables involved. The example in the first case study is a case of univariate analysis where we deal with one variable — volume of crime in England and Wales. Although we are interested in different types of crime in the study, we are not looking for or attempting to compare crime variable with any other variable. In a bivariate analysis, the focus is to examine the relationships between two variables: the explanatory and the outcome variable. The explanatory variable is the variable which is thought to be the variable of influence (it is also known as the independent, input or predictor variable). The outcome variable (also known as the dependent variable) is the one that we believe will be affected by the explanatory variable.
In a multivariate analysis, you can analyse more than two variables simultaneously. The techniques used to do this kind of analysis are quite advanced. While we have provided a general guide to analysing multivariate analysis in this chapter, you may need to consult books that deal specifically with quantitative analysis (see recommended reading list).
## Beyond descriptive analysis to inferential statistics
Using descriptive analysis, you will have described the data that you have collected and identified relationships between those variables. The next stage in analysis is to test to what extent the results of the data in your sample are generalisable to your sample’s population. These tests are called hypothesis tests or tests for statistical significance. The results from these tests tell you how confident you can be that the relationships observed in the sample are representative of that population.
You would use different hypothesis tests depending on the types of variables you are analysing. For example, if you were dealing with categorical explanatory and outcome variables and you require a chi-square test of association, then Phi or Cramer's V test will be appropriate. See Appendix 2 for a list of statistical tests and a brief notes on what they are designed to measure.
All of these tests will carry with them certain assumptions about your data. For example, they might require that your data be normally distributed, independent, continuous. However, in terms of hypothesis testing there is one assumption that is extremely important. The data needs to have been drawn from a random sample. Hypothesis tests are carried out when you want to know whether something found is a quirk of the data set or something that is a feature of the population.
A test carried out on a non-random sample cannot speak with confidence about generalising to the population. If your sample is not random, you would be advised to spend your time carrying out a thorough descriptive analysis and trying to interpret what is happening in the sample that you have collected. If you collect your own data for your undergraduate dissertation, you are unlikely to have a truly random sample large enough to analyse with inferential statistics. For this reason, we will not go too deep into inferential statistics in this book. If you are working with a random sample (if you are analysing data that has been collected by someone else as part of a much bigger survey, for example), you may look at some of the books in the list at the end of this chapter that will introduce you to some of the more sophisticated statistical techniques. You may also need to discuss analysis options with your dissertation supervisor.
## Analysing relationships – inferential statistics and hypothesis testing
Inferential statistics involves testing hypotheses in relation to the type of relationship that exist between variables. Part of this may involve measuring the degree of correlation.
Measuring correlation
Correlation is a statistical method for uncovering the nature and strength of relationships, if any, that exist between two or more variables. Not only will this technique tell you the kind of relationship that exists, but also enables you to evaluate, through hypothesis testing, the statistical validity of your result based on your sample data. You can carry out calculations to assess the degree or extent your two variables are related. The degree or association or correlation is determined by calculating the ‘correlation coefficients', and they are usually a value between zero and one.
Relationships between two events, or variables X and Y, could be described in two ways:
• • We could have an association between two variables where there is some kind of influence of one variable on the other, i.e. how X influences Y and vice versa;
• • We could have a case of a causal relationship where one variable X causes change to occur in the other variable Y.
A causal link exists if changes in event X triggers an action or reaction in event Y. This is often referred to as a ‘cause and effect’ relationship. The cause is often referred to as the independent variable; the variable that is affected is known as the dependent variable.
The correlation between two events or variables X and Y can be described as:
• • None (no correlation) — where changes in X has no effect on Y and vice versa;
• • Positive (positive correlation) — where an increase in one variable results in an increase in the other variable, or a decrease in one variable results in a decrease in the other variable);
• • Negative (negative correlation) — where an increase in one variable generates a decrease in the other.
Here are two common correlation coefficients:
• Pearson's r. This measures the relationship between two continuous variables. The value ranges from -1 (a perfect negative relationship) through 0 (no relationship) to +1 (perfect positive relationship). In order to conduct a Pearson’s r test, your data needs to meet certain assumptions:
• (a) The two variables need to have a normal distribution (i.e. the histogram would look like an upside-down bell); and
• (h) When the variables are plotted against each other in a scatterplot, there needs to be a linear relationship between them.
• Spearman’s rho. This test is similar to Pearson’s r but your data do not need to meet the same assumptions. In this test, variables are ranked. A ranking of +1 shows a perfect relationship. It is possible to use Spearman’s rho with both continuous and categorical data.
Case study 9.2 Analysing relationships using socioeconomic deprivation and crime data for English towns and cities 2015
Problem definition
A researcher interested in housing and socio-economic deprivation in English towns and cities obtained the data shown in Appendix 2. The objective of the study is to establish any connection between the degree of deprivation and crime in selected towns/ cities using appropriate statistical analysis.
(Data Source: Office for National Statistics licensed under the Open Government Licence.)
In order to test for statistical validity of any connection between deprivation and crime, the following hypotheses were posed:
The Null Hypothesis H„:
There is no statistically significant relationship between level of deprivation and crime rates in English towns and cities.
The Alternative Hypothesis Hp
There is a statistically relationship/connection between level of deprivation and crime rates in English towns and cities.
Given that the data set is ranked, it is appropriate to use Spearman's rho correlation technique to test whether or not there is a relationship between deprivation and crime and if any such relationship is statistically significant at 95% level of confidence.
Tips for IBM-SPSS procedure for Spearman's rho correlation analysis:
• • Log on and enter your data into IBM-SPSS.
• • Define your variables, e.g. index of multiple deprivation rank (IMD) and crime rank figures.
• • Analyse menu.
• • Correlation statistics.
• • Correlate > bivariate.
• • Move the two variables into the variable list box, e.g. IMD Rank and Crime Rank.
• • Select the appropriate correlation method e.g. Spearman.
• • Select the required test of significance, e.g. 2-tailed.
• • Check flag significant correlations box.
• • Click OK.
Tips for IBM-SPSS procedure for scatter plots and fitting regression line
• • From graphs menu, select scatter.
• • Choose simple scatterplot and click on define button.
• • Move the Crime Rank variable into the Y-axis box.
• • Move the IMD variable into the X-axis box.
• • Click on OK.
• • Double click the graph to open the chart editor.
• • Under chart menu, select options.
• • Check fit line box.
• • Check Display R-Square in legend.
• • Check include constant in equation.
• • Click continue.
• • Click OK.
Results:
The result shows that there is a strong connection between level of deprivation and level of crime in English towns and cities (See Table 9 6 and Figure 9-5).
Table 9.6 Correlation matrix of index of multiple deprivation (IMD) and level of crime in English towns and cities
Correlations Index of multiple deprivation Crime Spearman’s rho Index of multiple deprivation Correlation Coefficient 1.000 .687" Sig. (2-tailed) .000 N 109 109 Crime Correlation Coefficient .687" 1.000 Sig. (2-tailed) .000 N 109 109
** . Correlation is significant at the 0.01 level (2-tailed).
Source: IBM-SPSS Spearman Correlation, derived from Socio-economic deprivation in English towns and cities - 2015, Office of National Statistics.
Hypothesis testing offers the opportunity to establish whether there is strong enough evidence in the sample of data that you collected to infer or make a judgement on whether certain condition is true or false for the entire population to which your data relate. This is based on the nature and strength of connection between two variables (bivariate analysis). Inferential statistics require an understanding of key statistical concepts such as hypothesis formulation, correlation, cross-tabulation, bivariate analysis, confidence interval and statistical significance.
Figure 9-6 shows the scatterplots of index of multiple deprivation and crime based on the data in our case study.
In this example the relationship between the two variables is positive since an increase in the value of one variable show an increase in the other. The figure suggests a positive connection between crime and deprivation rank in English towns and cities.
Crime and deprivation - regression line
The line that runs through the middle of the scatter points in Figure 9-6 is known as the regression line. A relationship between two
Figure 9.6 Scatterplots of index of multiple deprivation and crime in English towns and cities
quantitative variables that can be represented by a straight line is called a linear relationship. One of the objectives of simple linear regression analysis is to help us determine the best line through the data points in a scatterplot. One common way of determining this line is through the method of least squares; therefore, the regression line is also known as the least square line. The underlying principle of the least square method is that the line of best fit through the scatter is such that the sum of the squares of the deviations from the points to the line is minimum. While a correlation statistic shows how closely the data points are distributed around a straight line, regression analysis involves calculating and fitting a line of best fit across the middle of the data points.
While the scatterplot provides a visual picture as to the nature of relationships between the two variables, Table 9-6 gives a much more detailed statistic with a correlation coefficient of 0.687 and significance level, p of 0.000. This means there is a strong positive correlation between crime and level of deprivation in our case study and the correlation is statistically significant.
In interpreting the result in relation to the stated hypotheses, we need to consider not only the correlation coefficient but also the level of significance, p. Since p < 0. 01, the Null hypothesis H„ in our case study can be rejected which means the Alternative hypothesis H! is true. Therefore, we can conclude that there is a statistically significant relationship between deprivation and crime in English towns and cities and we can make that conclusion at more than 99% level of confidence.
Crime and deprivation - linear regression model
In Figure 9 6, we can see the mathematical model or equation that defines the relationship between crime and deprivation. A linear relationship is usually represented by a linear equation.
The simple linear regression model is stated as:
Where Y is the dependent (or predicted) variable; (in our example — crime)
a is the intercept (the point at which the regression line touches the Y-axis)
x is the independent (or predictor) variable (IMD)
b is the slope or gradient of the regression line
e is the error term (that is the stochastic disturbance or the residuals).
Therefore, the mathematical model or equation that defines the relationship between crime and deprivation in English towns and cities can be stated as:
The R-Square value of 0.471 suggests that 47.1% of the variations in crime in English towns and cities can be explained by the level of multiple deprivation in those cities and towns.
Multivariate analysis and advanced analytical techniques
In our case study 9-2, we used only one explanatory variable, index of multiple deprivation (IMD) to explain the dependent variable (crime). You can use more than one independent variable to explain variations incrime. For example, the technique of multiple linear regression will enable you to use two or more explanatory variables in your analysis. You can learn more about multiple linear regression and other advanced statistical from relevant textbooks.
To help you along, we've included a number of variables in Appendix 2 that could help you with this. For example, you may wish to explore the relationship between crime and income deprivation, employment deprivation, health deprivation, education, skills and training deprivation, barriers to housing services and living environment deprivation. Using advanced statistical technique, you can determine the extent to which each of the factors listed above contribute to crime level in English towns and cities and the joint contribution of all the factors to the problem of crime in English towns and cities.
Besides multiple linear regression, there are a number of other advanced statistical and analytical techniques that are used in social sciences, such as modelling, hierarchical structure analysis, multinomial logistic techniques, factor analysis, etc., that are beyond the level expected of an undergraduate study. If you are thinking of doing postgraduate studies, it may well be that you are interested in developing your analytical skills further in these areas.
## Key messages
• • You need to understand the data that you are working with so that any calculations that you perform are valid.
• • Descriptive analysis allows you to examine the variables you have in your data set and to establish relationships between them.
• • There is no point conducting tests to establish the generalisability of your findings if you have a small sample, collected through convenience sampling.
• • If you are carrying out analysis of an existing data set, techniques of inferential statistics might well be appropriate.
• • Investigate computer packages that will help you with your analysis — investment of time in learning the software will pay dividends in terms of time spent analysing the data.
## Key questions
• • Does your university support a data analysis package and have you identified a package that can help you?
• • Do you know which variables you are working with?
• • Have you described your data using the appropriate techniques?
• • Have you used the right checks to establish relationships between your variables?
• • Does your sample comply with the assumptions necessary to carry out tests of statistical significance?
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Q: Find the difference: 48 – (–23)
A: 48 – (–23) = 48 + 23 = 71
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y=0
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,214 | 3,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-26 | latest | en | 0.863356 |
https://www.cfd-online.com/Forums/fluent/238614-what-settings-do-i-do-outlet-if-i-only-know-inlet-conditions.html | 1,656,694,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00793.warc.gz | 753,162,736 | 15,855 | # What settings do I do for the outlet if I only know the inlet conditions?
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September 22, 2021, 19:17
What settings do I do for the outlet if I only know the inlet conditions?
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What settings do I do for a density based solver if I only know the inlet conditions? I'm trying to do a validation study for a paper but I don't know how the authors did the outlet conditions. They only mention the boundary conditions for the air and hydrogen inlets.
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September 23, 2021, 00:58 #2 Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 4,844 Rep Power: 57 Does it even matter? I see that both inlets have a Mach number of 1 and 2, so downstream BC's shouldn't really affect the problem as long as you set a reasonable outlet pressure that maintains the sonic conditions at the inlets. Otherwise the inlets are overconstrained. Something has to be relaxed. If you really care about matching all the numbers at the inlet then one of the constraints needs to become a soft constraint. That is, you iterate something else until that constraint is satisfied. For example, you could use a pressure outlet BC and then iterate the outlet pressure until the inlet pressure matches the table.
September 23, 2021, 23:58
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Originally Posted by LuckyTran Does it even matter? I see that both inlets have a Mach number of 1 and 2, so downstream BC's shouldn't really affect the problem as long as you set a reasonable outlet pressure that maintains the sonic conditions at the inlets. Otherwise the inlets are overconstrained. Something has to be relaxed. If you really care about matching all the numbers at the inlet then one of the constraints needs to become a soft constraint. That is, you iterate something else until that constraint is satisfied. For example, you could use a pressure outlet BC and then iterate the outlet pressure until the inlet pressure matches the table.
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# Problem 44400. 二つのベクトルの要素ごとの積の平均を計算しよう
Solution 1934155
Submitted on 13 Sep 2019 by Paul Morant
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### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3]; w = [10 15 20]; y_correct = 100/3; assert(isequal(weighted_average(x,w),y_correct))
2 Pass
x = [1 2 3]; w = [10 15 20]; y_correct = 100/3; assert(isequal(weighted_average(x,w),y_correct)) | 165 | 477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-43 | latest | en | 0.614399 |
classes.css.wsu.edu | 1,406,242,814,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997892641.2/warc/CC-MAIN-20140722025812-00171-ip-10-33-131-23.ec2.internal.warc.gz | 53,422,458 | 8,418 | ### G. Campbell - Biophysical Measurements and Instrumentation
#### Chapter 4
No part of this manuscript can be reproduced without written permission of the author
Measuring Water Potential
In soil, and within living organisms, the moisture measurement which is relevant to water transport and growth is the water potential. The water potential is the potential energy per unit mass of water in a system. The potential energy of the water in an organism or soil is determined by the position (with respect to some reference level) of the water in a gravity field, the adsorptive forces binding water to a matrix, the concentration of dissolved substance in the water, and the hydrostatic or pneumatic pressure on the water. The total water potential is the sum of all these components, and is written as:
(1)
where the subscripts are for gravitational, matric, osmotic, and pressure potentials, respectively. In a given situation, some of the components of the water potential may not be important. For example, we normally do not consider the gravitational potential in calculation of water uptake by plants (except for tall trees) or the osmotic potential in water transport in soil. Careful attention needs to be given to the potentials acting under a given set of circumstances in order to decide which measurement is appropriate for describing the driving force for water flow.
Numerous methods have been devised for measuring water potential in soil and plant systems. We will discuss several of the most useful. The methods we will discuss are based on one of two principles. One class of instruments balances liquid water at a known water potential against water at an unknown potential across a semipermeable membrane. At equilibrium, the potential of the water in the measuring system is equal to the unknown water potential. Instruments which use this principle are the tensiometer for measuring matric potential of soils and other colloidal systems and the pressure bomb which is used to measure water potential of plant leaves.
The second class of instruments measures the relative humidity in equilibrium with a sample at unknown water potential. A relation derived from the first law of thermodynamics and the perfect gas law relates water activity, aw, to water potential:
(2)
where R is the gas constant (8.31 J mole-1 K-1), T is the Kelvin temperature, and Mw is the molecular weight of water. When the gas phase is in equilibrium with the liquid phase, then the relative humidity, hr, in the gas phase is equal to aw.
In principle, this measurement is simple, but in practice it requires measurement of humidity in the range of 0.95 to 1.0 with a precision of around 7 parts in 105. A special psychrometer is used for the measurement, and special precautions are required to obtain the needed precision. Instruments using this principle are called thermocouple psychrometers or thermocouple hygrometers depending on the way they are read.
Figure 1. Diagram of Pressure Bomb apparatus.
The pressure bomb is an instrument used for measuring the water potential of leaves. Figure l shows a typical pressure bomb arrangement. A leaf is severed from the plant and placed in the bomb with its petiole or stem protruding through the seal. Pressure is applied to the tissue until water just appears at the cut end of the petiole or stem. When water appears, the pressure applied to the tissue is equal to the negative of the leaf water potential. In terms of the principles mentioned earlier, consider the xylem of the leaf as being a continuous porous system, impermeable to air, but permeable to solutes and water. Since the xylem is vented to the atmosphere through the seal of the bomb, the potential of the water in the xylem, when water just appears at the cut surface, is equal to that of free water, or zero. The water potential of the cells is in equilibrium with the xylem potential so at the end point, the water potential of the cells is also zero. The water potential of the cells is the sum of the applied pneumatic pressure and the potential of the cell: P + = 0, so = -P.
It is worth noting that the tissue water potential is = p + o, where p is the turgor pressure and o is the osmotic potential of the cell sap. When the leaf wilts is zero, and the water potential of the leaf is equal to the osmotic potential of the cell sap. The pressure bomb can therefore be used for direct measurements of osmotic potential of wilted leaves. If several measurements are made at different leaf water contents, a pressure - volume curve can be constructed, from which osmotic potential at any leaf water content can be determined. More detail on this method is given in Campbell (1985).
Several precautions are necessary to assure accurate measurements with the pressure bomb. It is important to prevent water loss from the leaf during the time the measurement is being made. The entire range of water potentials from full to zero turgor for most leaves is covered with a change in water content of about l0%. Thus a loss of only a small fraction of the leaf water during the measurement can result in errors of several hundred joules per kilogram.
Another precaution is related to the rate of pressure increase in the bomb. The rate of pressure increase must be slow enough to give the water in the leaf time for equilibration. If the pressure is increased too rapidly the pressure reading will be too high. One method of determining whether the rate of pressure increase is too rapid is to take a measurement, then release the pressure until the water just disappears from the cut surface, then increase the pressure again to obtain a second reading. The two readings so obtained should be the same. If the second reading is lower, the rate of pressure increase was probably too high.
Another source of error results from the fact that air usually passes through some of the xylem vessels as the pressure is being increased, and fluids other than water in the phloem or xylem can coat the cut surface and cause bubbling which is often difficult to distinguish from the real end point. If one wipes the cut surface during the early stages of pressurization, it is usually possible to detect the proper end point. With twigs, it also helps to strip away the bark down to the cambium.
Some tissues have very porous xylem which fills with water as the measurement is being made. The storage of water in this tissue causes the pressure readings to be too high. Other leaves, such as sunflower or sugar beet have large fleshy petioles which are easily crushed and difficult to seal in the pressure bomb. With all of these, it is often best to cut a section from the leaf, leaving a length of mid-rib or other conducting tissue attached. Pass the conducting tissue through the seal to indicate the end point. Any xylem which is attached to the mesophyll should provide the necessary connection for detecting the end point.
The Tensiometer
The tensiometer consists of a water-filled porous ceramic cup attached to a vacuum gauge. Figure 2 shows a typical unit. Water is withdrawn from the cup by the matric forces in the soil until the pressure potential inside the tensiometer is equal to matric potential of the soil water. The cup is permeable to salts but not to air and soil colloids, so the osmotic potential of the soil solution has no effect on the reading. The useful range of the tensiometer is about 0 to -80 J/kg. Very accurate measurements are possible in this range, and this is the range in which most moisture flow occurs in soil. In theory, the tensiometer could be used to measure matric potentials much lower than -80 J/kg, since the adhesive and cohesive forces of water are sufficient to withstand tensions of many hundreds of joules per kilogram (c.f. in the xylem of plants), but impurities in the water and on the surfaces inside of the tensiometer, and dissolved gases in the water cause cavitation when the absolute water pressure approaches zero. Once a gas phase in present in the tensiometer, negative pressures are impossible.
Figure 2. A form of tensiometer for measuring soil water potential.
The Thermocouple Psychrometer
The thermocouple psychrometer measures the humidity of air in equilibrium with a sample of plant, soil, or solution. Equation 1 can then be used to infer the water potential of the sample from the humidity measurement. The humidity measurement is made by enclosing a small thermocouple in a sealed chamber with the sample, allowing time for vapor equilibrium, and measuring the wet bulb temperature depression. Two methods have been used to measure wet bulb depression. In one, the thermocouple has a small ceramic bead surrounding the measuring junction. This bead is dipped in water before the thermocouple is placed over the sample. When the wet junction is placed over the sample, it cools by evaporation, and the temperature of the wet ceramic (which is measured by the thermocouple) is read as the wet bulb temperature.
In the other method, the thermocouple is cooled using the Peltier effect. A current is passed through the junction for a specified time until water has condensed on it. The wet bulb reading is the temperature of the wet junction as the water evaporates.
The psychrometer equation could be used to find h, but it is usually easier and more accurate to relate cooling to the water potential by calibrating the psychrometer with salt solutions of known water potential. The water osmotic potentials of the salt solutions can be calculated using
(3)
where c is the concentration (moles/kg), R is the gas constant, T is kelvin temperature, n is the number of osmotically active particles per molecule of solute (2 for KCl and NaCl), and f is the osmotic coefficient. Values for f as a function of c are listed in Table 1.
Table 1. Osmotic coefficients (f ) of NaCl and KCl solutions
c NaCl KCl c NaCl KCl 0.1 0.932 0.927 1.6 0.962 0.904 0.2 0.925 0.913 2.0 0.983 0.912 0.3 0.922 0.906 3.0 1.045 0.937 0.4 0.920 0.902 4.0 1.116 0.965 0.5 0.921 0.899 5.0 1.192 0.7 0.926 0.897 6.0 1.271 1.0 0.936 0.897
The basic idea of the thermocouple psychrometer can be incorporated into a number of devices for measuring water potential or component potentials. A commercial sample chamber psychrometer is shown in Fig. 3.
Figure 3. C-52 Sample Chamber Psychrometer from Wescor, Inc., Logan, UT.
It is useful for measuring osmotic potential when solutions of unknown concentration are absorbed in filter paper disks and placed in the chamber. Osmotic potentials of soil solution, body fluids, and sap expressed from plant tissue can be determined in the manner. Various sizes of chamber can be used with the sample chamber psychrometer, so soil samples and leaf disks can be used. It should be pointed out, however, that equilibrium time with these samples is much longer than with osmotic samples. This causes no serious problems with soil samples, but excision and equilibration of leaves in the psychrometer chamber can cause errors as large as several hundred joules per kilogram due to changes which take place in the tissue during equilibration and adsorption of water vapor by the leaf surface (Campbell, 1985).
Figure 4. SC-10A Thermocouple Psychrometer Sample Changer from Decagon Devices, Inc., Pullman WA.
A sample changer which is better suited to soil samples is shown in Fig. 4. This instrument can equilibrate 9 samples at once, and uses the ceramic bead psychrometer, so that readings over dry samples can be obtained. It has also been successfully used to measure water activity of food and fiber samples.
The thermocouple psychrometer can also be placed in a porous ceramic cup and buried in the soil. Figure 5 shows a cutaway view of the Wescor soil psychrometer. This instrument measures the water potential of the soil in situ, and must therefore be calibrated at a number of temperatures to determine the appropriate calibration factor to use at a given soil temperature. These units have also been used to measure the water potential in the xylem of trees and in potato tubers.
A fourth version of the thermocouple psychrometer is designed into a special clamp which can be used on leaves. The leaf psychrometer is shown in Fig. 6. This instrument measures the water potential of the leaf in situ, and, when properly used, is the most accurate method of assessing leaf water potential. The chamber is sealed to the leaf, and an aluminum clamp assures that the system stays isothermal. A thin layer of Styrofoam insulation covered with aluminized mylar surrounding the unit helps keep the temperature stable.
Figure 5. Ceramic cup soil psychrometer from Wescor, Inc., Logan, UT
Figure 6. Wescor L-51 Thermocouple psychrometer and leaf clamp for measuring leaf water potential in situ.
Several precautions are necessary for accurate measurements with the psychrometer method for measuring water potential. The measurement is very sensitive to differences in temperature between the measuring junction and the sample. A 1C temperature difference between the measuring junction and the sample will result in an error of 12000 J/kg. We would like measurement errors to be smaller than about 10 J/kg, so temperature differences must be kept smaller than about 0.001C. The measurement is also very sensitive to contamination in the sample chamber. Contaminants take up water and slow or prevent equilibrium. At humidities near 1.0, many clean surfaces and almost any contamination on a surface will take up water. The importance of cleanliness cannot be over emphasized in the use of the psychrometer.
The theoretical range of the Peltier-cooled thermocouple psychrometer (the Wescor units) is 0 to around -5000 J/kg. They are limited on the dry end by the ability of the Peltier effect to cool the junction sufficiently for water to condense. The ceramic bead psychrometers (Decagon units) are not limited on the dry end because water is placed on the wet junction. On the wet end, the practical range without special precautions and techniques is around -50 J/kg for both types. Samples wetter than -50 J/kg are difficult to equilibrate.
The Hydraulic Press
The equipment discussed so far is quite slow and cumbersome. For many studies it would be desirable to have same portable and fast method for estimating water potential. A device constructed from a hydraulic jack is available from Decagon which presses leaf tissue between a rubber membrane and a Plexiglas plate. If the tissue is free to deform under pressure, then the pressure applied to the tissue will be transmitted to the water in the tissue, and water will be expressed from the tissue. The pressure required to express water from the tissue should be related to the water potential. Because of the irregular shape (on a micro-scale) of tissue, a small part of the tissue is always subjected to greater pressure than the average for the tissue, when pressure is first applied. A small amount of water is therefore expressed at low pressure, and the pressure required to do this is apparently uncorrelated with water potential. If one continues to increase the pressure until the intercellular spaces infiltrate with water and water comes out around the edge of the tissue, then the pressure is uniform on all parts of the tissue, and this can be used as the end point of the reading. The pressure at which the tissue starts to darken (cell wall air space fill with water) correlates well with the zero turgor water potential (measured with the pressure bomb). Additional study is necessary to finally establish a theoretical basis for the hydraulic press, but this correlation with daytime pressure bomb measurements allows one to use the press for quick (but relatively imprecise) estimates of osmotic potential or daytime water potential of plants. These measurements could be useful in screening breeding lines or assessing water potentials of tree seedlings before or after transplanting.
Summary
The various methods for measuring water potential, along with their precautions, ranges of operation, and principles of operation are summarized in Table 2.
EXPERIMENT
Use the tensiometer to measure soil water potential, the pressure bomb to measure leaf water potential, and the sample chamber psychrometer to measure osmotic potential. From the leaf water potential and osmotic potential measurements, calculate the turgor pressure of the leaf.
Questions:
1. Briefly describe the measurement principle for each measurement you made, and tell the precautions needed for accurate measurement.
2. Why is it necessary to freeze the leaf before expressing sap for the osmotic measurement?
References
Campbell, G. S. 1985. Instruments for measuring plant water potential and its components. in Marshall, B, and Woodward, F. I., Instrumentation for Environmental Physiology. Cambridge Univ. Press, Cambridge, p.193-209.
Campbell, G. S. 1988. Soil water potential measurement: an overview. Irrig. Sci. 9:265-273.
Table 2. Comparison of methods for measuring water potential
Method Measures Principle Range Precautions Tensiometer matric potential balances internal suction against external matric potential through a porous cup 0 to -80 J/kg difficult to keep system gas free Pressure bomb water potential of plant tissue (leaves); can also measure osmotic potential balances external pneumatic potential against cell water potential to produce zero potential in the xylem 0 to -6 kJ/kg sometimes difficult to see endpoint; use slow pressure increase (2 psi/s); prevent water loss from leaf; Hydraulic press (jack) probably measures osmotic potential; measures matric potential of frozen leaves principle not worked out. Calibration based on correlation with other methods. 0 to -6 kJ/kg Many endpoints visible. Care needed to choose the right one. Sample chamber psychrometer matric plus osmotic potential of soils, leaves, and solutions Measures hr of vapor equilibrated with sample. Uses eq. 2 to get potential 0 to -6 kJ/kg (some methods go to -100 kJ/kg) very sensitive to dirt and temperature fluctuations; has problems in the 0 to -100 J/kg range; requires frequent calibration in situ soil psychrometer matric plus osmotic potential in soil same as sample changer 0 to -5 kJ/kg same as sample changer in situ leaf psychrometer water potential of leaves same as sample changer 0 to -5 kJ/kg same as sample changer; should be shaded from direct sun; must have good seal to leaf | 3,929 | 18,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-23 | latest | en | 0.936589 |
https://codereview.stackexchange.com/questions/284539/a-recursive-sum-template-function-implementation-with-unwrap-level-in-c | 1,721,079,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00565.warc.gz | 155,229,092 | 45,599 | # A recursive_sum Template Function Implementation with Unwrap Level in C++
This is a follow-up question for A Summation Function For Various Type Arbitrary Nested Iterable Implementation in C++ and A recursive_transform_view Template Function Implementation. In the previous question, the implementation of recursive_sum template function performs summation operation on input container exhaustively. I am trying to make another version recursive_sum function which is used for dealing with nested iterables with unwrap level in C++.
The experimental implementation
• recursive_sum template function implementation:
template< std::size_t unwrap_level,
class T>
requires (unwrap_level <= recursive_depth<T>())
constexpr auto recursive_sum(const T& input)
{
if constexpr (recursive_depth<T>() - unwrap_level == 0)
{
return input;
}
else
{
return UL::recursive_transform<unwrap_level>(
input,
[](auto&& element){ return recursive_sum_all(element); }
);
}
}
Full Testing Code
The full testing code:
// A recursive_sum Template Function Implementation with Unwrap Level in C++
#include <algorithm>
#include <array>
#include <cassert>
#include <chrono>
#include <complex>
#include <concepts>
#include <deque>
#include <execution>
#include <exception>
#include <functional>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <mutex>
#include <numeric>
#include <optional>
#include <queue>
#include <ranges>
#include <stack>
#include <stdexcept>
#include <string>
#include <tuple>
#include <type_traits>
#include <utility>
#include <variant>
#include <vector>
// is_reservable concept
template<class T>
concept is_reservable = requires(T input)
{
input.reserve(1);
};
// is_sized concept, https://codereview.stackexchange.com/a/283581/231235
template<class T>
concept is_sized = requires(T x)
{
std::size(x);
};
template<typename T>
concept is_summable = requires(T x) { x + x; };
// recursive_depth function implementation
template<typename T>
constexpr std::size_t recursive_depth()
{
return 0;
}
template<std::ranges::input_range Range>
constexpr std::size_t recursive_depth()
{
return recursive_depth<std::ranges::range_value_t<Range>>() + 1;
}
// recursive_invoke_result_t implementation
template<typename, typename>
struct recursive_invoke_result { };
template<typename T, std::regular_invocable<T> F>
struct recursive_invoke_result<F, T> { using type = std::invoke_result_t<F, T>; };
template<typename F, template<typename...> typename Container, typename... Ts>
requires (
!std::regular_invocable<F, Container<Ts...>>&& // F cannot be invoked to Container<Ts...> directly
std::ranges::input_range<Container<Ts...>>&&
requires { typename recursive_invoke_result<F, std::ranges::range_value_t<Container<Ts...>>>::type; })
struct recursive_invoke_result<F, Container<Ts...>>
{
using type = Container<
typename recursive_invoke_result<
F,
std::ranges::range_value_t<Container<Ts...>>
>::type
>;
};
template<template<typename, std::size_t> typename Container,
typename T,
std::size_t N,
std::regular_invocable<Container<T, N>> F>
struct recursive_invoke_result<F, Container<T, N>>
{
using type = std::invoke_result_t<F, Container<T, N>>;
};
template<template<typename, std::size_t> typename Container,
typename T,
std::size_t N,
typename F>
requires (
!std::regular_invocable<F, Container<T, N>>&& // F cannot be invoked to Container<Ts...> directly
requires { typename recursive_invoke_result<F, std::ranges::range_value_t<Container<T, N>>>::type; })
struct recursive_invoke_result<F, Container<T, N>>
{
using type = Container<
typename recursive_invoke_result<
F,
std::ranges::range_value_t<Container<T, N>>
>::type
, N>;
};
template<typename F, typename T>
using recursive_invoke_result_t = typename recursive_invoke_result<F, T>::type;
// recursive_variadic_invoke_result_t implementation
template<std::size_t, typename, typename, typename...>
struct recursive_variadic_invoke_result { };
template<typename F, class...Ts1, template<class...>class Container1, typename... Ts>
struct recursive_variadic_invoke_result<1, F, Container1<Ts1...>, Ts...>
{
using type = Container1<std::invoke_result_t<F,
std::ranges::range_value_t<Container1<Ts1...>>,
std::ranges::range_value_t<Ts>...>>;
};
template<std::size_t unwrap_level, typename F, class...Ts1, template<class...>class Container1, typename... Ts>
requires ( std::ranges::input_range<Container1<Ts1...>> &&
requires { typename recursive_variadic_invoke_result<
unwrap_level - 1,
F,
std::ranges::range_value_t<Container1<Ts1...>>,
std::ranges::range_value_t<Ts>...>::type; }) // The rest arguments are ranges
struct recursive_variadic_invoke_result<unwrap_level, F, Container1<Ts1...>, Ts...>
{
using type = Container1<
typename recursive_variadic_invoke_result<
unwrap_level - 1,
F,
std::ranges::range_value_t<Container1<Ts1...>>,
std::ranges::range_value_t<Ts>...
>::type>;
};
template<std::size_t unwrap_level, typename F, typename T1, typename... Ts>
using recursive_variadic_invoke_result_t = typename recursive_variadic_invoke_result<unwrap_level, F, T1, Ts...>::type;
// https://codereview.stackexchange.com/a/253039/231235
template<template<class...> class Container = std::vector, std::size_t dim, class T>
constexpr auto n_dim_container_generator(T input, std::size_t times)
{
if constexpr (dim == 0)
{
return input;
}
else
{
return Container(times, n_dim_container_generator<Container, dim - 1, T>(input, times));
}
}
namespace UL // unwrap_level
{
template< std::ranges::input_range Container,
std::copy_constructible F>
requires (std::ranges::view<Container>&&
std::is_object_v<F>)
constexpr auto make_view(const Container& input, const F& f) noexcept
{
return std::ranges::transform_view(
input,
[&f](const auto&& element) constexpr { return recursive_transform(element, f ); } );
}
/* Override make_view to catch dangling references. A borrowed range is
* safe from dangling..
*/
template <std::ranges::input_range T>
requires (!std::ranges::borrowed_range<T>)
constexpr std::ranges::dangling make_view(T&&) noexcept
{
return std::ranges::dangling();
}
// clone_empty_container template function implementation
template< std::size_t unwrap_level = 1,
std::ranges::input_range Container,
std::copy_constructible F>
requires (std::ranges::view<Container>&&
std::is_object_v<F>)
constexpr auto clone_empty_container(const Container& input, const F& f) noexcept
{
const auto view = make_view(input, f);
recursive_variadic_invoke_result<unwrap_level, F, Container> output(std::span{input});
return output;
}
// recursive_transform template function implementation (the version with unwrap_level template parameter)
template< std::size_t unwrap_level = 1,
class T,
std::copy_constructible F>
requires (unwrap_level <= recursive_depth<T>()&& // handling incorrect unwrap levels more gracefully, https://codereview.stackexchange.com/a/283563/231235
std::ranges::view<T>&&
std::is_object_v<F>)
constexpr auto recursive_transform(const T& input, const F& f)
{
if constexpr (unwrap_level > 0)
{
auto output = clone_empty_container(input, f);
if constexpr (is_reservable<decltype(output)>&&
is_sized<decltype(input)>)
{
output.reserve(input.size());
std::ranges::transform(
input,
std::ranges::begin(output),
[&f](auto&& element) { return recursive_transform<unwrap_level - 1>(element, f); }
);
}
else
{
std::ranges::transform(
input,
std::inserter(output, std::ranges::end(output)),
[&f](auto&& element) { return recursive_transform<unwrap_level - 1>(element, f); }
);
}
return output;
}
else if constexpr(std::regular_invocable<F, T>)
{
return std::invoke(f, input);
}
else
{
static_assert(!std::regular_invocable<F, T>, "Uninvocable?");
}
}
/* This overload of recursive_transform is to support std::array
*/
template< std::size_t unwrap_level = 1,
template<class, std::size_t> class Container,
typename T,
std::size_t N,
typename F >
requires (std::ranges::input_range<Container<T, N>>)
constexpr auto recursive_transform(const Container<T, N>& input, const F& f)
{
Container<recursive_variadic_invoke_result_t<unwrap_level, F, T>, N> output;
std::ranges::transform(
input,
std::ranges::begin(output),
[&f](auto&& element){ return recursive_transform<unwrap_level - 1>(element, f); }
);
return output;
}
// recursive_transform function implementation (the version with unwrap_level, without using view)
template<std::size_t unwrap_level = 1, class T, class F>
requires (!std::ranges::view<T>)
constexpr auto recursive_transform(const T& input, const F& f)
{
if constexpr (unwrap_level > 0)
{
static_assert(unwrap_level <= recursive_depth<T>(),
"unwrap level higher than recursion depth of input"); // trying to handle incorrect unwrap levels more gracefully
recursive_variadic_invoke_result_t<unwrap_level, F, T> output{};
std::ranges::transform(
input, // passing a range to std::ranges::transform()
std::inserter(output, std::ranges::end(output)),
[&f](auto&& element) { return recursive_transform<unwrap_level - 1>(element, f); }
);
return output;
}
else
{
return std::invoke(f, input); // use std::invoke()
}
}
}
namespace NonUL
{
template< std::ranges::input_range Container,
std::copy_constructible F>
requires (std::ranges::input_range<Container>&&
std::ranges::view<Container>&&
std::is_object_v<F>)
constexpr auto make_view(const Container& input, const F& f) noexcept
{
return std::ranges::transform_view(
input,
[&f](const auto&& element) constexpr { return recursive_transform(element, f ); } );
}
/* Override make_view to catch dangling references. A borrowed range is
* safe from dangling..
*/
template <std::ranges::input_range T>
requires (!std::ranges::borrowed_range<T>)
constexpr std::ranges::dangling make_view(T&&) noexcept
{
return std::ranges::dangling();
}
/* Base case of NonUL::recursive_transform template function
https://codereview.stackexchange.com/a/283581/231235
*/
template< typename T,
std::regular_invocable<T> F>
requires (std::copy_constructible<F>)
constexpr auto recursive_transform( const T& input, const F& f )
{
return std::invoke( f, input );
}
/* The recursive case of NonUL::recursive_transform template function
https://codereview.stackexchange.com/a/283581/231235
*/
template< std::ranges::input_range Container,
std::copy_constructible F>
requires (std::ranges::input_range<Container>&&
std::ranges::view<Container>&&
std::is_object_v<F>)
constexpr auto recursive_transform(const Container& input, const F& f)
{
const auto view = make_view(input, f);
recursive_invoke_result_t<F, Container> output( std::ranges::begin(view), std::ranges::end(view) );
// One last sanity check.
if constexpr( is_sized<Container> && is_sized<recursive_invoke_result_t<F, Container>> )
{
assert( output.size() == input.size() );
}
return output;
}
/* The recursive case of NonUL::recursive_transform template function for std::array
https://codereview.stackexchange.com/a/283581/231235
*/
template< template<typename, std::size_t> typename Container,
typename T,
std::size_t N,
std::copy_constructible F>
requires std::ranges::input_range<Container<T, N>>
constexpr auto recursive_transform(const Container<T, N>& input, const F& f)
{
Container<recursive_invoke_result_t<F, T>, N> output;
std::ranges::transform( // Use std::ranges::transform() for std::arrays
input,
std::ranges::begin(output),
[&f](auto&& element){ return recursive_transform(element, f); }
);
// One last sanity check.
if constexpr( is_sized<Container<T, N>> && is_sized<recursive_invoke_result_t<F, Container<T, N>>> )
{
assert( output.size() == input.size() );
}
return output;
}
}
/* recursive_sum_all template function performs summation operation on input container exhaustively
*/
template<class T> requires is_summable<T>
auto recursive_sum_all(const T& input)
{
return input;
}
template<std::ranges::input_range T>
auto recursive_sum_all(const T inputArray)
{
typedef typename std::iterator_traits<typename T::iterator>::value_type
value_type;
decltype(recursive_sum_all(std::declval<value_type &&>())) sun_output{};
for (auto& element : inputArray)
{
sun_output += recursive_sum_all(element);
}
return sun_output;
}
template< std::size_t unwrap_level,
class T>
requires (unwrap_level <= recursive_depth<T>())
constexpr auto recursive_sum(const T& input)
{
if constexpr (recursive_depth<T>() - unwrap_level == 0)
{
return input;
}
else
{
return UL::recursive_transform<unwrap_level>(
input,
[](auto&& element){ return recursive_sum_all(element); }
);
}
}
template<class T>
requires (std::ranges::input_range<T>)
constexpr auto recursive_print(const T& input, const int level = 0)
{
T output = input;
std::cout << std::string(level, ' ') << "Level " << level << ":" << std::endl;
std::transform(input.cbegin(), input.cend(), output.begin(),
[level](auto&& x)
{
std::cout << std::string(level, ' ') << x << std::endl;
return x;
}
);
return output;
}
template<class T>
requires (std::ranges::input_range<T> &&
std::ranges::input_range<std::ranges::range_value_t<T>>)
constexpr T recursive_print(const T& input, const int level = 0)
{
T output = input;
std::cout << std::string(level, ' ') << "Level " << level << ":" << std::endl;
std::transform(input.cbegin(), input.cend(), output.begin(),
[level](auto&& element)
{
return recursive_print(element, level + 1);
}
);
return output;
}
void recursive_sum_tests()
{
auto test_vectors = n_dim_container_generator<std::vector, 4, int>(1, 3);
std::cout << "Play with test_vectors:\n\n";
std::cout << "recursive_sum_all function: \n";
auto recursive_sum_all_result = recursive_sum_all(test_vectors);
std::cout << recursive_sum_all_result << "\n\n";
std::cout << "unwrap_level = 4:\n";
auto test_output_1 = recursive_sum<4>(test_vectors);
recursive_print(test_output_1);
std::cout << "\n\n";
std::cout << "unwrap_level = 3:\n";
auto test_output_2 = recursive_sum<3>(test_vectors);
recursive_print(test_output_2);
std::cout << "\n\n";
std::cout << "unwrap_level = 2:\n";
auto test_output_3 = recursive_sum<2>(test_vectors);
recursive_print(test_output_3);
std::cout << "\n\n";
std::cout << "unwrap_level = 1:\n";
auto test_output_4 = recursive_sum<1>(test_vectors);
recursive_print(test_output_4);
std::cout << "\n\n";
std::cout << "unwrap_level = 0:\n";
auto test_output_5 = recursive_sum<0>(test_vectors);
std::cout << test_output_5 << "\n\n";
return;
}
int main()
{
recursive_sum_tests();
return 0;
}
The output of the test code above:
Play with test_vectors:
recursive_sum_all function:
81
unwrap_level = 4:
Level 0:
Level 1:
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 1:
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 1:
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
Level 2:
Level 3:
1
1
1
Level 3:
1
1
1
Level 3:
1
1
1
unwrap_level = 3:
Level 0:
Level 1:
Level 2:
3
3
3
Level 2:
3
3
3
Level 2:
3
3
3
Level 1:
Level 2:
3
3
3
Level 2:
3
3
3
Level 2:
3
3
3
Level 1:
Level 2:
3
3
3
Level 2:
3
3
3
Level 2:
3
3
3
unwrap_level = 2:
Level 0:
Level 1:
9
9
9
Level 1:
9
9
9
Level 1:
9
9
9
unwrap_level = 1:
Level 0:
27
27
27
unwrap_level = 0:
81
Godbolt link
All suggestions are welcome.
The summary information:
• Which question it is a follow-up to?
A recursive_transform_view Template Function Implementation
• What changes has been made in the code since last question?
I am trying to implement recursive_sum template function with unwrap level parameter in this post. In this example, I think that the usage of the version recursive_transform with unwrap level is necessary. If there is any misunderstanding, please let me know.
• Why a new review is being asked for?
Please review the experimental implementation of recursive_sum template function. The function perfroms the summation operation on specified level of container.
# Make it more generic
Summing is a very specific operation. What if you want to calculate the product of all elements instead? Or get the minimum or maximum value? Instead of hardcoding the operation, create a recursive_reduce() that works like std::reduce(). You can still have it sum by default if you don't specify which operation to perform.
# Associativity and initial values
Note that the order in which you perform operations matters. While operator+ is often associative, it doesn't have to be. That's why in C++23, std::ranges::fold_left() and std::ranges::fold_right() were introduced.
Furthermore, some value types might not have a default constructor, thus sun_output{} might not compile. This is why most algorithms in the STL that perform some kind of reduction take an initial value as a parameter. I recommend you add that here as well.
Sometimes you know the input is non-empty, and you want to avoid providing an initial values. C++23 introduced std::ranges::fold_left_first() and related functions for this purpose.
# Meaning of the unwrap level
I was surprised by the fact that your recursive_sum() uses recursive_transform() internally, and by the output from your example code. I would rather have expected the following:
std::vector<std::string> words = {"foo", "bar", "baz", "quux"};
std::cout << recursive_sum<2>(words) << '\n';
std::cout << recursive_sum<1>(words) << '\n';
To output:
:
foobarbazquux
Basically, the call to recursive_sum<2>(words) would unwrap two levels, so it would iterate over the characters in each string, whereas recursive_sum<1>(words) would unwrap only the vector, and add the strings together.
I think that would be more logical. Also consider math libraries that have vector and matrix types that can be iterated over; sometimes you want to sum the elements of a matrix, sometimes you want to sum matrices together, but your code will always sum the innermost elements.
Here is how I would implement a basic recursive_sum():
template<std::size_t unwrap_level,
class R,
class T = recursive_unwrap_type<unwrap_level, R>>
requires (unwrap_level <= recursive_depth<R>())
constexpr auto recursive_sum(const R& input, T init = {})
{
if constexpr (unwrap_level > 0) {
for (const auto& element: input) {
init = recursive_sum(element, init);
}
} else {
init += input;
}
return init;
}
Where recursive_unwrap_type<unwrap_level, R> would give you the type after unwrapping R for unwrap_level levels.
If you want your original behavior, the caller could then combine recursive_transform() and recursive_sum() themselves, like so:
auto test_output_3 = recursive_transform<2>(
test_vectors,
[](auto&& element){ return recursive_sum(element); }
);
• Thank you for answering. > I was surprised by the fact that your recursive_sum() uses recursive_transform() internally. Do you have any other idea for the implementation of recursive_sum function? Or there is any other better way to do this? Commented Apr 20, 2023 at 7:38
• Thank you for updating the answer. I understand the opinion you mentioned. :) Commented Apr 20, 2023 at 11:42 | 4,907 | 19,213 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.336074 |
https://betterlesson.com/lesson/reflection/14169/assessing-the-standard-quickly | 1,498,500,049,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320841.35/warc/CC-MAIN-20170626170406-20170626190406-00276.warc.gz | 724,881,792 | 23,196 | ## Reflection: Quizzes The Quilt Code: A Lesson in Triangles Through History - Section 5: Quick Assessment: Mastering the Standard
In this quick little quiz, I chose to add in a small short answer question, last minute. As I was roving the class, I decided that I wanted to know HOW they knew it was a right triangle and be able to explain that on this quick quiz. It was at first my intention to simply assess using a quilt block, whether or not they could pick out a right triangle. That would master the standard exactly. But, because I am committed to always holding them accountable to their thinking, I added in the writing piece at that bottom. It produced interesting answers that you can see in their samples.
Assessing the standard quickly
Quizzes: Assessing the standard quickly
# The Quilt Code: A Lesson in Triangles Through History
Unit 5: Geometry
Lesson 3 of 3
## Big Idea: Students learn to identify right triangles as they study the quilt blocks that were allegedly part of the controversial American Quilt Code abolitionists used to guide slaves to freedom.
Print Lesson
6 teachers like this lesson
Standards:
Subject(s):
Math, Social Studies, Civil War, Geometry, Symmetry, Language Arts, African American History, creativity, Art and Math, right triangle, Hands on Activity
53 minutes
### Mary Ellen Kanthack
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Environment: Urban | 511 | 2,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-26 | longest | en | 0.920435 |
https://www.answers.com/Q/How_big_is_the_engine_in_cubic_inches_in_the_2001_Chevy_1500 | 1,596,847,966,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00436.warc.gz | 597,760,465 | 33,232 | Late Model 1979-New Ford Mustangs
Math and Arithmetic
# How big is the engine in cubic inches in the 2001 Chevy 1500?
131415
###### 2005-10-14 23:20:36
If it is a 5.7 litre it is a 350 cubic inch, if it is a 5.3 litre it is a 327 cubic inch if it is a V8.
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## Related Questions
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1500 sq ft * 4 inches = 1500/9 sq yards * 1/9 yards = 18.518518... cubic yards.
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###### Math and ArithmeticCars & VehiclesFuel and EnginesUnits of MeasureChevy SilveradoChevy ExpressChevy CheyenneChevroletTiming and Firing OrdersAuto Body and InteriorsChevy 305Chevy S-10Flower GardeningCrankcases and PCV ValvesFuel FiltersChevy 350Chevy Blazer
Copyright ยฉ 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 608 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.906667 |
https://stats.stackexchange.com/questions/220304/manually-calculating-logistic-regression-coefficient/220306 | 1,585,567,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00000.warc.gz | 729,857,062 | 32,364 | # Manually calculating logistic regression coefficient
I am taking this short example from wiki:https://en.wikipedia.org/wiki/Logistic_regression
Hours | Pass
--------|-------
0.50 | 0
0.75 | 0
1.00 | 0
1.25 | 0
1.50 | 0
1.75 | 0
1.75 | 1
2.00 | 0
2.25 | 1
2.50 | 0
2.75 | 1
3.00 | 0
3.25 | 1
3.50 | 0
4.00 | 1
4.25 | 1
4.50 | 1
4.75 | 1
5.00 | 1
5.50 | 1
I am trying to manually calculate the intercept and coefficient.
How does the algorithm calculates the log of odds at each point to fit linear regression? If i try to calculate log of odds, I get into situation where it is log(0) or log(infinity). Can somebody please help here?
You are right that although you should be able to calculate the OLS coefficient estimate in logit space, you can't do it directly because the logit, $g(y) = \log \frac{p}{1-p}$, goes either to $-\infty$ for $y=0$ or $\infty$ for $y=1$. An added difficulty is that the variance in this model depends on $x$.
The likelihood for logistic regression is optimized by an algorithm called iteratively reweighted least squares (IRLS). There is a nice breakdown of this in Shalizi's Advanced Data Analysis from an Elementary Point of View, from which I have the details below:
• To deal with the infinite logit problem, make a first-order Taylor approximation to $g(y)$ around the point $p$ such that $g(y) \approx g(p) + (y − p)g'(p)$. Since $g(p)$ is by definition $\beta_0 + \beta x$, put that in there instead of $g(p)$ and say that your effective response is $z = \beta_0 + \beta x + (y-p)g'(p)$.
• Calculate the variance $V[Z|X=x] = V[(Y-p)g'(p) | X=x]=g'(p)^2V(p)$. Use this to weight your samples so that you can simply do a weighted regression of $z$ on $x$.
At this point you might ask yourself how you can use the regression coefficients you're trying to estimate to calculate your effective response, $z$. Of course you can't. And what is $p$ anyway? That's where the iterative part of IRLS comes in: you start with some guess at the $\beta$s, for instance to set them all to 0. From this you can first calculate the fitted probabilities $p$, and second use these fitted probabilities and your current coefficient estimates to calculate $z$.
All this and you get a new estimate for your $\beta$s, and it should be closer to the right one, but probably not the right one. So you iterate: use the new coefficients to calculate new fitted probabilities, calculate new effective responses, new weights, and go again. Sooner or later, unless you're unlucky, the $\beta$s will converge to a nice estimate. Says Shalizi:
The treatment above is rather heuristic, but it turns out to be equivalent to using Newton’s method, only with the expected second derivative of the log likelihood, instead of its actual value.
So in summary: you never use the logit directly because, as you point out, it's impractical. You can certainly calculate the logistic regression coefficients by hand, but it won't be fun. | 845 | 3,048 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-16 | latest | en | 0.879211 |
https://solutionsadda.in/topic/?tp=Graphs | 1,679,829,329,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00685.warc.gz | 608,035,551 | 39,651 | ## Graphs
Question 1
Let G be a simple undirected graph. Let TD be a depth first search tree of G. Let TB be a breadth first search tree of G. Consider the following statements.
(I) No edge of G is a cross edge with respect to TD.(A cross edge in G is between two nodes neither of which is an ancestor of the other in TD.)
(II) For every edge (u,v) of G, if u is at depth i and v is at depth j in TB, then ∣i−j∣ = 1.
Which of the statements above must necessarily be true?
A I only B II only C Both I and II D Neither I nor II
Data-Structures Graphs GATE 2018 Video-Explanation
Question 1 Explanation:
I. Undirected graph do not have cross edges in DFS. But can have cross edges in directed graph. Hence True.
II. Just draw a triangle ABC. Source is A. Vertex B and C are at same level at distance 1.
There is an edge between B and C too. So here |i - j| = |1 - 1| = 0. Hence, False.
Question 2
Let G be a weighted connected undirected graph with distinct positive edge weights. If every edge weight is increased by the same value, then which of the following statements is/are TRUE?
P: Minimum spanning tree of G does not change
Q: Shortest path between any pair of vertices does not change
A P only B Q only C Neither P nor Q D Both P and Q
Data-Structures Graphs GATE 2016 [Set-1] Video-Explanation
Question 2 Explanation:
Given undirected weighted graph with distinct positive edges.
Every edge weight is increased by the same value say,
P: Minimum Spanning Tree will not change ABC in both the cases.
Q: Shortest path will change because in 1st figure the path from A to C calculated as ABC but in fig.2, it is AC (Direct path).
Question 3
Consider the weighted undirected graph with 4 vertices, where the weight of edge {i,j} is given by the entry Wij in the matrix W.
The largest possible integer value of x, for which at least one shortest path between some pair of vertices will contain the edge with weight x is _________.
A 12 B 13 C 14 D 15
Data-Structures Graphs GATE 2016 [Set-1] Video-Explanation
Question 3 Explanation:
Let vertices be A, B, C and D.
x directly connects C to D.
The shortest path (excluding x) from C to D is of weight 12 (C-B-A-D).
Question 4
In an adjacency list representation of an undirected simple graph G = (V,E), each edge (u,v) has two adjacency list entries: [v] in the adjacency list of u, and [u] in the adjacency list of v. These are called twins of each other. A twin pointer is a pointer from an adjacency list entry to its twin. If |E| = m and |V| = n, and the memory size is not a constraint, what is the time complexity of the most efficient algorithm to set the twin pointer in each entry in each adjacency list?
A Θ(n2) B Θ(n+m) C Θ(m2) D Θ(n4)
Data-Structures Graphs GATE 2016 [Set-2] Video-Explanation
Question 4 Explanation:
Applying BFS on Undirected graph give you twin pointer.
Visit every vertex level-wise for every vertex fill adjacent vertex in the adjacency list. BFS take O(m+n) time.
Note:
Twin Pointers can be setup by keeping track of parent node in BFS or DFS of graph.
Question 5
Let G be a graph with n vertices and m edges. What is the tightest upper bound on the running time of Depth First Search on G, when G is represented as an adjacency matrix?
A θ(n) B θ(n+m) C θ(n2) D θ(m2)
Data-Structures Graphs GATE 2014 [Set-1]
Question 5 Explanation:
DFS visits each vertex once and as it visits each vertex, we need to find all of its neighbours to figure out where to search next. Finding all its neighbours in an adjacency matrix requires O(V ) time, so overall the running time will be O(V2).
Question 6
Consider the directed graph given below.
Which one of the following is TRUE?
A The graph does not have any topological ordering. B Both PQRS and SRQP are topological. C Both PSRQ and SPRQ are topological orderings. D PSRQ is the only topological ordering.
Data-Structures Graphs GATE 2014 [Set-1]
Question 6 Explanation:
There are no cycles in the graph, so topological orderings do exist.
We can consider P & S as starting vertex, followed by R & Q.
Hence, PSRQ & SPRQ are the topological orderings.
Question 7
Consider the tree arcs of a BFS traversal from a source node W in an unweighted, connected, undirected graph. The tree T formed by the tree arcs is a data structure for computing
A the shortest path between every pair of vertices. B the shortest path from W to every vertex in the graph. C the shortest paths from W to only those nodes that are leaves of T. D the longest path in the graph.
Data-Structures Graphs GATE 2014 [Set-2]
Question 7 Explanation:
One of the application of BFS algorithm is to find the shortest path between nodes u and v.
But in the given question the BFS algorithm starts from the source vertex w and we can find the shortest path from W to every vertex of the graph.
Question 8
Suppose depth first search is executed on the graph below starting at some unknown vertex. Assume that a recursive call to visit a vertex is made only after first checking that the vertex has not been visited earlier. Then the maximum possible recursion depth (including the initial call) is _________.
A 19 B 20 C 21 D 22
Algorithms Graphs GATE 2014 [Set-3]
Question 8 Explanation:
Note: We should not consider backtrack edges, it reduces recursion depth in stack.
So the maximum possible recursive depth will be 19.
Question 9
Consider a complete undirected graph with vertex set {0, 1, 2, 3, 4}. Entry Wij in the matrix W below is the weight of the edge {i, j}.
What is the minimum possible weight of a path P from vertex 1 to vertex 2 in this graph such that P contains at most 3 edges?
A 7 B 8 C 9 D 10
Data-Structures Graphs GATE 2010
Question 9 Explanation:
The minimum possible weight of a path p from vertex 1 to vertex 2 such that p contains atmost 3 edges,
= 1 + 4 + 3
= 8
Question 10
The most efficient algorithm for finding the number of connected components in an undirected graph on n vertices and m edges has time complexity
A θ(n) B θ(m) C θ(m+n) D θ(mn)
Data-Structures Graphs GATE 2008
Question 10 Explanation:
To find the number of connected components using either BFS or DFS time complexity is θ(m+n).
Suppose if we are using Adjacency matrix means it takes θ(n2).
Question 11
The Breadth First Search algorithm has been implemented using the queue data structure. One possible order of visiting the nodes of the following graph is
A MNOPQR B NQMPOR C QMNPRO D QMNPOR
Data-Structures Graphs GATE 2008
Question 11 Explanation:
Option C: QMNPRO
→ Queue starts with Q then neighbours of Q is MNP and it is matching with the given string .
→ Now , Next node to be considered is M . Neighbours of M are N, Q and R , but N and Q are already in Queue. So, R is matching with one given string
→ Now, next node to be considered is N. Neighbours of N are M, Q and O, but M and Q are already in Queue. So, O is matching with a given string.
Hence , Option C is Correct.
Similarly, check for option (D).
Question 12
G is a graph on n vertices and 2n–2 edges. The edges of G can be partitioned into two edge-disjoint spanning trees. Which of the following is NOT true for G?
A For every subset of k vertices, the induced subgraph has at most 2k–2 edges B The minimum cut in G has at least two edges C There are two edge-disjoint paths between every pair to vertices D There are two vertex-disjoint paths between every pair of vertices
Data-Structures Graphs GATE 2008
Question 12 Explanation:
→ In Spanning tree n nodes require n-1 edges. The above question they mentioned 2 disjoint spanning trees. So, it requires n-1 + n-1 = 2n-2 edges. Except option D everything is correct.
> Option A: True: Subgraph with k vertices here is no chance to get more than 2k−2 edges. Subgraph with n−k vertices, definitely less than 2n−2k edges.
-> Option B: True: Take any subgraph SG with k vertices. The remaining subgraph will have n−k vertices. Between these two subgraphs there should be at least 2 edges because we are taken 2 spanning trees in SG.
-> Option C: True: A spanning tree covers all the vertices. So, 2 edge-disjoint spanning trees in G means, between every pair of vertices in G we have two edge-disjoint paths (length of paths may vary).
Question 13
Consider the DAG with Consider V = {1, 2, 3, 4, 5, 6}, shown below. Which of the following is NOT a topological ordering?
A 1 2 3 4 5 6 B 1 3 2 4 5 6 C 1 3 2 4 6 5 D 3 2 4 1 6 5
Data-Structures Graphs GATE 2007
Question 13 Explanation:
The process to find topological order is,
(i) Go with vertex with indegree 0.
(ii) Then remove that vertex and also remove the edges going from it.
(iii) Repeat again from (i) till every vertex is completed.
Now we can see that in option (D), '3' is given first which is not possible because indegree of vertex '3' is not zero.
Hence option (D) is not topological ordering.
There are 13 questions to complete.
Register Now | 2,377 | 9,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-14 | latest | en | 0.900142 |
https://www.physicsforums.com/threads/man-standing-on-a-scale.899246/ | 1,531,884,065,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590046.11/warc/CC-MAIN-20180718021906-20180718041906-00307.warc.gz | 935,723,625 | 16,416 | # Homework Help: Man standing on a scale
1. Jan 5, 2017
### wah31
1. The problem statement, all variables and given/known data
Man standing on scale and pulling on a rope connected to two pulleys
2. Relevant equations
The mass of the man is: m=100 kg
The weight of the pulleys, cables, balance, and the frictions are not taken into account
g=10m x s-2
3. The attempt at a solution
the force that the man should draw in order to maitain balance:
T1=250 N
The weight displayed on the scale:
P= 125 kg
Other tensions:
T2=250 N
T3=500 N
T4=500 N
T5=1000 N
Do your agree with that ?
Thank you..
#### Attached Files:
• ###### Balance-page-001.jpg
File size:
5.7 KB
Views:
72
Last edited: Jan 5, 2017
2. Jan 5, 2017
### oz93666
You have not laid out the question properly ... what information is given in the question?
3. Jan 5, 2017
### wah31
You just have to validate or not thes values (T1, P, T2, T3, T4, T5)
4. Jan 5, 2017
### haruspex
I agree with all the tensions, but I do not understand how you got 125kg for the scale reading. Consider again which way the forces act on the man.
5. Jan 5, 2017
### wah31
I'm not convinced either
I thought about it..
6. Jan 5, 2017
### oz93666
Yes I agree there is something wrong
What do we know for sure ? ....T1= T2 and T3 =T4 ...... 4xT1 = 2xT4 = T5 and T5 is supporting all the mass = 100kg = 1000N ... so all the tensions are correct .
The weight indicated on the scale must be the man's weight , minus the the pull from T1 .... 1000 - 250 = 750N scale should indicate 75Kg not 125kg
7. Jan 5, 2017
### wah31
Yes you are right
Thank you.. it's cool
8. Jan 5, 2017
### haruspex
This is a homework forum. We point out errors, provide hints, explain principles.... We do not post solutions. | 553 | 1,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-30 | latest | en | 0.838961 |
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## ENGLISH CONTENT OF SITE
Jay Kappraff ANNE BULCKENS’ ANALYSIS OF THE PROPORTIONS OF THE PARTHENON AND ITS MEANINGS Based on the PhD thesis of Anne Bulckens, the proportional system of the Parthenon is examined. With the appropriate choice of a module and the length of a «Parthenon foot», all dimensions are shown to be integers. These integer values are shown to be related to the musical scale of Pythagoras. Among the many musical relationships expressed by this structure, this paper focuses on a pentatonic scale made up of lengths, widths and heights of the outer temple and the inner temple or cella. весь текст 28.08.2006
Eduard M Soroko GOLDEN CODE OF NEFERTITI`S IMAGE The logic by which the creator of the Nefertiti statuette was guided is clear and simple. Harmony is the prerogative of the divine order which dominates the Universe, and geometry is a means of its expression. The Queen is given to people by God. Hence, her image, personifyng the world’s wise tranquillity must be full of geometrical perfection to bear the stamp of high and irreproachable harmony, beauty and clearness. Such is, in fact, the whole philosophy of ancient Egyptian art, which glorified the eternal, the measured, the perfect in the ever changing being. весь текст 11.07.2006
Scott A. Olsen THE INDEFINITE DYAD AND THE GOLDEN SECTION:UNCOVERING PLATO'S SECOND PRINCIPLE Before starting, I offer the following overview, the details of which will be discussed in the rest of the article. An application of abductive reasoning to Plato's puzzles in the dialogues leads to the solution that the Divided Line in the Republic is constructed using a series of Golden Cuts (i.e., divisions in extreme and mean ratio). This leads to the discovery that there is a more primitive form than the √2 and √3 ratios (the roots inherent in the elementary triangles of the Timaeus), and that this form is based in the Golden Section. весь текст 07.07.2006
Аlexey Stakhov THE GOLDEN SECTION, SECRETS OF THE EGYPTIAN CIVILIZATION AND HARMONY MATHEMATICS The main goal of the present article is to consider the harmony mathematics from the point of view of the sacral geometry and to show how it can be used in this field. We also consider some secrets of the Egyptian civilization that have relation to the golden section and platonic solids. Briefly, this is considered to be the main concepts involved in harmony mathematics and its application to the sacral geometry. весь текст 05.05.2006
Аlexey Stakhov THE GENERALIZED GOLDEN PROPORTIONS, A NEW THEORY OF REAL NUMBERS, AND TERNARY MIRROR-SYMMETRICAL ARITHMETIC We consider two important generalizations of the golden proportion: golden p-proportions [Stakhov A.P.] and «metallic means» [Spinadel V.W.]. We develop a constructive approach to the theory of real numbers that is based on the number systems with irrational radices (Bergman's number system and Stakhov's codes of the golden p-proportions). It follows from this approach ternary mirror-symmetrical arithmetic that is the basis of new computer projects. весь текст 03.05.2006
Аlexey Stakhov, Boris Rozin ON A NEW CLASS OF HYPERBOLIC FUNCTIONS This article presents the results of some new research on a new class of hyperbolic functions that unite the characteristics of the classical hyperbolic functions and the recurring Fibonacci and Lucas series. The hyperbolic Fibonacci and Lucas functions, which are the being extension of Binet's formulas for the Fibonacci and Lucas numbers in continuous domain, transform the Fibonacci numbers theory into «continuous» theory because every identity for the hyperbolic Fibonacci and Lucas functions has its discrete analogy in the framework of the Fibonacci and Lucas numbers. весь текст 02.05.2006
Аlexey Stakhov, Boris Rozin THE GOLDEN SHOFAR The goal of the present article is to develop the «continues» approach to the recurrent Fibonacci sequence. The main result of the article is new mathematical model of a curve-linear space based on a special second-degree function named «The Golden Shofar». весь текст 28.04.2006
А. Stakhov, B. Rozin THE «GOLDEN» ALGEBRAIC EQUATIONS The special case of the (p+1)th degree algebraic equations of the kind xp+1 = xp + 1 (p = 1,2,3,...) is researched in the present article. For the case p = 1, the given equation is reduced to the well-known Golden Proportion equation x2 = x + 1. These equations are called the golden algebraic equations because the golden p-proportions τp, special irrational numbers that follow from Pascal's triangle, are their roots. A research on the general properties of the roots of the golden algebraic equations is carried out in this article. весь текст 24.04.2006
Аlexey P. Stakhov FIBONACCI MATRICES, A GENERALIZATION OF THE «CASSINI FORMULA», AND A NEW CODING THEORY We consider a new class of square Fibonacci (p + 1) × (p + 1)-matrices, which are based on the Fibonacci p-numbers (p = 0,1,2,3,...), with a determinant equal to +1 or — 1. This unique property leads to a generalization of the «Cassini formula» for Fibonacci numbers. An original Fibonacci coding/decoding method follows from the Fibonacci matrices. In contrast to classical redundant codes a basic peculiarity of the method is that it allows to correct matrix elements that can be theoretically unlimited integers. For the simplest case the correct ability of the method is equal 93.33% which exceeds essentially all well-known correcting codes. весь текст 21.04.2006
Аlexey Stakhov, Boris Rozin THE CONTINUOUS FUNCTIONS FOR THE FIBONACCI AND LUCAS P-NUMBERS The new continuous functions for the Fibonacci and Lucas p-numbers using Binet formulas are introduced. The article is of a fundamental interest for Fibonacci numbers theory and theoretical physics. весь текст 14.04.2006
Igor A. Melnik REMOTE ACTION OF ROTATION ON THE SEMICONDUCTOR DETECTOR OF GAMMA-RAY RADITION There have been obtained experimental results on remote effects of rotating objects on readings of semiconductor gamma – ray spectrometric equipment and devices. It is shown that the trace shift of statistic distribution is the function of velocity changes in semiconductor detector’s charges accumulation. In its turn, the accumulation rate is subject to a physical field of non-electromagnetic origin created in the result of rotation. весь текст 11.04.2006
Аlexey P. Stakhov, B. Rosin THEORY OF BINET FORMULAS FOR FIBONACCI AND LUCAS P-NUMBERS Modern natural science requires the development of new mathematical apparatus. The generalized Fibonacci numbers or Fibonacci p-numbers (p = 0,1,2,3,...), which appear in the "diagonal sums" of Pascal's triangle and are assigned in the recurrent form, are a new mathematical discovery. The purpose of the present article is to derive analytical formulas for the Fibonacci p-numbers. We show that these formulas are similar to the Binet formulas for the classical Fibonacci numbers. Moreover, in this article, there is derived one more class of the recurrent sequences, which is defined to be a generalization of the Lucas numbers (Lucas p-numbers). весь текст 07.04.2006
Alexey Stakhov THE GENERALIZED PRINCIPLE OF THE GOLDEN SECTION AND ITS APPLICATIONS IN MATHEMATICS, SCIENCE, AND ENGINEERING The «Dichotomy Principle» and the classical «Golden Section Principle» are two of the most important principles of Nature, Science and also Art. The Generalized Principle of the Golden Section that follows from studying the diagonal sums of the Pascal triangle is a sweeping generalization of these important principles. This underlies the foundation of «Harmony Mathematics», a new proposed mathematical direction. Harmony Mathematics includes a number of new mathematical theories: an algorithmic measurement theory, a new number theory, a new theory of hyperbolic functions based on Fibonacci and Lucas numbers, and a theory of the Fibonacci and «Golden» matrices. весь текст 31.03.2006
Alexey Stakhov FUNDAMENTALS OF A NEW KIND OF MATHEMATICSBASED ON THE GOLDEN SECTION As is well known, mathematics is one of the outstanding creations of the human intellect; a result of centuries of intensive and continuous creative work of man's geniuses. What is the goal of mathematics? The answer is not simple. Probably, the goal of mathematics is to discover «mathematical laws of the Universe» and to construct models of the physical world. It is clear that the progress of the human society depends on the knowledge of these laws. весь текст 29.03.2006
Haruo Hosoya SEQUENCES OF POLYOMINO AND POLYHEX GRAPHS WHOSE PERFECT MATCHING NUMBERS ARE FIBONACCI OR LUCAS NUMBERS: THE GOLDEN FAMILY GRAPHS OF A NEW CATEGORY A graph, G, is a mathematical object composed of vertices, j V}, and edges {E}, where an edge spans a pair of vertices (Harary, 1969). A matching of a graph is a set of edges of G such that no two of them share a vertex in common. If a graph with even n = |V| has a matching with л/2 edges it is called a perfect matching graph. The number of possible perfect matchings of G is the perfect matching number, or Kekule number, K(G). Although a tree graph has at most one Kekule structure, a number of interesting features have been found for the K{G) numbers of polycyclic graphs (HOSOYA, 1986). весь текст 22.03.2006
Haruo Hosoya SOME GRAPH-THEORETICAL ASPECTS OF THE GOLDEN RATIO: TOPOLOGICAL INDEX, ISOMATCHING GRAPHS, AND GOLDEN FAMILY GRAPHS It has already been shown by Lucas (1876) that the Fibonacci numbers can be obtained from the Pascal's triangle by rotation and addition in a certain way. The golden ratio can then be asymptotically obtained by taking the ratio of consecutive Fibonacci numbers, whose graph-theoretical aspects have been pointed out by the present author (Hosoya, 1971, 1973) in terms of the topological index, Z, which is the sum of the non-adjacent numbers for a given graph. Similar properties of the Lucas numbers related to the golden ratio have also been demonstrated. весь текст 16.03.2006
P. Gariaev (Ph.D., Dr.Sci., Acad.) PROSPECTS OF WAVE GENETICS. EXTENDED COMMENT We now have a situation in genetics, molecular biology, and medicine in general, that is simultaneously both paradoxical and promising. Long ago, science decided to investigate the genetic codes of human beings. Science has now completed the 10 years long effort of mapping the DNA sequences of human beings, called the Genome. All of the letters and sequences of the DNA codes of human beings are now known. весь текст 22.02.2006
Moshe Carmeli (Моше Кармели) DEAR DOCTOR SHIPOV I have followed the recent correspondence of several authors about your work, and I want to express my own opinion also. From these correspondents, which sometimes are scientific artiles by themselves, I can only suggest that your work, both in experiment and in theory, is of high interest and importance. I can only suggest that your work be kept going on in spite of the difficulties you are encountering and I personally have a high opinion on it. весь текст 30.12.2005
By Tim Ventura & Dr. Gennady Shipov, with revision by Paul A. Murad DR. GENNADY SHIPOV ON TORSION PHYSICS & BPP Dr. Gennady Shipov is one of the world’s leading physicists in Torsion-Physics research. He joins us to talk about the fundamentals of this emerging branch of scientific discovery, and provides some unique insight into how we can turn the fundamental forces of nature towards the goal of advanced propulsion… весь текст 12.12.2005
Murad A.P. TORSION PHYSICS. A VIEW FROM THE TRENCHES Gravity is nature's most mysterious force — or is it? Einstein's Relativity suggested that it's not a force at all, but instead a curvature of time & space. His later research into Unified Field Theory physics extended this notion with the concept of torsion, which many physicists believe has the power to «uncurve» space and make possible a new generation of advanced propulsion devices. весь текст 05.12.2005
R.M. Kiehn A STRONG EQUIVALENCE PRINCIPLE Consider Cn spaces where the exterior differentiation of a vector basis of C2 functions leads to a Cartan matrix of exterior differential 1-forms. A second exterior differentiation leads to a Cartan matrix of curvature 2-forms, which must evaluate to zero, by the Poincare lemma. The Cartan connection matrix can be decomposed in to two parts, one part based on a metric (Christoffel) connection, and the other part on a residue matrix of 1-forms such that [С] = [Γ] + [T]. The exterior differential of the composite connection must vanish such that the curvature 2-forms produced by the metric component must be balanced by all other matrices of 2-forms coming from the Residue connection and interaction 2-forms generated by exterior products of the [Γ] and [Γ]. весь текст 24.11.2005
Vera W. de Spinadel THE METALLIC MEANS FAMILY AND FORBIDDEN SYMMETRIES Abstract: In this paper, we present the Metallic Means Family (MMF), being the most paramount of its members, the Golden Mean f and in the second place, the Silver Mean s Ag. We call them a family because, aside from carrying the name of a metal – the Golden Mean, the Silver Mean, the Copper Mean, the Bronze Mean, the Nickel Mean – they enjoy common mathematical properties that confer them a fundamental importance in modern investigations. Among these applications, we have chosen the analysis of the forbidden symmetries in a quasicrystal. весь текст 18.11.2005
Stakhov A.P. CURRICULUM VITAE OF DR. VERA SPINADEL Dr. Vera W. de Spinadel has got her PhD in Mathematics at the University of Buenos Aires, Argentina, in 1958. At present, she is Full Consultant Professor of Mathematics at the University of Buenos Aires. This is her second period since she got this degree in 1995 until 2002 and it has been renewed until 2009. Since 1995, she is the Director of the research Center on Mathematics & Design MAyDI, having received many I&D grants and personal grants to develop the activitiesof many research groups. In April 2005, she has officially inaugurated the Mathematics & Design Laboratory at the Faculty of Architecture, Design and Urban Planning, University of Buenos. весь текст 25.10.2005
ADVISORY COUNCIL of the ACADEMY of TRINITARIZM INVITATION FOR IMPROVEMENT OF KNOWLEDGE OF THE SPACE LIFE HARMONY AND CREATION OF THE BEGINNINGS OF «HARMONY MATHEMATICS» ON THEIR BASE We invite philosophers, mathematicians, experts of all areas of science and education and also students of universities to take active part in the offered international action of scientific cooperation that is directed on the formation of harmonious mutual relations "Person-Society-Nature". The primary publication of materials concerning to the "Harmony Mathematics" and their discussion are realized on the site www.trinitas.ru and can be duplicated on any other sites and in different languages. весь текст 19.10.2005
G. Shipov DARK ENERGY IN THE THEORY OF PHYSICAL VACUUM Einstein believed that one of the main problems in unified field theory is the one of the geometrization of the energy-momentum tensor of matter on the right-hand side of his equations. This problem can be solved using the geometry of absolute parallelism and Cartan's structural equations in this geometry: весь текст 30.09.2005
G. Shipov DESCARTES MECHANICS: THE FOURTH GENERALIZATION OF NEWTON'S MECHANICS For 317 years we have been applying Newton's mechanics to explain non-relativistic mechanical experiments on the "bench table". Although Newton's mechanics has been generalized three times: the special relativity theory, general relativity theory, and quantum mechanics, there remains a possibility for its further generalization. весь текст 06.09.2005
Professor Moshe Carmeli PROFESSOR MOSHE CARMELI, ONE OF THE MOST SENIOR PHYSICISTS ON EARTH, COMMENTS ABOUT DR. G. SHIPOV BOOK «THE THEORY OF PHYSICAL VACUUM» Dr. Shipov has generalized the ordinary four-dimensional Relativity Theory. He showed that the right-hand sides of the Einstein field equations for gravity and the equations of general-relativistic electrodynamics can be geometrized successfully, if one uses not a Riemannian geometry but the geometry of absolute parallelism. The new field equations he suggests were written as весь текст 26.04.2005
Mikita Guriy WAVLET-SPECTRAL QUALITY MONITORING SOUNDINGS BELLS PRODUCTS IN VIEW OF FEATURES OF OBJECTS OF THE CONTROL Theoretical bases of a quality monitoring. Vibroacoustik the information from own vibrations bells products on the character carries relaccions character. At the description of this information in time it is possible to allocate the following periods of life Vibroacoustik information: весь текст 04.04.2005
Lobova M. AN OPEN LETTER TO WORLD STEERING COMMITTEE, WYP 2005 In response to your call for the ideas and information to promote the World Year of Physics, I wish to bring to your attention the revolutionary work of Dr. Gennady Shipov, Academician of the Russian Academy of Natural Science and Director of Science Center of Vacuum Physics. весь текст 04.03.2005
Lobova M. TO ACAD. E. P. KRUGLYAKOV, «PROF.A. KONKRETNY», ACAD. V.L. GINZBURG Since Russian Academy of Science's (RAS) «Commission on fight with Fraud in Science, Pseudoscientists and Torsion Fields» has launched a campaign against the work of Dr. G. Shipov, I would like to respond to your references in websites as well as to ask you several questions: весь текст 25.02.2005
Rainer W. Kühne POSSIBLE OBSERVATION OF A SECOND KIND OF LIGHT – MAGNETIC PHOTON RAYS Several years ago, I suggested a quantum field theory which has many attractive features. (1) It can explain the quantization of electric charge. (2) It describes symmetrized Maxwell equations. (3) It is manifestly covariant. (4) It describes local four-potentials. (5) It avoids the unphysical Dirac string. My model predicts a second kind of light, which I named «magnetic photon rays». Here I will discuss possible observations of this radiation by August Kundt in 1885, Alipasha Vaziri in February 2002, and Roderic Lakes in June 2002. весь текст [ Институт Физики Вакуума - Публикации ] 24.12.2004
Gubarev E.A. DYNAMICS OF AN ORIENTED POINT AND INERTIA Considered are the issues of modern physics concerned with the formulation of the equations of motion of a particle in conventional noninertial frames of reference. It is shown that in terms of G. Shipov's theory of physical vacuum a conventional noninertial frame can only be described as a frame undergoing a four-dimensional rotation in an absolute- parallelism space. весь текст [ Институт Физики Вакуума - Публикации ] 23.12.2004
G. Shipov GEOMETRICAL AND PHENOMENOLOGICAL TORSIONS IN RELATIVISTIC PHYSICS Продолжаем публикацию работ Г.И. Шипова на английском языке The Ricci and Cartan torsions are compared. Their common properties and distinctions are revealed. It is shown that Ricci torsion determines curvature and torsion in Frenet's equations and consequently the very torsion, instead of the Cartan one, can be connected with spin properties of matter. The theorems showing that in flat and curved spaces it is possible to present Frenet's curves as a first kind geodesic lines of space of absolute parallelism are proved. весь текст [ Институт Физики Вакуума - Теория ] 19.11.2004
G. Shipov UNIFICATION OF INTERACTIONS IN THE THEORY OF PHYSICAL VACUUM От Редакции. Теория физического вакуума имеет многочисленных сторонников и последователей во всем мире. Наш сайт посещают более чем из 100 стран мира, поэтому мы начинаем публикации работ Г.И.Шипова и сотрудников Института Физики Вакуума на английском языке. весь текст [ Институт Физики Вакуума - Теория ] 18.11.2004 | 4,742 | 19,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-25 | latest | en | 0.838564 |
https://gamedev.stackexchange.com/questions/168710/lock-enemys-y-axis-when-using-vector3-movetowards-to-follow-the-player/168713 | 1,631,810,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00189.warc.gz | 326,467,561 | 40,484 | # Lock enemy's y-axis when using Vector3.MoveTowards to follow the player
I am using the following code to make a zombie move towards me, but when playing the scene the zombie seems to be moving towards me, but also off the ground and not fixed on the y-axis
public GameObject ThePlayer;
public float TargetDistance;
public float AllowedRange = 10;
public GameObject TheEnemy;
public float EnemySpeed;
public int AttackTrigger;
public RaycastHit Shot;
void Update()
{
transform.LookAt(ThePlayer.transform);
if (Physics.Raycast(transform.position, transform.TransformDirection(Vector3.forward), out Shot))
{
TargetDistance = Shot.distance;
if (TargetDistance < AllowedRange)
{
EnemySpeed = 0.01f;
if (AttackTrigger == 0)
{
//TheEnemy.GetComponent<Animation>().Play("Walking");
transform.position = Vector3.MoveTowards(transform.position, ThePlayer.transform.position, EnemySpeed);
}
}
else
{
EnemySpeed = 0;
//TheEnemy.GetComponent<Animation>().Play("Idle");
}
}
if (AttackTrigger == 1)
{
EnemySpeed = 0;
//TheEnemy.GetComponent<Animation>().Play("Attacking");
}
}
void OnTriggerEnter()
{
AttackTrigger = 1;
}
void OnTriggerExit()
{
AttackTrigger = 0;
}
What can i do to ensure that it is sticking to the y axis at all times
The reason this is happening is because Vector3.MoveTowards will move all axes toward the target.
You can specifically exclude changes to the Y axis by creating a new target Vector3 having the Y axis set to the value it should stay at.
transform.position = Vector3.MoveTowards(
transform.position,
new Vector3(
ThePlayer.transform.position.x,
transform.position.y,
ThePlayer.transform.position.z
),
EnemySpeed
);
The above code creates a new Vector3 target where the Y axis is the same as the current position instead of the target, preventing Vector3.MoveTowards from changing the Y axis at all.
• This works perfectly thank you! But when the zombie gets into close contact with me it seems to lean back and fall into the floor (weird), is it possible to state when it gets within a certain distance of me, stop moving towards me? Mar 6 '19 at 16:24
• Sure, you can use Vector3.Distance(a, b) to check the distance between two objects. docs.unity3d.com/ScriptReference/Vector3.Distance.html Mar 6 '19 at 16:27
• Is there any benefit to this approach over calculating the move-towards vector as normal and then just maintaining the current Y position during the assignment? i.e. Vector3 moveTowards = Vector3.MoveTowards(transform.position, ThePlater.transform.position, EnemySpeed); transform.position = new Vector3(moveTowards.x, transform.position.Y, moveTowards.z); Mar 6 '19 at 23:41
• @Abion47 yes. Doing it as shown in EstelS's answer (if you apply appropriate frame time adjustment to the speed parameter not made explicit here), you can ensure that the speed of movement is exactly consistent no matter the origin/destination. If you compute the MoveTowards vector first, then override the y coordinate, you might shorten the total distance travelled - making the character move slower when seeking a target that's above/below them than when seeking directly sideways. In EstelS's answer, the shortening of the offset vector occurs before we enforce our consistent speed. Mar 13 '19 at 0:25
You can break out the Vector3 target parameter of Vector3.MoveTowards to prevent movement on the y-axis. Simply set the y value of the target position to the same y value as the current position. For example:
Vector3 currentPosition = transform.position;
Vector3 targetPosition = new Vector3(ThePlayer.transform.position.x, transform.position.y, ThePlayer.transform.position.z);
transform.position = Vector3.MoveTowards(currentPosition, targetPosition, EnemySpeed);
The other answers are great. But if you have a RigidBody attached, you can freeze the y rotation via constraints. Then no code change would be necessary.
• Except that the current code is moving the object via transform.position, which completely bypasses the Rigidbody and ignores any locking on its axes. Mar 7 '19 at 12:22
• @DMGregory indeed. Thank you. I'll delete this answer when I get to a computer
– Evorlor
Mar 7 '19 at 14:08
• Instead of deleting it, you could show how to move the Rigidbody with velocity instead of transform.position so the locked axes are respected. Mar 7 '19 at 20:02 | 960 | 4,310 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.744114 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-determine-if-the-series-ln-1-2-ln-1-3-ln-3-4-ln-k-k-1-converges | 1,723,664,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00891.warc.gz | 68,515,346 | 6,273 | # How do you determine if the series ln(1/2) + ln(1/3) + ln(3/4) + ... +ln[k/(k + 1)] + .... converges?
Mar 17, 2015
First of all I think there is a little mistake, the second term would be:
$\ln \left(\frac{2}{3}\right)$ instead of $\ln \left(\frac{1}{3}\right)$.
So it would be:
${\sum}_{k = 1}^{+ \infty} \ln \left(\frac{k}{k + 1}\right)$.
We can use an important logarithmic rule that says:
$\ln \left(\frac{a}{b}\right) = \ln a - \ln b$.
So:
${\sum}_{k = 1}^{+ \infty} \ln \left(\frac{k}{k + 1}\right) = \ln \left(\frac{1}{2}\right) + \ln \left(\frac{2}{3}\right) + \ln \left(\frac{3}{4}\right) + \ldots =$
$= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ldots + \ln k - \ln \left(k + 1\right) + \ldots$.
The first term is zero, all the other will cancel each other, but there is always another that survive. The bigger becomes $k$ the bigger becomes $\ln \left(k + 1\right)$, the smaller becomes $- \ln \left(k + 1\right)$. So the series will diverge to $- \infty$. | 379 | 986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.689305 |
https://ru.scribd.com/doc/205641101/ibps-po-2013-quantitative-aptitude-ebook-1-pdf | 1,571,810,019,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00334.warc.gz | 645,275,276 | 141,755 | Вы находитесь на странице: 1из 292
PREFACE ............................................................................................................................................. 6 IMPORTANT CHAPTER GUIDELINES ........................................................................................... 7 CHAPTER: SIMPLIFICATION & APPROXIMAITON ................................................................... 8
Solved Examples (Simplification & Approximation) ............................................................................................. 12 Practice Set-1 (Simplification & Approximation) .................................................................................................. 22 Practice Set-2 (Simplification & Approximaiton) .................................................................................................. 25 Simplification & Approximaiton Averages Practice Set-1 (Answers) .................................................................... 27 Simplification & Approximaiton Averages Practice Set-2 (Answers) .................................................................... 28
## CHAPTER: NUMBER SERIES......................................................................................................... 29
Solved Examples (Number Series)........................................................................................................................ 33 Practice Set (Number Series)................................................................................................................................ 38 Number Series Practice Set (Answers) ................................................................................................................. 41
## CHAPTER: NUMBER SYSTEM....................................................................................................... 42
Solved Examples (Number System)...................................................................................................................... 48 Practice Set-1 (Number System) .......................................................................................................................... 53 Practice Set-2 (Number System) .......................................................................................................................... 55 Number System Practice Set-1 (Answers) ............................................................................................................ 58 Number System Practice Set-2 (Answers) ............................................................................................................ 59
## CHAPTER: RATIO, PROPORTION & ALLIGATION .................................................................. 60
Solved Examples (Ratio, Proportion & Alligation) ................................................................................................ 66 Practice Set-1 (Ratio, Proportion & Alligation) ..................................................................................................... 70 Practice Set-2 (Ratio, Proportion & Alligation) ..................................................................................................... 74
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
Ratio, Proportion & Alligation Practice Set-1 (Answers)....................................................................................... 77 Ratio, Proportion & Alligation Practice Set-2 (Answers)....................................................................................... 78
## CHAPTER: AVERAGES ................................................................................................................... 79
Solved Examples (Averages) ................................................................................................................................ 80 Practice Set (Averages) ........................................................................................................................................ 86 Averages Practice Set (Answers) .......................................................................................................................... 90
## CHAPTER: PERCENTAGES, PARTNERSHIP AND SHARE ....................................................... 91
Solved Examples (Percentages, Partnership and Share) ....................................................................................... 95 Practice Set-1 (Percentages, Partnership and Share).......................................................................................... 101 Practice Set-2 (Percentages, Partnership and Share).......................................................................................... 104 Percentages, Partnership and Share Practice Set-1 (Answers) ........................................................................... 106 Percentages, Partnership and Share Practice Set-2 (Answers) ........................................................................... 107
## CHAPTER: PROFIT & LOSS ........................................................................................................ 108
Solved Examples (Profit & Loss) ......................................................................................................................... 112 Practice Set (Profit & Loss) ................................................................................................................................. 117 Profit & Loss Practice Set (Answers) .................................................................................................................. 121
## CHAPTER: SIMPLE INTEREST & COMPOUND INTEREST .................................................. 122
Solved Examples (Simple Interest & Compound Interest) .................................................................................. 126 Practice Set (Simple Interest & Compound Interest) .......................................................................................... 132 Simple Interest & Compound Interest Practice Set (Answers) ............................................................................ 136
## CHAPTER: TIME AND WORK .................................................................................................... 137
Solved Examples (Time and Work) ..................................................................................................................... 140 Practice Set (Time and Work)............................................................................................................................. 146 Averages Practice Set (Time and Work) ............................................................................................................. 151
## CHAPTER: SPEED DISTANCE & TIME ..................................................................................... 152
Solved Examples (Speed Distance & Time)......................................................................................................... 156 Practice Set (Speed Distance & Time) ................................................................................................................ 162 Speed Distance & Time Practice Set (Answers) .................................................................................................. 166
## CHAPTER: MENSURATION ........................................................................................................ 167
Solved Examples (Mensuration) ........................................................................................................................ 174 Practice Set (Mensuration) ................................................................................................................................ 178 Mensuration Practice Set (Answers) .................................................................................................................. 181
## CHAPTER: PERMUTATIONS, COMBINATIONS & PROBABILITY ..................................... 182
Solved Examples (Permutations, Combinations & Probability) .......................................................................... 188 Practice Set-1 (Permutations, Combinations & Probability) ............................................................................... 195 Practice Set-2 (Permutations, Combinations & Probability) ............................................................................... 197 Permutations, Combinations & Probability Practice Set-1 (Answers)................................................................. 199 Permutations, Combinations & Probability Practice Set-2 (Answers)................................................................. 200
CHAPTER: DATA INTERPRETATION ...................................................................................... 201 CHAPTER: DATA INTERPRETATION-TABLE CHART ......................................................... 204
Solved Examples (Data Interpretation-Table Chart) ........................................................................................... 206 Practice Set (Data Interpretation-Table Chart)................................................................................................... 213 Data Interpretation-Table Chart Practice Set (Answers) .................................................................................... 221
## CHAPTER: DATA INTERPRETATION-LINE GRAPHS ........................................................... 222
Solved Examples (Data Interpretation-Line Graphs) .......................................................................................... 225 Practice Set (Data Interpretation-Line Graphs) .................................................................................................. 229 Data Interpretation-Line Graphs Practice Set (Answers) .................................................................................... 233
CHAPTER: DATA INTERPRETATION-BAR GRAPHS............................................................ 234 4 IBPS BANK PO EXAM 2013 : Quantitative Aptitude
Solved Examples (Data Interpretation-Bar Graphs)............................................................................................ 237 Practice Set (Data Interpretation-Bar Graphs) ................................................................................................... 243 Data Interpretation-Bar Graphs Practice Set (Answers) ..................................................................................... 247
## CHAPTER: DATA INTERPRETATION-PIE DIAGRAM .......................................................... 248
Solved Examples (Data Interpretation-Pie Diagram) .......................................................................................... 250 Practice Set (Data Interpretation-Pie Diagram) .................................................................................................. 256 Data Interpretation- Pie Diagram Practice Set (Answers)................................................................................... 262
## CHAPTER: DATA INTERPRETATION-CASE LETS ................................................................ 263
Solved Examples (Data Interpretation-Caselets) ................................................................................................ 266 PracticeSet-(Data Interpretation- Caselets)........................................................................................................ 270 Data Interpretation-Caselets Practice Set (Answers).......................................................................................... 274 Solved Examples (Data Interpretation-Miscellaneous) ...................................................................................... 275 Practice Set (Data Interpretation-Miscellaneous) .............................................................................................. 285 Data Interpretation-Miscellaneous Practice Set (Answers) ................................................................................ 291
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
Preface
Jagranjoshs IBPS PO Exam 2013: Quantitative Aptitude e-Book is a one-stop solution for aspirants who endeavor to leave no stone unturned to score considerably in IBPS PO Written Examination 2013. The eBook is highly useful for all officers level Banking exams - IBPS Specialist officers Exam and SBI PO Exam. IBPS PO Exam 2013: Quantitative Aptitude e-Book is prepared by Jagranjoshs team of subject matter experts who worked up the best to come up with this all-inclusive preparation package for Quantitative Aptitude section of Officers level Banking Exams. This eBook is a perfect blend of chapter-wise basic concepts and questions regarding all the units included in the syllabus for Quantitative Aptitude Section of Officers level Banking Exams. The chapter-wise compilation of this e-Book makes the concept of Quantitative aptitudes easy to understand for students. It further includes previous year questions and model practice sets along with the importance factor of each and every chapter included out here. Our IBPS PO Exam 2013: Quantitative Aptitude e-Book will let students to practice for the QA section within the standard time limit set by the IBPS examination board. This will help them hone time management skills. The IBPS PO Exam 2013: Quantitative Aptitude e-Book includes: Chapter wise Basic Concepts & Questions IBPS PO Previous Year Questions Chapter wise Model Practice Set Important Chapter Guidelines Jagranjoshs IBPS PO Exam 2013: Quantitative Aptitude e-Book is a one-stop solution edition to help preparing for the QA Section of Officers level Banking Exams. All the chapters of this e-book are readerfriendly and easy to understand. Just prepare with it to score more. Our team at Jagranjosh.com wishes all the very best to the aspirants of Banking Exams. All the Best!
## Important Chapter Guidelines
Important Chapter Guidelines for IBPS PO 2013 : Quantitative Aptitude Chapter Number System Number Series Percentage Partnership and Share Simplification and Approximation Average Ratio ,Proportion & Alligation Time & Work Profit, Loss & Discount Simple Interest and Compound Interest Time, Distance and Speed Permutation and Combination & Probability Data Interpretation-Table Charts Data Interpretation-Bar Graph Data Interpretation-Line Graph Data Interpretation-Pie Charts Data Interpretation-Caselets Data Interpretation-Miscelleneous Graphs Mensuration Importance Very Important Very Important Very Important Very Important Important Important Important Very Important Important Important Very Important Very Important Very Important Very Important Very Important Very Important Very Important Important
## Chapter: Simplification & Approximaiton
Some Important Concepts BODMAS Rule: This BODMAS Rule shows the correct sequence of all the operations that are to be executed to find out the value of a given exp ression. In this rule B Stands for Bracket, O stands for of, D for Division, M for Multiplication, A for Addition and S for Subtraction. Therefore, the correct order to simplify an expression is: (a) (b) (c) (d) (e) (f) (g) (h) () {} [] of Division Multiplication Addition Subtraction
## Modulus of a Real Number: If the real number is r, then |r|= .
Example: What will be the value of x in the following equation? 5 + + x +2 = 9 . Solution: Simplifying the above equation x=9 - 5 2 x= -
x= x=
=1 .
Example: If
## Solution: Given expression:
( x/y =3/2)
Example: Arjun spends of his salary on house rent, of his salary on food and
of his salary
on conveyance. If he has Rs.2400 left with him, then find his expenditure on conveyance. Solution: Suppose Arjuns monthly salary is Rs. x Then, remaining part of his salary = x- ( + + Now, = 2400 x =Rs. 4500. )x=x( )x= = .
4500 ) = Rs.450.
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
Approximation In these types of questions there is no need to find out the exact values. The candidate is required to calculate the approximate value in the following manner: Step I: Round off the numbers given in the question. Step II: Simplify the value Step III: Round off the final simplified value Rounding off numbers: Rounding off the numbers given in the question can be done in the following manner: (a) Rounding off to the nearest 10: Example: Rounded off value of 56 is 60, because the digit at unit place is greater than 5, therefore, we will add 1 to the digit at tens place and replace the units digit by 0. Example: Rounded off value of 54 is 50, because the digit at unit place is less than 5, therefore, the value at tens place will remains the same and units digit will be replaced by 0. Example: Rounded off value of 35 is 40, because the digit at unit place is equal to 5, therefore, we will add 1 to the digit at tens place and replace the units digit by 0. (b) Rounding off to the nearest 100: Example: Rounded off value of 386 is 400, because the digit at tens place is greater than 5, therefore, we will add 1 to the digit at hundreds place and replace the digit at unit and tens places by 00. Example: Rounded off value of 741 is 700, because the digit at tens place is less than 5, therefore, the value at hundreds place will remains the same. Unit and tens digit will be replaced by 00. (c) Rounding off to the nearest 1000 Example: Rounded off value of 1963 is 2000, because the digit at hundreds place is greater than 5, therefore, we will add 1 to the digit at thousand place and replace the ones, tens and hundreds by 000. Although the numbers can be rounded off by the above procedures but it depends on the other numbers involved in the simplification. Rounding off a number to a decimal place:
10
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
The candidate is required to follow the following steps to round off a number to the nth decimal place: Step I: check the digit immediately, next right to the nth place. Step II: Add 1 to the digit in the nth place, if the next right digit is 5 or more, otherwise the digit will remains the same. Step III: Remove all the digits in places to the right of the nth place. Example: Rounded off value of 4.693 to the second place is 4.69, because next right digit to the second place is 3, therefore, the value will remain the same and the digit in place to the right of the second place will be removed.
11
## Solved Examples (Simplification & Approximation)
Directions (1-5) What will come in place of the question mark (?) in the following question? (IBPS PO/MT Exam 2012) 1. 4003 77 - 21015=? 116 (1) 2477 (2) 2478 (3) 2467 (4) 2476 (5) None of these Solution: ? 116 = 4003 77 21015 or, ? 116 = 308231 21015 = 287216 or, ? 116 = 287216 ?= Ans: (4) 2. (1) 143 (2) (3) 134 (4) (5) None of these Solution:
=
= 2476
- 361 - 361
12
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
3. (444440) + (64525) + (399126) =? (1) 280.4 (2) 290.4 (3) 295.4 (4) 285.4 (5) None of these Solution: (444440) + (64525) + (399126) = + +
## = 111.1 + 25.8 + 153.5 = 290.4 Ans: (2) 4. - (83)2 = (?)2 + (37)2
(1) 37 (2) 33 (3) 34 (4) 38 (5) None of these Solution: (?)2 + (37)2 = 2 2 2 or, (?) + (37) = 18251 - (83) or, (?)2 + 1369 = 9282-6889 =2393 or, (?)2 = 2393 - 1369 = 1024 ?= Ans: (5) 5. 5 4 11 + 2 =? = 32 - (83)2
(1) 303.75 (2) 305.75 (3) (4) (5) None of these 13 IBPS BANK PO EXAM 2013 : Quantitative Aptitude
Solution: ? = 5 = =
11 + 2 +
+ = 303 + =
## = 101 3 + = Ans: (2)
= 305.75
Direction (6-10): What approximate value should come in place of the question mark (?) in the following question? (Note: You are not excepted to calculate the exact value.) (IBPS PO/MT Exam 2012) 6. 8787 343 (1) 250 (2) 140 (3) 180 (4) 100 (5) 280 Solution: ? = 8787 343 = 25.61 7.07 = 181.09 Ans: (3) 7. (1) 48 (2) 38 (3) 28 (4) 18 (5) 58 Solution: 14 (3038) = (?)2 IBPS BANK PO EXAM 2013 : Quantitative Aptitude (3038) = (?)2 =?
## Ans: (2) 8. of 4011.33 + of 3411.22 =?
(1) 4810 (2) 4980 (3) 4890 (4) 4930 (5) 4850 Solution: ? = = + 4011.33 + 3411.22
= 2507.08 + 2387.854 = 2507 + 2387 = 4894 Ans: (3) 9. 23% of 6783 + 57% of 8431=? (1) 6460 (2) 6420 (3) 6320 (4) 6630 (5) 6360 Solution: ?= 23% of 6783 + 57% of 8431 6783 + = 23 = 1560.09 + 4805.67 = 6365.76 Ans: (5) 15 IBPS BANK PO EXAM 2013 : Quantitative Aptitude 8431
10. 335.01 244.9955=? (1) 1490 (2) 1550 (3) 1420 (4) 1590 (5) 1400 Solution: ? = 335.01 244.99 = 335 245 55 = 335 Ans: (1) Directions (Q.11-15): What will come in place of the question mark (?) in the following questions? (IBPS PO/MT Exam 2011) 11. 3463 295 - 18611 =? + 5883 (1) 997091 (2) 997071 (3) 997090 (4) 999070 (5) None of these Solution: 3463 x 295 -18611 =? + 5883 ? = 1021585 - 18611 - 5883 = 997091 Ans: (1) = = 1422.27 1490 55
16
## 12. (23.1)2 + (48.6)2 - (39.8)2 =? + 1147.69 (1) (13.6)2 (2)
(3) 163.84 (4) 12.8 (5) None of these Solution: 533.61 + 2361.96 - 1584.04 =? + 1147.69 or, ? = 1311.53 - 1147.69 = 163.84 Ans: (3) 13. + =?
1)
2) 0.75 3) 1
17
) X (8
+7
)]- 98 =?
## (5) None of these Solution: [ (3 +1) (8 + 7) ]- 98
= [4 = [60 Ans: (3) 15. (1) 3844 (2) 3721 (3) 3481 (4) 3638
15
] - 98
8] - 98 = 480 - 98 = 382
- (54)2 =
+ (74)2
or ,
- 2916 - 5476
## = 8453 - 2916 - 5476 = 61 18 IBPS BANK PO EXAM 2013 : Quantitative Aptitude
or , ? = (61)2 = 3721 Ans: (2) Directions ( 16-20): What approximate value should come in place of question mark (?) in the following questions? (Note: You are not expected to calculate the exact value] (IBPS PO/MT Exam 2011) 16. 39.897% of 4331 + 58.779% of 5003 =? (1) 4300 (2) 4500 (3) 4700 (4) 4900 (5) 5100 Solution: 40 + 59 = 1732 + 2950 = 4682 4700
Ans: (3) 17. 43931.03 / 2111.02 x 401.04 =? 1) 8800 2) 7600 3) 7400 4) 9000 5) 8300 Solution: 43931 or, ? = 44000 2111 401 =? 400
19
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
or, ? =
400 = 8800
Ans: (5) 18. 1) 3000 2) 2800 3) 2500 4) 3300 5) 2600 Solution: 34.993 = 80 x 35 = 2800 34.993 =?
Ans: (2) 19. 1) 7600 2) 7650 3) 7860 4) 7560 5) 7680 Solution: 17 + 349 =? or, 366 21 = ? 7680. 21 +349 =? 21.003
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
20. 59.88 1) 10 2) 50 3) 30 4) 70 5) 90
12.21
6.35 =?
12
6 = 30
21
## Practice Set-1 (Simplification & Approximation)
Directions (Q. 1-5): What will come in place of question mark (?) in the following questions? (IBPS RRB Grade A Officer Exam 2012) 1. (1) (2) (3) (4) (5) (2) (3) (4) (5) 5. (1) (2) (3) (4) (5) 6 2 4 0 None of these 64 8 7 9
2 16 8 None of these
2. 55% of (1) (2) (3) (4) (5) 3. (1) (2) (3) (4) 18 (5) 32 4. (1) 81 22 126.5 126.6 124.6 125.4 None of these
Directions (Q. 6-10). What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.) 6. 68% of 1288 + 26% of 734 215 =? (1) (2) (3) (4) (5) 7. (1) (2) (3) (4) (5) 670 530 420 780 960 620 930 540 850 710
## 8. 6578 67 15 =? 6 IBPS BANK PO EXAM 2013 : Quantitative Aptitude
(1) (2) (3) (4) (5) 9. (1) (2) (3) (4) (5) 10. (1) (2) (3) (4) (5)
## 200 250 150 100 300
(5) None of these 13. (0.0729 0.1)3 (0.081 10)5 (0.3 3)5 = (0.9)? + 3 (1) (2) (3) (4) (5) 14. (1) (2) (3) (4) (5) 1 2 4 7 None of these of 5) = 149.8 112
## 18 324 24 None of these
15. (27)2 6 9 + (7)3 + 71 = (?)3 431 (1) (2) (3) (4) (5) 11 (13)3 13 (11)2 None of these
Directions (Q.11 to 15): What will come in place of question mark (?) in the following questions? (RBI Grade B Officers Exam 2011) 11. [(3024 189)1/2 + (684 19)2] = (?)2 + 459 (1) (2) (3) (4) (5) -27 -29 31 841 1089 of 30% of 216 =?
Directions (Q. 16-20): What will come in place of question mark (?) in the following questions? (Corporation Bank PO Exam 2011) 16. 5907 1296 144 = ? 8 (1) (2) (3) (4) (5) 17. 726.75 767.25 737.25 676.75 None of these
12. 4.4 times (1) (2) (3) (4) 81.9 83.7 87.3 89.1 23
## IBPS BANK PO EXAM 2013 : Quantitative Aptitude
(1) (2) (3) 3 (4) 9 (5) -3 18. (1) (2) (3) (4) (5) 19. 11 5 7 9 None of these
(1) (2) 200 (3) (4) 100 (5) 10 20. (1) (2) (3) (4) (5) 169 13 14 196 None of these
24
## Practice Set-2 (Simplification & Approximaiton)
Directions (Q. 1-5): What approximate value will come in place of question mark (?) in the following questions? (Corporation Bank PO Exam 2011) (You are not expected to calculate the exact value.) 1. 3237 31 15 = ? 17 (1) (2) (3) (4) (5) 2. (1) (2) (3) (4) (5) 1760 1880 1950 1720 1650 (1) (2) (3) (4) (5) 90 100 110 120 80 (5) 160 5. 61% of 981 150.17 = ? 65% of 676 (1) (2) (3) (4) (5) 760 780 830 860 890
Directions-(Q. 6-10): What will come in place of question-mark (?) in the following questions? (Allahabad Bank Probationary Officers Exam 2011) 6. of 30% of 3420 = (?)2 x 2 (81)2 7 9 81 49
7. 1898 73 x 72 = (?)2 x 13 (1) (2) (3) (4) (5) 8. - 256 256 12 144 -16 =? 42 1024 1764 (1024)2 32
3. (34.34)2 + (5.96) 2 (23.09) 2 =? (1) (2) (3) (4) (5) 4. (1) (2) (3) (4) 200 180 120 140 25 510 540 620 680 650
## IBPS PO EXAM 2013 : Quantitative Aptitude
9. (081)2+ (0729)3 x (09)2 = (0.9)?-3 (1) (2) (3) (4) (5) 6 2 4 0 None of these x 5 =? + 154
12. 89988% of 6999 + 50002% of 99999 170015 =? (1) (2) (3) (4) (5) 13. (1) (2) (3) (4) (5) 760 800 690 870 780 990 900 920 960 860 =?
## 10. 65% of (1) (2) (3) (4) (5) 56 28 35 32 None of these
Directions-(Q. 11-15) What approximate value will come in place of question-mark (?) in the following questions? (Allahabad Bank Probationary Officers Exam 2011) (You are not expected to calculate the exact value.) 11. (1) (2) (3) (4) (5) 620 670 770 750 700 =? 8
14. 6999 70005 x 94998 =? x 19999 (1) (2) (3) (4) (5) 475 420 320 540 525
15. (4999)2 (89)2 (159)2 =? (1) (2) (3) (4) (5) 2165 2000 1965 1920 1885
26
## Simplification & Approximaiton Averages Practice Set -1 (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 5 1 1 4 3 4 2 2 5 3 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 2 4 1 5 1 3 2 4 5 2
27
## Simplification & Approximaiton Averages Practice Set -2 (Answers)
1) 2) 3) 4) 5) 6) 7) 8)
1 1 5 3 5 3 3 5
4 2 2 4 4 1 1
28
## Chapter: Number Series
In Number Series, questions are asked on the basis of relation between numbers given in a series. The questions asked can be divided into different types: Type I: In this type of questions, a series of numbers is given with one number missing represented by a question mark. Candidate has to select from the options choices to correct option in place of the question mark. The given sequences of numbers will be such that each number follows its predecessor in the same way, i.e., according to a particular pattern. Candidates are required to find out the correct ways in which the sequence is formed and there after find out the number to complete the series. 1. 30, 34, 43, 59, 84, 120,? (1) 169 (2) 148 (3) 153 (4) 176 (5) None of these Solution: (1) The given pattern is: +22, 32, +42, + 62, +72 So, missing term is 169=120 +72 2. 40, 54, 82,?, 180,250 (1) 142 (2) 124 (3) 136 (4) 163 (5) None of these Solution: (2) The pattern is: +14, + 28, + 42, + 52, + 70 29 IBPS PO EXAM 2013 : Quantitative Aptitude
So, missing term is 82 + 42=124 Type II: Here, we are given a sequence of number. Whole sequence, except the odd number follow a certain rule. You have to find that number which does not follow the rule. 1. 0, 1,3,8,18,35,264 (1) 62 (2) 35 (3) 18 (4) 8 (5) None of these Solution: (1) The pattern is +(02+1), +(12+1), + (22+1) ,+ (32+1), + (42+1), + (52+1) So, 62 is wrong and must be replaced by 35 + (52+1) = 62 2. 1, 9, 125, 49, 729, 121, 2147 (1) 2147 (2) 729 (3) 125 (4) 1 (5) None of these Solution: (1) Type III: In this type of questions, a number series is given. After the series is over, in the next line, a number is followed by (A), (B), (C), (D) and (E). The candidates have to complete the series starting with the number given following the sequence of the given series. 1. 2, 3 4, (A) 9, (B) 20, (C) 43, (D) 90 (E)
Which number will come in place of (D)? (1) 58 (2) 99 30 IBPS PO EXAM 2013 : Quantitative Aptitude
Similarly,
2. 3, 3
4, (A)
10, (B)
33, (C)
## 136, (D) (E)
Which number will come in place of (E)? (1) 1035 (2) 1165 (3) 1039 (4) 891 (5) None of these Solution: (2)
31
Similarly,
32
## Solved Examples (Number Series)
Directions: (1-5) In each of these questions a number series is given. In each series only one number is wrong. Find out the wrong number. (IBPS CWE PO MT 2012) 1. 5531 5506 5425 5304 5135 4910 4621 (1) 5531 (2) 5425 (3) 4621 (4) 5135 (5) 5506 Solution: The number should be 5555 in place of 5531. -72, -92, -112, -132, -152, -172... Ans: (1) 2. 6 7 9 13 26 37 69 (1) 7 (2) 26 (3) 69 (4) 37 (5) 9 Solution: The number should be 21 in place of 26. +1, +2, +4, +8, +16, +32 Ans: (2) 3. 1 3 10 36 152 760 4632 (1) 3 (2) 36 (3) 4632 (4) 760 33 IBPS PO EXAM 2013 : Quantitative Aptitude
(5) 152 Solution: The number should be 770 In place of 760. 1 +2, 2 +4, 3 +6, 4 + 8, 5 +10, 6 + 12, ... Ans: (4) 4. 4 3 9 34 96 219 435 (1) 4 (2) 9 (3) 34 (4) 435 (5) 219 Solution: The series is 02+ 4, 12+2, 32+0, 62-2, 102-4, 152- 6,212 - 8... Hence, 435 should be replaced with 433 Ans: (1) 5. 157.5 (1) 1 (2) 2 (3) 6 (4) 157.5 (5) 45 Solution: The number should be 2 in place of 1. 3.5, 3, 2.5, 2, 1.5, 1, ... Ans: (1) Directions (6-10): In the following number series only one number is wrong. Find out the wrong number. 6. 7 12 (1) 7 (2) 12 (3) 40 34 IBPS PO EXAM 2013 : Quantitative Aptitude 40 222 1742 17390 208608 45 15 6 3 2 1
(4) 1742 (5) 208608 Solution: The pattern of number series is as follows: 7 12 40 222 1744 2 - 2 = 12 4 - (2 + 6) = 48 - 8 = 40 6 - (8 + 10) = 240 - 18 = 222 8 - (18 + 14) = 1776 - 32 = 1744 1742
## 10 - (32 + 18) = 17440 - 50 = 17390
Ans: (4) 7. 6 (1) 91 (2) 70558 (3) 584 (4) 2935 (5) 35277 Solution: The pattern of number series is as follows: 6 91 582 2935 7 + 72 = 42 + 49 = 91 6 + 62 = 546 + 36 = 582 5+52 =2910 + 25=2935 4 + 42 = 11740 + 16 = 11756 584 91 584 2935 11756 35277 70558
11756 x 3 + 32 = 35268 + 9 = 35277 Ans: (3) 35 IBPS PO EXAM 2013 : Quantitative Aptitude
8. 9050 (1) 3478 (2) 1418 (3) 5675 (4) 2147 (5) 1077
5675
## 3478 2147 1418 1077 950
Solution: The pattern of number series is as follows: 9050 I53 = 9050 - 3375 = 5675 5675 - 133 = 5675 - 2197 = 3478 3478 - 113 = 3478 - 1331 = 2147 2147 - 93 = 2147 - 729 = 1418 1418 - 73 = 1418 - 343 = 1075 Ans: (5) 9. 1 4 25 256 (1)3125 (2) 823543 (3) 46656 (4) 25 (5) 256 Solution: The pattern of number series is as follows: 11 = 1;-22 = 4; 33 = 27 77 = 823543 Ans: (4) 25; 44 = 256; 55 = 3125; 66 = 46656; 3125 46656 823543 1077
36
## IBPS PO EXAM 2013 : Quantitative Aptitude
10. 8424 (1) 131.625 (2) 1051 (3) 4212 (4) 8424 (5) 263.25
## 4212 2106 1051 526.5
263.25 131.625
Solution: The pattern of number series is as follows: 8424 4212 2106 1053 2 = 4212 2 = 2106 2 = 1053 2 = 526.5 1051
37
## Practice Set (Number Series)
Directions (Q. 1-3): What will come in place of question mark (?) in the following number series? (IBPS RRB Grade Officer Exam 2012) 1. 987 587 331 187 123 (?) (1) 104 (2) 113 (3) 107 (4) 114 (5) None of these 2. 125 171 263 401 585 (?) (1) 835 (2) 815 (3) 792 (4) 788 (5) None of these 3. 121 132 167 226 309 (?) (1) 424 (2) 413 (3) 427 (4) 416 (5) None of these Directions (Q. 4-5): In the following number series, only one is wrong. Find out the wrong number. 4. 454 327 648 524 842 713 1036 (1) 327 (2) 648 (3) 521 (4) 842 (5) 713 5. 72.5 86 113 168 275 491 923 (1) 86 (2)113 (3)168 (4)275 (5)491 Directions (Q. 6-10): What will come in place of question mark (?) in the following number series? (RBI GradeB Officers Exam 2011) 6. 17 19 (1) (2) (3) (4) (5) 33 (?) 129 227
7. 35 256 38
## (1) (2) (3) (4) (5)
680 893 633 880 None of these 868 917 (?) 1051
## 8. 18 139 (1) (2) (3) (4) (5)
13. 342 337.5 328.5 315 297 (?) 265.5 274.5 270
1042 1036 942 996 None of these (?) 1162 874 730
## 9. 2890 658 (1) (2) (3) (4) (5)
260 None of these 14. 161 164 179 242 497 (?) (1) (2) (3) (4) (5) 1540 1480 1520 1440 None of these
## 1684 1738 1784 1672 None of these
10. 14 1004 1202 1251.5 1268 (?) (1) (2) (3) (4) (5) 1267.5 1276.25 1324.5 1367.25 None of these
15. 239 254 284 344 464 (?) (1) (2) (3) (4) (5) 726 716 724 714 None of these
Directions (Q. 11-15): What will come in place of question mark (?) in the following number series? (Corporation Bank PO 2011) 11. 8 11 20 47 128 (?) (1) (2) (3) (4) (5) 483 488 397 371 None of these
Directions-(Q. 16-20) What will come in place of question-mark (?) in the following number series? (Allahabad Bank Probationary Officers Exam 2011) 16. 958 833 733 658 608, (?) (1) (2) (3) (4) (5) 577 583 567 573 None of these
## IBPS PO EXAM 2013 : Quantitative Aptitude
17. 11 10 18 51 200, (?) (1) (2) (3) (4) (5) 885 1025 865 995 None of these
19. 14 24 43 71 108 (?) (1) (2) (3) (4) (5) 194 154 3) 145 4) 155 5) None of these
18. 25 48 94 186 370 (?) (1) (2) (3) (4) (5) 738 744 746 724 None of these
20. 144 173 140 169 136 (?) (1) (2) (3) (4) (5) 157 148 164 132 None of these
40
## Number Series Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 3 2 4 5 3 3 4 1 2 2 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 4 3 2 3 5 2 4 1 2 5
41
## Chapter: Number System
Introduction: Number is a symbol which represents quantity. There are three types of numbers: 1. Real Numbers: Real numbers are those numbers which can be easily indentify and quantify. For example: -10, -7.33, -1, 0, 1, 2, 5.77 etc. 2. Imaginary Numbers: Imaginary numbers are those numbers which we can just imagine but cannot physically perceive. For example: , etc, numbers are imaginary. is represented by i . Square root of all negative
3. Complex Number: Combination of Real and Imaginary number is called complex numbers. For example: (2+5i) , (1+3i) etc . Here we will only discuss about the real numbers . Types of Real Numbers: there are two types of real numbers 1. Rational numbers: - All that numbers which can be expressed in the form of where p & q are integers and q 0 are called rational numbers. =
For example: - -1, 2, , 0, 1 , 2.7 etc. , here -1 = Again Rational numbers are classified as:
(a) Integers: All rational numbers which do not have decimal or fractional parts are called integers. For example: -3, -1, 0 , 1 , 2 etc . Integers are of two type whole numbers and natural numbers. All the non negative integers are whole numbers , for example 0 , 1 , 2 , 3 etc and all the whole numbers except 0 are natural numbers , for example 1, 2 , 3 etc
42
## IBPS PO EXAM 2013 : Quantitative Aptitude
(b) Fractions: All rational numbers which are in the form of q 0 and p is not a multiple of q are called fractions. For example: - 1.2, , , 1.7, .3 etc. Fractions are of three types: Proper Improper Mixed fractions
## where p & q are integers and
A proper fraction is a fraction whose numerator is smaller than denominator, for example , etc. An improper fraction is a fraction whose numerator is equal to or greater than its , etc. and a mixed fraction is an integer plus a fraction, for
## denominator for example example , etc.
2. Irrational Numbers:- All that numbers which cannot be expressed in the form of are called irrational numbers. They have non-terminating and non-recurring decimal parts. For example: On the basis of origin: 1. Prime Numbers: All the natural numbers greater than 1 which are only divisible by 1 and the number itself are called prime numbers. For example: 2, 3, 5, 7, 11, 13, 17 etc. 2. Composite Numbers: All the natural numbers greater than 1 which are divisible by at least one more number other than 1 and the number itself are called composite numbers. For example: 4, 6, 8, 9, 10, 12 etc. Note: 1 is neither a prime number nor a composite. On basis of divisor: 1. Even Numbers: All the natural numbers which are multiple of 2 are called even numbers. For example: 2, 4, 6, 8, 10 etc. , , etc.
43
## IBPS PO EXAM 2013 : Quantitative Aptitude
Odd Numbers: All the natural numbers which are not a multiple of 2 are called odd numbers. They are denoted as 2k For example: 3, 5, 7, 9, 11 etc. To find whether a number is prime or not. Step 1: Find the approximate square root of a number . Step 2: Check if any prime number from 2 to that square root divides that number or not. Step 3: If none of those prime number divides the number than the number must be prime number. Example: Take 631 , the approx square root of 631 is 25 , now from 2 to 25 there are 2 , 3 ,5 , 7 , 11 , 13 , 17 , 19 and 23 prime number . Since none of these divides 631, so 631 must be a prime number. Conversion of recurring decimal into fraction. The form of purely recurring number = 1, where k is a natural number.
Let x = 0.77777.
10x = 7.77777
If x = 0.27272727 .
100 x = 27.272727
## The form of purely recurring number =
Let x= 0.143333333..
100x = 14.333333.
44
## Quotient and remainder: Dividend = (Divisor Quotient) + remainder.
For example: If dividend = 15968, Quotient = 89 and remainder = 37 then Divisor is? Divisor = =
45
## Divisibility Rules: For Number 2 Divisibility Rule Example Note
If the last digit of a number is 0,2,4,6,8 , then the number is divisible by 2 If the sum of all the digits of a number is divisible by 3 , then the number is divisible by 3
## 742 is divisible by 2 but 743 is not.
1458 (sum of digits = 18) is 766 (sum of digits divisible by 3, but 766 (sum of =19) the remainder digits =19) is not divisible by 3. when 19 is divided by 3 i.e. 1 will also be the remainder when 766 is divided by3 6732 is divisible by 4 as 32 is divisible by 4, but 2142 is not divisible by 4. Similarly the remainder when 42 is divided by 4 i.e 2 will also be the remainder when 2142 is divided by 4.
If the last two digits of a number are divisible by 4, then the number is also divisible by 4
If the last digits of a number 1465, 1320 are divisible by 5 are 0 and 5, then the as their last digit is 5 and 0 number is divisible by 5. respectively. If the number is divisible by 2 and 3 both, then it is also divisible by 6. 1452 is divisible by both 2 and 3 so it is divisible by 6 also, but 3362 is not divisible by 6 as it is not divisible by 3. If the number is divisible by 4 and 6 both, then it is not necessary that it is divisible by 24 (6 4).
If the last three digit of a number are divisible by 8 or are 000, then the number is divisible by 8 .
43102 and 13000 are divisible by 8 since 102 is divisible by 8 and 13000 have 000 as last three digits, but 2148 is not as
The remainder when 148 is divided by 8 i.e. 4 will also be the remainder of 2148
46
## IBPS PO EXAM 2013 : Quantitative Aptitude
148 is not divisible by 8 9 If the sum of all digits of a number is divisible by 9 , then the number is also divisible by 9. 25344 (sum of digits = 18) is divisible by 9 , 764 (sum of digits =17) is not.
when divided by 8. The remainder when 17 is divided by 9 i.e 8 will also be the remainder when 764 is divided by 9.
11
If the difference between the sum of the digits in the even places and the sum of the digits in the odd places is either 0 or is divisible by 11 , then the number is also divisible by 11.
9415956 is divisible by 11 as the difference of 9+1+9+6 =25 and 4+5+5 = 14 is 11, but 31872 is not as the sum of even places = 13 and sum of odd is 8 their difference is neither 0 nor 11.
47
## Solved Examples (Number System)
1. When X is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of X? (IBPS CWE PO MT 2012) (1) 8 (2) 6 (3) 4 (4) 5 (5) None of these Solution: Let be subtracted from the numbers 9, 15 and 27 we get continue proportion. Now, 9 : 15 : 27 b2 = ac (15 )2 = (9 ) (27 ) or, 225 30 + 2 = 243 + 2 36 or, 6 = 243 225 = 18 X=3 Hence number become 9 - x = 9-3=6 15 x = 15 -3 = 12 And 27-x = 27 -3 = 24 6 : 12 : 24 = 1: 2: 4 Thus 1: 2: 4 is continued proportion Ans: (5) 2. Sum of three consecutive numbers is 2262. What is 41% of the highest number? (IBPS CWE PO MT 2012) (1) 301.5.1 (2) 303.14 (3) 308.73 (4) 306.35 (5) 309.55 Solution: Let the three consecutive number be x, x + 1 and x + 2. Then, x + x + 1 + x + 2 =2262 or, 3x = 2262 3 = 2259 x= = 753 48 IBPS PO EXAM 2013 : Quantitative Aptitude
The Numbers are 753, 754, 755. The highest number is 755 41% of 755 = 755 = 41 7.55 = 309.55 Ans: (5) 3. Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy? (IBPS CWE PO MT 2012) (1) Either 12 or 13 (2) Either 11 or 12 (3) Either 10 or 11 (4) Either 9 or 11 (5) Either 9 or 10 Solution: Total number of items = 32 Maximum number of ice creams = 9 pastries> cookies> ice cream So, 13 10 9 12 11 9 Hence number of cookies is either 10 or 11. Number of pastries is either 13 or 12. Ans: (3) 4. The fare of a bus is Rs. x for the first five kilometres and Rs. 13/- per kilometre thereafter. If a passenger pays Rs. 2402/- for a journey of 187 kilometres, what is the value of X? (IBPS CWE PO MT 2012) (1) Rs. 29 (2) Rs. 39 (3) Rs. 36 (4) Rs. 31 (5) None of these Solution: Let the fare of first five kilometres be Rs. x. Total distance = 187 km Remaining distance = 187 - 5 = 182 km Now, x + 182 13 = 2402 49 IBPS PO EXAM 2013 : Quantitative Aptitude
x = 2402 - 2366= Rs.36 Ans: (3) 5. The product of three consecutive even numbers is 4032. The product of the first and the third number is 252. What is five times the second number? (1) 80 (2) 100 (3) 60 (4) 70 (5) 90 Solution: Let the three consecutive even numbers be 2x, 2x + 2 and 2x+4. Then, (2x) (2x + 2) (2x + 4) = 4032 ... (I) Again, product of first and third number 2x (2x + 4) = 252 ... (II) Putting the values of the product of first and third number in eqn (I), we have (2x + 2) 252 = 4032 or, 2x + 2 = = 16 x=7 Hence, first number = 7 8 = 14 Second number = 7 x 2 + 2 = 16 And third number = 7 x 2 + 4 = 18 Five times of second number = 5 x 16 = 80 Ans: (1) 6. Rubina could get equal number of Rs. 55, Rs. 85 and Rs. 105 tickets for a movie. She spends Rs. 2,940 for all the tickets. How many of each did she buy? (1) 12 (2) 14 (3) 16 (4) Cannot be determined (5) None of these Solution: Let she buy x tickets. Then total money spent = 55x + 85x + 105x or, 245x = 2940 or, x = 12 Ans: (1) 50 IBPS PO EXAM 2013 : Quantitative Aptitude
7. Seema bought 20 pens, 8 packets of wax colours, 6 calculators and 7 pencil boxes. The price of one pen is Rs. 7, one packet of wax colour is for Rs. 22, one calculator is for Rs. 175 and one pencil box costs Rs. 14 more than the combined price of one pen and one packet of wax colours. How much amount did Seema pay to the shopkeeper? (1) Rs. 1,491 (2) Rs. 1,725 (3) Rs. 1,667 (4) Rs. 1,527 (5) None of these Solution: Price of one pencil box = 14 + (Price of one pen + Price of one packet of wax colours)= 14+ (7+22) = Rs. 43 Total amount paid by Seema = {(20 7) + (8 22) + (6 175) + (7 43)} = Rs 1667 Ans: (3) 8. In a cricket match, Sachin and Viru scored a century each (more than 100 runs) and Yuvi and Gauti scored a half century each. Gauti scored 76 runs and Yuvi scored 12 runs less than Gauti. Viru scored 102 runs. The sum of the scores of all the four players is 442. How many runs did Sachin score? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) 200 210 180 160 None of these
Solution: Sachin's score = 442 76 (76 12) 102 = 200 Ans: (1) 9. The sum of seven consecutive even numbers of Set A is 378. What is the sum of a different set of four consecutive numbers whose lowest number is 23 less than the mean of Set A? (1) (2) (3) (4) (5) 132 146 156 124 None of these 51 IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: Mean of Set A Lowest number of Set B = (54 23) = 31 Sum of the four numbers of Set B = 31 + 32 + 33 + 34 = 130 Ans: (5) 10. The sum of six consecutive even numbers of Set-A is 402. What is the sum of another SetB of four consecutive numbers whose lowest number is 15 less than double the lowest number of set- A? (Allahabad Bank Probationary Officers Exam 2011) (1) (2) (3) (4) (5) 444 442 440 446 None of these = 67 - 1 = 66
## Solution: Third even number = smallest even number = 62
smallest number of set B = 2 x 62 - 15 = 109 required sum = 109 + 110 + 111+ 112 = 442 Ans: (2)
52
## Practice Set-1 (Number System)
1. A person on tour has Rs 360 for his daily expense. He decides to extend his tour programme by 4 days which leads to cutting down daily expense by Rs 3 a day. The number of days of his tour programme is (1) (2) (3) (4) 15 20 18 16
2. Two times a two-digit number is 9 times the number obtained by reversing the digits and sum of the digits is 9. The number is (1) (2) (3) (4) 72 54 63 81
3. If 5 students utilize 18 pencils in 9 days, how long at the same rate will 66 pencils last for 15 students? (1) (2) (3) (4) 10 days 12 days 11 days None of these
4. A man has certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcel, none is left over. What is the number of boxes he may have to pack? (1) (2) (3) (4) 106 301 309 400
5. Out of a group of swans, 7/2 times the square foot of the number are playing on the shore of the pond. The two remaining are inside the pond. What is the total number of swans? (1) 10 (2) 14 (3) 12 53 IBPS PO EXAM 2013 : Quantitative Aptitude
(4) 16 6. In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in the family? (1) (2) (3) (4) 4 3 2 5
7. A yearly payment to a servant is Rs. 90 plus one turban. The servant leaves the job after 9 months and received 65 and a turban. Then find the price of the turban. (1) (2) (3) (4) Rs 10 Rs 15 Rs 7.50 Cannot be determined
8. Mr. Mukherjee is 5 yr older to his wife and they have one son and one daughter. If the daughter is 23 yr younger to his mother, what is the difference in the ages of the brother and the sister? (1) (2) (3) (4) 7 yr 12 yr 4 yr 2 yr
9. The expenses of a hotel consist of two parts. The first one varies with the number of inmates, while the second one is fixed. When the numbers of inmates are 275 and 350, the expenses are Rs 1600 and Rs 1900 respectively. The expenses when the number of inmates is 315, are (1) (2) (3) (4) Rs 1760 Rs 1680 Rs 1780 Rs 1660
10. If the numerator of a fraction is doubled and the denominator is increased by 3, the new fraction is 3/5 what is the original fraction, if its denominator is more than twice the numerator by 1? (1) (2) (3) (4) 3/7 6/13 1/3 5/11 54 IBPS PO EXAM 2013 : Quantitative Aptitude
## Practice Set-2 (Number System)
1. The sum and product of two numbers are 5 and 6, respectively. The sum of reciprocals of their squares is (1) (2) (3) (4) 2. The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is (1) (2) (3) (4) 30 33 36 45
3. The sum of a natural number and its square equals the product of the first three prime numbers. The number is (1) (2) (3) (4) 2 3 5 6
4. 47 is added to the product of 71 and an unknown number. The new number is divisible by 7, giving the quotient 98. The unknown number is a multiple of (1) (2) (3) (4) 2 7 5 3
55
## IBPS PO EXAM 2013 : Quantitative Aptitude
5. Two natural numbers are in the ratio 3: 5 and their product is 2160. The smaller of the number is (1) (2) (3) (4) 36 24 18 12
6. The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers. (1) 1 (2) (3) (4) 7. The sum of the digits of a two digit numbers is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number. (1) (2) (3) (4) 81 46 64 60
8. The least number to be subtracted from 36798 to get a number which is exactly divisible by 78 is (1) (2) (3) (4) 18 60 38 68
9. The sum of first sixty number from one to sixty is divisible by (1) (2) (3) (4) 13 59 60 61
56
## IBPS PO EXAM 2013 : Quantitative Aptitude
10. Eight consecutive number are given. If the average of the two numbers that appear in the middle is 6, then the sum of the eight given numbers is (1) (2) (3) (4) 36 48 54 64
57
## Number System Practice Set-1 (Answers)
1 2 3 4 5 6 7 8 9 10 (2) (4) (3) (2) (4) (2) (1) (4) (1) (1)
58
1 (1)
(2)
(3)
(4)
(1)
(3)
(3)
(2)
(4)
10
(2)
59
## Chapter: Ratio, Proportion & Alligation
Introduction:Ratio is the relation which one quantity bears to another of the same kind. The ratio of two quantities a and b is the fraction and we write it as a: b. In the ratio a: b, we call a as the first term or antecedent and b, the second term or consequent. Note: The multiplication or division of each term of a ratio by the same non- zero number does not affect the ratio. Compound Ratio: - It is obtained by multiplying together the numerators for new numerator and denominators for new denominator. Example 1. If the ratios are 4:3, 15:20, 2:6 and 3:5 find the compound ratio? Sol 1. Required ratio = Duplicate ratio of a: b = Triplicate ratio of a: b = etc.
Example2. If we divide 4185 into two parts such that they are in ratio 7:2, then find the values of both the parts? Sol 2. Let the actual variable be 7x and 2x. Then 7x+2x = 4185 So, the 1st part = 7 The 2nd part = 2 Note: 60 IBPS PO EXAM 2013 : Quantitative Aptitude
The ratio of first , second and third quantities is given by ac : bc : bd If the ratio between first and second quantity is a:b and third and fourth is c:d. a: b c: d ac : bc : bd Similarly, the ratio of first, second, third and fourth quantities is given by ace : bce : bde : bdf If the ratio between first and second quantity is a: b and third and fourth is c:d .
a : b c : d e: f ace : bce : bde : bdf Example 3. If Savita has Rs 1880. How much money does Ravina have if the ratio of money with savita and Rita is 15: 7 and that with Rita and Ravina 16: 7? Solution3: Savita : Rita : Ravina 15 : 7 16 : 7 15 240 : 112 : 14 The ratio of money with Savita , Rita and Ravina is 240 : 112 : 14. We see that 240 x = 1880 Hence 14 61 IBPS PO EXAM 2013 : Quantitative Aptitude
Proportion
Introduction:Four quantities are said to be proportional if the two ratios are equal i.e. the A, B, C and
D are proportion. It is denoted by :: it is written as A : B : C : D where A and D are extremes and B and C are called means . Product of the extreme = Product of the means Direct proportion: - The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also increases (or decreases). Example 1. If 5 pens cost Rs 10 then 15 pen cost? Sol 1. It is seen that if number of pens increases then cost also increases. So, 5 pens: 15 pens:: Rs 10 : required cost Required cost = Inverse proportion: - The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also decreases (or increases). Example 2.If 10 men can do a work in 20 days then in how many days 20 men can do that work? Sol 2. Here if men increase then days should decrease, so this is a case of inverse proportion, so 10 men: 20 men :: required days : 20 days Required days =
Rule of three: It Is the method of finding 4th term of a proportion if all the other three are given, if ratio is a:b :: c:d then , d= Compound proportion: - Lets take an example to explain this. 62 IBPS PO EXAM 2013 : Quantitative Aptitude
Example3. If 9 men can do a piece of work in 40 days of 10 hours each, how many men will it take to do 12 times the amount of work if they work for 30 days of 9 hours? Solution 3: Step 1. Days: . : . Hours: . : .. Work: . : .. Step 2. 1. Compare days with men : to do the work in less days we will need more men , so it is the case of inverse proportion hence , 30 : 40 :: 9 : required no. of men 2. Compare hours with men : to do the work in less hours we will need more men , so it is the case of inverse proportion hence , 9 : 10 :: 9 : required no. of men 3. Compare work with men : to do the more work we will need more men , so it is the case of direct proportion hence , 1: 12:: 9: required no. of men Put all the values in step 1, 30: 40 9: 10 1: 12 Now, :: 9: required no. of men :: 9: required no. of men
63
## IBPS PO EXAM 2013 : Quantitative Aptitude
ALLIGATION Introduction:The word allegation means linking. It is used to find: 1. The proportion in which the ingredients of given price are mixed to produce a new mixture at a given price. 2. The mean or average value of mixture when the price of the two or more ingredients and the proportion in which they are mixed are given. Mathematical Formula: For two ingredient:-
Mean Price
## (d-m) Then (cheaper quantity): (dearer quantity) = (d-m): (m-c)
(m-c)
Example 1: If the rice at Rs 3.20 per kg and the rice at Rs 3.50 per kg be mixed then what should be their proportion so that the new mixture be worth Rs 3.35 per kg ? Sol 1: CP of 1 kg of cheaper rice 320 paisa CP of 1 kg of dearer rice 350 paisa
Mean Price (m) 335 paisa 15 64 15 IBPS PO EXAM 2013 : Quantitative Aptitude
By allegation: Hence they must be mixed in equal proportion i.e. 1:1 For three ingredient:1. The price of the ingredient should be reduced to one denomination and then place them in ascending order under one another. 2. After that place the mean prices to the left of all the price 3. Then pair the price so that price less than and greater than the mean prices go together. 4. Then find out the difference between mean price and each price and place it opposite to the price with which it is linked. 5. These difference will help to find out the given answer and similarly it will work for four ingredients . Example 2: Find out the ratio of new mixture so that it will cost Rs 1.40 per kg from the given three kinds of rice costing Rs 1.20, Rs 1.45 and Rs 1.74? Sol 2: 1st rice cost = 120, 2nd rice cost = 145 and 3rd rice cost = 174 paisa. From the above rule: we have, 120 140 145 174 5+34 [(145-140) + (174-140)] 20 20 (140-120) (140-120)
Therefore, three rice must be mixed in 39: 20: 20 ratios to have a new mixture of rice.
65
## Solved Examples (Ratio, Proportion & Alligation)
1. A certain amount was to be distributed among A, B and C in the ratio 2: 3: 4 respectively, but was erroneously distributed in the ratio 7: 2: 5 respectively. As a result of this, B got Rs 40 less. What is the amount? (IBPS CWE PO MT 2012) (1) Rs. 210 (2) Rs. 270 (3) Rs. 230 (4) Rs. 280 (5) None of these Solution: Let the amount be x. B's share = Due to error B's share = Difference in B's share due to error = 40 or, or, 24x = 40 x= Ans: (1) 2. Rs.73,689/- are divided between A and B in the ratio 4: 7. What is the difference between thrice the share of A and twice the share of B? (IBPS CWE PO MT 2012) (1) Rs. 36,699 (2) Rs. 46,893 (3) Rs. 20,097 (4) Rs. 26,796 (5) Rs. 13,398 Solution: Let As share be 4x and Bs share be 7x. 4x +7x = 73689 or, 11x = 73689 66 IBPS PO EXAM 2013 : Quantitative Aptitude = 40 = 40 126 = Rs. 210
x=
= 6699
As share = 6699 4 = 26796 Bs share = 6699 7 = 44893 Thrice the share of A = 26796 3 = 80388 Twice the share of B = 46893 2= 93786 Difference = 93786 - 80388 = Rs. 13398 Ans: (5) 3. The ratio of the present age of Manisha and Deepali is 5:X. Manisha is 9 years younger than Parineeta. Parineetas age after 9 years will be 33 years. The difference between Deepali's and Manisha's age is the same as the present age of Parineeta. What should come in place of X? (1) 23 (2) 39 (3) 15 (4) Cannot be determined (5) None of these Solution: Parineetas present age = (33 - 9 =) 24 yrs. Manisha's present age = (24 - 9 =) 15 yrs. Deepalis present age = IS + 24 = 39 yrs. Ratio of the present age of Manisha and Deepali = 15 : 39 = 5 : 13 X = 13 Ans: (5) 4. The ratio between the three angles of a quadrilateral is 3 : 5 : 9. The value of the fourth angle of the quadrilateral is 71. What is the difference between the largest and the smallest angles of the quadrilateral? (IBPS RRB Group A Officers Exam 2012) (1) 82 (2) 106 (3) 102 (4) 92 (5) None of these 67 IBPS PO EXAM 2013 : Quantitative Aptitude
Solution. Let the quadrilateral angles be 3x, 5x, 9x and 71. Total sum of angles = 3x + 5x + 9x + 71 = 360 or, 17x = 360 71 = 289 x = 17 Hence angles are 51, 85, 153, and 71. Difference = 153 51 = 102. Ans: (3) 5. The second largest and the smallest angles of a triangle are in the ratio of 6 : 5. The difference between the second largest angle and the smallest angle 'of the triangle is equal to 9. What is the difference between the smallest and the largest angles of the triangle? (IBPS RRB Group A Officers Exam 2012) (1) 36 (2) 24 (3) 12 (4) 18 (4) None of these Solution. Let the second largest angle of the triangle be 6x and the smallest angle 5x. Now, 6x - 5x = 9 or, x = 9 Second largest angle = 54 Smallest angle = 45 Sum of angles of a triangle = 180 largest angle = 180 - 99 = 810 Difference = 81 - 45 = 36. Ans: 1 6. The ratio between the speed of a bus and train is 15 : 27 respectively. Also, a car covered a distance of 720 km in 9 hours. The speed of the bus is three- fourth the speed of the car. How much distance will the train cover in 7 hours? (Allahabad Bank Probationary Officers Exam 2011) (1) 760 km (2) 756 km (3) 740 km 68 IBPS PO EXAM 2013 : Quantitative Aptitude
(4) Cannot be determined (5) None of these Solution: Speed of the car = Speed of the bus = Speed of the train = 60km/hr 108km/hr = = 80hm/hr
69
## Practice Set-1 (Ratio, Proportion & Alligation)
1. A container has 80 L of milk. From this container 8 L of milk was taken out and replaced by water. The process was further repeated twice. The volume of milk in the container after that is [SSC Quantitative Aptitude (Arihant)] a) 58.23 L b) 85.23 L c) 58.32 L d) 85.32 L 2. A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 L of mixture is drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. Litres of liquid A contained by the can initially was a) 10 b) 20 c) 21 d) 25 3. What number should be added to or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2? a) 5 is subtracted b) 10 is added c) 7 is added d) 10 is subtracted 4. The ratio of weekly incomes of A and B is 9 : 7 and the ratio of their 70 expenditures is 4 : 3. If each saves Rs. 200 per week, then the sum of their weekly incomes is a) Rs. 3200 b) Rs. 4200 c) Rs. 4800 d) Rs. 5600 5. The ratio of alcohol and water in 40 L of mixture is 5 : 3.8 L of the mixture is removed and replaced with water, Now, the ratio of the alcohol and water in the resultant mixture is a) 1 : 2 b) 1 : 1 c) 2 : 1 d) 1 : 3 6. Rama's expenditure and savings are in the ratio 3 : 2. His income increases by 10%. His expenditure also increases by 12%. His saving increases by a) 7% b) 10% c) 9% d) 13% 7. Three numbers are in the ratio 3 : 4 : 5. The sum of the largest and the smallest equals the sum of the second and 52. The smallest number is a) 20
## IBPS PO EXAM 2013 : Quantitative Aptitude
b) 27 c) 39 d) 52 8. The ratio of the ages of Ram and Rahim 10 yr ago was 1 : 3. The ratio of their ages 5 yr hence will be 2 : 3. Then, the ratio of their present ages is a) 1 : 2 b) 3 : 5 c) 3 : 4 d) 2 : 5 9. The ratio of milk and water in mixtures of four container are 5 : 3, 2 : 1, 3 : 2 and 7 : 4, respectively. In which container is the quantity of milk, relative to water, minimum? a) First b) Second c) Third d) Fourth 10. In a mixture of 25 L, the ratio of acid to water is 4 : 1. Another 3 L of water is added to the mixture. The ratio of acid to water in the new mixture is a) 5 : 2 b) 2 : 5 c) 3 : 5 d) 5 : 3 11. A shopkeeper buys two varieties of tea, the price of the first being twice the second. He sells the mixture at Rs 36 per kilogram and makes a profit of 20%. If the ratio of quantities of the first and second 71
variety in these mixture is 3 : 4, then what is the cost price of each variety of tea ? [Mission MBA MAT (Arihant)] a) Rs 21, 42 b) Rs 15, 30 c) Rs 16.5, 33 d) Rs 17, 34 12. Two liquids are mixed in the ratio 3 : 5 and the mixture is sold at Rs 120 with a profit of 20%. If the first liquid is costlier than the second by Rs 2 per litre, find the cost of the costlier liquid per litre. a) Rs 92.30 b) Rs 74.10 c) Rs 101.25 d) Rs 99.25 13. A grocer buys two kinds of rice at Rs 1.80 and Rs 1.20 per kg respectively. In what proportion should these be mixed, so that by selling the mixture at Rs 1.75 per kg, 25% may be gained? a) 2 : 1 b) 3 : 2 c) 3 : 4 d) 1 : 2 14. A jar full of whisky contains 40% of alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26. The quantity of whisky replaced is a)
## IBPS PO EXAM 2013 : Quantitative Aptitude
b) c) d) 15. A container contains 240 L of wine. 80 L is taken out of the container everyday and an equal quantity of water is put into it. Find the quantity of the wine that remains in the container at the end of the fourth day. a) 39.2 L b) 32 L c) 42.5 L d) 47.40 L 16. A tea trader mixed two varieties of tea, one costing Rs 3.50 per kg and the other costing Rs 4 per kg and sells 40 kg of the mixture to a vendor at Rs 4.50 per kg and makes a profit of 20%. How much of each variety did the vendor mix? a) 30 kg, 10 kg b) 20 kg, 20 kg c) 10 kg, 30 kg d) None of these 17. A vessel contains 50 L milk. The milkman delivers 10 L to the first house and adds an equal quantity of water. He does exactly the same at the second and third house. What is the ratio of milk and water when he has finished delivering at the third house? a) 61 : 64 72
b) 27 : 37 c) 16 : 19 d) None of these 18. Prabhu purchased 30 kg of rice at the rate of Rs 17.50 per kg and another 30 kg rice at a certain rate. He mixed the two rice and sold the entire quantity at the rate of Rs 18.60 per kg and made 20% over all profit. At what price per kg did he purchase the lot of another 30 kg rice? a) Rs 14.50 b) Rs 12.50 c) Rs 15.50 d) Rs 13.50 19. A person has a chemical of Rs 50 per litre. In what ratio should water be mixed in that chemical so that after selling the mixture at Rs 40 per litre he may get a profit of 50%. a) 8 : 7 b) 9 : 8 c) 10 : 7 d) 4 : 3 20. A trader has 50 kg of pulses, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. What is the quantity sold at 18% profit? a) 30 kg b) 35 kg c) 40 kg d) None of these
## IBPS PO EXAM 2013 : Quantitative Aptitude
21. A container of capacity 120 L is filled with milk and water. 80% of milk and 40% of water is taken out of vessel. It is found that the vessel is vacated by 65%. What is the ratio of milk to water? a) 5 : 3 b) 6 : 5 c) 3 : 5 d) 4 : 3
73
## Practice Set-2 (Ratio, Proportion & Alligation)
1. An employer reduces the number of employees in the ratio 8 : 5 and increases their wages in the ratio 7 : 9. As a result, the overall wages bill is a) increased in the ratio 56 : 69 b) decreased in the ratio 56 : 45 c) increased in the ratio 13 : 17 d) decreased in the ratio 17 : 13 2. A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 L of mixture is drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. Litres of liquid A contained by the can initially was e) 10 f) 20 g) 21 h) 25 3. What number should be added to or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2? e) 5 is subtracted f) 10 is added 74 g) 7 is added h) 10 is subtracted 4. The ratio of weekly incomes of A and B is 9 : 7 and the ratio of their expenditures is 4 : 3. If each saves Rs. 200 per week, then the sum of their weekly incomes is e) Rs. 3200 f) Rs. 4200 g) Rs. 4800 h) Rs. 5600 5. Rama's expenditure and savings are in the ratio 3 : 2. His income increases by 10%. His expenditure also increases by 12%. His saving increases by e) 7% f) 10% g) 9% h) 13% 6. If A : B is 2 : 3, B : C is 6 : 11, then A : B : C is a) 2 : 3 : 11 b) 4 : 6 : 22 IBPS PO EXAM 2013 : Quantitative Aptitude
c) 4 : 6 : 11 d) 2 : 6 : 11 7. The ratio of the ages of Ram and Rahim 10 yr ago was 1 : 3. The ratio of their ages 5 yr hence will be 2 : 3. Then, the ratio of their present ages is e) 1 : 2 f) 3 : 5 g) 3 : 4 h) 2 : 5 8. If the annual income of A, B and C are in the ratio 1 : 3 : 7 and the total annual income of A and C is Rs. 800000, then the monthly salary of B is a) Rs. 20000 b) Rs. 25000 c) Rs. 30000 d) Rs. 15000 9. Two numbers are such that the ratio between them is 4 : 7. If each is increased by 4 the ratio becomes 3 : 5. The larger number is a) 36 b) 48 c) 56 75
d) 64 10. Acid and water are mixed in a vessel A in the ratio of 5 : 2 and in the vessel B in the ratio of 8 : 5. In what proportion should quantities be taken out from the two vessels, so as to form a mixture in which the acid and water will be in the ratio of 9 : 4? a) 7 : 2 b) 2 : 7 c) 7 : 4 d) 2 : 3 11. In an alloy, Zinc and Copper are in the ratio 1 : 2. In the second alloy, the same elements are in the ratio 2 : 3. If these two alloys be mixed to form a new alloy in which two elements are in the ratio 5 : 8, the ratio of these two alloys in the new alloy is a) 3 : 10 b) 3 : 7 c) 10 : 3 d) 7 : 3 12. A boy has a few coins of denominations 50 paise, 25 paise and 10 paise in the ratio 1 : 2 : 3. If the total amount of the coins is Rs. IBPS PO EXAM 2013 : Quantitative Aptitude
6.50, the number of 10 paise coins is a) 5 b) 10 c) 15 d) 20 13. The sum of the ages of a father and his son is 100 yr now. 5 yr ago their ages were in the ratio of 2 : 1. The ratio of the ages of father and son after 10 yr will be a) 5 : 3 b) 4 : 3 c) 10 : 7 d) 3 : 5 14. In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the
school, the ratio of boys and girls becomes a) 12 : 7 b) 10 : 7 c) 8 : 7 d) 4 : 3 15. A box contains Rs. 1,50 paise and 25 paise coins in the ratio 8 : 5 : 3. If the total amount of money in the box is Rs. 112.50, the number of 50 paise coins is a) 80 b) 50 c) 30 d) 42
76
## Ratio, Proportion & Alligation Practice Set-1 (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11)
c c d a b a c b c a a
12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22)
c d c d b a d a a a b
77
## Ratio, Proportion & Alligation Practice Set-2 (Answers)
1) 2) 3) 4) 5) 6) 7) 8)
b c d a a c b b
c a a c a d b
78
## IBPS PO EXAM 2013 : Quantitative Aptitude
Chapter: Averages
The term Average refers to the sum of all observations divided by the total number of observations. Average is used quite regular in our day to day life. For example to calculate the average marks of the students, Average height of a particular group etc. The term average is also referred to as Mean. Basic formula to calculate the average is as follows: Average = ( Example. What is the average of First 10 Prime numbers? Solution: First 10 Prime number are 2,3,5,7,11,13,17,19,23,29. Hence, Average = {2+3+5+7+11+13+17+19+23+29} / 10 = 129 / 10 = 12.90 )
## So, Average of First 10 Prime numbers is 12.90.
Example. The total number of sales visits made by a Salesman in the month of June is 90. What is the Average visit he makes per day? Solution: Number of days in the month of June are 30 Hence, Average Visit per day = Number of total visits / Number of total days = 90 / 30 =3 So, the salesman makes 3 visits per day.
79
## Solved Examples (Averages)
1. Ramola's monthly income is three times Ravina's monthly income. Ravina's monthly income is fifteen per cent more than Ruchira's monthly income. Ruchira's monthly income is Rs. 32,000. What is Ramola's annual income? (1) Rs. 1,10.400 (2) Rs. 13,24,800 (3) Rs. 36,800 (4) Rs. 52,200 (5) None of these Solution: Ravina's monthly income = 32000 = Rs. 36800
## 110400 = Rs. 1324800
Ans: (2) 2. The average marks in English of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 in lieu of the actual marks 48, 59 and 67 respectively, then what would be the correct average? (1) 56.5 (2) 59 (3) 57.5 (4) 58 (5) None of these Solution: Total marks = 24 56 = 1344 Total of actual marks = 1344 - (44 + 45 + 61) + (48 + 59 + 67) = 1368 Actual Average = = 57
80
## IBPS PO EXAM 2013 : Quantitative Aptitude
Ans: (5) 3. In a test, a candidate secured 468 marks out of maximum marks' A:. Had the maximum marks' A' converted to 700, he would have secured 336 marks. What was the maximum marks of the test? (1) 775 (2) 875 (3) 975 (4) 1075 (5) None of these Solution: Converted maximum marks = 700
## 468 is 48 % of maximum marks 'A' A=
Ans : (3)
100 =975
4. The ratio between the speed of a truck, car and train is 3: 8: 12. The car moved uniformly and covered a distance of 1040 km in 13 hours. What is the average speed of the truck and the train together? (IBPS RRB Group A Officers Exam 2012)
(1) 75km/hr (2) 60 km/hr (3) 48 km/hr (4) Cannot be determined (5) None of these
Solution: Speed of car Ratio of speed of truck, car and train = 3 : 8 : 9 Now 8x = 80 81 IBPS PO EXAM 2013 : Quantitative Aptitude
x = 10 Hence truck = 30 kmph. Train = 90 kmph. Average speed of truck and train together
Ans: 2 5. The circumference of a circle is twice the perimeter of a rectangle. The area of the circle is 5544 sq cm. What is the area of the rectangle if the length of the rectangle is 40cm? (IBPS RRB Group A Officers Exam 2012) (1) 1120 sq cm (2) 1020 sq cm (3) 1140 sq cm (4) 1040 sq cm (5) None of these Solution: Area of circle
r = 42 Circumference of circle = 2 perimeter of rectangle Or, Or, perimeter of rectangle = 132 cm Or, 2(l + b) = 132 l + b = 66 b = 66 40 = 26 82 IBPS PO EXAM 2013 : Quantitative Aptitude
Area of rectangle = 40 26 = 1040 cm2 = 1040 sq.cm. Ans: 4 6. Among three numbers the first is twice the second thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is (1) 54 (2) 28 (3) 39.5 (4) 41.5 Solution: Let first number = x Then, second number = and third number = According to given condition, X + + = 3 49.5
= 148.5
11x = 6 148.5
x =
= 81
83
## IBPS PO EXAM 2013 : Quantitative Aptitude
7. 10 yrs ago, the average age of P and Q was 20 yrs. Average age of P, Q is 30 yrs now. After 10 yr, the age of R will be (1) (2) (3) (4) 35 yr 40 yr 30 yr 45 yr
Solution: 10 yr ago, total age P and q = 20 2 = 40 yr Present total age of P and Q = 40+210= 60 yr Present total age of P and Q and R=303= 90 yr Rs age = 90-60 = 30 yr After 10 yr Rs age = (30+10) yr = 40 yr Ans: (2) 8. In an examination, the average of marks was found to be 50. For 100 students, marks was computed wrongly as 90 instead of 60.For deducting marks for computational errors, the average of marks came down to 45. The total number of candidate, who appeared at the examination, was (1) (2) (3) (4) 600 300 200 150
Solution: Let the total number of candidate be x. By given condition. 50x- 90x100+ 60x100= 45x
## 50x 45x = 3000
x= Ans: (1) 84 IBPS PO EXAM 2013 : Quantitative Aptitude = 600.
9. The average of 25 observations 13. It was later found that an observation 73 was wrongly entered as 48. The new average is (1) (2) (3) (4) 12.6 14 15 13.8 Required new average = = Ans: (2) 10. A tabulator while calculating the average marks of 100 students of an examination, by mistake enters 68, instead of 86 and obtained the average marks of those students are (1) (2) (3) (4) 58.8 57.82 58.81 57.28 = = 14
Solution:
Solution: Actual total number of 100 students = 5800 + (86 - 68) = 5818 Required actual average = Ans: (1) = 58.18
85
## Practice Set (Averages)
1. The average age of a jury of 5 is 40. If a member aged 35 resigns and a managed 25 becomes a member, then the average age of the new jury is a) 30 yr b) 38 yr c) 40 yr d) 42 yr 2. The average of the runs made by Raju, Shyam and Hari is 7 less than that made by Shyam, Hari and Kishore. If the number of Kishore's run is 35, what is Raju's run? a) 21 b) 35 c) 7 d) 14 3. The mean of 50 numbers is 30. Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean. a) 36.12 b) 30.66 86 IBPS PO EXAM 2013 : Quantitative Aptitude c) 29.28 d) 38.21 4. One third of a certain journey is covered at the rate of 25 km/h, one-fourth at the rate of 30 km/h and the rest at 50 km/h. The average speed for the whole journey is a) 35 km/h b) c) 30 km/h d) 5. The average of 5 numbers is 140. If one number is excluded, the average of the remaining 4 numbers is 130. The excluded number is a) 135 b) 134 c) 180 d) 150
6. The average weight of 5 persons sitting in a boat is 38 kg. The average weight of the boat and the persons sitting in the boat is 52 kg. What is the weight of the boat? a) 228 kg b) 122 kg c) 232 kg d) 242 kg 7. There are 50 students in a class. Their average weight is 45 kg. When one student leaves the class the average weight reduces by 100 g. What is the weight of the student who left the class? a) 45 kg
9. In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score for the whole class is 71.8. Find the percentage of girls a) 40% b) 50% c) 55% d) 60% 10. The average of the first 100 positive integers is a) 100 b) 51 c) 50.5 d) 49.5
b) 47.9 kg c) 49.9 kg d) 50.1 kg 8. Out of 4 numbers, whose average is 60, the first one is one fourth of the sum of the last three. The first number is a) 15 b) 45 c) 48 d) 60 11. Out of the three numbers, the first number is twice of the second and the second is thrice of the third number. If the average of these 3 numbers is 20, then the sum of the largest and smallest numbers is a) 24 b) 42 c) 54 d) 60
87
## IBPS PO EXAM 2013 : Quantitative Aptitude
12. The average age of 40 students of a class is 18 yr. When 20 new students are admitted to the same class, the average age of the students of the class is increased by 6 months. The average age of newly admitted students is a) 19 yr b) 19 yr 6 months c) 20 yr d) 20 yr 6 months 13. The average age of 40 students of a class is 15 yr. When 10 new students are admitted, the average age is increased by 0.2 yr. The average age of new students is a) 15.2 yr b) 16 yr c) 16.2.yr d) 16.4 yr 14. The average of 6 observations is 45.5. If one new observation is added to the previous observation, then the new average becomes 47. The new observation is a) 58 b) 56 c) 50 88
d) 46 15. The average of marks scored by the students of a class is 68. The average of marks of the girls in the class is 80 and that of boys is 60. What is the percentage of boys in the class? a) 40 b) 60 c) 65 d) 70 16. The average age of 30 boys in a class is 15 yr. One boy aged 20 yr, left the class but two new boys came in his place whose ages differ by 5 yr. If the average age of all the boys now in the class still remains 15 yr, the age of the younger newcomer is a) 20 yr b) 15 yr c) 10 yr d) 8 yr
## IBPS PO EXAM 2013 : Quantitative Aptitude
17. Out of 10 teachers of a school, one teacher retires and in his place a new teacher of age 25 yr joins. As a result of it, the average age of the teachers is reduced by 3 yr. The age of the retired teacher is a) 60 yr b) 58 yr c) 56 yr d) 55 yr 18. The mean weight of 34 students of a school is 42 kg. If the weight of the teacher be included, the mean rises by 400 g. Find the weight of the teacher (in kg). a) 66
b) 56 c) 55 d) 57 19. Ram aims to score an average of 80 marks in quarterly and half yearly exams. But his average in quarterly is 3 marks less than his target and that in half yearly is 2 marks more than his aim. The difference between the total marks scored in both the exams is 25. Total marks aimed by Ram is a) 380 b) 400 c) 410 d) 430
89
## IBPS PO EXAM 2013 : Quantitative Aptitude
1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
b d c b c b c c a c
## 11) 12) 13) 14) 15) 16) 17) 18) 19)
b b b b b b d b a
90
## Chapter: Percentages, Partnership and Share
INTRODUCTION The word per cent means per hundred. Thus 19 parts out of 100 parts. This can also be written as . Fraction Equivalents of important Percentages.
PER CENT TO FRACTION To convert a per cent to a fraction, divide it by 100 and delete the % sign. Example: 2% can be converted to a fraction as PER CENT OF A NUMBER Per cent of a number is the product of equivalent fraction (of rate per cent) and the number. Example: To find out 25% of 500 91 IBPS PO EXAM 2013 : Quantitative Aptitude
## Solution: Required value = 25% of 500
equivalent fraction for 25% = 125 Example: 9% of what number is 36? Solution: the required number (base number)
= 400 Example: If 30% of a number is 48, then what is 40% of the number? Solution: Here, unitary method can be used to save the time. 30% 48 Hence, the required value is 64 Calculating % EXCESS OR % SHORTNESS 1% 40%
92
## IBPS PO EXAM 2013 : Quantitative Aptitude
i.e. B is more than A by Example: If the income of Ram is more than that of Mohan by 25% then by much percentage Mohan's income is than that of Ram?
Solution: Required % shortness (less) income of Mohan = 20% Therefore, income of Mohan is 20% less than that of Ram. Partnership and Shares Meaning: When two or more than two persons run a business jointly, they are called partners in that business and the deal between them is known as partnership. There are two types of partners in the business 1. Working Partner: A person who manages the business is known as working partner. 2. Sleeping Partner: A person who simply invests the money is known as sleeping partner. Some Important Formulae: Suppose two persons P and Q invests Rs. X and Rs. Y respectively for a year in a business, then their share of profit or loss at the end of the year: = Suppose two persons P and Q invests Rs. X for m month and Rs. Y for n months respectively, then = Example1: P, Q and R started a business by investing Rs. 1, 50,000, Rs. 2, 50,000 and 3, 50,000 respectively. At the end of the year, out of an annual profit of Rs.30, 000 find the share of P, Q and R respectively. Solution: Ratio of shares of P, Q and R respectively = Ratio of their investments 93 IBPS PO EXAM 2013 : Quantitative Aptitude
## = 1,50,000 : 2,50,000 : 3,50,000 = 3: 5:7
Share of Ps profit = Rs. (30,000 Share of Qs profit = Rs. (30,000 Share of Rs profit = Rs. (30,000
## ) = Rs. 6,000. ) = Rs. 10,000. ) = Rs. 14,000.
Example2: Ram started a business investing Rs. 50,000.After 4 months, Shyam joined him with a capital of Rs. 30,000 .After another 2 months, Mohan joined them with a capital of Rs. 60,000.At the end of the year, they made a profit of Rs. 24,000.Find the share of profits of Ram, Shyam and Mohan. Solution: According to the given problem, it is clear that Ram invested his capital for 12 months, Shyam invested for 8 months and Mohan invested for 6 months. Then, ratio of their Capitals = (50,000 12) : (30,000 8) : (60,000 6) = 60:24:36 = 5:2:3
Share of Rams profit = (24,000 Share of Shyams profit = (24,000 Share of Mohans profit = (24,000
= Rs.12,000;
) = Rs.4,800; ) = Rs.7,200.
94
## Solved Examples (Percentages, Partnership and Share)
1. Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many marks did he score in subject C? (IBPS CWE PO MT 2012) (1) 84 (2) 86 (3) 79 (4) 73 (5) None of these Solution: Akash scored in subject A = 73 marks Subject B = = 84 marks Total marks Akash got in all the three subjects together = 54 4.5 = 243 marks Let Akash's marks in subject C be X. A + B + C = 243 or, A + B + X = 243 or, X = 243- (84 +73) = 243 -157= 86 marks Ans: (2) 2. An HR Company employs 4800 persons, out of which 45 per cent are males and 60 per cent of the males are either 25 years or older. How many males are employed in that HR Company who are younger than 25 years? (1) 2640 (2) 2160 (3) 1296 (4) 864 (5) None of these Solution: Total number of persons = 4800 95 IBPS PO EXAM 2013 : Quantitative Aptitude
## Number of males = 45% of 4800 =
= 2160
Now, according to the question, Number of males who are younger than 25 years = (100-60 =) 40% of 2160 = 864
Ans: (4) 3. Six-elevenths of a number is equal to 22 per cent of the second number. The second number is equal to one- fourth of the third number. The value of the third number is 2400. What is 45% of the first number? (1) 109.8 (2) 111.7 (3) 117.6 (4) 123.4 (5) None of these Solution: According to the question, x First number = 22% of second number Second number = x Third number or, Second number = x 2400 = 600 or, First number = = 242 required answer = 45% of 242 = 96 = 108.9
## IBPS PO EXAM 2013 : Quantitative Aptitude
Ans: (5) 4. In an entrance examination, Ritu scored 56 per cent marks, Smita scored 92 per cent marks and Rina scored 634 marks. The maximum marks of the examination is 875. What is the average marks scored by all the three girls together? (1) 1929 (2) 815 (3) 690 (4) 643 (5) None of these Solution: Ritu's marks = 875 x Smita's, marks = 875 x Rina's marks = 634 Total marks = 490 + 80S + 634 = 1929 Average = Ans: (4) 5. If twenty five per cent of three-sevenths of twenty six per cent of a number is 136.5, what is the number? (IBPS RRB Group A Officers Exam 2012) (1) 6300 (2) 5600 (3) 4800 (4) 4900 (5) None of these Solution: Let the number be x. Then, = 643 = 805 = 490
## Ans: (4) 97 IBPS PO EXAM 2013 : Quantitative Aptitude
6. Two-thirds of Ranjits monthly salary is equal to Ramans monthly salary. Ramans monthly salary id thirty per cent more than pawans monthly salary. Pawans monthly salary is Rs. 32000. What is Ranjits monthly salary? (IBPS RRB Group A Officers Exam 2012) (1) Rs. 64200 (2) Rs. 62500 (3) Rs. 64500 (4) Rs. 62400 (5) None of these Solution: Pawans monthly salary = Rs. 32,000 Ramans monthly salary Ranjits monthly salary Ans: (4) 7. In a class there are 60 students, out of whom 15 per cent are girls. Each girls monthly fee is Rs. 250 and each boys monthly fee is 34 per cent more than a girl. What is the total monthly fees of girls and boys together? (IBPS RRB Group A Officers Exam 2012) (1) Rs. 19335 (2) Rs. 18435 (3) Rs. 19345 (4) Rs. 19435 (5) None of these Solution: Number of girls Total monthly fee of girls = 250 9 = Rs. 2250 Number of boys = 60 9 = 51 Monthly fee of one boy Total monthly fees of boys = 51 335 = Rs. 17085 Sum = 17085 + 2250 = Rs. 19,235 Ans: (1)
98
## IBPS PO EXAM 2013 : Quantitative Aptitude
8. Ravi scored 225 marks in a test and failed by 15 marks. If the passing percentage of the test is 25 per cent, what is the maximum marks of the test? (1) (2) (3) (4) (5) 860 840 920 960 None of these
Solution: Passing marks = 225 + 15 = 240 Maximum marks Ans: (4) 9. Ravi scored 225 marks in a test and failed by 15 marks. If the passing percentage of the test is 25 per cent, what is the maximum marks of the test? (1) (2) (3) (4) (5) 860 840 920 960 None of these
Solution: Passing marks = 225 + 15 = 240 Maximum marks Ans: (4) 10. In a school, there are 800 students out of whom 12 per cent are girls. Each boy's monthly fee is Rs. 220 and each girl's monthly fee is 25 per cent less than that of a boy. What is the total monthly fee of all the girls and boys together? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) Rs. 1,72,020 Rs. 1 ,80,780 Rs. 1,70,720 Rs. 1,80,600 None of these
99
## IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: Total number of girls Total number of boys = 800 96 = 704 Each girls monthly fee Total monthly fee of girls and boys together = 96 165 + 704 220 = 15840 + 154880 = Rs. 170720 Ans: (3)
100
## Practice Set-1 (Percentages, Partnership and Share)
1. The number of seats in an auditorium is increased by 25%. The price of a ticket is also increased by 12%. Then, the increase in revenue collection will be a) 40% b) 35% c) 45% d) 48% 2. Two numbers are 30% and 40% more than the third number, respectively. The first number is x % of the second. Then, x is equal to a) b) c) d) 3. The price of cooking oil has increased by 25%. The percentage of reduction that a family should effect in the use of cooking oil, so as not to increase the expenditure on this account is a) 15% b) 20% c) 25% d) 30% 4. In an examination, a student had to obtain 33% of the maximum marks 101 to pass. He got 125 marks and failed by 40 marks. The maximum marks were a) 500 b) 600 c) 800 d) 1000 5. In an office 40% of the staff is female, 40% of the female and 60% of the male voted for me. The percentage of votes I got was a) 24% b) 42% c) 50% d) 52% 6. If A's income is 50% less than that of B's, then B's income is what per cent more than that of A? a) 125% b) 100% c) 75% d) 50% 7. In an examination, 35% of the candidates failed in Mathematics and 25% in English. If 10% failed in both Mathematics and English, then how much per cent passed in both the subjects? a) 50% b) 55% c) 57% d) 60%
## IBPS PO EXAM 2013 : Quantitative Aptitude
8. The price of sugar rise by 25%. If a family wants to keep their expenses on sugar the same as earlier, the family will have to decrease its consumption of sugar by a) 25% b) 20% c) 80% d) 75% 9. If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction becomes . The original fraction is a) b) c) d) 10. If the price of tea is increased by 20%, by how much per cent the consumption of tea be reduced, so that there is no increase in the expenditure on it? a) b) 20% c) d)
11. A number reduced by 25% becomes 225. What per cent should it be increased, so that it becomes 375? a) 25% b) 30% c) 35% d) 75% 12. 25% of the candidates who appeared in an examination failed to qualify and only 450 candidates qualified. The number of candidates, who appeared in the examination was a) 700 b) 600 c) 550 d) 500 13. A worker suffers a 20% cut in his wages. He may regain his original wages by obtaining a rise of a) 27.5% b) 25.0% c) 22.5% d) 20.0% 14. A certain company has 80 engineers. If the engineers constitute 40% of its workers, then the number of people employed in the company is a) 150 b) 800 c) 200 d) 3200
102
## IBPS PO EXAM 2013 : Quantitative Aptitude
15. A saves 20% of his monthly salary. If his monthly expenditure is Rs. 6000. Then, his monthly saving is a) Rs. 1200 b) Rs. 4800
103
## Practice Set-2 (Percentages, Partnership and Share)
1. A starts business with Rs. 7000 and after 5 months, B joined as a partner. After a year the profit is divided in the ratios 2: 3. The capital of B is (a) Rs 10000 (b) Rs 6500 (c) Rs 18000 (d) Rs 9000 2. A, Band C started a business by investing Rs 40500, Rs 45000 and Rs 60000, respectively. After 6 months C withdrew Rs 15000 while A invested Rs 4500 more. In annual profit of Rs 56100 the share of C will exceed that of A by (a) Rs 900 (b) Rs 1100 (c) Rs 3000 (d) Rs 3900 3. A, B and C entered into partnership in a business. A got of the profit and B and C 6. A began a business with Rs 10500 and is joined afterwards by B with Rs 18000. After how many months did B join, if the profit at the end of year is divided equally? (a) 5 months (b) 15 months (c) 10 months IBPS PO EXAM 2013 : Quantitative Aptitude (d) Rs 800 4. A began a business with Rs 2250 and was joined afterwards by B with Rs 2700. If the profits at the end of the year were divided in the ratio of 2: 1, after how much time B joined the business? (a) 5 months (b) 6 months (c) 3 months (d) 7 months 5. Amit and Brijesh started a business with initial investments in the ratio of 12: 11 and their annual profits were in the ratio of 4: 1. If Amit invested the money for 11 months, then for what time Brijesh invested the money? (a) 9 months (b) 3 months (c) 5 months (d) 10 months distributed the remaining profit equally. If C got Rs 400 less than A, the total profit was (a) Rs 1600 (b) Rs 1200 (c) Rs 1000
104
(d) 9 months 7. A and B started a business by investing Rs 35000 and Rs 20000, respectively. After 5 months B left the business and C joined the business with a sum of Rs 15000. The profit earned at the end of year is Rs 84125. What is the share of B in profit?
105
## Percentages, Partnership and Share Practice Set-1 (Answers)
1) 2) 3) 4) 5) 6) 7) 8)
a d b a d b a b
c c a b b c c
106
## Percentages, Partnership and Share Practice Set-2 (Answers)
1) 2) 3) 4) 5) 6) 7)
c d c d b a c
107
## Chapter: Profit & Loss
Profit and Loss is an extension of the chapter of percentages. It is a very important branch of basic Mathematics. This branch deals with the study of Profit and loss made in any commercial transaction. The entire economy and the concept of capitalism is based on the so called Profit Motive. Some basic terms used in Profit and loss are: Cost price The price, at which an article is purchased, is called Cost price and it is abbreviated by C.P. Selling Price The price, at which an article is sold, is called its selling price and it is abbreviated by S.P. Profit If S.P. > C.P., then seller is said to have a profit. Loss If SP < CP, Then seller is said to have incurred a loss. Formulae Profit or Gain = S.P. C.P. Loss = C.P. S.P. Gain % =
Loss % =
S.P. =
C.P.
S.P. =
C.P.
C.P =
S.P.
108
## IBPS PO EXAM 2013 : Quantitative Aptitude
C.P =
S.P.
Example: 100 apples are bought at the rate of Rs. 500 and sold at the rate of Rs. 84 per dozen. What will be the percentage of profit and loss? Solution: We will solve this in steps Step I: Given that C.P. of 100 apples = 500 Then, C.P. of 1 apple = =5 Step II: Also given that per dozen S.P. of apples = 84 Then, S.P. of 1 apple = =7 Step III: Now, we know that Gain % = = = = = 40% Therefore, there is a profit of 40% in the whole selling process. If a person sells two similar items, one at a gain of A%, and the other at a loss of A%, then the seller always incurs a loss. This loss can be calculated by:
Loss % =
109
## IBPS PO EXAM 2013 : Quantitative Aptitude
Example : A man sold two plots for Rs. 15, 00,000 each. On one he gains 25% while on the other he loses 25%. How much does he gain or loss in the whole transaction. Solution: In such a case there is always a loss
Loss % =
=6.25%.
If an article sold at two different selling price . On one is made and on the other is made then:
= Profit calculation on the basis of equating the Amount Spent and the Amount Earned: If the person is going through the transaction has got back all the money that he has spent, but has ended up with some amount of goods left over after the transaction. % Profit = 100
Example: A fruit vendor recovers the cost of 15 oranges by selling 10 oranges. Find his percentage profit. Solution: Here the money spent is equal to the money earned the percentage profit is given by % Profit = Discount: Discount is the reduction offered amount on the market price. Therefore, SP = Where, MP = Market Price of the product 110 IBPS PO EXAM 2013 : Quantitative Aptitude MP 100 = 5 100/10 =50%.
d = discount in percentage on the market price Equivalent Single discount for successive discounts a% and b% = (a+b )%
Example: What are the successive discounts of 10 %, 12 % and 15% amount to a single discount? Solution: Suppose the marked price = Rs. 100 The, S.P. = 85% of 88% of 90% of Rs. 100 = Rs. 67.32. Therefore, the single discount = (100 67.32) % = 32.68 %.
111
## Solved Examples (Profit & Loss)
1. An article was purchased for Rs. 78,350/-. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price? (IBPS CWE PO MT 2012) (1) 4 (2) 7 (3) 5 (4) 3 (5) 6 Solution: Cost price = Rs. 78350 Marked price = 78350 Selling price = 101855 = Rs. 101855 = Rs. 81484
Profit = 81484 -78350 = 3134 Reqd % profit = Ans: (1) 2. The cost of 8 kg of almonds is equal to the cost of 50 kg of apples. The cost of 19 kg of mangoes is Rs. 456. The cost of 1 kg of apples is twice the cost of 2 kg of mangoes. What is the total cost of 3 kg of almonds and 4 kg of apples together? (1) Rs. 2,168 (2) Rs. 2,248 (3) Rs. 2,184 (4) Rs. 2,264 (5) None of these Solution: Cost of 1 kg of mangoes Cost of 1 kg of apples = 2 48 = Rs. 96 Cost of 1 kg of almonds 112 IBPS PO EXAM 2013 : Quantitative Aptitude 100 = 4%
Cost of 3 kg of almonds and 4 kg of apples = 3 600 + 4 96 = Rs. 2184 Ans: (3) 3. Meena purchases 1500 ml of milk every day. If the cost of one liter of milk is Rs. 44, how much amount will she pay in 20 days? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) Rs. 1,340 Rs. 1,320 Rs. 1,280 Rs. 1,260 None of these
Solution: Amount paid in 20 days = 1.5 20 44= Rs. 1320 Ans: (2) 4. Kamya purchased an item of Rs. 46,000 and sold it at loss of 12 per cent. With that amount she purchased another item and sold it at a gain of 12 per cent. What was her overall gain/loss? (Allahabad Bank Probationary Officers Exam 2011) (1) (2) (3) (4) (5) Loss of Rs 662.40 Profit of Rs 662.40 Loss of Rs 642.80 Profit Rs 642.80 None of these = Rs 40480 = Rs 45337.6
## Solution: First S.P. = Second S.P =
Loss = Rs (46000-45337.6) = RS 662.4 Ans: (1) 5. A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights. His total profit percentage is (1) (2) 21% 23% 113 IBPS PO EXAM 2013 : Quantitative Aptitude
(3) (4)
25% 20%
Ans: (1) Solution: Here, x = 10%, y = 10% Total profit percentage = = 21%
6. When the price of sugar decreases by 10%, a man could buy 1 kg more for Rs 270. Then, the original price of sugar per kg is (1) (2) (3) (4) Rs 25 Rs 30 Rs 27 Rs 32
## According to the give condition,
=1 =1 =1 = Rs 30 per kg
x=
114
## IBPS PO EXAM 2013 : Quantitative Aptitude
7. For a certain article, if discount is 25% then profit is 25%. If the discount is 10% then the profit is (1) 50% (2) 40% (3) 30% (4) %
x =
= Rs 510 = Rs
115
## IBPS PO EXAM 2013 : Quantitative Aptitude
8. A man bought orange at the rate of 8 for Rs 34 and sold them at the rate of 12 for Rs 57. How many orange should be sold to earn a net profit of RS 45? (1) (2) (3) (4) 90 100 135 150
Ans: (1) Solution: Cost price of one orange = and selling price of one orange = Profit of one orange = = = =
116
## Practice Set (Profit & Loss)
1. A man makes a profit of 20% on the sale by selling 20 articles for Rs 1, the number of articles he bought by Rs 1 is (a) 20 (b) 24 (c) 25 (d) 30 2. A man buys one table and one chair for Rs 500. He sells the table at a loss of 10% and the chair at a gain of 10%. He still gains Rs 10 on the whole. The cost price of the chair is (a) Rs 300 (b) Rs 350 (c) Rs 200 (d) Rs 250 3. What single discount is equivalent to two successive discounts of 20% and 15%? (a) 35% (b) 32% (c) 34% (d) 30% 4. If the selling price of 10 articles is equal to the cost price of 11 articles, then the gain per cent is (a) 10% 117 IBPS PO EXAM 2013 : Quantitative Aptitude (c) loss 8 % (d) gain 8 % 6. A shopkeeper makes a profit of 20% even after giving a discount of 10% on the marked price of an article. If marked price is Rs 500, then the cost price of the article is (a) Rs 350 (b) Rs 375 (c) Rs 425 (d) Rs 475 (b) 11% (c) 15% (d) 25% 5. Krishna purchased a number of articles at Rs 10 for each and the same number for Rs 14 each. He mixed them together and sold them for Rs 13 each. Then, his gain or loss per cent is (a) loss 8 % (b) gain 8 %
7. A fruit seller bought 240 bananas at the rate of Rs 48 per dozen. He sells of them
equals the cost price of the article. The cost price of the article is (a) Rs 90
at the rate of Rs 5 per banana. th of the remaining are 6 found to be rotten. The price per banana at which he has to sell the remaining bananas to get a profit of 25% on his entire investment is (a) Rs 5.5 (b) Rs 6.0 (c) Rs 5.0
(b) Rs 8O (c) Rs 75 (d) Rs 60 11. When the price of cloth was reduced by 25%, the quantity of cloth sold increased by 20%. What was the effect on gross receipt of the shop? (a) 5% increase
(d) Rs 6.5 (b) 5% decrease 8. The difference between the selling price and cost price of an article is Rs 210. If the profit per cent is 25, then the selling price of the article is (a) Rs 950 (b) Rs 1050 (c) Rs 1150 (d) Rs 1250 9. A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, his advertised price of the item, is (a) Rs 820 (b) Rs 780 (c) Rs 790 (d) Rs 8OO 10. By selling an article for Rs 144, a person gained such that the percentage gain 118 IBPS PO EXAM 2013 : Quantitative Aptitude (c) 10% increase (d) 10% decrease 12. If an article is sold at 200% profit, then the ratio of its cost price to its selling price will be (a) 1:2 (b) 2:1 (c) 1:3 (d) 3:1
13. If an electricity bill is paid before due date one gets a reduction of 4% on the amount of the bill. By paying the bill before due date a person got a reduction of Rs 13. The amount of his electricity bill was (a) Rs 125 (b) Rs 225 (c) Rs 325 (d) Rs 425 14. A shopkeeper marks an article at (. 60 and sells it at a discount of 15%. He also gives a gift worth Rs 3. If he still makes 20% profit, the cost price, (in Rs) is (a) 22 (b) 32 (c) 40 (d) 42 15. A shopkeeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book is (a) 45:56 (b) 45:51 (c) 47: 56 (d) 47: 51 16. A reduction of 10% in the price of sugar enables a housewife to buy 6.2 kg more for Rs 1116. The reduced price per kg is (a) Rs 12 119
(b) Rs 14 (c) Rs 16 (d) Rs 18 17. A man buys a certain number of oranges at 20 for Rs 60 and an equal number at 30 for Rs 60. He mixes them and sells them at 25 for Rs 60. What is gain or loss per cent (a) Gain of 4% (b) Loss of 4% (c) Neither gain nor loss (d) Loss of 5% 18. The marked price of a shirt and trousers are in the ratio 1:2. The shopkeeper gives 40% discount on the shirt. If the total discount on the set of the shirt and trousers is 30%, the discount offered on the trousers is (a) 15% (b) 20% (c) 25% (d) 30%
## IBPS PO EXAM 2013 : Quantitative Aptitude
19. The percentage of loss when an article is sold at Rs 50 is the same as that of the profit when it is sold at Rs 70. The above mentioned percentage of profit or loss on the article is (a) 10% (b) 16 % (c) 20% (d) 22 % 20. I purchased 120 exercise books at the rate of Rs 3 each and sold of them at the rate of Rs 4 each, of them at the rate of Rs 5 each and the rest at the cost price. My profit per cent was (a) 44 % (b) 44 % (c) 44 % (d) 45%
120
## Profit & Loss Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
b a b a d b b b d b
11) 12) 13) 14) 15) 16) 17) 18) 19) 20)
d c c c a d b c b b
121
## Chapter: Simple Interest & Compound Interest
Definition: If a person X borrows some money from another person Y for a certain period, then after that specified period, X (borrower) has to return the borrowed money with some additional money. This additional money that X (borrower) has to pay is called Interest. The actual borrowed money is called Principal or Sum. The Principle and interest together is called amount, and the time for which X the borrower has been used the borrowed money is called the time. The interest that X has to pay for every 100 rupees each year is called rate percent per annum. If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called Simple Interest and it is denoted by S.I. FORMULAE: Let Principle = P, Rate =R% per annum, and Time = T years. Then S.I. = ( Or P= or R= or T= Now, Simple Interest + Principle = Amount If we denote the amount by A, then ) )
122
## Simple Interest = A P S.I. = A P = A=P (1+ ) = SI (1+ )
Two different cases can be compared by using the following formula = Example: What will be the simple interest on Rs. 78,000 at 10% per annum for 9 years? Solution: Here, given that Principal (P) =78,000 Rate (R) = 10% Time (T) = 9 years Now, we know that S.I. = ( S.I. = ( S.I. = Rs. 70,200 Therefore, the simple interest on Rs. 78,000 at 10% per annum for 9 years will be Rs. 70, 200.
123
## IBPS PO EXAM 2013 : Quantitative Aptitude
Compound Interest When the borrower X and the lender Y agrees to fix up a certain time for example yearly, half yearly or quarterly to settle the previous money, then the difference between the amount and the money borrowed is said to be the Compound Interest and it denoted by C.I. In these calculations, principal for the second unit of time is the amount of first unit of time and so on. Some important facts and formulae Let Principal = P, Time = n years (i) If interest is compounded annually, then Amount = P (ii) If interest is compounded half- yearly, then Amount = P (iii) If interest is compounded Quarterly, then Amount = P (iv) If time is in fractions and the interest is compounded yearly, say 2 then
Amount = P
(1+
Example: What will be the Compound Interest on Rs. 5000 at 5% per annum for 3 years, compounded annually? Solution: Amount = Rs. [5,000 ] = Rs. [5,000 = Rs.5788.125 ] = Rs. [5,000 ]
124
## IBPS PO EXAM 2013 : Quantitative Aptitude
Example: What is the compound interest on Rs. 12000 in 4 years at 20 % per annum, the interest being compounded half yearly. Solution: Given that, Principal = Rs. 12000, Rate = 20% per annum, Time = 4 years
125
## Solved Examples (Simple Interest & Compound Interest)
1. What is the difference between the simple and compound interest on Rs. 7,300/- at the rate of 6 p.c.p.a. in 2 years? (IBPS CWE PO MT 2012) (1) Rs. 29.37 (2) Rs. 26.28 (3) Rs. 31.41 (4) Rs. 23.22 (5) Rs. 21.34 Solution: SI = CI = 7300 [(1 7300 [( = = 876 )2 - 1] ) = 902.28
## Difference = 902.28 876 = 26.28 Quick Method: CI = - (6 + 6)
= 12.36 -12 = 0.36% = 0.36 per cent of 7300 = 26.28 Ans: (2) 2. The simple interest accrued on an amount of Rs. 22,500 at the end of four years is Rs. 10,800. What would be the compound interest accrued on the same amount at the same rate of interest at the end of two years? (1) 16,908 (2) 5,724 (3) 28,224 (4) 8,586 (5) None of these 126 IBPS PO EXAM 2013 : Quantitative Aptitude
2
= 12% 22500
## - 22500 = 28224 22500 = 5724
3. The simple interest accrued on a sum of a certain principal is Rs. 35,6727 in seven years at the rate of 8 pcpa. What would be the compound interest accrued on that principal at the rate of 2 pcpa in 2 years? (A) Rs. 2573.48 (B) Rs. 2564.86 (C) Rs. 2753.86 (D) Rs. 2654.48 (E) None of these Solution. Principal
CI Quicker Method:
Ans: (1)
127
## IBPS PO EXAM 2013 : Quantitative Aptitude
4. What will be the simple interest on Rs. 78,000 at 10% per annum for 9 years? Solution: Here, given that Principal (P) =78,000 Rate (R) = 10% Time (T) = 9 years Now, we know that S.I. = ( S.I. = ( S.I. = Rs. 70,200 Therefore, the simple interest on Rs. 78,000 at 10% per annum for 9 years will be Rs. 70, 200 5. What will be the simple interest earned on an amount of Rs. 18,000 in 6 months at the rate of 25% p.a.? (a) Rs.2250.50 (b) Rs.2350.50 (c) Rs.2, 250 (d) Rs 2,400 Solution : P= 18,000, R = 25%, T = 6/12 year S.I. = = = Rs.2, 250.
6. What will be the simple interest on Rs. 2,400 at 4 % per annum for the period from 1st Feb, 2005 to 15th April, 2005? (a) Rs. 20 (b) Rs.20.5 (c) Rs. 22 (d) Rs.25 Solution: Principal (P) = Rs. 2,400, Rate (R) = = year 128 IBPS PO EXAM 2013 : Quantitative Aptitude , Time (T) = (27+31+15) days =73 days = year
Now, S.I. = (
S.I. = (2,400
S.I. = 20.
Therefore, the simple interest on Rs. 2,400 at 4 % per annum for the period from 1 st Feb, 2005 to 15th April, 2005 will be Rs. 20. NOTE: The day on which money is withdrawn is counted while the day on which money is deposited is not counted.
7. A sum of Rs. 600 amounts to Rs. 900 in 5 years at simple interest. What would be the amount if the interest rate is increased by 5%? (a) Rs.1, 000 (b) Rs 1,230 (c) Rs.1, 050 (d) Rs.1, 125 Solution : According to the given situation a sum of Rs. 600 amounts to Rs. 950 in 5 years, then S.I. = Rs. 900 Rs.600 = Rs.300 P = Rs. 600, T = 5 Years Therefore, R= R= % = 10%
If the interest rate is increased by 10 %, then New Rate = (10+5) % =15%, New S.I. =Rs. Then, New Amount = Rs.600 + Rs.450 = Rs.1050 8. What will be the compound interest on a sum of Rs 8000 at the rate of 15% per annum after 2 years? (a) Rs.2400 (b) Rs.2450 (c) Rs.2580 (d) Rs.2650 129 IBPS PO EXAM 2013 : Quantitative Aptitude = Rs.450
## Solution : Amount = Rs. [8000 Rs. [8000 = Rs. 10,580
C.I. = Rs. [10580 8000] = Rs. 2,580 9. What is the difference between the compound interests on Rs. 10000 for 2 years at 4%pe annum yearly and half yearly? (a) Rs.8 (b) Rs.8.49 (c) Rs.7 (d) Rs. 10 Solution: C.I. , When interest is compounded yearly
## Amount = Amount = = Rs.11,032.32
C.I. = 11032.32 10000 = 1032.32 C.I. When interest is compounded half yearly Amount = Rs [10000 ] = Rs. 11040.80803
130
## IBPS PO EXAM 2013 : Quantitative Aptitude
10. A sum of money tripled itself at compound interest in 10 years. In how many years will it become 27 times. (a) (b) (c) (d) 35 years 31 years 30 years 32 years = 3P
Solution : P
131
## Practice Set (Simple Interest & Compound Interest)
1. A sum of money placed at compound interest doubles itself in 4 yr. In how many years will it amount to four times itself? a) 12 yr b) 13 yr c) 8 yr d) 16 yr 2. The difference between the compound interest and simple interest on Rs. 10000 for 2 yr is Rs. 25. The rate of interest per annum is a) 5% b) 7% c) 10% d) 12% 3. The simple interest on a sum of money is of the principal and d) 4. Ratio of the principal and the amount after 1 yr is 10 : 12. Then, the rate of interest per annum is a) 12% b) 16% c) 18% d) 20% 5. At some rate of simple interest, A lent Rs. 6000 to B for 2 yr and Rs. 1500 to C for 4 yr and received Rs. 900 as interest from both of them together. The rate of interest per annum was a) 5% b) 6% c) 8% d) 10% 6. What annual payment will discharge a debt of Rs. 6450 due in 4 yr at 5% per annum simple interest? a) Rs. 1400 b) Rs. 1500 c) Rs. 1550 c) 132 d) Rs. 1600 IBPS PO EXAM 2013 : Quantitative Aptitude
the number of years is equal to the rate per cent per annum. The rate per cent per annum is equal to a) 3% b)
7. In how many years will a sum of money double itself at interest per annum? a) 24 yr b) 20 yr c) 16 yr d) 12 yr 8. In how many years will a sum on Rs. 800 at 10% per annum compound interest, compounded semi-annually becomes Rs. 926.10? a) simple
a) b) c) d) 11. In what time Rs. 8000 will amount to Rs. 9261 at 10% per annum compound interest, when the interest is compounded half-yearly? a) b)
b) c) c) d) d) 9. A certain sum amount to Rs. 5832 in 2 yr at 8% per annum compound interest, then the sum is a) Rs. 5000 b) Rs. 5200 c) Rs. 5280 d) Rs. 5400 10. In what time will Rs. 10000 amount to Rs. 13310 at 20% per annum compounded half-yearly? c) d) 12. Simple interest on Rs. 500 for 4 yr at 6.25% per annum is equal to the simple interest on Rs. 400 at 5% per annum for a certain period of time. The period of time is a) 4 yr b) 5 yr
133
## IBPS PO EXAM 2013 : Quantitative Aptitude
13. If the difference between the compound and simple interests on a certain sum of money for 3 yr at 5% per annum is Rs. 15.25, then the sum is a) Rs. 2000 b) Rs. 1000 c) Rs. 1500 d) Rs. 2500 14. If the simple interest for 6 yr be equal to 30% of the principal, it will be equal to the principal after a) 20 yr b) 30 yr c) 10 yr d) 22 yr 15. The effective annual rate of interest, corresponding to a nominal rate of 6% per annum payable half-yearly is a) 6.06% b) 6.07% c) 6.08% d) 6.09% 16. What annual instalment will discharge a debt of Rs. 6450 due in 4 yr at 5% simple interest? a) Rs. 1500 b) Rs. 1835 c) Rs. 1935 134
d) Rs. 1950 17. At what rate per cent per annum will the simple interest on a sum of money be of the amount in 10 yr? a) 4% b) 6% c) d) 18. The difference between the simple and compound interest on a certain sum of money for 2 yr at 4% per annum is Rs. 4. The sum is a) Rs. 2500 b) Rs. 2400 c) Rs. 2600 d) Rs. 2000 19. The compound interest on a certain sum of money at a certain rate for 2 yr is Rs. 40.80 and the simple interest on the same sum is Rs. 40 at the same rate and for the same time. The rate of interest per annum is a) 2% b) 3% c) 4% d) 5%
## IBPS PO EXAM 2013 : Quantitative Aptitude
20. A sum of money becomes eight times of itself in 3 yr at compound interest. The rate of interest per annum is a) 100% b) 80% c) 20% d) 10%
135
## Simple Interest & Compound Interest Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) c a d d a b c a a a 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) b c a a d a d a c a
136
## Chapter: Time and Work
Introduction: Pipes and cisterns problems are similar to time and work problem, the only difference is that pipes and cisterns problems have outlets and inlets . 1. Inlet is a pipe connected with a tank which fills it 2. Outlet is a pipe connected with a tank to empty it. Mathematical use :1. If pipe can fill (or empty) a tank in x hours, then the part filled (or emptied) in 1 hour = . 2. If a pipe P fills a tank in x hours and another pipe Q empties the full tank in y hours and if both the pipes are opened then the net part filled in 1 hour = ( ).
Example 1: How much time will it take to fill the tank if a pipe A fill it in 30 hours and another pipe B empties it in 40 hours? Sol 1: Net part filled in 1 hour = Hence the time taken to fill the tank = 12 hours. Direct method:The time taken to fill the tank = .
3. If two pipes fills the tank in x and y hours respectively then the net part filled in 1 hours , when both the pipes are opened = ( ).
137
## IBPS PO EXAM 2013 : Quantitative Aptitude
Example 2: How much time will it take to fill the tank if two pipes A and B fill it in 20 hours and 30 hours respectively and both the pipes are opened? Sol 2: Net part filled by both pipes A and B together in 1 hour = Hence the time taken to fill the tank = 12 hours. Direct method:The time taken to fill the tank = .
4. If two pipes fills the tank in x and y hours respectively and a third pipe emptied the full tank in z hours , then the net part filled in 1 hours , when all the pipes are opened = ( ).
Example3: How much time will it take to fill the tank if two pipes A and B fill it in 10 hours and 20 hours respectively and a third pipe C empties it in 40 hours? Sol 3: Net part filled in 1 hour = Hence the time taken to fill the tank = 8 hours. Direct method:The time taken to fill the tank = 5. If a pipe fills the tank in x hours but due to leakage in bottom it is filled in y hours , then the time taken by leak to empty the tank if the tank is full = .
138
## IBPS PO EXAM 2013 : Quantitative Aptitude
Example 4: How much time will the leak take to empty the tank if a pipe A fill it in 10 hours but due to leak in the bottom it is filled in 15 hours ? Sol 4: Tank empty due to leak in 1 hour = ( Hence the leak will empty the full tank in 30 hours Direct method:The required time =
139
## Solved Examples (Time and Work)
1. A and B together can complete a task in 20 days. Band together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone? (IBPS CWE PO MT 2012) (1) 2: 5 (2) 2: 7 (3) 3: 7 (4) 1: 5 (5) 3: 5 Solution: A and B can finish the work in 20 days. A and Bs one days work = B and C and finish the work in 30 days. B and Cs one days work = A and C can finish the work in 40 days A and Cs one day's work = Adding we get 2(A + B + C)s one days work = + + = =
140
## Cs one day work =
C alone can finish the work in 240 days. Reqd ratio = Ans: (4) 2. Seven girls can do a piece of work in 13 days, six boys can do the same piece of work in 12 days, nine men can do the same piece of work in nine days and six women can do the same piece of work in 14 days. Who are the most efficient? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) Boys Girls Women Men Both men and women =1:5
Solution: One girl can complete the work in 7 13 = 91 days. One boy can complete the work in 6 12 = 72 days. One man can complete the work in 9 9 = 81 days. One woman can complete the work in 6 14 = 84 days. Hence, boys are the most efficient. Ans: (1) 3. A and B together can complete a work in 12 days. A alone can complete in 20 days. If B does the work only half a day daily, then in how many days A and B together will complete the work. (1) (2) (3) (4) 10 days 20 days 11 days 15 days
141
## IBPS PO EXAM 2013 : Quantitative Aptitude
As 1 day work = Bs 1 day work = = = B does the work only half day. Bs 1 days work = = + = = =
## Now, A+B)s day work = =
Hence, A and B together will complete the work in 15 days. 4. A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do same work is (1) 3 (2) (3) 5 (4) Ans: (3) Solution: A can do a piece of work in 70 days while B us 40% more efficient than A. B do the same work in
142
## IBPS PO EXAM 2013 : Quantitative Aptitude
= 70 = 70 = 50 days
5. A and B working separately can do a piece of work in 9 and 12 days, respectively. If they work for a day alternately with A beginning, the work would be completed in (1) 28 days (2) 14 days (3) 5 days (4) Ans: (3) Solution: One day work of A = One day work of B = Two days work of (A+B) = + = 10 days = 5 2 days work of (A + B) = Remaining work = 1Now, As turn. A, will complete the work in 9 = day = 5 = = days
143
## IBPS PO EXAM 2013 : Quantitative Aptitude
6. A and B can do a work in 72 days. B and C can do it in 120 days. The number of days needed for A to the work alone is (1) (2) (3) (4) 20 22 33 44
Ans: (3) Solution: Work done by (A+B) in 1 day = Work done by (B+C) in 1 day = Work done by (C+A) in 1 day =
## Work done by 2 (A+B+C) in 1 day =
= Work done by (A+B+C) in 1 day = As 1 day work = = Hence, A do the work alone in 120 days. 7. A, B and C individually can do a work in 10 days, 12 days and 15 days, respectively. If they start working together, then the number of days required to finish the work is (1) (2) (3) (4) 16 days 8 days 4 days 2 days 144 IBPS PO EXAM 2013 : Quantitative Aptitude =
## Ans: (3) Solution: Given, x = 10, y = 12 and z = 15 Required number of days =
= = 4 days 8. A cistern has two pipes. One can fill it with water in 8 h and other can empty it in 5 h. In how many hours will the cistern be emptied, if both the pipes are opened together when the cistern is already full of water? (1) h (2) 10 h (3) 6 h (4) h of
145
## Practice Set (Time and Work)
1. Two men A and B are started a job in which A was thrice as good as B and therefore took 60 days less than B to finish the job. How many days will they take to finish the job, if they start working together? (a) 15 days (b) 20 days (c) 22 days (d) 25 days 2. A and B can separately do a piece of work in 20 and 15 days, respectively. They worked together for 6 days, after which B was replaced by C. The work as finished in next 4 days. The number of days in which C alone could do the work is (a) 30 days (b) 45 days (c) 40 days (d) 35 days 3. A can do a work in 12 days. When he had worked for 3 days, B joined him. If they complete the work in 3 more days, in how many days can B alone finish the work? (a) 6 days (b) 12 days 146 (a) 3 (d) 5 days 6. A and B can do a job alone in 12 days and 15 days, respectively. A starts the work and after 6 days B also joins to finish the work together. For how many days B actually worked on the job? (c) 4 days (d) 8 days 4. A and B can complete a piece of work in 8 days, B and C can do it in 12 days, C and A can do it in 8 days. A, Band C together can complete it in (a) 4 days (b) 5 days (c) 6 days (d) 7 days 5. X is 3 times as fast as Yand is able to complete the work in 40 days less than Y. Then, the time in which they can complete the work together is (a) 15 days (b) 10 days (c) 7 days
## IBPS PO EXAM 2013 : Quantitative Aptitude
(b) 9 (c) 5 (d) 6 7. Two pipes can fill a cistern separately in 24 min and 40 min, respectively. A waste pipe can drain off 30 L/min. If all the there pipes are opened, the cistern fills in one hour. The capacity of the cistern is (a) 800 L (b) 400 L (c) 600 L (d) 500 L 8. A cistern is normally filled in 8 h but takes another 2 h longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in (a) 16 h (b) 20 h (c) 25 h (d) 40 h 9. Two pipes, P and Q can fill a cistern in 12 and 15 min, respectively. Both are opened together but at the end of 3 min, P is turned off. In how many more minutes will Q fill the cistern? (a) 7min (b) 7 min 147
(c) 8 min (d) 8 min 10. Tapas works twice as fast as Mihir. If both of them together complete a work in 12 days. Tapas alone can complete it in (a) 15 days (b) 18 days (c) 20 days (d) 24 days 11. One man and one woman together can complete a piece of work in 8 days. A man alone can complete the work in 10 days. In how many days can one woman alone complete the work? (a) (b) 30 (c) 40 (d) 42 12. A and B together can do a work in 10 days. Band C together can do the same work in 6 days. A and C together can do the work in 12 days. Then, A, Band C together can do the work in (a) 28 days (b) 14 days (c) 5 days (d) 8 days IBPS PO EXAM 2013 : Quantitative Aptitude
13. A is thrice as good a workman as Band therefore is able to finish a job in 40 days less than B. Working together, they can do it in (a) 14 days (b) 13 days (c) 20 days (d) 15 days 14. 45 men can complete a work in 16 days. Four days after they started working, 36 more men joined them. How many days will they now take to complete the remaining work? (a) 6 days (b) 8 days (c) 6 day (d) 7 days 15. A can do a work in 5 days less than the time taken by B to do it. If both of them together take 11 days, then the time taken by B alone 9 to do the same work (in days) is (a) 15 (c) 25 (b) 20 (d) 30
16. A can complete of a work in 5 days and B, of the work in 10 days. In how many days both A and B together can complete the work? (a) 10 (b) 9 (c) 8 (d) 7 17. A boy and girl together fill a cistern with water. The boy pours 4 L of water every 3 min and the girl pours 3 L every 4 min. How much time will it take to fill 100 L of water in the cistern? (a) 36 min (b) 42 min (c) 48 min (d) 44 min 18. If 28 men complete of a piece of work in a week, then the number of men, who must be engaged to get the remaining work completed in another week, is (a) 5 (b) 6 (c) 4 (d) 3
148
## IBPS PO EXAM 2013 : Quantitative Aptitude
19. While working 7 h a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, working 8 h a day? (a) 3 days (b) 4 days (c) 2.5 days (d) 3.6 days 20. A and B can complete a piece of work in 12 and 18 days, respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in (a) 14 days
22. A pipe can empty a tank in 40 min. A second pipe with diameter twice much as that of the first is also attached with the tank to empty it. The two pipes together can empty the tank in (a) 8 min (b) 13 min (c) 30 min (d) 38 min 23. A and B can do a work in 45 days and 40 days, respectively. They began the work together but A left after sometime and B completed the remaining work in 23 days. After how many days of the start of the work did A leave? (a) 10 days
## (b) 9 days (c) 8 days (d) 5 days
days
21. A man and a boy received Rs 800 as wages for 5 days for the work they did together. The man's efficiency in the work was three times that of the boy. What are the daily wages of the boy? (a) Rs 76 (b) Rs 56 (c) Rs 44 (d) Rs 40 149
24. A and B working separately can do a piece of work in 10 days and 15 days, respectively. If they work on alternate days beginning with A, in how many days will the work be completed? (a) 18 days (b) 13 days (c) 12 days (d) 6 days
## IBPS PO EXAM 2013 : Quantitative Aptitude
25. Two pipes can fill a tank separately in 20 min and 30 min, respectively. If both the pipes are opened simultaneously, then the tank will be filled in (a) 10 min (b) 12 min (c) 15 min (d) 25 min
150
## 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13)
c c a c a a c d d b c c d
14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25)
c c b c c a a d b b c b
151
## Chapter: Speed Distance & Time
Speed, Distance and time is one of the most important chapter for the purpose of the maths section in aptitude exams. Time and Distance Formulae relates Time, Distance and Speed. These relationships have many practical applications. The basic concepts of Speed, Distance and time are used in solving questions based on relative motion, circular motion, problem based on trains, problem based on boats, races, etc. The questions asked in IBPS Bank PO is very diverse in nature, therefore this chapter is very important for Bank PO aspirants. For example if you know the speed of any vehicle and the distance covered by that vehicle, then we can easily calculate the time taken in whole journey by using the formula of Time and Distance. Important Formulae (i) Speed = (ii) S km/hr =(s (iii) S m/sec =(s Points to remember: 1. Read the units of time speed and distance carefully. 2. If the distance is given in km and the speed is in m/s then always convert the units. According to the demand of the question you can change the kilometer in to meter or m/s in to km /hr. Example1. A Chennai Express travelling at of its actual speed and covers 42 km in 1 hr 40 ,Distance= Speed Time, Time= )m/sec )km/hr
min 48 sec, find the actual speed of the Chennai Express. Solution: Suppose the actual speed of Chennai Express = S km/hr Then new speed = S
152
## IBPS PO EXAM 2013 : Quantitative Aptitude
= 1hr+40
hr+48
[Because 1hr = 60 minutes, 1 minute = 60 Second, & 1 minute = 1 Second = hr, minute
= 1hr + hr + = hrs
hr
Now, according to the formula New Speed Time taken by Chennai Express with new speed = Distance covered by Chennai Express S = 42
S= S =55km/hr Hence, the actual speed of the Chennai Express is 55 km/hr TRAINS, BOATS AND STREAMS IMPORTANT POINTS ON TRAINS: 1.) Time taken by a train x metres long to pass a single post or a pole or a standing man = Time taken by the train to cover x metres. 2.) Time taken by a train x metres long to pass a stationary object of length y metres = Time taken by the train to cover (x+y) meters. 3.) Suppose two trains or two bodies are moving in the same direction at u kmph and v kmph such that u > v, then their relative speed = (u-v) kmph.
153
## IBPS PO EXAM 2013 : Quantitative Aptitude
4.) If 2 trains of lengths x km and y km are moving in the same direction at u kmph and v kmph , where u > v, then time taken by faster train to cross the slower train = hrs.
5.) Suppose two trains or two bodies are moving in opposite directions at u kmph and v kmph . Then their relative speed = (u+v) kmph 6.) If two trains of lengths x km and y km are moving in opposite directions at u kmph and v kmph , then time taken by the trains to cross each other = hrs.
7.) If two trains start at the same time from two points A and B towards each other and after crossing they take a and b hours in reaching B and A respectively. Then, As speed: Bs speed = (b: a). 8.) x kmph = 9.) y metre/sec = m/sec km/hr EXAMPLES Example 1: Find the time taken by a train 180 m long, running at 72 kmph, in crossing an electric pole. Solution: speed of the train = (72 5/18) m/sec = 20 m/sec Distance moved in passing the pole = 180m Required time taken = (180/20)sec = 9 sec. Example 2: Two trains 137m and 163m in lengths are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmph. In what time will they be clear of each other from the moment they meet? Solution: Relative speed of the trains = (42+48) kmph = 90 kmph = (905/18)m/sec= 25m/sec. Time taken by the trains to pass each other = Time taken to cover (137+163)m at 25m/sec = (300/25)sec = 12 sec. IMPORTANT POINT ON BOATS AND STREAMS: 1.) In water, the direction along the stream is called downstream and the direction against the stream is called upstream. 2.) If speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then Speed downstream = (u + v) km/hr Speed upstream = (u v) km/hr 3.) If the speed downstream is a km/hr and the speed upstream is b km/hr, then : 154 IBPS PO EXAM 2013 : Quantitative Aptitude
Speed in still water = (a + b)/2 km/hr Rate of stream = (a b)/2 km/hr EXAMPLES Example 1: A man can row upstream at 7 kmph and downstream at 10 kmph. Find mans rate in still water and the rate of current. Solution: Rate in still water = (10 + 7)/2 km/hr= 8.5 km/hr Rate of current = (10 7)/2 km/hr = 1.5km/hr Example 2: A man can row 8 kmph in still water and the river is running at 2 kmph. If the man takes 1 hour to row to a place and back, how far is the place? Solution: Mans rate downstream = (8+2) kmph = 10 kmph Mans rate upstream = (8-2) kmph = 6 kmph
155
## Solved Examples (Speed Distance & Time)
1. A 476-metre-Iong moving train crosses a pole in 14 seconds. The length of a platform is equal to the distance covered by tile train in 20 seconds. A man crosses the same platform in 7 minutes and 5 seconds. What is the speed of the man in metre/second? (IBPS RRB Group A Officers Exam 2012) (1) 1.8 m/s (2) 1.4 m/s (3) 1.6 m/s (4) 2 m/s (5) 1.2 m/s Solution: Speed of train Length of platform = 34 20 = 680 metre. ( 7 minute 5 second = 7 60 + 5 = 425 second) Speed of man Ans: 3 2. Train A crosses a pole in 20 seconds and Train B crosses the same pole in one minute. The length of Train A is half the length of Train B. What is the ratio of the speed of Train A to that of Train B? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) 3:2 3:4 4:3 Cannot be determined None of these
Solution: Let the length of Train B be 2x and that of Train A be x. Speed of Train A 156 IBPS PO EXAM 2013 : Quantitative Aptitude
Speed of Train B Ratio: Ans: 1 3. A 280 metre long train moving with an average speed of 108 km/hr crosses a platform in 12 seconds. A man crosses the same platform in 10 seconds. What is the speed of the man? (Allahabad Bank Probationary Officers Exam 2011) (1) (2) (3) (4) (5) 5 m/s 8 m/s 12 m/s Cannot be determined None of these m/sec
Solution: Speed of the train = 108 km/hr = If the length of the platform be x metres, then = 30
x + 280 = 360
x = (360-280 =) = 80 metres
mans speed = Ans: 2 4. A student goes to school at the rate of 2 km/h and reaches 6 min late. If he travels at the speed of 3 km/h, he is 10 min early. The distance (in km) between the school and his house is (1) (2) (3) (4) 5 4 3 1 = 8 m/sec
## Ans: (2) 157 IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: Let the distance between school and his house = x km According to the given condition. - =
- = = =
x = 4 km 5. The Speeds of two trains are in the ratio 6 : 7. If the second train runs 364 km in 4 h, then the speed of first train is (1) (2) (3) (4) 60 km/h 72 km/h 78 km/h 84 km/h
Ans: (3) Solution: Let the speeds of two rain be 6x km/h and 7x km/h, respectively. By given condition Speeds of second train = 7x = 91 x= = 13 km/h = 91 km/h.
158
## IBPS PO EXAM 2013 : Quantitative Aptitude
6. If a man walks 3 km/h, he is late to his office by 20 min. If he increases his speed to 6 km/h, he reaches the office 30 min early. The distance of his office from the starting place is (1) (2) (3) (4) 6 km 5 km 5.5 km 4 km
Ans: (2) Solution: Let the distance of his office from the starting place is x km. By given condition, = + =
x =
= 5 km
7. In covering a distance of 30 km, Abhay takes 2 h more than Sameer. If Abhay doubles his speed, then he would take 1 h less than Sameer. Abhay's speed (in km/h) is (1) (2) (3) (4) 5 6 6.25 7.5
Ans: (1) Solution: Let the speeds of Abhay be x km/h and Sameer be y km/h respectively. Then, And = 2 ..(i) = 1 ..(i)
On adding Eqs. (i) and (ii), we get 159 IBPS PO EXAM 2013 : Quantitative Aptitude
=3
30 = 6x x = 5 km/h
8. A man completed a certain journey by a car. If he covered 30% of the distance at the speed of 20 km/h and the remaining distance at 10 km/h, his average speed for the whole journey was (1) (2) (3) (4) 25 km/h 28 km/h 30 km/h 33 km/h
Ans: (1) Solution: Let the total be 100 km. Average Speed = = =
= = 25 km/h 9. Two trains started at the time one from A to B and other from B to A. If they arrived at B and A, respectively 4 h and 9 h after they passed each other, the ratio of the speeds of the two trains was (1) (2) (3) (4) 2:1 3:2 4:3 5:4
Ans: (2) Solution: Required ratio of the speeds of two trains 160 IBPS PO EXAM 2013 : Quantitative Aptitude
10. A man takes 6 h 15 min in walking a distance and ridding to the starting place. He could walk both ways in 7h 45 min. The time taken by him to ride both ways, is (1) (2) (3) (4) 4h 4 h 30 min 4 h 45 min 5h
Ans: (3) Solution: Time taken in walking both ways = 7 h 45 min (i)
Time taken in walking one way and riding back = 6 h 15 min (ii)
On multiplying by 2 in Eq. (ii) and then subtracting Eq . (i) from Eq. (ii), we get Time taken by the man to ride both ways = 12 h 30 min 7h 45 min = 7 h 45 min
161
## Practice Set (Speed Distance & Time)
1. A person travels 285 km in 6 h. In the first part of the journey, he travels at 40 km/h by bus. In the second part, he travels at 55 km/h by train. The distance travelled by train is (a) 165 km (b) 615 km (c) 561 km (d) 156 km 2. With average speed of 40 km/h, a train reaches its destination at time. If it goes with an average speed of 35 km/h, it is late by 15 min. The total journey is (a) 30 km (b) 40 km (c) 70 km (d) 80 km 3. If a train runs at 40 km/h, it reaches its destination late by 11 min but, if it runs at 50 km/h, it is late by 5 min only. The correct time (in min) for the train to complete the journey is (a) 13 (b) 15 (c) 19 (d) 21 (d) 38 km/h 4. Two trains, 80 m and 120 m long, are running at the speed of 25 km/h and 35 km/h, respectively in the same direction on parallel tracks. How many seconds will they take to pass each other? (a) 48 s (b) 64 s (c) 70 s (d) 72 s 5. A train, 300 m long, passed a man, walking along the line in the same direction at the rate of 3 km/h in 33 s. The speed of the train is (a) 30 km/h (b) 32 km/h (c) 32 km/h
6. By walking at of his usual speed, a man reaches his office 20 min later than his usual time. The usual time taken by him to reach his office is (a) 75 min (b) 60 min (c) 40 min
162
## IBPS PO EXAM 2013 : Quantitative Aptitude
(d) 30 min 7. In a 100 m race, Kamal defeats Bimal by 5 s. If the speed of Kamal is 18 km/h, then the speed of Bimal is (a) 15.4 km/h (b) 14.5 km/h (c) 14.4 km/h (d) 14 km/h 8. A train, 240 m long, crosses a man walking along the line in opposite direction at the rate of 3 km/h in 10 s. The speed of the train is (a) 63 km/h (b) 75 km/h (c) 83.4 km/h (d) 86.4 km/h 9. A boy is late by 9 min, if he walks to school at a speed of 4 km/h. If he walks at the rate of 5 km/h he arrives 9 min early. The distance to his school is (a) 9 km (b) 5km (c) 4 km (d) 6 km 10. Two towns A and B are 500 km apart. A train starts at 8 am from A towards B at a speed of 70 km/h. At 10 am, another train starts from B towards A at a speed of 110 km/h. When will the two trains meet? (a) 1 pm 163
(b) 12 noon (c) 12: 30 pm (d) 1: 30 pm 11. A train 150 m long passes a pole in 15 s and another train of the same length travelling in the opposite direction in 12 s. The speed of the second train is (a) 45 km/h (b) 48 km/h (c) 52 km/h (d) 54 km/h 12. A, B and C start together from the same place to walk round a circular path of length 12 km. A walks at the rate of 4 km/h, B 3 km/h and C km/h. They will
meet together at the starting place at the end of (a) 10 h (b) 12 h (c) 15 h (d) 24 h 13. A train travelling at 48 km/h crosses another train, having half its length and travelling in opposite direction at 42 km/h in 12 s. It also passes a railway platform in 45 s. The length of the railway platform is (a) 200 m (b) 300 m (c) 350 m IBPS PO EXAM 2013 : Quantitative Aptitude
(d) 400 m 14. In a race of one kilometer, A gives B a start of 100 m and still wins by 20 s but, if A gives B a start of 25 s, B wins by 50 m. The time taken by A to run one kilo metre is (a) 17 s (b) (c) (d) s
(b) 27 km/h (c) 25 km/h (d) 24 km/h 17. A and B run a kilometre and A wins by 25 s. A and C run a kilometre and A wins by 275 m. When Band C run the same distance, B wins by 30 s. The time taken by A to run a kilometre is (a) 2 min 25 s
s s
(b) 2 min 50 s (c) 3 min 20 s (d) 3 min 30 s 18. A train passes a man standing on a platform in 8 s and also crosses the platform which is 264 m long in 20 s. The length of the train is (a) 188 m (b) 176 m (c) 175 m (d) 96 m
15. A student walks from his house at a speed of 2 km/h and reaches his school 6 min late. The next day he increases his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house? (a) km (b) km (c) km (d) km
19. A walks at a uniform rate of 4 km/h and 4 h after his start, B bicycles after him at the uniform rate of 10 km/h. How far from the starting point will B catch A? (a) 16.7 km (b) 18.6 km
16. A train passes two persons walking in the same direction at a speed of 3 km/h and 5 km/h respectively, in 10 sand 11 s, respectively. The speed of the train is (a) 28 km/h 164
## IBPS PO EXAM 2013 : Quantitative Aptitude
20. A constable follows a thief who is 200 m ahead of the constable. If the constable and the thief run at speeds of 8 km/h and 7 km/h respectively, the constable would catch the thief in (a) 10 min (b) 12 min (c) 15 min (d) 20 min
165
## Speed Distance & Time Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) a c c d d b c c d b 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) d d d b b c a b d b
166
## IBPS PO EXAM 2013 : Quantitative Aptitude
Chapter: Mensuration
Mensuration is a branch of Mathematics which deals with lengths of lines, areas of surfaces and volume of solids. Mensuration may be divided into two parts: (i) (ii) Plane mensuration which deals with the sides, perimeters and areas of plane figures of different shapes. Solid mensuration which deals with the areas and volumes of solid objects.
## Perimeter and Area of Triangles:
1.) If a, b, c are three sides of a triangle; then i) Perimeter = a+b+c ii) Area = {s(s-a)(s-b)(s-c)} , where s= semi-perimeter of triangle i.e., s= (a+b+c)/2 2.) If base and the corresponding altitude of triangle are known, then Area of triangle = base height 3.) If a is side of an equilateral triangle then i.) Perimeter = 3a ii.) Area = (3/4) a2 Perimeter and area of rectangles, square:
l b a a
1.) If l and b are length and breadth of rectangle, Then i.) Perimeter = 2(l + b) ii.) Area = lb iii.) Diagonal= (l2 + b2) 167 IBPS PO EXAM 2013 : Quantitative Aptitude
2.) If a is side of square , then i.) Perimeter = 4a ii.) Area = a2 iii.) Diagonal , d = a2 + a2 = a2 iv.) Side = area Area of Trapezium
Two sides of trapezium are parallel. If a and b are parallel sides and h is distance (i.e. height) between them; then Area of trapezium = (a + b) h Parallelogram
Area of parallelogram = bh where: b = base of the parallelogram h=height of the parallelogram Rhombus
Rhombus has all the sides equal and diagonals of rhombus bisect each other at right angles. If a is the side of rhombus and d1 , d2 are diagonals ; then Perimeter of rhombus = 4a Area of rhombus = d1 d2
168
## IBPS PO EXAM 2013 : Quantitative Aptitude
Circle
If r is radius of circle, then diameter= 2r, circumference of circle = 2r, area of circle = r2 The Greek letter (pronounced as Pie) = 22/7 = 3.14 EXAMPLES: 1) Find the area of triangle whose sides are 10cm, 24cm and 26cm. Solution: a= 10cm b= 24cm c=26cm s= (a+ b+ c)/2= (10+ 24+ 26)/2 =30 area of triangle = {s(s-a)(s-b)(s-c)} = {30(30-10)(30-24)(30-26)} =120cm2 2) If two sides of a triangle are 6cm and 8cm and height of the triangle corresponding to 6cm side is 4cm; find the area of triangle: Solution: area of triangle= baseheight = 64 =12cm2 3.) Find the length and perimeter of the rectangle whose: Area = 120cm2 and breadth= 8cm Solution: area of rectangle =120cm2 breadth b=8cm Area = lb l 8 =120 l = 15cm Perimeter = 2 (l + b) = 2 (15 + 8) = 223 = 46cm 4.) The parallel sides of a trapezium are 8.4cm and 12.3cm. If its height is 7.2 cm, find its area. 169 IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: sum of parallel sides = 8.4+ 12.3 = 20.7cm Height = 7.2cm Area of trapezium = (sum of parallel sides) height = 20.77.2 = 74.52cm2 3-D FIGURES: CUBOID: h lllllllll b l Lateral / curved surface area = 2(l+b)h Total surface area = 2(lb+bh+lh) Volume = lbh Diagonal of cuboid= (l2+b2+h2) Where: l= length, b= breadth, h = height CUBE: l l
Lateral/curved surface area = 4l2 Total surface area = 6l2 Volume = l3 Diagonal of cube = l3 Where: l = edge of cube
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## RIGHT CIRCULAR CYLINDER:
Lateral/curved surface area= 2rh Total surface area= 2r(r+h) Volume = r2h Where: r = radius h = height RIGHT CIRCULAR CONE:
Lateral/curved surface area = rl Total surface area = r(l+r) Volume = (1/3)r2h Where: r=radius of base h=height l=slant height = (r2+h2)
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## IBPS PO EXAM 2013 : Quantitative Aptitude
SPHERE:
Lateral/curved surface area = 4r2 Total surface area = 4r2 Volume = (4/3)r3 ,Where: r= radius of sphere HEMISPHERE:
Lateral/curved surface area = 2r2 Total surface area = 3r2 Volume = (2/3)r3 where: r= radius of hemisphere SPHERICAL SHELL:
Total surface area = 4(R2+r2) Volume = (4/3)(R3-r3) Where: R= external radius r = internal radius 172 IBPS PO EXAM 2013 : Quantitative Aptitude
EXAMPLES: 1.) Find the volume of right circular cylinder which has a height of 21cm, and the base radius 5cm. Also find the curved surface area of the cylinder. Solution: Given h=21cm, r=5cm Volume of cylinder=r2h = (22/7) 2521 = 1650cm3 Curved surface area = 2rh= 2 (22/7) 5 21= 660cm2 2.) A rectangular sheet of paper 44cm 18cm is rolled along its length and a cylinder is formed. Find the volume of cylinder.(Use =22/7) Solution: The sheet of paper is rolled along the length. Then, height of the cylinder= 18cm and the circumference of base of the cylinder= 44cm Let the radius of the base of the cylinder. 2 (22/7) r=44 r = 7cm and height = 18cm Volume of the cylinder = r2h = (22/7)7718 = 2772cm3 3.) A beam 9m long, 40cm wide and 20cm deep is made up of iron which weighs 50kg per cubic meters. Find the weight of the beam. Solution: Volume of the beam = lbh = 9 (4/100) (20/100)m3 = 72/100 m3 Since the beam is made up of iron which weighs 50kg per cubic metre. Weight of the beam = (72/100) 50 = 36kg 4.) A rectangular reservoir is 120m long and 75m wide. At what speed per hour must water flow into it through a square pipe of 20cm side so that the water rises by 24m in 18 hours? Solution: Volume of accumulated in 18 hours = (1207524)cu.m. Volume of water accumulated in one hour = (1207524)/18 m3 Area of the end of the square pipe = (20/100)2 = (1/25) m2 = 0.4 m2 Speed of the water per hour = (1207524)/(180.4) = 30,000 m = 30 km Hence water must flow at 30km/hour into the reservoir.
173
## Solved Examples (Mensuration)
1. The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law, the sum of their ages is 92. Assuming that there has been on other change in the family structure and all the people-are alive, what is the difference in the age of the daughter and the daughter-in-law? (IBPS CWE PO MT 2012) (1) 22 years (2) 11 years (3) 25 years (4) 19 years (5) 15 years Solution: There are four members in a family. Five years ago the sum of ages of the family members = 94 years Now, sum of present ages of family members = 94 + 5 4 = 114 years Daughter is replaced by daughter-in-law. Thus, sum of family member's ages becomes 92 years. Difference = 114 - 92 = 22 years Ans: (1) 2. The area of a square is 1444 square meters. The breadth of a rectangle is 1/4th the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle? (IBPS CWE PO MT 2012) (1) 1152.38 sq mtr (2) 1169.33 sq mtr (3) 1181.21 sq mtr (4) 1173.25 sq mtr (5) None of these Solution: Area of square = 1444 Let the side of square be a. So a2 = 1444 a= = 38 m Breadth of rectangle = 38 = 9.5 metres Length = 3 x 9.5 = 28.5 m 174 IBPS PO EXAM 2013 : Quantitative Aptitude
Area of rectangle = 28.5 x 9.5 = 270.75 m2 Difference = 1444 - 270.75 = 1173.25 sq m Ans: (4) 3. The side of a square is half the diameter of a circle. The area of the square is 1225 sq cm. What is the area of the circle? (1) 962.5 sq cm (2) 3850 sq cm (3) 15400 sq cm (4) 15600 sq cm (5) None of these Solution: Side of the square Radius of the circle = 35 cm Area of the circle Ans: (2) 4. The adjacent angles of a parallelogram are in the ratio of 5 : 7. The larger angle of the parallelogram is equal to the largest angle of a triangle. The smallest angle of the triangle is 40 less than the smaller angle of the parallelogram. What is the measure of the second largest angle of the triangle? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) 30 40 55 35 None of these
Solution: Sum of adjacent angles of a parallelogram = 180 Larger angle of the parallelogram Smaller angle of the parallelogram = 180 105 = 75 Second-largest angle of the triangle = 180 105 (75 40) = 40 175 IBPS PO EXAM 2013 : Quantitative Aptitude
Ans: (2) 5. The side of a square is half the diameter of a circle. The area of the square is 1225 sq cm. What is the area of the circle? (Corporation Bank PO 2011) (1) (2) (3) (4) (5) 962.5 sq cm 3850 sq cm 15400 sq cm 15600 sq cm None of these
Solution: Side of the square Radius of the circle = 35 cm Area of the circle Ans: (2) 6. The sum of the circumference of a circle and the perimeter a square is equal to 272 cm. The diameter of the circle is 56 cm. What is the sum of the area of the circle and the area of the square? (Allahabad Bank Probationary Officers Exam 2011) (1) (2) (3) (4) (5) 2464 sq cm 2644 sq cm 3040 sq cm Cannot be determined None of these
176
## Area of the circle = r2 =
x 28 x 28 = 2464 sq cm
required sum = (576 + 264) sq cm = 3040 sq cm Ans: (3) 7. The ratio of the three angles of a quadrilateral is 13: 9: 5 respectively. The value of the fourth angle of the quadrilateral is 36. What is the difference between the largest and the second smallest angles of the quadrilateral? (Allahabad Bank Probationary Officers Exam 2011) (1) (2) (3) (4) (5) 104 108 72 96 None of these
Solution: Let the three angles of the quadrilateral be13xo, 19xo and 5x o respectively. Now, according to the question; 13x + 9x + 5x = 360 - 36 = 324
27x = 324
x= = 12
177
## Practice Set (Mensuration)
1. The diameter of garden roller is 1.4 m and it is 2 m long. The area covered by the roller in 5 revolutions is a) 4.4 m2 b) 44 m2 c) 16.8 m2 d) 8.8 m2 2. In a cylindrical vessel of diameter 24 cm filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Then, the increase in height of water level is a) 1.5 cm b) 2 cm c) 3 cm d) 4.2 cm 3. If the length of the diagonal of a cube is , then its surface area is a) 192 cm2 b) 512 cm2 c) 768 cm2 d) 384 cm2 4. A solid sphere of radius 1 cm is melted to convert into a wire of length 100 cm. The radius of the wire (take ) is a) 0.08 cm 178 IBPS PO EXAM 2013 : Quantitative Aptitude b) 0.09 cm c) 0.16 cm d) 0.11 cm 5. A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a corner of the field and the earth taken out is evenly spread over the remaining area of the field. The level of the field raised is a) 12 cm b) 14 cm c) 16 cm d) 18 cm 6. A right circular cylinder, a hemisphere and a right circular cone stand on the same base and have the same height. The ratio of the volumes is a) 3 : 6 : 1 b) 3 : 4 : 1 c) 3 : 2 : 1 d) 4 : 3 : 1 7. The base of a right pyramid is an equilateral triangle of side 4 cm. The height of the pyramid is half of its slant height. Its volume is a) b)
c) d) 8. A tent is of the shape of a right circular cylinder upto a height of 3 m and then becomes a right circular cone with maximum height of 13.5 m above the ground. If the radius of the base is 14 m, the cost of painting the inner side of the tent at the rate Rs. 2 per sq m is a) Rs. 2050 b) Rs. 2060 c) Rs. 2068 d) Rs. 2080 9. A solid cone of height 9 cm with diameter of its base 18 cm is cut out from a wooden solid sphere of radius 9 cm. The percentage of wood wasted is a) 25% b) 30% c) 50% d) 75% 10. The number of spherical bullets that can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being of 4 cm diameter, is a) 2541 b) 2451 c) 2514 d) 2415 11. From a solid cylinder whose height is 12 cm and diameter 10 cm, a 179
conical cavity of same height and same diameter of the base is hollowed out. The volume of the remaining solid is approximately
a) 942.86 cm3 b) 314.29 cm3 c) 628.57 cm3 d) 450.76 cm3 12. The base of a right pyramid is a square of side 16 cm long. If its height be 15 cm, then the area of the lateral surface (in cm 2) is a) 136 b) 544 c) 800 d) 1280 13. The curved surface area of a cylindrical pillar is 264 sq m and its volume is 924 cu m. The ratio of its diameter to height is a) 3 : 7 b) 7 : 3 c) 6 : 7 d) 7 : 6 14. The ratio of the volume of a cube and of a solid sphere is 363 : 49. The ratio of an edge of the cube and the radius of the sphere is
a) b) c) d)
7 : 11 22 : 7 11 : 7 7 : 22
## IBPS PO EXAM 2013 : Quantitative Aptitude
15. The base of a solid right prism is a triangle whose sides are 9 cm, 12 cm and 15 cm. The height of the prism is 5 cm. Then, the total surface area of the prism is a) 180 cm2 b) 234 cm2 c) 288 cm2 d) 270 cm2 16. Volume of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5. The ratio of their heights is a) 1 : 5 b) 5 : 4 c) 5 : 16 d) 25 : 64 17. Water is being pumped out through a circular pipe whose internal diameter is 7 cm, if the flow of water is 12 cm/s then how many litres of waters being pumped out in one hour? a) 1663.2 L b) 1500 L c) 1747.6 L d) 2000 L 18. A right circular cylinder of height 16 cm is covered by a rectangular tin foil of size 16 cm 22 cm. The volume of the cylinder is
a) 352 cm3 b) 308 cm3 c) 176 cm3 d) 616 cm3 19. The volume of a cube (in cm3), whose diagonal measures cm is a) 16 b) 27 c) 64 d) 8 20. Some solid metallic right circular cones, each with radius of the base 3 cm and height 4 cm, are melted to form a solid sphere of radius 6 cm. The number of right circular cones is a) 12 b) 24 c) 48 d) 6 21. A cuboidal water tank contains 216 L of water. Its depth is of its length and breadth is of of the difference between length and depth. The length of the tank is a) 72 dm b) 18dm c) 6 dm d) 2 dm
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## IBPS PO EXAM 2013 : Quantitative Aptitude
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11)
b b d d c c c c d a c
12) 13) 14) 15) 16) 17) 18) 19) 20) 21)
b b b c d a d c b b
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## Chapter: Permutations, Combinations & Probability
Permutation The arrangement of a number of things taking some or all of them at a time is called permutation. If there are n number of things and we have to select r things at a time then the total number of permutation is denoted by n =
For example if there are 3 candidates A ,B and C for the post of president and vice president of a college , since we have to select only 2 candidates , it can be done in 3! Ways. i.e. (A, B) (B, C) (A, C) (B, A) (C, B) and (C, A). Here order of arrangement matters. Restricted Permutation: Sometimes we have to find out the number of permutation keeping few specific objects at specific places. In this case, we find out the number of permutation of filling remaining vacant places by the remaining objects. If r objects are taken out of n dissimilar objects (i) A specific object is taken each time: if there are n objects Suppose that is taken each time. If .
takes first place then the remaining (n-1) can take any place so number of
## objects can be arranged in n-1
ways. Since
permutation is r n-1 . (ii) Specific object never taken: then r objects are taken out of (n-1) objects, so number of permutation is n-1 . {Note: n = n-1 + r n-1 }
Permutation of things when some are identical: If we have n things in which p are exactly of one kind , q of second kind , r of third kind and the rest are different then the number of permutation of n things taken all at a time n 182 IBPS PO EXAM 2013 : Quantitative Aptitude =
Example: In how many ways can the letters of the word LEADER be arranged? Solution: The word LEADER contains total 6 letters namely 1L, 2E, 1A, 1D, 1R Therefore, the number of ways to arrange the letters of the word LEADER = = 360.
Repetition of things: The number of permutation formed by taking r things at a time out of n things in any object arrangement such that each object can be taken any number of time is Circular permutation: If we fix one of the objects around the circumference of a circle then number of permutation of n different thing taken all at a time is (n-1)! Ways. It will be same as by putting (n-1) objects at (n-1) places. But if we do not consider the direction i.e. clockwise and anticlockwise then the number of permutation is Combination From a given group of object each of the number of groups which are formed by taking some objects or all objects at a time without caring about the sequence of the objects is called combination. The number of combination formed by taking r objects at a time out of n object is denoted by n n = where C expresses combination. . .
For example if we have 3 objects A , B and C , 2 objects are taken out at a time then 3 combination are formed AB , BC and CA. Note: If r= 0 , then n = =1
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## IBPS PO EXAM 2013 : Quantitative Aptitude
If r= 1 , then n If r= n , then n n =n
= =
=n =1
## . = = 100 {Because 1! =1}
The combination of r object out of n objects on which p specific objects: 1. Are always included is n-p remaining (n-p). 2. Are never included is n-p . Since p specific object are never included we have to form the combination taking r obects out of (n-p) objects. The number of ways to select some or all thing out of any number of given thing: There are 2 ways to select anything i.e. either it will be selected or not. Therefore number of ways to select n things is 2 For non- empty selection is Note: n +n 2 -1. = -1. n times = . In these empty selection is also include. . We have to keep aside p specific objects and to select
+. + n
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## IBPS PO EXAM 2013 : Quantitative Aptitude
Difference between permutation and combination: Suppose there are 5 objects out of which 2 have to be chosen. Permutation Number of required way = = = = = 5 4 = 20 = = 10 Combination
So it is clear that in permutation order matters while in combination order does not matter. Probability The mathematical measure of the uncertainty is called probability. For example, consider the following questions: (a) (b) (c) (d) Will it rain today? Which of the three candidates will win? On throwing a dice, the number obtained will be even or odd? On tossing a coin, head will occur or tail will occur?
The answer to all these question is not sure i.e. there is uncertainty .We study the uncertainty of the result of such question in the theory of probability , which may not have one result but more than one result are possible . Random experiment: The experiments in which the outcomes cannot be predicted before hand is called random experiments. When these kind of experiment are repeated under identical condition, they do not produce the same outcome every time and there may be many possible outcome which depends upon chance and cannot be predicted. For example, on tossing a coin either the head will come up or the tail will come up, we cannot predict it. This is an example of random event. Sample Space: The set of all possible outcomes of experiments is called the sample space and it is denoted by S. And the subset of a sample space is called an event. That is, every subset A of the sample 185 IBPS PO EXAM 2013 : Quantitative Aptitude
space S is an event of that random experiment. For example, in an experiment of tossing a coin, if h is obtained then it is a random event, since here S = {H, T} and {H} S Now, the probability of any event A can be defined as the ratio between the number of favourable outcomes to the event A and the number of total equiprobable outcomes, that is P(A) = Here it should be noted that the probability of a certain or sure event is 1 and that of impossible event is 0. Now, since the probability of an event to occur is =
So the probability of an event A not to occur is = 1 Mutually Exclusive events: Two events A and B are said to be mutually exclusive if they cannot occur together, that is simultaneously. For example , on throwing a dice , the events A = { 2,4,6 } and B = { 1,3,5 } are mutually exclusive events , i.e. A B = . In term of probability if A and B are mutually exclusive events, then P (A B) = P (A) + P (B) and, P (S) = P (A) + P (A) = 1 where A is Complement of A. Example: If a dice is thrown once then the probability of the number appearing on dice is more than 2? (a) (b) (c) (d) 1/3 1/2 2/3 4.3/4
Solution: As there are 6 faces on a dice, So the total number of possible events are 1, 2 , 3 . 6 , that is = 6 Now the number more than are 3, 4 , 5 and 6 So total number of favourable events =
186
## IBPS PO EXAM 2013 : Quantitative Aptitude
Probability of an event =
Required probability = = Example : An urn contains 3 green, 6 red, and 4 black balls. 3 balls are drawn. Find the probability that all 3 balls are of same colour? (a) 3/44 (b) 25/286 (c) 15/286 (d) 5/286
Solution: Total number of balls in an urn is 13. Number of ways 3 balls can be drawn out of 13 balls = Numbers of ways 3 green balls are drawn = Numbers of ways 6 red balls are drawn = Numbers of ways 4 black balls are drawn = Now, The required probability = + + = = = = =1 = 20 =4 = = 286
187
## Solved Examples (Permutations, Combinations & Probability)
1. In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together? (IBPS CWE PO MT 2012) (1) 720 (2) 1440 (3) 5040 (4) 3600 (5) 4800 Solution: Total number of letters is 7, and these letters can be arranged in 7! ways . = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 ways There are seven letters in the word THERAPY including 2 vowels. (E, A) and five consonants. Consider two vowels as one letter. We have 6 letters which can be arranged in 6P6 = 6 ways. But vowels can be arranged in 2! ways. Hence, the number of ways, all vowels will come together = 6! x 2! = 1 x 2 x 3 x 4 x 5 x 6 x 2 = 1440 Total number of ways in which vowels will never come together = 5040 - 1440 = 3600 Ans: (4) 2. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour? (IBPS CWE PO MT 2012) (1) (2) (3) (4)
188
## IBPS PO EXAM 2013 : Quantitative Aptitude
(5) Solution: Total number of balls = 13 + 7 = 20 Number of sample space = n(S) = 20C2 = 190 Number of events = n(E) = 13C2 + 7C2 = 78 + 21 = 99 P(E) = Ans: (4) Directions (1-5): Study the given information carefully to answer the questions that follow. An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles. 3. If two marbles are drawn at random, what is the probability that both are red or at least one is red? (1) (2) (3) (4) (5) None of these Solution: Total number of marbles in the urn = 4 + 5 + 2 + 3 = 14 Total number of possible outcomes = Selection of 2 marbles out of 14 marbles = 14C2= 91 =
Total number of favourble cases = 2C2 + 2C1 + 12C1 = 1 + 2 x 12 = 25 189 IBPS PO EXAM 2013 : Quantitative Aptitude
required probability = Ans: (5) 4. If three marbles are drawn at random, what is the probability that at least one is yellow? (1)
(2)
(3)
(4)
(5) None of these Solution: Total number of possible outcomes = 14C3 = = 364
## When no marbles is yellow, favourable number of cases = 14C3 = = 165
Probability that no marble is yellow = required probability = (Probability that at least one is yellow) = (1 - Probability that no marble is yellow) 1Ans: (3) = =
190
## IBPS PO EXAM 2013 : Quantitative Aptitude
5. If eight marbles are drawn at random, what is the probability that there are equal numbers of marbles of each colour? (1) (2) (3) (4) (5) None of these Solution: Total possible outcomes = 14C8 = 14C6 [ = 3003 Total number of favourable cases = 4C2 x 5C2 x 2C2 X 3C2 = 6 x 10 x 1 x 3 = 180 = 3003 required probability = Ans: (3) 6. If three marbles are drawn at random, what is the probability that none is green? (1) =
n
Cr = nCn-r]
(2)
(3)
(4)
191
(5)
## Solution: Total number of possible outcomes = 14C2 = = 364
Now, according tot the question, no marble should be green. Total number of favourable outcomes = Selection of 3 marbles out of 5 blue, 2 red and 3 yellow marbles = 10C3 = = 120 =
## required probability = Ans: (5)
7. If three marbles are drawn at random, what is the probability that two are blue and two are red? (1)
(2) (3)
(4) (5) None of these Solution: Total number of possible outcomes = 14C4 = = 1001
Total number of favourable cases 192 IBPS PO EXAM 2013 : Quantitative Aptitude
= 5C2 x 22C = 10 x 1 = 10 required probability = Ans: (1) Directions (Q. 8-10): Study the given information carefully and answer the questions that follow: A basket contains 4 red, 5 blue and 3 green marbles. 8. If three marbles are picked at random, what is the probability that either all are green or all are red? (1) (2) (3) (4) (5) None of these Solution: P(All Green) + P(All Red) = 3C3 / 12C3 + 4C3 / 12C3 = 1/44 Ans : (4) 9. If two marbles are drawn at random, what is the probability that both are red? (1) (2) (3) (4) (5) None of these 193 IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: 4C2 / 12C2 = 1/11 Ans: (5) 10. If three marbles are picked at random, what is the probability that at least one is blue? (1) (2) (3) (4) (5) None of these Solution: 1 P (None Blue) = 1- (7C3 / 12C3) = 37/44 Ans: (2)
194
## Practice Set-1 (Permutations, Combinations & Probability)
1. In how many ways can six different rings be worn on four fingers of one hand? a) 10 b) 12 c) 15 d) 16 2. There are three prizes to he distributed among five students. If no student gets more than one prize, then this can be done in a) 10 ways b) 30 ways c) 60 ways d) 80 ways 3. In a hockey championship, there were 153 matches played. Every two teams played one match with each other. The number of teams participating in the championship is a) 18 b) 19 c) 17 d) 16 4. In an examination paper, there are two groups each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected? 195 a) 24 b) 48 c) 96 d) None of these 5. After a get together every person present shakes the hand of every other person. If there were 105 hands shakes in all, how many persons were present in the party? a) 14 b) 13 c) 15 d) 16 6. There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is a) 1048 b) 1072 c) 1024 d) none of these 7. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is a) 140 b) 280 c) 196 d) 346
## IBPS PO EXAM 2013 : Quantitative Aptitude
8. After a get-together every person present shakes the hand of every other person. If there were 105 hands shakes in all, how many persons were present in the party? a) 14 b) 13 c) 15 d) 16 9. Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side. a) 864 b) 865 c) 863 d) 1728
196
## Practice Set-2 (Permutations, Combinations & Probability)
1. Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number? a) b) c) a) d) 2. Two dice are tossed. The probability that the total score is a prime number is a) b) c) d) 3. A bag contains 3 white balls and 2 black balls. Another bag contains 2 white balls and 4 black balls. A bag and a ball are picked at random. The probability that the ball will be white is 197 a) b) c) d) b) c) d) 5. A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is probability that the balls drawn contain no blue ball? a) b) c) d) 4. Suppose six coins are flipped. Then the probability of getting at least one tail is
## IBPS PO EXAM 2013 : Quantitative Aptitude
6. I forgot the last digit of a 7-digit telephone number. If I randomly dial the final 3 digits after correctly dialling the first four, then what is the chance of dialling the correct number? a) b) c) d) 7. A box contains 6 white balls and 7 black balls. Two balls are drawn at random. What is the probability that both are of the same colour? a) b) c) d) 8. A brother and sister appear for an interview against two vacant posts in an office. The probability of the brother's selection is and that of
the sister's selection is . What is the probability that one of them is selected? a) b) c) d) 9. A room has 3 lamps. From a collection of 10 light bulbs of which 6 are not good, a person selects 3 at random and puts them in a socket. The probability that he will have light, is a) 5/6 b) 1/2 c) 1/6 d) none of the above 10. Four boys and three girls stand in queue for an interview. The probability that they will stand in alternate positions is a) 1/34 b) 1/35 c) 1/17 d) 1/68
198
## Permutations, Combinations & Probability Practice Set-1 (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) c a a b c d a c d
199
## Permutations, Combinations & Probability Practice Set-2 (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
c b d c b d b b d a
200
## Chapter: Data Interpretation
Introduction: Analyzing data is the major part of our daily routine. Financial data as in P & L sheet, marketing and sales data, data on productivity, data on performance appraisal, data on each and every thing that we can imagine. In some companies entire department generates and manages every conceivable data that we can imagine. To collect information from all the amount of data it needs to be presented in a lucid and concise manner. Therefore we use data representation as it immediately provides the overall scenario and it is also sufficient to compute any detailed information. Strategies: While studying the DI section one should follow few strategies given below: 1. Solve the graph with which you are most comfortable for example some are more comfortable with line graph while the other may be with pie chart. 2. The problems with numbers with 2 or 3 digits like 82, 114 etc are easier to solve then the problem with number of 4 or 5 digits like 3457, 52468. 3. In pie charts if sectors are 15% 20% etc then it will be easier calculation then the sectors like 17.5%, 23.6 % etc. 4. We must also look at the number in answer option, if the answers are 12.5 5 6.33% etc then it will be easier to arrive at. 5. Wider are the choices in answer, easier will be the elimination process and lesser will be the calculation. 6. The answer choice Cannot be determined makes the question much easier as it becomes data sufficiency question and if the question can be answered then only four options are left.
201
## IBPS PO EXAM 2013 : Quantitative Aptitude
7. Pick the alternative which is the middle one and check if your answer is less than, equal to or greater than this value. In this way only one iteration will give you the correct answer. 8. Te answer choice None of these on the other hand make it more difficult as we have to calculate the question to the exact value. 9. It is better to solve a line chart with two lines and 5 points than to solve a table with 5 rows and 6 columns. But this may not be the case always looking at the other factors. Growth rate and Growth: Growth and Growth rate are two different cases. Growth refers to just increase in the underlying value, while the Growth rate refers to the percentage increase. For example following table shows the sales and profit of a company A in Rs. Lakhs
## 1999 Sales Profit Here, 240 50
2000 290 55
2001 320 70
2002 350 50
2003 380 40
Growth during the period from 2000 to 2002 is simple the difference of sales = 350-290 = 60 But the Growth rate during this period is the percentage of growth = Profit Percentages: The profit percentage is not but it is . For example in above .
202
## as cost price of this year is 190
Also the percentage change in profit percentage is percent of profit percent of two years with base as previous year profit percent. For example the percentage change in profit percentage in the year 2002 over that in year 2001 is:
= =
But in line graph growth rate is related to slope of line. There are few points to understand: 1. The slope of each segment is same across all years of company B but still the growth rate is not the same in all years. The same slope simply means the sales grow by a constant amount each year. But the growth rate depends on the base value so in company B its decreasing from 2000 to 2004. 2. The line representing sales of A in 2002-03 is steeper but 2003 is not the year with highest growth rate of sales of A . The highest growth rate of sale of A is in year 2001.
203
## Chapter: Data Interpretation-Table Chart
In studying problems on statistics, the data collected by the investigator are arranged in systematic form, called the tabular form. In order to avoid some heads again and again, we make tables, consisting of horizontal lines called rows and vertical lines called columns with distinctive heads, known as captions. Units of measurements are given along with the captions. Example: The table given below shows the population, literates and illiterates (in thousands) and the percentage of literacy in 3 states, in a year: State Population Literates Illiterates Percentage of literacy .. 16.1
49342 . 60314
6421 4068
.. 16790
After reading the table, mark a tick () against the correct answer in each question given below and hence complete the table. 1) Percentage of literacy in Madras is (a) 14.9% (b) 13.01% (c) 12.61% (d) 15.04% 2) Percentage of literacy in Bombay is (a) 19.5% (b) 16.7% (c) 18.3% (d) 14.6%
204
## IBPS PO EXAM 2013 : Quantitative Aptitude
3) Number of literates in Bengal (in thousands) is: (a) 50599 (b) 9715 (c) 76865 (d) 9475 Solution: 1) (b) percentage of literacy in Madras = (6421/49342)100% = 13.01% 2) (a) Population of Bombay = (4068+16790) thousands = 20858 thousands. Therefore, percentage of literacy in Bombay = (4068/20858)100% = 19.5% 3) (b) Number of literates in Bengal =(16.1/100)60314 = 9715 thousands
205
## Solved Examples (Data Interpretation-Table Chart)
Direction (1-5): Study the table carefully to answer the questions that follow :- (IBPS PO Exam 2011) Number of people visiting six different Super-markets and the percentage of Men, Women and Children visiting those Super-markets Names of the Total Number of Supermarkets Percentage of
People
Men
Women
Children
34560
35
55
10
65900
37
43
20
45640
35
45
20
55500
41
26
33
42350
06
70
24
59650
24
62
14
206
## IBPS PO EXAM 2013 : Quantitative Aptitude
1. The Number of men visiting Super-market D forms approximately what percent of the total number of people visiting all the Super-markets together? (1) 11 (2) 5.5 (3) 13 (4) 9 (5) 7.5 Solution: Number of men visiting supermarket D = 41% of 55500 = = 4155500 = 22755
Total number of people visiting all the supermarkets together = 34560 + 65900 + 45640 + 55500 + 42350 + 59650 = 303600 required probability = Ans: (5) 2. The Number of children visiting Super-market C forms what percent of number of children visiting Super- market F? (rounded off to two digits after decimal) (1) 91.49 (2) 49.85 (3) 121.71 (4) 109.30 (5) None of these Solution: Number of children visiting supermarkets C = 20% of 45640 = 20 x 45640 = = 9128 100 = 7.5% (Aprox)
Number of children visiting supermarket F = 14% of 59650 = 207 = 8351 IBPS PO EXAM 2013 : Quantitative Aptitude
## required percentage = Ans: (4)
x 100 = 109.30%
3. What-is the total number of children visiting Super-markets B and D together? (1) 18515 (2) 28479 (3) 31495 (4) 22308 (5) None of these Solution: Total number of children visiting supermarket B and D together = 20% of 65900 + 33% of 55500 = +
= 13180 + 18315 = 31495 Ans: (3) 4. What is the average number of women visiting all the Super-markets together? (1) 24823.5 (2) 22388.5 (3) 26432.5 (4) 20988.5 (5) None of these Solution: Total number of women = 55% of 34560 + 43% of 65900 + 45% of 45640 + 26% of 55500+ 70% of 42350 + 62% of 59650 = 19008 +'28337 + 20538 + 14430 + 29645 + 36983 = 148941 required average = Ans: (1) = 24823.5
208
## IBPS PO EXAM 2013 : Quantitative Aptitude
5. What is the ratio of number of women visiting Super-markets A to that visiting Supermarket C? (1) 35: 37 (2) 245: 316 (3) 352: 377 (4) 1041: 1156 (5) None of these Solution: Required ratio = 19008: 20538 = 1056: 1141 Ans : (3) (Directions Q.6-10): Study the table carefully to answer the questions that follow :- (IBPS PO Exam 2011)
Percentage SUBJECTS (Maximum Marks) of Marks Obtained by Students Strategic Brand Compensation Consumer Service Training & Different Students in Management Management Management Behaviour Marketing Development Different (150) (100) (150) (125) (75) (50) Subject of MBA Anushka 66 75 88 56 56 90 ,
## Archit Arpan Garvita Gunit Pranita
82 76 90 64 48
76 66 88 70 56
84 78 96 68 50
96 -88 76 7.2 64
92 72 84 68 64 ,
88 70 86 74 58 1;;' ~ ,'_ ,-
6. How many marks did Anushka get in all the Subjects together? (1) 369 (2) 463 (3) 558 (4) 496 (5) None of these Solution: Total marks of Anuska = 209 + 75 + + + + 45
## IBPS PO EXAM 2013 : Quantitative Aptitude
= = 99 + 75 + 132 + 70 + 42 + 45 = 463 Ans: (2) 7. The Marks obtained by Garvita in Brand Management are what percent of marks obtained by Archit in the same Subject? (rounded off to two digits after decimal) (1) 86.36 (2) 101.71 (3) 115.79 (4) 133.33 (5) None of these Solution: Marks obtained by Garvita in Brand Management = 88% of 100 = 88 Marks obtained by Archita in Brand Management = 76% of 100 = 76 required percentage = Ans: (3) 8. What is the average marks obtained by all students together in Compensation Management? (1) 116 (2) 120 (3) 123 (4) 131 (5) None of these Solution: Average marks obtained by all students together in Compensation Management x 100 115.79%
210
## IBPS PO EXAM 2013 : Quantitative Aptitude
150=116
Ans: (1) 9. Who has scored the highest total marks in all the subjects together? (1) Archit (2) Gunit (3) Pranita (4) Garvita (5) Arpan Solution: Total obtained in all the subjects together by Arapn: 76% of 150 + 66% of 100 + 78% of 150 + 88% of 125 + 72% of 75 + 70% of 50 = + + + + +
= 114 + 66 + 117 + 110 + 54 + 35 = 496 Archit: 82% of 150 + 76% of 100 + 84% of 150 + 96% of 125 + 92% of 75 + 88% of 50 = + + + + +
= 123 + 76 + 126 + 120 + 69 + 44 = 558 Garvita: 90% of 150 + 88% of 100 + 96% of 150 + 76% of 125 + 84% of 75 + 86% of 50 . = 135 + 88 + 144 + 95 + 63 + 43 = 568 = + + + + +
= 135 + 88 + 144 + 95 + 63 + 43 = 568 Gunit: 64% of 150 + 70% of 100 + 68% of 150 + 72% of 125 + 68% of 75 + 74% of 50 = + + + + +
= 96 + 70 + 102 + 90 + 51 + 37 = 446 Pranita: 48% of 150 + 56% of 100 + 50% of 150 + 64% of 125 + 64% of 75 + 58% of 50 211 IBPS PO EXAM 2013 : Quantitative Aptitude
= 72 + 56 + 75 + 80 + 48 + 29 = 360 Clearly, Garvita scored the highest total marks in all the subjects together. Ans: (4) 10. How many Students have scored the highest marks in more than one Subject? (1) three (2) two (3) one (4) none (5) Now of these Solution: Archit (consumer behaviour and service marketing) and Garvita (strategic management, brand management and compensation management). Ans: (2)
212
## Practice Set (Data Interpretation-Table Chart)
Directions (Q. 1-5): Study the table carefully to answer the questions that follow Number of cars (in thousand) of two models (Basic and Premium) produced by five different companies in five different years (IBPS RRB Grade Officer Exam 2012) Compan y Year 2006 2007 2008 2009 2010 Basi c Premiu m Basi c Premiu m Basi c Premiu m Basi c Premiu m Basi c Premiu m A B C D E
## 5.1 5.5 11.5 12.8 12.2
1. The number of cars of premium model produced by Company D in the year 2009 was approximately what per cent of the total number of cars (both models) produced by Company C in the year 2007? (1) (2) (3) (4) (5) 70 51 56 61 66 213 IBPS PO EXAM 2013 : Quantitative Aptitude
2. What was the approximate percentage decrease in the number of cars of basic model produced by Company B in the year 2009 as compared to the previous year? (1) 15 (2) 20 (3) 10 (4) 80 (5) 85 3. What was the average number of cars of premium model produced by Company A over all the years together? (1) (2) (3) (4) (5) 9000 8000 6000 48000 None of these
4. In which year was the difference between the basic model and the premium model of cars produced by Company E the second highest? (1) (2) (3) (4) (5) 2010 2006 2007 2008 2009
5. In which company did the production of cars of premium model consistently increase from the year 2006 to the year 2010? (1) (2) (3) (4) (5) Both C and E Both C and d C only D only E only
214
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 6-10): Study the table carefully to answer the questions that follow. Number of animals in grasslands of four different countries in five different years (RBI GradeB Officers Exam 2011) Country Year South Africa Tiger 1990 1995 2000 2005 2010 145 134 120 110 160 Lion 156 165 135 184 224 Bear 250 354 324 285 264 Tiger 320 445 583 466 411 China Lion 346 256 325 475 535 Bear 436 542 454 322 534 Sri Lanka Tiger 280 354 433 343 535 Lion 468 354 345 324 532 Bear 255 343 545 546 453 Tiger 423 368 354 562 349 England Lion 342 136 267 235 345 Bear 234 345 456 567 324
6. What is the average of the number of tigers in the grassland of Sri Lanka over all the years together? (1) (2) (3) (4) (5) 386 389 369 276 None of these
7. What is the difference between the total number of lions and bears in the grassland of England in the year 2005 and the number of tigers in the grassland of South Africa in the year 1995? (1) (2) (3) (4) (5) 597 558 677 668 None of these 215 IBPS PO EXAM 2013 : Quantitative Aptitude
8. The total number of animals together in the grassland of China in the year 1990 is approximately what per cent of the total number of bears in the grassland of Sri Lanka over all the years together? (1) (2) (3) (4) (5) 44% 56% 41% 47% 51%
9. If 35 per cent of the total number of animals in the grassland of China in the year 2010 died due to an epidemic, how many animals remained in the grassland of China in the year2010? (1) (2) (3) (4) (5) 976 952 986 962 None of these
10. What is three-fourths of the total number of lions in the grasslands of all the four countries in the year 2000? (1) (2) (3) (4) (5) 848 868 804 824 None of these
216
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 11-5): Study the table carefully to answer the questions that follow: Number of girls and boys (in hundreds) in six different years in five different schools School Years 2005 2006 2007 2008 2009 2010 A B C D E
Boys Girls Boys Girls Boys Girls Boys Girls Boys Girls 3.3 6.6 9.3 5.4 3.6 4.2 6.9 9.6 5.2 4.9 4.7 6.3 7.5 9.8 3.1 2.2 4.2 5.4 5.9 4.4 5.5 6.9 5.8 6.6 8.7 4.5 3.3 4.9 5.2 2.4 4.4 6.4 5.3 1.4 6.5 2.3 5.5 3.3 2.7 5.4 5.4 6.6 3.6 2.4 5.7 6.5
## 11.7 4.2 12.2 9.4 10.8 12.7
11. What is the approximate percentage decrease in the number of boys in School D in the year 2008 as compared to that in the previous year? (1) (2) (3) (4) (5) 17 12 9 5 23
12. The number of girls in School B in the year 2009 is approximately what per cent of the total number of students (both boys and girls) in School E in the year 2006? (1) (2) (3) (4) (5) 46 52 70 58 65
13. What is the average number of girls in School A in all the years taken together? (1) 760 (2) 800 (3) 860 217 IBPS PO EXAM 2013 : Quantitative Aptitude
(4) 600 (5) None of these 14. What is the ratio of the number of boys in School C in the year 2009 to the number of girls in School A in the year 2009? (1) (2) (3) (4) (5) 29 : 41 36 : 11 29 : 43 36: 13 None of these
15. In which year is the total number of students (both girls and boys together) the third highest in School E? (1) (2) (3) (4) (5) 2006 2007 2008 2005 2010
218
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions-(Q. 16-20) Study the table carefully to answer the questions that follow: Number of Athletes (in hundreds) who participated in a Sports Event from Five Different Countries over the years (Allahabad Bank Probationary Officers Exam 2011) Countries Year 2005 2006 2007 2008 2009 2010 A B C D E
Male Fem Male Fem Male Fem Male Fem Male Fem 4.4 66 46 9.6 11.8 82 3.3 42 18 49 64 52 6.3 84 74 11.4 106 64 4.2 62 48 84 52 72 4.5 69 48 66 79 108 3.1 33 28 42 63 69 5.6 84 93 126 14.4 156 4.1 63 73 94 4.7 78 87 89 2.1 52 65 58 92 98
## 102 118 121 136
16. In which of the following years was the total number of participants (athletes) second highest from Country C? (1) (2) (3) (4) (5) 2005 2006 2007 2008 None of these
17. What was the average number of female athletes who participated from Country B over all the years together? (1) (2) (3) (4) (5) 1200 400 600 1800 3600
219
## IBPS PO EXAM 2013 : Quantitative Aptitude
18. What was the approximate percentage decrease in the number of male athletes who participated from Country C in the year 2007 as compared to the previous year? (1) (2) (3) (4) (5) 21 30 35 39 25
19. The Number of female athletes who participated from Country E in the year 2009 was approximately what percentage of the 42 total number of athletes who participated from Country-B in the year 2008? (1) (2) (3) (4) (5) 40 46 50 56 60
20. In which of the following country is the difference between the number of male and female participants second highest in the year 2006? (1) (2) (3) (4) (5) A B C D E
220
## Data Interpretation-Table Chart Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 5 2 5 5 3 2 4 5 4 3 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 1 5 3 3 5 5 3 2 2 5
221
## Chapter: Data Interpretation-Line Graphs
Line graphs of a frequency distribution is obtained from the histogram of the frequency distribution by joining the mid points of respective tops of the rectangles in a histogram. To complete the line graphs, the mid-points at each end are joined to the immediately lower or higher mid-points (as the case may be) at zero frequency. Example: Study the following graph and answer the following questions:
1.) The total expenditure of which of the following pairs of years was equal to the income in 1992? (a) 1987 and 1988 (b) 1987 and 1989 (c) 1988 and 1989 (d) 1988 and 1990 (e) none of these 2.) What was the percentage decrease in expenditure from 1988 and 1989? 222 IBPS PO EXAM 2013 : Quantitative Aptitude
(a) 80 (b) 50 (c) 40 (d) 10 (e) none of these 3.) In how many of the given years was the expenditure more than the average expenditure of the given years? (a) 4 (b) 3 (c) 1 (d) 5 (e) none of these 4.) In which of the following years was the percentage of expenditure to income, the highest? (a) 1987 (b) 1988 (c) 1989 (d) 1991 (e) none of these 5.) What was the approximate percentage increase in income from 1991 to 1992? (a) 35 (b) 40 (c) 20 (d) 15 (e) 25 Solution: 1.) (c) : income in 1992= 475 crores Total expenditure in 1988 and 1989 = Rs.( 250+225) crores =Rs. 475 crores. 2.) (d) : expenditure in 1988 = Rs. 250 crores Expenditure in 1989 = Rs. 225 crores.
Decrease % =
= 10%
3.) (b): average expenditure =Rs. =Rs. 287.5 The expenditure is greater than the average expenditure during the years 1987, 1990 and 1992. Required no. of years =3 4.) (e): the required percentage : In 1987 is (300100/450)% = 66.66% In 1988 is (250100/400)% = 62.5% In 1989 is (225100/350)% = 64.29% 223 IBPS PO EXAM 2013 : Quantitative Aptitude
In 1990 is (375100/425)% = 88.24% In 1991 is (175100/375)% = 46.6% In 1992 is (400 100/475)% = 84.21% Clearly the percentage is highest in 1990. 5.) (e): income in 1991 = 375 crores Income in 1992 = 475 crores Therefore increase % = (100100/375)% = 26.6% = 25% nearly
224
## Solved Examples (Data Interpretation-Line Graphs)
Directions-(Q.1-5): Study the following graph carefully to answer the questions that follow: (Allahabad Bank Probationary Officers Exam 2011)
1. In which state was the total number of trees planted by NGO A and NGO B together second lowest? (1) (2) (3) (4) (5) Bihar Punjab Haryana Assam Tamil Nadu
Solution: Number of tree planted by NGO-A and NGO-B together in Bihar: 100 + 60 = 160 Punjab: 120 + 80 = 200 Haryana: 80 + 140 = 220 Assam: 150 + 160 = 310 225 IBPS PO EXAM 2013 : Quantitative Aptitude
Tamil Nadu: 140 + 180 = 320 Ans: (2) 2. What was the difference between the trees planted by NGO A in Haryana and the number of trees planted by NGO C in Tamil Nadu? (1) (2) (3) (4) (5) 90 60 120 100 None of these
Solution: Required difference = 160 80 = 80 Ans: (5) 3. What was the average number of trees planted in Haryana by all the NGOs together? (1) (2) (3) (4) (5) 420 140 120 390 None of these = 129
## Solution: Required average = Ans: (5)
4. The total number of trees planted by NGO A and NGO B together in Bihar was approximately what per cent of the total number of trees planted by NGO-B and NGO-C together in Punjab? (1) (2) (3) (4) (5) 85 90 105 110 95 100 = 100 95%
226
## IBPS PO EXAM 2013 : Quantitative Aptitude
5. What was the ratio of the number of trees planted by NGO B in Tamil Nadu, number of trees planted by NGO C in Assam and the number of trees planted by NGO A in Assam? (1) (2) (3) (4) (5) 5: 3 : 6 5: 6 : 3 6: 4 : 5 6: 5 : 3 None of these
Solution: Required ratio: 180: 120 + 150 = 6: 4: 5 Ans: (3) Direction (Q. 6 10): Study the graph carefully and answer the questions that follow: Per cent profit made by two companies over the years
6. If in the year 2006 the expenditures incurred by company P and Q were same, what was the ratio of the income of company Q to that of company P in that year? (a) 26 : 27 (b) 27 : 26 (c) 24 : 25 (d) 25 : 24 (e) None of these Answer: (c) 227 IBPS PO EXAM 2013 : Quantitative Aptitude
7. If the amount of profit earned by company Q in the year 2007 was Rs. 2.4 lakhs, what was its expenditure in that year? (a) Rs. 13 lakhs (b) Rs. 15 lakhs (c) Rs. 24 lakhs (d) Rs. 16 lakhs (e) Rs. 20 lakhs Answer: (d) 8. What is the average per cent profit earned by company P over all the years together? (a) 30 (b) 25 (c) 40 (d) 33 (e) None of these Answer: (b) 9. If in the year 2009, the incomes of both the companies P and Q were same, what was the ratio of the expenditure of company P to the expenditure of company Q in the same year? (a) 26 : 23 (b) 23 : 26 (c) 24 : 25 (d) 25 : 24 (e) None of these Answer: (a) 10. What is the ratio of the amount of profit earned by company A to that by company B in the year 2010? (a) 27 : 24 (b) 24 : 27 (c) 23 : 24 (d) 24 : 23 (e) None of these Cannot be determined Answer: (e)
228
## Practice Set (Data Interpretation-Line Graphs)
Directions (1-5): Study the following graph carefully and answer the questions given below: (Allahabad Bank PO Exam: 2010) Profit earned by Three Companies over the years (Rs. in crores)
1. What was the average profit earned by all the three companies in the year 2008? (a) Rs. 300 crore (b) Rs. 400 crore (c) Rs. 350 crore (d) Rs. 520 crore (e) None of these 2. In which of the following years was the difference between the profits earned by company B and company A the minimum? (a) 2003 (b) 2004 229 IBPS PO EXAM 2013 : Quantitative Aptitude
(c) 2005 (d) 2008 (e) None of these 3. In which of the following years was the total profit earned by all three comapnies together the highest? (a) 2004 (b) 2007 (c) 2008 (d) 2009 (e) None of these 4. What was the approximate percentage increase in the profit earned by Company A from 2006 to 2007? (a) 36 (b) 24 (c) 40 (d) 20 (e) 54 5. What was the difference between the profit earned by company A in 2004 and the profit earned by company C in 2009? (a) Rs. 50 crore (b) Rs. 1 crore (c) Rs. 100 crore (d) Rs. 200 crore (e) None of these
230
## IBPS PO EXAM 2013 : Quantitative Aptitude
Direction (Q. 6 10): Study the given graph carefully and answer the questions that follow: The line diagram shows the cost of production and profit of six companies for the year 201112. (The figures are in 'Lakhs'). Revenue = Cost of Production + Profit.
6. The ratio of profits of company B and D to the profits of A and E is: (a) (b) (c) (d) (e) 2:3 10 : 9 3:2 10 : 7 None of these
7. The profit of company C is what percentage of the revenue of company F? (a) (b) (c) (d) (e) 20% 25% 30% 35% None of these
8. The revenue of company C is how many times of company E's profit? (a) (b) (c) (d) (e) 5.5 5.25 5.75 5 None of these
231
## IBPS PO EXAM 2013 : Quantitative Aptitude
9. Which company has the maximum percentage of profit? (a) (b) (c) (d) (e) C D E F None of these
10. What is the average profit of the last five companies (B, C, D, E and F)? (a) (b) (c) (d) (e) Rs. 500 Rs. 5,000 Rs. 50,000 Rs. 4,66,667 None of these
232
## Data Interpretation-Line Graphs Practice Set (Answers)
1. 2. 3. 4. 5. 6. 7. 8. 9. (b) (e) (d) (a) (c) (d) (a) (c) (b)
10. (e)
233
## Chapter: Data Interpretation-Bar Graphs
In a bar diagram, information is presented by means of rectangles, whose lengths indicate the quantity of the variable which the bar is representing. The following points are important: 1) All bars are in the form of rectangles and the width of the bars is uniform throughout the diagram. 2) The height of each bar is proportional to the frequency of the variable. 3) The gap between various bars is uniform. 4) The base line of all the bars is the same. 5) The bars can be either horizontal or vertical depending on the space available. Example: The expenditure of a company under different heads (in thousands of rupees) is given below: Head Expenditure(in thousands of rupees) 400 100 150 200 300
234
## IBPS PO EXAM 2013 : Quantitative Aptitude
Example: The following bar diagram represents the percentages of total expenditure incurred by a state during the years 1981- 90 for different items. In each bar the blue portion stands for the expenditure during the first five years and the red portion stands for the next five years. Study the graph and answer questions 1-5.
1) Which of the items listed below accounts for the maximum expenditure during the year 1981 to 1985? 235 IBPS PO EXAM 2013 : Quantitative Aptitude
(a) Communication (b) education (c) health (d) housing 2) Which of the items listed below accounts for the maximum expenditure during 1986 to 1990? (a) Agriculture (b) communication (c) education (d) health 3) The amount of expenditure on Agriculture is approximately what proportion of that on industry during the year 1986-90? (a) 1/5 (b) (c) 1/3 (d) data inadequate 4) If the total expenditure on housing is Rs. 610 crores during 1981-85, the total expenditure on industry during the same period would (approximately) (a) Rs 2440 crores (b) Rs 1220 crores (c) Rs 4620 crores (d) none of these 5) Out of every 10,000 rupees spent during 1981-90 approximately, how much was spent during the years 1981-85 on housing? (a) Rs 1400 (b) Rs 700 (c) Rs 1000 (d) Rs 2800 Answer: 1) (d) out of the items listed in the question, clearly maximum expenditure during 1981-85 is on housing. 2) (a) out of the items listed in the question, clearly the maximum expenditure during 1986-90 is on agriculture. 3) (a) expenditure on agriculture during 1986-90 a. = (15-10)% of total expenditure = 5x/100 = x/20 b. Expenditure on industry during 1986-90 c. = (52.5-27.5)% of total expenditure = 25x/100 =x/4. d. Required ratio = x/20 : x/4 = 1:5 4) (d) expenditure on housing during 1981-85 = 10% of total expenditure. a. Let the total expenditure be Rs. x. b. Then , 105 of x = 610 crores or 10x/100 =610 crores c. x=6100 crores d. total expenditure on industry during 1981-85 = 25% of 6100 crores e. = Rs 1525 crores. 5) (c) 20% of total expenditure during 1981-90 was spent on housing. a. Expenditure on housing during 1981-90 for a total expenditure of Rs 10000 = (2010000/100) = Rs 2000 b. Ratio of expenditure on housing during 1981-85 and that during 1986-90= 10%/(20-10)% = 1/1 c. Expenditure on housing during 1981-85 = Rs 1000
236
## Solved Examples (Data Interpretation-Bar Graphs)
Directions (Q. 1-5): Study the following graph and answer the questions given below:
1. Out of the total number of students who opted for the given three subjects, in the year 2009, 38% were girls. How many boys opted for Mathematics in the same year? (1) 1322 (2) 1332 (3) 1312 (4) Cannot be determined (5) None of these Solution: Number of students who opted for all three subjects in 2009 = (20 + 20 + 5) thousand = 45000 Number of boys = = 27900
Since, we do not know the number of girls in Mathematics, number of boys opted for Mathematics cannot be determined. Ans : (4)
237
## IBPS PO EXAM 2013 : Quantitative Aptitude
2. If the total number of students in the University in the year 2007 was 455030, then, the total number of students who opted for the given three subjects were approximately what percent of the total students? (1) 19 (2) 9 (3) 12 (4) 5 (5) 23 Solution: Required percentage = = Ans: (2) 3. What is the total number of students who opted for Hindi and who opted for Mathematics in the years 2006, 2007 and 2009 together? (1) 97000 (2) 93000 (3) 85000 (4) 96000 (5) None of these Solution: Required number of students = (5 + 35 + 15 + 15 + 20 + 5) x 1000 = 95 x 1000 = 95000 Ans: (5) 4. The total number of students who opted for Mathematics in the years 2005 and 2008 together are approximately what percent of the total number of students who opted for all three subjects in the same years? (1) 38 (2) 28 (3) 42 (4) 32 (5) 48 100 9 100
238
## Solution: Required percentage = 100
= Ans: (4)
100 =
100
32
5. What is the respective ratio between the number of students who opted for English in the years 2006 and 2008 together and the number of students who opted for Hindi in the year 2005 and 2009 together? (1) 11: 5 (2) 12: 7 (3) 11: 7 (4) 12:5 (5) None of these Solution: Required ratio = (25 + 30): (5+20) = 55:25 = 11:15 Ans: (1) Directions (Q. 6-10): Study the following graph carefully to answer the questions that follow: Monthly income (Rs in thousand) of three different persons in six different years (IBPS RRB Grade A Officers Exam 2012)
239
## IBPS PO EXAM 2013 : Quantitative Aptitude
6. What was the difference between the total monthly salary of Arun in all the years together and Suman's monthly income in the year 2007? (1) (2) (3) (4) (5) Rs. 1.24 Iakh Rs. 1.14 Iakh Rs. 11.4 lakh Rs. 12.4 lakh None of these
## Solution. Arun monthly income in all year together
Suman's monthly income in the year 2007 = 15 thousand Difference = 129 15 = 114 = 114 1000 = 114000 lakh Ans: 2 7. What is the ratio of Arun's monthly income in the year 2006, Suman's monthly income in the year 2007 and Jyoti's monthly income in the year 2005? (1) (2) (3) (4) (5) 6:3:5 6:4:5 5:6:4 5:4:7 None of these
240
## IBPS PO EXAM 2013 : Quantitative Aptitude
8. In which year was the difference between Jyoti's and Arun's monthly income the second highest? (1) (2) (3) (4) (5) 2005 2006 2007 2009 2010
Solution. Difference in 2005 14 9 = 5 2006 18 10 = 8 2007 23 18 = 5 2008 27 21 = 6 2009 27 26 = 1 2010 Ans: 2 9. The monthly income of Suman in the year 2009 was approximately what percentage of the monthly income of Jyoti in the year 2010? (1) (2) (3) (4) (5) 72 89 83 67 95 35 26 = 9
Solution. Monthly income of Suman in 2009 = 29000 Monthly income of Jyoti in 2010 = 35000
Ans: 3
241
## IBPS PO EXAM 2013 : Quantitative Aptitude
10. What was the percentage increase in the monthly income of Jyoti in the year 2008 as compared to the previous year? (1) (2) (3) (4) (5) 50 150 160 60 None of these
Solution. Ans: 1
242
## Practice Set (Data Interpretation-Bar Graphs)
Directions (Q. 1-5): Study the following graph carefully to answer the questions that follow: (Corporation Bank PO 2011)
1. What is the percentage increase in the number of runs scored by Team B in Match 4 as compared to that in the previous match (Match 3)? (1) (2) (3) (4) (5) 40 30 20 25 None of these 243 IBPS PO EXAM 2013 : Quantitative Aptitude
2. What is the ratio of the number of runs scored by Team A in Match 2 to the number of runs scored by Team C in Match 6? (1) (2) (3) (4) (5) 5:4 2:5 2:3 3:4 None of these
3. What is the average number of runs scored by Team B in all the matches together? (1) (2) (3) (4) (5) 250 275 200 300 225
4. The number of runs scored by all the teams together in Match 3 is approximately what percentage of the total runs scored by Team C in all the matches together? (1) (2) (3) (4) (5) 37 57 52 47 42
5. In which match is the total runs scored by all the teams together the second highest? (1) (2) (3) (4) (5) Match 2 only Match 6 only Match 4 only Both Match 2 and Match 6 Both Match 2 and Match 4
244
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 6-10): Study the given graph carefully to answer the questions that follow:
6. What is the average number of people using mobile service M for all the years together? 1) 16
2) 14444
3) 16666
4) 14 5) None of these 7. The total number of people using all the three mobile services in the year 2007 is what per cent of the total number of people using all the three mobile services in the year 2008? (rounded off to two digits after decimal) 1) 89.72 2) 93.46 3) 88.18 245 IBPS PO EXAM 2013 : Quantitative Aptitude
4) 91.67 5) None of these 8. The number of people using mobile service N in the year 2006 forms a proximately what per cent of the total number of people using all the three mobile services in that year? 1) 2) 3) 4) 5) 18 26 11 23 29
9. What is the ratio of the number of people using mobile service L in the year 2005 to that of those using the same service in the year 2004? 1) 2) 3) 4) 5) 8: 7 3:2 19: 13 15: 11 None of these
10. What is the total number of people using mobile service M in the years 2008 and 2009 together? 1) 2) 3) 4) 5) 35,000 30,000 45,000 25,000 None of these
246
## Data Interpretation-Bar Graphs Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
3 3 1 5 4 5 4 1 2 3
247
## Chapter: Data Interpretation-Pie Diagram
In a pie chart, the values of different components of a frequency distribution are represented by the sectors of a circle. These sectors are so constructed that the area of each sector is proportional to the corresponding value of the component. Since the sum of all the central angles is 360 degrees, we have Central angle of a component = degrees
Example: The following pie diagram shows the expenditure incurred on the preparation of a book by a publisher, under various heads. A: Paper 20, B: Printing 25%, C: Binding, Canvassing, Designing etc 30% D: Miscellaneous 10% E: Royalty 15% Study the diagram carefully and answer the questions 1-5:
1) What is the angle of pie diagram showing the expenditure incurred on paying the royalty? (a) 15 degrees (b) 24 degrees (c) 48 degrees (d) 54 degrees
248
## IBPS PO EXAM 2013 : Quantitative Aptitude
2) The marked price of a book is 20% more than the C.P. If the marked price of the book be Rs 30, what is the cost of paper used in a single copy of the book? (a) Rs. 6 (b) Rs. 5 (c) Rs 4.50 (d) Rs 6.50 3) Which two expenditures together will form an angle of 108 degrees at the centre of the pie diagram: (a) A and E (b) B and E (c) A and D (d) D and E 4) If the difference between two expenditures be represented by 18 degrees in the piediagram, these expenditures are : (a) B and E (b) A and C (c) B and D (d) none of these
1) (d) Angle representing royalty D = (15360/100)degrees = 54 degrees 2) (b) C.P. of a book = Rs ( 10030/120) = Rs 25 Cost of paper = Rs (2025/100) = Rs 5 3) (c) angle A = (20360/100)degrees =72 degrees Angle B = (25360/100) degrees = 90 degrees Angle C = (30360/100)degrees = 108 degrees Angle D = (10360/100) degrees = 36 degrees Angle E = (15360/100) degrees = 54 degrees Thus, A and D together will form an angle of 108 degrees. 4) (d) These expenditures are A and B; Band C; D and E; and A and E.
249
## Solved Examples (Data Interpretation-Pie Diagram)
Direction (Q. 1-5): Study the following pie-chart and answer the questions given below: Preferences of students among six beverages in terms of degree of angle in the pie-chart Total No. of students = 6800
1. What is the difference between the total number of students who prefer Beverage A and C together and the total number of students who prefer beverage D and F together? (1) 959 (2) 955 (3) 952 (4) 954 (5) None of these Solution: Difference of corresponding angles = (122.4 + 21.6)0 - (79.2 + 14.4)0 = 50.40 required difference = Ans: 3 x 6800 = 952
250
## IBPS PO EXAM 2013 : Quantitative Aptitude
2. What is the ratio of the number of students who prefer beverage F and the number of students who prefer beverage A? (1) 3: 11 (2) 3: 13 (3) 6: 11 (4) 5: 11 (5) None of these Solution: Required Ratio = 21.6: 79.2 = 3: 11 Ans: 1 3. The number of student who prefer beverage E and F together is what per cent of the total of student? (1) 18 (2) 14 (3) 26 (4) 24 (5) None of these Solution: Required percentage = ( Ans: 4 4. The number of students who prefer beverage C are approximately what percent of the number of students who prefer Beverage D? (1) 7 (2) 12 (3) 18 (4) 22 (5) 29 Solution: Required percentage = Ans: 2 5. How many students prefer beverage B and beverage E together? (1) 2312 (2) 2313 (3) 2315 (4) 2318 251 IBPS PO EXAM 2013 : Quantitative Aptitude 100 = 11.76 12% ) 100 =24%
(5) None of these Solution: Number of students who prefer beverages B and E together = Ans: 1 Directions (Q. 6-10): Study the following pie-chart and answer the following questions. (IBPS RRB Group A Officers Exam 2012) Percentagewise distribution of teachers in six different universities. Total number of teachers = 6400 68000 = = 2312
6. The number of teachers in University B is approximately what per cent of the total number of teachers in University D and University E together? (1) (2) (3) (4) (5) 55 59 49 45 65
252
## Number of teachers in University D Number of teachers in University E Required percentage
Ans: 3 7. If twenty five per cent of the teachers in University C are females, what is the number of male teachers in University C? (1) (2) (3) (4) (5) 922 911 924 912 None of these
## Solution: Number of teachers in University C Number of female teachers in University C
Number of male teachers in University C = 1216 304 = 912 Ans: 4 8. The difference between the total number of teachers in University A, University B and University C together and the total number of teachers in University D, University E and University F together is exactly equal to the number of teachers of which University? (1) (2) (3) (4) University A University B University C University D 253 IBPS PO EXAM 2013 : Quantitative Aptitude
(5) University F Solution: Number of teachers in University A Number of teachers in University B Number of teachers in University C Number of teachers in University D Number of teachers in University E Number of teachers in University F Difference = 3392 3008 = 384 Quicker method: Difference = (D + E + F)% (A + B + C)% = (53 47) = 6% 6% of 6400 = 384 Hence, University of D is equal to 6%. Ans: 4 9. If one-thirty sixth of the teachers from University F are professors and the salary of each professor is Rs. 96000, what will be the total salary of all the professors together from University F? (1) (2) (3) (4) (5) Rs. 307.2 lakh Rs. 32.64 lakh Rs. 3.072 lakh Rs. 3.264 lakh None of these
Solution: Number of teachers in University F 254 IBPS PO EXAM 2013 : Quantitative Aptitude
Number of professors in University F Total Salary of professors in University F = 32 96000 = 30.72 lakh Ans: 5 10. What is the average number of teachers in University A, University C, University D and University F together? (1) (2) (3) (4) (5) 854 3546 3456 874 None of these
255
## Practice Set (Data Interpretation-Pie Diagram)
Directions (1-5): Study the following pie-charts carefully and answer the questions given below: Discipline-wise Breakup of the Number of candidates appeared in Interview and Disciplinewise Break up of the Number of candidates selected by an organisation Discipline-wise Breakup of Number of candidates appeared in Interview Total Number of Candidates Appeared In the Interview = 25780 Percentage Distribution
Discipline-wise Break-up of Number of candidates selected after Interview by the organization Total Number of Candidates selected After Interview = 7390 Percentage Distribution
256
## IBPS PO EXAM 2013 : Quantitative Aptitude
1. What was the ratio of the number of candidates appeared . in interview from other disciplines and the number of candidates selected from Engineering discipline respectively (rounded off to the nearest integer)? 1) 2) 3) 4) 5) 3609: 813 3094: 813 3094: 1035 4125: 1035 3981: 767
2. The total number of candidates appeared in interview from Management and other discipline was what percentage of number of candidates appeared from Engineering discipline? 1) 2) 3) 4) 5) 50 150 200 Cannot be determined None of these
3. Approximately what was the difference between the number of candidates selected from Agriculture discipline and number of candidates selected from Engineering discipline? 1) 2) 3) 4) 5) 517 665 346 813 296
4. For which discipline was the difference in number of candidates selected to number of candidates appeared in interview the maximum? 1) 2) 3) 4) 5) Management Engineering Science Agriculture None of these
5. Approximately what was the total number of candidates selected from Commerce and Agricultural discipline together? 1) 1700 257 IBPS PO EXAM 2013 : Quantitative Aptitude
2) 3) 4) 5)
## 1800 2217 1996 1550
Directions (Q.6-10): Study the following pie-chart carefully to answer these questions. (Central Bank of India (PO) 2010) Percentagewise Distribution of teachers who teach six different subjects Total number of Teachers = 1800 Percentage of teachers
6. If two-ninths of the teachers who teach Physics are female, then the number of male Physics teachers is approximately what percentage of the total number of teachers who teach Chemistry? 1) 57 2) 42 3) 63 4) 69 5) 51 7. What is the total number of teachers teaching Chemistry, English and Biology? 1) 1, 226 258 IBPS PO EXAM 2013 : Quantitative Aptitude
2) 1, 116 3) 1, 176 4) 998 5) None of these 8. What is the difference between the total number of teachers who teach English and Physics together and the total number of teachers who teach Mathematics and Biology together? 1) 352 2) 342 3) 643 4) 653 5) None of these 9. What is the ratio of the number of teachers who teach Mathematics to the number of teachers who teach Hindi? 1) 13:7 2) 7:13 3) 7: 26 4) 8: 15 5) None of these 10. If the percentage of Mathematics teachers is increased by 50 per cent and the percentage of Hindi teachers decreased by 25 per cent then what will be the total number of Mathematics and Hindi teachers together? 1) 390 2) 379 3) 459 4) 480 5) None of these 259 IBPS PO EXAM 2013 : Quantitative Aptitude
Direction (Q. 11 15): Study the given pie-charts carefully and answer the questions that follow: Discipline-wise breakup of the number of candidates appeared in Interview and Discipline-wise breakup of the candidates selected by and organisation. Total number of candidates appeared in the interview = 25,600 and total number of candidates selected after interview = 7,500.
11. What was the ratio of the number of candidates appeared in interview from other disciplines and the number of candidates selected from art disciplines? (a) 256 : 125 (b) 125 : 256 (c) 125 : 216 (d) Cannot be determined (e) None of these 12. The total number of candidates appeared in interview from Management and Art disciplines was what per cent of the number of candidates from Engineering discipline? (a) 66.67 260 IBPS PO EXAM 2013 : Quantitative Aptitude
(b) 75 (c) 80 (d) 120 (e) 150 13. What was the difference between the number of candidates selected from Science discipline and the number of candidates selected from Commerce discipline? (a) 1,000 (b) 1,100 (c) 1,200 (d) 1,250 (e) None of these 14. From which discipline was the difference in number of candidates selected to number of candidates appeared in interview the maximum ? (a) Management (b) Engineering (c) Commerce (d) Science (e) Art 15. What was the total number of candidates selected from Commerce and Art discipline together? (a) 1,800 (b) 1,950 (c) 2,100 (d) 2,250 (e) 2,400 261 IBPS PO EXAM 2013 : Quantitative Aptitude
## Data Interpretation- Pie Diagram Practice Set (Answers)
1. 2. 3. 4. 5. 6. 7. 8. 2 2 5 3 1 1 2 2 9. 10. 11. 12. 13. 14. 15. 5 3 a e c d b
262
## Chapter: Data Interpretation-Case lets
Direction for questions 1 to 5: Answer these questions on the basis of the following information. Shekhar bought 10 acres of land for Rs.250000 in 2011. That year he cultivated Sugarcane and Soya bean in the 10 acres with the ratio of area under Sugarcane and Soya bean being 5:4. The profit obtained from Sugarcane and Soya bean was in the ratio 3:2 with the total profit being Rs.58500. This was 15% of the amount he invested in cultivation that year. The next year he again cultivated Sugarcane and Soya bean, with the areas being same as before and reaped a profit of Rs.66000 in total with that from Sugarcane and Soya bean being in the ratio 8:7 but his return on his investment that year was only 14%. 1. What is the amount invested by Shekhar for cultivation in 2011? (a) Rs.356000 (b) Rs.374800 (c) Rs.380000 (d) Rs. 390000 Solution: (d) Shekhar had a profit of Rs.58500 and this profit was 15% of the money he invested, his investment was
2. What is the profit obtained by Shekhar by cultivating Sugarcane in 2011? (a) Rs.43800 (b) Rs.35100 (c) Rs.36200 (d) None of these Solution: (b) The profit obtained by Shekhar by cultivating Sugarcane in 2012 =
263
## IBPS PO EXAM 2013 : Quantitative Aptitude
3. What is the profit obtained by cultivating Soya bean in 2012? (a) Rs.30800 (b) Rs.36100 (c) Rs.24200 (d) None of these Solution: (a) The profit obtained by cultivating Soya bean in 2012 = 4. What is the ratio of the profit obtained from Sugarcane and Soya bean in the two years together? (a) 89 : 79 (b) 167 : 211 (c) 703 : 542 (d) None of these Solution: (c) The profit obtained in 2011 from Sugarcane = Rs.35100 The profit obtained in 2012 from Sugarcane = 66000 - 30800 = Rs.35200 Total profit from Sugarcane = Rs.70300 Total profit in the two years = Rs.58500 + Rs.66000 = Rs.124500 Therefore, Profit from Soya bean Rs.54200. Now, the required ratio = 70300: 54200 = 703 : 542.
264
## IBPS PO EXAM 2013 : Quantitative Aptitude
5. What is the approximate amount invested by Shekhar for cultivation in 2012? (a) Rs.428500 (b) Rs.471400 (c) Rs.495300 (d) Rs.518650 Solution: (b) Here the profit of 66,000 is 14% of the amount invested, therefore, the invested =
265
## Solved Examples (Data Interpretation-Caselets)
Directions: (1-5) Study the following information and answer the questions that follow: (IBPS CWE PO MT 2012) The premises of a bank are to be renovated. The renovation is in terms of flooring. Certain areas are to be floored either with marble or wood. All rooms/halls and pantry are rectangular. The area to be renovated comprises of a hall for customer transaction measuring 23 m by 29 m, branch managers room measuring 13 m by 17 m, a pantry measuring 14 m by 13 m, a record keeping cum server room measuring 21 m by 13 m and locker area measuring 29 m by 21 m. The total area of the bank is 2000 square meters. The cost of wooden flooring is f 170/- per square meter and the cost of marble flooring is Rs. 190/- per square meter. The locker area, record keeping cum server room and pantry are to be floored with marble. The branch manager's room and the hall for customer transaction are to be floored with wood. No other area is to be renovated in terms of flooring. 1. What is the respective ratio of the total cost of wood en flooring to the total cost of marble flooring? (1) 1879: 2527 (2) 1887: 2386 (3) 1887: 2527 (4) 1829: 2527 (5) 1887: 2351 Solution: Total flooring area with marble = locker area + record keeping + pantry = 182+273 +609 = 1064 sqm Cost of flooring = 1064 190 Total flooring area with wooden = Branch Manager room + Hall = 221 + 667 = 888 sqm Cost of flooring = 888 170 Ratio= 888 170: 1064 190 = 888 17: 1064 19 = 15096 : 20216 = 1887: 2527 Ans: 3 266 IBPS PO EXAM 2013 : Quantitative Aptitude
2. If the four walls and ceiling of the branch managers room (The height of the room is 12 meters) are to be painted at the cost off 190/- per square meter, how much will be the total cost of renovation of the branch manager's room including the cost of flooring? (1) Rs. 1, 36,800/(2) Rs. 2, 16,660/(3) Rs. 1, 78,790/(4) Rs. 2, 11,940/(5) None of these Solution: Cost of flooring of branch manager room =221 170= Rs. 37570 Cost of painting = [2(17 12+ 13 12)+ 13 x 17] 190 = [2( 204 + 156) + 221] 190= (2 360 + 221) 190 = (720 + 221) 190 = 941 190 = Rs. 178790 Total cost = 178790 + 31570 = Rs.216360 Ans: (5) 3. If the remaining area of the bank is to be carpeted at the rate of Rs. 110/- per square meter, how much will be the increment in the total cost of renovation of bank premises? (1) Rs. 5,820/(2) Rs. 4,848/(3) Rs. 3,689/(4) Rs. 6,890/(5) None of these Solution: Total area of bank = 2000 sqm Total flooring area = 1952 sqm Remaining area = 2000 - 1952 = 48 sqm Cost of carpeting = 48 110 = Rs.5280 Ans: (5) 4. What is the percentage area of the bank that is not to be renovated? (1) 2.2 % (2) 2.4 % (3) 4.2 % (4) 4.4 % 267 IBPS PO EXAM 2013 : Quantitative Aptitude
(5) None of these Solution: Area not to be renovated = 48 sq m Reqd % = l00 = 2.4% Ans: 2 5. What is the total cost of renovation of the hall for customer transaction and the locker area? (1) Rs. 2, 29,100 (2) Rs. 2, 30,206 (3) Rs. 2, 16,920 (4) Rs. 2, 42,440 (5) None of these Solution: Cost of renovation of hall + locker area = 667 170 + 609 190 = 113390 + 115710= Rs. 229100 Ans: (1) Directions (Q. 6 10): Study the given information carefully to answer the questions that follow: An organization consists of 3500 employees working in different departments, viz HR, Marketing, IT, Production and Accounts. The ratio of male to female employees in the organisation is 3 : 2. 8% of the males work in the HR department. 22% of the female work in the account department. The ratio of males to females working in the HR department is 3 : 5. Oneseventh of the females work in the IT department. 46% of the males work in the Production department. The number of females is one-sixth of the males working in the same. The remaining females work in the Marketing department. The total number of employees working in the IT department is 375. 22% of the males work in the Marketing department and remaining work in the Account department. 6. The number of males working in the Account department forms approximately what per cent of the total number of males in the organisation? (a) 6 (b) 8 (c) 10 268 IBPS PO EXAM 2013 : Quantitative Aptitude
(d) 11 (e) 12 Answer: (a) 7. How many females work in Production department? (a) 140 (b) 200 (c) 180 (d) 160 (e) None of these Answer: (e) 8. The total number of employees working in the Account department forms approximately what per cent of the total number of female employees in the organisation? (a) 28 (b) 32 (c) 29 (d) 31 (e) 30 Answer: (d) 9. The ratio of the numbers of females working in IT department to the numbers of males working in the same department is (a) 15 : 8 (b) 1 : 2 (c) 8 : 15 (d) 2 : 1 (e) 7 : 11 Answer: (c) 10. What is the total number of employees working in the Marketing and Production departments together? (a) 1900 (b) 2040 (c) 2020 (d) 2031 (e) 2042 Answer:(b)
269
## PracticeSet-(Data Interpretation- Caselets)
Directions (Q. 1-5): Study the information carefully to answer the questions that follow. A company produced five different products, viz mobile phone, pen drive, calculator, television and washing machine. The total number of all the five products is 1650.24% of the total number of products is mobile phones. One-sixth of the total number of products is pen drives. 14% of the total number of products is calculators. Remaining products are either television or washing machine. The number of washing machines is 50 more than the number of televisions produced. (IBPS RRB Grade Officer Exam 2012) 1. What is the ratio of the number of washing machines to the number of calculators produced by the company? (1) (2) (3) (4) (5) 17:11 19:11 11:17 19: 13 None of these 3. The number of televisions produced is approximately what per cent of the total number of calculators and washing machines produced together? (1) (2) (3) (4) (5) 63 55 59 51 67
4. What is the difference between the total number of televisions and mobile phones together and the number of calculators produced? (1) 534 (2) 524 (3) 511 (4) 523 (5) None of these 5. What is the total number of pen drives, calculators and washing machines produced by the company? (1) 907 (2) 917 (3) 925 (4) 905 (5) None of these
2. If 24 per cent of the pen drives are defective, what is the number of pen drives which are not defective? (1) (2) (3) (4) (5) 209 215 219 225 None of these 270
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 6-10): Study the following information carefully to answer the questions that follow: (RBI GradeB Officers Exam 2011) There are two trains, Train A and Train B. Both trains have four different types of coaches, viz General, Sleeper, First Class and AC. In Train A, there are total 700 passengers. Train B has thirty per cent more passengers than Train A. Twenty per cent of the passengers of Train A are in General Coach. One-fourth of the total number of passengers of Train A are in AC coach. Twenty three per cent of the passengers of Train A are in Sleeper Coach. Remaining passengers of Train A are in First Class Coach. The total number of passengers in AC Coach in both the trains together is 480. Thirty per cent of the number of passengers of Train B are in Sleeper Coach. Ten per cent of the total passengers of Train B are in First Class Coach. The remaining passengers of Train B are in General Coach. 6. What is the ratio of the number of passengers in First Class Coach of Train A to the number of passengers in Sleeper Coach of Train B? (1) (2) (3) (4) (5) 13 : 7 7 : 13 32 : 39 Data Inadequate None of these
(3) 435 (4) 445 (5) None of these 8. What is the difference between the number of passengers in the AC Coach of Train A and the total number of passengers in Sleeper and First Class Coach together of Train B? (1) (2) (3) (4) (5) 199 178 187 179 None of these
9. The total number of passengers in General Coaches of both the trains together is approximately what percentage of the total number of passengers in Train B? (1) (2) (3) (4) (5) 35 42 46 38 31
10. If the cost per ticket of First Class coach is Rs.450, what will be the total amount generated from First Class Coach of Train A? (1) (2) (3) (4) (5) Rs.1, 00, 080 Rs.1, 08, 000 Rs.1, 00, 800 Rs.10, 800 None of these
7. What is the total number of passengers in the General Coach of Train A and the AC Coach of Train B together? (1) 449 (2) 459 271
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 11-15): Study the information carefully to answer the questions that follow: A company produces five different products, viz Television, Refrigerator, Mobile Phone, Oven and Water Heater. The total number of all the five products manufactured is 1200. 15 per cent of the total number of products are Televisions. Three-tenths of the total number of products are Refrigerators. The number of Mobile Phones manufactured is 40 more than the number of Televisions. 22 per cent of the total number of products is Oven and the remaining number of products is Water Heaters. (Corporation Bank PO 2011) 11. What is the total number of Refrigerators and Water Heaters? (1) (2) (3) (4) (5) 436 476 576 536 None of these
## Rs. 21,68,400 Rs. 2,16,48,000 Rs. 2,16,840 None of these
14. What is the difference between the number of Refrigerators and the number of Mobile Phones manufactured? (1) (2) (3) (4) (5) 160 140 120 130 None of these
15. If 25 per cent of the number of Ovens are defective, what is the number of nondefective Ovens? (1) 176 (2) 188 (3) 198 (4) 186 (5) None of these
12. The number of Mobile Phones is approximately what per cent of the number of Televisions? (1) 115 (2) 140 (3) 135 (4) 130 (5) 120 13. If the cost of one Oven is Rs. 8,200, what is the cost of all the Ovens manufactured by the company? (1) Rs. 2,16,480 272 IBPS PO EXAM 2013 : Quantitative Aptitude
Directions-(Q. 16-20): Study the information carefully to answer the questions that follows: In a ship there are 1200 passengers. 18 per cent of the total number of passengers is from Britain. Two- fifth of the total number of passengers is from South Africa 6 per cent of the total number of passengers is from Madagascar. Remaining number of passengers is from India. 25 per cent of the number of passengers from Britain is females. Half the numbers of passengers from South Africa are male. There is no female passenger from Madagascar. Twothird of the number of passengers from India are females. (Allahabad Bank Probationary Officers Exam 2011) 16. What is the ratio of the number of passengers from Madagascar, number of female passengers from South Africa and the total number of passengers from India? (1) (2) (3) (4) (5) 2: 5 : 18 3: 10 : 18 3: 11 : 18 2: 18 : 5 None of these
## 111 115 120 125 131
18. What is the average number of male passengers from all the four countries? (1) (2) (3) (4) (5) 154.5 164.5 145 164 None of these
19. What is the difference between the number of male passengers from Madagascar and the number of male passengers from India? (1) (2) (3) (4) (5) 64 82 74 72 None of these
20. What is the total number of male from Britain passengers female passengers from India together? (1) (2) (3) (4) (5) 340 420 350 460 None of these
17. The number of male passengers from South Africa is approximately what percentage of the total number of passengers from Britain?
273
## Data Interpretation-Caselets Practice Set (Answers)
1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 2 1 2 5 4 3 4 5 2 3 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 4 5 5 2 3 2 1 1 4 5
274
## Solved Examples (Data Interpretation-Miscellaneous)
Directions: (1-5) Study the following graph and table carefully and answer the questions given below: Time Taken To Travel (In Hours) By Six Vehicles On Two Different Days
## Distance covered (in kilometers) by six vehicles on each day
(IBPS CWE PO MT 2012) 1. Which of the following vehicles travelled at the same speed on both the days? (1) Vehicle A (2) Vehicle C (3) Vehicle F 275 IBPS PO EXAM 2013 : Quantitative Aptitude
(4) Vehicle B (5) None of these Solution: The speed of Vehicle B on both the days is 43 km/hr. Ans: (4) 2. What was the difference between the speed of vehicle A on day 1 and the speed of vehicle C on the same day? (1) 7 km/hr (2) 12 km/hr (3) 11 km/hr (4) 8 km/hr (5) None of these Solution: Speed of A on 1st day = 52 km/hr Speed of C on 1st day = 63 km/hr Difference = 65 - 52 = 11 km/hr Ans: 2 3. What was the speed of vehicle C on day 2 in terms of meters per second? (1) 15.3 (2) 12.8 (3) 11.5 (4) 13.8 (5) None of these Solution: Speed of Vehicle C on 2nd day = 45 km/hr = 45 = 2.5 5 = 12.5 m/sec Ans: (5) 4. The distance travelled by vehicle F on day 2 was approximately what percent of the distance travelled by it on day 1? (1) 80 (2) 65 (3) 85 276 IBPS PO EXAM 2013 : Quantitative Aptitude
(4) 95 (5) 90 Solution: Reqd % = Ans: (5) 5. What is the respective ratio between the speeds of vehicle 0 and vehicle E on day 2? (1) 15: 13 (2) 17: 13 (3) 13: 11 (4) 17: 14 (5) None of these Solution: Reqd Ratio = = Ans: (2) = = 17:13 100 = 90.46 90%
277
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (6-10) Study the following pie-chart and table carefully and answer the questions given below: Percentagewise distribution of the number of mobile phones sold a shopkeeper during six months Total number of mobile phones sold = 45000
The respective ratio between the numbers of mobile phones sold of company A and company B during six months
6. What is the respective ratio between the number of mobile phones sold of company B during July and those sold during December of the same company? (1) 119: 145 (2) 116: 135 (3) 119: 135 (4) 119: 130 (5) None of these Solution: Total number of mobiles sold in the month of July 278 IBPS PO EXAM 2013 : Quantitative Aptitude
= 45000
= 7650
Mobile phones sold by Company B in the month of July = 7650 = 3570 Total number of mobile phones sold in the month of December = 45000 = 7200 Mobile phones sold by Company B in the month of December = 7200 = 4050 Reqd ratio = Ans: (3) 7. If 35% of the mobile phones sold by company A during November were sold at a discount, how many mobile phones of company A during that month were sold without a discount? (1) 882 (2) 1635 (3) 1638 (4) 885 (5) None of these Solution: Number of mobile phones sold in the month of November = 45000 = 5400 Number of mobile phones sold by Company A in the month of November = 5400 = 2520 Number of mobile phones without discount in the month of November by Company A = 2520 = 2520 0.65 = 1638 Ans: 3 8. If the shopkeeper earned a profit of Rs. 433/- on each mobile phone sold of company B during October, what was his total profit earned on the mobile phones of that company during the same month? (1) Rs. 6,49,900 (2) Rs. 6,45,900 (3) Rs. 6,49,400 (4) Rs 6,49,500 (5) None of these 279 IBPS PO EXAM 2013 : Quantitative Aptitude = = = 119: 135
Solution: Number of mobile phones sold in the month of October = 45000 = 3600 Number of mobile phones sold by Company B in the month of October = 3600 = 1500 Total profit earned by Company B in the month of October = 1500 433 = 649500 Ans: (4) 9. The number of mobile phones sold of company A during July is approximately what percent of the number of mobile phones sold of company A during December? (1) 110 (2) 140 (3) 150 (4) 105 (5) 130 Solution: Number of mobile phones sold in the month of July = 45000 = 7650 Number of mobile phones sold by Company A in the month of July = 7650 = 4080 Number of mobile phones sold in the month of December = 45000 = 7200 Number of mobile phones sold by Company A in the month of December = 7200 = 3150 Reqd % = Ans: (5) 10. What is the total number of mobile phones sold of company B during August and September together? (1) 10000 (2) 15000 (3) 10500 (4) 9500 (5) None of these = 129.52 130
280
## IBPS PO EXAM 2013 : Quantitative Aptitude
Solution: Number of mobile phones sold in the month of August = 45000=9900 Number of mobile phones sold in the month of September = 45000 = 45000 = 11250 Number of mobile phones sold by Company B in the month of August = 9900 = 5500 Number of mobile phones sold by Company B in September= 11250 = 4500 Total number of mobile phones sold in August and September by Company B = 5500+4500= 10000 Quicker Method: Total number of mobile phones sold by Company B in August and September = 10000 Ans: (1) .
281
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions: (11-15) Study the following information and answer the questions that follow: The graph given below represents the production (in tonnes) and sales (in tonnes) of company a from 2006-2011.
The table given below represents the respective ratio of the production (in tonnes) of Company A to the production (in tonnes) of Company B, and the respective ratio of the sales (in tonnes) of Company A to the sales (in tonnes) of Company B.
11. What is the approximate percentage increase in the production of Company A (in tonnes) from the year 2009 to the production of Company A (in tonnes) in the year 2010? (1) 18% (2) 38% (3) 23% (4) 27% (5) 32% Solution: Production of Company A in year 2009 = 550 Production of Company A in year 2010 = 700
282
## IBPS PO EXAM 2013 : Quantitative Aptitude
Reqd % = = = 27.27
100 = 27%
100
Ans: (4) 12. The sale of Company A in the year 2009 was approximately what percent of the production of Company A in the same year? (1) 65% (2) 73% (3) 79% (4) 83% (5) 69% Solution: Sales of Company A in year 2009 = 400 Production of Company A in year 2009 = 550 Reqd % = Ans: (2) 13. What is the average production of Company B (in tonnes) from the year 2006 to the year 2011? (1) 574 (2) 649 (3) 675 (4) 593 (5) 618 Solution: Average production of Company B = = = 675 100 = = 72.72 73%
Ans: (3)
283
## IBPS PO EXAM 2013 : Quantitative Aptitude
14. What is the respective ratio of the total production (in tonnes) of C ompany A to the total sales (in tonnes) of Company A? (1) 81: 64 (2) 64: 55 (3) 71: 81 (4) 71: 55 (5) 81: 55 Solution: = = = 81: 55
Ans: (5) 15. What is the respective ratio of production of Company B (in tonnes) in the year 2006 to production of Company B (in tonnes) in the year 2008? (1) 2: 5 (2) 4: 5 (3) 3: 4 (4) 3: 5 (5) 1: 4 Solution: Production of Company B in the year 2006 = 150 = 600 Production of Company B in the year 2008 = 200 Ratio = = 3: 4 Ans: (3)
284
## Practice Set (Data Interpretation-Miscellaneous)
Directions (Q. 1-5): Study the radar-graph carefully and answer the questions that follow. Monthly salary (in thousands) of five different persons in three different years (RBI GradeB Officers Exam 2011)
1. What is the average of the monthly salary of Sumit in the year 2008, Anil in the year 2009 and Jyoti in the year 2010? (1) (2) (3) (4) (5) Rs.20, 000 Rs.5, 000 Rs.45, 000 Rs.15, 000 None of these
2. The total monthly salary of Arvind in all years together is what per cent of the total monthly salary of all the five persons together in the year 2008? (1) (2) (3) (4) (5) 55% 60% 75% 70% None of these 285 IBPS PO EXAM 2013 : Quantitative Aptitude
3. Among the five persons, whose earning per month over all the years together is the second lowest? (2) (3) (4) (5) (6) Sumit Anil Jyoti Arvind Poonam
4. What is the per cent decrease in the monthly salary of Poonam in the year 2009 as compared to her monthly salary in the previous year? (1) (2) (3) (4) (5) 40% 10% 20% 80% None of these
5. If Jyoti's monthly salary in the year 2010 was increased by 30 per cent what would her monthly salary be in that year? (1) (2) (3) (4) (5) Rs.36, 000 Rs.39, 000 Rs.45, 000 Rs.42, 000 None of these
286
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions (Q. 6-10): Study the following pie-chart and bar-graph and answer the following questions. (Corporation Bank PO 2011) Percentage wise distribution of teachers in six different districts Total number of Teachers = 4500 Percentage of Teachers
287
## IBPS PO EXAM 2013 : Quantitative Aptitude
6. What is the total number of male teachers in District F, female teachers in District C and female teachers in District B together? (1) (2) (3) (4) (5) 1080 1120 1180 1020 None of these
7. The number of female teachers in District D is approximately what per cent of the total number of teachers (both male and female) in District A? (1) (2) (3) (4) (5) 70 75 80 95 90
8. In which district is the number of male teachers more than the number of female teachers? (1) (2) (3) (4) (5) B only D only Both B and E Both E and F None
9. What is the difference between the number of female teachers in District F and the total number of teachers (both male and female) in District E? (1) (2) (3) (4) (5) 625 775 675 725 None of these
10. What is the ratio of the number of male teachers in District C to the number of female teachers in District B? (1) (2) (3) (4) (5) 11 : 15 15 : 11 15 : 8 30 : 13 None of these
288
## IBPS PO EXAM 2013 : Quantitative Aptitude
Directions-(Q. 11-15) Study the following pie-chart and bar chart and answer the following questions. (Allahabad Bank Probationary Officers Exam 2011) Percentagewise Distribution of students in six different Schools Total number of Students = 6000 Percentage of Students
289
## IBPS PO EXAM 2013 : Quantitative Aptitude
11. What is the sum of the number of girls in School C. Number of girls in School-E and the number of boys in school-D together? (1) (2) (3) (4) (5) 1700 1900 1600 1800 None of these
12. What is the respective ratio of the number of boys in School-C, number of girls in School B and total number of students in School E? (1) (2) (3) (4) (5) 45: 7: 97 43: 9: 97 45: 7: 87 43: 9: 87 None of these
13. What is the difference between the total number of students in School-P and the number of boys in School E? (1) (2) (3) (4) (5) 820 860 880 900 None of these
14. In which school the total numbers of students (both boys and girls) together are equal to the number of girls in School E? (1) (2) (3) (4) (5) A B C D F
15. The Number of girls in School A is approximately what per cent of total number of students in School B? (1) 55 (2) 50 (3) 35 (4) 45 (5) 4 290 IBPS PO EXAM 2013 : Quantitative Aptitude
## Data Interpretation-Miscellaneous Practice Set (Answers)
1 2 3 4 5 6 7 8 (1) (2) (4) (4) (2) 13 (1) 14 (5) 15 (3) (5) (2) (5) 9 10 11 12 (2) (3) (4) (3)
291
## IBPS PO EXAM 2013 : Quantitative Aptitude
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http://www.energeticforum.com/renewable-energy/12466-konstatin-meyls-mistake-theory-practice-2.html | 1,386,605,502,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163988740/warc/CC-MAIN-20131204133308-00013-ip-10-33-133-15.ec2.internal.warc.gz | 318,969,923 | 17,687 | THIS IS THE NEW BOOK THAT GOES WITH THE VIDEO
Energetic Forum Konstatin Meyl's mistake in Theory and Practice
Homepage Energetic Science Ministries Register FAQ Calendar Search Today's Posts Mark Forums Read
Renewable Energy Discussion on various alternative energy, renewable energy, & free energy technologies. Also any discussion about the environment, global warming, and other related topics are welcome here.
10-21-2012, 02:11 AM
Farmhand Senior Member Join Date: Jun 2010 Location: Australia Posts: 3,214
Since this is all Tesla related I've been looking for references made by Tesla
to a longitudinal wave ect. I admit I haven't read much as yet but I see no
reference by Tesla as so far to longitudinal waves/electricity. But I did find
this quote below. Which clearly states the the energy of the currents he
passes into the Earth are stored as "Electromagnetic momentum of the vibrations. To me that says LC tank.
Nikola Tesla On His Work With Alternating Currents -- Chapter IV
Quote:
Counsel That is what I want you to explain. I must be mistaken, because my conception does not fit in with your statements. Tesla All right, I will explain that. In my first efforts, of course I simply contemplated to disturb effectively the earth, sufficiently to operate instruments. Well, you know you must first learn how to walk before you can fly. As I perfected my apparatus, I saw clearly that I can recover, of that energy which goes in all directions, a large amount, for the simple reason that in the system I have devised, once that current got into the earth it had no chance of escaping, because my frequency was low; hence, the electro-magnetic radiation was low. The potential, the electric potential, is like temperature. We might as well call potential electric temperature. The earth is a vast body. The potential differences in the earth are small, radiation is very small. Therefore, if I pass my current into the earth, the energy of the current is stored there as electromagnetic momentum of the vibrations and is not consumed until I put a receiver at a distance, when it will begin to draw the energy and it will go to that point and nowhere else.
I have my doubts he actually described a longitudinal wave/component in
relation to these devices.
Could someone provide a reference where Tesla clearly refers to or describes
this longitudinal component ?
Cheers
Last edited by Farmhand : 10-21-2012 at 02:13 AM.
10-21-2012, 07:44 AM
Ernst Senior Member Join Date: Jul 2012 Posts: 288
@ Farmhand,
Have a look here
This is a pretty complete overview (I believe) of Tesla's work on various forms of radiation.
Since at that time (and in fact still today) there was little known about the nature of the various forms of radiation, you will not hear Tesla speak in todays terminology. But as you read this paper you will realise that Tesla was way ahead of his contemporaries on this subject as well.
These are photons,... no,... waves,... no,... photons,... erhm, wavylike particles that have no mass when at rest. But they never are at rest, they always move at the speed of light. Hence their 0-mass is multiplied by infinity, to give a small (lucky us) mass so they can have momentum.
Therefor they are also attracted by gravitation, which alters their speed, which is constant, especially near objects like black holes. Then they also interact with eachother creating interference patterns where it looks like 2 photons annihilate leaving no trace of their energy nor momentum. Which of course is impossible so we come up with photons in superposition; traveling more than 1 path at the same time.
This may sound a little bit far-fetched-ish, but this is actually todays science.
I had seen a video on youtube about 2 years ago giving an alternative explanation for experimental data without needing to resort to these (or other) obscure particles. I remember it was pretty convincing, something about treating an EM wave as a string of capacitors.
Ernst.
Last edited by Ernst : 10-21-2012 at 09:13 AM. Reason: typo
10-21-2012, 08:59 AM
Farmhand Senior Member Join Date: Jun 2010 Location: Australia Posts: 3,214
I was actually looking for direct quotes from Tesla in relation to the
transmission of energy with regard to his improved transmitter using
longitudinal waves or whatever.
Many things are surmised or confusingly lumped in together, Tesla showed
two ways of wireless transmission, by atmosphere and by the ground. And
from a transmitter.
I disagree with this statement below, the idea for making and breaking the
circuit in a radiant energy collector is to allow the energy to build before
applying it, in most cases. Otherwise the current is feeble. Although a radiant
energy receiver can in fact supply energy to a transmitter I don't see how it
could effectively transmit as an oscillator and at the same time collect from
the elevated terminal.
Just because he did "other" experiments at Colorado Springs doesn't mean
those other experiments were related to the actual transmission of energy.
Quote from near the end.
Quote:
"He has operated his radiations receiver as an oscillator with interrupter circuitry. And if one is only interested to collect free charges one does not continuously break the (direct) current flow. This alternating oscillation induction in the receiver does only make sense if one is interested in resonance coupling!"
http://www.andre-waser.ch/Publicatio...CosmicRays.pdf
Cheers
10-21-2012, 09:36 AM
Ernst Senior Member Join Date: Jul 2012 Posts: 288
I think this Andre gives a very good overview of all instances where Tesla says something about the nature of radiation. He distinquishes a few forms.
When it comes to transmission of energy or electricity, he always says that 'Herzian waves' can not be used for this purpose.
Herzian waves are those that we today would call transverse EM radiation.
So it will have to be another form of radiation that he uses when he is talking about energy transfer through the air.
(in most cases, I believe, Tesla uses the earth to transfer the energy. And I have read somewhere that the ionossphere mirrors earths potential fluctuations, and vice versa. This would make it possible to transfer energy through capacitive coupling. But I forgot where I read this.)
Tesla clearly mentions that particle beams/radiation can be used for efficient energy transfer, but I believe in most cases he did not use this.
From this point on, we can only guess what kind of radiation he was using. But by reconstructing his work we should be able to do some educated guess work.
Ernst.
10-21-2012, 10:46 AM
Ernst Senior Member Join Date: Jul 2012 Posts: 288
Forgot....
Tesla does mention that the waves he uses for energy transmission resemble sound waves.
This is probably the statement that comes closest to what you are looking for?
Sound waves are longitudinal as opposed to 'Herzian waves'.
Ernst.
10-21-2012, 12:46 PM
Farmhand Senior Member Join Date: Jun 2010 Location: Australia Posts: 3,214
Yes well what of the quote I gave above, he clearly states the currents he
passes into the Earth are stored there as electromagnetic momentum, is this
not analogous to a tank circuit ? Are the oscillations in a tank circuit "hertz" waves ?
Or are the oscillations in a tank circuit EM radiation ? Or are they
electromagnetic momentum of vibrations ? I don't think they are radiation.
Resembling sound waves is "not" sound waves, just "like" sound waves. But I
don't see how longitudinal waves are not recognized if sound waves are
longitudinal. Sound waves resonate and reverberate and such whereas
radiation does not. So any resonance in a tank is "like" sound waves.
Seems like good Logic to me.
Anyway the directed energy is a different concept to the transmission of
energy with the patented transmitters. I think he just intended using the transmitters
to transmit the energy to the energy directing device and/or control them (remotely)
and/or to produce the enormous potential to run the energy directing
apparatus..
The atmospheric transmissions he explains that the currents are transmitted
by conduction.
Nikola Tesla On His Work With Alternating Currents -- Chapter IV
By direct quotes I mean to words written by Tesla or recorded from what he actually said, in context.
Cheers
10-21-2012, 02:42 PM
Ernst Senior Member Join Date: Jul 2012 Posts: 288
Ok Farmhand,
Quote:
I disagree with this statement below, the idea for making and breaking the circuit in a radiant energy collector is to allow the energy to build before applying it, in most cases.
I agree with you, then.
Quote:
Just because he did "other" experiments at Colorado Springs doesn't mean those other experiments were related to the actual transmission of energy.
Absolutely.
Concerning the document I refered you to; I do not agree with the author of that document on many points. But when you are looking for quotes from Tesla on radiation then this document is a very good starting point.
Quote:
Yes well what of the quote I gave above, he clearly states the currents he passes into the Earth are stored there as electromagnetic momentum, is this not analogous to a tank circuit ?
I would say you are correct again.
Quote:
Resembling sound waves is "not" sound waves, just "like" sound waves. But I don't see how longitudinal waves are not recognized if sound waves are longitudinal.
I understand the first line, and agree there, but I do not understand the second line.
Quote:
Sound waves resonate and reverberate and such whereas radiation does not.
Not sure on this one. I think various forms of EM radiation are known to resonate.
Quote:
So any resonance in a tank is "like" sound waves. Seems like good Logic to me.
True.
Quote:
The atmospheric transmissions he explains that the currents are transmitted by conduction.
and true again. Though I do not remember having read that he actually uses this type of transmission in a (real) experiment. Whenever he does transmit power, he mostly does so through the earth. I will have to think hard to see if I can remember one of his experiments where he does not use the earth.
I'll try to find something for you. I feel you are trying to say that Tesla never used longitudinal electricity through the air, he was using longitudinal electricity through the earth... Is that right?
Ernst.
10-21-2012, 04:49 PM
Farmhand Senior Member Join Date: Jun 2010 Location: Australia Posts: 3,214
Hi Ernst, I'm not looking for quotes from Tesla on radiation in particular, I'm
looking for quotes by Tesla on Longitudinal "waves" or such with respect
directly to the operation of the Energy Transmitters patented for the
transmission of energy by the atmospheric conduction or the ground currents.
Nikola Tesla On His Work With Alternating Currents -- Chapter IV
Demonstration for the patent examiner.
Quote:
Having discovered that, I established conditions under which I might operate in putting up a practical commercial plant. When the matter came up in the patents before the Examiner, I arranged this experiment [shown in Fig. 78] for him in my Houston Street laboratory. I took a tube 50 feet long, in which I established conditions such as would exist in the atmosphere at a height of about 4 1/2 miles, a height which could be reached in a commercial enterprise, because we have mountains that are 5 miles high; and, furthermore, in the mountainous regions we have often great water power, so that the project of transmitting it, if the plan was rational, would be practicable. Then, on the basis of the results I had already obtained, I established those conditions, practically, in my laboratory. I used that coil which is shown in my patent application of September 2, 1897 (Patent No. 645,576 of March 20, 1900), the primary as described, the receiving circuit, and lamps in the secondary transforming circuit, exactly as illustrated there. And when I turned on the current, I showed that through a stratum of air at a pressure of 135 millimeters, when my four circuits were tuned, several incandescent lamps were lighted.
Quote:
and true again. Though I do not remember having read that he actually uses this type of transmission in a (real) experiment. Whenever he does transmit power, he mostly does so through the earth. I will have to think hard to see if I can remember one of his experiments where he does not use the earth. I'll try to find something for you. I feel you are trying to say that Tesla never used longitudinal electricity through the air, he was using longitudinal electricity through the earth... Is that right?
Not sure, I'm not a learned guy, I have to try to make my own conclusions.
The Earth currents travel by conduction as well, obviously, I just don't think
there needs to be a second path back to the transmitter because of the
terminal reference.
Here is the quote on the process through the atmosphere it has a return
through the Earth, I assume, it's not clearly stated.
Quote:
The earth is 4,000 miles radius. Around this conducting earth is an atmosphere. The earth is a conductor; the atmosphere above is a conductor, only there is a little stratum between the conducting atmosphere and the conducting earth which is insulating. Now, on the basis of my experiments in my laboratory on Houston Street, the insulating layer of air, which separates the conducting layer of air from the conducting surface of the earth, is shown to scale as you see it here. Those [radii lines] are 60 of the circumference of the earth, and you may notice that faint white line, a little bit of a crack, that extends between those two conductors. Now, you realize right away that if you set up differences of potential at one point, say, you will create in the media corresponding fluctuations of potential. But, since the distance from the earth's surface to the conducting atmosphere is minute, as compared with the distance of the receiver at 4,000 miles, say, you can readily see that the energy cannot travel along this curve and get there, but will be immediately transformed into conduction currents, and these currents will travel like currents over a wire with a return. The energy will be recovered in the circuit, not by a beam that passes along this curve and is reflected and absorbed, because such a thing is impossible, but it will travel by conduction and will be recovered in this [emphasis in original] way. Had I drawn this white line to scale on the basis of my Colorado experiments, it would be so thin that you would have to use a magnifying glass to see it.
For the wireless transmission of energy by the atmosphere he used two conductors,
the atmosphere and the Earth. For the wireless transmission of energy not
using the atmosphere he used only one conductor, the Earth. At least that is
how I see it in general.
Oh I am taking your hint about the vector calculus and intend to begin study
on some things.
Cheers
P.S. Now I forgot something. While it may be possible for radiation to be
reflected/deflected, when we talk in terms of radiation leaving a
system/device or radiation from a system/device I think it's gone by definition,
I think true radiation only travels in one direction and is lost to the system
which it originated from. Terms can be confusing, radiations can be emitted
and collected but radiation goes in all directions. Directed beams of energy or
particles are not radiations in my opinion. It's about control and intent.
If there is an effective intentional direction of waves I don't think it is radiation. Radiation - radii.
..
Last edited by Farmhand : 10-21-2012 at 05:16 PM.
10-22-2012, 03:07 AM
Ernst Senior Member Join Date: Jul 2012 Posts: 288
Quote:
I'm not looking for quotes from Tesla on radiation in particular, I'm looking for quotes by Tesla on Longitudinal "waves" or such with respect directly to the operation of the Energy Transmitters patented for the transmission of energy by the atmospheric conduction or the ground currents.
Yes, Farmhand, I think I understand what you want. But what I am trying to say is:
- At Tesla's time there was less knowledge about the nature of the various forms of radiation. Today we distinquish transverse, longitudinal and particle (beta: electrons, alfa: protons, for example), but you can not expect someone in Tesla's time to do the same.
- Tesla does clearly distinquish various forms of radiation, though.
- Tesla does mention which forms of radiation can NOT be used for the transmission of energy
- Tesla does describe various experiments in which energy is transmitted
I do not think that a direct quote such as you are looking for exists. The best thing we can do is look at what we DO have and reconstruct it in todays terminology.
On the experiment that you quote... That is a simulation experiment, that is why I said
Quote:
I do not remember having read that he actually uses this type of transmission in a (real) experiment.
All actual experiments that I can now think of, I believe he uses the earth to conduct the electric energy or particle beams. But there may be some experiment that either I do not know of or that escapes my memories at this time.
On vector calculus: this
was the (first of three) video that I thought is very helpful for visualising the concepts of div, grad and curl. Once you can visualise this, try to imagine the universe filled with ether and currents running through wires dragging ether outside the wire along. You can then see the vortices of magnetism appear. And of course the other way around.
It will also become clear why a constant current induces a constant magnetic field, while a constant magnetic field does not induce potential differences and therefor no current.
You need a change in magnetic 'speed' (either faster or slower) in order to create a potential difference.
Then if you want more try to find a copy of "Div, grad, curl and all that". It is easy to read and more or less "all there is to it".
Good luck!
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September 16, 2009, 16:09 udf define property on a wall surface #1 New Member beibei Join Date: Aug 2009 Posts: 6 Rep Power: 9 I need to define aborption coefficient as a function of position x on a wall surface (pipe). FLUENT didn't seen to render it even though it could interprete. What could it be wrong? Couldn't fluent render udf on a surface? Thanks! Here's my code: #include"udf.h" DEFINE_PROPERTY(absorption_coefficient, c, t) { real abs_coeff; real x[ND_ND]; C_CENTROID(x,c,t); if (x[0] > 0 && x[0] < 0.2) abs_coeff = 593.9187; elseif (x[0] > 0 && x[0] < 0.2) abs_coeff = 197.6521; elseif (x[0] > 0 && x[0] < 0.2) abs_coeff = 177.0521; elseif (x[0] > 0 && x[0] < 0.2) abs_coeff = 140.5854; else abs_coeff = 140.5854; Message("Abs coefficient for this cell %e = %e", x[0],abs_coeff); return abs_coeff; }
November 27, 2011, 12:10 udf #2 New Member arash Join Date: Jun 2011 Posts: 7 Rep Power: 8 hi, can i modelling asphaltene precipitation that exist in crude oil by uusing ceramic filter monolit by defult formolation in fluent? i want to modelling separation of asphaltene from crude oil by 19-channel ceramic filter,but i think it is not define in FLUENT that precipitate asphaltene on surface of ceramic channel base on diffrences between pore size. i must write UDF for this porpuse or not? most of oil flow must pass through the membrane wall and most of asphaltene must form a gel-layer and precipitation on inner surface od channel. please help me as soon as possible thank you so much
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12th October 2003, 01:19 AM #1 tubetvr diyAudio Member Join Date: Jan 2003 Location: Sweden How much output offset voltage is too much? Hi I am experimenting with different amplifier circuits and have experienced that output offset voltage can be quite different in different circuits, also some circuits drift with temperature so even if offset is very low initially it can increase quite a lot when the amplifier gets warm Now my question, how much offset is too much? It is obvious that if offset is very high, i.e. several volts most electrodynamic speakers will react and the speaker membrane will be in a offset position either out or in from the resting position. In this case either the voice coil will go to the end position or the speaker will give excessive distorsion as the voice coil is not centered in the magnet gap. So how much offset is tolerable? 50mV, 100mV, 200mV, 500mV or 1V or even more? Regards Hans
12th October 2003, 01:42 AM #2 Tube_Dude diyAudio Member Join Date: Mar 2002 Location: Aveiro-Portugal Hi Hans... For me , my target is less than 50 mV at the output...with the normal +- 5 Ohms speaker DC resistence ...10 mA is the of set current circulating in the voice coil... But some people can be less stringent... __________________ Jorge
12th October 2003, 01:29 PM #3 Circlotron diyAudio Member Join Date: Jun 2002 Location: Melbourne, Australia My class A has a constant 325mV at the output. No problemo. Zero. __________________ Best-ever T/S parameter spreadsheet. http://www.diyaudio.com/forums/multi...tml#post353269
12th October 2003, 04:45 PM #4 Richard C diyAudio Member Join Date: Jul 2003 Location: Nottingham, England I'd go for less than 50mV if possible. Moving coil loudspseakers are only reasonably linear over a few milimeters of cone travel so it pays to operate with as little offset as possible.
fmak
diyAudio Member
Join Date: May 2001
Location: London UK
Quote:
Originally posted by Circlotron My class A has a constant 325mV at the output. No problemo. Zero.
--------------------------------------------------
This is way too high and biases your speaker cone popsition.
----------------------------------------------------
I'd go for less than 50mV if possible. Moving coil loudspseakers are only reasonably linear over a few milimeters of cone travel so it pays to operate with as little offset as possible. [/QUOTE]
----------------------------------------------------
I agree; I'd go for <25mV and my systems have less than 10mV.
12th October 2003, 05:57 PM #6 peranders Electrons are yellow and more is better! diyAudio Member Join Date: Apr 2002 Location: Göteborg, Sweden I think this question has a rather floating answer. Take a small 10 W speaker and a huge PA speaker. I'll guess that over 1 volt isn't good if we talk distortion but under 100 mV is good I think and normal also. __________________ /Per-Anders (my first name) or P-A as my friends call me Group buy: SSR03 Super Regulator Power Supply. Sign up HERE 53 pcb's in interest so far.
EchoWars
diyAudio Member
Join Date: Dec 2002
Location: Left of the Dial
Quote:
Originally posted by peranders I think this question has a rather floating answer.
Probably sums it up. It's a personal thing. On amps I'm repairing, anything over 50mV warrrants investigation. On my personal amps, anything over 15mV needs attention. I like to have them at <2mV.
12th October 2003, 11:31 PM #8 markp diyAudio Member Join Date: Oct 2003 Location: L.A., CA My amps are within 5mv without a servo. If you have lots of offset then try to match your transistors better or build a servo to control it. __________________ If it sounds good... it is good!
13th October 2003, 12:09 AM #9 EchoWars diyAudio Member Join Date: Dec 2002 Location: Left of the Dial I'm with you Mark. But the problem with matching is that they aren't going to stay matched. Check your amp in a few years. I have only recently been introduced to the whole servo thing...I like. Long-term stability, no muss, no fuss. I do believe I am a convert.
fmak
diyAudio Member
Join Date: May 2001
Location: London UK
Quote:
Originally posted by EchoWars Probably sums it up. It's a personal thing. On amps I'm repairing, anything over 50mV warrrants investigation. On my personal amps, anything over 15mV needs attention. I like to have them at <2mV.
--------------------------------------------------
I agree with your numbers. However, this is more than a personal thing and doesn't dpend on the size of the speakers but on sensitivity and cone displacement.
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All times are GMT. The time now is 06:01 PM. | 1,676 | 7,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-47 | latest | en | 0.887277 |
https://wiki.lesswrong.com/index.php?title=Logical_uncertainty&diff=next&oldid=15358 | 1,558,432,329,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00208.warc.gz | 669,838,939 | 7,603 | # Difference between revisions of "Logical uncertainty"
Logical uncertainty applies the rules of probability to logical facts which are not yet known to be true or false.
Is the googolth digit of pi odd? The probability that it is odd is, intuitively, 0.5. Yet we know that this is definitely true or false by the rules of logic, even though we don't know which. Formalizing this sort of probability is the primary goal of the field of logical uncertainty.
The problem with the 0.5 probability is that it gives non-zero probability to false statements. If I am asked to bet on whether the googolth digit of pi is odd, I can reason as follows: There is 0.5 chance that it is odd. Let P represent the actual, unknown, parity of the googolth digit (odd or even); and let Q represent the other parity. If Q, then anything follows. (By the Principle of Explosion, a false statement implies anything.) For example, Q implies that I will win \$1 billion. Therefore the value of this bet is at least \$500,000,000, which is 0.5 * \$1,000,000, and I should be willing to pay that much to take the bet. This is an absurdity. Only expenditure of finite computational power stands between the uncertainty and 100% certainty.
Logical uncertainty is closely related to the problem of counterfactuals. Ordinary probability theory relies on counterfactuals. For example, I see a coin that came up heads, and I say that the probability of tails was 0.5, even though clearly, given all air currents and muscular movements involved in throwing that coin, the probability of tails was 0.0. Yet we can imagine this possible impossible world where the coin came up tails. In the case of logical uncertainly, it is hard to imagine a world in which mathematical facts are different.
## References
Questions of Reasoning Under Logical Uncertainty by Nate Soares and Benja Fallenstein. | 416 | 1,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-22 | longest | en | 0.941058 |
https://nl.mathworks.com/matlabcentral/answers/1642475-how-to-find-the-nearest-value-if-a-value-is-already-provided-from-a-given-array | 1,660,402,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571959.66/warc/CC-MAIN-20220813142020-20220813172020-00189.warc.gz | 395,914,126 | 24,221 | how to find the nearest value (if a value is already provided) from a given array?
1 view (last 30 days)
Reana Taylor on 3 Feb 2022
Answered: David Hill on 3 Feb 2022
I want to find the nearest value of x in the array m
x=0.61;
m=linspace(0,1,9);
meshs = 1×9
0 0.1250 0.2500 0.3750 0.5000 0.6250 0.7500 0.8750 1.0000
We know the nearest value is 0.6250, but i would like a code to do this for me, for any value of x.
Yongjian Feng on 3 Feb 2022
Try this:
x=0.61;
m=linspace(0,1,9);
[a, b] = min(abs(m-x));
m(b)
David Hill on 3 Feb 2022
[~,idx]=min(abs(m-x));
M=m(idx);
R2020b
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Start Hunting! | 256 | 703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-33 | latest | en | 0.801522 |
https://im.kendallhunt.com/K5_ES/teachers/kindergarten/unit-4/lesson-16/preparation.html | 1,718,906,828,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00518.warc.gz | 272,531,487 | 26,485 | Lesson 16
Encontremos el valor de expresiones
Lesson Purpose
The purpose of this lesson is for students to find the value of addition and subtraction expressions in a way that makes sense to them.
Lesson Narrative
In previous lessons, students interpreted expressions and connected expressions to story problems and drawings. This is the first lesson where students begin by working with only expressions. Because students have matched expressions to drawings in previous lessons, students may create a drawing to find the value of the expression. Students may also use their fingers or objects to represent the expression and count to find the total or difference.
• Action and Expression
• MLR8
Learning Goals
Teacher Facing
• Find the value of addition and subtraction expressions.
Student Facing
• Encontremos el valor de expresiones.
Required Materials
Materials to Gather
Materials to Copy
• Roll and Add Stage 2 Recording Sheet
• Number Mat 1–5
Required Preparation
Activity 3:
• Gather materials from:
• Roll and Add, Stages 1 and 2
• Shake and Spill, Stages 1-3
• Number Race, Stage 1
• Math Stories, Stages 1 and 2
Lesson Timeline
Warm-up 10 min Activity 1 10 min Activity 2 10 min Activity 3 20 min Lesson Synthesis 5 min Cool-down 5 min
Teacher Reflection Questions
In a previous unit, students represented numbers in multiple ways, including using their fingers, objects, and drawings. How did students work with representing numbers prepare them to find the value of expressions in this lesson? | 332 | 1,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.864124 |
http://blogs.oregonstate.edu/geog566spatialstatistics/2018/05/10/multivariate-analysis-location-behavior-changes/ | 1,558,434,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00299.warc.gz | 31,374,950 | 8,881 | # GEOG 566
May 10, 2018
### Multivariate analysis of the location of behavior changes
Filed under: Exercise/Tutorial 2 2018 @ 1:25 pm
Question:
Because the spatial distribution of behavior changes doesn’t change with boom angle, the results of Exercise 1 imply that hydraulic conditions do not drive fish behavior in this experiment (Figure 1). However, because analogous research and common sense implies that thresholds of hydraulics indeed affect fish behavior, a statistical analysis is necessary to determine if environmental and/or internal factors did affect the location of behavior changes we observed. Five hydraulic variables – water speed (m/s), turbulent kinetic energy (or TKE, m2/s2), TKE gradient (m2/s2/m), velocity gradient (m/s/m, or s-1), and acceleration (m/s2) – were drawn from the locations of every behavior change in Exercise 1. Then, the locations of behavior changes are compared with channel hydraulics, boom angle, and a fish’s visual fitness (as measured by an optomotor assay) using three methods: 1) multivariate regression analysis, 2) principal component analysis (PCA), and 3) a partial least squares regression. With this analysis, we hope to answer the question: do any of 5 hydraulic variables, the geometry of the channel, or the visual fitness of a fish correlate well with the location of its behavior change?
##### Figure 1. The results of Exercise 1 indicate that despite differences in hydraulics created by varying boom angle, the spatial distribution of location changes is not observed the change between boom angles. Contour lines show two-dimensional 95% confidence intervals.
Method and steps for analysis:
Three methods of regression analysis were used to answer the question above. Although previous work was conducted in Python, R is better suited for complex statistical analyses. First, a multivariate regression examined the correlation between the locations of behavior change and the independent variables. A multivariate regression enables the analysis of more than one dependent variable – in this case, the X- and Y-coordinates of a behavior change. This is not to be confused with multiple linear regression, which only analyzes the correlation of predictor variables with one dependent variable (i.e. just X or just Y). Interestingly, angle, water speed, and velocity gradient show significant, positive correlations with location (where positive locations are downstream and against the left channel wall; Figure 2). Acceleration and TKE gradient, on the other hand, show significant negative correlations with location of behavior change. TKE (red box) shows no significant correlation, nor does visual fitness (blue box). Clearly, the high correlation of hydraulics warrants further analyses to better understand which hydraulic variable, if any, truly influences behavior changes.
##### Figure 2. The results of a multivariate regression analysis of hydraulic, geometric, and visual variables on the locations of behavior change.
Principal component analysis is one method of identifying important variables (in the form of components) from a larger set of variables, especially when they’re highly correlated. A principal component is a linear combination of input variables that explains the variation in the original data. By quantifying the amount of variation of each principal component, an idea of influential variables can be grasped. In the case of X (the downstream position of a behavior change) the first principal and second principal components account for over 85% of the variation observed (Figure 3). Within Principal Components 1 and 2, boom angle, water speed, and velocity gradient influence the dependent variable, X, to the greatest degree, indicated by the length of arrow in Figure 4. Again, visual fitness, TKE, and now TKE gradient lack influence on X. Although promising, the results of PCA are limited by the analysis of only one dependent variable. A partial least squares regression allows a similar investigation into the principal components of behavior changes locations in the X and Y dimensions.
##### Figure 4. The contribution (as indicated by length of arrows) of independent variables to Principal Components 1 and 2. Variables near the bottom, top, and left of the graph have more influence than those near the apex of the arrows.
Partial least squares regression, or PLS regression, combines principal components of PCA and linear regression of multivariate regression. It benefits our data because multiple dependent variables may be analyzed for principal components of many correlated predictor variables. However, the results of PLS perhaps realize our original fear – that no consistent hydraulic variable emerges as a strong predictor of the location of behavior change in our experiment (Figure 5). Instead, TKE gradient now dominates Axis 1 (analogous to Principal Component 1), while velocity gradient and angle largely influence Axis 2. A PLS regression within boom angle failed to identify consistent hydraulic or visual variables that dominate the analysis’s axes.
##### Figure 5. Partial least squares regression of independent hydraulic, visual, and geometric variabls on X and Y, the locations of behavior change. TKE gradient now dominates Axis 1, while velocity gradient and boom angle influence Axis 2 the greatest.
Results:
No hydraulic or visual variable measured in this experiment consistently predicted the locations of behavior changes observed in this experiment. If anything, boom angle most consistently dominates the axes, principal components, and correlations of PLS, PCA, and multivariate regressions. This implies channel geometry, independent of the hydraulics it created, had the largest influence on fish behaviors as we observed them. Taken at face value, this result seems illogical. However, it may indicate that a bias existed in the behavior changes we observed.
Critique of methods:
Multivariate regression analysis easily identifies correlations of independent variables with more than one response variable. However, highly correlated independent variables require PCA or PLS to explain variation amongst many variables. Although powerful, these analyses are more difficult to interpret (in the case of PCA and PLS) and unable to investigate more than 1 response variable (in the case of PCA). | 1,229 | 6,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-22 | latest | en | 0.9006 |
https://www.physicsforums.com/threads/speed-of-light.510850/ | 1,527,424,465,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00536.warc.gz | 787,740,981 | 17,319 | # Speed of light .
1. Jun 30, 2011
### quantizedzeus
Speed of light.....
It makes no sense to me that if the speed of light is always c......then in different mediums why this speed changes......?? I may have got it in a wrong way.....But can anyone help me out in detail......(thanks)
2. Jun 30, 2011
### Staff: Mentor
Re: Speed of light.....
The speed of light in a vacuum is always c.
3. Jun 30, 2011
### phinds
Re: Speed of light.....
Does it puzzle you that the speed of sound is different in air than in water?
4. Jun 30, 2011
### ZapperZ
Staff Emeritus
Re: Speed of light.....
Zz.
5. Jun 30, 2011
### srivi
Re: Speed of light.....
The speed of light is constant in vacuum and it changes in different mediums.
when light enters a denser medium (like from air to glass) the speed and wavelength of the
light wave decrease while the frequency stays the same.
How much light slows down depends on the new medium's refractive index, n.
6. Jul 3, 2011
### kevinfrankly
Re: Speed of light.....
speed of light (c = 3 x 10^8) is an constant while on vacuum
but c vary in other medium depend on its density. => n = c/v while n = refractive index.
higher refractive index means less density.
CMIIW
7. Jul 3, 2011
### danR
Re: Speed of light.....
Seems perfectly natural to me. But don't just say the speed changes. It only gets slower in a medium.
But the fact is that light is constantly absorbed and re-emitted in a medium. The light that goes out isn't the light that came in.
When a light-particle bangs into the electron cloud in an atom in a medium, it dies and is disappeared, and an impostor steals its clothes and runs out the same direction a bit later. Everyone is fooled. Even the cops.
8. Jul 5, 2011
### Keyur
Re: Speed of light.....
Speed of light is slower in all mediums except VACUUM.
v=frequency x wavelength
v changes but wavelength doesn't.
as v changes this causes refraction when light enters a different medium.
9. Jul 6, 2011
### danR
Re: Speed of light.....
I wouldn't include a vacuum as a medium, though. And strictly speaking, does c remain as a constant even with a medium, between re-emissions and re-absorptions?
10. Jul 6, 2011
### danR
Re: Speed of light.....
After reading the FAQ on transmission through a medium, I have to abandon the simple adsorbtion, re-emission model, although the wording at a key point in that article is a bit confusing.
11. Jul 8, 2011
### kevinfrankly
Re: Speed of light.....
so the light particle that come in will stay there ?? or what ?
12. Jul 17, 2011
### piroman
Re: Speed of light.....
If the light bulb is in the midle of the room and you turn it on the light reaches south wall same time as north wall. Does that mean you multiply the speed of light times 2?
13. Jul 17, 2011
### Staff: Mentor
Re: Speed of light.....
No, it doesn't. Speed is measured from one object to another. Not from one object to two others at the same time. That's not speed.
14. Jul 18, 2011
### Drakkith
Staff Emeritus
Re: Speed of light.....
You need to read up on what light is. Hit up wikipedia and the FAQ here on PF. | 831 | 3,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-22 | longest | en | 0.90813 |
https://www.physicsforums.com/threads/basis-nullspace-linear-transformtion.648059/ | 1,508,276,306,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00124.warc.gz | 959,096,004 | 16,419 | # Basis, Nullspace, Linear transformtion
1. Oct 29, 2012
### pyroknife
The problem is attached, I did parts 1-3, but I am having trouble with part 4.
This is what i was planning on doing for part 4 (my teacher said this wasn't the correct method):
set T(v)=0
and solve the augmented matrix
1 0 -1 1 0
2 1 -2 4 0
3 1 -1 7 0
rref gives
1 0 0 2 0
0 1 0 2 0
0 0 1 1 0
So N(T)=span{[-2 -2 -1 1]^t}
So shouldn't a basis for N(T) be [-2 -2 -1 1]^t?
I don't understand how this is wrong.
#### Attached Files:
• ###### Untitled.png
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Views:
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Last edited: Oct 29, 2012
2. Oct 30, 2012
### Fredrik
Staff Emeritus
I don't see anything wrong with it either. What you're doing is to solve the equation
$$T\begin{bmatrix}a\\ b\\ c\\ d\end{bmatrix}=0.$$ Maybe your teacher just thought that this was a bad idea compared to solving
$$T\begin{bmatrix}2b-2c\\ b\\ c\\ -c\end{bmatrix}=0,$$ where two of the unknowns have already been eliminated because we're using what we know about the null space of A. But I get the same result with both methods, so I can't say that the second one is significantly better.
3. Oct 30, 2012
### pyroknife
I talked to him again and he said "the way I did it, I was finding all of the vectors in R4 such that T(x)=0. The question is asking to find all vectors in V such that T(x)=0 which is a different question."
The basis found in part 1 is {[2 1 0 0] [2 0 -1 1]}.
The linear transformation of these two vectors is
[2 5 7]^t and [4 10 14]^t respectively.
Te matrix for this is
2 4
5 10
7 14
Rref is
1 2
0 0
0 0
The coordinate vector that spans the null space is [2 -1]^t
So the basis is 2[2 1 0 0] + 1[2 0 -1 1]= [-2 -2 -1 1]
Wow that got really tricky a d conplicated
4. Oct 31, 2012
### Fredrik
Staff Emeritus
Right, that's what I meant. The thing is, both methods give us the same result, so what you did can't be completely wrong. I think the only problem is this: When you do it your way, you don't automatically know that the vector you find is actually a member of V (= the null space of A), so you also have to verify that it is.
5. Oct 31, 2012
### Fredrik
Staff Emeritus
I don't understand exactly what you're doing here. I think it may be an advantage for me that I don't remember the fancy ways of doing these things. All I did was this: You know from part 1 that any member of V can be written as a linear combination of the two basis vectors you found: a[2 1 0 0]T+b[2 0 -1 1]T =[2a+2b a -b b]T So now you just need to solve T[2a+2b a -b b]T=0. This gives you a relationship between a and b. Use that relationship and set b=1, or whatever is convenient. | 862 | 2,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.949977 |
http://blog.csdn.net/wu_yihao/article/details/17115211 | 1,505,945,406,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687484.46/warc/CC-MAIN-20170920213425-20170920233425-00424.warc.gz | 42,708,830 | 16,664 | 547人阅读 评论(0)
# charge-station
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 876 Accepted Submission(s): 459
Problem Description
There are n cities in M^3's empire. M^3 owns a palace and a car and the palace resides in city 1. One day, she wants to travel around all the cities from her palace and finally back to her home. However, her car has limited energy and can only travel by no more than D meters. Before it was run out of energy, it should be charged in some oil station. Under M^3's despotic power, the judge is forced to build several oil stations in some of the cities. The judge must build an oil station in city 1 and building other oil stations is up to his choice as long as M^3 can successfully travel around all the cities.
Building an oil station in city i will cost 2i-1 MMMB. Please help the judge calculate out the minimum cost to build the oil stations in order to fulfill M^3's will.
Input
There are several test cases (no more than 50), each case begin with two integer N, D (the number of cities and the maximum distance the car can run after charged, 0 < N ≤ 128).
Then follows N lines and line i will contain two numbers x, y(0 ≤ x, y ≤ 1000), indicating the coordinate of city i.
The distance between city i and city j will be ceil(sqrt((xi - xj)2 + (yi - yj)2)). (ceil means rounding the number up, e.g. ceil(4.1) = 5)
Output
For each case, output the minimum cost to build the oil stations in the binary form without leading zeros.
If it's impossible to visit all the cities even after all oil stations are build, output -1 instead.
Sample Input
3 3
0 0
0 3
0 1
3 2
0 0
0 3
0 1
3 1
0 0
0 3
0 1
16 23
30 40
37 52
49 49
52 64
31 62
52 33
42 41
52 41
57 58
62 42
42 57
27 68
43 67
58 48
58 27
37 69
Sample Output
11
111
-1
10111011
Hint
In case 1, the judge should select (0, 0) and (0, 3) as the oil station which result in the visiting route: 1->3->2->3->1. And the cost is 2^(1-1) + 2^(2-1) = 3.
Source
Recommend
zhoujiaqi2010 | We have carefully selected several similar problems for you: 4812 4811 4810 4809 4808
#include <cstdio>
#include <cstring>
#include <cmath>
int que[2000000];
int dist[140];
bool vis[140];
bool charge[140];
int x[140];
int y[140];
int n;
int D;
int Dist(int a,int b)
{
return ceil(sqrt(double(((x[a]-x[b])*(x[a]-x[b]) + (y[a]-y[b])*(y[a]-y[b])))));
}
int min(int a,int b)
{
return a<b?a:b;
}
bool bfs()
{
int l = 0;
int r = 0;
for (int i=1;i<=n;i++)
{
vis[i] = false;
if (charge[i]) dist[i] = 0;
else dist[i] = 0x3f3f3f3f;
}
r ++;
que[r] = 1;
vis[1] = true;
while (l < r)
{
l ++;
int u = que[l];
for (int v=1;v<=n;v++)
{
int d = Dist(u,v);
if (!vis[v] && d<=D)
{
dist[v] = min(dist[v],dist[u]+d);
if (charge[v])
{
vis[v] = true;
r ++;
que[r] = v;
}
}
}
}
for (int i=1;i<=n;i++)
{
if (charge[i] && !vis[i]) return false;
if (!charge[i] && dist[i]*2>D) return false;
}
return true;
}
int main()
{
while (scanf("%d%d",&n,&D)==2)
{
for (int i=1;i<=n;i++)
charge[i] = true;
for (int i=1;i<=n;i++)
{
scanf("%d%d",x+i,y+i);
}
if (!bfs()) {printf("-1\n");continue;}
for (int i=n;i>1;i--)
{
charge[i] = false;
if (!bfs()) charge[i] = true;
}
int i = n;
while (i>1 && !charge[i]) i--;
while (i > 0)
{
printf("%d",int(charge[i]));
i --;
}
printf("\n");
}
return 0;
}
0
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最新评论 | 1,229 | 3,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-39 | longest | en | 0.925169 |
https://backroadsofamericanmusic.com/book/4109-geometry-chapter-5-resource-book-lesson-53-practice-a-answers-558-218.php | 1,606,910,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00511.warc.gz | 172,631,895 | 10,223 | # Geometry chapter 5 resource book lesson 5.3 practice a answers
8.26 · 8,476 ratings · 775 reviews
## Holt geometry lesson practice a answers
Sometimes students want an alternative explanation of an idea along with additional practice problems. The Parent Guide resources are arranged by chapter and topic. The format of these resources is a brief restatement of the idea, some typical examples, practice problems, and the answers to those problems. Download the entire Parent Guide Full Version. Professional Learning Shop. Back Why CPM?
File Name: geometry chapter 5 resource book lesson 5.3 practice a answers.zip
Size: 75345 Kb
Published 16.06.2019
## Lesson 5.3 Estimate Quotients
All Images Videos News. Period: All.
## Search results
Answers will vary. Yes; Converse of the Angle Bisector Theorem 6! Solutions to Geometry :: Free Homework Professional Learning Shop.
SAS Congruence Postulate6. View Notes - 5! YouTube - Aug 09, - All Images Videos News.
Holt geometry lesson 5. EnClave Documents. Section folders have the Powerpoint lesson notes, Lesson Practice homework. Chapter Circles and Conditional Probability. Answers will vary.
Introduction to Geometry 1. Proofs and Reasoning. The Chapter 5 Resource Mastersincludes the core materials needed for Chapter 5. These materials include worksheets, extensions, and assessment options. Which is a possible value of x? Based on the diagram, which is a true statement Use the figure shown and the given information.
### Updated
Sample answer: The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. LaRella Documents. The Triangle Inequality Theorem states that no side of a triangle ggeometry be longer than the sum of the lengths of the other two sides. Category Documents.
In order for the incenter to lie at a vertex, makingan angle of 08 which does not make a triangle. Chapter 3: Justification and Similarity. Refl exive Property of Congruence 6. Hi everyone.
Make your selection below 5. Geometry 3. All Images Videos News. Geometry Module 1, Lesson 7.
Chapter 8: Polygons and Circles. Find the value of x. Chapter 3: Justification and Similarity! All r ight s re serv ed.
## Reginald hill books dalziel and pascoe
358 books — 18 voters
Schools around the world book pdf
638 books — 57 voters
## 5 thoughts on “Holt geometry lesson practice a answers”
1. Travers P. says:
No; you need to know that the congruent segments are to the rays. Linear Pair supplementary. Statements Reasons 1. Lesson 5.
2. Christelle V. says: | 601 | 2,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-50 | latest | en | 0.826044 |
https://www.teacherspayteachers.com/Product/Graphs-and-Data-Posters-for-5th-Grade-421071 | 1,542,362,131,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00236.warc.gz | 1,001,617,183 | 19,471 | # Graphs and Data Posters for 5th Grade
Subject
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Product Rating
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PDF (Acrobat) Document File
2 MB|14 pages
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Product Description
This is a set of colorful, child-friendly, posters that you can put on your Math bulletin board or Word Wall. These posters are designed to fit the NGSSS curriculum, but will align with many Common Core standards. This set is designed to align with the series Go Math! for 5th grade, the division unit.
Personally, I keep these up for the entire unit. I start by putting the posters up several days before we start the unit. This gives students a chance to preview the content. They know to expect that I'll ask questions regarding the posters on the first day of the unit. They often refer to these (and are encouraged to do) regularly during math lessons.
The "Chapter 5" page in the PDF corresponds with Go Math. However, I added two pages at the end. One page states the learning goal as "Analyzing Graphs and Data". The last page just says "Learning Goal" and leaves space for you to write in your own.
Posters inform on:
- types of graphs: double bar, line, venn diagrams
- continuous versus discrete data
- parts of a graph: TLC (title, labels, create a scale)
- types of venn diagrams
- ordered pairs ("you learn to crawl before you walk")
- midpoint
- describing relationships
In case you are curious about the types of venn diagrams mentioned in my posters, here is an explanation. In my class, we talk about venn diagrams in rather unique ways. There is the 'Mickey Mouse' and the 'Caterpillar'. We discuss that if a word problem gives you a LOT of numbers and mentions that x-amount of people are involved in "ALL" three categories, you make a "Mickey Mouse", as described on the posters. Reason being: this diagram looks very much like our mousey friend. =) This allows you to place a number in the part of the diagram that overlaps all three categories.
On the other hand, when a word problem only uses the word "BOTH", you draw a caterpillar, allowing you to place a number in the part of the diagram that overlaps the two categories. Seems silly, but it does wonders for their understanding of which to use. They learned to circle the key words in the word problems, and immediately set up their work to reflect the appropriate diagram.
Feel free to use the pages you see fit. One size may not fit all. =)
Total Pages
14 pages
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\$4.00 | 565 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-47 | latest | en | 0.937517 |
http://excelmathmike.blogspot.com/2012/03/happy-birthday-arizona100-years-of.html | 1,508,331,418,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00736.warc.gz | 122,032,265 | 22,040 | ## Monday, March 12, 2012
### Happy Birthday, Arizona—100 Years of Statehood
Happy Birthday! This year the Oreo cookie turns 100, and Arizona celebrates its centennial as part of the United States of America.
Here's a brief timeline of some of the major events from Arizona's history. For a more complete look at the past and a schedule of centennial events, visit Arizona Highways.
In Excel Math, we teach students how to read timelines and number lines so they can avoid bread lines later in life. Let's take a look at a Projectable Lesson from Excel Math First Grade. In this example, number lines are used to help students solve subtraction problems:
Excel Math First Grade Student Sheet Lesson 79
Did you figure out how to solve number 2 on the number line?
Saguaro Cactus
Now back to our timeline and some fun facts about Arizona. It became a state in 1912, and this year Arizona celebrates 100 years of statehood. Do know its nickname? How about its state flower? (the white saguaro cactus flower, the largest cactus in the United States)
The saguaro is found in the Sonoran Desert, which is in the southwestern part of Arizona. The saguaro cactus first blooms when it is 50 to 75 years old. These cacti grow very slowly at first. It takes them about ten years to grow to be just an inch tall. The Tohono O'odham people use long sticks to pick the fruit from the saguaro. They make the fruit into jelly. (Yum!) The best place to see saguaros is in Saguaro National Park near Tucson.
Arizona is also called the Grand Canyon State. The Grand Canyon is a mile deep and more than 200 miles long. At its widest point, it is 18 miles wide.
President Theodore Roosevelt called it "the one great sight which every American should see." This is a view of the Grand Canyon from the South Rim:
Grand Canyon
The canyon was formed by the Colorado River. This photo shows some of the many rock layers carved out over the years by erosion from rain, snow, and streams:
Grand Canyon
Another beautiful area of Arizona, Sedona is often called "red rock country." The red sandstone and manzanita trees are two of its signature landmarks. Sedona is surrounded by red-rock monoliths named Coffeepot, Cathedral and Thunder Mountain. At the north end of the city is the stunning Oak Creek Canyon, a breathtaking chasm.
Sedona
Arizona has 20 Indian reservations that cover about one-fourth of the state. The Navajo Reservation, in the northeastern part of the state, is the largest in the country. This is a vista from an overlook on one of the Navajo Reservations located outside Grand Canyon National Park:
Navajo Reservation
Excel Math is alive and well in Arizona. The latest Arizona Mathematics Standards include the Common Core State Standards plus Arizona additions. Excel Math is fully correlated to the Common Core Standards and to the Arizona State Standards. Here's what one principal from Arizona has to say about Excel Math:
"The teachers are very pleased with the Excel Math program and with what they are able to learn about their students’ math abilities. We are very encouraged about the benefits of using this program."
—Susan, Principal from Phoenix, Arizona
With Excel Math, students learn higher-order thinking skills beyond what is required of these standards. Download the Arizona correlations (or those of your state) to see how the Excel Math Lessons, Stretches, Activities, and Exercises for each grade level correlate to those educational standards. At the bottom of each correlation, we list additional concepts covered by Excel Math. You can use these additional concepts to provide accelerated learning for your students who are ready for more.
Happy Birthday, Arizona! Here's to another 100 candles on your birthday cake. | 807 | 3,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-43 | longest | en | 0.949774 |
https://codegolf.stackexchange.com/questions/57617/is-this-number-a-prime?page=4&tab=scoredesc | 1,718,991,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00424.warc.gz | 138,919,960 | 88,897 | # Is this number a prime?
Believe it or not, we do not yet have a code golf challenge for a simple primality test. While it may not be the most interesting challenge, particularly for "usual" languages, it can be nontrivial in many languages.
Rosetta code features lists by language of idiomatic approaches to primality testing, one using the Miller-Rabin test specifically and another using trial division. However, "most idiomatic" often does not coincide with "shortest." In an effort to make Programming Puzzles and Code Golf the go-to site for code golf, this challenge seeks to compile a catalog of the shortest approach in every language, similar to "Hello, World!" and Golf you a quine for great good!.
Furthermore, the capability of implementing a primality test is part of our definition of programming language, so this challenge will also serve as a directory of proven programming languages.
Write a full program that, given a strictly positive integer n as input, determines whether n is prime and prints a truthy or falsy value accordingly.
For the purpose of this challenge, an integer is prime if it has exactly two strictly positive divisors. Note that this excludes 1, who is its only strictly positive divisor.
Your algorithm must be deterministic (i.e., produce the correct output with probability 1) and should, in theory, work for arbitrarily large integers. In practice, you may assume that the input can be stored in your data type, as long as the program works for integers from 1 to 255.
### Input
• If your language is able to read from STDIN, accept command-line arguments or any other alternative form of user input, you can read the integer as its decimal representation, unary representation (using a character of your choice), byte array (big or little endian) or single byte (if this is your languages largest data type).
• If (and only if) your language is unable to accept any kind of user input, you may hardcode the input in your program.
In this case, the hardcoded integer must be easily exchangeable. In particular, it may appear only in a single place in the entire program.
For scoring purposes, submit the program that corresponds to the input 1.
### Output
Output has to be written to STDOUT or closest alternative.
If possible, output should consist solely of a truthy or falsy value (or a string representation thereof), optionally followed by a single newline.
The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation.
• This is not about finding the language with the shortest approach for prime testing, this is about finding the shortest approach in every language. Therefore, no answer will be marked as accepted.
• Submissions in most languages will be scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8.
The language Piet, for example, will be scored in codels, which is the natural choice for this language.
Some languages, like Folders, are a bit tricky to score. If in doubt, please ask on Meta.
• Unlike our usual rules, feel free to use a language (or language version) even if it's newer than this challenge. If anyone wants to abuse this by creating a language where the empty program performs a primality test, then congrats for paving the way for a very boring answer.
Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.
• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Headsecks or Unary), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.
• Built-in functions for testing primality are allowed. This challenge is meant to catalog the shortest possible solution in each language, so if it's shorter to use a built-in in your language, go for it.
• Unless they have been overruled earlier, all standard rules apply, including the http://meta.codegolf.stackexchange.com/q/1061.
As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalog as complete as possible. However, do primarily upvote answers in languages where the author actually had to put effort into golfing the code.
### Catalog
The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.
## Language Name, N bytes
where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:
## Ruby, <s>104</s> <s>101</s> 96 bytes
If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:
## Perl, 43 + 2 (-p flag) = 45 bytes
You can also make the language name a link which will then show up in the snippet:
## [><>](http://esolangs.org/wiki/Fish), 121 bytes
<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 57617; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>
• Is there a reason for the full program requirement, rather than allowing the full range of default input types? E.g. answering with a function that takes its input as an argument, is currently disallowed? codegolf.meta.stackexchange.com/questions/2447/… Commented Dec 12, 2017 at 6:21
• @LyndonWhite This was intended as a catalog (like “Hello, World!”) of primality tests, so a unified submission format seemed preferable. It's one of two decisions about this challenge that I regret, the other being only allowing deterministic primality tests. Commented Dec 12, 2017 at 12:51
• Could a case be made for locking this challenge and posting a new, less restrictive one? Commented Jun 25, 2018 at 12:59
• @Shaggy Seems like a question for meta. Commented Jun 25, 2018 at 13:44
• Yeah, that's what I was thinking. I'll let you do the honours, seeing as it's your challenge. Commented Jun 25, 2018 at 13:45
# Logicode, 601511 467 bytes
circ o(n)->cond n<->0+n/o(n>)
circ p(n)->[
cond n->var c=~((~(o(n)))>)/var c=0
cond (~n)<->var d=p(c)+0/var d=c+1
d
]
circ q(n)->[
cond n->var e=~((~(o(n)))>)/var e=0
cond (~n)<->var f=e+0/var f=q(e)+1
f
]
circ r(a,b)->cond *a&*b->r(q(a),q(b))/a
circ s(a,b)->!(*(r(b,a)))
circ t(a,b)->cond b->t(q(a),q(b))/a
circ u(a,b)->cond s(a,b)->u(t(a,b),b)/!(*a)
circ v(a)->[
var j=p(j)
cond s(a,j)->[
var k=k+u(a,j)
var l=v(a)
]/[
var k=k
]
*((~k)>)
]
var j=1
var k
out v(binp)
Oh my, that is a lot of code.
Here's a rundown of what each circuit does:
• o is a trimmer, and it strips the extra 0's at the start (this is used for p and q.
• p is a successor, and q is a decrement.
• r is a preliminary bit to s (lessthan).
• s is the lessthan (which uses the remainder checker).
• t is a subtractor, which calculates a - b for any two positive integers a and b (in binary).
• u is a mod checker, which returns 1 if a%b is not 0, and 0 if it is 0.
• v is the actual prime checker, which returns 1 if the number is not prime, and 0 if the number is.
• The j and k at the bottom are the divisor (b in the mod checker) and the un-bool'd output respectively.
The final line is the "input" bit, which asks for user input in binary (any other character that is not 0 or 1 in the input will be ignored), and returns 1 if the result is not prime, and 0 if the result is.
## Dyalog APL, 10 9 bytes
2=∘≢∘∪⍳∨⊢
I don't think this one has been posted.
# Tcl, 78 75 bytes
Thanks to sergiol
if $argv<2 {exit 1} incr d while {[incr d]<$argv} {if $argv%$d==0 {exit 1}}
Original version
if {$argv<2} {exit 1} incr d while {[incr d]<$argv} {if {$argv%$d==0} {exit 1}}
Works for all integer values (both negative and arbitrarily large). However, as this aims to be short and not efficient, it is written with a simple divisor={2,3,4,...} loop, so you'll begin getting noticeable lag around eight digit numbers (n ≥ 108).
The input is taken on the command-line; the output is an exit code: 0 for prime and 1 for not prime.
On Windows you can use the following batch file to test it:
@echo off
tclsh a.tcl %1
if ERRORLEVEL 1 (
echo not prime
) else (
echo prime
)
Use it as:
C:\foo> run.bat 2017
prime
On *nixen you can use the following bash script to test it:
#! /bin/sh
tclsh a.tcl $1 if [$? -eq 0 ]
then
echo prime
else
echo not prime
fi
Use it as:
% ./run.sh 2017
prime
Enjoy!
• You can save some bytes: tio.run/##K0nO@f8/… Commented Oct 23, 2017 at 0:40
• Please be careful to test stuff before suggesting edits. ! has higher precedence than %, so the ==0 remains... Other edits made from your suggestion. Thanks! Commented Oct 23, 2017 at 1:49
• I tested it on my local machine and it seemed to work Commented Oct 23, 2017 at 9:18
• Nobody else is testing the validity of the inputs, so you can remove the 1. line. Commented Nov 18, 2023 at 19:41
# Symbolic Python, 48 bytes
___=_
__("__=_/_"+";_=-~_;__*=_*_"*~-_)
_=__%___
Try it online!
Uses Wilson's theorem, i.e. that $$\(n-1)!^{2} \% n = 1\$$ if $$\n\$$ is a prime, otherwise it equals $$\0\$$.
### Explanation:
___=_ # Save the value of input to ___
__( ) # Evaluate string
"__=_/_" # Initialise __ as 1
+"; "*~-_ # Then repeat n-1 times
_=-~_; # Decrement n
__*=_*_ # Multiply __ by n squared
_=__%___ # Output the value of __ modulo ___
# Alchemist, 126 bytes
_->In_a
a->b+c
0e+b+c->g+h
f+0c+b->e+b
e+0h+c->f+c
f+0b+c->c
0e+0f+g->b
0f+h+0s->c
0f+0a+0g+0h+c->f
f+0b+0c+h->s
s+0h->Out_"1"
Try it online!
Outputs 1 for primes, and nothing for composite numbers. Here's a script that tests the program against numbers below 100.
This is rather inefficient, as I removed a restriction that prevented one rule undoing another rule for a sweet one byte save (worth it!). To be much more efficient, we can replace the 0e in the third rule with f and add f to the other side too. Try it online!
### Explanation:
_->In_a # Create the input number of a atoms
a->b+c # Convert all them to b and c atoms
# b will serve as a permanent save of the input
# c will be a counter starting at the input
0f+0a+0g+0h+c->f # Decrement c and start the main loop by setting the f flag
# Note that f and 0e are mostly interchangeable, and the same with 0f and e
0e+b+c->g+h # Subtract the counter from the input, keeping a copy of both
f+0c+b->e+b # When the counter runs out, set the e flag
0f+h+0s->c # Move the temporary counter atoms back to the main counter
e+0h+c->f+c # Once done, set the f flag again
f+0b+c->c # If the counter isn't 0 when the input reaches 0
# Then the input is not divisible by the counter
0e+0f+g->b # Convert the temporary input atoms back to the main input
# Reuse the temp counter to main counter rule again
# Once both are done, start the loop over again, decrementing the counter
# This continues until:
f+0b+0c+h->s # If both the counter and the input reach 0 at the same time
# Then the input is currently divisible by the counter
# Decrement the temporary counter atom
s+0h->Out_"1" # If the temp counter is 0 after that (i.e. the highest factor is 1)
# Print 1
# Java, 108 bytes
interface P{static void main(String[]a){long l=new Long(a[0]),i=1;for(;0<l%++i%l;);System.out.print(l==i);}}
Try it online!
Port of my Ink answer. Would beat all existing answers in C#, Python (2 and 3), and possibly other languages if ported.
# Ungolfed
interface PrimeChecker {
// Unlike members of classes, members of interfaces are public by default.
static void main(String[] args) {
long input = new Long(args[0]),
div = 1;
for (; 0 < ( input % (++div) // Trial division, finish when div divides input
% input /* If input is at least 2, this has no effect.
But (1 % n) = 1 for any n > 1
so without this, the loop would never end
when the input is 1 */
);
) { /* The loop body is empty, we're just using it to set div */ }
// div is now the lowest number greater than 1 that divides the input
// (or 2, if the input is 1)
// The input is prime iff that number is equal to the input
System.out.println(input == div);
}
}
• Note for those who, like me, didn't understand the existence of the last %l at first, it's to shortcut the loop when the input is 1. Commented May 16, 2019 at 15:42
# Taxi, 1238 bytes
Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.Pickup a passenger going to Cyclone.Pickup a passenger going to The Underground.Go to The Underground:n 2 r 2 r.Switch to plan q i.[l]Pickup a passenger going to Cyclone.Go to Zoom Zoom:n 3 l 2 r.Go to Cyclone:w.Pickup a passenger going to Divide and Conquer.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to Divide and Conquer.Go to Sunny Skies Park:n 1 r.Go to Divide and Conquer:n 1 r 1 r 2 r 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:e 1 l 1 l 2 l.Pickup a passenger going to The Underground.Pickup a passenger going to Equal's Corner.Pickup a passenger going to Trunkers.Go to Trunkers:s 1 l.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:w 1 l.Switch to plan c i.Go to The Underground:n 3 r 1 r 2 l.Switch to plan p i.[q]0 is waiting at Writer's Depot.[p]1 is waiting at Writer's Depot.Go to Writer's Depot:n 3 l 2 l.Pickup a passenger going to Post Office.Go to Post Office:n 1 r 2 r 1 l.[c]Go to Sunny Skies Park:n.Pickup a passenger going to Cyclone.Go to The Underground:n 1 r 1 r 2 r.Switch to plan l.
Try it online!
With many ideas from this answer, but 1 character shorter using the same shortening techniques.note This takes a slightly different approach so that “The Underground” is the return point for each iteration instead of “Zoom Zoom”. Even though this ends up only 1 character shorter, it's actually one whole instruction shorter! The reduction is relatively little because there's 2 more turns and “The Underground” is mentioned more.
• Removing all quotes around plans and strings. As long as there is no space or other special character they are not needed.
• For the Switch to plan "name" if no one is waiting the only check is whether there is any text after "name". This means that the if no one is waiting part can be replaced with i.
• If we did use quotes like the original answer we could also remove the spaces after closing quotes.
This is the code ungolfed with some comments.
[Read number and convert to int]
Go to Post Office: west 1st left, 1st right, 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left, 1st right.
[Duplicate to get numerator and denominator]
Pickup a passenger going to Cyclone.
Go to Cyclone: north 1st left, 1st left, 2nd right.
[Decrement denominator and duplicate both]
Pickup a passenger going to Cyclone.
Pickup a passenger going to The Underground.
Go to The Underground: north 2nd right, 2nd right.
Switch to plan "not prime" if no one is waiting.
[loop]
Pickup a passenger going to Cyclone.
Go to Zoom Zoom: north 3rd left, 2nd right.
Go to Cyclone: west.
[Pickup for division...]
Pickup a passenger going to Divide and Conquer.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to Divide and Conquer.
Go to Sunny Skies Park: north 1st right.
Go to Divide and Conquer: north 1st right, 1st right, 2nd right, 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: east 1st left, 1st left, 2nd left.
[Copy result to check if integer, first pickup ]
Pickup a passenger going to The Underground.
Pickup a passenger going to Equal's Corner.
Pickup a passenger going to Trunkers.
Go to Trunkers: south 1st left.
Pickup a passenger going to Equal's Corner.
[Check if integer]
Go to Equal's Corner: west 1st left.
Switch to plan "continue" if no one is waiting.
Go to The Underground: north 3rd right, 1st right, 2nd left.
Switch to plan "prime" if no one is waiting.
[Print whether it's a prime or not]
[not prime]
0 is waiting at the Writer's Depot.
[prime]
1 is waiting at the Writer's Depot.
Go to Writer's Depot: north 3rd left, 2nd left.
Pickup a passenger going to Post Office.
Go to Post Office: north 1st right, 2nd right, 1st left.
[Pickup numerator and decrement denominator for checking lower numbers]
[continue]
Go to Sunny Skies Park: north.
Pickup a passenger going to Cyclone.
Go to The Underground: north 1st right, 1st right, 2nd right.
Switch to plan "loop".
Try it online!
# Vyxal, 3 bytes
KḢ₃
Try it Online!
I thought something non trivial would be nice for a change. æ does the job for one byte but where's the fun in that?
-1 thanks to EmanresuA
## Explained
KḢ₃
KḢ # factors(input)[1:]
₃ # len(^) == 1 // prime numbers have only [1, n] as factors...other numbers have 1 or 3+ factors
• This can be KḢ₃ Commented Nov 1, 2021 at 8:57
• Also, 5 bytes version with no factorisation - 4 with r if you remove the $ Commented Nov 1, 2021 at 9:03 ## <>^v, 184 bytes Prints 1 if number is prime, otherwise 0. >]v2i>IT%0=vIT[vv > vv 1 I TI T T ) T = i t > ^ v , ^ < v >"1"; > >"0"; ^@| #### Explanation @ Pointer starts here, goes right | Mirror — reverse pointer direction @ No-op because the program is already running ^ Send instruction pointer up , Read number from stdin and push to stack t Pop and store in variable t T Push value of variable t T Same 1 Push 1 > Send instruction pointer right ] If top element of stack is greater or equal to the second element of the stack v Send instruction pointer down > Send instruction pointer right Go to [not prime] Else 2 Push 2 i Pop and store into variable i [loop start] > Send instruction pointer right I Push value of variable i T Push value of variable t % Set top of stack to top of stack modulo second element of stack 0 Push 0 = If the top two elements of the stack are equal v Send instruction pointer down I Push value of variable i T Push value of variable t = If the top two elements of the stack are equal > Send instruction pointer right ^ Send instruction pointer up > Send instruction pointer right Go to [prime] Else [not prime] > Send instruction pointer right "0" Push "0" ; Print top of stack ';' was the last character of this row, program exits Else I Push value of variable i T Push value of variable t If the first element of the stack (t) is smaller or equal to the second element of the stack (i) v Send instruction pointer down [prime] v Send instruction pointer down T Push value of T v Send instruction pointer down (there only to ensure the row is long enough) > Send instruction pointer right "1" Push "1" ; Print top of stack Since ';' was the last character of the row, exit v Send instruction pointer down v Idem I Push value of variable i ) Increment top of stack i Pop stack and store in variable i v Send instruction pointer down < Send instruction pointer left ^ Send instruction pointer up Go to [loop start] More simply, it first checks if the number is lower or equal to 1. If so, it prints 0 and exits. Then, it sets the variable i (current divisor) to 2. Then the program enters a loop. In that loop: • If the number modulo i is 0: • If i == number, print 1 and exit (prime). Else, print 0 and exit (has found a divisor, number isn't prime). • Then it increments i, and goes back to the beginning of the loop. run online # Google Sheets, 29 bytes =mod(fact(A1-1),A1)=mod(-1,A1 Try it on my Google Sheet It uses Wilson’s theorem, which, in modular arithmetic, is: (n − 1)! ≡ −1 (mod n) read: n − 1 factorial is congruent modulo n to negative 1 However, it is important to maintain the distinction between (mod n) notation and and the modulo operator. We could rewrite Wilson’s theorem using more familiar mod operators: (n − 1)! % n = -1 % n and that’s what i wrote into google sheets. Google Sheets also does parenthesis autocompletion, so i omitted the final parentheses. # Trilangle 1.3, 31 29 bytes <'?<#2%._zS<.>(>.,)2-/\\_/!@@ Reads a single integer from STDIN, and prints 0 iff it's prime. Try it on the online interpreter! ## TL;DR Keeps a running counter that starts at 2, and increments it until (input % counter) == 0. Then, it prints '0' if input == counter. ## What is Trilangle? Trilangle is a 2-D stack-based programming language. Program flow can be redirected with "mirrors" /|\_ or with branches <^7>vL. The branches can split, merge, or reflect control flow, depending on how they're used. If the IP walks off the board, it continues one row or diagonal to its left. ## Code Explanation Unfolds to this example from the README: < ' ? < # 2 % . _ z S < . > ( > . , ) 2 - / \ \ _ / ! @ @ . . . . . . . Equivalent to this C code: #include <stdio.h> // Get an integer from stdin. Implementation provided by the interpreter. extern int getint(void); int main() { // RED PATH int input = getint(); int counter = 2; while (input % counter) // GREEN PATH ++n; // BLUE PATH if (input == counter) // YELLOW PATH return 0; // MAGENTA PATH puts("0"); return 0; } This uses the new 2DUP instruction z to make stack management easier: it copies both the input and counter so that they may be operated on non-destructively. The older version below used a combination of j (indexed read), 2 (duplicate), and S (swap) to achieve a similar effect. ## Older answer: Trilangle 1.0, 42 41 40 bytes '2.?..<[email protected]'2,<|>(%!.\S)S,,)S<.....@>- # Ruby, 16 + 6 = 22 bytes [*$<];p$..prime? or equivalently p$<.count.prime?
Rules abuse! Kind of, anyway. This answer requires that the input be in unary, and that the character used for input be a newline. Invoke like
ruby -rprime prime_test.rb input
Where input is a file containing n newlines.
I calculate this at 22 bytes: 6 for "rprime" and 16 for the code. However, I also calculate manatwork's answer at 22 bytes if you golf the command line invocation (7 for 'nrprime' and 15 for the code).
• Are you sure that Ruby's prime? doesn't use a probabilistic test? Commented Sep 14, 2015 at 10:23
• Yes, it loops through a pseudoprime generator (which returns a superset of the primes) and checks for divisibility of each number less than it. Commented Sep 14, 2015 at 12:07
# Ruby, 21 + 1 = 22 19 + 1 = 20 bytes
Also throwing my hat into the Ruby battle using the regex approach (and improved using histocrat's suggestion):
p !/^(11+)\1+$|^1$/
Takes input as a unary string and invoked using the n flag:
$ruby -ne 'p !/^(11+)\1+$|^1$/' <<< 11111 true • Ooh, you can drop two in fact. p !/^(11+)\1+$|^1$/ Commented Sep 11, 2015 at 18:25 • @histocrat Interesting. I didn't know about that. However, it gives me warning: regex literal in condition. Since the rules state "if possible, output should consist solely of the string representation of a truthy or falsy value," I'd need to add the W0 flag, so it's a wash. Thanks for the tip, though. Commented Sep 11, 2015 at 18:52 • @histocrat A-ha, it only gives that warning when I run it from a file rather than as a one-liner. So thank you indeed for that. Commented Sep 11, 2015 at 20:45 # O, 23 bytes j.1>\J2/{Jn%0={0}{}?}dp Take 2! This one uses trial by division. # Python 2, 466552 51 bytes 1. Went up to 65 bytes - Fix made thanks to feersum & Mauris 2. Went down to 52 bytes - Suggestions made by kirbyfan64sos and accepts in input from STDIN to complete the code. 3. Went down to 51 bytes - Suggestion made by kirbyfan64sos (thanks again!) to remove the space between 1 and and. Apparently if you have a letter that follows a number, a space isn't needed... how weird, but cool! This finds the remainder / modulus of the input integer n divided by every number from 2 up to n-1. If there is at least one number in this sequence that has no remainder, or results in 0, this means that the number is not prime. If every value in this sequence is non-zero, the value is prime. Also by definition, 1 isn't prime and so that has to be taken care of separately. n=input();print n!=1and all(n%i for i in range(2,n)) # Example Runs I ran this in IPython: In [10]: n=input();print n!=1and all(n%i for i in range(2,n)) 1 False In [11]: n=input();print n!=1and all(n%i for i in range(2,n)) 2 True In [12]: n=input();print n!=1and all(n%i for i in range(2,n)) 3 True In [13]: n=input();print n!=1and all(n%i for i in range(2,n)) 4 False In [14]: n=input();print n!=1and all(n%i for i in range(2,n)) 5 True In [15]: n=input();print n!=1and all(n%i for i in range(2,n)) 6 False In [16]: n=input();print n!=1and all(n%i for i in range(2,n)) 7 True In [17]: n=input();print n!=1and all(n%i for i in range(2,n)) 10 False In [18]: n=input();print n!=1and all(n%i for i in range(2,n)) 15 False In [19]: n=input();print n!=1and all(n%i for i in range(2,n)) 17 True In [20]: n=input();print n!=1and all(n%i for i in range(2,n)) 20 False In [21]: n=input();print n!=1and all(n%i for i in range(2,n)) 30 False • It has to return false for 1. Commented Sep 12, 2015 at 0:40 • You made it return False for every possible input but 1, which should be False :) – lynn Commented Sep 12, 2015 at 1:05 • @Mauris - Oops. I didn't fix that properly lol. Was on mobile. I'm on my computer now. Fixing. Commented Sep 12, 2015 at 1:07 • @Mauris - Fixed now. Thanks. Commented Sep 12, 2015 at 1:12 • Wow, this is almost the exact same code I wrote :) You can save a char by using > instead of !=. Commented Sep 21, 2015 at 15:14 # C, 61 bytes r;main(i,j){r=(--i>1);for(j=i-1;j>1;)r*=!!(i%j--);return r;} Uses unary representation - the number is encoded in the number of commandline arguments. The return value is initialized to 1 and then, it's multiplied by the sign of remainder of division of the tested number and numbers 2..i-1 in a loop. So it will be zeroed when any non-trivial divisor found. The result is returned to the system as the exit code. Test: echo 'r;main(i,j){r=(--i>1);for(j=i-1;j>1;)r*=!!(i%j--);return r;}' > main.c gcc -o main main.c ./main 1 ; echo$?
./main 1 1 ; echo $? ./main 1 1 1 ; echo$?
./main 1 1 1 1 ; echo $? ./main 1 1 1 1 1 ; echo$?
./main 1 1 1 1 1 1 ; echo $? ./main 1 1 1 1 1 1 1 ; echo$?
./main 1 1 1 1 1 1 1 1 ; echo $? ./main 1 1 1 1 1 1 1 1 1 ; echo$?
./main 1 1 1 1 1 1 1 1 1 1 ; echo $? ./main 1 1 1 1 1 1 1 1 1 1 1 ; echo$?
# Bash+coreutils, 29 bytes
echo $[factor$1|wc -w==2]
Test:
echo 'echo $[factor$1|wc -w==2]' > primetest.sh
chmod +x primetest.sh
./primetest.sh <NUMBER_TO_TEST>
## Fortran 90, 66 55 bytes
Using trial division.
read*,i
do1 j=2,i
1 if(mod(i,j)==0)exit
print*,i==j
end
Saved 11 bytes thanks to sigma.
A do loop with labelled statement (not an enddo or continue) is still valid in Fortran 90, but obsolete.
Alternative using Wilson's theorem (56 bytes):
read*,j
k=1
do1 i=2,j-1
1 k=mod(k*i,j)
print*,k==j-1
end
• Aww you beat me, now I'll have to come up with a different Fortran solution! Some tips though: the program statement is not necessary, which will save you 9 bytes, and you can also get away with writing do1 instead of do 1. Commented Sep 13, 2015 at 20:43
# Smalltalk, 47 characters
Obviously no competition for isPrime if the Smalltalk dialect has it, but not all do. For example, the one used in Coding Ground (GNU Smalltalk v3.2.5) does not have it.
I'm relying on the observation that GCD((n - 1)!, n) = 1 for prime n which I haven't seen used very often. Ridiculously bad algorithm, but Smalltalk has no problem working with large integers. Replace the 2 with whatever you want to test:
|n|n:=2.((n>1)&((n-1)factorial gcd:n)=1)inspect
It does not consider 1 as prime as required by the OP. However, one widely accepted definition of a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. So by this definition, 1 shouldn't be a candidate for primality testing, any more than 0, ½, i, e or π should be.
Note that in some situations, -1 is considered prime, because -1 = 1 × -1, and is used as such in some factorization algorithms.
## J, 4 bytes
1&p:
This is essentially a built-in function for J as p: provides several different prime-related functions, depending on the left argument (attached here with 1&).
As with J in general, it is incredibly array-friendly:
1&p: 0 1 2 3 4 5 6 7 8 9 10
0 0 1 1 0 1 0 1 0 0 0
# C#, 156133 130 bytes
(second attempt after the first didn't exactly work)
using System;class C{static void Main(string[]a){int i=2,n=int.Parse(a[0]),b=n-1;for(;i<=n/2;)b*=n%i++>0?1:0;Console.Write(b>0);}}
(and thanks to Dennis for helping me out with my first code golf :) )
• Welcome to Programming Puzzles & Code Golf! In its current form, your answer simply prints the result of the last divisibility test. You can fix that by initializing b=1 and multiplying b by the results (b*=...). Commented Sep 15, 2015 at 16:05
• I'm not a C# expert, but since C# doesn't have implicit int-to-bool conversion, I don't think 1/0 satisfy the definition of truthy/falsy. using System;class C{static void Main(string[] a){int i=2,n=Int32.Parse(a[0]),b=n-1;for(;i<=n/2;)b*=n%i++>0?1:0;Console.Write(b>0);}} produces the proper output (and is 23 bytes shorter). Commented Sep 15, 2015 at 16:14
• Ah, I see. I would have said that 1/0 does satisfy the condition, but I'll stick with your definition. :) Commented Sep 15, 2015 at 16:16
• For most languages, it does. However, our definition counts 1 as truthy if if(1){...} executes the block, which doesn't seem to wok in C# (again, not an expert). Commented Sep 15, 2015 at 16:20
• You can remove the space between string[] and a. Also, you can change Int32 to just int. Commented Sep 19, 2015 at 15:53
# Perl, 25 bytes
#!perl -p
$_=2==grep$'%$_<1,//..$_
Counting the shebang as one. Input is taken, in decimal, from stdin.
Sample Usage:
$echo 101101 | perl isprime.pl$ echo 101107 | perl isprime.pl
1
Perl, 24 bytes
#!perl -p
$_=3>grep$'%$_<1,//..$_
One byte can be shaved by replacing 2== with 3>, however, it will incorrectly identify both 0 and 1 as prime.
## Ouroboros, 39 bytes
Sr0s1(
)S1+.@@.@@%!Ms+S.@@.@>6*(6s2=n1(
Each line of code in an Ouroboros program represents a snake eating its tail.
Snake 1
S switches to the shared stack; r0 reads a number from input and pushes a 0. Then s1( switches back to snake 1's stack, pushes a 1, and eats that many characters of the tail. The instruction pointer is on a character that gets eaten, so the snake dies.
Snake 2
Here the magic happens. We check every number from 1 up through n, adding 1 to a tally if it divides our input number. At the end we check whether the number of factors equals 2 and print 1 or 0 accordingly.
) is a no-op the first time through. S switches to the shared stack. We then push a 1 (just after the first snake pushes its 0) and add. The stack now contains the input number and the factor we're testing for divisibility.
.@@.@@%! makes copies of both numbers, takes the modulus, and negates (1 if it is a factor, 0 if not). M moves that result to snake 2's stack, where we're storing the tally of factors; then s+S switches to that stack, adds the top two numbers, and switches back to the shared stack.
Next, .@@.@>6* makes copies of both numbers and tests whether the input number is greater than the test factor, pushing 6 if so and 0 if not. ( then eats that many characters from the end of the snake.
• If the number is still greater than the factor, the uneaten code now ends after (6. This pushes a 6 and wraps execution back to the beginning. There ) regurgitates the 6 characters we just ate. S does nothing because we're already on the shared stack. 1+ then increments the test factor, and we go through the loop again.
• When the number is no longer greater than the factor, nothing gets eaten and execution continues. We push a 6 but then switch to snake 2's stack, where the number of factors is sitting. 2=n tests whether it's 2 and outputs the result (1 or 0) as a number. Finally, 1( eats the last character and dies.
Try it out
// Define Stack class
function Stack() {
this.stack = [];
this.length = 0;
}
Stack.prototype.push = function(item) {
this.stack.push(item);
this.length++;
}
Stack.prototype.pop = function() {
var result = 0;
if (this.length > 0) {
result = this.stack.pop();
this.length--;
}
return result;
}
Stack.prototype.top = function() {
var result = 0;
if (this.length > 0) {
result = this.stack[this.length - 1];
}
return result;
}
Stack.prototype.toString = function() {
return "" + this.stack;
}
// Define Snake class
function Snake(code) {
this.code = code;
this.length = this.code.length;
this.ip = 0;
this.ownStack = new Stack();
this.currStack = this.ownStack;
this.alive = true;
this.wait = 0;
this.partialString = this.partialNumber = null;
}
Snake.prototype.step = function() {
if (!this.alive) {
return null;
}
if (this.wait > 0) {
this.wait--;
return null;
}
var instruction = this.code.charAt(this.ip);
var output = null;
console.log("Executing instruction " + instruction);
if (this.partialString !== null) {
// We're in the middle of a double-quoted string
if (instruction == '"') {
// Close the string and push its character codes in reverse order
for (var i = this.partialString.length - 1; i >= 0; i--) {
this.currStack.push(this.partialString.charCodeAt(i));
}
this.partialString = null;
} else {
this.partialString += instruction;
}
} else if (instruction == '"') {
this.partialString = "";
} else if ("0" <= instruction && instruction <= "9") {
if (this.partialNumber !== null) {
this.partialNumber = this.partialNumber + instruction; // NB: concatenation!
} else {
this.partialNumber = instruction;
}
next = this.code.charAt((this.ip + 1) % this.length);
if (next < "0" || "9" < next) {
// Next instruction is non-numeric, so end number and push it
this.currStack.push(+this.partialNumber);
this.partialNumber = null;
}
} else if ("a" <= instruction && instruction <= "f") {
// a-f push numbers 10 through 15
var value = instruction.charCodeAt(0) - 87;
this.currStack.push(value);
} else if (instruction == "") { // Toggle the current stack if (this.currStack === this.ownStack) { this.currStack = this.program.sharedStack; } else { this.currStack = this.ownStack; } } else if (instruction == "s") { this.currStack = this.ownStack; } else if (instruction == "S") { this.currStack = this.program.sharedStack; } else if (instruction == "l") { this.currStack.push(this.ownStack.length); } else if (instruction == "L") { this.currStack.push(this.program.sharedStack.length); } else if (instruction == ".") { var item = this.currStack.pop(); this.currStack.push(item); this.currStack.push(item); } else if (instruction == "m") { var item = this.ownStack.pop(); this.program.sharedStack.push(item); } else if (instruction == "M") { var item = this.program.sharedStack.pop(); this.ownStack.push(item); } else if (instruction == "y") { var item = this.ownStack.top(); this.program.sharedStack.push(item); } else if (instruction == "Y") { var item = this.program.sharedStack.top(); this.ownStack.push(item); } else if (instruction == "\\") { var top = this.currStack.pop(); var next = this.currStack.pop() this.currStack.push(top); this.currStack.push(next); } else if (instruction == "@") { var c = this.currStack.pop(); var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(c); this.currStack.push(a); this.currStack.push(b); } else if (instruction == ";") { this.currStack.pop(); } else if (instruction == "+") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(a + b); } else if (instruction == "-") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(a - b); } else if (instruction == "*") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(a * b); } else if (instruction == "/") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(a / b); } else if (instruction == "%") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(a % b); } else if (instruction == "_") { this.currStack.push(-this.currStack.pop()); } else if (instruction == "I") { var value = this.currStack.pop(); if (value < 0) { this.currStack.push(Math.ceil(value)); } else { this.currStack.push(Math.floor(value)); } } else if (instruction == ">") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(+(a > b)); } else if (instruction == "<") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(+(a < b)); } else if (instruction == "=") { var b = this.currStack.pop(); var a = this.currStack.pop(); this.currStack.push(+(a == b)); } else if (instruction == "!") { this.currStack.push(+ !this.currStack.pop()); } else if (instruction == "?") { this.currStack.push(Math.random()); } else if (instruction == "n") { output = "" + this.currStack.pop(); } else if (instruction == "o") { output = String.fromCharCode(this.currStack.pop()); } else if (instruction == "r") { var input = this.program.io.getNumber(); this.currStack.push(input); } else if (instruction == "i") { var input = this.program.io.getChar(); this.currStack.push(input); } else if (instruction == "(") { this.length -= Math.floor(this.currStack.pop()); this.length = Math.max(this.length, 0); } else if (instruction == ")") { this.length += Math.floor(this.currStack.pop()); this.length = Math.min(this.length, this.code.length); } else if (instruction == "w") { this.wait = this.currStack.pop(); } // Any unrecognized character is a no-op if (this.ip >= this.length) { // We've swallowed the IP, so this snake dies this.alive = false; this.program.snakesLiving--; } else { // Increment IP and loop if appropriate this.ip = (this.ip + 1) % this.length; } return output; } Snake.prototype.getHighlightedCode = function() { var result = ""; for (var i = 0; i < this.code.length; i++) { if (i == this.length) { result += '<span class="swallowedCode">'; } if (i == this.ip) { if (this.wait > 0) { result += '<span class="nextActiveToken">'; } else { result += '<span class="activeToken">'; } result += escapeEntities(this.code.charAt(i)) + '</span>'; } else { result += escapeEntities(this.code.charAt(i)); } } if (this.length < this.code.length) { result += '</span>'; } return result; } // Define Program class function Program(source, speed, io) { this.sharedStack = new Stack(); this.snakes = source.split(/\r?\n/).map(function(snakeCode) { var snake = new Snake(snakeCode); snake.program = this; snake.sharedStack = this.sharedStack; return snake; }.bind(this)); this.snakesLiving = this.snakes.length; this.io = io; this.speed = speed || 10; this.halting = false; } Program.prototype.run = function() { this.step(); if (this.snakesLiving) { this.timeout = window.setTimeout(this.run.bind(this), 1000 / this.speed); } } Program.prototype.step = function() { for (var s = 0; s < this.snakes.length; s++) { var output = this.snakes[s].step(); if (output) { this.io.print(output); } } this.io.displaySource(this.snakes.map(function (snake) { return snake.getHighlightedCode(); }).join("<br>")); } Program.prototype.halt = function() { window.clearTimeout(this.timeout); } var ioFunctions = { print: function (item) { var stdout = document.getElementById('stdout'); stdout.value += "" + item; }, getChar: function () { if (inputData) { var inputChar = inputData[0]; inputData = inputData.slice(1); result = inputChar.charCodeAt(0); } else { result = -1; } var stdinDisplay = document.getElementById('stdin-display'); stdinDisplay.innerHTML = escapeEntities(inputData); return result; }, getNumber: function () { while (inputData && (inputData[0] < "0" || "9" < inputData[0])) { inputData = inputData.slice(1); } if (inputData) { var inputNumber = inputData.match(/\d+/)[0]; inputData = inputData.slice(inputNumber.length); result = +inputNumber; } else { result = -1; } var stdinDisplay = document.getElementById('stdin-display'); stdinDisplay.innerHTML = escapeEntities(inputData); return result; }, displaySource: function (formattedCode) { var sourceDisplay = document.getElementById('source-display'); sourceDisplay.innerHTML = formattedCode; } }; var program = null; var inputData = null; function showEditor() { var source = document.getElementById('source'), sourceDisplayWrapper = document.getElementById('source-display-wrapper'), stdin = document.getElementById('stdin'), stdinDisplayWrapper = document.getElementById('stdin-display-wrapper'); source.style.display = "block"; stdin.style.display = "block"; sourceDisplayWrapper.style.display = "none"; stdinDisplayWrapper.style.display = "none"; source.focus(); } function hideEditor() { var source = document.getElementById('source'), sourceDisplay = document.getElementById('source-display'), sourceDisplayWrapper = document.getElementById('source-display-wrapper'), stdin = document.getElementById('stdin'), stdinDisplay = document.getElementById('stdin-display'), stdinDisplayWrapper = document.getElementById('stdin-display-wrapper'); source.style.display = "none"; stdin.style.display = "none"; sourceDisplayWrapper.style.display = "block"; stdinDisplayWrapper.style.display = "block"; var sourceHeight = getComputedStyle(source).height, stdinHeight = getComputedStyle(stdin).height; sourceDisplayWrapper.style.minHeight = sourceHeight; sourceDisplayWrapper.style.maxHeight = sourceHeight; stdinDisplayWrapper.style.minHeight = stdinHeight; stdinDisplayWrapper.style.maxHeight = stdinHeight; sourceDisplay.textContent = source.value; stdinDisplay.textContent = stdin.value; } function escapeEntities(input) { return input.replace(/&/g, '&').replace(/</g, '<').replace(/>/g, '>'); } function resetProgram() { var stdout = document.getElementById('stdout'); stdout.value = null; if (program !== null) { program.halt(); } program = null; inputData = null; showEditor(); } function initProgram() { var source = document.getElementById('source'), stepsPerSecond = document.getElementById('steps-per-second'), stdin = document.getElementById('stdin'); program = new Program(source.value, +stepsPerSecond.innerHTML, ioFunctions); hideEditor(); inputData = stdin.value; } function runBtnClick() { if (program === null || program.snakesLiving == 0) { resetProgram(); initProgram(); } else { program.halt(); var stepsPerSecond = document.getElementById('steps-per-second'); program.speed = +stepsPerSecond.innerHTML; } program.run(); } function stepBtnClick() { if (program === null) { initProgram(); } else { program.halt(); } program.step(); } function sourceDisplayClick() { resetProgram(); } .container { width: 100%; } .so-box { font-family:'Helvetica Neue', Arial, sans-serif; font-weight: bold; color: #fff; text-align: center; padding: .3em .7em; font-size: 1em; line-height: 1.1; border: 1px solid #c47b07; -webkit-box-shadow: 0 2px 2px rgba(0, 0, 0, 0.3), 0 2px 0 rgba(255, 255, 255, 0.15) inset; text-shadow: 0 0 2px rgba(0, 0, 0, 0.5); background: #f88912; box-shadow: 0 2px 2px rgba(0, 0, 0, 0.3), 0 2px 0 rgba(255, 255, 255, 0.15) inset; } .control { display: inline-block; border-radius: 6px; float: left; margin-right: 25px; cursor: pointer; } .option { padding: 10px 20px; margin-right: 25px; float: left; } h1 { text-align: center; font-family: Georgia, 'Times New Roman', serif; } a { text-decoration: none; } input, textarea { box-sizing: border-box; } textarea { display: block; white-space: pre; overflow: auto; height: 50px; width: 100%; max-width: 100%; min-height: 25px; } span[contenteditable] { padding: 2px 6px; background: #cc7801; color: #fff; } #stdout-container, #stdin-container { height: auto; padding: 6px 0; } #reset { float: right; } #source-display-wrapper , #stdin-display-wrapper{ display: none; width: 100%; height: 100%; overflow: auto; border: 1px solid black; box-sizing: border-box; } #source-display , #stdin-display{ font-family: monospace; white-space: pre; padding: 2px; } .activeToken { background: #f93; } .nextActiveToken { background: #bbb; } .swallowedCode{ color: #999; } .clearfix:after { content:"."; display: block; height: 0; clear: both; visibility: hidden; } .clearfix { display: inline-block; } * html .clearfix { height: 1%; } .clearfix { display: block; } <!-- Designed and written 2015 by D. Loscutoff Much of the HTML and CSS was taken from this Befunge interpreter by Ingo Bürk: http://codegolf.stackexchange.com/a/40331/16766 --> <div class="container"> <textarea id="source" placeholder="Enter your program here" wrap="off">Sr0s1( )S1+.@@.@@%!Ms+S.@@.@>6*(6s2=n1(</textarea> <div id="source-display-wrapper" onclick="sourceDisplayClick()"><div id="source-display"></div></div></div><div id="stdin-container" class="container"> <textarea id="stdin" placeholder="Input" wrap="off">5</textarea> <div id="stdin-display-wrapper" onclick="stdinDisplayClick()"><div id="stdin-display"></div></div></div><div id="controls-container" class="container clearfix"><input type="button" id="run" class="control so-box" value="Run" onclick="runBtnClick()" /><input type="button" id="pause" class="control so-box" value="Pause" onclick="program.halt()" /><input type="button" id="step" class="control so-box" value="Step" onclick="stepBtnClick()" /><input type="button" id="reset" class="control so-box" value="Reset" onclick="resetProgram()" /></div><div id="stdout-container" class="container"><textarea id="stdout" placeholder="Output" wrap="off" readonly></textarea></div><div id="options-container" class="container"><div class="option so-box">Steps per Second: <span id="steps-per-second" contenteditable>20</span></div></div> # Vitsy, 2 bytes Yes, it's that Simple...x. pN Implicit grab of STDIN as number, if possible. p If it's prime, push 1 to the stack. Else, push 0. N Output as number. Interestingly, adding an i to this will also find prime characters. ipN For input % (ASCII character 37) will output 1. • Haha, so punny! +1 from me. Commented Nov 7, 2015 at 19:00 # Jelly, 4 bytes ’!²% Try it online! Jelly has a built-in for primality testing (ÆP, 2 bytes), but it uses a probabilistic method. This answer uses Wilson's theorem instead. For input x, it calculates (x - 1)!² % x, which yields 1 if x is a prime number and 0 if not. Input: x ’ Decrement; compute x - 1. ! Apply factorial atop the previous result. Yields (x - 1)!. ² Apply square atop the previous result. Yields (x - 1)!². % Do a modulus hook; compute (x - 1)!² % x. # Par, 9 bytes Counted using its own, non-UTF-8 encoding. ✶↓″↑p~1=* Explanation: ✶ Parse the input (which is implicitly on the stack). n ↓ Subtract one. (n-1) ″ Duplicate. (n-1) (n-1) ↑ Add one. (n-1) n p Prime divisors. For 1, this strangely returns (1). (n-1) np ~ Length. (n-1) np~ 1= Is the length one? This is iff n isn't composite. (n-1) noncomposite(n) * Multiply the top of stack. (n-1)*noncomposite(n) # Binary-Encoded Golfical, 13+1 (-x flag) = 14 bytes This can converted to the standard graphical version using the included Encoder utility, or run directly by adding the -x flag. Hex dump: 00 40 02 15 14 49 1b 00 00 00 01 17 17 The original image: Zoomed in 100x with color value lables: Explanation: 10,0,0->Input number 14,3,0->Turn right if prime 11,0,1->Go east 0,0,0->Set to 0 0,0,1->Set to 1 10,1,0->Print number # Arcyóu, 42 7 bytes Note: I added a builtin for primality testing after I submitted this answer. In the interest of completeness, I have left the old answer below, but this is the new official answer: (p?(#(l Primality check on a line of input casted to int. ### Old answer: ((F(n)(?([ n)(&(f x(_ 2 n)(% n x)))f))(#(l Arcyóu is a LISP-like golfing language of my own devising. Explanation: ((F(n) ; Anonymous function taking one argument n (? ([ n) ; If-statement with condition n-1 (handling the special case) (& ; & is both bitwise AND and an 'all' function (f x (_ 2 n) ; For loop iterating over a range from 2 to n (% n x))) ; n mod x f)) ; If n did equal 1, return false (# (l ; Now call the function on a line of input casted to int The interpreter allows you to leave off final close-parens, since adding them back is trivial. # MediaWiki templates with ParserFunctions, 101 + 1 = 102 bytes (for title) {{#ifexpr:{{{n}}} mod {{{f|(n-1)}}}==0|false|{{#ifexpr:{{{f}}}==1|true|{{:p|n=n|f={{#expr:f-1}}}}}}}} Ungolfed: {{#ifexpr:{{{n}}} mod {{{f|(n-1)}}}==0|false| {{#ifexpr:{{{f}}}==1|true| {{p|n=n|f={{#expr:f-1}}}} }} }} This recursive trial division method theoretically works, but to determine the primality of a positive integer n, wgMaxTemplateDepth (in the MediaWiki config) must exceed n - 2.
• So, in practice, this works for no single input? According to the rules, it must work at least for integers 1 to 255... Commented Dec 21, 2015 at 20:25
• This solution is valid for wikis that allow at least a recursive depth of 253. Commented Dec 21, 2015 at 20:29
# Befunge 93, 44 bytes
&:v>0.@ @.-1<
03<_v#%p03+1:g03::_^#g
This works by trial division.
There's a hidden unprintable character between the v> on the first line; it's the character whose value is 2. The base64 of the file is as follows:
Jjp2Aj4wLkAgICAgICAgQC4tMTwKMDM8X3YjJXAwMysxOmcwMzo6X14jYGc=
Opening it as hex in Sublime Text looks like this (newline confusion, though):
263a 7602 3e30 2e40 2020 2020 2020 2040
2e2d 313c 0d0a 3033 3c5f 7623 2570 3033
2b31 3a67 3033 3a3a 5f5e 2360 670d 0a
` | 15,320 | 54,320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.894125 |
https://tableaumagic.com/shaped-bar-charts-3d-in-tableau/ | 1,611,391,055,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00270.warc.gz | 602,891,991 | 37,924 | We previously had an article about creating Shaped Bar Charts in Tableau; in that previous tutorial, we used data densification to create a different look and feel for our bar charts. In this tutorial, we are going to explore using 3D shapes to create 3D Bar Charts in Tableau.
Note: 3D data visualisation is not considered best practice for data visualisations, however, I have seen these used to create interesting and engaging infographics or used for infotainment.
## Data
``````Country,Value
United Kingdom,100
United States,80
France,60
Germany,50
Italy,30
Australia,70
India,50``````
Once your data is loaded into Tableau, right-click on the data source and click on Edit Data Source… with the Data Source Editor open, paste the following:
``````Path
0
100``````
You should get an error as there is no joining column, however, click on Add new join clause, go to Create Join Calculation, type 1 and click OK. Do this for the right-hand side as well. Ensure that you have Inner join selected and you should see the following:
Note: we need additional records as we are going to be drawing shapes and using densification to get more points on our canvas. For more information, check out our article on Data Densification.
## Calculated Fields
With our data set loaded into Tableau, we are going to create the following Bin and Calculated Fields:
Path (bin)
• Right-click on Path, go to Create and select Bins…
• In the Edit Bins dialogue window:
• Set New field name to Path (bin).
• Set Size of bins to 1.
• Click Ok.
Index
``100-INDEX()``
TC_Value
``WINDOW_MAX(MAX([Value]))``
Rows
``````IF [Index] <= [TC_Value] THEN
[Index]
ELSE
NULL
END``````
With this done, let us start creating our data visualisation.
## Worksheet
We will now build our first worksheet:
• Change the Mark Type to Shape.
• Drag Country onto the Columns Shelf.
• Drag Country onto the Marks Shelf.
• Drag Path (bin) onto Columns Shelf.
• Right-click on this object and ensure that Show Missing Values is selected.
• Drag this object onto the Detail Mark.
• Drag the Rows Object onto the Rows Shelf.
• Right-click on this object, go to Compute Using and select Path (bin).
If all goes well, you should see the following:
We have now created our Shaped Bar Chart, but now we will need to download, unzip, and add copy the following shapes to our Shapes directory. If you have not done this before, check out our article on Custom Shapes.
Note: These are cube objects that I have created and colored using Material Design colors.
• Set Grid Lines to None.
• Set Zero Lines to None.
• Set Axis Rulers to None.
• Assign our Cube Shapes to each Country.
• Click on the 254 nulls and select Filter.
If all goes well, you should now have the following final data visualisation:
Exercise: Explore with our own objects, and let me know what you come up with.
and boom, we are done! I hope you enjoyed this tutorial, and as always, you can find this data visualisation on Tableau Public at https://public.tableau.com/profile/toan.hoang#!/vizhome/ShapedBarCharts3D/ShapedBarChart3D
## Summary
I hope you all enjoyed this article as much as I enjoyed writing it and as always do share the love. Do let me know if you experienced any issues recreating this Visualisation, and as always, please leave a comment below or reach out to me on Twitter @Tableau_Magic. Do also remember to tag me in your work if you use this tutorial.
If you like our work, do consider supporting us on Patreon, and for supporting us, we will give you early access to tutorials, exclusive videos, as well as access to current and future courses on Udemy:
Also, do be sure to check out our various courses:
Toan Hoang, Tableau Zen Master 2020, has over 15 years of experience in Business Intelligence, Data Management, Big Data, Data Lakes, Internet of Things (IoT), Data Visualisation and the Data Analytics space; the last six years has been dedicated to delivering end-to-end solutions using Tableau.
1. Hi Tableau Magic, I applied the same steps on sample super store data where I am using sales and subcategory to create 3D bar chart but in my view all the bars are same in height..how can we fix this ?
• Hi Imran, it sounds like you are not setting the Table Calculations correctly. Right-click on the Y axis pill and go to Edit Table Calculations and adjust accordingly.
• Imran, actully the thing is if u use sales by subcategory it wont as any of the sub category have more than 100 then it will appear as full bar, so that that u need to use this ({fixed: sum (sale )}/sum(sales))*100. To make each category as percentage number which will be below 100 for sure
2. Hello Toan, thanks for the all the helpful posts. was trying the 3d bar chart with a different data set where I am showing “Aging groups”. However, bars are not starting from the x axis. I see floating cubes, any idea why?
3. hi Toan can we do this 3 D charts only if source data is excel? or it possible with other sources too, I am using a BI report which is a hyperlink. when I try the first step at joins its failing.
4. I am using Tableau 2020.2.7. In this version there is no concept of Join. But there is a relationship. and 1 =1 , in a relationship does not work. Can you help me here, Super urgent
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,278 | 5,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.807248 |
https://de.mathworks.com/matlabcentral/cody/problems/44261 | 1,713,453,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00806.warc.gz | 180,539,188 | 19,065 | # Problem 44261. Multivariate polynomials - sort monomials
In Problem 44260, multivariate polynomials were defined as a sum of monomial terms using|exponents|, a matrix of integers, and|coefficients|, a vector (follow the above link for an explanation). It can be useful to order the monomials. But first we need to define the total degree of a monomial as the sum of the exponents. For example, the total degree of 5*x is 1 and the total degree of x^3*y^5*z is 9.
Write a function
```function [coeffs,exponents] = sortMonomials(coeffs,exponents)
```
to sort the monomials. Sort them first by descending total degree, and then for a given total degree, by lexicographical order of the exponents (by the first exponent, then the second, and so on, each in descending order). The coefficients should be sorted so they stay with the correct monomial.
Example: Consider the polynomial p(x,y,z) = 3*x - 2 + y^2 +4*z^2, which is represented as:
```exponents = [1 0 0; 0 0 0; 0 2 0; 0 0 2], coefficients = [3; -2; 1; 4]
```
The sorted version is
```exponents = [0 2 0; 0 0 2; 1 0 0; 0 0 0], coefficients = [1; 3; 1; 4].
```
You can assume that a given combination of exponents is never repeated.
### Solution Stats
37.5% Correct | 62.5% Incorrect
Last Solution submitted on Jul 26, 2023
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 414 | 1,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.851689 |
https://resources.altium.com/p/how-total-harmonic-distortion-affects-your-power-system | 1,716,802,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00282.warc.gz | 418,955,428 | 24,636 | # Total Harmonic Distortion In Power Systems
Zachariah Peterson
| Created: July 27, 2021
It would be nice if, at least in the US, the power that came from the wall was truly noise free. Unfortunately, this is not the case, and although a power system can appear to output a clean sine wave, zooming into an oscilloscope trace or using an FFT will tell you a different story. When you take dirty power, put it through rectification, and then pass it through a switching regulator, you introduce additional noise into the system that further degrades power quality. If you’re a power supply or power systems designer, then you know the value of supplying your devices with clean, noise-free power.
Multiple disciplines in engineering use total harmonic distortion as a metric for quantifying the quality of power supplied to your electronics. So where does this harmonic content come from, and what can you do about it in your design? The key is to control or compensate for nonlinearity in your circuit design, and following some basic layout practices can help ensure power supply noise is contained.
## Diagnosing Harmonics in Power Systems
When a power supply or power converter draws power from AC mains or a generator, it’s common for the waveform to be a bit distorted when viewed on an oscilloscope. The distorted voltage/current that gets measured from mains or generator power deviates from an ideal sine wave in terms of its magnitude, although slight variations in frequency and phase can also occur. In power systems engineering, THD and power factor are most commonly used to quantify the extent to which an AC waveform deviates from an ideal sinusoidal waveform. The latter can be defined in terms of the former and can be measured in the time domain.
Thanks to Fourier analysis, we can write the voltage (or current) of an AC waveform as a sum of some fundamental frequency (50 or 60 Hz in power systems) and all of its harmonics.
Whenever the harmonics (components with n > 1) are non-zero, you have harmonic distortion. Harmonics in power systems can have very small amplitudes, reaching -30 dB or less than the fundamental frequency, but the power they carry adds up over hundreds of harmonics. So, we need some way to diagnose the total amount of distortion in our waveform.
### Total Harmonic Distortion
The AC waveform in power systems will always have some harmonic content that creates distortion, but what’s important is whether this distortion has a noticeable effect on efficiency and signal quality at the system output. Total harmonic distortion is defined as the sum of squared peak voltages contained in the harmonics to the voltage of the fundamental waveform:
Standards like IEEE 519-2014, IEC 61000-3-6, and NEMA IS07 P1-2019 specify acceptable limits on harmonic content in power systems with THD targets. The above equation is the standard metric used to define THD, although other definitions include noise in the numerator (THD + N) or harmonics in the denominator as a point of reference. The ideal AC waveform will have a THD value of zero (i.e., no harmonics).
One thing that is important to note is that harmonic content in a power system, whether coming from the grid or generated by a regulator, can be measurable up to very high harmonics. In signal integrity, we usually only talk up to the 5th or 7th harmonic, but power systems can have noise spanning dozens of harmonics.
### Relation to Power Factor
The main concern regarding harmonics in a power system is power efficiency in the power regulator/converter section, which is a measure of efficiency. This is quantified using the power factor, or the ratio of the voltage (or current) in the fundamental frequency to the total power carried by all frequencies. This can be rewritten in terms of THD and phase angle between the mains voltage and current.
Power factor is a measure of efficiency. When the phase angle above is zero, the power is dissipated at the load as real power, rather than as reactive power. The phase angle will depend on both the load impedance (any reactance) as well as the THD value of the input power. At some point, you will only get the phase difference so small; ideally, you want to get THD as small as possible to prevent low power factor.
### Identifying Total Harmonic Distortion
The simplest way to identify harmonic content in your power system is with an oscilloscope that can perform an FFT. As an example, consider the waveform shown below. This is part of a 60 Hz 110 VAC sinusoid. Some glitches can be seen along the crest of the waveform, equating to about 5 V of dropout.
The harmonic content can be seen in an FFT of the above waveform:
If we can remove this harmonic content, we can clean up this AC wave and ensure higher efficiency power conversion.
## Reducing Total Harmonic Distortion in a PCB
There are some steps that can be taken in circuit design and in the PCB layout to remove power system interharmonics that produce harmonic distortion. From the circuit design perspective, you have two options:
• Filtering: The optimal path forward here is to shunt high frequency noise on the input to ground with a large capacitor. Make sure you’ve set the band edge between the first and second harmonic. Some systems may use a pi filter, which along with the output inductor in a switching converter, will form a 4th order filter with high rolloff.
• PFC circuit: This circuit is used in high-current switching converters to compensate for the pulsed current drawn by a switching regulator. It effectively involves recreating the sinusoidal current draw by switching a MOSFET in continuous conduction mode, so the current draw will closely match the voltage supplied by the system.
### Filtering Component Placement
Where you place filtering elements in your regulator section will affect the effectiveness of your THD reduction strategy. The first point where EMI filtering should be placed is at the main power input. For AC/DC or DC/DC systems, this will provide low-pass filtering, so it will remove some differential and common-mode harmonic content, thus reducing total harmonic distortion.
Since circuits like PFCs and high power converters use switching elements, these can generate new harmonic content that also needs to be filtered alongside harmonics in the input power. Placement of capacitors near these elements is an important aspect of ensuring filtering with low inductance. Proper placement of capacitors for filtering ensures you do not unintentionally add a pole to the PFC transfer function, which will prevent any gain and oscillation in the circuit at high frequencies.
To do this, you’ll need to place input capacitors with tight loops so that there is low parasitic inductance on the signal path. This will create an effective high-pass filter back to ground. Remember the 3-capacitor guideline used for bypass/decoupling capacitors on high-speed digital ICs? A similar idea applies here to provide a low-pass filter for sourcing power to MOSFETs.
### What About Parallel MOSFETs?
Be careful when working in power systems that use MOSFETs in parallel as they have their own placement strategies needed to prevent oscillations during driving. This is normally handled by adding a small amount of resistance to provide damping, rather than trying to filter such an oscillation and any harmonics it generates with filtering. You can read more about placement near parallel MOSFETs in this article.
When you need to design power systems with low total harmonic distortion, low EMI, and safe placement of components, make sure you use the best set of PCB layout features in Altium Designer®. When you need to examine conducted EMI from power systems that contain harmonics or interharmonics, you can use the integrated SPICE package or import your design into a field solver to perform a range of SI/PI/EMI simulations. When you’ve finished your design, and you want to release files to your manufacturer, the Altium 365™ platform makes it easy to collaborate and share your projects.
We have only scratched the surface of what’s possible with Altium Designer on Altium 365. Start your free trial of Altium Designer + Altium 365 today. | 1,708 | 8,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.934873 |
http://mathoverflow.net/revisions/18238/list | 1,369,342,391,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703788336/warc/CC-MAIN-20130516112948-00097-ip-10-60-113-184.ec2.internal.warc.gz | 165,198,717 | 6,203 | 2 slightly changed question to fit the context given by answers
On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.
I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?
I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).
So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.)
EDIT: Thanks, Petya, I guess I need to restrict to smooth singular chains. Also thanks Gonçalo for pointing out that I appear to really be talking about currents - I will take a look at the book! My remaining question: first of all, it seems to me like in the context of a compact Riemannian manifold, the space of k-currents is naturally identified with the space of k-forms via the inner product. So in this space, is the set of currents given by integration over a smooth k-submanifold a proper subspace? I understand that the point of currents is that in the general case they are broader, but in the compact case it seems like maybe that doesn't happen, and I'm having difficulty understanding the statements about the mass norm that seem to concern this question.
1
# Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds
On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.
I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?
I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).
So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.) | 1,271 | 5,433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2013-20 | latest | en | 0.916292 |
http://www.eprisner.de/MAT109/Analysis51.html | 1,679,973,377,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00510.warc.gz | 69,490,858 | 8,568 | # Analysis Stratego
Precision, put in 100 or 1000 or 10000
1 2 3 4 5 1 2 3 4 5
x1: x2:
#### The game where higher cards beat lower cards, except 1 against 5, which is a tie (both cards are replaced)
Solution of the 1,2,4,5 Game: In the first move, play 1 with probability 0.3268, play never 2, Play 4 with probability 0.2693, and play 5 with probability 0.4039.
For the 1,2,3,4,5 game we need games beyond 3*n. This is done, using an online applet. Here is the data:
1234 1235 1245 1345 2345 1234 0 -0.25 -0.431 -0.5 -0.75 1235 0.25 0 -0.573 -0.643 -0.318 1245 0.431 0.573 0 -0.454 -0.161 1345 0.5 0.643 0.454 0 0 2345 0.75 0.318 0.161 0 0
1234 1235 1245 1345 2345 12345 1 0.749 0.602 0.494 1
1 2 3 4 5 1 0 -1 -1 -1 0.75 2 1 0 -0.494 -0.494 -0.494 3 1 0.494 0 -0.602 -0.602 4 1 0.494 0.602 0 -0.749 5 -0.75 0.494 0.602 0.749 0
Solution of the 1,2,3,4,5 Game: At the beginning, play 5 with probability 0.4, 1 or 4 both with probability 0.3, but never play 2 or 3 at the beginning.
May 23-25, 2007
### Complete Strategy:
Ann Beth 1: 2: 3: 4: 5: --- 1: 2: 3: 4: 5: --- exp 12345 12345 0.3 0.3 0.4 0.3 0.3 0.4 0 12345 1235 0.15 0.59 0.26 1: 2: 3: 4: 5: 0.749 12345 1245 0.65 0.03 0.32 1: 2: 3: 4: 5: 0.602 12345 1345 0.31 0.31 0.38 1: 2: 3: 4: 5: 0.494 12345 125 0.2 0.58 0.22 1: 2: 3: 4: 5: 0.893 12345 135 0.25 0.52 0.23 1: 2: 3: 4: 5: 0.85 12345 145 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: 0.72 12345 15 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: 1
Your Hand Opponent 1: 2: 3: 4: 5: --- 1: 2: 3: 4: 5: --- exp 1234 1234 1 1 0 1234 2345 0.25 0.75 0.75 0.25 -0.75 1235 1235 0.3268 0.2693 0.4039 0.3268 0.2693 0.4039 0 1235 1245 0.2542 0.5702 0.1756 0.6184 0.0403 0.3413 -0.573 1235 1345 0.2417 0.4463 0.3121 0.0631 0.6241 0.3128 -0.643 1235 2345 0.1478 0.0418 0.6626 0.1478 0.4289 0.1067 0.1235 0.3409 -0.318 1245 1245 0.3268 0.2693 0.4039 0.3268 0.2693 0.4039 0 1245 1345 0.2561 0.5313 0.0154 0.1973 0.2676 0.062 0.2604 0.41 -0.453 1245 2345 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp 1345 1345 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp 1345 2345 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp 2345 2345 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp
Your Hand Opponent 1: 2: 3: 4: 5: --- 1: 2: 3: 4: 5: --- exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp You Opponent 1: 2: 3: 4: 5: 1: 2: 3: 4: 5: exp | 1,703 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-14 | latest | en | 0.519079 |
http://www.edugeek.net/forums/office-software/print-109759-counting-coloured-cells-ms-excel-2010-a.html | 1,418,924,645,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767301.77/warc/CC-MAIN-20141217075247-00005-ip-10-231-17-201.ec2.internal.warc.gz | 532,380,825 | 3,347 | # Counting coloured cells in MS Excel 2010
• 4th March 2013, 04:01 PM
QWERTY72
Counting coloured cells in MS Excel 2010
Hi
I was wondering if anybody could give me a helping hand please. I am exporting data from our SIMS into Excel and rather than counting *, =, +, - etc. I would like to count the coloured background of those cells. I can filter the data by the coloured background, but I don't know how to count them (other than manually, obviously).
Grateful for any help!
Thanks
• 4th March 2013, 04:06 PM
If it's only a quick count, after filtering if you highlight all the cells it should give a summary of stats including count.
If you want it permanently some form of COUNTIF() could be used.
• 4th March 2013, 04:12 PM
QWERTY72
I was going for the COUNTIF, but when when I tell Excel to look for a = (equal sign) within the designated range, it won't accept that. Now if I could tell Excel to count the number of red coloured cells, that would solve my problem.
• 4th March 2013, 04:14 PM
VeryPC_Tom_M
the only real way is to get VBA involved..
Here's a website with a bunch of colour functions for Excel 2003 - Color Functions In Excel
• 4th March 2013, 04:19 PM
LosOjos
Quote:
Originally Posted by QWERTY72
I was going for the COUNTIF, but when when I tell Excel to look for a = (equal sign) within the designated range, it won't accept that. Now if I could tell Excel to count the number of red coloured cells, that would solve my problem.
Don't overcomplicate things, count the symbols if you have them.
Code:
```To count "=": =COUNTIF(RANGE,"==") To count "+": =COUNTIF(RANGE,"=+") To count "-": =COUNTIF(RANGE,="=-")```
• 4th March 2013, 04:20 PM
QWERTY72
I don't know where to start with VBA - it looks terribly complicated.I'm just using the standard colours red, yellow, dark and light green.
• 4th March 2013, 04:31 PM
LosOjos
Quote:
Originally Posted by QWERTY72
I don't know where to start with VBA - it looks terribly complicated.I'm just using the standard colours red, yellow, dark and light green.
If you must count by colour (and I don't recommend it, see my previous post for Excel formulas that will count the symbols), then you'll want to add this code to a module in your workbook:
Code:
```Public Function cell_colour(Cell_Check As Range) As Long cell_colour = Cell_Check.Interior.Color End Function Public Function count_colour(Count_Range As Range, Colour As Long) As Long Dim x As Range count_colour = 0 For Each x In Count_Range If x.Interior.Color = Colour Then count_colour = count_colour + 1 End If Next x End Function```
Then you can use those functions in your worksheet like any other excel function, like so:
Code:
`=COUNT_COLOUR(RANGE, CELL_COLOUR(CELL))`
The range is the range of cells to count, and using the cell_colour function and highlighting a cell that is the same colour as that which you wish to count will get the colour code (simpler than working it out every time)
• 4th March 2013, 05:45 PM
QWERTY72
This works fine, thank you! But I will try the colour counting as well - been curious about it for quite some time. | 823 | 3,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-52 | latest | en | 0.965383 |
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# Miscellany1 - by choosing an orthonormal basis e 1 e N for...
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Analysis 2 Miscellaneous Problems 1 1. Let H be a Hilbert space, in which v is a point and X a closed subspace. Prove that min { k v - x k : x X } = max { |h v | y i| : y X , k y k = 1 } . 2. Fix a 2 . Prove that the following subset of 2 is compact: A = { z = ( z n ) n = 1 : n > 1 ⇒ | z n | 6 | a n |} . 3. Let X be an N -dimensional subspace of C [0 , 1] with inner product given by h f | g i = Z 1 0 f ( t ) g ( t )d t . Define K : [0 , 1] × [0 , 1] C
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Unformatted text preview: by choosing an orthonormal basis { e 1 , . . . , e N } for X and setting K ( s , t ) = N X n = 1 e n ( s ) e n ( t ) . Prove that K is choice-independent and has the (‘reproducing’) property that if f ∈ X then Z 1 K ( s , t ) f ( t )d t = f ( s ) . 1...
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## This note was uploaded on 07/08/2011 for the course MHF 3202 taught by Professor Larson during the Spring '09 term at University of Florida.
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https://m.ebrary.net/206264/computer_science/floating_point_arithmetic | 1,669,826,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00782.warc.gz | 428,488,183 | 7,557 | Home Computer Science
# Floating-Point Arithmetic
Let X and Y be two floating-point numbers, be expressed as (Xs, XJ;) and (Ys, YE), respectively. Therefore, the numerical value of X is Xs x BXE and that of Y is Ys x BYE. To explainthis, some realistic assumptions in this respect are needed to be made, which are as follows: • Xs is an ns-bit two's complement or sign-magnitude binary fraction; • X£ is an nE-bit integer in excess 2'V1 code, implying an exponent bias of 2V1;}} [1]
TABLE 7.1
Rules for Basic Operations Involving Floating-Point Operands
Operation Respective Rules Add/Subtract 1. Checking of zeros for both the numbers X and У. 2. To equalize the exponent of the two input numbers X and Y, choose the number with the smaller exponent and shift its significand right a number of steps equal to the difference in exponents of two numbers, X and Y. 3. Set the exponent of the number after shift equal to the largest exponent. 4. Perform addition/subtraction on the significands and determine the sign of the result. 5. Normalize the result value, if necessary. Multiply In multiplication and division operations, no alignment of mantissas is needed. 1. Add the exponents and subtract 127. 2. Multiply the significands, and determine the sign of the result. 3. Normalize the resulting value, if necessary. Divide 1. Subtract the exponents and addl27. 2. Divide the mantissas, and determine the sign of the result. 3. Normalize the resulting value, if necessary. *In multiply and divide rules, 127 is added or subtracted. This is due to using the excess-127 notation for exponents.
Basic operations: General methods for floating-point addition, subtraction, multiplication, and division are given in Table 7.1. For addition and subtraction, it is necessary to ensure that both operands have the same exponent value. This may require shifting the radix point on one of the operands to realize alignment. Multiplication and division are relatively simple, because the significands (mantissas) and exponents can be processed independently. The floating-point operations normally produce the usual expected expressible results, but at times, they may give rise to one of these situations, such as:
i. Significand underflow is observed while aligning significands that digits may flow off the right end of the significand. To cope with this situation, some form of rounding-off is required, to be explained later in section;
ii. significand overflow occurs when the addition of two significands of the same sign may result in a carry-out of the most significant bit. This can be fixed by realignment, which will be explained later;
iii. Exponent underflow happens when a negative exponent becomes less than the minimum possible exponent value (e.g. -145 is less than -127) in the prescribed format. This means that the number is too small to be represented, and thus may be considered to be equal to 0; and
iv. exponent overflow happens when a positive exponent exceeds the maximum possible exponent value defined in the prescribed format. This may be designated in some systems as +~ or -°°.
Floating-point addition and subtraction are relatively complex since the exponents of the two input operands must be made equal before the corresponding significands can be added or subtracted. Following the floating-point format as already described, the two operands must be placed in the respective registers within the ALU to execute the required operation. The floating-point includes an implicit bit in the significand, but that bit must be made explicit at the time of executing the operation. The procedures being followed to perform addition and subtraction, however, are explained in Table 7.1.
During addition/subtraction, if the signs of two numbers are the same, there also exists the possibility of significand overflow, the rectification of which, in turn, may invite exponent overflow. Whatever be it is, the appropriate actions would then be taken with suitable intimation, and possibly the operation is to be halted, and the subsequent needful actions are then required. After addition/subtraction, the result may be required to be normalized, which may invite exponent underflow. Again, suitable actions should be taken to resolve the situation.
A typical flowchart for performing addition/subtraction incorporating all the activities as mentioned in Table lalong with a solved example is given in the website: http:// routledge.com/9780367255732.
### Implementation: Floating-Point Unit
A floating-point arithmetic unit can be built up by connecting two loosely coupled fixed- point arithmetic circuits, one to be used as an exponent unit and the other as a significand (mantissa) unit. As the significand unit is required to perform all four basic arithmetic operations on the significands, a conventional fixed-point arithmetic circuit (already described earlier) can be used for this purpose. The exponent unit, however, is implemented by a relatively simpler circuit, capable of only adding, subtracting, and comparing exponents of the input operands. Comparison of exponents can be made by a comparator or by subtracting the exponents. With this idea, a schematic structure of a floating-point unit can be built up on the lines of the illustration shown in Figure 7.18.
The exponents of the input operands are loaded in registers £1 and E2, which are connected to an adder that computes El + E2. The comparison of exponents required for addition and subtraction is made by computing El - E2 (i.e. El + (-E2), essentially is an addition) and placing the result in a counter E. The larger exponent is then determined from the sign of E. The bit-shift of one of the significands (mantissas) required before the addition/subtraction of the significands can be controlled by E. The magnitude of E is sequentially decremented to zero. After each such decrement, the corresponding significand located in the significand unit is shifted one-digit position. After the needed alignment of the respective significand
FIGURE 7.18
Schematic block diagram of a floating-point arithmetic unit.
(equalizing the exponent, i.e. when £ becomes 0, of the two input numbers X and Y), they are processed in the usual manner depending on the type of arithmetic operation being required. The exponent of the result is also computed and is placed in E.
All the computers have the fixed-point arithmetic instructions as well as the floatingpoint instructions; it is, hence, always desirable to have a single unit within the ALU to execute both these types of instructions. But, as the sophisticated, faster, and also cheaper electronic technology is now readily available in abundance, it is almost common nowadays in most of the computer systems to incorporate separate units: one dedicated for fixed-point integer (FXU) and another for floating-point arithmetic operations (FPU). Separation of these two individual units located within the architecture of ALU facilitates the execution of fixed-point and floating-point instructions to continue in parallel.
## Multiplication and Division
Multiplication and division are relatively simpler and somewhat easier than addition and subtraction, in that no alignment of significand (equalization of the exponents) is needed. As usual, the input operands here are represented in 2's (two's) complementary form.
In multiplication, if either operand is 0, the result is automatically declared as 0. The next step is to add the exponents. If the exponents are stored in biased form, the sum of the exponents would then contain double the bias value. Hence, the bias value must be subtracted from the sum. The result may sometimes give rise to a situation of exponent overflow or underflow which must be intimated with the termination of the process. However, if the exponent of the product (result) lies within the specified range, the next step is to multiply the significands of the input operands, taking into account their signs, as is done for integer multiplication (already described earlier). The product (result) will be double the length of the multiplier or multiplicand, which one is larger. The extra bits may be lost due to rounding-off the result. After obtaining the product, the result as usual needs to be normalized, and rounded-off, if required. The action of normalization may sometimes lead to a situation of exponent underflow. Appropriate actions should then be taken to resolve the situation.
Division is performed almost on the same lines as multiplication. Here too, the testing of 0 is to be carried out first. If the divisor is 0, an error is to be declared, or the result may be set to infinity, as per the guidelines of the particular implementation. But, for having a dividend of 0, the final result will be 0. The next step is to subtract the divisor exponent from the dividend exponent. This subtraction removes the bias, which must be added back in, but this addition may result in exponent overflow. However, appropriate tests are then made to inspect exponent underflow or overflow, if any, and a befitting test report can then be accordingly issued. The next action is to divide the significands of the two input operands. Finally, the result as obtained will go through the usual process of normalization, and rounding, if needed.
Two typical flowcharts for separately performing multiplication and division incorporating all the activities as described in Table 1, respectively, are shown in the website: http://routledge.com/9780367255732.
### Implementation: Floating-Point Multiplication
A multiplier circuit can be implemented using a multistage CSA circuit (already described earlier). This circuit is popularly known as a Wallace tree after the name of its inventor (Wallace 1964). The inputs to the adder tree are n terms of the form M, = x, Y 2k. Here, M, represents the multiplicand Y multiplied by the ;'th multiplier bit weighted by the appropriate power of 2. Suppose M, is 2n-bit long, and that a full double-length product is required,
ti-i
the desired product P is ^ M,. This sum is computed by the CSA tree that produces a
• (=0
• 2n-bit sum and a 2n-bit carry word. The final carry assimilation is then usually performed by a fast adder, a CLA, for instance, with normal internal carry propagation.
A brief detail of this topic along with a befitting figure is given in the website: http:// routledge.com/9780367255732.
• [1] В is the base w'hich here is equal to 2. The above assumptions also hold good for Y. In addition, it is also assumed that thefloating-point numbers are stored in their normalized form only, and thus, the final resultof each floating-point operation should also be normalized.
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Machines and Mechanisms Applied Kinematic Analysis
# TEXTBOOK SOLUTIONS FOR Machines and Mechanisms Applied Kinematic Analysis 4th Edition
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PROBLEM
Chapter: Problem:
Graphically position the links for the rear-wiper mechanism shown in Figure. Then reposition the links as the 2-in. crank is rotated 50° clockwise. Determine the resulting angular displacement of the wiper arm and the linear displacement at the end of the wiper blade.
FIGURE
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Follow the steps to graphically position the links of rear wiper mechanism:
1) Locate the two points apart and mark them as and
2) Draw an arc of radius from and position the crank at an angle of in clockwise direction with respect to horizontal.
3) Draw arcs of radius and from points and respectively, intersecting below horizontal at point . Join and .
4) Draw an arc of radius from and position the wiper arm at an angle with respect to in clockwise direction.
• Chapter , Problem is solved.
Corresponding Textbook
Machines and Mechanisms Applied Kinematic Analysis | 4th Edition
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Finding the Lexicographically K-th Combinatorial Object | Reply NameFinding the Lexicographically K-th Combinatorial ObjectProblemFrom all combinatorial objects of length n generated by specified rules you need to find the lexicographically k-th one.SolutionLet k be 0-based. Then let’s suppose am,p to be a number of combinatorial objects of length m satisfying certain condition marked as p. A totality of all p’s for some m describes a condition which suffixes of length m of any our combinatorial objects of length n must satisfy (and every such suffix must satisfy exactly one of those p’s).So the general solution is like this pseudo-code:```class CombinatorialObject { void getMthSymbol (int m, int p, int k, vector &ans) { if (m==0) return; ans.push_back(symbol(m,p,k)); getMthSymbol(m-1,newP(m,p,k),newK(m,p,k),ans); } }; ```Where function symbol(m,p,k) returns an appropriate symbol, function newP(m,p,k) returns new condition for a suffix of length m-1, and function newK(m,p,k) returns a number of appropriate suffix under this condition. These two functions may rely on am,p. To obtain the answer, we should call getMthSymbol with arguments n, condition p that actually means no condition in addition to problem description, k, and an empty array ans.But there are three important questions: how we should calculate a(m,p), newP(m,p,k), and newK(m,p,k). If only there was a universal method, we would describe it here, but modes of calculating these vary too much for different problems, so we explain them on examples.DiscussionIf k is small enough, you may use brute force algorithm generating combinatorial objects in lexicographical order and simply take the k-th one. Our sophisticated approach is for cases when there are too many objects to iterate throw for passing time limit.I couldn’t find any SRM problems requiring this algorithm easier than Division One – Level Three, so I suppose it will be more understandable if I give some other examples.In simple cases you can calculate am,p directly and find k using a formula. Let’s suppose we need to find k-th lexicographically array of n integers from 0 to n-1, inclusive. This is basically the task of writing down number k in n-based notation, and can be solved easier, but we'll handle it in a manner that illustrates our idea.In this simple case there are no constraints p on suffixes depending on previously set symbols, and am,p (actually simply am) is just a number of arrays of length m where each element is an integer between 0 and n-1, inclusive. am=mn, so we precompute it in array pow[m] before calling recursion.```#include using namespace std; class Arrays { private: vector pow; void getMthSymbol(int m, int k, vector &ans) { if (m==0) return; ans.push_back(k/pow[m-1]); getMthSymbol(m-1,k%pow[m-1],ans); } public: vector getKth(int n, int k) { pow.resize(n); int i; pow[0]=1; for (i=1; i ans(0); getMthSymbol(n,k,ans); return ans; } }; ```
Re: Finding the Lexicographically K-th Combinatorial Object (response to post by Ferlon) | Reply
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Converting Start and End Day/Time Fields into Number Rounded
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6 - Interface Innovator
Hi everyone! I have done several searches and cannot find exactly what I am trying to do. I am relatively new to all of this and my ADHD brain struggles with all of the technical stuff at first, so I apologize if this is very obvious.
I want to simply have a “total hours” field that takes the start and end date/time fields and puts it in hours rounded to the nearest quarter. 2/27/2020 8:00AM to 2/27/2020 9:45 would automatically compute to 1.75.
This is for a timesheet :slightly_smiling_face:
Thanks!
1 Solution
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18 - Pluto
A duration rounded to the nearest quarter hour
Existing fields:
• {Start} a date/time field for when the task starts
• {End} a date/time field for when the task ends
The goal is a field showing the duration between the start and end times
• in hours
• in decimal format
• rounded to the nearest quarter of an hour
Getting the duration between two times
You want a formula field that uses the `DATETIME_DIFF` function. It and all the other functions in this post are documented in the formula field reference. In addition to the fields with the start and end times, the function takes a unit specifier. The unit specifier for hours is `h`.
`DATETIME_DIFF({End}, {Start}, 'h')`
If you do not need rounding, you can just use this formula. In order to see the decimal part, set the number of decimals in the formatting tab when defining the field.
If you do need rounding to the nearest quarter hour, read on …
The logic for rounding a duration to a quarter hour
The rounding functions in Airtable round to a specific decimal place, not to a quarter of an hour. Thus we need figure out the logic behind what functions and operators to combine.
There are many ways to go about this task, including this formula with 8 lines and this formula with 5 lines. I also created several possibilities (including one with 15 lines) before settling on this one:
1. Calculate the number of minutes
2. Round the number of minutes to the nearest multiple of 15
3. Convert minutes to hours
Calculating the number of minutes
Use the same function, but change the unit specifier.
`DATETIME_DIFF({End}, {Start}, 'm')`
Rounding the number of minutes
An internet search for how to round a number {n} to the nearest multiple of another number {m} produced this formula:
`ROUND({n}/{m})*{m}`
Thus, to round the number of minutes to the nearest multiple of 15:
1. Divide the minutes remaining by 15
`DATETIME_DIFF({End}, {Start}, 'm') / 15`
2. Round the result to the nearest whole number
`ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15)`
3. Multiply times 15
`ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) * 15`
Convert minutes to hours
Divide by 60 to convert minutes to hours
``````ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) * 15 / 60
``````
Notice that the last two operations are multiplying by 15 and dividing by 60. When combined those two operations are the same as simply dividing by 4.
``````ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) / 4
``````
Make sure you set the number of decimal places in the formatting tab when defining the field, or you won’t see the decimal part.
Tidying things up
The above formula will produce an `#ERROR!` if there is no start or end time.
To avoid the unsightly error, wrap the formula in an `IF` function that tests for the existence of values in both of the date/time fields.
Here is the final formula.
``````IF( AND({End}, {Start}),
ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) / 4
)
``````
18 - Pluto
A duration rounded to the nearest quarter hour
Existing fields:
• {Start} a date/time field for when the task starts
• {End} a date/time field for when the task ends
The goal is a field showing the duration between the start and end times
• in hours
• in decimal format
• rounded to the nearest quarter of an hour
Getting the duration between two times
You want a formula field that uses the `DATETIME_DIFF` function. It and all the other functions in this post are documented in the formula field reference. In addition to the fields with the start and end times, the function takes a unit specifier. The unit specifier for hours is `h`.
`DATETIME_DIFF({End}, {Start}, 'h')`
If you do not need rounding, you can just use this formula. In order to see the decimal part, set the number of decimals in the formatting tab when defining the field.
If you do need rounding to the nearest quarter hour, read on …
The logic for rounding a duration to a quarter hour
The rounding functions in Airtable round to a specific decimal place, not to a quarter of an hour. Thus we need figure out the logic behind what functions and operators to combine.
There are many ways to go about this task, including this formula with 8 lines and this formula with 5 lines. I also created several possibilities (including one with 15 lines) before settling on this one:
1. Calculate the number of minutes
2. Round the number of minutes to the nearest multiple of 15
3. Convert minutes to hours
Calculating the number of minutes
Use the same function, but change the unit specifier.
`DATETIME_DIFF({End}, {Start}, 'm')`
Rounding the number of minutes
An internet search for how to round a number {n} to the nearest multiple of another number {m} produced this formula:
`ROUND({n}/{m})*{m}`
Thus, to round the number of minutes to the nearest multiple of 15:
1. Divide the minutes remaining by 15
`DATETIME_DIFF({End}, {Start}, 'm') / 15`
2. Round the result to the nearest whole number
`ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15)`
3. Multiply times 15
`ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) * 15`
Convert minutes to hours
Divide by 60 to convert minutes to hours
``````ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) * 15 / 60
``````
Notice that the last two operations are multiplying by 15 and dividing by 60. When combined those two operations are the same as simply dividing by 4.
``````ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) / 4
``````
Make sure you set the number of decimal places in the formatting tab when defining the field, or you won’t see the decimal part.
Tidying things up
The above formula will produce an `#ERROR!` if there is no start or end time.
To avoid the unsightly error, wrap the formula in an `IF` function that tests for the existence of values in both of the date/time fields.
Here is the final formula.
``````IF( AND({End}, {Start}),
ROUND(DATETIME_DIFF({End}, {Start}, 'm') / 15) / 4
)
`````` | 1,718 | 6,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | longest | en | 0.884658 |
http://www.nektar.info/wiki/3.2/Reference/XmlFileFormat/Geometry?version=1 | 1,369,277,639,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702749808/warc/CC-MAIN-20130516111229-00007-ip-10-60-113-184.ec2.internal.warc.gz | 615,479,642 | 4,385 | Version 1 (modified by ccantwel, 14 months ago) (diff)
--
# XML Geometry definition
This section defines the mesh. It specifes a list of vertices, edges (in two or three dimensions) and faces (in three dimensions) and how they connect to create the elemental decomposition of the domain. It also defines a list of composites which are used in the Expansions and Conditions sections of the file to describe the polynomial expansions and impose boundary conditions.
The GEOMETRY section is structured as
``` <GEOMETRY DIM="2" SPACE="2">
<VERTEX>
...
</VERTEX>
<EDGE>
...
</EDGE>
<FACE>
...
</FACE>
<ELEMENT>
...
</ELEMENT>
<CURVED>
...
</CURVED>
<COMPOSITE>
...
</COMPOSITE>
<DOMAIN> ... </DOMAIN>
</GEOMETRY>
```
It has two attributes:
• DIM: specifies the dimension of the expansion elements.
• SPACE: specifies the dimension of the space in which the elements exist.
These attributes allow, for example, a two-dimensional surface to be embedded in a three-dimensional space. The FACES section is only present when DIM=3. The CURVED section is only present if curved edges or faces are present in the mesh.
## Vertices
Vertices have three coordinates. Each has a unique vertex ID. They are defined in the file within VERTEX subsection as follows:
``` <VERTEX>
<V ID="0"> 0.0 0.0 0.0 </V>
...
</VERTEX>
```
VERTEX subsection has three optional attributes: XSCALE, YSCALE and ZSCALE. They specify scaling factors to corresponding vertex coordinates. For example, the following snippet
``` <VERTEX XSCALE="5">
<V ID="0"> 0.0 0.0 0.0 </V>
<V ID="1"> 1.0 2.0 0.0 </V>
</VERTEX>
```
defines two vertices with coordinates . Values of XSCALE, YSCALE and ZSCALE attributes can be arbitrary analytic expressions depending on pre-defined constants, parameters defined earlier in the XML file and mathematical operations/functions of the latter. If omitted, default scaling factors 1.0 are assumed.
## Edges
Edges are defined by two vertices. Each edge has a unique edge ID. They are defined in the file with a line of the form
``` <E ID="0"> 0 1 </E>
```
## Faces
Faces are defined by three or more edges. Each face has a unique face ID. They are defined in the file with a line of the form
``` <T ID="0"> 0 1 2 </T>
<Q ID="1"> 3 4 5 6 </Q>
```
The choice of tag specified (T or Q), and thus the number of edges specified depends on the geometry of the face (triangle or quadrilateral).
## Element
Elements define the top-level geometric entities in the mesh. Their definition depends upon the dimension of the expansion. For two-dimensional expansions, an element is defined by a sequence of three or four edges. For three-dimensional expansions, the element is defined by a list of faces. Elements are defined in the file with a line of the form
``` <T ID="0"> 0 1 2 </T>
<H ID="1"> 3 4 5 6 7 8 </H>
```
Again, the choice of tag specified depends upon the geometry of the element. The element tags are:
• S: Segment
• T: Triangle
• A: Tetrahedron
• P: Pyramid
• R: Prism
• H: Hexahedron
## Curved Edges and Faces
For mesh elements with curved edges and/or curved faces, a separate entry is used to describe the control points for the curve. Each curve has a unique curve ID and is associated with a predefined edge or face. The total number of points in the curve (including end points) and their distribution is also included as attributes. The control points are listed in order, each specified by three coordinates. Curved edges are defined in the file with a line of the form
``` <E ID="3" EDGEID="7" TYPE="PolyEvenlySpaced" NUMPOINTS="3">
0.0 0.0 0.0 0.5 0.5 0.0 1.0 0.0 0.0
</E>
```
## Composites
Composites define collections of elements, faces or edges. Each has a unique composite ID associated with it. All components of a composite entry must be of the same type. The syntax allows components to be listed individually or using ranges. Examples include
``` <C ID="0"> T[0-862] </C>
<C ID="1"> E[68,69,70,71] </C>
```
## Domain
This tag specifies composites which describe the entire problem domain. It has the form of
``` <DOMAIN> C[0] </DOMAIN>
``` | 1,073 | 4,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.812472 |
https://stage.geogebra.org/m/F3EnbEQe | 1,695,376,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00286.warc.gz | 604,987,753 | 10,788 | # Parallelograms Investigation
## The object is a parallelogram.
After you complete each step, feel free to undo each measurement so you can start fresh at each step. This might be easier, or else you'll have a complicated diagram at the end of the investigation.
1. Use the distance feature (located under the angle icon [4th from the right]) to measure all the sides. What do you notice about the opposite sides?
2. Use the angle measure feature (located under the angle feature icon [4th from the right]) to measure each angle of the parallelogram. What do you notice about opposite angles?
3. Find the sum of two consecutive angles (like angles A and B). Then find the sum of two different consecutive angles. What do you notice about the sums of consecutive angles?
• Using the segment feature (under the line line feature icon [3rd from the left]), draw the diagonals of the parallelogram.
• Use the intersection feature (located under the point icon [2nd from the left]) to find the intersection point of the parallelogram's diagonals.
• Use the distance feature to measure the distance from each vertex to the intersection point.
• What is true about the intersection point?
You're all done with your parallelogram investigation! Good work!! | 270 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.896907 |
https://punctualessays.com/2021/06/05/the-following-table-shows-the-age-and-annual-salary-in-thousands/ | 1,701,756,590,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00136.warc.gz | 555,862,673 | 11,659 | The following table shows the age and annual salary (in thousands
According to Advertising Age’s annual salary review, Mark Hurd, the 49-year-old chairman, president, and CEO of Hewlett-Packard Co., received an annual salary of \$817,000, a bonus of more than \$5 million, and other compensation exceeding \$17 million. His total compensation was slightly better than the average CEO total pay of \$12.4 million. The following table shows the age and annual salary (in thousands of dollars) for Mark Hurd and 14 other executives who led publicly held companies. Executive Title Company Age Salary (1000s) Charles Prince,Chmn/CEO,Citigroup,56,1000 Harol McGraw III,Chmn/Pres/CEO,MCGraw-Hill Cos.,57,1172 James Dimon,Pres/CEO,JP Morgan Chase & Co,50,1000 K. Rupert Murdoch, Chmn/CEO,News Corp.,75,4509 Kenneth D. Lewis,Chmn/Pres/CEO,Bank of America,58,1500 Kenneth I. Chenault,Chmn/CEO,American Express Co,54,1092 Louis C. Camilleri,Chmn/CEO, Altria Group,51,1663 Mark V Hurd,Chmn/Pres/CEO,Hewlett-Packard Co,49,817 Martin S. Sorrell,CEO,WPP Group,61,1562 Robert L Nardelli, Chmn/Pres/CEO, Home Depot, 57, 2164 Samuel J Palmisano ,Chmn/Pres/CEO, IBM Corp, 55, 1680 David C Novak, Chmn/Pres/CEO, Yum Brands, 53, 1173 Henry R Silverman, Chmn/CEO, Cendant Corp, 65, 3300 Robert C Wright, Chmn/CEO, NBC Universal, 62, 2500 Sumner Redstone, Exec Chmn/Founder, Viacom, 82, 5807 a)Develop a scatter diagram for these data with the age of the executive as the independent variable. b) What does the scatter diagram developed in part (a) indicate about the relationship between the two variables? c) Develop the least squares estimated regression equation. d) Suppose Bill Gustin is the 72-year-old chairman, president, and CEO of a major electronics company. Predict the annual salary for Bill Gustin.
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# Most economists in the United States seem captivated by the
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Joined: 28 Dec 2005
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Most economists in the United States seem captivated by the [#permalink]
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Updated on: 12 Oct 2017, 14:24
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Part of New RC Series- Please check this link for more questions
Most economists in the United States seem captivated by the spell of the free market. Consequently, nothing seems good or normal that does not accord with the requirements of the free market. A price that is determined by the seller or, for that matter, established by anyone other than the aggregate of consumers seems pernicious. Accordingly, it requires a major act of will to think of price-fixing (the determination of prices by the seller) as both “normal” and having a valuable economic function. In fact, price-fixing is normal in all industrialized societies because the industrial system itself provides, as an effortless consequence of its own development, the price-fixing that it requires. Modern industrial planning requires and rewards great size. Hence, a comparatively small number of large firms will be competing for the same group of consumers. That each large firm will act with consideration of its own needs and thus avoid selling its products for more than its competitors charge is commonly recognized by advocates of free-market economic theories. But each large firm will also act with full consideration of the needs that it has in common with the other large firms competing for the same customers. Each large firm will thus avoid significant price-cutting, because price-cutting would be prejudicial to the common interest in a stable demand for products. Most economists do not see price-fixing when it occurs because they expect it to be brought about by a number of explicit agreements among large firms; it is not.
Moreover, those economists who argue that allowing the free market to operate without interference is the most efficient method of establishing prices have not considered the economies of non-socialist countries other than the United states. These economies employ intentional price-fixing, usually in an overt fashion. Formal price-fixing by cartel and informal price-fixing by agreements covering the members of an industry are commonplace. Were there something peculiarly efficient about the free market and inefficient about price-fixing, the countries that have avoided the first and used the second would have suffered drastically in their economic development. There is no indication that they have.
Socialist industry also works within a framework of controlled prices. In the early 1970’s, the Soviet Union began to give firms and industries some of the flexibility in adjusting prices that a more informal evolution has accorded the capitalist system. Economists in the United States have hailed the change as a return to the free market. But Soviet firms are no more subject to prices established by a free market over which they exercise little influence than are capitalist firms; rather, Soviet firms have been given the power to fix prices.
1. The primary purpose of the passage is to
(A) refute the theory that the free market plays a useful role in the development of industrialized societies
(B) suggest methods by which economists and members of the government of the United States can recognize and combat price-fixing by large firms
(C) show that in industrialized societies price-fixing and the operation of the free market are not only compatible but also mutually beneficial
(D) explain the various ways in which industrialized societies can fix prices in order to stabilize the free market
(E) argue that price-fixing, in one form or another, is an inevitable part of and benefit to the economy of any industrialized society
Spoiler: :: OA
E
2. The passage provides information that would answer which of the following questions about price-fixing?
I. What are some of the ways in which prices can be fixed?
II. For what products is price-fixing likely to be more profitable that the operation of the free market?
III. Is price-fixing more common in socialist industrialized societies or in non-socialist industrialized societies?
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Spoiler: :: OA
A
3. The author’s attitude toward “Most economists in the United States” (line 1) can best be described as
(A) spiteful and envious
(B) scornful and denunciatory
(C) critical and condescending
(D) ambivalent but deferential
(E) uncertain but interested
Spoiler: :: OA
C
4. It can be inferred from the author’s argument that a price fixed by the seller “seems pernicious” (line 7) because
(A) people do not have confidence in large firms
(B) people do not expect the government to regulate prices
(C) most economists believe that consumers as a group should determine prices
(D) most economists associate fixed prices with communist and socialist economies
(E) most economists believe that no one group should determine prices
Spoiler: :: OA
C
5. The suggestion in the passage that price-fixing in industrialized societies is normal arises from the author’s statement that price-fixing is
(A) a profitable result of economic development
(B) an inevitable result of the industrial system
(C) the result of a number of carefully organized decisions
(D) a phenomenon common to industrialized and non-industrialized societies
(E) a phenomenon best achieved cooperatively by government and industry
Spoiler: :: OA
B
6. According to the author, price-fixing in non-socialist countries is often
(A) accidental but productive
(B) illegal but useful
(C) legal and innovative
Spoiler: :: OA
E
7. According to the author, what is the result of the Soviet Union’s change in economic policy in the 1970’s?
(A) Soviet firms show greater profit.
(B) Soviet firms have less control over the free market.
(D) Soviet firms have some authority to fix prices.
(E) Soviet firms are more responsive to the free market.
Spoiler: :: OA
D
8. With which of the following statements regarding the behavior of large firms in industrialized societies would the author be most likely to agree?
(A) The directors of large firms will continue to anticipate the demand for products.
(B) The directors of large firms are less interested in achieving a predictable level of profit than in achieving a large profit.
(C) The directors of large firms will strive to reduce the costs of their products.
(D) Many directors of large firms believe that the government should establish the prices that will be charged for products.
(E) Many directors of large firms believe that the price charged for products is likely to increase annually.
Spoiler: :: OA
A
9. In the passage, the author is primarily concerned with
(A) predicting the consequences of a practice
(B) criticizing a point of view
(C) calling attention to recent discoveries
(D) proposing a topic for research
(E) summarizing conflicting opinions
Spoiler: :: OA
B
Originally posted by pmenon on 07 Jun 2009, 20:52.
Last edited by Gnpth on 12 Oct 2017, 14:24, edited 2 times in total.
Economist GMAT Tutor Discount Codes Optimus Prep Discount Codes e-GMAT Discount Codes
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Joined: 16 Apr 2009
Posts: 275
Re: Most economists in the United States seem captivated by the [#permalink]
### Show Tags
13 Jun 2009, 15:12
1
1-E
2-A
3-B
4-D
5-b
6-E
7-D
8-C
9-B
_________________
Director
Joined: 23 May 2008
Posts: 731
Re: Most economists in the United States seem captivated by the [#permalink]
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16 Jun 2009, 15:51
tough passage
I got
SVP
Joined: 04 May 2006
Posts: 1751
Schools: CBS, Kellogg
Re: Most economists in the United States seem captivated by the [#permalink]
### Show Tags
01 Jul 2009, 09:34
1 C 08:22
2 A 00:53
3 C 00:36
4 C 02:07
5 D 01:21
6 E 00:17
7 B 00:25
8 A 00:39
9 B 00:18
At the noise!
_________________
Manager
Joined: 22 Jul 2009
Posts: 170
Location: Manchester UK
Re: Most economists in the United States seem captivated by the [#permalink]
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03 Dec 2009, 12:49
Man....I got just 3 correct. This seemed to be one of the toughest. Hope not to get this kind on D-Day.
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Joined: 13 Aug 2009
Posts: 179
Schools: Sloan '14 (S)
Re: Most economists in the United States seem captivated by the [#permalink]
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09 Dec 2009, 05:29
1. E
2. A
3. C
4. C
5. B
6. E
7. d
8. a
9. b
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18 Dec 2009, 06:38
Here are the OAs.
EACCBEDAB
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Re: Most economists in the United States seem captivated by the [#permalink]
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03 Jan 2010, 08:31
pmenon wrote:
Most economists in the United States seem captivated by the spell of the free market. Consequently, nothing seems good or normal that does not accord with the requirements of the free market. A price that is determined by the seller or, for that matter (for that matter: so far as that is concerned), established by anyone other than the aggregate of consumers seems pernicious. Accordingly, it requires a major act of will to think of price-fixing (the determination of prices by the seller) as both “normal” and having a valuable economic function. In fact, price-fixing is normal in all industrialized societies because the industrial system itself provides, as an effortless consequence of its own development, the price-fixing that it requires. Modern industrial planning requires and rewards great size. Hence, a comparatively small number of large firms will be competing for the same group of consumers. That each large firm will act with consideration of its own needs and thus avoid selling its products for more than its competitors charge is commonly recognized by advocates of free-market economic theories. But each large firm will also act with full consideration of the needs that it has in common with the other large firms competing for the same customers. Each large firm will thus avoid significant price-cutting, because price-cutting would be prejudicial to the common interest in a stable demand for products. Most economists do not see price-fixing when it occurs because they expect it to be brought about by a number of explicit agreements among large firms; it is not.
Moreover, those economists who argue that allowing the free market to operate without interference is the most efficient method of establishing prices have not considered the economies of non-socialist countries other than the United states. These economies employ intentional price-fixing, usually in an overt fashion. Formal price-fixing by cartel and informal price-fixing by agreements covering the members of an industry are commonplace. Were there something peculiarly efficient about the free market and inefficient about price-fixing, the countries that have avoided the first and used the second would have suffered drastically in their economic development. There is no indication that they have.
Socialist industry also works within a framework of controlled prices. In the early 1970’s, the Soviet Union began to give firms and industries some of the flexibility in adjusting prices that a more informal evolution has accorded the capitalist system. Economists in the United States have hailed the change as a return to the free market. But Soviet firms are no more subject to prices established by a free market over which they exercise little influence than are capitalist firms; rather, Soviet firms have been given the power to fix prices.
1. The primary purpose of the passage is to
(A) refute the theory that the free market plays a useful role in the development of industrialized societies
(B) suggest methods by which economists and members of the government of the United States can recognize and combat price-fixing by large firms
(C) show that in industrialized societies price-fixing and the operation of the free market are not only compatible but also mutually beneficial
(D) explain the various ways in which industrialized societies can fix prices in order to stabilize the free market
(E) argue that price-fixing, in one form or another, is an inevitable part of and benefit to the economy of any industrialized society
2. The passage provides information that would answer which of the following questions about price-fixing?
I. What are some of the ways in which prices can be fixed?
II. For what products is price-fixing likely to be more profitable that the operation of the free market?
III. Is price-fixing more common in socialist industrialized societies or in non-socialist industrialized societies?
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
3. The author’s attitude toward “Most economists in the United States”(line 1) can best be described as
(A) spiteful and envious
(B) scornful and denunciatory
(C) critical and condescending
(D) ambivalent but deferential
(E) uncertain but interested
4. It can be inferred from the author’s argument that a price fixed by the seller “seems pernicious” (line 7) because
(A) people do not have confidence in large firms
(B) people do not expect the government to regulate prices
(C) most economists believe that consumers as a group should determine prices
(D) most economists associate fixed prices with communist and socialist economies
(E) most economists believe that no one group should determine prices
5. The suggestion in the passage that price-fixing in industrialized societies is normal arises from the author’s statement that price-fixing is
(A) a profitable result of economic development
(B) an inevitable result of the industrial system
(C) the result of a number of carefully organized decisions
(D) a phenomenon common to industrialized and non-industrialized societies
(E) a phenomenon best achieved cooperatively by government and industry
6. According to the author, price-fixing in non-socialist countries is often
(A) accidental but productive
(B) illegal but useful
(C) legal and innovative
7. According to the author, what is the result of the Soviet Union’s change in economic policy in the 1970’s?
(A) Soviet firms show greater profit.
(B) Soviet firms have less control over the free market.
(D) Soviet firms have some authority to fix prices.
(E) Soviet firms are more responsive to the free market.
8. With which of the following statements regarding the behavior of large firms in industrialized societies would the author be most likely to agree?
(A) The directors of large firms will continue to anticipate the demand for products.
(B) The directors of large firms are less interested in achieving a predictable level of profit than in achieving a large profit.
(C) The directors of large firms will strive to reduce the costs of their products.
(D) Many directors of large firms believe that the government should establish the prices that will be charged for products.
(E) Many directors of large firms believe that the price charged for products is likely to increase annually.
9. In the passage, the author is primarily concerned with
(A) predicting the consequences of a practice
(B) criticizing a point of view
(C) calling attention to recent discoveries
(D) proposing a topic for research
(E) summarizing conflicting opinions
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14 mins
A
A
C
C
B
A
D
C
B
OE plz!!
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Re: Most economists in the United States seem captivated by the [#permalink]
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30 Jul 2010, 13:52
Most economists in the United States seem captivated by the spell of the free market. Consequently, nothing seems good or normal that does not accord with the requirements of the free market. A price that is determined by the seller or, for that matter, established by anyone other than the aggregate of consumers seems pernicious. Accordingly, it requires a major act of will to think of price-fixing (the determination of prices by the seller) as both “normal” and having a valuable economic function. In fact, price-fixing is normal in all industrialized societies because the industrial system itself provides, as an effortless consequence of its own development, the price-fixing that it requires. Modern industrial planning requires and rewards great size. Hence, a comparatively small number of large firms will be competing for the same group of consumers. That each large firm will act with consideration of its own needs and thus avoid selling its products for more than its competitors charge is commonly recognized by advocates of free-market economic theories. But each large firm will also act with full consideration of the needs that it has in common with the other large firms competing for the same customers. Each large firm will thus avoid significant price-cutting, because price-cutting would be prejudicial to the common interest in a stable demand for products. Most economists do not see price-fixing when it occurs because they expect it to be brought about by a number of explicit agreements among large firms; it is not.
Moreover, those economists who argue that allowing the free market to operate without interference is the most efficient method of establishing prices have not considered the economies of non-socialist countries other than the United states. These economies employ intentional price-fixing, usually in an overt fashion. Formal price-fixing by cartel and informal price-fixing by agreements covering the members of an industry are commonplace. Were there something peculiarly efficient about the free market and inefficient about price-fixing, the countries that have avoided the first and used the second would have suffered drastically in their economic development. There is no indication that they have.
Socialist industry also works within a framework of controlled prices. In the early 1970’s, the Soviet Union began to give firms and industries some of the flexibility in adjusting prices that a more informal evolution has accorded the capitalist system. Economists in the United States have hailed the change as a return to the free market. But Soviet firms are no more subject to prices established by a free market over which they exercise little influence than are capitalist firms; rather, Soviet firms have been given the power to fix prices.
3. The author’s attitude toward “Most economists in the United States”(line 1) can best be described as
(A) spiteful and envious
(B) scornful and denunciatory
(C) critical and condescending
(D) ambivalent but deferential
(E) uncertain but interested
4. It can be inferred from the author’s argument that a price fixed by the seller “seems pernicious” (line 7) because
(A) people do not have confidence in large firms
(B) people do not expect the government to regulate prices
(C) most economists believe that consumers as a group should determine prices
(D) most economists associate fixed prices with communist and socialist economies
(E) most economists believe that no one group should determine prices
8. With which of the following statements regarding the behavior of large firms in industrialized societies would the author be most likely to agree?
(A) The directors of large firms will continue to anticipate the demand for products.
(B) The directors of large firms are less interested in achieving a predictable level of profit than in achieving a large profit.
(C) The directors of large firms will strive to reduce the costs of their products.
(D) Many directors of large firms believe that the government should establish the prices that will be charged for products.
(E) Many directors of large firms believe that the price charged for products is likely to increase annually.
..............................
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Re: Most economists in the United States seem captivated by the [#permalink]
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Updated on: 01 Aug 2010, 06:31
gauravnagpal wrote:
c d b
Btw you got the last 2 wrong
Originally posted by goalsnr on 31 Jul 2010, 06:45.
Last edited by goalsnr on 01 Aug 2010, 06:31, edited 1 time in total.
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Re: Most economists in the United States seem captivated by the [#permalink]
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31 Jul 2010, 09:41
1
3) c.
4) E. => "A price that is determined by the seller or, for that matter, established by anyone other than the aggregate of consumers seems pernicious" in the first paragraph
8) b => "Modern industrial planning requires and rewards great size."
What is the OA?
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Re: Most economists in the United States seem captivated by the [#permalink]
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31 Jul 2010, 22:04
C,C,A IMO
Not too sure about the last one. Just guessed it.
OA?
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Re: Most economists in the United States seem captivated by the [#permalink]
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01 Aug 2010, 06:33
bibha wrote:
C,C,A IMO
Not too sure about the last one. Just guessed it.
OA?
What is the reasoning behind the answer choices?
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Re: Most economists in the United States seem captivated by the [#permalink]
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01 Aug 2010, 07:29
2. A price that is determined by the seller or, for that matter, established by anyone other than the aggregate of consumers seems pernicious.
This means that price should be determined by the consumer group.
3.Each large firm will thus avoid significant price-cutting, because price-cutting would be prejudicial to the common interest in a stable demand for products
As i said, that was a guess....i don't have any reason...Just saw the word "demand" haha
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Re: Most economists in the United States seem captivated by the [#permalink]
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01 Aug 2010, 18:33
1
C,C,B
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Re: Most economists in the United States seem captivated by the [#permalink]
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26 Jan 2012, 00:38
How do you know if there's no OA ?
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Re: Most economists in the United States seem captivated by the [#permalink]
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22 Feb 2012, 11:30
1c - not right
2a - right
3c - right
4e - not right
5b - right
6a - not right
7d - right
8a - right
9b - right
Total time: 14 min 30 seconds...
was a fun challenge. Got 6/9. Need to start shoring in the time it took to read the passage. Almost took me 6 minutes!
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Re: Most economists in the United States seem captivated by the [#permalink]
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05 Aug 2012, 02:06
Can anybody explain these questions?
Question 1 (between choices A and E)
Question 2 (why is I correct and III incorrect?)
Question 3 (between choices B and C)
Question 8 (between choice A and C)
I took 19 mins and got 5/9.
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Re: Most economists in the United States seem captivated by the [#permalink]
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22 Feb 2013, 08:37
3 : either B or C ( I chose B ) . Wanted to match Pernicious . wasn't sure how condescendin does that.
4 : C
Satement says
A price that is determined by the seller or, for that matter, established by anyone other than the aggregate (group) of consumers seems pernicious.
E is close but C ooked better
8.
C (POE)
Eliminate
A - talks @ Demand
B - talks @ Profit
D and E not inferable.
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24 Feb 2013, 02:49
smayekar wrote:
3 : either B or C ( I chose B ) . Wanted to match Pernicious . wasn't sure how condescendin does that.
4 : C
Satement says
A price that is determined by the seller or, for that matter, established by anyone other than the aggregate (group) of consumers seems pernicious.
E is close but C ooked better
8.
C (POE)
Eliminate
A - talks @ Demand
B - talks @ Profit
D and E not inferable.
Agree ... C C C (No reason for the last one ... other options were worse)
... OAs?
Re: Most economists in the United States seem captivated by the &nbs [#permalink] 24 Feb 2013, 02:49
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https://en.wikipedia.org/wiki/Random_forests | 1,701,439,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00170.warc.gz | 257,003,281 | 57,663 | # Random forest
(Redirected from Random forests)
Random forests or random decision forests is an ensemble learning method for classification, regression and other tasks that operates by constructing a multitude of decision trees at training time. For classification tasks, the output of the random forest is the class selected by most trees. For regression tasks, the mean or average prediction of the individual trees is returned.[1][2] Random decision forests correct for decision trees' habit of overfitting to their training set.[3]: 587–588
The first algorithm for random decision forests was created in 1995 by Tin Kam Ho[1] using the random subspace method,[2] which, in Ho's formulation, is a way to implement the "stochastic discrimination" approach to classification proposed by Eugene Kleinberg.[4][5][6]
An extension of the algorithm was developed by Leo Breiman[7] and Adele Cutler,[8] who registered[9] "Random Forests" as a trademark in 2006 (as of 2019, owned by Minitab, Inc.).[10] The extension combines Breiman's "bagging" idea and random selection of features, introduced first by Ho[1] and later independently by Amit and Geman[11] in order to construct a collection of decision trees with controlled variance.
## History
The general method of random decision forests was first proposed by Ho in 1995.[1] Ho established that forests of trees splitting with oblique hyperplanes can gain accuracy as they grow without suffering from overtraining, as long as the forests are randomly restricted to be sensitive to only selected feature dimensions. A subsequent work along the same lines[2] concluded that other splitting methods behave similarly, as long as they are randomly forced to be insensitive to some feature dimensions. Note that this observation of a more complex classifier (a larger forest) getting more accurate nearly monotonically is in sharp contrast to the common belief that the complexity of a classifier can only grow to a certain level of accuracy before being hurt by overfitting. The explanation of the forest method's resistance to overtraining can be found in Kleinberg's theory of stochastic discrimination.[4][5][6]
The early development of Breiman's notion of random forests was influenced by the work of Amit and Geman[11] who introduced the idea of searching over a random subset of the available decisions when splitting a node, in the context of growing a single tree. The idea of random subspace selection from Ho[2] was also influential in the design of random forests. In this method a forest of trees is grown, and variation among the trees is introduced by projecting the training data into a randomly chosen subspace before fitting each tree or each node. Finally, the idea of randomized node optimization, where the decision at each node is selected by a randomized procedure, rather than a deterministic optimization was first introduced by Thomas G. Dietterich.[12]
The proper introduction of random forests was made in a paper by Leo Breiman.[7] This paper describes a method of building a forest of uncorrelated trees using a CART like procedure, combined with randomized node optimization and bagging. In addition, this paper combines several ingredients, some previously known and some novel, which form the basis of the modern practice of random forests, in particular:
1. Using out-of-bag error as an estimate of the generalization error.
2. Measuring variable importance through permutation.
The report also offers the first theoretical result for random forests in the form of a bound on the generalization error which depends on the strength of the trees in the forest and their correlation.
## Algorithm
### Preliminaries: decision tree learning
Decision trees are a popular method for various machine learning tasks. Tree learning "come[s] closest to meeting the requirements for serving as an off-the-shelf procedure for data mining", say Hastie et al., "because it is invariant under scaling and various other transformations of feature values, is robust to inclusion of irrelevant features, and produces inspectable models. However, they are seldom accurate".[3]: 352
In particular, trees that are grown very deep tend to learn highly irregular patterns: they overfit their training sets, i.e. have low bias, but very high variance. Random forests are a way of averaging multiple deep decision trees, trained on different parts of the same training set, with the goal of reducing the variance.[3]: 587–588 This comes at the expense of a small increase in the bias and some loss of interpretability, but generally greatly boosts the performance in the final model.
### Bagging
The training algorithm for random forests applies the general technique of bootstrap aggregating, or bagging, to tree learners. Given a training set X = x1, ..., xn with responses Y = y1, ..., yn, bagging repeatedly (B times) selects a random sample with replacement of the training set and fits trees to these samples:
For b = 1, ..., B:
1. Sample, with replacement, n training examples from X, Y; call these Xb, Yb.
2. Train a classification or regression tree fb on Xb, Yb.
After training, predictions for unseen samples x' can be made by averaging the predictions from all the individual regression trees on x':
${\displaystyle {\hat {f}}={\frac {1}{B}}\sum _{b=1}^{B}f_{b}(x')}$
or by taking the plurality vote in the case of classification trees.
This bootstrapping procedure leads to better model performance because it decreases the variance of the model, without increasing the bias. This means that while the predictions of a single tree are highly sensitive to noise in its training set, the average of many trees is not, as long as the trees are not correlated. Simply training many trees on a single training set would give strongly correlated trees (or even the same tree many times, if the training algorithm is deterministic); bootstrap sampling is a way of de-correlating the trees by showing them different training sets.
Additionally, an estimate of the uncertainty of the prediction can be made as the standard deviation of the predictions from all the individual regression trees on x':
${\displaystyle \sigma ={\sqrt {\frac {\sum _{b=1}^{B}(f_{b}(x')-{\hat {f}})^{2}}{B-1}}}.}$
The number of samples/trees, B, is a free parameter. Typically, a few hundred to several thousand trees are used, depending on the size and nature of the training set. An optimal number of trees B can be found using cross-validation, or by observing the out-of-bag error: the mean prediction error on each training sample xi, using only the trees that did not have xi in their bootstrap sample.[13] The training and test error tend to level off after some number of trees have been fit.
### From bagging to random forests
The above procedure describes the original bagging algorithm for trees. Random forests also include another type of bagging scheme: they use a modified tree learning algorithm that selects, at each candidate split in the learning process, a random subset of the features. This process is sometimes called "feature bagging". The reason for doing this is the correlation of the trees in an ordinary bootstrap sample: if one or a few features are very strong predictors for the response variable (target output), these features will be selected in many of the B trees, causing them to become correlated. An analysis of how bagging and random subspace projection contribute to accuracy gains under different conditions is given by Ho.[14]
Typically, for a classification problem with p features, p (rounded down) features are used in each split.[3]: 592 For regression problems the inventors recommend p/3 (rounded down) with a minimum node size of 5 as the default.[3]: 592 In practice, the best values for these parameters should be tuned on a case-to-case basis for every problem.[3]: 592
### ExtraTrees
Adding one further step of randomization yields extremely randomized trees, or ExtraTrees. While similar to ordinary random forests in that they are an ensemble of individual trees, there are two main differences: first, each tree is trained using the whole learning sample (rather than a bootstrap sample), and second, the top-down splitting in the tree learner is randomized. Instead of computing the locally optimal cut-point for each feature under consideration (based on, e.g., information gain or the Gini impurity), a random cut-point is selected. This value is selected from a uniform distribution within the feature's empirical range (in the tree's training set). Then, of all the randomly generated splits, the split that yields the highest score is chosen to split the node. Similar to ordinary random forests, the number of randomly selected features to be considered at each node can be specified. Default values for this parameter are ${\displaystyle {\sqrt {p}}}$ for classification and ${\displaystyle p}$ for regression, where ${\displaystyle p}$ is the number of features in the model.[15]
### Random forests for high-dimensional data
The basic Random Forest procedure may not work well in situations where there are a large number of features but only a small proportion of these features are informative with respect to sample classification. This can be addressed by encouraging the procedure to focus mainly on features and trees that are informative. Some methods for accomplishing this are:
- Prefiltering: Eliminate features that are mostly just noise. [16] [17]
- Enriched Random Forest (ERF): Use weighted random sampling instead of simple random sampling at each node of each tree, giving greater weight to features that appear to be more informative. [18] [19]
- Tree Weighted Random Forest (TWRF): Weight trees so that trees exhibiting better accuracy are assigned higher weights. [20] [21]
## Properties
### Variable importance
Random forests can be used to rank the importance of variables in a regression or classification problem in a natural way. The following technique was described in Breiman's original paper[7] and is implemented in the R package randomForest.[8]
#### Permutation Importance
The first step in measuring the variable importance in a data set ${\displaystyle {\mathcal {D}}_{n}=\{(X_{i},Y_{i})\}_{i=1}^{n}}$ is to fit a random forest to the data. During the fitting process the out-of-bag error for each data point is recorded and averaged over the forest (errors on an independent test set can be substituted if bagging is not used during training).
To measure the importance of the ${\displaystyle j}$-th feature after training, the values of the ${\displaystyle j}$-th feature are permuted in the out-of-bag samples and the out-of-bag error is again computed on this perturbed data set. The importance score for the ${\displaystyle j}$-th feature is computed by averaging the difference in out-of-bag error before and after the permutation over all trees. The score is normalized by the standard deviation of these differences.
Features which produce large values for this score are ranked as more important than features which produce small values. The statistical definition of the variable importance measure was given and analyzed by Zhu et al.[22]
This method of determining variable importance has some drawbacks.
• For data including categorical variables with different number of levels, random forests are biased in favor of those attributes with more levels. Methods such as partial permutations[23][24][25] and growing unbiased trees[26][27] can be used to solve the problem.
• If the data contain groups of correlated features of similar relevance for the output, then smaller groups are favored over larger groups.[28]
• Additionally, the permutation procedure may fail to identify important features when there are collinear features. In this case permuting groups of correlated features together is a remedy.[29]
#### Mean Decrease in Impurity Feature Importance
This feature importance for random forests is the default implementation in sci-kit learn and R. It is described in the book "Classification and Regression Trees" by Leo Breiman.[30] Variables which decrease the impurity during splits a lot are considered important:[31]
${\displaystyle {\text{unormalized average importance}}(x)={\frac {1}{n_{T}}}\sum _{i=1}^{n_{T}}\sum _{{\text{node }}j\in T_{i}|{\text{split variable}}(j)=x}p_{T_{i}}(j)\Delta i_{T_{i}}(j),}$
where ${\displaystyle x}$ indicates a feature, ${\displaystyle n_{T}}$ is the number of trees in the forest, ${\displaystyle T_{i}}$ indicates tree ${\displaystyle i}$, ${\displaystyle p_{T_{i}}(j)={\frac {n_{j}}{n}}}$ is the fraction of samples reaching node ${\displaystyle j}$, ${\displaystyle \Delta i_{T_{i}}(j)}$ is the change in impurity in tree ${\displaystyle t}$ at node ${\displaystyle j}$. As impurity measure for samples falling in a node e.g. the following statistics can be used:
The normalized importance is then obtained by normalizing over all features, so that the sum of normalized feature importances is 1.
The sci-kit learn default implementation of Mean Decrease in Impurity Feature Importance is susceptible to misleading feature importances:[29]
• the importance measure prefers high cardinality features
• it uses training statistics and therefore does not "reflect the ability of feature to be useful to make predictions that generalize to the test set"[32]
### Relationship to nearest neighbors
A relationship between random forests and the k-nearest neighbor algorithm (k-NN) was pointed out by Lin and Jeon in 2002.[33] It turns out that both can be viewed as so-called weighted neighborhoods schemes. These are models built from a training set ${\displaystyle \{(x_{i},y_{i})\}_{i=1}^{n}}$ that make predictions ${\displaystyle {\hat {y}}}$ for new points x' by looking at the "neighborhood" of the point, formalized by a weight function W:
${\displaystyle {\hat {y}}=\sum _{i=1}^{n}W(x_{i},x')\,y_{i}.}$
Here, ${\displaystyle W(x_{i},x')}$ is the non-negative weight of the i'th training point relative to the new point x' in the same tree. For any particular x', the weights for points ${\displaystyle x_{i}}$ must sum to one. Weight functions are given as follows:
• In k-NN, the weights are ${\displaystyle W(x_{i},x')={\frac {1}{k}}}$ if xi is one of the k points closest to x', and zero otherwise.
• In a tree, ${\displaystyle W(x_{i},x')={\frac {1}{k'}}}$ if xi is one of the k' points in the same leaf as x', and zero otherwise.
Since a forest averages the predictions of a set of m trees with individual weight functions ${\displaystyle W_{j}}$, its predictions are
${\displaystyle {\hat {y}}={\frac {1}{m}}\sum _{j=1}^{m}\sum _{i=1}^{n}W_{j}(x_{i},x')\,y_{i}=\sum _{i=1}^{n}\left({\frac {1}{m}}\sum _{j=1}^{m}W_{j}(x_{i},x')\right)\,y_{i}.}$
This shows that the whole forest is again a weighted neighborhood scheme, with weights that average those of the individual trees. The neighbors of x' in this interpretation are the points ${\displaystyle x_{i}}$ sharing the same leaf in any tree ${\displaystyle j}$. In this way, the neighborhood of x' depends in a complex way on the structure of the trees, and thus on the structure of the training set. Lin and Jeon show that the shape of the neighborhood used by a random forest adapts to the local importance of each feature.[33]
## Unsupervised learning with random forests
As part of their construction, random forest predictors naturally lead to a dissimilarity measure among the observations. One can also define a random forest dissimilarity measure between unlabeled data: the idea is to construct a random forest predictor that distinguishes the "observed" data from suitably generated synthetic data.[7][34] The observed data are the original unlabeled data and the synthetic data are drawn from a reference distribution. A random forest dissimilarity can be attractive because it handles mixed variable types very well, is invariant to monotonic transformations of the input variables, and is robust to outlying observations. The random forest dissimilarity easily deals with a large number of semi-continuous variables due to its intrinsic variable selection; for example, the "Addcl 1" random forest dissimilarity weighs the contribution of each variable according to how dependent it is on other variables. The random forest dissimilarity has been used in a variety of applications, e.g. to find clusters of patients based on tissue marker data.[35]
## Variants
Instead of decision trees, linear models have been proposed and evaluated as base estimators in random forests, in particular multinomial logistic regression and naive Bayes classifiers.[36][37][38] In cases that the relationship between the predictors and the target variable is linear, the base learners may have an equally high accuracy as the ensemble learner.[39][36]
## Kernel random forest
In machine learning, kernel random forests (KeRF) establish the connection between random forests and kernel methods. By slightly modifying their definition, random forests can be rewritten as kernel methods, which are more interpretable and easier to analyze.[40]
### History
Leo Breiman[41] was the first person to notice the link between random forest and kernel methods. He pointed out that random forests which are grown using i.i.d. random vectors in the tree construction are equivalent to a kernel acting on the true margin. Lin and Jeon[42] established the connection between random forests and adaptive nearest neighbor, implying that random forests can be seen as adaptive kernel estimates. Davies and Ghahramani[43] proposed Random Forest Kernel and show that it can empirically outperform state-of-art kernel methods. Scornet[40] first defined KeRF estimates and gave the explicit link between KeRF estimates and random forest. He also gave explicit expressions for kernels based on centered random forest[44] and uniform random forest,[45] two simplified models of random forest. He named these two KeRFs Centered KeRF and Uniform KeRF, and proved upper bounds on their rates of consistency.
### Notations and definitions
#### Preliminaries: Centered forests
Centered forest[44] is a simplified model for Breiman's original random forest, which uniformly selects an attribute among all attributes and performs splits at the center of the cell along the pre-chosen attribute. The algorithm stops when a fully binary tree of level ${\displaystyle k}$ is built, where ${\displaystyle k\in \mathbb {N} }$ is a parameter of the algorithm.
#### Uniform forest
Uniform forest[45] is another simplified model for Breiman's original random forest, which uniformly selects a feature among all features and performs splits at a point uniformly drawn on the side of the cell, along the preselected feature.
#### From random forest to KeRF
Given a training sample ${\displaystyle {\mathcal {D}}_{n}=\{(\mathbf {X} _{i},Y_{i})\}_{i=1}^{n}}$ of ${\displaystyle [0,1]^{p}\times \mathbb {R} }$-valued independent random variables distributed as the independent prototype pair ${\displaystyle (\mathbf {X} ,Y)}$, where ${\displaystyle \operatorname {E} [Y^{2}]<\infty }$. We aim at predicting the response ${\displaystyle Y}$, associated with the random variable ${\displaystyle \mathbf {X} }$, by estimating the regression function ${\displaystyle m(\mathbf {x} )=\operatorname {E} [Y\mid \mathbf {X} =\mathbf {x} ]}$. A random regression forest is an ensemble of ${\displaystyle M}$ randomized regression trees. Denote ${\displaystyle m_{n}(\mathbf {x} ,\mathbf {\Theta } _{j})}$ the predicted value at point ${\displaystyle \mathbf {x} }$ by the ${\displaystyle j}$-th tree, where ${\displaystyle \mathbf {\Theta } _{1},\ldots ,\mathbf {\Theta } _{M}}$ are independent random variables, distributed as a generic random variable ${\displaystyle \mathbf {\Theta } }$, independent of the sample ${\displaystyle {\mathcal {D}}_{n}}$. This random variable can be used to describe the randomness induced by node splitting and the sampling procedure for tree construction. The trees are combined to form the finite forest estimate ${\displaystyle m_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})={\frac {1}{M}}\sum _{j=1}^{M}m_{n}(\mathbf {x} ,\Theta _{j})}$. For regression trees, we have ${\displaystyle m_{n}=\sum _{i=1}^{n}{\frac {Y_{i}\mathbf {1} _{\mathbf {X} _{i}\in A_{n}(\mathbf {x} ,\Theta _{j})}}{N_{n}(\mathbf {x} ,\Theta _{j})}}}$, where ${\displaystyle A_{n}(\mathbf {x} ,\Theta _{j})}$ is the cell containing ${\displaystyle \mathbf {x} }$, designed with randomness ${\displaystyle \Theta _{j}}$ and dataset ${\displaystyle {\mathcal {D}}_{n}}$, and ${\displaystyle N_{n}(\mathbf {x} ,\Theta _{j})=\sum _{i=1}^{n}\mathbf {1} _{\mathbf {X} _{i}\in A_{n}(\mathbf {x} ,\Theta _{j})}}$.
Thus random forest estimates satisfy, for all ${\displaystyle \mathbf {x} \in [0,1]^{d}}$, ${\displaystyle m_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})={\frac {1}{M}}\sum _{j=1}^{M}\left(\sum _{i=1}^{n}{\frac {Y_{i}\mathbf {1} _{\mathbf {X} _{i}\in A_{n}(\mathbf {x} ,\Theta _{j})}}{N_{n}(\mathbf {x} ,\Theta _{j})}}\right)}$. Random regression forest has two levels of averaging, first over the samples in the target cell of a tree, then over all trees. Thus the contributions of observations that are in cells with a high density of data points are smaller than that of observations which belong to less populated cells. In order to improve the random forest methods and compensate the misestimation, Scornet[40] defined KeRF by
${\displaystyle {\tilde {m}}_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})={\frac {1}{\sum _{j=1}^{M}N_{n}(\mathbf {x} ,\Theta _{j})}}\sum _{j=1}^{M}\sum _{i=1}^{n}Y_{i}\mathbf {1} _{\mathbf {X} _{i}\in A_{n}(\mathbf {x} ,\Theta _{j})},}$
which is equal to the mean of the ${\displaystyle Y_{i}}$'s falling in the cells containing ${\displaystyle \mathbf {x} }$ in the forest. If we define the connection function of the ${\displaystyle M}$ finite forest as ${\displaystyle K_{M,n}(\mathbf {x} ,\mathbf {z} )={\frac {1}{M}}\sum _{j=1}^{M}\mathbf {1} _{\mathbf {z} \in A_{n}(\mathbf {x} ,\Theta _{j})}}$, i.e. the proportion of cells shared between ${\displaystyle \mathbf {x} }$ and ${\displaystyle \mathbf {z} }$, then almost surely we have ${\displaystyle {\tilde {m}}_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})={\frac {\sum _{i=1}^{n}Y_{i}K_{M,n}(\mathbf {x} ,\mathbf {x} _{i})}{\sum _{\ell =1}^{n}K_{M,n}(\mathbf {x} ,\mathbf {x} _{\ell })}}}$, which defines the KeRF.
#### Centered KeRF
The construction of Centered KeRF of level ${\displaystyle k}$ is the same as for centered forest, except that predictions are made by ${\displaystyle {\tilde {m}}_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})}$, the corresponding kernel function, or connection function is
{\displaystyle {\begin{aligned}K_{k}^{cc}(\mathbf {x} ,\mathbf {z} )=\sum _{k_{1},\ldots ,k_{d},\sum _{j=1}^{d}k_{j}=k}&{\frac {k!}{k_{1}!\cdots k_{d}!}}\left({\frac {1}{d}}\right)^{k}\prod _{j=1}^{d}\mathbf {1} _{\lceil 2^{k_{j}}x_{j}\rceil =\lceil 2^{k_{j}}z_{j}\rceil },\\&{\text{ for all }}\mathbf {x} ,\mathbf {z} \in [0,1]^{d}.\end{aligned}}}
#### Uniform KeRF
Uniform KeRF is built in the same way as uniform forest, except that predictions are made by ${\displaystyle {\tilde {m}}_{M,n}(\mathbf {x} ,\Theta _{1},\ldots ,\Theta _{M})}$, the corresponding kernel function, or connection function is
${\displaystyle K_{k}^{uf}(\mathbf {0} ,\mathbf {x} )=\sum _{k_{1},\ldots ,k_{d},\sum _{j=1}^{d}k_{j}=k}{\frac {k!}{k_{1}!\ldots k_{d}!}}\left({\frac {1}{d}}\right)^{k}\prod _{m=1}^{d}\left(1-|x_{m}|\sum _{j=0}^{k_{m}-1}{\frac {(-\ln |x_{m}|)^{j}}{j!}}\right){\text{ for all }}\mathbf {x} \in [0,1]^{d}.}$
### Properties
#### Relation between KeRF and random forest
Predictions given by KeRF and random forests are close if the number of points in each cell is controlled:
Assume that there exist sequences ${\displaystyle (a_{n}),(b_{n})}$ such that, almost surely,
${\displaystyle a_{n}\leq N_{n}(\mathbf {x} ,\Theta )\leq b_{n}{\text{ and }}a_{n}\leq {\frac {1}{M}}\sum _{m=1}^{M}N_{n}{\mathbf {x} ,\Theta _{m}}\leq b_{n}.}$
Then almost surely,
${\displaystyle |m_{M,n}(\mathbf {x} )-{\tilde {m}}_{M,n}(\mathbf {x} )|\leq {\frac {b_{n}-a_{n}}{a_{n}}}{\tilde {m}}_{M,n}(\mathbf {x} ).}$
#### Relation between infinite KeRF and infinite random forest
When the number of trees ${\displaystyle M}$ goes to infinity, then we have infinite random forest and infinite KeRF. Their estimates are close if the number of observations in each cell is bounded:
Assume that there exist sequences ${\displaystyle (\varepsilon _{n}),(a_{n}),(b_{n})}$ such that, almost surely
• ${\displaystyle \operatorname {E} [N_{n}(\mathbf {x} ,\Theta )]\geq 1,}$
• ${\displaystyle \operatorname {P} [a_{n}\leq N_{n}(\mathbf {x} ,\Theta )\leq b_{n}\mid {\mathcal {D}}_{n}]\geq 1-\varepsilon _{n}/2,}$
• ${\displaystyle \operatorname {P} [a_{n}\leq \operatorname {E} _{\Theta }[N_{n}(\mathbf {x} ,\Theta )]\leq b_{n}\mid {\mathcal {D}}_{n}]\geq 1-\varepsilon _{n}/2,}$
Then almost surely,
${\displaystyle |m_{\infty ,n}(\mathbf {x} )-{\tilde {m}}_{\infty ,n}(\mathbf {x} )|\leq {\frac {b_{n}-a_{n}}{a_{n}}}{\tilde {m}}_{\infty ,n}(\mathbf {x} )+n\varepsilon _{n}\left(\max _{1\leq i\leq n}Y_{i}\right).}$
### Consistency results
Assume that ${\displaystyle Y=m(\mathbf {X} )+\varepsilon }$, where ${\displaystyle \varepsilon }$ is a centered Gaussian noise, independent of ${\displaystyle \mathbf {X} }$, with finite variance ${\displaystyle \sigma ^{2}<\infty }$. Moreover, ${\displaystyle \mathbf {X} }$ is uniformly distributed on ${\displaystyle [0,1]^{d}}$ and ${\displaystyle m}$ is Lipschitz. Scornet[40] proved upper bounds on the rates of consistency for centered KeRF and uniform KeRF.
#### Consistency of centered KeRF
Providing ${\displaystyle k\rightarrow \infty }$ and ${\displaystyle n/2^{k}\rightarrow \infty }$, there exists a constant ${\displaystyle C_{1}>0}$ such that, for all ${\displaystyle n}$, ${\displaystyle \mathbb {E} [{\tilde {m}}_{n}^{cc}(\mathbf {X} )-m(\mathbf {X} )]^{2}\leq C_{1}n^{-1/(3+d\log 2)}(\log n)^{2}}$.
#### Consistency of uniform KeRF
Providing ${\displaystyle k\rightarrow \infty }$ and ${\displaystyle n/2^{k}\rightarrow \infty }$, there exists a constant ${\displaystyle C>0}$ such that, ${\displaystyle \mathbb {E} [{\tilde {m}}_{n}^{uf}(\mathbf {X} )-m(\mathbf {X} )]^{2}\leq Cn^{-2/(6+3d\log 2)}(\log n)^{2}}$.
While random forests often achieve higher accuracy than a single decision tree, they sacrifice the intrinsic interpretability present in decision trees. Decision trees are among a fairly small family of machine learning models that are easily interpretable along with linear models, rule-based models, and attention-based models. This interpretability is one of the most desirable qualities of decision trees. It allows developers to confirm that the model has learned realistic information from the data and allows end-users to have trust and confidence in the decisions made by the model.[36][3] For example, following the path that a decision tree takes to make its decision is quite trivial, but following the paths of tens or hundreds of trees is much harder. To achieve both performance and interpretability, some model compression techniques allow transforming a random forest into a minimal "born-again" decision tree that faithfully reproduces the same decision function.[36][46][47] If it is established that the predictive attributes are linearly correlated with the target variable, using random forest may not enhance the accuracy of the base learner.[36][39] Furthermore, in problems with multiple categorical variables, random forest may not be able to increase the accuracy of the base learner.[48]
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48. ^ Piryonesi, Sayed Madeh (November 2019). Piryonesi, S. M. (2019). The Application of Data Analytics to Asset Management: Deterioration and Climate Change Adaptation in Ontario Roads (Doctoral dissertation) (Thesis). | 10,200 | 37,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 93, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.957544 |
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# What is |3x+8| =0 ?
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The equation |3x+8| = 0 has no solution because the absolute value of any number is always non-negative. Since the absolute value of 3x+8 is equal to 0, it means that 3x+8 must be equal to 0. However, there is no value of x that will make 3x+8 equal to 0. Therefore, the equation has no solution.
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## Basic Information
Lecturers: Gerhard Schmidt (lecture) and Marco Gimm (exercise) Room: KS2/Geb.F - SR-II Language: English Target group: Students in electrical engineering and computer engineering Prerequisites: Basic (bachelor) knowlegde about signals and systems Contents: This course teaches advanced knowledge in signal systems theory for electrical engineering and information technology. Topic overview: Introduction Discrete signals and random processes Spectra Idealized linear, shift-invariant systems Hilbert transform State-space description and system realizations Generalizations for signals, systems and spectra References: A. V. Oppenheim, R. W. Schafer: Discrete-Time Signal Processing, 2nd edition, Prentice Hall, 1999J. G. Proakis, D. K. Manolakis: Digital Signal Processing, 4th edition, Prentice Hall, 2006S. Haykin, B. Van Veen: Signals and Systems, 2nd edition, Wiley, 2002A. Papoulis: Probability, Random Variables, and Stochastic Processes, McGraw Hill, 1965
## News
Start of exercise:
Date: Thursday, 02.11.2017
Time: 11:15 h - 12:45 h
Exam Results and Exam Inspection:
The results of the winter semester's exam of xx.xx.xxxx are now online in the QIS-system. Please note the date of the exam inspection:
Date: Weekday, xx.xx.xxxx
Time: xx:xx h - xx:xx h
Room: Library of the DSS-group (D-037 A)
## Lecture slides
Slides of the lecture "Introduction" (Contents, literature, notation) Slides of the lecture "Discrete Signals and Random Processes"(Definitions, stochastic processes, examples) Slides of the lecture "Spectra"(Fourier transform, z transfrom, discrete Fourier transform)) Slides of the lecture "Discrete Systems"(Descriptions, classifications, responses to deterministic and stochastic signals) Slides of the lecture "Idealized Linear, Shift-invariant Systems"(Ideal transmission systems, amplitude and phase distortions) Slides of the lecture "Hilbert Transform"(Frequency and impulse response, analytic signal) Slides of the lecture "State-space description and system realizations" (State-space description and signal-flow graphs) Slides of the lecture "Extensions"(All-pass filters)
## Exercises
Exercises can be found on an extra web page.
-
## Previous Exams
Exam of the summer term 2017 Corresponding solution Exam of the winter term 2016/2017 Corresponding solution Exam of the summer term 2016 Corresponding solution Exam of the winter term 2015/2016 Corresponding solution Exam of the summer term 2015 Corresponding solution
### Website News
01.10.2017: Started with a Tips and Tricks section for KiRAT.
01.10.2017: Talks from Jonas Sauter (Nuance) and Vasudev Kandade Rajan (Harman/Samsung) added.
13.08.2017: New Gas e.V. sections (e.g. pictures or prices) added.
05.08.2017: The first "slide carousel" added.
### Recent Publications
J. Reermann, P. Durdaut, S. Salzer, T. Demming, A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241, https://doi.org/10.1016/j.measurement.2017.09.047, 2017
S. Graf, T. Herbig, M. Buck, G. Schmidt: Low-Complexity Pitch Estimation Based on Phase Differences Between Low-Resolution Spectra, Proc. Interspeech, pp. 2316 -2320, 2017
### Contact
Prof. Dr.-Ing. Gerhard Schmidt
E-Mail: gus@tf.uni-kiel.de
Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory
Kaiserstr. 2
24143 Kiel, Germany
## Recent News
Jens Reermann Defended his Dissertation with Distinction
On Friday, 21st of June, Jens Reermann defended his research on signals processing for magnetoelectric sensor systems very successfully. After 90 minutes of talk and question time he finished his PhD with distinction. Congratulations, Jens, from the entire DSS team.
Jens worked for about three and a half years - as part of the collaborative research center (SFB) 1261 - on all kinds of signal ... | 1,020 | 4,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-43 | latest | en | 0.754809 |
http://slideplayer.com/slide/4645704/ | 1,638,639,756,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362999.66/warc/CC-MAIN-20211204154554-20211204184554-00336.warc.gz | 67,950,222 | 20,707 | # Warm-up Graph: 4x – 6y = 18 Write the equation of the line with slope ½ and through the point (4, -5). Use point slope form. Write it in slope intercept.
## Presentation on theme: "Warm-up Graph: 4x – 6y = 18 Write the equation of the line with slope ½ and through the point (4, -5). Use point slope form. Write it in slope intercept."— Presentation transcript:
Warm-up Graph: 4x – 6y = 18 Write the equation of the line with slope ½ and through the point (4, -5). Use point slope form. Write it in slope intercept form
3.6 Slopes of Parallel and Perpendicular Lines
10/31 with practice on 11/1
Learning Targets I can tell if two lines are parallel or perpendicular by finding their slopes. I can write equations of parallel and perpendicular lines.
Parallel Lines have equal slopes (and different y intercepts)
Line 1 contains A(-4, 2) and B(3, 1) Line 2 contains C(-4, 0) and D(8, -2) Are Lines 1 and 2 Parallel? Hint, find the slopes
Are Lines 4y – 12x = 20 and y = 3x -1 Parallel?
Are lines y = x + 6 and 2x + 4y = 20 Parallel?
Write an equation for the line parallel to y = -4x + 3 and through the point (1, -2)
Perpendicular Lines have slopes that are negative reciprocals of each other (their product is -1)
What is a negative reciprocal?
Line 1 goes through (-2, 3) and (6, -3)
Are lines 1 and 2 Perpendicular? Hint: find the slopes
Are the following lines perpendicular?
2x – 7y = -42 4y = -7x – 2
Are the following lines perpendicular?
2x + 3y = 6 6x – 4y = 24
Write an equation of a line Perpendicular to y = -3x + 5 and through the point (-3, 7)
Write the equation for a line that contains (1, 8) and is perpendicular to y = ¾ x + 1
No homework, but be prepared to practice this further.
Day 3 Review (11/1) Practice worksheet with follow up quiz on Monday
Monday 11/4 Review, go over hmwk wkst, review problems
Quiz with mins left
Homework p. 314 #2-36 even
Download ppt "Warm-up Graph: 4x – 6y = 18 Write the equation of the line with slope ½ and through the point (4, -5). Use point slope form. Write it in slope intercept."
Similar presentations | 613 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-49 | latest | en | 0.906711 |
https://scholar.archive.org/work/dacnb5nswrewbc5oht4katzlxe | 1,620,524,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00076.warc.gz | 518,037,975 | 6,167 | ### Almost-Dedekind rings
E. W. Johnson
1994 Glasgow Mathematical Journal
Throughout we assume all rings are commutative with identity. We denote the lattice of ideals of a ring R by L(R), and we denote by L(R)* the subposet L(R) -R. A classical result of commutative ring theory is the characterization of a Dedekind domain as an integral domain R in which every element of L(R)* is a product of prime ideals (see Mori [5] for a history). This result has been generalized in a number of ways. In particular, rings which are not necessarily domains but which otherwise
more » ... hich otherwise satisfy the hypotheses (i.e. general ZPI-rings) have been widely studied (see, for example, Gilmer [3]), as have rings in which only the principal ideals are assumed to satisfy the hypothesis (i.e. ^-rings). General ZPI-rings and ^-rings can both be thought of as "almost Dedekind". In both cases, one gets a representation as the finite direct product of integral domains of the same type (Dedekind domains in the first case, ^-domains in the second case) and quotients of discrete (rank one) valuation rings (i.e. special principal ideal rings-or SPIRS as they have come to be called). Note that ZPI-rings are rings in which every ideal in L(R)* satisfies the "product of prime ideals" condition, whereas only the principal ideals of a ;r-ring are assumed to satisfy this condition. This naturally raises consideration of rings in which every ideal of L(R)* generated by n elements is a product of prime ideals. Any UFD is a Jr-ring; so a jr-ring need not be a general ZPI-ring. In this regard, Levitz [4, 5] has obtained the very interesting result that wrings are the single exception. If every doubly generated ideal in L(R)* is the product of prime ideals, then every ideal in L(R)* is. Butts and Gilmer [3] have characterized ZPI-rings in a somewhat different manner. They have shown that ZPI-rings are characterized by the property that every ideal in L(R)* is a finite intersection of powers of prime ideals. In this paper, we obtain the analogue of Levitz's theorem for the Butts-Gilmer characterization of general ZPI-rings. That is, we show that, once again, two elements suffice: if R is a ring in which every double generated ideal in L(R)* is the intersection of powers of prime ideals, then every ideal in L(R)* is. For convenience, we will say that a ring R satisfies "Property D" if every doubly generated ideal in L(R)* is the intersection of powers of prime ideals. We begin with a simple but useful observation. L E M M A 1. Let (R,M) be a quasi-local ring satisfying Property D. If x , yeM then there are only a finite number of primes minimal over (x,y). n Proof (x,y) is the finite intersection of powers of prime ideals, say (x,y) = P) Pf. Then any prime minimal over (x,y) is one of the primes P t . ' = l We also note the following. LEMMA 2. If R satisfies Property D and if S is a multiplicatively closed subset of R, then R s satisfies Property D. Proof. (a/s l ,b/s)R s = (a,b)R s . Glasgow Math. J. 36 (1994) 131-134. https://www.cambridge.org/core/terms. https://doi. | 764 | 3,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.949297 |
https://inches-to-cm.appspot.com/1/et/4540-tolli-et-sentimeetrit.html | 1,722,832,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00039.warc.gz | 246,099,211 | 6,352 | Tolli Et Sentimeetrit
# 4540 in cm4540 Tolli Sentimeetrit
in
=
cm
## Kuidas teisendada 4540 tolli sentimeetrit?
4540 in * 2.54 cm = 11531.6 cm 1 in
## Convert 4540 in ühistele pikkused
MõõtühikPikkus
nanomeeter1.15316e+11 nm
Mikromeeter115316000.0 µm
Millimeeter115316.0 mm
Sentimeeter11531.6 cm
Toll4540.0 in
Jalg378.333333333 ft
Õu126.111111111 yd
Meeter115.316 m
Kilomeeter0.115316 km
Miil0.0716540404 mi
Meremiil0.0622656587 nmi
## Alternatiivsed õigekirja
4540 Toll cm, 4540 in Sentimeetrit, 4540 in cm, 4540 Tolli Sentimeeter, 4540 in Sentimeeter, 4540 Tolli cm, 4540 Toll Sentimeeter | 265 | 599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.240505 |
http://jwilson.coe.uga.edu/EMAT6680/Evans/Assignment%209/Assignment9WriteUp.htm | 1,544,893,986,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826892.78/warc/CC-MAIN-20181215152912-20181215174912-00609.warc.gz | 154,015,457 | 1,864 | Assignment 9:
Pedal Triangles
The purpose of this assignment is to construct pedal triangles using a point P. We will investigate what happens when point P is:
• Any point in the plane of triangle ABC
• The Centroid of triangle ABC
• The Incenter of triangle ABC
• The Orthocenter of triangle ABC (Even if outside of triangle ABC)
• The Circumcenter of triangle ABC (Even if outside of triangle ABC)
• On a side of the triangle ABC
• One of the vertices of triangle ABC
How do you Construct a Pedal Triangle?
First, construct a triangle and a point P. Then extend the sides of the triangle by drawing a parallel line constructed by selecting a side of the triangle and a point on that side. Next, construct perpendicular lines through point P to all of the three sides of the triangle. The vertices of the pedal triangle are the intersection point of the extended parallel lines and the perpendicular lines.
What if Point P is Any point in the plane of triangle ABC?
The pedal triangle constructed from a point P in the plane of the triangle ABC is the blue triangle DEF.
Click here to see what happens when you move point P around in the plane of triangle ABC.
What if Point P is the Centroid of triangle ABC?
The pedal triangle constructed from point P being the centroid is the purple triangle DEF.
Click here to see what happens to the pedal triangle when you change the shape of the triangle ABC
What if Point P is the Incenter of Triangle ABC?
The pedal triangle constructed from point P being the incenter is the purple triangle DEF.
Click here to see what happens to the pedal triangle when you change the shape of the triangle ABC.
What if Point P is the Orthocenter of Triangle ABC?
Even if Outside of Triangle ABC?
The pedal triangle constructed above from point P being the orthocenter INSIDE triangle ABC is the purple triangle DEF.
The pedal triangle constructed above from point P being the orthocenter OUTSIDE triangle ABC is the purple triangle DEF.
Click here to see what happens to the pedal triangle when you change the shape of the triangle ABC.
What if Point P is the Circumcenter of Triangle ABC?
Even if Outside of Triangle ABC?
The pedal triangle that has been constructed above from point P being the circumcenter INSIDE triangle ABC is the purple triangle DEF.
The pedal triangle constructed above from point P being the circumcenter OUTSIDE triangle ABC is the purple triangle DEF.
Click here to see what happens to the pedal triangle when you change the shape of the triangle ABC.
What if Point P is on a Side of the Triangle ABC?
The pedal triangle that has been constructed from point P being on a side of triangle ABC is the purple triangle DEF above.
Click here to see what happens to the pedal triangle when you change the shape of the triangle ABC. Notice that one of the vertices of the pedal triangle is on the side of the triangle ABC.
What if Point P is One of the Vertices of Triangle ABC?
The pedal triangle constructed from point P being one of the vertices of triangle ABC is the purple segment above.
Click here to see what happens to the segment when you change the shape of the triangle ABC. | 654 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-51 | latest | en | 0.870082 |
https://uk.pokernews.com/poker-strategy-theory/when-weak-means-weak-analysing-clear-bluffing-opportunity-21665.htm | 1,638,657,461,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00398.warc.gz | 662,413,324 | 17,989 | # When Weak Means Weak: Analysing a Clear Bluffing Opportunity
I recently played a hand that required me to move in on the turn against an opponent who had taken the lead on multiple streets. Although my thinking in the hand was straightforward, it led to a shove that many players might miss, and the hand contains a couple of worthwhile lessons.
I was playing a tournament with 50,000-chip starting stacks. The blinds were 800/1,600 with a 200 ante, and the action started when a player with roughly 40,000 limped in from under the gun. It folded to me on the button where I had been dealt . I raised to 5,500, the blinds folded, and the UTG player (whom I had covered) called.
This is a small raise, especially given that we were in the ante levels, but it was enough to fold out the blinds if they didn't have much. It was also enough to start building a pot against UTG (or even to get him to fold if he was limping very weak).
The flop came , giving me a Broadway gutshot and an overcard. The UTG player, to my surprise, led out for the 1,600 minimum into a pot of roughly 15,000. The key to this hand is interpreting this leading bet. Although we can't be completely sure what it means, these very small bets out of position often indicate that a player wants to "set his price" either with a draw or with a weak made hand.
The bad news was I had no pair, couldn't beat even a weak ten, and was vulnerable against a draw. The good news was that I was ahead of a draw and drawing live against a weak made hand. So I raised to roughly 6,700. Again, I chose a very small bet size, but it was enough to get a fold from his weakest hands, and — crucially — it left him with enough of a stack that I could get him to fold later.
He called, which put the pot at roughly 28,000. At this point he had just about the same amount behind.
The turn came the , and UTG led out for 1,600 again.
His making another weak lead made my decision very clear, because it gave even more evidence of my initial read that he had either a draw or a weak made hand.
If he had a draw, jamming would be a good play. Ace-high would probably be best in this case, and I would rather get the draw to fold in this situation than fail to charge it. If he had a weak made hand, jamming would again be a good play. Most players who think to make a weak lead do not want to call a nearly pot-sized raise all-in with just a weak pair. Finally, in the unlikely event that my read was incorrect, my hand would still have some value.
I moved in and he folded.
In truth this is a relatively simple bluff, but the hand contains a couple of useful lessons.
First, sometimes things are exactly as they seem. Weak bets don't always indicate weakness, but often they do.
Second, sometimes all roads lead to the same conclusion. Here, the question whether UTG had a strong draw, a weak draw, or a weak made hand made little practical difference, as the correct response was to jam in any of these cases.
Your time and attention are limited at the table. Don't spend it on questions the answers to which don't change your best play.
* * * * *
The Kindle edition of Nate Meyvis's new book, Thinking Tournament Poker, Volume 2, is now available on Amazon. Be sure also to check out Nate and Andrew Brokos on the Thinking Poker Podcast, and for more from Nate visit his blog at natemeyvis.com. | 786 | 3,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.994181 |
https://unicourse.co/ders/bilkent-universitesi-econ102spring23-final | 1,716,332,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00729.warc.gz | 508,441,547 | 25,345 | # ECON 102 (Spring 24) • Introduction to Economics II • Final
Online Economics School / Gökhan Işık.
Bilkent biz geldik! Bu dersimizde Makroekonominin en temel kavramları olan; GDP,GNP,Unemployment,Inflation,Interest Rate,Fiscal Policy, Monetary Policy kavramlarını detaylıca öğrenecek, sonrasında çıkmış eski sınav sorularını çözerek sınava eksiksiz gireceksin. Bu dersimizde sunduğumuz içerikler sırasıyla
Ders Tanıtımı
Determining Equilibrium Output – 1
Ücretsiz
Determining Equilibrium Output – 2
Determining Equilibrium Output – 3
Determining Equilibrium Output - Exam Type Question
Ücretsiz
Determining Equilibrium Output - Exam Type Question 2
Ücretsiz
Multiplier Effect(Government Expenditure Multiplier & Fiscal Policy) Part1
Multiplier Effect(Government Expenditure Multiplier & Fiscal Policy) Part2
Output Gap
Planned Aggregate Expenditure and Keynesian Cross
Question 1
Question 2
Question 3
Question 4
Banks and Money Creation (Money Supply, Money Multiplier...)
Quantity Theory of Money (Money Supply, Velocity...)
Money Supply, Money Demand and Equilibrium Interest Rate
Ücretsiz
Tools of Monetary Policy
Exam Like Question 1
Exam Like Question 2
Exam Like Question 3
Money Supply, Money Demand, Equilibrium Interest Rate and Exam Like Question
Money Supply, Money Demand, Inflation Rate and Velocity of Money and Exam Like Question
Exam Like Question 1: Ms, Md, Inflation, Velocity of Money
Exam Like Question 2 - Part 1: Ms, Md, Inflation, Velocity of Money
Exam Like Question 2 - Part 2: Ms, Md, Inflation, Velocity of Money
IS-LM-FE Model
IS-LM-FE Model: Numerical Exam Like Question
AD & AS - Part 1
AD & AS - Part 2
Shifts of AD & AS and Exam Like Question
AD & AS Exam Like Question
AD & AS Exam Like Question 2
Philips Curve Part 1
Phillips Curve Part 2 ( Phillips Curve'ü Shift Ettiren Unsurlar)
NAIRU
Exam Type Question
Exchange Rates: Nominal Vs. Real
Long-Run Exchange Rate Determination: Purchasing Power Parity-Exam Type Question
Exchange Rate-Exam Type Question
Supply and Demand for Foreign Exchange
Saving,Investment,Net Export,Net Foreign Investment
## Eğitmen
Gökhan Işık
Eğitmen
Gökhan Işık olarak yıllardır üniversitelerin yabancı dilde eğitim veren Ekonomi ve İşletme fakültelerindeki öğrencilere yardımcı olmaktayım. Kendimize has anlatım teknikleri ile her üniversite ve ders hocasını ayrı ayrı ele alarak sınavlarda doğrudan hedefe yönelik çalışmalar yapmaktayız.
1299 TL
Hemen Al | 634 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.505833 |
http://www.coderanch.com/t/556027/java/java/parentheses | 1,467,180,010,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397567.28/warc/CC-MAIN-20160624154957-00090-ip-10-164-35-72.ec2.internal.warc.gz | 462,023,727 | 11,423 | Win a copy of Mesos in Action this week in the Cloud/Virtualizaton forum!
# If and parentheses
Sam Samson
Ranch Hand
Posts: 63
Hi
Could someone tell me how parentheses are virtually set in this example? Is there a simple rule like 'if there are no parentheses, the first to checks build a pair'?
I mean, will z > 1 && y >6 be together or y > 6 || z == 2?
greez
Sam
Bear Bibeault
Author and ninkuma
Marshal
Posts: 64838
86
It all has to do with the precedence of the operators.
Java operator precedence
Sam Samson
Ranch Hand
Posts: 63
Thank you, that was quick
So in my example
z > 1 && y >6 is a pair
and in:
y > 6 | z == 2 would be a pair, right?
Bear Bibeault
Author and ninkuma
Marshal
Posts: 64838
86
The easiest way to figure it all out is to use the operator precedence and start adding parentheses around the highest to lowest terms.
For example, if you started with 2 + 3 * 4, you'd end up with (2 + (3 * 4)). Give it a try.
fred rosenberger
lowercase baba
Bartender
Posts: 12124
30
• 1
And this is why you should always use parens. really, there is zero cost to adding them, and having them saves the next person reading it (which could be yourself) the trouble of having to remember the correct operator precedence.
Henry Wong
author
Marshal
Posts: 21117
78
• 1
Bear Bibeault wrote:The easiest way to figure it all out is to use the operator precedence and start adding parentheses around the highest to lowest terms.
To add to that, if you encounter two operators with the same precedence, then you use the associativity. Most go from left to right, but some goes right to left.
Henry
Bear Bibeault
Author and ninkuma
Marshal
Posts: 64838
86
P.S. I agree with fred -- for any but the most trivial of expressions, I use parens to make the meaning of the expression crystal clear.
Winston Gutkowski
Bartender
Posts: 10417
63
Sam Samson wrote:if(z > 1 && y > 6 | z == 2)
It may also be worth pointing out that a construct like y > 6 | z == 2 is extremely rare except when dealing with bit manipulation; it's almost always better to use '||'.
Winston
Stephan van Hulst
Bartender
Posts: 5811
61
All the point are very valid, but personally I also find it handy to remember a small list so you don't get too surprised when you're reading other people's code. From highest to lowest priority:
1. Dereference (array[i] and obj.foo)
2. Unary (includes casts)
3. Arithmetic (includes shifts)
4. Comparison (includes instanceof)
5. Bitwise
6. Logical
7. Assignment
For instance, this is one I frequently bump my head into: (Something)something.getFoo(); when they mean: ((Something)something).getFoo(), because dereferences (. or []) have a higher precedence than casts. | 723 | 2,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-26 | latest | en | 0.891263 |
https://sqlpete.wordpress.com/2017/02/12/upper-bound-of-the-information-value-statistic/comment-page-1/ | 1,498,713,965,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00505.warc.gz | 826,700,206 | 37,809 | ## Upper bound of the Information Value statistic
### Information Value
Despite having worked with it for years, it has always irked me that I don’t know the derivation of the Information Value (IV) statistic.
It’s used liberally throughout credit risk work, but the background to its invention seems somewhat hazy. Clearly it’s related to Shannon Entropy, via the $\sum p \log(p)$ construct. In Naeem Siddiqi’s well-known book Credit Risk Scorecards, he writes “Information Value, […] comes from information theory” and references Kulback’s 1959 book Information Theory and Statistics, which I don’t have. Someone else suggested that it stems from the work of I.J. Good, but I can’t find an explicit definition in any of his papers I’ve managed to look at. (I bought his book Good Thinking, about the foundations of probability and statistical interference, but it’s waaaay too complex for me!)
The Information Value (IV) is defined as:
$\mathrm{IV} = \sum_{i=1}^{k} (g_{i} - b_{i}) \log_e (g_{i} / b_{i})$
, where $g_{i}$ is the number of ‘goods’ in category i, and $b_{i}$ is the number of ‘bads’.
In his book, Siddiqi gives the following rule of thumb regarding the value of IV:
< 0.02 unpredictive 0.02 to 0.1 weak 0.1 to 0.3 medium 0.3 to 0.5 strong 0.5+ “should be checked for over-predicting”
For an independent variable with an IV over 0.5, it might be somehow related to the dependent variable, and you might want to consider leaving it out. (If you build a scorecard that has a bureau score as one of your variables, then you’ll almost certainly see this.)
[See these two links for more about Information Value, and an example or two of its use: All about “Information Value” and Information Value (IV) and Weight of Evidence (WOE).]
### Upper Bound
The lower bound of the IV is fairly obviously zero: if $g_{i} \equiv b_{i}$ for all the categories, then the difference is zero, so their sum is zero times $\log_{e}(1)$, which is also zero. But what about the upper bound?
I’ve put together this small PDF document: Upper bound of the Information Value (IV), in which (I think!) I show that the upper bound is very close to $\log_{e}(N_{G}) + \log_{e}(N_{B})$, where $N_G$ is the total number of goods, and $N_B$ is the total number of bads.
Of course, it’s wise to at least check the result with some code — so in R, let’s create a million tables at random, and look at the actual figures that are produced:
Z <- 1000000; # number of iterations
IV <- rep(0, Z); # array of IVs
lGB <- rep(0, Z); # array of (log(n_g) + log(n_b))
for (i in 1:Z)
{
k <- sample(2:20, 1); # number of categories
g <- sample(1:100, k, replace=T); # good
b <- sample(1:100, k, replace=T); # bad
ng <- sum(g);
nb <- sum(b);
IV[i] <- sum( ((g/ng)-(b/nb)) * log((g/ng)/(b/nb)) );
lGB[i] <- log(ng) + log(nb);
}
plot(IV, lGB, xlab="IV", ylab="log(N_G)+log(N_B)",
main="IV vs log(N_G)+log(N_B)", pch=19,col="blue",cex=0.5);
abline(a=0,b=1,col="red",lwd=2); # draw the line x=y
As you can see, there are no points below the red ‘x=y’ line; in other words, the IV is always less than $\log_{e}(N_{G}) + \log_{e}(N_{B})$. There are a few points that are close; the closest is:
min(lGB-IV)
[1] 0.2161227
I know that $\log_{e}(N_{G}) + \log_{e}(N_{B})$ is not the best possible upper bound — a closer, but more complex answer is reasonably obvious from the document — but “log(number of goods) plus log(number of bads)” is (a) memorable, and (b) close enough for me!
1. #1 by Michael Hore on March 25, 2017 - 11:55 am
Interesting topic. I have been tempted to investigate.
I think I found the answer to where this blasted stat was derived! It’s in the first few pages of that information theory book you linked to which is available on Google, in the section on Divergence – this is what would apparently now be called symmetric kullback-leibler divergence according to wikipedia.
Lets say we’re doing WOE/IV on age binning and I will summarize my possibly incorrect understanding. We have our distributions b(age) and g(age) which give the probability that a sample falls in a given age bucket given that they are bad or good respectively. | 1,164 | 4,144 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-26 | longest | en | 0.910032 |
https://www.gradesaver.com/textbooks/math/algebra/differential-equations-and-linear-algebra-4th-edition/chapter-3-determinants-3-5-chapter-review-additional-problems-page-243/12 | 1,576,217,560,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00290.warc.gz | 723,330,227 | 12,001 | ## Differential Equations and Linear Algebra (4th Edition)
$-6$.
If $A$ and $B$ are $n\times n$ matrices, then $det(AB)=det(A)det(B)$. Also $det(A^T)=det(A)$ and if $det(A)\ne0$, then $det(A^{-1})=\frac{1}{det(A)}$ Hence here $det(AB)=det(A)det(B)=-2\cdot3=-6$. | 104 | 262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.261102 |
https://www.statisticssolutions.com/reliability-analysis/ | 1,713,187,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00264.warc.gz | 917,117,138 | 22,506 | # Reliability Analysis
Quantitative Results
Statistical Analysis
Reliability analysis refers to the fact that a scale should consistently reflect the construct it is measuring. There are certain times and situations where it can be useful.
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An aspect in which the researcher can use reliability analysis is when two observations under study that are equivalent to each other in terms of the construct being measured also have the equivalent outcome.
There is a popular technique called the split half reliability. This method splits the data into two parts. The score for each participant in the analysis is then computed on the basis of each half of the scale. In that type of reliability analysis, if the scale is very reliable, then the value of the person’s score on one half of the scale would be equivalent to the score on the other half. In this type of reliability analysis, the previous fact should remain true for all the participants.
The major problem is that there are several ways in which a set of data can be divided into two parts, and therefore the outcome could be numerous.
In order to overcome this problem, Cronbach (1951) introduced a measure that is common in reliability analysis. This measure is loosely equivalent to the splitting of the data in two halves in every possible manner and further computing the correlation coefficient for each split. The average of these values is similar to the value of Cronbach’s alpha.
There are basically two versions of alpha in reliability analysis. The first version is the normal version. The second version is the standardized version.
The normal version of alpha is applicable when the items on a scale are summed to produce a single score for that scale. The standardized version of alpha is applicable when the items on a scale are standardized before they are summed up.
According to Kline (1999), the acceptable value of alpha in reliability analysis is 0.8 in the case of intelligence tests, and the acceptable value of alpha in reliability analysis is 0.7 in the case of ability tests.
There are certain assumptions that are assumed.
While conducting reliability analysis in SPSS, the researcher should click on “Tukey’s test of additivity” as additivity is assumed.
Independence within the observations is assumed. However, it should be noted by the researcher that the test retest type of reliability analysis involves the correlated data between the observations which do not pose a statistical problem in assessing the reliability.
It is assumed that the errors are uncorrelated to each other. This means that no association exists among the errors and therefore all the errors are different.
To attain reliability in the data, the coding done by the researcher should be consistent. This means that the high values must be coded consistently, such that they have the same meaning across the items.
In the split half type of reliability analysis, the random assignment of the subjects is assumed. Generally, the odd numbered items fall in one category and the even numbered items fall in the other. | 718 | 3,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | longest | en | 0.933088 |
https://www.robinstewart.com/blog/2010/09/math-in-ancient-india/ | 1,685,909,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00162.warc.gz | 1,071,199,748 | 6,879 | # Math in Ancient India
I just checked my web server statistics and found that part of my high school research paper on the history of mathematics is getting well over a thousand requests a month.
The topic of that paper is “the usually unrecognized achievements of Vedic and Hindu mathematicians from 2000-300 B.C.” When writing it, I was surprised at how hard it was to find good research on the topic:
In Mathematical Thought from Ancient to Modern Times, [a “comprehensive” summary of] the history of mathematics, [author Morris Kline] included only half a chapter (out of 50 chapters total) on Indian math. Everything he said seemed to sneer at them, put them down, and belittle their accomplishments.
Despite this dearth of understanding, the facts were clear:
Besides using simple arithmetic operations like addition, subtraction, etc., Indians invented the decimal system and the idea of positional notation, both of which are still in use today. They also used the “Pythagorean” theorem and “Pascal’s” triangle long before either of those men were born!
Today, it turns out that if you google “Sulva Sutras” (the title of the web page getting most of the visits), my research paper is the first result, above Wikipedia and everything else! If you search for “mathematics in vedas”, I’m the third result.
True, this seeming popularity may have something to do with spelling inconsistencies — the Wikipedia article uses “Shulba Sutras” and is the first result if you search with that spelling. It’s also true that my paper was written in 1999 and has been on the web since 2003, so has had time to gather links from other websites (which influence Google’s ranking).
But we’re talking about the founding documents of mathematics! The origin of zero! The “Pythagorean” theorem, recorded hundreds of years before Pythagoras! And the most relevant article was written by a fifteen-year-old?
In my research paper’s conclusion (which is mostly too embarrassing to quote), I wrote, “I find it unbelievable how little work has been done in the field…. The vast majority of the work has been done only by Indians. Most of the books on the subject are written in Sanskrit or Hindi [and are ignored by] eurocentric scholars.”
Ten years later, my incredulity lives on. | 491 | 2,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.960623 |
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