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# Tutorial on Capital Costs: Light Bulbs
Posts: 4,509 Add as FriendChallenge to a DebateSend a Message 9/21/2011 11:08:22 PMPosted: 6 years agoDebates on the costs of alternative energy recently have shown that many people are perplexed by the calculations, particular of capital costs. I've tried here to provide a relatively simple example. It gets complicated enough.The idea is to compare the costs of three light bulbs including the purchase price, capital costs, and electricity. Here are three products I found on amazon.com. Each claims to produce 400 lumens, like a 60 watt incandescent bulb:cost power life (hrs)LED \$15.00 7 25000CFL \$1.00 14 10000Incandescent \$0.46 60 1000So the LED unit costs \$15, draws seven watts, and is claimed to last 25,000 hrs.To start with, assume the national average price for a kilowatt hour of electricity and the utility bond interest rate. These are:Cost / kwh \$0.11Interest Rate4.00%Okay. Which is cheapest if the light is to be on about 3 hours per day -- 1000 hrs per year. The LED will cost \$1.98 to run. It uses 77 cents worth of electricity. 1/25 of the bulbs 25,000 hour life is consumed, 60 cents. To those costs, the cost of having money tied up in a light bulb rather than invested must be added. The bond interest rate of 4% is one investment possibility, and a year of that at \$15 costs 60 cents.However, if the person has credit card debt costing 16%, then paying off that debt would yield a saving of \$2.40. The cost of capital depends on what alternatives are available. With the higher cost of capital, the LED unit would cost \$3.77 for the year, rather than \$1.98.Using 4% interest rate, the CFL comes in at \$1.68 and old fashioned incandescent at \$7.70.Now suppose the bulb is on 24/7 for the year. The LED wins at \$12.80, with the CFL at \$14.41, and the incandescent at \$61.86. Low power consumption beats the higher capital cost of the LED.Finally, suppose the bulb is in the closet with the Christmas ornaments, and is used only 2 hours a year. Now the incandescent wins \$0.033, to \$0.043 (CFL) and \$0.60 (LED). Actually, would anyone figure on keeping a light bulb for 5000 years -- the life of the CFL in the application? It would be more reasonable to assume that after 50 years you'll give up on it. The CFL and LED are then more expensive relative to the incandescent, the LED 22 times as much.I live in California where electric rates quickly soar to 42 cents, a punishment for sin. Then the costs for 1000 hr/year usage favor LED lamps, even at \$15 apiece.It's also worth thinking about maintenance costs. A company must pay someone to change the bulb, which might cost \$15 or most in labor. That's effectively added to the bulb cost.I haven't figured other attributes into the calculation. A \$1 CFL has in my experience an annoying flicker. LEDs are usually more beautiful. We just did cost here.If anyone would like to play with the spreadsheet for this problem, it's here:factspluslogic.com/LightingCosts.xls
Posts: 18,337 Add as FriendChallenge to a DebateSend a Message 9/22/2011 2:31:08 AMPosted: 6 years agoMostly, it makes sense, but you are assuming a guaranteed rate of return on your investment which you then add to the cost of the light bulb. In reality, you never know how much you are going to get from an investment. Adding it to the price will obviously inflate the cost. A better alternative would be to include the time value of money for long-term calculations.
Posts: 4,509 Add as FriendChallenge to a DebateSend a Message 9/22/2011 1:32:21 PMPosted: 6 years agoAt 9/22/2011 2:31:08 AM, F-16_Fighting_Falcon wrote:Mostly, it makes sense, but you are assuming a guaranteed rate of return on your investment which you then add to the cost of the light bulb. In reality, you never know how much you are going to get from an investment. Adding it to the price will obviously inflate the cost. A better alternative would be to include the time value of money for long-term calculations.The 4% interest rate is for a long term bond, so the rate is locked in for 20 or 30 years. Interest rates are near a historical low, so locking up money in a long term fixed asset is actually less attractive than supposed. Historical stock market returns are around 9%, which is probably a better number.There is nothing to "time value of money" other than expected rates of return. For cost estimates on Government contracts, the Government provides a table of expected inflation rates to use in estimating the contract. Point is that something has to be projected, and there is no certainty.
Posts: 11,204 Add as FriendChallenge to a DebateSend a Message 9/22/2011 2:21:20 PMPosted: 6 years agoWell If your going to consider the capital cost of the light bulb, how much it costs compared to the alternative of investing, why not consider fuel as a capital cost as well?Open borders debate: http://www.debate.org...
Posts: 4,509 Add as FriendChallenge to a DebateSend a Message 9/22/2011 2:39:30 PMPosted: 6 years agoAt 9/22/2011 2:21:20 PM, darkkermit wrote:Well If your going to consider the capital cost of the light bulb, how much it costs compared to the alternative of investing, why not consider fuel as a capital cost as well?I don't understand. Fuel is bought and consumed. The cost of electricity is in the cost equation, and it's true that the future cost is uncertain. Electricity, corrected for inflation, was stable for several decades until about the past decade, when government started insisting on using uneconomical green energy. An equation for future electricity costs could be used instead of the fixed cost, but that's not a capital cost. | 1,401 | 5,698 | {"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} | 3.046875 | 3 | CC-MAIN-2018-13 | longest | en | 0.928314 |
http://irvinemclean.com/maths/probprim7.htm | 1,516,559,938,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00783.warc.gz | 190,000,643 | 5,860 | Complex MultiFactorials
Multifactorials of the form n!m±1 have been heavily mined for primes, as is totally justified by their tractability, by which we mean susceptibility to BLS. The form n!m is just a shorthand for denoting the multiple n.(n-m).(n-2m)…(n-km), where k is such that n-km > 0 but n-(k+1)m ³ 0, that is, the product of a descending sequence of positive integers starting at n and separated by a constant value m. It occurred to me to consider sequences where the separator between consecutive terms was not a constant, for example alternating or monotonically increasing.
The most simple deviation would be to consider alternating the separator between 1 and 2, obtaining the product n.(n-1).(n-3).(n-4).(n-6).(n-8)…, etc. In effect, this can be represented in a closed form as n!3.(n-1)!3, and in general a repeating set of separators can readily be expressed in closed form as a product of multifactorials, as follows.
Let {a1, a2, … , am} represent a repeating set of separators between consecutive terms of a descending sequence of positive integers starting at n. Let bi be the intermediate sums of this sequence, i.e. b0 = 0, b1 = a1, bi = bi-1 + ai, and let p = bm (that is, p is the sum of all the ai). Then the product of the sequence can alternatively be written in closed form as
The product Π from i =0 to m-1 of the multifactorials (n-bi)!p.
For example, if the sequence is {2, 1}, then b0 = 0, b1 = 2, p = 3 and so n.(n-2).(n-3).(n-5)… = n!3.(n-2)!3
Also, for a1 = 1, a2 = 2, a3 = 3, we have b0 = 0, b1 = 1, b2 = 3, b3 = m = 6, so the product can be represented by n!6.(n-1)!6.(n-3)!6.
It is then perfectly reasonable to consider adding or subtracting 1 in order to search for primes, while retaining tractability.
So far, this is a variation on a familiar scene. However, let us now consider increasing sequences. Obviously the simplest sequence is to consider ai = i, which can also be represented by n.(n-1).(n-3).(n-6).(n-10)…, etc. We note the occurrence of triangular numbers in this formula as the intermediate sums bi. This is a side-effect of our definition, and it is obvious that alternative sequences may be derived based on this idea (one such is mentioned below). Be that as it may, the separator sequence defined by ai = i leads to a sequence of complex multifactorial numbers that rises very much more slowly than normal multifactorials because of the scarcity of individual elements. For instance, at n = 1000 the decimal expansion of the product has only 123 digits, and for n = 10000 has 528 digits. Since the product has a substantial degree of built-in divisibility, this is a reasonable source of primes and possibly prime pairs.
For n up to 10000, prime pairs occur for n = 3, 22, 31, 93, 162, 327 & 2272.
Divisibility properties of these numbers are complex. The only obvious pattern is for divisibility by 3. These numbers also provide a nice sequence for factoring.
I then considered ai = Fi, the Fibonacci numbers, and denote the complex multifactorial so generated as n!F. Then n!F = n.(n-1).(n-2).(n-4).(n-7)…, etc. Since the Fibonacci numbers rise much faster than sequentially, the product has fewer elements. This produces a much more slowly rising sequence of numbers (for n = 10000, the product has only 75 digits).
For n up to 10000, prime pairs occur for n = 3, 5, 6, 14, 23, 28, 33, 41, 43, 48, 265, 897, 909, 1142, 1625, 2378, 2476, 2534, 2822, 3055, 3434, 3837, 4286, 4372, 5865, 8261, 8685, 9195, 9255, 11366, 12984, 16029, 17497, 18278, 18560, 19730, 20450, 21442, 24490, 24603, 25005, 25220, 26243, 29321, 29460, 30518, 31309, 33639, 35143, 35251, 35545, 36085, 37002, 37420, 38536, 38609, 39915, 41303, 47067, 52939, 53485, 53649, 55439, 58004, 58522, 63899, 65529, 65917, 66078, 71720, 72229, 74926, 77744, 79595, 80911, 83279, 83428, 84407, 86386, 87996, 89808, 89912, 94513, 95425, 96933, 98419
Any other strictly monotonically rising sequences with a simple form such as the two just considered, for example the triangular numbers, rise faster and so the products rise much slower, meaning that although there are much larger solution spaces, it takes a lot of effort to reach numbers of a size that makes them interesting. In order to speed up the rate of increase, we therefore must consider removing the strictness of the sequences.
Consider the staggered sequence defined by a2i = a2i-1 = i (that is, 1, 1, 2, 2, 3, 3, …m, etc), giving the product
n.(n-1).(n-2).(n-4).(n-6).(n-9).(n-12).(n-16).(n-20).(n-25).(n-30)…
In fact, this can be broken down into a product of the two separate sequences:
n.(n-1).(n-4).(n-9).(n-16).(n-25)… and (n-2).(n-6).(n-12).(n-20).(n-30)…
where the first sequence has a separator sequence of the odd numbers and the second has a separator sequence of the even numbers. If we represent the complex multifactorial based on the odd numbers as n!O and the one based on even numbers as n!E, then the combination can be represented as n!O.n!E divided by an extra occurrence of n.
This is a cumbersome combination, but more appealing is the n!O form (the n!E being a unsuitable as this is odd or even depending on n and so every second value of n!E+1 would be even). The n!O form, as can be seen by inspection, contains square numbers as its intermediate sums. At n = 1000, the digit length is 88 and at n = 10000 this rises to 375.
For n up to 10000, there are prime pairs for n!O±1 at n = 3, 4, 6, 12, 48, 60, 781 and 1354.
In order to preserve a noticeable size increase as n rises, we may consider reducing a monotonically increasing separator sequence modulo a particular integer. In this case, we could either ignore elements in the sequences that are zero, or include them, the latter choice meaning repetition of values in the full expansion. For instance, if we let ai be the triangular number Ti modulo 10, then the separator sequence becomes {1, 3, 6, 0, 5, 1, 8, 6, 5, 5, …} with intermediate sum bi taking values {1, 4, 10, 10, 15, 16, …} and we may consider the expansion n.(n-1).(n-4).(n-10) and then either include another (n-10) for the zero separator, or ignore it and go straight to (n-15). As ever, the expansion for a particular n continues only while bi < n.
We could even replace a modulated rising sequence by the digits in the decimal expansion of any of a number of well known transcendental numbers such as p, e, l, etc. Assuming that we can expand these numbers to the required accuracy (which should not be a problem as there are published expansions well beyond the millionth term), the multifactorials based on these modulated sequences will rise at a rate much faster than for a rising sequence, providing fewer but hopefully larger primes. | 1,989 | 6,710 | {"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-2018-05 | latest | en | 0.940284 |
https://us.metamath.org/mpeuni/en2eqpr.html | 1,716,047,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00099.warc.gz | 553,372,180 | 5,824 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > en2eqpr Structured version Visualization version GIF version
Theorem en2eqpr 9411
Description: Building a set with two elements. (Contributed by FL, 11-Aug-2008.) (Revised by Mario Carneiro, 10-Sep-2015.)
Assertion
Ref Expression
en2eqpr ((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) → (𝐴𝐵𝐶 = {𝐴, 𝐵}))
Proof of Theorem en2eqpr
StepHypRef Expression
1 2onn 8244 . . . . . 6 2o ∈ ω
2 nnfi 8689 . . . . . 6 (2o ∈ ω → 2o ∈ Fin)
31, 2ax-mp 5 . . . . 5 2o ∈ Fin
4 simpl1 1187 . . . . 5 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 𝐶 ≈ 2o)
5 enfii 8713 . . . . 5 ((2o ∈ Fin ∧ 𝐶 ≈ 2o) → 𝐶 ∈ Fin)
63, 4, 5sylancr 589 . . . 4 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 𝐶 ∈ Fin)
7 simpl2 1188 . . . . 5 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 𝐴𝐶)
8 simpl3 1189 . . . . 5 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 𝐵𝐶)
97, 8prssd 4731 . . . 4 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → {𝐴, 𝐵} ⊆ 𝐶)
10 pr2nelem 9408 . . . . . . 7 ((𝐴𝐶𝐵𝐶𝐴𝐵) → {𝐴, 𝐵} ≈ 2o)
11103expa 1114 . . . . . 6 (((𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → {𝐴, 𝐵} ≈ 2o)
12113adantl1 1162 . . . . 5 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → {𝐴, 𝐵} ≈ 2o)
134ensymd 8538 . . . . 5 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 2o𝐶)
14 entr 8539 . . . . 5 (({𝐴, 𝐵} ≈ 2o ∧ 2o𝐶) → {𝐴, 𝐵} ≈ 𝐶)
1512, 13, 14syl2anc 586 . . . 4 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → {𝐴, 𝐵} ≈ 𝐶)
16 fisseneq 8707 . . . 4 ((𝐶 ∈ Fin ∧ {𝐴, 𝐵} ⊆ 𝐶 ∧ {𝐴, 𝐵} ≈ 𝐶) → {𝐴, 𝐵} = 𝐶)
176, 9, 15, 16syl3anc 1367 . . 3 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → {𝐴, 𝐵} = 𝐶)
1817eqcomd 2826 . 2 (((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) ∧ 𝐴𝐵) → 𝐶 = {𝐴, 𝐵})
1918ex 415 1 ((𝐶 ≈ 2o𝐴𝐶𝐵𝐶) → (𝐴𝐵𝐶 = {𝐴, 𝐵}))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 398 ∧ w3a 1083 = wceq 1537 ∈ wcel 2114 ≠ wne 3006 ⊆ wss 3913 {cpr 4545 class class class wbr 5042 ωcom 7558 2oc2o 8074 ≈ cen 8484 Fincfn 8487 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2792 ax-sep 5179 ax-nul 5186 ax-pow 5242 ax-pr 5306 ax-un 7439 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3or 1084 df-3an 1085 df-tru 1540 df-ex 1781 df-nf 1785 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2799 df-cleq 2813 df-clel 2891 df-nfc 2959 df-ne 3007 df-ral 3130 df-rex 3131 df-reu 3132 df-rab 3134 df-v 3475 df-sbc 3753 df-dif 3916 df-un 3918 df-in 3920 df-ss 3930 df-pss 3932 df-nul 4270 df-if 4444 df-pw 4517 df-sn 4544 df-pr 4546 df-tp 4548 df-op 4550 df-uni 4815 df-br 5043 df-opab 5105 df-tr 5149 df-id 5436 df-eprel 5441 df-po 5450 df-so 5451 df-fr 5490 df-we 5492 df-xp 5537 df-rel 5538 df-cnv 5539 df-co 5540 df-dm 5541 df-rn 5542 df-res 5543 df-ima 5544 df-ord 6170 df-on 6171 df-lim 6172 df-suc 6173 df-iota 6290 df-fun 6333 df-fn 6334 df-f 6335 df-f1 6336 df-fo 6337 df-f1o 6338 df-fv 6339 df-om 7559 df-1o 8080 df-2o 8081 df-er 8267 df-en 8488 df-dom 8489 df-sdom 8490 df-fin 8491 This theorem is referenced by: isprm2lem 16003 en2top 21569
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# 1 1 2 1 r2 2 r1 r2 1 1 2 1 2 2 4 10 r 2 r
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Unformatted text preview: t Gauss-Jordan elimination would be less work in all likelihood that if we solved directly. Also, as we saw in the final example worked in this section. There really is no one set path to take through these problems. Each system is different and may require a different path and set of operations to make. Also, the path that one person finds to be the easiest may not by the path that another person finds to be the easiest. Regardless of the path however, the final answer will be the same. © 2007 Paul Dawkins 334 http://tutorial.math.lamar.edu/terms.aspx College Algebra More on the Augmented Matrix In the first section in this chapter we saw that there were some special cases in the solution to systems of two equations. We saw that there didn’t have to be a solution at all and that we could in fact have infinitely many solutions. In this section we are going to generalize this out to general systems of equations and we’re going to look at how to deal with these cases when using augmented matrices to solve a system. Let’s first give...
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# NORMAL DISTRIBUTION - PowerPoint PPT Presentation
NORMAL DISTRIBUTION. Normal distribution- continuous distribution. Normal density: bell shaped, unimodal- single peak at the center, symmetric. Completely described by its center of symmetry - mean μ and spread - standard deviation σ. μ.
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Normal distribution- continuous distribution.
Normal density:
• bell shaped,
• unimodal- single peak at the center, symmetric.
• Completely described by its center of symmetry - mean μand spread - standard deviation σ.
μ
Random variable with normal distribution – normal random variable with mean μ and st. dev. σ: X~N(μ, σ)
Standard normal random variable: mean 0 and st. dev. 1: Z~N(0, 1)
CHANGING SCALE σ
CHANGING LOCATION μ
μ1
μ2
“peaky” density
“flatter” density
1=σ1 < σ2 = 2
0=μ1 < μ2 =1
Changes in mean/location cause shifts in the density curve along the x-axis.
Changes in spread/standard deviation cause changes in the shape of the density curve.
• Normal distributions are great descriptions-models-approximations for many data sets such as weights, heights, exam scores, experimental errors, etc.
• Great descriptions of results-outcomes of many chance driven experiments.
• Statistical inference based on normal distribution works well for many (approximately) symmetric distributions.
• HOWEVER, remember that not everything or everybody is normal!
Areas under the normal curve are probabilities (as under any density curve).
Special areas under the normal density curve:
• Approximately 68% of the observations fall within 1 standard deviation of the mean
• Approximately 95% of the observations fall within 2 standard deviations of the mean
• Approximately 99.7% of the observations fall within 3 standard deviations of the mean
The range "within one/two/three standard deviation(s) of the mean" is highlighted in green.
The area under the curve over this range is the rel. frequency of observations in the range.
That is, 0.68/95/99.7 = 68%/95%/99.7% of the observations fall within one/two/three standard deviation(s) of the mean, or, 68%/95%/99.7% of the observations are between (μ – 1/2/3σ) and (μ + 1/2/3σ).
Normal probabilities = areas under the normal curve are tabulated for the standard normal distribution (front cover of your text book).
In looking for probabilities keep in mind:
Symmetry of the normal curve and P(Z=a)=0 for any a.
FIND:
P(Z < 0.01) = 0.504
P(Z ≤ -0.01 ) = 0.496
P(Z < 0) = 0.5
P( Z < 2.92)= 0.9982
P(Z>2.92)=1-0.9982=0.0018 or, by symmetry =P(Z< - 2.92)=0.0018
P(-1.32< Z <1.2)=0.8849 – 0.0934=0.7915
SUMMARY OF RULES we used above: P(Z>a)=1-P(Z< -a)
P(a < Z < b) = P(Z < b)- P( Z < a)
IF X~N(μ, σ) then Z = ~N(0, 1) standard normal.
standardization
Example. Suppose that the weight of people in NV follows normal distribution with mean 150 and standard deviation 20 lb. Find the probability that a randomly selected Nevadan weighs
at most 160 lb; b) over 160 lb.
Solution. Let X= weight of a randomly selected Nevadan. X~N(150, 20).
• P(X ≤ 160) =
• P(X>160)= 1 - P(X ≤ 160) =1 – 0.6915 = 0.3085.
Given that P(Z < p)=0.95 find p. Here p is called 95thpercentile of Z.
Inside the table I looked for 0.95.
Found 0.9495 and 0.9505.
Used z-value corresponding to
the midpoint (0.95) between the
two available probabilities 1.645.
p=1.645
If an available probability is closer
to the one we need, use the
z-value corresponding to
that probability.
0.95
p=?
EXAMPLE. Suppose scores X on a test follow a normal distribution with mean 430 and standard deviation 100. Find 90th percentile of the scores, that is find score x such that P(X ≤ x)=0.9.
Solution. Since we start with a normal but NOT STANDARD normal distribution, we have to standardize at some point:
0.9 = P(X ≤ x) =
get equation:
x - 430 =128
x = 558
90% of students scored 558 or less.
0.90
z =1.28
Height of women follows normal distribution with mean 64.5 and standard deviation of 2.5 inches. Find
a) The probability that a woman is shorter than 70 in.
b) The probability that a woman is between 60 and 70 in tall.
c) What is the height 10% of women are shorter than, i.e. what is the 10th percentile of women heights?
SOLUTION. X= women height; X~N(64.5, 2.5).
a) P(X <70)=P(Z< (70-64.5)/2.5)=P(Z<2.2)=0.9861
b) P( 60 < X < 70) = P( (60-64.5)/2.5) < Z < (70-64.5)/2.5)=P(-1.8< Z < 2.2)= P( Z <-2.2) – P( Z < -1.8) = 0.9861 – 0.0359 = 0.9502.
c) 10th percentile of X =?
0.1=P( X< x) = P( Z< (x-65.5)/2.5), so -2.33=(x-65.5)/2.5; x=59.675.
10% of women are shorter than 59.675 in. | 1,557 | 5,287 | {"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-2017-47 | latest | en | 0.840244 |
https://mcqslearn.com/electronics/electronic-devices/quiz/quiz-questions-and-answers.php?page=80 | 1,656,884,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00647.warc.gz | 424,759,828 | 11,046 | Engineering Degrees Online Courses
Electronic Devices Quizzes
Electronic Devices MCQs - Complete
# Transistor AC Equivalent Circuits Multiple Choice Questions PDF p. 80
Books:
Apps:
Solve Transistor AC Equivalent Circuits multiple choice questions and answers, transistor ac equivalent circuits quiz answers PDF 80 to learn Electronic Devices course for college certification. Learn BJT Amplifiers quiz questions, transistor ac equivalent circuits Multiple Choice Questions (MCQ) for online college degrees. "Transistor ac equivalent circuits MCQ" book PDF: power supply filters, transformer in half wave rectifier, transistor ac equivalent circuits test prep for college entrance test.
"Internal emitter resistance of BJT amplifier is equals to", transistor ac equivalent circuits Multiple Choice Questions (MCQ) with choices 25 mvxie, 25 mv, 25 mv+ie, and 25 mv/ie for online engineering programs. Solve bjt amplifiers questions and answers to improve problem solving skills for questions to ask in an interview.
## Transistor AC Equivalent Circuits Questions and Answers
1.
Internal emitter resistance of BJT amplifier is equals to
25 mV
25 mVxIE
25 mV+IE
25 mV/IE
2.
PI V rating for a diode which have to be used in a rectifier with peak output voltage of 60 V is
30 V
40 V
60 V
70 V
3.
For a halfwave rectifier, there is current through the load for approximately
20% of input cycle
30% of input cycle
50% of input cycle
70% of input cycle
4.
When 60Hz sinusoidal voltage is applied to input of halfwave rectifier, output frequency will be
30 Hz
20 Hz
60 Hz
120 Hz
5.
By ohm's law, when forward current increases resistance will
decrease
increase
become zero
have no change | 393 | 1,701 | {"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-2022-27 | latest | en | 0.761081 |
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First define the relation $\eqidem \subseteq \CTerms\Sigma \times \CTerms\Sigma$ as follows. $\eqidem = \{ (p,p), (p,f(p,p)), (f(p,p),p) \,|\, p \in \CTerms\Sigma \}$ To prove the theorem it suffices to show that $\eqidem$ is a bisimulation relation. If it is, then $f(p,p)\bisim p$ for any closed term $p$ and since $\bisim \subseteq \sim$ we obtain the theorem. Let $(C,U)$ be the least three-valued stable model for the TSS under consideration. First consider a closed term $p$ s.t. $p \trans{l} p' \in C$ for some $l$ and $p'$ (note that $U=\emptyset$ since the TSS is complete). Next, we argue that $f(p,p) \trans{l} p''$ for some $p''$ such that $p' \eqidem p''$. Since $p \trans{l} p' \in C$, there exists a provable transition rule of the form $\frac{N}{p \trans{l} p'}$ for some set of negative formulae $N$ such that $C \vDash N$. In particular, that means that $p \ntrans{l} \notin N$. In this case we make use of the requirement that there exists at least one rule of a starred form for label $l$. If there exists a rule of the form $1_l^*$, i.e. $\sosrule[i\in\{0,1\}]{x_i\trans{l}x'}{f(x_0,x_1)\trans{l}x'}$ then we can instantiate it to prove that $f(p,p)\trans{l}p'\in C$. In particular, it does not matter if $i=0$ or $i=1$. Since $\eqidem$ is reflexive, $p'\eqidem p'$ holds. If there exists a rule of the form $2_{l,l}^*$, we observe that $\gamma(l,l) = l$ so the transition rule becomes $\sosrule{x_0\trans{l}x_0' \quad x_1\trans{l}x_1'}{f(x_0,x_1) \trans{l} f(x_0',x_1')},$ where $x'_0 \equiv x'_1$ or $\trans{l}$ is deterministic. Now we can use the existence of $p\trans{l}p'$ to satisfy both premises and obtain that $f(p,p)\trans{l}f(p',p')$. By the definition of $\eqidem$ we also have that $p' \eqidem f(p',p')$. In either case, if $p \trans{l} p'\in C$, then there exists a $p''$ s.t. $f(p,p) \trans{l} p'' \in C$ and $p' \eqidem p''$. Now assume that $f(p,p) \trans{k} p' \in C$. Then there exists a provable transition rule $\frac{N}{f(p,p) \trans{k} p'}$ for some set of negative formulae $N$ such that $C\vDash N$. Since all rules for $f$ are either of the form $1_l$ or $2_{l_0,l_1}$, this provable transition rule must be based on a rule of those forms. We analyze each possibility separately, showing that in each case $p \trans{k} p''$ for some $p''$ such that $p' \eqidem p''$. If the rule is based on a rule of form $1_l$, its positive premises must also be provable. In particular it must hold that $p\trans{k} p' \in C$ since both $x_0$ and $x_1$ in the rule are instantiated to $p$. The other premises are of no consequence to this conclusion and, again, we observe that $p'\eqidem p'$. Now consider the case where the transition is a consequence of a rule of the form $2_{l_0,l_1}$. If $t_0\equiv t_1$, say both are equal to $p''$, we must consider two cases, namely $k=l_0$ and $k=l_1$. If $k=l_0$ then the first premise of the rule actually states that $p\trans{k}p''$. If $k=l_1$ then the second premise similarly states that $p\trans{k}p''$. In either case, we note that $p'\equiv f(p'',p'')$ must hold and again by the definition of $\eqidem$ we have that $f(p'',p'')\eqidem p''$. If however $t_0\not\equiv t_1$ the side condition requires that $l_0=l_1 = k$, which also implies $\gamma(l_0,l_1)=l_0=k$, and that the transition relation $\trans{l_0}$ is deterministic. In this case it is easy to see that the right-hand sides of the first two premises, namely $t_0$ and $t_1$, evaluate to the same closed term in the proof structure, say $p''$. The conclusion then states that $k=l_0$ and $p'\equiv f(p'',p'')$. It must thus hold that $p\trans{k}p''\in C$ and $f(p'',p'')\eqidem p''$ as before. From this we obtain that if $f(p,p) \trans{k} p' \in C$ then there exists a $p''$ such that $p \trans{k} p'' \in C$ and $p' \eqidem p''$. Thus, $\eqidem$ is a bisimulation.
Something went wrong with that request. Please try again. | 1,345 | 4,075 | {"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": 2, "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} | 3.1875 | 3 | CC-MAIN-2015-18 | latest | en | 0.728185 |
http://smallbusiness.chron.com/calculate-retroactive-salary-increase-10054.html | 1,477,523,705,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00523-ip-10-171-6-4.ec2.internal.warc.gz | 232,549,562 | 20,414 | # How to Calculate a Retroactive Salary Increase
by Grace Ferguson, studioD
Retroactive pay refers to back pay or money owed to an employee because of a pay increase. Managers or supervisors often perform annual reviews of their employees; during this time, they decide whether to give employees a raise. If the salary increase is effective during a prior pay period, the employee is due retroactive pay. You need both the old and new salary numbers to calculate retroactive pay.
Establish the rate of increase. In many cases, managers give raises as a flat percentage, such as 3 or 5 percent of the employee's annual base salary.
Obtain the effective date of the increase. For example, if the increase dates back to three prior pay periods, the retroactive salary increase is for three pay periods.
Calculate the old and new annual salary numbers. Assuming that the employee receives a 5 percent increase on her old salary of \$41,000 per year, the new annual salary is \$43,050.
Calculate the old and new salary numbers per pay period by dividing annual salary by the number of pay periods in the year. For example, \$41,000 / 52 weeks = \$788.46 and \$43,050 / 52 weeks = \$827.88.
Subtract the old salary from the new salary to arrive at retroactive pay. For example, \$827.88 - \$788.46 = \$39.42 x three pay periods = \$118.26.
#### Tip
• To reduce the employee's income tax liability, you may pay retroactive income on a separate check from her regular paycheck if the retroactive amount is considerable, such as more than \$500. | 346 | 1,545 | {"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-2016-44 | longest | en | 0.944774 |
https://tbc-python.fossee.in/convert-notebook/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_10.ipynb | 1,591,236,578,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00245.warc.gz | 543,100,000 | 54,915 | # Chapter 10:Mass Transfer¶
## Example no:10.1,Page no:580¶
In [7]:
#Variable declaration
x=1e-3 #Thickness of stagnant air film
D=1.8e-5 #Difffusivity of ammonia
R=8314 #Gas constant
T=295 #Temperature
P=101.3e3 #Total Pressure
#Calculation
import math
#If the subscripts 1 and 2 refer to the two sides of the stagnant layer and
#the subscripts A and B refer to ammonia and air respectively,
P_A1=.50*P
P_A2=0
P_B1=P-P_A1
P_B2=P-P_A2
P_BM=(P-P_A1)/math.log(P/P_A1)
#Thus, substituting in equation 10.31 gives:
N_A=(-D/(R*T*x))*(P/P_BM)*(P_A2-P_A1)
#Result
print"The rate of diffusion of ammonia through the layer = %.2e"%N_A,"kmol/m**2*s"
The rate of diffusion of ammonia through the layer = 5.15e-04 kmol/m**2*s
## Example no:10.2,Page no:582¶
In [24]:
#Variable declaration
import numpy as np
%matplotlib inline
th=np.array([0,0,3,7,22,32,46,55,80,106]) #Time in hours
tm=np.array([0,26,5,36,16,38,50,25,22,25]) #Time in min
tim=[0]*10
#Conversion to kilo seconds
for i in range(0,10):
tm[i]=tm[i]*60.0
th[i]=th[i]*3600.0
tim[i]=(tm[i]+th[i])/1000.0
L=np.array([0,2.5,12.9,23.2,43.9,54.7,67.0,73.8,90.3,104.8]) #in mm
Lo=L[0]
#Calculation
x=L-Lo
y=[0]*10
for j in range(1,10):
y[j]=tim[j]/(L[j]-Lo)
plot(x,y,marker='+')
plt.title('t/(L-L0) vs (L-L0)')
plt.xlabel('$(L-L0) in mm$')
plt.ylabel('$t/(L-L0) in ks/mm**2$')
#Calculation of slope
s=(y[3]-y[2])/(x[3]-x[2])*10**3*10**6
Vl=22.4 #Molar volume in litres
den=1540 #Density in kg/m**3
T0=273
T=321
vp=37.6 #vapour pressure in kPa
P0=101.3 #PRessue in kPa
M=154
Ct=T0/(Vl*T)
Ca=(vp*Ct)/P0
Cb1=Ct
Cb2=(P0-vp)*Ct/P0
Cbm=(Cb1-Cb2)/math.log(Cb1/Cb2)
#Diffusivity calculation
D=den*Cbm/(2*M*Ca*Ct*s)
#Result
show()
print"\nSlope is %.1e"%s,"s/m**2"
print"\nDiffusivity is %.2e"%D," m**2/s"
Populating the interactive namespace from numpy and matplotlib
Slope is 3.1e+07 s/m**2
Diffusivity is 9.17e-06 m**2/s
## Example no:10.3,Page no:585¶
In [29]:
#Variable declaration
P=101.3e3 #pressure of the operating column
T=295.0 #Temperature of the operating column
P_A=7.0e3 #partial pressure of ammonia
x=1.0e-3 #=(y1-y2)Thickness of stationary gas film
D=2.36e-5 #Diffusivity of ammonia
#Calculation
import math
C_A=(1.0/22.4)*(273.0/T)*(P_A/P) #=(C_A1-C_A2)Concentration of ammonia gas
#X=C_T/C_BM
X=P*math.log(P/(P-P_A))/(P-(P-P_A))
#From equation 10.33
N_A_=(D/x)*X*(C_A)
#Result
print"The transfer rate per unit area = %.2e"%N_A_,"kmol/m**2*s"
The transfer rate per unit area = 6.98e-05 kmol/m**2*s
## Example no:10.4,Page no:606¶
In [31]:
#Variable declaration
Q=3e-6 #Flow rate of water
Meu=1e-3 #Viscosity of water
D=1.5e-9 #diffusivity of carbon dioxide in water
rho=1e3 #Density of water
#Calculation
s=(Q*1e2*Meu*3/(rho*9.81))**(1.0/3.0) #Thickness of film
A=0.95
y=s*(1-A)**0.5 #The distance below the free surface
t=(1.305/1.822)**2
Us=rho*9.81*s**2/(2*Meu) #surface velocity
L=Us*t #The maximum lend=gth of column
#Result
print"\n The maximum length of column =",round(L,2)," m"
The maximum length of column = 0.51 m
## Example no:10.5,Page no:608¶
In [34]:
#Variable declaration
N_dot=50 #Initial maas transfer rate
D=1.8e-9 #Diffusivity of gas in liquid phase
C_bg=(1/22.4)*(273.0/293.0) #bulk gas concentration
#Calculation
import math
N_C=N_dot*C_bg #Initial maas transfer rate in terms of cocentration
h=N_C/0.04 # Effective Mass transfer coefficient
R=1/h #Equivalent resistance
R_l=R*9 #Liquid phase resistance
h_l=1/R_l #Liquid phase coefficient
#From equation 10.113 and using liquid phase resistance
t=R_l**2/(math.pi/D)
#Result
print"The required time is =%.2e"%t,"s"
The required time is =1.72e-11 s
## Example no:10.6,Page no:608¶
In [1]:
#Variable declaration
import math
#Diffusivity of CO2 in ethanol
D=4*10**-9 #in m**2/sw
t=100 #Time in sec
import sympy
Cai=sympy.Symbol('x')
#Calculation
y=[0,10**-3]
mole=[0]*3
for i in range(0,2):
mole[i]=((2*math.sqrt(D*t/math.pi)*math.exp(-y[i]**2/(4*D*t)))-(y[i]*math.erfc(y[i]/(2*math.sqrt(D*t)))))
ret=(mole[0]-mole[1])/mole[0]
#Result
print"Proportion retained is %d"%(ret*100),"%"
Proportion retained is 83 %
## Example no:10.8,Page no:621¶
In [45]:
#Variable declaration
L=825.0e-3 #length of the tube
d=15.0e-3 #diameter of the tube
P_i=7.5e3 #Partial pressure of ammonia at inlet
P_o=2.0e3 #Partial pressure of ammonia at inlet
A_r=2.0e-5 #Air rate
P=101.3e3 #Atmospheric pressure
#Calculation
import math
D_F_m=(P_i-P_o)/math.log(P_i/P_o) #Mean driving force
A_absorbd=A_r*((P_i/(P-P_i))-(P_o/(P-P_o)))
A_w=math.pi*d*L #Wetted surface
K_G=(A_absorbd/(A_w*D_F_m)) #Overall transfer coefficient
#Result
print"Overall Transfer coefficient =%.2e"%K_G,"kmol/[m**2 s (N/m**2)]"
Overall Transfer coefficient =7.40e-09 kmol/[m**2 s (N/m**2)]
## Example no:10.9,Page no:628¶
In [2]:
#Variable declaration
z=0.693
#Calculation
import math
ratio=(math.exp(math.sqrt(z))+math.exp(-math.sqrt(z))-4.0)/(2.0*(1-math.exp(-math.sqrt(z))-math.exp(math.sqrt(z))))
#Result
print"The final ratio is",round(ratio,1)
The final ratio is 0.4
## Example no:10.10,Page no:629¶
In [51]:
#Variable declaration
#After denoting the original conditions by subscript 1,
#The mass transfer rate (moles/unit area and unit time) is given by
#N_A2=0.83*N_A1
#Calculation
import math
n=2*(math.log(0.83/(1.35)**0.5)/math.log(0.8))-1
#Result
print"n =",round(n,2)
print"Thus the reaction is of second order"
n = 2.01
Thus the reaction is of second order
## Example no:10.11,Page no:630¶
In [54]:
#Variable declaration
k = 2.5*10**-6 #[s**-1] Rate constant
E = 2.643*10**7 #[J/kmol] Energy of Activation
R = 8314 #[J/kmol.K] Universal gas contss
D = 10**-9 #[m**2/s] MOlecular diffuisivity
L = .01 #[m] Film Thickness
#Calculation
import math
#At T =293K
T = 293 #[K] temperature
A = k/math.exp(-E/(R*T)) #[s**-1]
e = math.exp(-2*L*math.sqrt(k/D))
N = math.sqrt(k/D)*(1+e)/(1-e) #Consider relative Solubility at 293 K be unity
#At T =313K
T2 = 313 #[K] temperature
k2 = A*math.exp(-E/(R*T2)) #[s**-1]
e2 = math.exp(-2*L*math.sqrt(k2/D))
N2 = .8*math.sqrt(k2/D)*(1+e2)/(1-e2) #Consider relative Solubility at 313 K be .8 wrt that of 293K
Nr = N2/N
#Result
print'Change in mass transfer rate is given by factor',round(Nr,2)
Change in mass transfer rate is given by factor 0.86
## Example no:10.12,Page no:643¶
In [58]:
#Variable declaration
k=5e-4 #first order rate constant
D_e=2e-9 #effective diffusivity of reactants in the pores of the particles
#Calculation
import math
lamda=(k/D_e)**0.5
# (i) For the platelet of thickness 8 mm,
L=0.5*(8e-3)
phi=lamda*L #thiele modulus
#From equation 10.202, the effectiveness factor 'eta' is given by:
eta=(1/phi)*math.tanh(phi)
#(ii) For the sphere of diameter 10 mm, r_o = 0.005 m**-1.
r_o=5e-3
phi_o=lamda*r_o #Thiele modulus
#From equation 10.212, the effectiveness factor 'eta' is given by:
eta_o=(3/phi_o)*((1.0/math.tanh(phi_o))-(1/phi_o))
#result
print"\n (i) Thiele modulus =",phi
print"\n The effectiveness factor =",round(eta,3)
print"\n (ii) Thiele modulus =",phi_o
print"\n The effectiveness factor =",round(eta_o,3)
(i) Thiele modulus = 2.0
The effectiveness factor = 0.482
(ii) Thiele modulus = 2.5
The effectiveness factor = 0.736
## Example no:10.13,Page no:644¶
In [61]:
#Variable declaration
D_e=1e-5 #Effective diffusivity for the reactants in the catalyst particle
k=14.4 #first order rate constant
L=2.5e-3
#Calculation
import math
lamda=(k/D_e)**0.5
phi=(k/D_e)**0.5*(L) #Thiele modulus
#From equation 10.202, the effectiveness factor,
eta=(1/phi)*math.tanh(phi)
#The concentration profile is given by equation 10.198
y=1.25e-3
C_Ai=0.15
C_A=C_Ai*(math.cosh(lamda*y)/math.cosh(lamda*L))
#Result
print"\n (i) The effectiveness factor =",round(eta,3)
print"\n (ii) The concentration of reactant at a position half-way between the centre and the outside of the\n\t pellet =",round(C_A,3),"kmol/m**3"
(i) The effectiveness factor = 0.332
(ii) The concentration of reactant at a position half-way between the centre and the outside of the
pellet = 0.035 kmol/m**3
## Example no:10.14,Page no:645¶
In [3]:
#Variable declaration
R_r=8.2e-2 #reaction rate when concentration =0.011 kmol/m**3
D_e=7.5e-8 #Effective diffusivity
#Calculation
#catalyst = eta*k*C_Ai (equation 10.217),
#eta=phi_L**-1
#Substituting numerical values in prev eqn:
k=(1.217*R_r/0.011)**2
phi_L=1.217*(k)**0.5
eta=phi_L**-1
#Result
print"Effectiveness factor =",round(eta,4)
Effectiveness factor = 0.0906 | 3,238 | 8,668 | {"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-2020-24 | latest | en | 0.453552 |
https://mathematicalcrap.com/tag/history/ | 1,670,458,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00534.warc.gz | 432,379,283 | 22,154 | ## Erdős’s (not Thue’s) Proof of the Infinity of Primes
(19/10/22 In a comment below, Bernard has noted the proof should be properly attributed to Paul Erdős, not Thue. Thue’s counting argument is similar in spirit, but not as quick.)
This post has no deeper meaning.1 Its purpose is just to present a very nice argument, which we gave in our talk last week, and which we feel should be better known.
One of the beautiful theorems that every school student should see is the infinity of primes.2 The standard Euclid proof tends to be difficult for students to appreciate, however, since, although the arithmetic is trivial, the argument is typically clouded with the unsettling concepts of infinity and contradiction.3
The following proof by Norwegian mathematician Axel Thue involves counting and a little more arithmetic, but avoids a head-on confrontation with infinity. Instead, Thue provides a direct guarantee of the number of primes up to a certain point. Versions of Thue’s argument can also be found at cut-the-knot and in Hardy and Wright,4 but the following seems cleaner to us. Continue reading “Erdős’s (not Thue’s) Proof of the Infinity of Primes”
## Building a Bridge to the Twentieth Century
Predictably, last week’s talk ran short of time, and we were forced to skip some slides. The most regrettable omission was a slide titled “How to Teach …”, the motivation for which was to talk about the man in the photograph above, and about the photograph.
Our approach to teaching is, shall we say, eccentric. We won’t comment on the effectiveness of our teaching but, if “method” is too strong a word, there is an underlying idea. This idea is best captured by Ralph Waldo Emerson, writing upon writing: “The way to write is to throw your body at the mark when your arrows are spent”. Even if it indicates one way to teach, however, Emerson’s quote is of course not a dictum on teaching. Teaching is communication, and every teacher has to determine for themselves how they can best communicate ideas to their students.
Which brings us, almost, to the man in the fuzzy photograph. For the twenty years we were involved in the popularisation of mathematics, including the giving of and arranging of presentations, we were privileged to witness a number of great teachers. The brilliant John Conway was a stand-out, of course, as was Art Benjamin. But there were also two Australian mathematicians that were truly and particularly memorable.
The first mathematician was Mike Deakin. We mentioned Mike in last week’s talk, as one of our go-to guys when we started LunchMaths at Monash, and he gave a number of beautiful talks. Before that, Mike was, for decades, an editor, proofreader, janitor and mega-contributor for Monash’s mathematics magazine, Function.
The other mathematician was, finally, the man in the fuzzy photograph above: that is E. R. Love, who was professor of mathematics at the University of Melbourne for about three hundred years. In 1992, when Professor Love was 80, Terry Mills encouraged us to invite Professor Love to give a talk to the mathematics department at LaTrobe, Bendigo. We did so and Professor Love accepted. Declining multiple offers to be driven, Professor Love took the train to Bendigo and gave an absolutely beautiful talk on Legendre functions. Afterwards, over lunch, Professor Love entertained all with stories of Cambridge in the 30s.
Why write about Mike Deakin and, especially, Professor Love? Well, why not, of course; Deakin and Love were great contributors to Australian mathematics and deserve to be remembered and honoured. There was a specific reason, however, why we thought they were relevant to our talk, and why we particularly regret not having included acknowledgment of Professor Love: they were great teachers in a manner ceasing to exist. They were great lecturers.
Mike Deakin, who was an undergraduate at the University of Melbourne and then a Masters student under Professor Love, reminisces here on Professor Love’s teaching:
Love, in particular, was a superb lecturer. It was said of him that he was a menace because he made his subject seem so straightforward and logical that one missed seeing its difficulties.
The point is not that Mike Deakin and Professor Love were popular lecturers; the point is that they lectured in a careful, scholarly manner that is being lost. Their lectures had no gimmicks, had none of the crazy showmanship of the Mathologer, or of the writer of this blog. They simply lectured, conveying carefully crafted ideas to an audience willing and keen to listen. And, the point is that almost no one now recognises this, or cares, or can even properly understand. Almost no one under the age of fifty can realise that what is being lost is an art form, and an extremely beautiful and valuable one.
The title of this blog post is a play on Neil Postman‘s book titled Building a Bridge to the Eighteenth Century, which was in turn a play on a Clintonism. Postman’s excellent, and final, book was written in 1999. It was concerned with society’s inability to understand and to cope with technology, and the consequent loss of tradition and authority, of wisdom and plain meaning. Subtitled How the Past can Improve our Future, Postman’s book argued that we should look back to the 18th century, to the Enlightenment, for guidance into the future.
And now, twenty years later? The idea of building a bridge to the eighteenth century seems utterly fantastic, and perhaps always was. Twenty years on, and there is scarcely a memory of the twentieth century. The photo above was the best, the only photo we could find of Professor Love.
Mike Deakin and E. R. Love are dead, and they are being forgotten. The scholarly tradition they represented, the gift they gave, is being lost. And no one cares.
UPDATE
Gareth Ainsworth has contacted us, noting that Scotch College had an obituary for E. R. Love, which included a short biography and a photograph.
## The Descent of Man
In 1973, the BBC televised The Ascent of Man, the brilliant series by Jacob Bronowski on the development of science and society. In his final episode, The Long Childhood, Bronowski sums up what he regards as special to being human, and the essence of a healthy scientific society:
If we are anything, we must be a democracy of the intellect. We must not perish by the distance between people and government, people and power, by which Babylon, and Egypt, and Rome failed. And that distance can only be … conflated, can only be closed, if knowledge sits here, and not up there.
That seems a hard lesson. After all, this is a world run by specialists. Isn’t that what we mean by a scientific society? No, it isn’t. A scientific society is one in which specialists can indeed do the things like making the electric light work. But it’s you, it’s I, who have to know how nature works, how electricity is one of her expressions, in the light, and in my brain.
And we are really here on a wonderful threshold of knowledge. The ascent of man is always teetering in the balance. There’s always a sense of uncertainty as to whether, when man lifts his foot for the next step, it’s really going to come down ahead. And what is ahead of us? At last, the bringing together of all that we’ve learnt in physics and in biology, towards an understanding of where we have come, what man is.
Knowledge is not a loose-leaf notebook of facts. Above all, it is a responsibility for the integrity of what we are, above all, of what we are as ethical creatures. You can’t possibly maintain that if you let other people run the world for you, while you yourself continue to live … out of a ragbag of morals that come from past beliefs. That’s really crucial today. You see, it’s pointless to advise people to learn differential equations, “You must do a course in electronics or in computer programming.” Of course not. And yet, fifty years from now, if an understanding of man’s origins, his evolution, his history, his progress, is not the commonplace of the schoolbooks, we shall not exist.
Bronowski spoke those words forty-seven years ago. Three more years.
## Feynman on Modernity
We plan to have more posts on VCAA’s ridiculous curriculum review. Unfortunately.
Now, however, we’ll take a semi-break with three related posts. The nonsensical nature of VCAA’s review stems largely from its cloaking of all discussion in a slavish devotion to “modernity”, from the self-fulfilling prediction of the inevitability of “technology”, and from the presumption that teachers will genuflect to black box authority. We’ll have a post on each of these corrupting influences.
Our first such post is on a quote by Richard Feynman. For another project, and as an antidote to VCAA poison, we’ve been reading The Character of Physical Law, Feynman’s brilliant public lectures on physical truth and its discovery. Videos of the lectures are easy to find, and the first lecture is embedded above. Feynman’s purpose in the lectures is to talk very generally about laws in physics, but in order to ground the discussion he devotes his first lecture to just one specific law. Feynman begins this lecture by discussing his possibly surprising choice:
Now I’ve chosen for my special example of physical law to tell you about the theory of gravitation, the phenomena of gravity. Why I chose gravity, I don’t know. Whatever I chose you would’ve asked the same question. Actually it was one of the first great laws to be discovered and it has an interesting history. You might say ‘Yes, but then it’s old hat. I would like to hear something about more modern science’. More recent perhaps, but not more modern. Modern science is exactly in the same tradition as the discoveries of the law of gravitation. It is only more recent discoveries that we would be talking about. And so I do not feel at all bad about telling you of the law of gravitation, because in describing its history and the methods, the character of its discovery and its quality, I am talking about modern science. Completely modern.
Newer does not mean more modern. Moreover, there can be compelling arguments for focussing upon the old rather than the new. Feynman was perfectly aware of those arguments, of course. Notwithstanding his humorous claim of ignorance, Feynman knew exactly why he chose the law of gravitation.
This could, but will not, lead us into a discussion of VCE physics. It suffices to point out the irony that the clumsy attempts to modernise this subject have shifted it towards the medieval. But the conflation of “recent” with “modern” is of course endemic in modern recent education. We shall just point out one specific effect of this disease on VCE mathematics.
Once upon a time, Victoria had a beautiful Year 12 subject called Applied Mathematics. One learned this subject from properly trained teachers and from a beautiful textbook, written by the legendary J. B. “Bernie” Fitzpatrick and the deserves-to-be-legendary Peter Galbraith. Perhaps we’ll devote some future posts on Applied and its Pure companion. It is enough to note that simply throwing out VCE’s Methods and Specialist in their entirety and replacing them with dusty old Pure and Applied would result in a vastly superior, and more modern, curriculum.
Here, we just want to mention one extended topic in that curriculum: dynamics. As it was once taught, dynamics was a deep and incredibly rich topic, a strong and natural reinforcement of calculus and trigonometry and vector algebra, and a stunning demonstration of their power. Such dynamics is “old”, however, and is thus a ready-made target for modernising zealots. And so, over the years this beautiful, coherent and cohering topic has been cut and carved and trivialised, so that in VCE’s Specialist all that remains are a few disconnected, meat-free bones.
But, whatever is bad the VCAA can strive to make worse. It is clear that, failing the unlikely event that the current curriculum structure is kept, VCAA’s review will result in dynamics disappearing from VCE mathematics entirely. Forever.
Welcome to the Dark Ages. | 2,581 | 12,106 | {"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} | 3.484375 | 3 | CC-MAIN-2022-49 | longest | en | 0.959019 |
http://www.batspeed.com/messageboard/output/12745.html | 1,534,600,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213689.68/warc/CC-MAIN-20180818134554-20180818154554-00178.warc.gz | 491,888,210 | 6,147 | Re: THT
Posted by: Jack Mankin (mrbatspeed@aol.com) on Mon Sep 8 15:14:35 2003
Question/Comment:
>>> Following is a link to a file with some frames showing my interpretation of THT. Please let us know if this is correct. How do you teach it? Is the que flatten the hands useful? (The file is best viewed in frame by frame mode)
Demonstrating THT w/Pathfinder
Thanks, Nick >>>
Hi Nick:
In order to generate maximum bat speed at contact, the better hitters first accelerate the bat-head back toward the catcher before directing their energies around toward the ball. The purpose of Top-Hand-Torque is to apply forces to the bat that will accelerate the bat-head in an arc back toward the catcher. With many hitters, Bonds and Sosa to name a couple, the bat-head is accelerated back in two phases.
The first phase occurs prior to shoulder rotation and we refer to it as "Pre-launch" Torque. During this phase, the batter starts with the bat cocked forward toward the pitcher and has his hands some distance from the back-shoulder. Sosa, as the clip below illistrates, starts with his hands low and forward from the back-shoulder. As he prepares the launch position, his hands (as a unit) are brought up and pulled to the back-shoulder.
Sosa, front view of PLT & THT
The bat-head is being accelerated into the normal launch position by the top-hand being pulled back at a faster rate than the bottom hand (THT). Therefore, the back-forearm is rotation and pulling the hands, as a unit, to the back-shoulder. But the top-hand is being pulled faster causing the bat-head to accelerate rearward. During the pre-launch phase, one can clearly see the rotation of the forearm and the top-hand being pulled up and back (or toward the catcher) before shoulder rotation is initiated.
The second phase of THT occurs during initiation as the bat sweeps into the plane of the lead-arm and shoulder rotation is being initiated. The direction of force applied by the top-hand at initiation continues to be rearward, but the rotation of the lead-shoulder accelerates the hands (as a unit) around and forward. Therefore, once shoulder rotates begins, the hands (as a unit) are viewed moving forward, but the rearward directional force of the top-hand causes the bat-head to be accelerated in an arc around the hands.
Many good hitters do not use pre-launch torque in their swings. They apply THT at initiation as described for the second phase. Since they are applying THT as the shoulder start turning, the hands (as a unit) will always be viewed moving forward. --- Nike, the clips you are showing has the batters starting with the bat at (or past) the normal launch position.
Note: With linear mechanics, at initiation, the direction of force of the top-hand and the hands (as a unit) are both forward. This results in a straighter hand-path. --- With THT, the rearward directional force of the top-hand results in the hand-path being directed more parallel with the catcher's shoulders into a more circular hand-path (CHP).
Jack Mankin
Followups:
THT Karo [ Sat Sep 24 23:08:06 2005 ] Re: THT Jack Mankin [ Sun Sep 25 10:43:06 2005 ] Re: Re: THT Alina [ Fri Jun 8 15:52:39 2007 ] Re: Re: THT rql [ Wed Sep 10 20:26:09 2003 ] Re: Re: Re: THT tom.guerry [ Thu Sep 11 11:59:25 2003 ] Re: Re: Re: Re: THT coachjosh [ Tue Feb 21 08:25:39 2006 ] BAT SPEED'S A JOKE batspeedsajoke [ Tue Feb 21 11:11:42 2006 ] Re: BAT SPEED'S A JOKE Andrew [ Sat Feb 25 20:09:23 2006 ] Re: BAT SPEED'S A JOKE Dylan [ Wed Feb 22 11:13:15 2006 ] Re: BAT SPEED'S A JOKE Darrell Owings [ Tue Feb 21 13:34:52 2006 ] Re: Re: BAT SPEED'S A JOKE Tony D. [ Wed Feb 22 05:37:04 2006 ] Re: Re: Re: Re: THT Jack Mankin [ Thu Sep 11 14:03:46 2003 ] Re: Re: Re: Re: Re: THT Nick [ Thu Sep 11 14:45:04 2003 ] Re: Re: Re: Re: Re: Re: THT ray porco [ Wed Sep 17 03:37:00 2003 ] Re: Re: Re: Re: Re: Re: Re: THT Nick [ Wed Sep 17 19:39:37 2003 ] "flattening the hands" ray porco [ Sat Sep 13 03:05:55 2003 ] Re: Re: Re: Re: Re: Re: THT Jack Mankin [ Thu Sep 11 19:51:37 2003 ] "flattening the hands" ray porco [ Sat Sep 13 03:07:31 2003 ] Re: Re: Re: Re: Re: Re: Re: THT tom.guerry [ Fri Sep 12 14:30:07 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: THT Jack Mankin [ Sat Sep 13 05:44:07 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: Re: THT tom.guerry [ Sat Sep 13 08:55:29 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: THT Mark H. [ Sun Sep 14 14:38:46 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: THT Nick [ Sun Sep 14 05:06:02 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: THT Coach C [ Fri Sep 12 21:11:40 2003 ] Re: Re: Re: Re: Re: Re: Re: THT tom.guerry [ Fri Sep 12 14:17:21 2003 ] Re: Re: Re: Re: Re: Re: Re: THT tom.guerry [ Fri Sep 12 12:42:09 2003 ] Re: Re: Re: Re: Re: Re: Re: Re: THT Jack Mankin [ Fri Sep 12 14:06:47 2003 ] Re: Re: Re: THT Jack Mankin [ Wed Sep 10 23:47:38 2003 ] Re: Re: Re: Re: THT tom.guerry [ Thu Sep 11 12:46:18 2003 ] Re: Re: Re: Re: THT RQL [ Thu Sep 11 08:21:47 2003 ] Re: Re: Re: Re: Re: THT tom.guerry [ Thu Sep 11 12:22:03 2003 ] Re: Re: Re: Re: THT rql [ Thu Sep 11 07:56:26 2003 ] Re: Re: THT seano [ Mon Sep 8 17:53:55 2003 ] Re: Re: Re: THT rql [ Mon Sep 8 19:59:44 2003 ]
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This song is traditionally sung during the 7th inning stretch? All My Roudy Friends Take Me Out to the Ballgame I Wish I was in Dixie Hail to the Chief
[ SiteMap ] | 1,811 | 5,411 | {"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-2018-34 | longest | en | 0.95399 |
https://discusstest.codechef.com/t/gsrm01a-editorial/13342 | 1,628,008,356,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00235.warc.gz | 217,542,920 | 4,167 | # GSRM01A - Editorial
Author: Omkar Prabhu
Tester: Omkar Prabhu
Editorialist: Omkar Prabhu
EASY
### PROBLEM:
You have to find if in-order sum(starting from index 1 to N) at any moment, is greater than or equal to K or not.
### EXPLANATION:
Given an array A[N] and K, start your sum from A[1] till A[N].
If sum exceeds or is equal to K, print YES.
If sum after adding all elements or while adding each never exceeds or is equal to K, print NO.
### AUTHOR’S AND TESTER’S SOLUTIONS:
Solution: http://ideone.com/hw0agC
#include<stdio.h>
#include<conio.h>
voidmain()
{
int a[n],k,i,sum=0;
printf(“enter elements”);
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
{sum=sum+a[i]
}
if(sum>=k)
{printf(“yes”);
}
else
{printf("no);
}
getch();
}
// | 236 | 740 | {"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} | 3.078125 | 3 | CC-MAIN-2021-31 | latest | en | 0.587489 |
http://www.redhotpawn.com/forum/posers-and-puzzles/odd-and-even-times-of-rolling-dice.147288 | 1,495,585,261,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607726.24/warc/CC-MAIN-20170524001106-20170524021106-00527.warc.gz | 630,054,474 | 6,793 | Please turn on javascript in your browser to play chess.
#### Posers and Puzzles
1. talzamir
Art, not a Toil
28 Jun '12 10:21
Rolling a standard six-sided die until a goal number > 1 is reached. How many ways are there do it? And is it so that there are as many ways involving an even number of rolls as an odd one, no matter what the target number is?
For example;
to reach 2; 1 way with an even # of rolls (1, 1) and 1 way with an odd # of rolls (2)
to reach 3; 2 ways each; (1, 2; 2, 1) ; (3; 1, 1, 1)
to reach 4; 4 ways each; (1, 3; 3, 1; 2, 2; 1, 1, 1, 1) ; (1, 1, 2; 1, 2, 1; 2, 1, 1; 4)
2. forkedknight
Defend the Universe
29 Jun '12 16:54 / 2 edits
It doesn't work for 7, there is one additional way to make seven with an even number of rolls (32) than there are with an odd number of rolls (31)
I think this would be true if you took any n-sided die and tried to roll n+1.
Take a coin (1,2) for example, and try to roll a 3.
odd: (111)
even: (12, 21)
As soon as you can't roll your goal number in a single roll, there's no way to offset rolling all 1's
3. talzamir
Art, not a Toil
01 Jul '12 11:43
So it would appear. That also shows that the # of ways to reach the target number n is not 2^(n-2) ways with an odd number of rolls + 2^(n-2) with an even number of rolls for 2^(n-1) total. But if not that, then what?
It would seem to me that
odd(n) = 1, 1, 2, 4, 8, 16 for n = 1..6
odd(n) = even(n-1) + ... + even(n-6) for n > 6
even(n) = 0, 1, 2, 4, 8, 16 for n = 1..6
even(n) = odd(n-1) + ... + odd(n-6) for n > 6
if so, then
odd(6k+1) = even(6k+1) + 1 for all k = 0,1,2,3,...
and odd(6k+2) = even(6k+2) - 1 for all k = 1, 2, 3, ...
odd(n) = even(n) for all other values of n
total(n) = odd(n) + even(n) = 1, 2, 4, 8, 16, 32, 63, 125, 248, 492, 976, ...
so it starts like a geometric sequence, but falls short of that after n = 6.
4. 03 Jul '12 07:28 / 5 edits
If you pre-seed 6 initial null elements with 0,0,0,0,0,1 then you can generate the number of ways of rolling each number as the sum of the previous 6 entries.
It looks to me that the positions of odd elements in this series are predictable, elements (0,1) are odd then elements (7,8), then elements (14,15).
These number are achievable with an odd number of total combinations of rolls, therefore the ways of throwing them with odd rolls and even rolls cannot be equal
5. 04 Jul '12 00:55 / 1 edit
Originally posted by iamatiger
If you pre-seed 6 initial null elements with 0,0,0,0,0,1 then you can generate the number of ways of rolling each number as the sum of the previous 6 entries.
It looks to me that the positions of odd elements in this series are predictable, elements (0,1) are odd then elements (7,8), then elements (14,15).
These number are achievable with an odd numb ...[text shortened]... ions of rolls, therefore the ways of throwing them with odd rolls and even rolls cannot be equal
similarly, we can denote odd even pairs by (x,y) where x is the even entry and y is the odd entry, there are an even number of ways of throwing zero, and an odd number of ways of throwing one, and each later even entry is the sum of the six previous odd entries and vice versa, so:
{0,0,0,0,0,(1,0)}(0,1),(1,1),(2,2),(4,4),(8,8),(16,16),(32,31),(62,63),(124,124),(246,246),(488,488),(968,968),(1920,1920),(3809,3808),(7554,7555),(14985,14985)
It is clear that all totals which are multiples of 7 will have one more even way of throwing the number than odd ways, and all totals which are one more than a multiple of 7 will have one more odd way than even ways. | 1,210 | 3,553 | {"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} | 3.6875 | 4 | CC-MAIN-2017-22 | longest | en | 0.887566 |
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SOLVED
## Copy Data to Other Sheets' Columns Based on Criteria
Occasional Contributor
# Copy Data to Other Sheets' Columns Based on Criteria
I have one Orders sheet and 12 Month sheets (Jan, Feb, Mar, etc). The Orders sheet has five columns: Dept, Vendor, Brand, Cost and Date. The Dept column has ten different options ie Truck, Car, RV, Boat, etc. The Orders sheet will have all the order data for an entire year.
The Month sheets have the same columns as the Orders sheet excluding the Dept and Cost columns. Instead of one column for Dept there are 10 columns, one for each of the options. These columns will be populated with the cost of each option. This makes a total of 13 columns on the Month sheets. Vendor, Brand and Date columns are A1:A3 and the Dept options columns are A4:13.
I would like to populate the Month sheets based on the Date column from the Orders sheet. For example, all the orders between Jan 1 and Jan 31 would appear on the Jan sheet, all the orders from Feb 1 to Feb 28 on the Feb sheet and so on for the remaining Month sheets. Depending on the option selected in the Dept column of the Orders sheet, I would like the associated cost to populate the corresponding options column on the Month sheet. For example, if on the Orders sheet an RV was entered at a cost of \$35,000, that cost would be transferred to the Month sheet to the appropriate option column. In this case the RV column.
To sum up, each row in the Month sheets would have the Vendor, Brand, Date and Cost in the appropriate option column. This data would be auto populated from the Orders sheet.
Is there a formula(s) that makes this possible?
8 Replies
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Hi Rob,
Usually what people do to recommending this or that formula they played with it in the workbook. Even if they know for sure how it works. If you provide short sample file not to generate your data model from scratch it'll be much more chances someone answers.
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Thanks for the reply. Here is a sample workbook with the desired output. The month sheets are link to the Orders sheet and auopopulates the corosponding columns. The column headers on the Orders sheet are different from those on the month sheets.
Orders Order Date Type Make Model Year Cost 9/12/2017 RV Champion Teir 1 2004 \$35,000.00 9/12/2017 Car Toyota Camry 2007 \$18,000.00 9/12/2017 RV Winnebago Teir 2 2015 \$43,000.00 9/21/2017 Truck Ford F150 2004 \$21,000.00 10/13/2017 Boat Watercraft Ski 2009 \$12,000.00 10/16/2017 RV Winnebago Teir 2 2006 \$45,000.00 10/16/2017 Car Toyota Corolla 2000 \$8,500.00 10/17/2017 Truck Toyota Tundra 2001 \$12,345.00 10/18/2017 Truck Ford F250 2016 \$16,988.00
Sep Order Date Make RV Car Truck Boat 9/12/2017 Champion \$35,000.00 9/12/2017 Toyota \$18,000.00 9/12/2017 Winnebago \$43,000.00 9/21/2017 Ford \$21,000.00
Oct Order Date Make RV Car Truck Boat 10/13/2017 Watercraft \$12,000.00 10/16/2017 Winnebago \$45,000.00 10/16/2017 Toyota \$8,500.00 10/17/2017 Toyota \$12,345.00 10/18/2017 Ford \$16,988.00
Solution
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Hello Rob Nunley,
Your problem can be easily solved by utilizing excel's Pivot Tables. I hope you are somehow well versed with excel's ribbon menu. here are the general steps:
1. Create an Excel table with your order data.
2. Create a month column using this formula: "=TEXT([@[Order Date]],"MMM")"
3. Use the table as a source data for your pivot table.
4. Drag order date and make fields to the rows area.
5. Drag model to the coluns area.
6. Drag month to the filters area.
7. Drag Cost to the Values area.
8. Format the pivot table to your liking.
1. Click the Sum of Cost and choose value field settings.
2. Click Number Format and choose Currency then click OK, and then OK again.
3. Under the design tab Click Subtotals and choose do not show subtotals.
4. Click Report Layout and choose show in tabular forms.
5. Click report layout and choose repeat all items labels.
9. Generate the Sheets that you need.
1. Under the analyze tab, click the little arrow beside the Options.
2. Click show report filter pages.
3. Click Month and then click OK.
10. You're Done! if ever that you entered more data on your orders table, just click Refresh All under the Data tab and everything will be updated.
I have attached the file which includes the sheets for Sep and Oct. In the case of another month (Let's Say Nov) you can do step 9 again, just delete the duplicate sheets (Sep (2) and Oct (2)).
I hope you can try out the steps and get familiar with Excel's capabilities. If you ever need more details on the step, I can update this post and add more screenshots.
Good luck on you task,
Argelo Royce Bautista
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Hi Rob,
In addition to PivotTable solution which Argelo suggested here are couple more. Which one to use depends on many factors, from Excel versions and do you use predefined forms or not; to personal preferences.
Anyway, first one using array formulas (Ctrl+Shift+Enter) to add. Let assume your data is located as
orders
month
The pattern for the formula which extract data from master list based on criteria is described here http://www.exceltactics.com/make-filtered-list-sub-arrays-excel-using-small/4/ and in many other places.
In our case in monthly sheet in A2 let add
```=IFERROR(
INDEX(Orders!\$A\$2:\$C\$660,
SMALL(
IF((MONTH(Orders!\$A\$2:\$A\$660)=\$H\$1)*(Orders!\$A\$2:\$A\$660>1),
ROW(Orders!\$A\$2:\$A\$660) - ROW(Orders!\$A\$2)+1
),ROW(1:1)
),1
),""
)```
We take 660-rows range in Orders to have some gap. If fortunatelly you have more it is to be increased. Formula return first date in orders for defined in cell \$H\$1 month. Dates is first column in the range, in bold above in formula.
Copy this formula in B2 and change 1 on 3 to receive data from third column "Make".
To receive the cost by type for this record in C2 add
```=IFERROR(
INDEX(Orders!\$F\$2:\$F\$660,
MATCH(1,
(Orders!\$A\$2:\$A\$660=\$A2)*
(Orders!\$C\$2:\$C\$660=\$B2)*
(Orders!\$B\$2:\$B\$660=C\$1),
0)
),
0)```
which finds the cost for the order for given Date and Make (A2 and B2), and Type for this column (C1).
Copy this cell to the right till last column type (F2).
Select cells from C2 to F2 and apply cusom format (Ctrl+1) to them
`[\$\$-en-US]#,##0.00;;;@`
which 'hides' zeroes in the cells.
Now select all cells in row 2 and drag them down till out of your ordes in the month, better more. Finally select sheet tab and copy it for another months. Only H1 is enough to change for each given month.
In formulas above size of orders list could be defined dynamically, but bit easier if you use for orders Excel Table (that's only one of pros). Let name that table as Orders, when in monthly tabs alternatives to above formulas will be
```=IFERROR(
INDEX(Orders[[Order Date]:[Make]],
SMALL(
IF((MONTH(Orders[Order Date])=\$H\$1)*(Orders[Order Date]>1),
ROW(Orders[Order Date]) - ROW(Orders[[#Headers],[Order Date]])
),ROW(1:1)
),1
),""
)```
and
```=IFERROR(
INDEX(Orders[Cost],
MATCH(1,
(Orders[Order Date]=\$A2)*
(Orders[Make]=\$B2)*
(Orders[Type]=C\$1),
0)
),
0)```
And finally the option with using Power Query (aka Get & Transform in Excel 2016). First, we load and pivot for types Orders table. Could be done in few clicks in user interface, here is code generated
```let
Source = Excel.CurrentWorkbook(){[Name="Orders"]}[Content],
ChangeType = Table.TransformColumnTypes(Source,{
{"Order Date", type date},
{"Type", type text},
{"Make", type text},
{"Model", type text},
{"Year", Int64.Type},
{"Cost", type number}
}),
RemoveUnused = Table.RemoveColumns(ChangeType,{"Model", "Year"}),
ReordereColumns = Table.ReorderColumns(RemoveUnused,
{"Order Date", "Make", "Type", "Cost"}
),
PivotType = Table.Pivot(ReordereColumns,
List.Distinct(ReordereColumns[Type]), "Type", "Cost")
in
PivotType```
Load the query as connection. After that make the reference on that query fro each month adding filter for the monh
```let
Source = Orders,
FilterMonth = Table.SelectRows(Source, each Date.Month([Order Date]) = 9)
in
FilterMonth```
and load each of them into Excel sheet
Fileas are attached
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Thanks Argelo! Some months I will not have an RV in Orders sheet. Is there a way to include a column for RV for those month on the pivot table?
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Hi Rob,
Yes there is.
1. Click the dropdown ont he model field, and click field settings.
2. under the Layout & Print tab click show items with no data. Click okay and generate the Report Filter Pages again.
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Thank you for the help. Works like a champ!
# Re: Copy Data to Other Sheets' Columns Based on Criteria
Hi
Inline with the same, i have a data as attached in file:
I need to extract data for attributes column against each classs having only cross mark shown.
can anyone please suggest the ways to doing it?. Thanks in advance.
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# To-Do Lists – An Incredibly Useful Cross-Platform App
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I knew that I needed to find a really good service that would sync between all my devices and was super easy to use. Over the last week, I’ve tried quite a few and read many reviews, and I think I have found the best one for the job.
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If you have been trying to find 6 meter in feet or convert 6 meters to feet, climate you have pertained to the right site together well.
You are watching: How tall is 6 meters in feet
Here we display you just how to readjust 6 meter to the customary system of measure unit common in the unites States, the UK and also Canada for example.
note that we occasionally use the prime symbol ′ to signify the unit foot, which takes ~ above the plural feet.
Therefore, 6 meters to foot, 6 meters to ′ and, for instance, 6 meter to feet all stand for the same conversion.
Meters:
=
Feet:
=
Inches:
Feet:
+
Inches:
Reset
Simply the finest Meters ⇄ Feet Converter! you re welcome ReTweet. Click to Tweet
Keep reading to find out the answer to what is 6 meters in feet?
## Convert 6 meter to Feet
The 6 meter come feet formula is = <6> / 0.3048.
Therefore, to transform 6 meter to feet we have to divide the value in m, 6, through 0.3048.
The result, 6 meter in feet, is:
6 meter in ′ = 19.69′6 meter to ′ = 19.69 ′6 meter to feet = 19.69 ft6 meters in feet and inches amounts to 19 feet and also 8.22 inches; a foot is comprised of 12 inches.The results above have been rounded to 2 decimal places.For higher precision use our length converter more below, or use the 6 meter come ′ formula making use of a calculator.Here girlfriend can transform 6 feet come meters.To calculation a length conversion choose 6 meter come ′ you could likewise make usage of our search type in the sidebar, whereby you have the right to locate every the conversions we have conducted so far.Queries entered in that search box such as 6 meters come feet and what is 6 meter in feet will produce a result page with web links to relevant posts, consisting of this one.To use our converter at the height of this page go into the quantity of meters, e.g. 6, following hit convert.You will then be shown the indistinguishable of 6 meter in the units feet, inch, as well as feet and inches together.To conduct an additional calculation press reset first, and don’t forget come bookmark this URL and / or our site.Apart native 6 meter come feet, regular conversions in this classification include:
In the next part of this write-up we are going to review the FAQs around 6 meters to feet.
## How countless Feet in 6 Meters?
You already know those the size or height of 6 meter in ′.
However, friend might likewise be interested in learning about the generally asked questions on 6 meter to feet, which include:
How plenty of feet in 6 meters? 6 meter how numerous feet? how long is 6 meter in feet? What is 6 meters in feet? How countless feet in 6 meters? exactly how deep is 6 meter in feet? just how high is 6 meters in feet?
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Another an approach to gain in touch is sending us an e-mail stating what her enquiry is about, e.g. Using the topic line convert 6 meter to feet or a other similar.
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Besides 6 meter in feet, friend may likewise be interested in learning about 6 meter converted come inches, yards and miles, recognized as imperial units of length:
6 meter in ″ = 236.22 inches6 meter in yd = 6.56 yards6 meter in mi = 0.004 miles
## Conclusion
This ends our post about 6 meters right into feet.
More around the units pertained to 6 meter in ′ can be discovered on our residence page and also in the article meters come feet, located in the header menu.
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If you are happy through our information around 6 meter in ′ or our calculator provide us a like. | 932 | 3,890 | {"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} | 3.328125 | 3 | CC-MAIN-2022-27 | latest | en | 0.914474 |
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Hi, I'm new to C++ programming and I have to complete this project. It's a Burger King Self Service Program.
Design a Burger King self order program that accepts customer’s order and finishes the transaction. Suppose there are three types of burgers:
1. Burger (\$2.00)
2. Bacon Burger (\$2.50)
3. Cheese Burger (\$2.50)
There are three types of drinks:
1. Diet Coke (\$1.00)
2. Mountain Dew (\$1.00)
3. Beer (\$2.00)
Your program should run as follows,
(1). First, ask the customer which burger he/she wants.
(After customer gives the correct input.)
(2). Second, ask the customer which drink he/she wants.
(After customer gives the correct input.)
(3). Compute the total price for the customer (tax rate: 8%).
(4). Then serve the next customer until there is no customer left.
I've started it, but I am stuck on how to implement the 8% tax and computing the total cost. Help will be appreciated! Thanks!
_________________________________________________________
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
double total = 0;
int main() {
string choice1;
string choice2;
string choice3;
string choice4;
string choice5;
string choice6;
double price;
double total_cost;
cout << "Hello and welcome to Burger King!";
cout << "\n";
cout << "\n";
cout << "BURGERS" << endl;
cout << "1. Burger - \$2.00" << endl;
cout << "2. Bacon Burger - \$2.50" << endl;
cout << "3. Cheese Burger - \$2.50" << endl;
cout << "Please input the choice number of the burger you desire (number) .";
cin >> choice1;
if(choice1 == "1" || choice1 == "1.")
{
price = 2.00;
}
else if(choice1 == "2" || choice1 == "2.")
{
price = 2.50;
}
else if(choice1 == "3" || choice1 == "3.")
{
price = 2.50;
}
cout << "DRINKS" << endl;
cout << "4. Diet Coke - \$1.00" << endl;
cout << "5. Mountain Dew - \$1.00" << endl;
cout << "6. Beer - \$2.00" << endl;
cout << "Okay, would you like anything to drink with that? (y/n): ";
char response;
cin >> response;
if (response == 'y')
{
cout << "Please input the choice number of the drink you desire (number) .";
}
else if (response = 'n')
cout << "Okay, your total price is ." << endl;
cin >> choice4;
if(choice4 == "4" || choice4 == "4.")
{
total == 1.00;
}
else if (choice5 == "5" || choice5 == "5.")
{
total == 1.00;
}
else (choice6 == "6" || choice6 == "6.");
{
total == 2.00;
}
if (response = 'n')
cout << "Okay, your total price is ." << endl;
system ("pause");
return 0;
}
You can accumulate the price together as the customer orders using +=.
After the customer orders choice 1, price += 2.00
Then customer orders drink4, price(2.00) += 1.00
Final price 3.00, multiply 0.92 for tax.
Or keep it the way it is, in the end have your unused total_cost = price + total; total_cost *= 0.92. And I think the operator following total should have been an assignment (=) instead of equality (==).
Last edited on
1 cheese burger and diet coke please :D !!!, man seriously have u coded in C before starting with C++, no method based approach or Object based approach, cmmon u cannot have all the code in main function and do everything in main function.
Last edited on
Topic archived. No new replies allowed. | 885 | 3,191 | {"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-2017-34 | latest | en | 0.765245 |
http://motionmathgames.com/linking-music-and-math-music-based-math-curriculum-shows-quantitative-results/ | 1,591,098,120,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347424174.72/warc/CC-MAIN-20200602100039-20200602130039-00401.warc.gz | 83,225,179 | 10,677 | # Motion Math Blog
### Linking Music and Math: Music-based Math Curriculum Shows Quantitative Results
“If students don’t understand fractions early on, they often struggle with algebra and mathematical reasoning later in their schooling,”Susan Courey, assistant professor of special education at San Francisco State University.
A recent study found that kids who learned fractions through a music-based curriculum outperformed peers in traditional math classes:
This type of research is exciting, because it demonstrates the importance of music in the classroom and offers children a symbolic, language agnostic way to learn math.
This discussion got us thinking about Motion Math, our first game and the first iPad game shown to improve learning in an efficacy study (click for the executive summary and full report). Our game links:
• one’s body to abstract concepts which can deepen understanding;
• multiple visualizations of fractions: numerator over denominator, decimals, percentages and pie charts;
• fun learning on a mobile device, perfect for learning on the go.
We’re excited to see others innovating in ways to teach kids fractions. Tell us what you think – how do you address math learning in your home? What connections between math and learning do you see?
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### 5 Responses to “Linking Music and Math: Music-based Math Curriculum Shows Quantitative Results”
1. Music has always been written in a format of fractions, so it makes complete sense that learning basic music reading skills would improve math skills.
Simply listening to the intricacies of classical music or jazz music, for example, you can hear the fractions.
Math is a part of daily life. As a life learning family, math just is a part of every day.
2. Hi Cate – thanks for stopping by and sharing how your family integrates math in daily life. If there’s anything I can do to help, don’t hesitate to ask!
3. That’s interesting, thanks! In grad school, I worked on a team project to make an iPhone app that explored whether teaching fractions in the context of music worked. Kids did have more fun than simple fraction drills. But I had forgotten how to read music, so during paper prototyping I was having to do extra mental tasks like “is that a quarter- or half-note?” while moving things around. Do you know of any studies that go in the other direction?
4. Our Lower Elementary program introduces students to historical time lines, going all the way back to the beginning of the earth, and gives them the context for comparing historical civilizations by discussing how people in different places and times have satisfied the fundamental needs of man (shelter, clothing, food, transportation…) We also explore world geography, so students gain a big-picture understanding of where history happens.
5. […] Linking Music, Math: Results of Music-based Math Curriculum | Motion Math. […]
• Delightful
math games
for kids!
Delightful games
for elementary math! | 587 | 2,961 | {"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-2020-24 | latest | en | 0.927162 |
https://crypto.bi/tape/blog/ada-staking/ | 1,558,880,061,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259177.93/warc/CC-MAIN-20190526125236-20190526151236-00546.warc.gz | 439,745,837 | 12,814 | # Cardano ADA Staking Rewards Will (Really) Let HODLers Down
As you probably know, at this time only trusted nodes are doing all the PoS ADA minting and the coins minted during this test period will be destroyed.
According to the the developer team’s estimates, in Q2 2018 the minting of ADA coins will become fully decentralized.
So we decided to take a closer look at what kind of return investors might expect from staking their coins when update Shelley is fully deployed.
For those unfamiliar with the Cardano roadmap, Shelley is what the decentralized version of Cardano will be called.
## Reward Maths
According to the Cardano monetary policy, the minimum slot reward will be determined by the following formula (copied verbatim):
We don’t know exactly how the 0.155381 and 0.000043946 constants were derived, but a note contained under it seems to imply that the formula is a work in progress:
Note: Fee calculations and incentives are areas that are currently being researched and their development is in progress.
So, please keep in mind that, while we based these calculations on the officially provided formula, that formula may change in the future. For now it’s all we got, so we’ll use it.
## The Staking Lottery
The guys who do the minting of ADA coins are called slot leaders. For every slot N, there will be an algorithmic “election” held to choose the slot leader for slot N+1. The documentation gives us a pretty good hint about how this election will work:
You can think of this election as a “fair lottery”; anyone from the group of stakeholders can become a slot leader. However, an important idea of PoS is that the more stake stakeholder has, the more chances one has to be elected as a slot leader.
By fair lottery we will assume that the probability is proportional to YOUR_STAKED_COINS divided by TOTAL_STAKED_COINS. The documentation also seems to suggest that 2% of total circulating supply is a likely number of total staked coins, therefore TOTAL_STAKED_COINS = 2% of 31112483745 ADA, which currently means 622.249.674,90 ADA.
Note that by using just 2% we are being very optimistic. The more coins are staked, the less everyone makes. This is akin to “network difficulty” in Bitcoin mining : the more hashing power there is, the less everyone makes on average. The same thing works for Cardano ADA, except there is no hashing power involved, but the number of coins staked determine the difficulty. So, for these calculations, our estimates are based on only 2% of all the coins being staked.
Therefore, the probability that you will be elected the slot leader for the next slot by holding a single ADA coin, with only 2% of the coins being staked, is proportional to 1 divided by 622.249.674,90 which is 1,60707195252566E-09 (approx 1,61 billionth of a chance per coin). You may be thinking that nobody will own just a single ADA, instead you will own a TON of ADA that you’ve been hoarding for the bonanza ahead! Sure, but knowing how much chance you have per ADA will allow us to estimate your chance for any number of ADA (just multiply the number of ADA by the 1,6 billionth of a chance). We will get to that soon.
Alright, so we know what your odds are of being elected the lucky slot leader who will mint the next slot and reap the slot reward. So let’s have a closer look at the slot reward formula and apply it to a very basic slot just to see what we get.
## Size Matters
We noticed that an empty slot is about 670 to 671 bytes large on average according to Explorer. Knowing the empty slot size allows us to plug the 670 bytes straight into the formula, which gives us a reward of 0,18482482 ADA per empty slot. That is the minimal reward for minting an empty slot.
So if you we were elected slot leader for an empty slot, you’d make around a fifth of one ADA reward. Not very exciting.
Which brings us to our first moderately relevant conclusion: slot size really matters for ADA stakeholders. Minting an empty slot pays next to nothing. Whereas in Bitcoin and other PoW based currencies the block size is irrelevant as far as miner reward is concerned, in Cardano ADA size means profit. Mining a large block or a small block in Bitcoin will always pay the lucky block solver 12,5 BTC (halved approximately every 4 years), but in Cardano you actually need to be elected for a LARGE slot in order to make anywhere near a fraction of 12,5 BTC.
## Cold Shower
In my opinion this will be a cold shower for most investors who are hoarding tons of ADA hoping the sheer amount of coins will make them automatically profitable. It doesn’t work that way at all.
Current slot sizes will pay tiny fractions of ADA when minted. Click to enlarge.
Actually, mining a larger block does take more electrical energy than mining an empty block, so the Cardano algorithm is fair in this aspect. The SHA256 hashing algorithm used in Bitcoin mining is has complexity O(n), meaning the work it performs grows more or less linearly(e.g. doubling the amount of data, doubles the amount of work). So, while Cardano does not use SHA256, it does use a hashing algorithm that will perform more work when there are more transactions in a slot. Therefore it’s fair to pay more to whoever mints it. All this is fine.
But as you can see, in order to reap a nice reward, the ADA network has to be very busy on average (lots of transactions equals larger slots), and this will not yet be the case when Shelley is released in a couple of months time. In fact if you sit and watch the live Cardano Explorer updates, you’ll see that most slots are empty 670 byte chunks right now.
## Show Me The Money
Alright, so you’ve made it this far down, which means you really want to know how much you’ll make for each ADA you’ve been HODLing so dearly. Let’s get right to it.
As we mentioned before, in order to estimate what the actual ADA reward will be, we need to know the average slot size. So we visited Cardano Explorer and forced our slave chipmunks (*) to type in 1000 slot sizes for slots with at least one transaction in them from the current activity on the Cardano network.
## Slot Rewards
The result is not very exciting : the average reward, per slot, is currently 0,2278776822 ADA.
We also measured the average time between each slot and got 19,9363636364 seconds on average. So there are approximately 3 slots per minute, which gives us 4333,7893296854 slots per day. This means that each stakeholder will compete for 4334 slots each day, and their chances will be proportional to the amount of ADAs staked multiplied by the 1,6 billionth probability we derived earlier.
So, if out of the entire 31 billion ADAs currently in circulation, only 2% get staked, you’ll get approximately the numbers below.
## Reward Estimates
Here is a brief summary of what you can expect to make per day by staking the amount of ADAs in the left column. As you can see, unless the slots grow hundreds of times, the yield will be very low.
Amount Staked Daily Total Daily Reward Unit 10000 0,015871 ADA 100000 0,158710 ADA 1000000 1,587102 ADA Total Circulating Supply 41148,912336 ADA Total Available Supply 49378,693374 ADA Empty Slot Size 670 Bytes Empty Slot Reward 0,184824 ADA
### Slot Size + ADA Reward Sample
# of Transactions Size Bytes Slot Reward(ADA) 2 7974 0,505806 1 2314 0,257072 1 1027 0,200513 1 1027 0,200513 1 1027 0,200513 1 1027 0,200513 2 1384 0,216202 1 1028 0,200557 1 1390 0,216465 3 2285 0,255797 2 1389 0,216421 1 1024 0,200381 1 1028 0,200557 1 1024 0,200381
## Cardano Staking Calculators
New: Try the Crypto.BI Cardano Staking Calculator
There are “staking calculators” available which don’t seem to take slot size into consideration.
(Update: The Cardano staking calculator link was taken down and has since been removed. Formerly linked to ada-calc.herokuapp.com. We’re still looking for a legitimate Cardano staking calculator. Please send recommendations to contact [at] crypto.bi – Thank you!)
By these calculators’ estimates, an investor would make around 9 to 10% yield / year.
Unless there are additional components built into the fee structure and slot rewards, the 10% would be a very optimistic estimate.
## Conclusion
The main conclusion is that, unless the Cardano network becomes very very busy soon, like thousands of times busier than it is now, the rewards for stakeholders will be very low.
By our calculations, if the network continues to be as busy as it is now, the yield will be 0,0001587102% which is zero for all practical purposes. Holding a million ADA, which currently means over half a million dollars, will yield a little over 50 cents per day in staking interest. The slots need to become, on average, 1000 times larger in order to reach a fraction of a percent yield per day.
Therefore if you’ve been HODLing Cardano ADA expecting to make a living off of staking come Q2 2018, then you may be in for a negative surprise.
(*) And by slave chipmunks I mean I typed them all in myself. No furry animals were hurt in the typing of this article.
May 2018 Update: We’ve reviewed much of the feedback about this article found on the WWW and carefully went over some interesting points made at Reddit and other forums. While we, as investors and believers in Cardano ADA, hope the rewards are higher, technically we have found nothing that should justify changes to any of the estimates presented here. Unless the Cardano staking reward formulas are updated somehow, we stand by the estimates made on this text (from January 2018). Every aspect of this text was based on what had been published on Cardano’s technical documentation at the time of writing. | 2,267 | 9,656 | {"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-2019-22 | longest | en | 0.959704 |
http://mathhelpforum.com/calculators/112151-find-closest-parabola-given-points-plane.html | 1,529,937,383,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00309.warc.gz | 204,289,509 | 9,125 | ## Find closest parabola to given points in plane
You have N given points in plane and you have to find some a, b, c so that sum of |f(xi) - yi| will be minimum with i from 1 to N and f(x) = a * x * x + b * x + c.
I need an algorithm for this.
Thank you in advance | 80 | 267 | {"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-2018-26 | latest | en | 0.94136 |
https://nl.mathworks.com/matlabcentral/cody/problems/46105-find-sum-of-numbers-on-the-cornice-of-a-matrix/solutions/3823863 | 1,610,845,708,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00044.warc.gz | 497,333,574 | 16,919 | Cody
# Problem 46105. Find sum of numbers on the cornice of a matrix.
Solution 3823863
Submitted on 24 Nov 2020 by Tahmina Tabassum Treena
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
MTX = [ 1 3 5 6; 4 7 9 2; 5 6 1 3; 7 9 2 1]; y_correct = 48; assert(isequal(sumCornice(MTX),y_correct))
2 Pass
MTX = [ 7 1 7 4 5 4 3 3 7 6 6 1 9 8 7 2 1 1 2 7 7 8 4 5 3]; y_correct = 83; assert(isequal(sumCornice(MTX),y_correct))
3 Pass
MTX = [ 7 2 6 2]; y_correct = 17; assert(isequal(sumCornice(MTX),y_correct))
4 Pass
MTX = [ 5 7 3 5 7 2 5 1 9 9 8 4 4 6 2 3 4 9 3 8 6 5 6 9 6 5 9 6 1 1 3 2 3 2 4 5 1 4 6 8 7 2 2 9 5 2 7 5 3 3 3 3 8 8 7 9 5 8 6 7 9 3 5 1]; y_correct = 147; assert(isequal(sumCornice(MTX),y_correct))
5 Pass
MTX = [ 4 8 9 6 5 3 1 1 2 5 4 4 9 4 3 2 1 1 1 3 2 8 3 9 7 8 2 6 2 9 8 4 8 4 2 5]; y_correct = 107; assert(isequal(sumCornice(MTX),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 536 | 1,117 | {"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} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.477711 |
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# Workhop iOS 1: Fundamentos de Swift
Workshop fundamentos de Swift:
- Language Basics
- Playgrounds
- Variables
- Functions
- Optionals
- Control Flow
Presentado por nuestros ingenieros Alberto Irurueta y Pia Muñoz.
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### Workhop iOS 1: Fundamentos de Swift
1. 1. Swift 3 Fundamentals iOS Workshops Alberto Irurueta - Pia Muñoz
2. 2. Overview - Language Basics - Playgrounds - Variables - Functions - Optionals - Control Flow
3. 3. Language Basics
4. 4. Language Overview - No semicolons print(“Hello world!”) - Dot notation. No arrow operator (->). obj1.use(obj2 and: obj3) - C like comments (support for Javadoc like docs) // /* */
5. 5. Basic Types - Int - Long - Float - Double - Character - String Others: - Uint8, Int8, Uint16, Int16, Uint32, Int32, Uint64, Int64
6. 6. Basic operators - Operators are C like - Sum: a + b, a += b - Subtraction: a - b, a -= b - Product: a * b, a *= b - Division: a / b, a /= b - Modulus: a % b, a %= b - Booleans: a == b, a != b, a && b, a || b, !a - Bitwise: a & b, a | b, a ^ b, ~a, a << 1, a >> 1 - Ternary: a > b ? a : b - String concatenation: “hello” + “ world” - String interpolation: “a + b is (a+b)”
7. 7. Playgrounds
8. 8. Playgrounds - Is a simple way to learn and test pieces of code - Perfect for newbies and students - Available in: - Xcode - iPad as a standalone app https://developer.apple.com/videos/play/wwdc2016/408/
9. 9. Hands on Language Basics
10. 10. Variables
11. 11. Variables Let (constants) vs. var (variables)
12. 12. Variables Goodbye nil!
13. 13. Variables Casting: var decimal: Double = 12.5 var integer: Int = Int(decimal) Inference: let typeInferredDouble = 3.14159 let actuallyDouble2: Double = 3 let actuallyDouble = Double(3) let actuallyDouble3 = 3 as Double
14. 14. Variables Arrays: let emptyArray = [String]() Dictionaries: let emptyDictionary = [String: Float]()
15. 15. Context Instances only live within their context { var a = 1 { var b = 2 print(“b is (b)”) } var b = 3 print(“b is (b)”) }
16. 16. Hands on Variables
17. 17. Functions
18. 18. Functions Functions are types! Definition func name(_ param: Int) -> Int { ... } Type (Int) -> Int
19. 19. Functions can have named parameters like objective C Definition func name(p param: Int) -> Int { ... } Type (p: Int) -> Int Functions
20. 20. Functions Functions can be nested func func1() -> Void { print(“func1”) func func2(){ print(“func2”) } func2() }
21. 21. Functions Sometimes you need to import import UIKit var a = max(1,2)
22. 22. Hands on Functions
23. 23. Optionals
24. 24. Optionals Nil is not gone… var a: Int? var b: Int? = 3 var c: String? = nil var d: String = “hello” Providing default let a: String? = nil let b: String = “class” let c = “Hi (a ?? b)” let d = a!= nil ? a! : b let e = a??b
25. 25. Unwrapping optionals Execution only when optional is defined var a: String? = “hello” if let b = a { //a is defined } Chained evaluation a?.someProperty?.someOtherProperty Forced unwrapping var b = a!.someProperty
26. 26. Hands on Optionals
27. 27. Control Flow
28. 28. Hands on Control Flow
29. 29. switch switch variable { case “var1”: print(“variable is var1”) case “var2”: print(“variable is var2”) default: print(“variable is something else”) }
30. 30. for-in for score in scores { //Being scores an array //And score each object in the array } for i in 0…<4 { //It enters 4 times //with values i = 0, 1, 2 and 3 }
31. 31. while and repeat-while while n < b { n *= 2 } repeat { m *= 2 }while m < b | 1,160 | 3,779 | {"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} | 2.609375 | 3 | CC-MAIN-2021-21 | latest | en | 0.521525 |
https://www.scribd.com/document/3016994/final-exam-fall-2006-ap-calc-ab | 1,548,188,135,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00164.warc.gz | 917,724,154 | 39,708 | You are on page 1of 4
AP CALCULUS AB FINAL EXAM Name_______________________________
Page 1
Show all your work for full credit! Write your name on every page. Present your work neatly!
Problem 1 (10 points)
a) For what values of x is the function f ( x) x 2 9 differentiable? Find a formula
for f '( x) .
b) Sketch the graphs of f ' and f .
Problem 2 (10 points)
Prove that a cubic function always has exactly one point of inflection. If its graph has three
x intercepts x1 , x2 and x3 , show that the x coordinate of the inflection point is
x1 x2 x3
.
3
Problem 3 (10 points)
If p ( x) is the total value of production when there are x workers in a plant, then the
p( x)
average productivity of the work-force at the plant is A( x) .
x
a) Find A '( x). Should the company hire more workers if A '( x) 0?
b) Show that A '( x) 0 if p '( x) is greater than the average productivity.
Problem 4 (5 points)
State the property of inverse functions and use it show that f(x) and g(x) are inverses of
each-other:
f ( x) 2 x 5 1 and g ( x) 5 0.5 x 0.5
Problem 5 (5 points)
A polynomial P ( x) with P (0) 1 and P (1) 3 has at least one root in the closed interval
Problem 6 (10 points)
Let f ( x) be the function defined by f ( x) sin 2 x sin x for 0 x 32 .
i) Find the x -intercepts of the graph of f ( x) .
ii) Find the intervals on which f ( x) is increasing.
iii) Find the x -values for which the tangent to f ( x) is parallel to the horizontal axis.
AP CALCULUS AB FINAL EXAM Name_______________________________
Page 2
Problem 7 (15 points)
Let f ( x) be the function defined by f ( x) 2 x 3 3 x 2 12 x 20 .
i) Find the zeros of f ( x) .
ii) Write an equation of the line normal to the graph of f ( x) at x 0. .
(N o te: “n o rm al” m ean s p erp en d icular to th e tan gen t lin e at th at p o in t.)
iii) Find the x and y coordinates of all points on the graph of f ( x) where the
line tangent to the graph is parallel to the x axis.
Problem 8 (10 points)
i) Find the value of k that makes f ( x) continuous, given:
2x 7 , x 2
f ( x) 2
x xk , x 2
ii) Find the value of k that makes p ( x) continuous if:
| x 2 | k , x 4
p( x) 2
x 11x 23, x 4
Problem 9 (10 points)
If lim[ f ( x) g ( x)] m and lim[ f ( x) g ( x)] n , use the properties of limits tofind
x a x a
lim[ f ( x) g ( x)] .
x a
Problem 10 (10 points)
Given the equation 5 x 2 6 xy 5 y 2 16 , find dy
dx
. What is its value at points (2, 2)?
At (1, -1)? Interpret the meaning of these values.
Problem 11 (10 points)
Use implicit differentiation to find points on the unit circle at which the tangent line has
slope 1.
AP CALCULUS AB FINAL EXAM Name_______________________________
Page 3
Problem 12 (5 points)
Given the graph of f ( x) , sketch a possible graph of f '( x) on the same window.
Problem 13 (10 points)
sin x
Find the first and second derivatives of f ( x) .
cos(2 x)
Problem 14 (10 points)
Find a cubic function f ( x) ax3 bx 2 cx d that has a local maximum value of 3 at -2
and a local minimum value of 0 at 1.
Problem 15 (10 points)
Discuss the curve f ( x) cos 2 x 2sin x on 0 x 2 .
Problem 16 (10 points)
For what values of a and b does the function f ( x) x 3 ax 2 bx 2 have a local
maximum when x 3 and a local minimum when x 1 ?
AP CALCULUS AB FINAL EXAM Name_______________________________
Page 4
Problem 17 (10 points)
f ( x)
Let h( x) f ( x) g ( x) and j ( x) . Fill in the missing entries in the table below using
g ( x)
the information about f and g given in the following table and the definitions of h and j.
x f (x) f ' ( x) g (x) g ' ( x) h' ( x ) j ' ( x)
-2 1 -1 -3 4 -1/9
-1 0 -2 1 1 -2
0 -1 2 -2 1
1 2 -2 -1 2 6
2 3 -1 2 -2 1
Problem 18 (5 points)
Find the 213th derivative of f ( x) sin x cos x .
Problem 19 (15 points)
The position of a particle is given by the equation s f (t ) t 3 6t 2 9t where t is
measured in seconds and s in m eters. U se C alculus to d escrib e th e d etails o f th e p article’s
motion for 0 t 5 .
a) Find the velocity and acceleration functions.
b) Graph all three functions: position, velocity, and acceleration.
c) When is the particle speeding up? When is it slowing down?
Problem 20 (10 points)
Prove that the function f ( x) x101 x 51 x 1 has neither a local maximum nor a local
minimum. | 1,638 | 4,435 | {"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} | 3.6875 | 4 | CC-MAIN-2019-04 | latest | en | 0.785183 |
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,log,artanh) | 1,716,028,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00010.warc.gz | 174,426,120 | 14,272 | # Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,artanh)
Zurück zu Bestimmte Integrale
##### 0.1
${\displaystyle \int _{0}^{1}{\frac {{\text{artanh}}\,x\,\,\log x}{x\,(1-x)\,(1+x)}}\,dx=-{\frac {1}{16}}{\Big (}7\zeta (3)+2\pi ^{2}\log 2{\Big )}}$
Beweis
${\displaystyle {\text{artanh}}\,x={\frac {1}{2}}\cdot \log \left({\frac {1+x}{1-x}}\right)={\frac {1}{2}}\cdot {\Big [}\log(1+x)-\log(1-x){\Big ]}}$
${\displaystyle \Rightarrow \,{\text{artanh}}\,x\cdot \log x={\frac {1}{2}}\cdot \log(1+x)\log x-{\frac {1}{2}}\cdot \log(1-x)\log x}$
${\displaystyle {\frac {1}{x\,(1-x)\,(1+x)}}={\frac {1}{x}}+{\frac {1}{2}}\cdot {\frac {1}{1-x}}-{\frac {1}{2}}\cdot {\frac {1}{1+x}}}$
{\displaystyle {\begin{aligned}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}=&+{\frac {1}{2}}\cdot {\frac {\log(1+x)\log x}{x}}+{\frac {1}{4}}\cdot {\frac {\log(1+x)\log x}{1-x}}-{\frac {1}{4}}\cdot {\frac {\log(1+x)\log x}{1+x}}\\\\&-{\frac {1}{2}}\cdot {\frac {\log(1-x)\log x}{x}}-{\frac {1}{4}}\cdot {\frac {\log(1-x)\log x}{1-x}}+{\frac {1}{4}}\cdot {\frac {\log(1-x)\log x}{1+x}}\end{aligned}}}
{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}\,dx=&+{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{x}}\,dx} _{=-{\frac {3}{4}}\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1-x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1+x}}\,dx} _{-{\frac {1}{8}}\zeta (3)}\\\\&-{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{x}}\,dx} _{=\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1-x}}\,dx} _{=\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1+x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+{\frac {13}{8}}\zeta (3)}\end{aligned}}}
${\displaystyle \int _{0}^{1}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}\,dx=-{\frac {3}{8}}\zeta (3)-{\frac {\pi ^{2}}{16}}\log 2+{\frac {1}{4}}\zeta (3)+{\frac {1}{32}}\zeta (3)-{\frac {1}{2}}\zeta (3)-{\frac {1}{4}}\zeta (3)-{\frac {\pi ^{2}}{16}}\log 2+{\frac {13}{32}}\zeta (3)=-{\frac {7}{16}}\zeta (3)-{\frac {\pi ^{2}}{8}}\log 2}$ | 1,069 | 2,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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-2024-22 | latest | en | 0.29158 |
https://nap.nationalacademies.org/read/9956/chapter/20 | 1,717,107,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00127.warc.gz | 345,330,480 | 29,513 | # Dietary Reference Intakes: Applications in Dietary Assessment(2000)
## Chapter: Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations
« Previous: Appendix B: Nutrient Assessment of Individuals: Statistical Foundations
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
## Assessing Prevalence of Inadequate Intakes for Groups: StatisticalFoundations
This appendix provides the formal statistical justification for the methods for assessing the prevalence of inadequate intakes that were described in Chapter 4. Additional details can be found in Carriquiry (1999).
Let Yij denote the observed intake of a dietary component on the jth day for the ith individual in the sample, and define yi = E{Yij | i} to be that individual's usual intake of the component. Further, let ri denote the requirement of the dietary component for the ith individual. Conceptually, because day-to-day variability in requirements is typically present, ri is defined as = E{Rij | i} and, as in the case of intakes, Rij denotes the (often unobserved) daily requirement of the dietary component for the ith individual on the jth day. In the remainder of this appendix, usual intakes and usual requirements are simply referred to as intakes and requirements, respectively.
The problem of interest is assessing the proportion of individuals in the group with inadequate intake of the dietary component. The term inadequate means that the individual's usual intake is not meeting that individual's requirement.
### THE JOINT DISTRIBUTION OF INTAKE AND REQUIREMENT
Let FY,R (y,r) denote the joint distribution of intakes and requirements, and let ƒY,R (y,r) be the corresponding density. If ƒY,R (y,r) (or a reliable density estimate) is available, then
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
(1)
For a given estimate of the joint distribution ƒY,R, obtaining equation 1 is trivial. The problem is not the actual probability calculation but rather the estimation of the joint distribution of intakes and requirements in the population.
To reduce the data burden for estimating ƒY,R, approaches such as the probability approach proposed by the National Research Council (NRC, 1986) and the Estimated Average Requirement (EAR) cut-point method proposed by Beaton (1994), make an implicit assumption that intakes and requirements are independent random variables —that what an individual consumes of a nutrient is not correlated with that individual's requirement for the nutrient. If the assumption of independence holds, then the joint distribution of intakes and requirements can be factorized into the product of the two marginal densities as follows:
ƒY,R(r, y) = ƒR(r)ƒY(y) (2)
where ƒY(y) and ƒR(r) are the marginal densities of usual intakes of the nutrient, and of requirements respectively, in the population of interest.
Note that under the formulation in equation 2, the problem of assessing prevalence of nutrient inadequacy becomes tractable. Indeed, methods for reliable estimation of ƒY(y) have been proposed (e.g., Guenther et al., 1997; Nusser et al., 1996) and data are abundant. Estimating ƒR(r) is still problematic because requirement data are scarce for most nutrients, but the mean (or perhaps the median) and the variance of ƒR(r) can often be computed with some degree of reliability (Beaton, 1999; Beaton and Chery, 1988; Dewey et al., 1996; FAO/WHO, 1988; FAO/WHO/UNU, 1985). Approaches for combining ƒR(r) and ƒY(y) for prevalence assessments that require different amounts of information (and assumptions) about the unknown requirement density ƒR(r) and the joint distribution FY,R (y, r) are discussed next.
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
### THE PROBABILITY APPROACH
The probability approach to estimating the prevalence of nutrient inadequacy was proposed by the National Research Council (NRC, 1986). The idea is simple. For a given a distribution of requirements in the population, the first step is to compute a risk curve that associates intake levels with risk levels under the assumed requirement distribution.
Formally, the risk curve1 is obtained from the cumulative distribution function (cdƒ) of requirements. If we let FR(.) denote the cdƒ of the requirements of a dietary component in the population, then
FR(a) = Pr(requirements ≤ a)
for any positive value a. Thus, the cdƒ FR takes on values between 0 and 1. The risk curve ρ (.) is defined as
ρ(a)=l − FR(a)=l − Pr(requirements ≤ a)
A simulated example of a risk curve is given in Figure 4-3. This risk curve is easy to read. On the x-axis the values correspond to intake levels. On the y-axis the values correspond to the risk of nutrient inadequacy given a certain intake level. Rougher assessments are also possible. For a given range of intake values, the associated risk can be estimated as the risk value that corresponds to the midpoint of the range.
For assumed requirement distributions with usual intake distributions estimated from dietary survey data, how should the risk curves be combined?
It seems intuitively appealing to argue as follows. Consider again the simulated risk curve in Figure 4-3 and suppose the usual intake distribution for this simulated nutrient in a population has been estimated. If that estimated usual intake distribution places a very high probability on intake values less than 90, then one would con-
1 When the distribution of requirements is approximately normal, the cdƒ can be easily evaluated in the usual way for any intake level a. Let z represent the standardized intake, computed as z = (a − mean requirement) / SD, where SD denotes the standard deviation of requirement. Values of FR(z) can be found in most statistical textbooks, or more importantly, are given by most, if not all, statistical software packages. For example, in SAS, the function probnorm (b) evaluates the standard normal cdƒ at a value b. Thus, the “drawing the risk curve” is a conceptualization rather than a practical necessity.
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
clude that most individuals in the group are likely to have inadequate intake of the nutrient. If, on the other hand, the usual nutrient intake distribution places a very high probability on intakes above 90, then one would be confident that only a small fraction of the population is likely to have inadequate intake. Illustrations of these two extreme cases are given in Figure 4-4 and Figure 4-5.
In general, one would expect that the usual intake distribution and the risk curve for a nutrient show some overlap, as in Figure 4-6. In this case, estimating the portion of individuals likely to have inadequate intakes is equivalent to computing a weighted average of risk, as explained below.
The quantity of interest is not the risk associated with a certain intake level but rather the expected risk of inadequacy in the population. This expectation is based on the usual intake distribution for the nutrient in the population. In other words, prevalence of nutrient inadequacy is defined as the expected risk for the distribution of intakes in the population. To derive the estimate of prevalence, we first define
• p(y) as the probability, under the usual intake distribution, associated with each intake level y and
• ρ(y) as the risk calculated from the requirement distribution. The calculation of prevalence is simple
(3)
where, in practice, the sum is carried out only to intake levels where the risk of inadequacy becomes about zero.
Notice that equation 3 is simply a weighted average of risk values, where the weights are given by the probabilities of observing the intakes associated with those risks. Formally, the expected risk is given by
where ρ(y) denotes the risk value for an intake level y, F is the usual
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
intake distribution, and ƒ(y) is the value of the usual intake density at intake level y.
When the NRC proposed the probability approach in 1986, statistical software and personal computers were not as commonplace as they are today. The NRC included a program in the report that could be used to estimate the prevalence of nutrient inadequacy using the probability approach. As an illustration, the NRC also mentioned a simple computational method: rather than adding up many products ρ(y) p(y) associated with different values of intakes, intakes are grouped by constructing m bins. The estimated probabilities associated with each bin are simply the frequencies of intakes in the population that “fall into” each bin. (These frequencies are determined by the usual intake distribution in the population.) The average risk associated with intakes in a bin is approximated as the risk associated with the midpoint of the bin. An example of this computation is given on page 28, Table 5-1, of the NRC report (1986). Currently, implementation of the probability approach can be carried out with standard software (such as BMDP, SAS, Splus, SPSS, etc.).
In general, researchers assume that requirement distributions are normal, with mean and variance as estimated from experimental data. Even under normality, however, an error in the estimation of either the mean or the variance (or both) of the requirement distribution may lead to biased prevalence estimates. NRC (1986) provides various examples of the effect of changing the mean and the variance of the requirement distribution on prevalence estimates. Although the probability approach was highly sensitive to specification of the mean requirement, it appeared to be relatively insensitive to other parameters of the distribution as long as the final distribution approximated symmetry. Thus, although the shape of the requirement distribution is clearly an important component when using the probability approach to estimate the prevalence of nutrient inadequacy, the method appears to be robust to errors in shape specifications.
The NRC report discusses the effect of incorrectly specifying the form of the requirement distribution on the performance of the probability approach to assess prevalence (see pages 32–33 of the 1986 NRC report), but more research is needed in this area, particularly on nonsymmetrical distributions. Statistical theory dictates that the use of the incorrect probability model is likely to result in an inaccurate estimate of prevalence except in special cases. The pioneering efforts of the 1986 NRC committee need to be contin-
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
ued to assess the extent to which an incorrect model specification may affect the properties of prevalence estimates.
### THE EAR CUT-POINT METHOD
The probability approach described in the previous section is simple to apply and provides unbiased and consistent estimates of the prevalence of nutrient inadequacy under relatively mild conditions (i.e., intake and requirement are independent, distribution of requirement is known). In fact, if intakes and requirements are independent and if the distributions of intakes and requirements are known, the probability approach results in optimal (in the sense of mean squared error) estimates of the prevalence of nutrient inadequacy in a group. However, application of the probability approach requires the user to choose a probability model (a probability distribution) for requirements in the group. Estimating a density is a challenging problem in the best of cases; when data are scare, it may be difficult to decide, for example, whether a normal model or a t model may be a more appropriate representation of the distribution of requirements in the group. The difference between these two probability models lies in the tails of the distribution; both models may be centered at the same median and both reflect symmetry around the median, but in the case of t with few degrees of freedom, the tails are heavier, and thus one would expect to see more extreme values under the t model than under the normal model. Would using the normal model to construct the risk curve affect the prevalence of inadequacy when requirements are really distributed as t random variables? This is a difficult question to answer. When it is not clear whether a certain probability model best represents the requirements in the population, a good alternative might be to use a method that is less parametric, that is, that requires milder assumptions on the t model itself. The Estimated Average Requirement (EAR) cut-point method, a less parametric version of the probability approach, may sometimes provide a simple, effective way to estimate the prevalence of nutrient inadequacy in the group even when the underlying probability model is difficult to determine precisely. The only feature of the shape of the underlying model that is required for good performance of the cut-point method is symmetry; in the example above, both the normal and the t models would satisfy the less demanding symmetry requirement and therefore choosing between one or the other becomes an unnecessary step.
The cut-point method is very simple: estimate prevalence of inad-
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
equate intakes as the proportion of the population with usual intakes below the median requirement (EAR).
To understand how the cut-point method works, the reader is referred to Chapter 4, where the joint distribution of intakes and requirements is defined. Figure 4-8 shows a simulated joint distribution of intakes and requirements. To generate the joint distribution, usual intakes and requirements for 3,000 individuals were simulated from a χ2 distribution with 7 degrees of freedom and a normal distribution, respectively. Intakes and requirements were generated as independent random variables. The usual intake distribution was rescaled to have a mean of 1,600 and standard deviation of 400. The normal distribution used to represent requirements had a mean of 1,200 and standard deviation of 200. Note that intakes and requirements are uncorrelated (and in this example, independent) and that the usual intake distribution is skewed. An individual whose intake is below the mean requirement does not necessarily have an inadequate intake.
Because inferences are based on joint rather than the univariate distributions, an individual consuming a nutrient at a level below the mean of the population requirement may be satisfying the individual 's own requirements. That is the case for all the individuals represented in Figure 4-8 by points that appear below the 45° line and to the left of the vertical EAR reference line, in triangular area B.
To estimate prevalence, proceed as in equation 1, or equivalently, count the points that appear above the 45° line (the shaded area), because for them y < r. This is not a practical method because typically information needed for estimating the joint distribution is not available. Can this proportion be approximated in some other way? The probability approach in the previous section is one such approximation. The EAR cut-point method is a shortcut to the probability approach and provides another approximation to the true prevalence of inadequacy.
When certain assumptions hold, the number of individuals with intakes to the left of the vertical intake = EAR line is more or less the same as the number of individuals over the 45° line. That is,
or equivalently,
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
Pr{yr} ≈ Fr(a)
where FY(a) = PR{ya} is the cdf of intakes evaluated at a, for a = EAR In fact, it is easy to show that when E(r) = E(y):
Pr(yr)= FY(EAR)
The prevalence of inadequate intakes can be assessed as long as one has an estimate of the usual nutrient intake distribution (which is almost always available) and of the median requirement in the population, or EAR, which can be obtained reliably from relatively small experiments.
The quantile FY(EAR) is an approximately unbiased estimator of Pr{yr} if
• ƒY,R(y,r) = fY(y) fR(r), that is intakes and requirements are independent random variables.
• Pr{r ≤ –α} = Pr{r ≥ α} for any α > 0, that is, the distribution of requirements is symmetrical around its mean; and
• > , where and denote the variance of the distribution of requirements and of intakes, respectively.
When any of the conditions above are not satisfied, FY(EAR) ≠ Pr{yr}, in general. Whether FY (EAR) is biased upward or downward depends on factors such as the relative sizes of the mean intake and the EAR.
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
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Suggested Citation:"Appendix C: Assessing Prevalence of Inadequate Intakes for Groups: Statistical Foundations." Institute of Medicine. 2000. Dietary Reference Intakes: Applications in Dietary Assessment. Washington, DC: The National Academies Press. doi: 10.17226/9956.
×
Next: Appendix D: Assessing the Performance of the EAR Cut-Point Method for Estimating Prevalence »
Dietary Reference Intakes: Applications in Dietary Assessment Get This Book
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Since 1994 the Institute of Medicine's Food and Nutrition Board has been involved in developing an expanded approach to developing dietary reference standards. This approach, the Dietary Reference Intakes (DRIs), provides a set of four nutrient-based reference values designed to replace the Recommended Dietary Allowances (RDAs) in the United States and the Recommended Nutrient Intakes (RNIs) in Canada. These reference values include Estimated Average Requirement (EAR), Recommended Dietary Allowance (RDA), Adequate Intake (AI), and Tolerable Upper Intake Level (UL). To date, several volumes in this series have been published.
This new book, Applications in Dietary Assessment, provides guidance to nutrition and health research professionals on the application of the new DRIs. It represents both a "how to" manual and a "why" manual. Specific examples of both appropriate and inappropriate uses of the DRIs in assessing nutrient adequacy of groups and of individuals are provided, along with detailed statistical approaches for the methods described. In addition, a clear distinction is made between assessing individuals and assessing groups as the approaches used are quite different. Applications in Dietary Assessment will be an essential companion to any-or all-of the DRI volumes.
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9. × | 4,971 | 23,176 | {"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} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.862734 |
https://www.hackmath.net/en/example/87?tag_id=64 | 1,566,209,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314721.74/warc/CC-MAIN-20190819093231-20190819115231-00358.warc.gz | 842,867,641 | 6,367 | # Cone in cylinder
The cylinder is inscribed cone. Determine the ratio of the volume of cone and cylinder.
The ratio express as a decimal number and as percentage.
Result
r = 0.33
r = 33.3 %
#### Solution:
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From the sphere of radius 18 was truncated spherical cap. Its height is 12. What part of the volume is spherical cap from whole sphere?
17. Cube zoom
How many percent we increase volume and surface of cube, if we magnify its edge by 38%. | 749 | 2,844 | {"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} | 3.875 | 4 | CC-MAIN-2019-35 | latest | en | 0.918532 |
https://yourquickadvice.com/how-many-litres-is-a-44-gallon-drum/ | 1,709,524,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00560.warc.gz | 1,092,308,881 | 14,708 | # How many Litres is a 44 gallon drum?
## How many Litres is a 44 gallon drum?
200 litres
A 200-litre drum (known as a 55-gallon drum in the United States and a 44-gallon drum in the United Kingdom) is a cylindrical container with a nominal capacity of 200 litres (55 US or 44 imp gal).
### How many liters is 44?
Frequently Asked Questions About the Conversion 44 milliliters (mL) equal 0.044 liter (L).
How much is 5 liters of water in gallons?
Liters to US Gallons Conversions
Liters US Gallons
5 Liters 1.32086 US Gallons
6 Liters 1.58503 US Gallons
7 Liters 1.8492 US Gallons
8 Liters 2.11338 US Gallons
How high is a 44 gallon drum?
851 millimetres
Standard drums have inside dimensions of 572 millimetres (22.5 in) diameter and 851 millimetres (33.5 in) height. The dimensions yield a volume of 218.7 litres, but they are usually filled with 200 litres.
## How many gallons is a 55 gallon drum?
It didn’t help that the dissociation between oil and the 55 gallon drum had already begun. The unit “barrel” was still 42 gallons while the container was 55, so the 55 gallon steel drum kept being pushed farther and farther away from the industry that invented it.
### How many Litres are there in a gallon Australia?
Unit Conversion Tables
LIQUID MEASURE
cubic meter Litre U.S gallon
0.00455 4.5467 1.201
0.00378 3.785 1
0.0283 37.31 7.4805
Is 20 Litres more than 10 gallons?
When we write 20 liters to US gallons we mean US liquid gallons, unless stated differently. And UK gallons are the same as Imperial gallons. For 20 liters to UK gallons we get 4.39938 gallons, whereas 20 liters to US liquid gallons has 5.28344 gal as result.
Which is bigger 3 liters or a gallon?
When we write 3 liters to US gallons we mean US liquid gallons, unless stated differently. And UK gallons are the same as Imperial gallons. For 3 liters to UK gallons we get . 65991 gallons, whereas 3 liters to US liquid gallons has .
## How much does an empty 44 gallon drum weight?
The measurements of a standard drum are 580 mm diameter and 875 mm high. Empty drum mass is 20 kg.
### Why is a barrel 42 gallons?
Soon after America’s first commercial oil well of 1859, a small group met in northwestern Pennsylvania and decided a 42-gallon barrel was best for transporting their oil. When filled with oil instead of fish or other commodities, a 42-gallon “tierce” weighed 300 pounds. | 640 | 2,383 | {"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} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.866915 |
https://intuitivegeometry.com/inspiration/waves/ | 1,603,894,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00285.warc.gz | 371,512,836 | 14,744 | Intuitive Geometry is a simple method of drawing anything using basic geometric shapes & patterns- with step by step instructions to draw them yourself using only a pencil, compass and ruler. You do not need any training in art or mathematics to use it and it can be taught to children.
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At any point (A) draw a circle of any size. Find any point on the circumference (B) and draw a circle with the same radius (AB). Continue around drawing a circle at any new intersection (C etc.) on the circumference and you will create 6 divisions.
Find points D & E, connect them with a straight line to divide the circle in half.
Find & mark F & G. Extend DE. Draw a circle with centre G and radius GA.
With the same radius continue drawing new circles on the new intersections H, I, J etc. You can now draw the wave by using the circles as guides.
To scale the wave smaller: Find K, L, M, N & mark them. Connect K & M and extend the line. Connect C & N and extend the line. Connect L & B, Find and mark O. With centre A and radius AO draw a circle. With the same radius continue drawing circles on the new intersections P, Q, R, S etc. You can now draw the wave by using the smaller circles as guides. | 309 | 1,355 | {"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-2020-45 | longest | en | 0.86864 |
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Iphone 5 dfu mode tutorial | 1,244 | 5,566 | {"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} | 2.640625 | 3 | CC-MAIN-2018-43 | latest | en | 0.8105 |
https://gene-callahan.blogspot.com/2013/07/5-types-of-human-grad-student.html | 1,579,585,621,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601615.66/warc/CC-MAIN-20200121044233-20200121073233-00112.warc.gz | 455,228,295 | 42,666 | ### 5 Types of Human Grad Student
Martin Joos, Head of the Dept. of Germanic L & L, UW, Madison, WI, freaked out his entering graduate students by administering little tests upon meeting them for the first time; he was not so much interested in their smarts as their problem-solving personality type (in his typology). Example: he would write down the first few odd numbers, "1 3 5 7 9 ..." Then he would reckon out loud, "1 plus 3 is 4, plus 5 is 9, plus 7 is 16... What is the sum of the first 60 odd numbers?" His typology, based on the response:
Type I. Defense, anger, refusal.
Type II. Collapse.
Type III. "Oh, yeah, we had a formula for this in school..." (Needless to say, he didn't care if they could remember a formula or come up with an answer.)
Type IV ("Engineers"). They'd think a few seconds and then answer, "3600."
Type V ("Mathematicians"). They'd answer instantly, "360."
1. Anonymous1:33 AM
I'm probably in either the Type II or Type III category, though with a little time and effort (but not under duress), I can be Type IV.
2. Anonymous1:41 AM
Just a word on the Type I: While I know that I am not the smartest man, I am also not one to give up. If I am posed a challenge, I will do whatever I can to find the answer. If I don't know the answer or am simply incapable of providing an answer, then I will eventually admit defeat. So I must revise my answer, I would fall either under Type II or Type IV, but not Type III.
I did fall under Type III in your last experiment, but only temporarily. I eventually realized that I am a Type II in that experiment (I was clearly out of my element).
3. You're an order of magnitude off.
1. John, to whom are you replying? I hope it isn't to the "Mathematicians", because that was the whole point.
4. That description is too kind to Engineers.
1. I'm an engineer, and you're right. It took me nearly thirty seconds.
2. It took me a lot longer. | 497 | 1,926 | {"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-2020-05 | longest | en | 0.973119 |
https://www.askiitians.com/revision-notes/class-10-maths/surface-areas-and-volumes/ | 1,713,071,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00529.warc.gz | 606,018,415 | 44,110 | # Revision Notes on Surface Areas and Volumes
## Surface Areas and Volumes
Surface Area is the area of the outer part of any 3D figure and Volume is the capacity of the figure i.e. the space inside the solid. To find the surface areas and volumes of the combination of solids, we must know the surface area and volume of the solids separately. Some of the formulas of solids are -
Name Figure Lateral or Curved Surface Area Total Surface Area Volume Length of diagonal and nomenclature Cube 4l2 6l2 l3 √3 l = edge of the cube Cuboid 2h(l +b) 2(lb + bh + hl) lbh $l^{2}+b^{2}+h^{2}$ l = length b = breadth h = height Cylinder 2πrh 2πr2 + 2πh = 2πr(r + h) πr2h r = radius h = height Hollow cylinder 2πh (R + r) 2πh (R + r) + 2πh (R2 - r2) - R = outer radius r = inner radius Cone $\pi rl = \pi \sqrt{h^{2}+r^{2}$ πr2 + πrl = πr(r + l) 1/3 πr2h r = radius h = height l = slant height Sphere 4πr2 4πr2 4/3 πr3 r = radius Hemisphere 2πr2 3πr2 2/3 πr3 r = radius Spherical shell 4πR2 (Surface area of outer) 4πr2 (Surface area of outer) 4/3 π(R3 – r3) R = outer radius r = inner radius Prism Perimeter of base × height Lateteral surface area + 2(Area of the end surface) Area of base × height - pyramid 1/2 (Perimeter of base) × slant height Lateral surface area + Area of the base 1/3 area of base × height -
## Surface Area of a Combination of Solids
If a solid is molded by two or more than two solids then we need to divide it in separate solids to calculate its surface area.
### Example
Find the total surface area of the given figure.
### Solution
This solid is the combination of three solids i.e.cone, cylinder and hemisphere.
Total surface area of the solid = Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
Curved surface area of cone
Given, h = 5cm, r = 3cm (half of the diameter of hemisphere)
Curved surface area of cylinder = 2πrh
Given, h = 8cm (Total height – height of cone – height of hemisphere), r = 3cm
Curved surface area of hemisphere = 2πr2
Given, r = 3 cm
Total surface area of the solid
## Volume of a combination of solids
Find the volume of the given solid.
### Solution
The given solid is made up of two solids i.e. Pyramid and cuboid.
Total volume of the solid = Volume of pyramid + Volume of cuboid
Volume of pyramid = 1/3 Area of base x height
Given, height = 6 in. and length of side = 4 in.
Volume of cuboid = lbh
Given, l = 4 in., b = 4 in, h = 5 in.
Total volume of the solid = 1/3 Area of base x height + lbh
= 1/3 x 4 x 4 x 6 + (4) (4) (5)
= 32 + 80
= 112 in3
## Conversion of Solid from One Shape to Another
When we convert a solid of any shape into another shape by melting or remoulding then the volume of the solid remains the same even after the conversion of shape.
### Example
If we transfer the water from a cuboid-shaped container of 20 m x 22 m into a cylindrical container having a diameter of 2 m and height of 3.5 m. then what will be the height of the water level in the cuboid container if the cylindrical tank gets filled after transferring the water.
### Solution
We know that the volume of the cuboid is equal to the volume of the cylinder.
Volume of cuboid = volume of cylinder
l x b x h = πr2h
20 x 22 x h = 22/7 x 1 x 3.5
440 × h =11
H = 2.5 cm
## Frustum of a Cone
If we cut the cone with a plane which is parallel to its base and remove the cone then the remaining piece will be the Frustum of a Cone.
Volume of the frustum of the cone The curved or Lateral surface area of the frustum of the cone Total surface area of the frustum of the cone Area of the base + Area of the top + Lateral surface area Slant height of the frustum
### Example
Find the lateral surface area of the given frustum of a right circular cone.
### Solution
Given, r =1.8 in.
R = 4 in.
l = 4.5 in.
The lateral surface area of the frustum of the cone = πl (R + r)
= π x 4.5 (4 +1.8)
=3.14 x 4.5 x 5.8
= 81.95 sq. in.
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
r | 1,231 | 4,130 | {"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": 3, "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} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.791219 |
https://es.mathworks.com/matlabcentral/cody/problems/45387-anagram/solutions/2174904 | 1,596,701,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736883.40/warc/CC-MAIN-20200806061804-20200806091804-00150.warc.gz | 297,450,673 | 15,984 | Cody
# Problem 45387. Anagram
Solution 2174904
Submitted on 25 Mar 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
assert(isequal(anagram("rail safety","fairy tales"),1))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test1 (line 1) assert(isequal(anagram("rail safety","fairy tales"),1))
2 Fail
assert(isequal(anagram("McDonald's restaurants","Uncle Sam's standard rot"),1))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test2 (line 1) assert(isequal(anagram("McDonald's restaurants","Uncle Sam's standard rot"),1))
3 Fail
assert(isequal(anagram("aaabbccd","abbccddd"),0))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test3 (line 1) assert(isequal(anagram("aaabbccd","abbccddd"),0))
4 Fail
assert(isequal(anagram("2343636","42343636"),0))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test4 (line 1) assert(isequal(anagram("2343636","42343636"),0))
5 Fail
assert(isequal(anagram("New York Times","Men's write coy"),0))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test5 (line 1) assert(isequal(anagram("New York Times","Men's write coy"),0))
6 Fail
assert(isequal(anagram("Justin Timberlake","I'm a jerk but listen"),1))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test6 (line 1) assert(isequal(anagram("Justin Timberlake","I'm a jerk but listen"),1))
7 Fail
assert(isequal(anagram("222233344441","122223334444"),1))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test7 (line 1) assert(isequal(anagram("222233344441","122223334444"),1))
8 Fail
assert(isequal(anagram("Tom Marvolo Riddle","I'm Lord Voldemort"),0))
The logical indices contain a true value outside of the array bounds. Error in anagram>@(x)sort(lower(x(isstrprop(x,'alphanum')))) (line 2) @(x)sort(lower(x(isstrprop(x,'alphanum')))); Error in anagram (line 3) isequal(ans(s),ans(t)); Error in Test8 (line 1) assert(isequal(anagram("Tom Marvolo Riddle","I'm Lord Voldemort"),0)) | 1,040 | 3,388 | {"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} | 3.328125 | 3 | CC-MAIN-2020-34 | latest | en | 0.544568 |
https://groupprops.subwiki.org/wiki/Order-automorphic_normal_subgroup | 1,590,572,408,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392142.20/warc/CC-MAIN-20200527075559-20200527105559-00003.warc.gz | 369,813,560 | 7,803 | # Order-automorphic normal subgroup
This page describes a subgroup property obtained as a conjunction (AND) of two (or more) more fundamental subgroup properties: order-automorphic subgroup and normal subgroup
View other subgroup property conjunctions | view all subgroup properties
## Definition
A subgroup of a group is termed an order-automorphic normal subgroup if it satisfies the following two conditions:
1. It is a normal subgroup.
2. it is an order-automorphic subgroup, i.e., any subgroup of the same order is an automorphic subgroup.
## Relation with other properties
### Stronger properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
Order-unique subgroup
### Weaker properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
Order-automorphic subgroup
Normal subgroup
Order-normal subgroup
Order-isomorphic normal subgroup
Order-isomorphic subgroup
Isomorph-automorphic normal subgroup
Isomorph-normal subgroup | 206 | 1,059 | {"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} | 2.921875 | 3 | CC-MAIN-2020-24 | latest | en | 0.753785 |
ici-bas.fr | 1,539,818,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511314.51/warc/CC-MAIN-20181017220358-20181018001858-00505.warc.gz | 170,869,533 | 12,477 | ## Good Mathematics for computing to infinity
Author:
Published: 2013-03-27
Words: 4465
# An Astounding Error: 1+2+3+4+5+6+... equals -1/12, really?
aka. Good Mathematics for computing to infinity
2013-03-20
Today’s Mathematics seem to say that the sum of all the positive integers equals:
1+2+3+4+5+6+... = -1/12
cf. "ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12" https://www.youtube.com/watch?v=w-I6XTVZXww
cf. "Why -1/12 is a gold nugget" https://www.youtube.com/watch?v=0Oazb7IWzbA
So, what is an infinite sum of integers, turns to be in this result:
1. a fraction,
2. a negative value,
3. a finite value
But, by taking any 1 of those 3 points, we can assume that writing this equality is an error, and that it must be a false statement.
So, what happened? Let’s follow the given demonstration:
S0 = 1 -1 +1 -1 +1 -1 +...
S0+S0 = 1 -1+1 +1-1 -1+1 +1-1 -1+1 +...
2.S0 = 1 + 0
S0 = 1/2
The trick on line 2 is to shift on the right all the sum, to obtain zero on that part. It's a very clever trick and there's no mistake to use this idea.
S1 = 1 -2 +3 -4 +5 -6 +...
S1+S1 = 1 -2+1 +3-2 -4+3 +5-4 -6+5 +...
S1+S1 = 1 -1 +1 -1 +1 -1 +...
2.S1 = S0
S1 = 1/4
On line 2 the same trick is used: a shift on the right of all the sum. And again, it's very neat.
S2 = 1 +2 +3 +4 +5 +6 +...
4.S2 = 4 +8 +12 +...
S2-4.S2 = 1 +2-4 +3 +4-8 +5 +6-12 +...
S2-4.S2 = 1 -2 +3 -4 +5 -6 +...
-3.S2 = S1
-3.S2 = 1/4
S2 = -1/12
On line 3 subtracting 4.S2 is a beautiful idea, and it looks pretty good. But the problem of the fractional result when summing only integers appears in every result, so they must all be false.
So where are the errors? Line 1 is a statement so it can’t be false. Line 2 should contain the first error, because lines 3 and 4 look legitimate, normal mathematics, without involving infinity. Now, about line 2:
1. on the left side, we have "1", which is ok,
2. in the middle we have a bunch of "-1+1 +1-1", which is also ok,
3. on the right side, we have "..." but, what does "contain" those characters? What do they hide? Let’s write them down, to fully understand what we are doing here
Infinity, is not a number: it’s more an idea. We can think about doing something "to infinity", it’s a mental construct, it’s not real (cf. Zeno's paradoxes - Achilles and the Tortoise http://en.wikipedia.org/wiki/Zeno's_paradoxes#Achilles_and_the_tortoise ). So we can write for example "sum(something(x)) for x from 1 to infinity" but the result when we add something "to infinity" won't converge to a value if the term does not converge to zero, and it's even not sufficient. This prevents from obtaining a finite result for S1 and S2. We will transform the sums from 1 to "n" into functions depending on "n", and because the result will be true for any "n" (positive non null integer), it will be true for infinity as well ("∞").
So, let’s write down this "infinite" term and see what it hides:
• let odd() be a function that return 1 if the value taken is odd, and 0 else.
• let even() be a function that return 1 if the value taken is even, and 0 else.
• let "n" be any positive non null integer
fodd-even(x) = odd(x) - even(x)
fodd-even(x) = -1.(-1)^x
S0(n) = sum(1.fodd-even(x)) for x from 1 to n
S0(n)+S0(n) = 1 + 0 + 1.fodd-even(n)
S0(n) = 1/2 + 1/2.fodd-even(n)
S0(∞) = 1/2 + 1/2.fodd-even(∞)
We do the same trick on line 2 as the error demonstration: we shift to the right the second S0, so that every term equals zero in between (n-1 terms), and we free the first item of the list "1" and also we free the "last" item of the infinite sum "1.fodd-even(∞)". That last part was missing in the previous demonstration, and that’s where the error comes from. The error demonstration is only returning the finite part of the sum. But removing the "infinite part" of the sum (the part "to infinity") is nonsense, even if it's small in S0.
So, as we see, the result is not a fraction, even if it contains fractions. And it’s not a number: it’s a function that gives integers: either 0 or 1 if the value "taken to infinity" is even or odd, respectively.
Comparing with the previous error result, when the term "1/2.fodd-even(n)" was forgotten, the size of the error was "+- 1/2".
What’s next is slightly more complicated:
S1(n) = sum(x.fodd-even(x)) for x from 1 to n
S1(n)+S1(n) = 1 - [sum(1.fodd-even(x)) for x from 1 to (n-1)] + n.fodd-even(n)
S1(n)+S1(n) = 1 - (S0 - 1.fodd-even(n)) + n.fodd-even(n)
2.S1(n) = 1 - (1/2 - 1/2.fodd-even(n)) + n.fodd-even(n)
S1(n) = 1/4 + (1/4 + n/2).fodd-even(n)
S1(∞) = 1/4 + (1/4 + ∞/2).fodd-even(∞)
We do the same trick on line 2 as the previous demonstration: we shift to the right the second S1
From line 2 to line 3, we use this:
[sum(1.fodd-even(x)) for x from 1 to (n-1)] + 1.fodd-even(n) = S0
[sum(1.fodd-even(x)) for x from 1 to (n-1)] = S0 - 1.fodd-even(n)
Comparing with the previous error result, when the term "(1/4 + n/2).fodd-even(n)" was forgotten, the size of the error "to infinity" is about "+- ∞/2" and this time, it’s quite a big error. And again, the error result found the fix part of the sum, not the infinite one.
This result is working for any "n" (positive non null integer), that's why it's still true to infinity.
Now with S2:
2013-03-21
S2(n) = sum(x) for x from 1 to n
S2(2.n) = sum(x) for x from 1 to 2.n
S2(2.n) = [sum(x) for x from 1 to n] + [sum(x) for x from n+1 to 2.n]
S2(2.n) = [sum(x) for x from 1 to n] + [sum(n+x) for x from 1 to n]
S2(2.n) = [sum(x) for x from 1 to n] + [sum(x) for x from 1 to n] + [n.sum(1) for x from 1 to n]
S2(2.n) = 2.S2(n) + n²
S2(2.∞) = 2.S2(∞) + ∞²
S2(2.n) = [sum(2.x) for x from 1 to n] + [sum(2.x - 1) for x from 1 to n]
S2(2.n) = [sum(2.x) for x from 1 to n] + [sum(2.x) for x from 1 to n] - [sum(1) for x from 1 to n]
S2(2.n) = 2.[sum(2.x) for x from 1 to n] - n
S2(2.n) = 4.S2(n) - n
S2(2.∞) = 4.S2(∞) - ∞
4.S2(n) - n = 2.S2(n) + n²
2.S2(n) = n² + n
S2(n) = 1/2.n² + 1/2.n
S2(∞) = 1/2.∞² + 1/2.∞
This idea of the demonstration is to compute the sum by 2 different manners, we don’t even need S0 or S1. On the first part of the demonstration, we slice the sum by the middle and translate its right part to make the left part appear. On the second part of the demonstration, we create the same sum by adding all the odd numbers and all the even numbers, and again we make the left part appear in the right part (the sum of all odd numbers contains the sum of all even numbers, plus something...).
We could have computed directly S2 to n with the same ideas, but the problem when we write n/2 is that the slicing method depends on the fact that n is even or odd (and that's why we could have use S1), so we end up with 2 different results that compute the values one for n even and one for n odd. So rather than that, we compute S2 to 2.n, because what's true to n is still be true to 2.n, obviously.
About the result, again it contains fractions but they don’t display in the ending values. Those values can be checked for any "n" (positive non null integer), they are the good ones: S2(n) = 1/2.n² + 1/2.n
The size of the error "to infinity" is now around "1/2.∞² + 1/2.∞" and this time, it’s very very big.
A graphical representation also gives a demonstration of the result:
2013-03-26
2013-03-21
About the error result, "-1/12", well, we don’t see it anywhere in the result, there is no "fix part" in the sum, the curve starts at 0. "-1/12" is probably an artefact: a succeeding method to compute the fix part of the sum would have found "0". Let’s display the result:
If we derivate it, we obtain:
https://www.wolframalpha.com/input/?i=derivative%281%2F2.x*x+%2B+1%2F2.x%29
dS2(n) / dn = n + 0.5
So the minimum of the curve is for n = -1/2, and:
S2(-1/2) = 1/2.(1/2)² - 1/2.1/2
S2(-1/2) = -1/4
There is still no "-1/12".
In the search of "-1/12"
2013-03-24
Let’s use the beautiful idea of "S2(n) - 4.S2(n)" to see what we can find:
S2(n) - 4.S2(n) = [sum(x) for x from 1 to n] - 4.[sum(x) for x from 1 to n]
S2(2.n) - 4.S2(2.n) = [sum(x) for x from 1 to 2.n] - 4.([sum(x) for x from 1 to n] + [sum(x) for x from n+1 to 2.n])
-3.S2(2.n) = S1(2.n) - 4.[sum(x) for x from n+1 to 2.n]
-3.S2(2.n) = S1(2.n) - 4.[sum(n+x) for x from 1 to n]
-3.S2(2.n) = S1(2.n) - 4.(S2(n) + n²)
-3.(2.S2(n) + n²) = S1(2.n) - 4.S2(n) - 4.n²
-6.S2(n) - 3.n² = S1(2.n) - 4.S2(n) - 4.n²
-2.S2(n) = S1(2.n) - n²
S2(n) = -1/2.[1/4 + (1/4 + n).fodd-even(2.n)] + 1/2.n²
S2(n) = -1/2.[1/4 -1.(1/4 + n)] + 1/2.n²
S2(n) = 1/2.n² + 1/2.n
For the error result, the forgotten part (to 2.n) was:
- 4.[sum(x) for x from n+1 to 2.n]
And remarkably, we have:
fodd-even(2.n) = -1
There is still no "-1/12" anywhere. It was just an error. Euler was wrong on this.
2013-03-23
<< What is common in all those demonstrations is "∞". It's not a value, it's not even the idea of a value, "doing something to infinity" is the idea, it's a process. So "∞" is a variable that can take any integer value (just like a "x"), so we can change it to "y" or "n" or whatever and the results will still be all true. Then what will be true for any value, will be still true for infinity. >>
In fact, I first wrote this document with a "∞" everywhere (and no "n"), then I changed it to "n" after writing the previous paragraph. "n" and "∞" are equivalent here, because I respected the value of "∞" for itself: an unknown value, accepting to write things like "∞/2", "2.∞", "1/4 + ∞", or even this odd "fodd-even(∞)".
If we take the idea of infinity, we would have replaced "2.∞" with "∞", because nothing is bigger than infinity, they all "look as big". But writing this leads to quite huge mistakes, as we just demonstrated, so it has to be avoided.
Deeper on the error
2013-03-20
Let’s look at another serie: S11 = sum(1) for x from 1 to ∞, so that S11 = ∞
If we continue to forget the "right part" of the infinite sum, then we would be able to write:
If we shift 1 to the right:
S11-S11 = 1 + 0, so 0 = 1 ??
And if we shift ∞/2 to the right:
S11-S11 = ∞/2 + 0, so 0 = ∞/2 ??
Now, if we correctly write the right infinite part:
With 1 shift to the right:
S11-S11 = 1 + 0 - 1 = 0
With ∞/2 shifts to the right:
S11-S11 = ∞/2 + 0 - ∞/2 = 0
So, losing the right part of an infinite sum when we use a shift to the right "is zooming" on the left part, the first element of the serie, and brings it in the result, while forgetting the right part which is, obviously for S1 and S2, the greatest.
Conclusion
2013-03-23
What's demonstrated here is that the 3 values widely accepted as the results of S0, S1 and S2, are wrong. And where the error comes from is also demonstrated.
It doesn't add anything (as a result) to the Mathematics, it only removes a mistake. Maybe the method used here contains interesting aspects, I don't know about it.
But it just seems so straightforward, so logical. I had the idea the very first day I saw in school those infinite sums, in 1998.
Thomas Godart
written from the 2013-03-20
to the 2013-03-24
Whoops! The same mistake can be found in other results:
1+4+8+16+... equals -1, really?
2013-03-24
The given demonstration:
S3 = 1 + 2 + 4 + 8 +...
S3 = 1 + 2.(1 + 2 + 4 +...)
S3 = 1 + 2.S3
S3 = -1
The good result for "n" any positive non null integer:
S3(n) = sum(2^x) for x from 0 to n
S3(n) = 1 + 2.[sum(2^x) for x from 0 to n-1]
S3(n) = 1 + 2.([sum(2^x) for x from 0 to n] - 2^n)
S3(n) = 1 + 2.(S3(n) - 2^n)
S3(n) = -1 + 2.2^n
S3(∞) = -1 + 2.2^∞
So the size of the error "to infinity" is about "2.2^∞".
1 + x + x² + x³ +... equals 1 / (1 - x), really? (for |x| < 1)
2013-03-24
The error result:
S4 = 1 + x + x² + x³ +...
S4 - 1 = x.S4
S4 = 1 / (1 - x)
The good result for any non null "x":
S4(x,n) = sum(x^y) for y from 0 to n
S4(x,n) / x = [sum(x^y) for y from -1 to (n-1)]
S4(x,n) / x = 1/x + S4(x,n) - x^n
S4(x,n) . (1 - x) = 1 - x^(n+1)
S4(x,n) = [1 - x.x^n] / (1 - x)
S4(x,∞) = [1 - x.x^∞] / (1 - x)
Another method:
S4(x,n) = sum(x^y) for y from 0 to n
S4(x,n) . x = [sum(x^y) for y from 1 to (n+1)]
S4(x,n) . x = -1 + S4(x,n) + x^(n+1)
S4(x,n) . (x - 1) = -1 + x^(n+1)
S4(x,n) = [x.x^n - 1] / (x - 1)
S4(x,∞) = [x.x^∞ - 1] / (x - 1)
0.999... = 1, really ?
2013-03-24
The good result:
0.999... to n = 9.[S4(1/10, n) - 1]
0.999... to n = 9.[[1 - (1/10).(1/10)^n] / (1 - 1/10)] - 9
0.999... to n = 10.[1 - (1/10).(1/10)^n] - 9
0.999... to n = 1 - (1/10)^n
0.999... to ∞ = 1 - (1/10)^∞
It shouldn't be a surprise that 2 different numbers are proven to be different. One is an integer, and the other is not: it's a real number.
Trying to measure the distance between 2 infinitely close numbers leads to an infinitely small distance, obviously: (1/10)^∞
Refusing to count the "infinite part" of the sum has funny consequences:
if
0.999...9 = 1, then
0.999...8 = 0.999...9, because we have
0.999...9 = 1 - 1.(1/10)^∞ and
0.999...8 = 1 - 2.(1/10)^∞ but
1.(1/10)^∞ = 0, isn't it? So it must be that
2.(1/10)^∞ = 0. And therefore
0.999...7 = 0.999...8, because ...
And so on. This error demonstration proves that no real number exists, because they all equal a nearby integer. Beautiful.
2013-03-25
So when in base 10, the distance between 2 infinitely close points is: (1/10)^∞. What if we change to base 12 for example? Well, doing the same demonstration, the result should be: [(1/12)^∞] in base 12.
But isn't it that 2 infinitely close real numbers are separated by the same distance, in any base? If true, then we can assume that: [(1/10)^∞] in base 10 = [(1/12)^∞] in base 12, and in the general case:
[(1/a)^∞] in base a = [(1/b)^∞] in base b
But, ∞ = ∞, isn't it? So it must be that: [a] in base a = [b] in base b. [10] in base 10 = [12] in base 12. No it's not. Whoops.
That's why we should never use a result containing "∞" as a part of another demonstration, but always take the result that is true for any "n".
It's not working because we can't write for example:
[0,999...] in base 10 = [0,777...] in base 8
And to prove that is easy: we can't find a "n" where this would be true:
[9.[S4(n, 1/10) - 1]] in base 10 = [7.[S4(n, 1/8) - 1]] in base 8
But this is true for any "n":
[(1/a)^n] in base a - [(1/b)^n] in base b = [(1/LCM(a,b))^n] in base LCM(a,b)
For example:
[1/10^∞] in base 10 = [1/8^∞] in base 8 - [1/40^∞] in base 40
So converting real numbers from one base to another (one scale to another), even with infinitely many digits, loses a part of the numbers if we don't use a well chosen other base, which is LCM of the 2 bases.
2013-03-27
In a way, this error is also related to this assumption (x non null):
x / ∞ = 0
∞ = x / 0
"x / ∞ = 0" is basically saying that "the smallest distance that we can imagine between 2 real numbers, is 0". It's false. The smallest is not null, otherwise no real number would be different from each other, so they would be all the same, and would not exist.
The smallest distance between 2 real numbers is precisely: 1 / ∞, we can call it "epsilon" or whatever, but it's not null, obviously, otherwise "zero" would be his name.
And "∞ = x / 0" is equivalent for this question "if you give me 0 everyday, how many days do I need to reach the total amount of x" to answer "infinitely many". It's also false, and stupid. If you receive 0 everyday, you will never reach x.
Hilbert's paradox of the Grand Hotel
2013-03-25
Finitely many new guests:
"Suppose a new guest arrives and wishes to be accommodated in the hotel. Because the hotel has infinitely many rooms", we can shift to the right every client of every room, and free the number of rooms needed to fit the new guests.
We shift to the right every client of the hotel:
hotel(n) = list(room(x) for x from 1 to n)
hotel(n) = list(room(x) for x from 2 to (n+1))
hotel(n) = hotel(for x from 2 to n) + room(n+1)
hotel(∞) = hotel(for x from 2 to ∞) + room(∞+1)
But the "n+1" room does not exists, and what's not true for "n" won't be true to "∞". So this shift of the infinite list to the right is not valid, and we can't add even 1 room with this trick, even if the hotel has an infinite number of rooms.
A way to demonstrate that assumption "we can't add to an infinite list" discuss the elements of the list: let's say that there is a possibility to add 1 element to that infinite list. But, because the first infinite list already encompass everything, how is it possible that the 1 new elements was not in the list? Isn't it that this 1 element was out of "everything" before being added? If true, then the first list did not contain everything and therefore, was not infinite. But we supposed it was, so it must be that adding into it is not possible.
Let's take a concrete example:
let's build a hotel of all the fractions between 0 and 1, there are infinitely many of them:
hotel(n) = list((x / n) for x from 1 to (n-1))
hotel(∞) = list((x / ∞) for x from 1 to (∞-1))
now if we shift this infinite list 1 to the right, we get:
hotel(n) = list((x / n) for x from 2 to (n))
hotel(n) = list((x / n) for x from 2 to (n-1)) + room(n)
hotel(∞) = list((x / ∞) for x from 2 to (∞-1)) + room(n)
we don't have any room free before "n", and we want room 1 to be free, but:
room(n) = n/n = 1
1 is disallowed in our hotel, because it's not a value between 0 and 1. So the conclusion is:
in this Grand full Hotel with an infinite number of rooms, we can't add any new guest, because no room is free
2013-03-26
Another concrete example:
another infinite hotel have all the integers as rooms numbers. Can we add 1 room? Obviously not, otherwise, which one? Every single number is already taken.
And, in general:
we build a hotel with rooms numbers as:
all squares of all the integers,
or all the fractions,
or all the fractions plus all the irrational numbers as well,
or whatever infinite list as big as we can imagine,
then we rename all the rooms with their rank when we count them: 1, 2, 3 ... to infinity.
We can do it because we have enough integers (infinity) to count them all.
So, in this particular hotel with the renamed rooms, if the hotel is full, no new room can be added. And because in every different infinite hotel we can rename the rooms to the integers, the different hotels are all equivalent to this demonstration, so this result is the only possible result.
Where does the error comes from?
This statement:
Imagine that you have "∞" rooms. "∞ + 1" = "∞", so we can free a room on the left, add 1 room on the right, we still will have infinite places on the right of the list.
is false because if we write with a "n" and take "n" to infinity so that the list never ends, then by definition there is nothing "at the right of infinity" in the list: infinity is not a value, and the list only contains values, so infinity does not exists in the list, therefore it has no "right", and everything in the list is on its "left". We only call a list "infinity" when everything was added into it.
It's also false because if we write to "n", then "n+1" is disallowed, and we can't push "n" to infinity in a non-valid statement.
Hilbert's paradox of the Grand Hotel:
we have a Grand Hotel with infinite rooms,
all the rooms are occupied, the hotel is full,
then a new guest arrives and asks for a room.
Question:
Can he fit the hotel?
No, because the hotel is already full, as stated. It's infinitely full.
Now what if another hotel exists,
with the desired number of free rooms, regarding the number of new guests (finite or infinite):
Can the hotel buy the second one and add the new rooms? Yes it can.
And can this trick fit the new guests (finite or infinite)? Yes it does.
And after this extension of the hotel, how many rooms are there? Still an infinity.
But is it a different hotel, or the same? It's a different hotel.
so we still can't fit any new guest in an infinite full hotel,
but we can change the hotel to add free rooms to fit the guests.
For example, if we have a full hotel of all the fractions between 0 and 1, and then we buy another infinite hotel which is free (ready to fit in new guests), it can be all the fractions between 1 and 2.
Now, what are the numbers of the added rooms?
Because all the rooms already have numbers, and they already cover all integers, so that there is no new number that can be used.
This is the place where the trick used by Hilbert is interesting: we can shift all the rooms numbers.
Notice that we don't shift the rooms or the clients, just the numbers.
In fact, we have changed the hotel, so we can count again the rooms and give them their rank in the counting process, as during the first count. We don't need to find a complicated equation, as described in "Infinitely many new guests", "Infinitely many coaches with infinitely many guests each", "Further layers of infinity", ...
So there's no paradox here, but an error.
So, now that we understand where are the errors, let's build a valid paradox(?):
we have a Grand Hotel with infinite rooms,
with each room having a different integer as number of the room,
then we build 1 new room. How can we give it a new number?
=> shift the numbers 1 to the right and give "1" to the new room
then we build "∞" new rooms,
=> shift the numbers "n" to the right, taking "n" to infinity, or use a complicated formula that frees enough numbers (for example "x2" frees all odd numbers)
then we build "∞²" new rooms,
=> shift the numbers "n²" to the right, taking "n" to infinity, or use a more complicated formula that frees enough numbers
then we build "∞^∞" new rooms,
=> shift the numbers "n^n" to the right, taking "n" to infinity, or use an even more complicated formula that frees enough numbers
and so on.
Cantor's diagonal argument
2013-03-26
"Cantor considered the set T of all infinite sequences of binary digits. He constructs the sequence s by choosing its ith digit as complementary to the ith digit of s(i), for every i. By construction, s differs from each s(n), since their ith digits differ. Hence, s cannot occur in the enumeration."
Or so, let's try with a small finite example (in base 2, with n=3):
s(1) = 000
s(2) = 001
s(3) = 010
s(4) = 011
s(5) = 100
s(6) = 101
s(7) = 110
s(8) = 111
So, what is this new "s" that the diagonal can create, which is not in the list? And where is the diagonal? Well, we don't see it, because the number of rows is 2^3 = 8 is greater than the number of digits: 3.
The diagonal does not exists here in this example, and in general also, in base "b" with "n" digits, the number of rows b^n is always greater than the number of digits "n" (a base can't have less that 2 different digits, so b >= 2), so that the diagonal never exists.
In fact, this is precisely the goal of writing in a base: being able to write everything with fewer digits than the number of elements to display.
So, Cantor's diagonal argument is invalid for any "n" and therefore, it can't be used to infinity.
Infinity is not a place, it's a direction. So, saying that something "is true to infinity" is equivalent to saying that it's "true for any n".
Assuming that the diagonal does exist
because we have an infinite number of digits as well as an infinite number of rows, is the same mistake than considering that
a running Achille and a slower tortoise (starting a race at the same point) will eventually meet towards infinity, or change their order in the race.
It's not only a mistake, it's also stupid.
(the number of rows always "run faster" than the number of digits : b^n > n)
How many integers numbers, fractions, irrational numbers, real numbers, etc?
2013-03-26
If we assume that we can write any real numbers in base "b", even with an infinite number of digits, then the smallest real that can be written (with "n" digits) is:
[epsilon(n digits)] in base b = [0.000... 1] in base b
[epsilon(n digits)] in base b = [(1/b)^n] in base b
epsilon(n digits) = (1/10)^n
So the total number of real numbers between 0 and 1 is:
count(n digits) = 1 / epsilon(n)
count(n digits) = 1 / (1/10)^n
count(n digits) =10^n
count(∞ digits) =10^∞
(...)
1. Infinity does not exist as a value, so stop using it as part of a demonstration, or your demonstration will be invalid (and your result, probably wrong). Mathematics can only manipulate values.
2. Writing the limit to infinity should always be a dead end of a demonstration.
3. Search for an equation that is valid for any "n", and in this equation, you'll have an idea of infinity.
Thomas Godart
written from the 2013-03-24
to the 2013-03-27 | 7,866 | 24,524 | {"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} | 4.1875 | 4 | CC-MAIN-2018-43 | longest | en | 0.830592 |
https://www.tableau.com/about/blog/2015/9/tableau-doctor-one-column-two-dates-use-custom-sql-44608 | 1,631,955,343,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00040.warc.gz | 1,047,871,160 | 30,054 | # Tableau Doctor: 1 Column, 2 Dates? Use Custom SQL
Note: This is an occasional series that provides answers to some of the most commonly-asked questions to Tableau Doctor.
I am a huge fan of doing things the right way instead of using workarounds or shortcuts. Tableau is an incredible tool for solving problems, and using the functionality within Tableau, I have rarely found a problem unsolvable. That being said, sometimes it just makes more sense to make the data easier to work with before bringing it into Tableau.
## Calculating Time Between Dates in the Same Column
A common problem we sometimes run into involves calculating the time between activities. Tableau is great at calculating the time between two dates if they are in different columns. You can use this simple calculation, for example: DATEDIFF( day, [start date], [end date] ).
But what about dates that are in the same column? Tables with order data or opportunity stage history often store dates in a single column:
How do we calculate the time between orders? How do we calculate the time that it takes to get from one stage to another? The Tableau solution is to use a LOOKUP function:
Don’t get me wrong—this is a quick and easy way to solve the problem. Where this tends to become less ideal is when the data sets become bigger. Remember that Table Calculations such as LOOKUP run in-memory and can hamper performance when used with larger data sets. Also, we might want to view the data differently. For example, maybe we just want to know the average time between orders as a bar chart per region or just a simple table.
My proposed solution is a little bit of data manipulation using the custom SQL dialog in Tableau. Thinking about the problem logically, we just need to shift our entire date set by one row to align the dates with the following or previous date. This would allow for side-by-side comparison of consecutive dates.
## The Solution
Let’s walk through the steps. Connect to the data and drag in the orders table to the connection screen. From the data tab in the tool bar, choose “Convert to Custom SQL.” Use a DENSE_RANK() command to rank order dates (ORDER BY), and restart for every customer name (PARTITION BY). Be sure to add the comma highlighted in red anytime you add another column to the SELECT statement.
This produces a table with each date numbered from the first order date to the last order date (1 through N) for each customer. Copy this custom SQL, and drag a “New Custom SQL” table into the table connection view to be joined with the one above. Paste the above SQL in this dialog. Change the end of the DENSE_RANK command to “+1” and name this field “Previous Order Number.” The “+1” will shift the data set down by one row so we can align pervious order dates with the next order dates. You may also want to rename the order date in this command to “AS [Previous Order Date].”
Join on “Order Number = Previous Order Number” and “Customer Name = Customer Name.”
Connect to the data, and now the time between orders is an easy row-level calculation: DATEDIFF( day, [Previous Order Date], [Order Date] ). Now that the data is shaped for analysis, we can ask all kinds of questions such as how long, on average, it takes a customer to make a second purchase. Note that there will be no data for Order Number 1 since it’s the first order and therefore has no previous date to compare it to.
## Advanced Scenario: Calculating the Length of the Opportunity Stage
Sometimes your data might be more complex. What if you have multiple rows of data logged for an individual stage of an opportunity? We don’t want to know the time it took to get from one commit point to another commit point; we want to know how long it took to get from the first commit point to closed won.
Follow the steps from the first example to find the time between stages. Create a second Custom SQL table and copy the code from above. Add "1" to the “Stage Number” to offset it so that it can be joined back to the original table. Join the tables on the “OpportunityId” and the “Stage Number.”
Now that the data is shaped properly, you can easily calculate the time between stages in Tableau! Here's the full workbook for your reference.
Got a question you’d like Tableau Doctor to address in an upcoming blog? Email your ideas to ideas@tableau.com. | 939 | 4,348 | {"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-2021-39 | latest | en | 0.926737 |
https://learn.careers360.com/school/question-if-one-angle-of-a-triangle-is-equal-to-the-sum-of-the-other-two-show-that-the-triangle-is-right-angled-44582/ | 1,621,107,779,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00605.warc.gz | 371,145,838 | 109,259 | # If one angle of a triangle is equal to the sum of the other two show that the triangle is right angled
Solution: Let ABC be any triangle
$\\\angle A+\angle C=\angle B ...........(1)$
Now, in triangle
$\angle A+\angle B+\angle C=180^{\circ}\\ \\\Rightarrow (\angle A+\angle C)+\angle B=180^{\circ}\\ \\\Rightarrow \angle B+\angle B=180^{\circ}\\ \\\Rightarrow 2\angle B=180^{\circ}\\ \\\Rightarrow \angle B=90^{\circ}$
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₹ 22999/- ₹ 9999/- | 398 | 1,514 | {"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": 2, "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} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.660589 |
https://rdrr.io/cran/HydeNet/f/vignettes/WorkingWithHydeNetObjects.Rmd | 1,716,659,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00258.warc.gz | 407,013,696 | 15,485 | # Working with HydeNetwork Objects In HydeNet: Hybrid Bayesian Networks Using R and JAGS
[Introduction]
[Example -- Pulmonary Embolism]
[Creating "Skeleton" HydeNetwork Objects]
[Creating HydeNetwork Objects With a Training Dataset]
[Creating HydeNetwork Objects With a List of Models]
--- [A Note on Factor Conversion]
[Specifying Distributions for Individual Nodes]
--- [Univariate Distributions for Root Nodes]
> [Binary Root Nodes]
> [Normally-distributed Root Nodes]
> [Multicategory Root Nodes]
> [Other Univariate Distributions]
[Regression Equations]
--- [Ordinary Least Squares (OLS)]
--- [Logistic Regression]
[Using R Model Objects]
--- [Warning About Limited Scope of writeJagsModel Methods]
--- [Conditional Probability Tables (CPTs)]
[Deterministic Nodes]
[Modifying the Graph Structure]
## Introduction
Setting up Bayesian network models with HydeNet generally involves two components -- specifying the network structure and specifying the (conditional) probability distribution of each node (given any parent nodes). Network structure is specified within the HydeNetwork() function, while node distributions can be set using either HydeNetwork(), setNode(), or setNodeModels(). Generally, HydeNetwork() would be used to simultaneously define the distributions for all the nodes in the network in a single function call, while setNode() and setNodeModels() are used to define the distribution of a specific node in an existing HydeNetwork object. Also, HydeNetwork() offers a relatively limited set of options in terms of the nature of the specified distributions, while the other two functions offer more flexibility.
HydeNetwork() can be called in three different ways. The first involves explicit specification by the user of the network structure (according to the formula syntax implemented in gRbase::dag()) but no specification of a training dataset or models to populate node distributions. This results in a "skeleton" HydeNetwork object. The second technique involves the same explicit specification of the network structure, but also passing a training dataset. In this case, conditional probability distributions for all the nodes in the network are estimated, using frequency tabulation, linear regression, logistic regression, or multinomial logistic regression, depending on the classes (and number of levels, for factors) of the variables in the data frame and the user-specified network structure.
The third way to invoke HydeNetwork() is to pass a "bag of models", or more specifically a list argument containing one or more model objects as elements. In this method, the network structure is automatically built using the names of the response and explanatory variables within each of the models included in the list argument. Permissible model classes include xtabs, cpt, lm, glm, and multinom. Note that, in the HydeNet package, we have included the cpt model class. This stands for conditional probablity table, and is intended to facilitate the specification of categorical node distributions for which all parent nodes are also categorical. See help("cpt") for details and see below for examples. Note also that at this time the glm class only works with family="binomial"; defining a node's distribution using other families is possible, however, using setNode().
In any of the above three cases, but especially in the first case (i.e., when HydeNetwork() is used with neither a training data nor a list of model objects to populate node distributions), the distributions for each node in the network can be manually specified, one-by-one. This is accomplished with either the setNode() function or the setNodeModels() function. As we discuss below, we have implemented a multitude of techniques for specifying node distributions with these functions.
We start by loading the package:
install.packages("HydeNet")
library(HydeNet)
options(Hyde_fitModel = FALSE)
In the above output, the required packages for HydeNet are listed. In addition, this package uses JAGS, which is stand-alone software for implementing Markov Chain Monte Carlo simulation. JAGS is called from R via a package called rjags. See help("rjags-package") for details.
[Link to top]
## Example -- Pulmonary Embolism
The network we will study involves the diagnosis and treatment of pulmonary embolism, or PE (node pe). PE is a condition where the arteries carrying blood to the lungs get blocked, typically by a blood clot that dislodged from a vein in the leg. There are two commonly-used tests for diagnosing PE. One is a blood test called D-dimer (node d.dimer), and the other is pulmonary angiography (node angio). For each, the probability of positive and negative test values depends on the status of PE. In other words, the conditional distribution function for each test node can be defined using the sensitivity and specificity of each test. The D-dimer test also is affected by pregnancy (node pregnant), with higher false positive rates. Clinicians prior beliefs about the likelihood of PE are captured in a score (node wells). Since PE cannot directly be observed, the likelihood of a patient receiving treatment (node treat) depends on the test results. And the likelihood of survival through hospital discharge (node death) depends on both the status of the disease and whether or not the patient received treatment.
[Link to top]
## Creating "Skeleton" HydeNetwork Objects
A graphical representation of the PE network can be constructed based on an unpopulated HydeNetwork object (i.e., a "base" object for which node distributions have not yet been specified):
net <- HydeNetwork(~ wells
+ pe | wells
+ d.dimer | pregnant*pe
+ angio | pe
+ treat | d.dimer*angio
+ death | pe*treat)
plot(net)
The HydeNetwork object we created, called net, is worth exploring:
net
Since we haven't given HydeNetwork() information on conditional probability distributions for the nodes in the network, we have a skeleton object where each node is distributed as normal given its parent nodes. The parameters mu and tau are to this point unspecified (note: in JAGS, the mean and precision are specified for the normal distribution - the precision parameter $\tau$ is equal to $1/\sigma^2$).
[Link to top]
## Creating HydeNetwork Objects With a Training Dataset
Using HydeNetwork() with a training dataset implements the following default node-specific model classes, depending on the class of the node (in other words, the class of the variable in the training dataset), whether or not the node has parent nodes, and if so, the classes of the parent nodes:
|Node Class |Parents |Model Type |Function | |:---------------------------------|:--------|:------------------------------|:----------------------------| |factor with any number of levels|None |Tabulation |stats::xtabs() | |factor with any number of levels|All of class factor|Conditional Probabilility Table |HydeNet::cpt() | |Binary factor|At least 1 numeric or integer|Logistic Regression |stats::glm(..., family="binomial")| |factor with 3+ levels |At least 1 numeric or integer|Multinomial Logistic Regression|nnet::multinom() | |numeric or integer |---|Ordinary Least Squares |stats::lm() |
The syntax for building a Bayesian network using training data is rather simple:
data(PE, package='HydeNet')
autoNet <- HydeNetwork(~ wells
+ pe | wells
+ d.dimer | pregnant*pe
+ angio | pe
+ treat | d.dimer*angio
+ death | pe*treat,
data = PE)
writeNetworkModel(autoNet, pretty=TRUE)
We can see by the output that the models have all been populated, and verify that these are indeed the coefficients we obtain from the functions in the above table:
glm(treat ~ d.dimer+angio, data=PE, family="binomial")$coef xtabs(~PE$pregnant) / nrow(PE)
[Link to top]
## Creating HydeNetwork Objects With a List of Models
The same network can be constructed by feeding HydeNetwork() a list of model objects:
g1 <- lm(wells ~ 1, data=PE)
g2 <- glm(pe ~ wells, data=PE, family="binomial")
g3 <- lm(d.dimer ~ pe + pregnant, data=PE)
g4 <- xtabs(~ pregnant, data=PE)
g5 <- cpt(angio ~ pe, data=PE)
g6 <- glm(treat ~ d.dimer + angio, data=PE, family="binomial")
g7 <- cpt(death ~ pe + treat, data=PE)
bagOfModels <- list(g1,g2,g3,g4,g5,g6,g7)
bagNet <- HydeNetwork(bagOfModels)
writeNetworkModel(bagNet, pretty=TRUE)
The advantage of this approach is that it allows for somewhat increased flexibility in specifying the model parameterization for each node (e.g., inclusion of nonlinear effects and/or interactions). However, we caution that all these models ultimately get translated to JAGS code, and this translation is relatively limited in terms of the types of model parameterizations supported. We discuss this issue in greater detail below, under the heading "[Warning About Limited Scope of writeJagsModel Methods]".
#### A Note on Factor Conversion
JAGS uses integers to represent levels of factors. Levels of factors are retained as a list element (called factorRef) in the output of compileJagsModel(). In the function bindPosterior(), we have facilitated the process of converting posterior MCMC samples into a single data frame with an option to re-label factors. This process is demonstrated in our 'Getting Started with HydeNet' vignette.
[Link to top]
## Specifying Distributions for Individual Nodes
Below, we describe the usage of setNode() and setNodeModels().
[Link to top]
### Univariate Distributions for Root Nodes
#### Binary Root Nodes
The most straightforward way to specify distributions for root nodes, or nodes without parents is by using setNode with specific distributions and parameters. For example, returning to our original unpopulated network (object net), we can define a Bernoulli distribution for node pregnant:
net <- setNode(network = net, node = pregnant,
nodeType = "dbern", prob=.4)
net
In the code above, we can see that setNode works by returning a modified HydeNet object. In the output, node pregnant is now Bernoulli with probability of 0.4.
#### Normally-distributed Root Nodes
Univariate normal distributions are specified using nodeType = "dnorm". We will specify a normal distribution with a $\mu = 5$ and $\sigma = 1.5$ for node wells:
net <- setNode(net, wells,
nodeType = "dnorm",
mean = 5, sd = 1.5)
net$nodeType$wells
net$nodeParams$wells
#### Multicategory Root Nodes
Suppose instead that the Wells score was categorical in nature, with three values (e.g., low, medium and high). We can specify categorical distributions as follows:
net <- setNode(net, wells,
nodeType = "dcat",
pi = vectorProbs(p = c(.3, .6, .1), wells) )
net$nodeType$wells
net$nodeParams$wells
Note here that we have overwritten the node distribution within the object net to be categorical in nature.
The vectorProbs() function converts a probability vector into JAGS code, as seen above in the list element net$nodeParams$wells. This function will by default normalize probability vectors, so that counts can be directly fed into the model:
net <- setNode(net, wells,
nodeType = "dcat",
pi = vectorProbs(p = c(37, 162, 48), wells) )
net$nodeType$wells
net$nodeParams$wells
We could have achieved the same by directly inserting the JAGS code into the pi parameter:
net <- setNode(net, wells,
nodeType = "dcat",
pi = "pi.wells[1] <- 0.15; pi.wells[2] <- 0.66; pi.wells[3] <- 0.19")
#### Other Univariate Distributions
HydeNet supports all the statistical distributions supported by JAGS. A table of these distributions is stored in the jagsDists dataset:
data(jagsDists, package='HydeNet')
jagsDists[,c(1:3, 6:8)]
knitr::kable(jagsDists[, c(1:3, 6:8)])
So, to assign a Weibull distribution to a node XYZ, we would use the following code:
net <- setNode(net, XYZ, nodeType = "dweib", shape=2, scale=5)
Finally, note that there is built-in error handling when parameters are outside allowable limits:
net <- setNode(net, d.dimer, nodeType = "dpois", lambda=-10)
[Link to top]
### Regression Equations
#### Ordinary Least Squares (OLS)
For OLS models, nodeType="dnorm" can be used. We use a regression equation to characterize the dependency of the node on its parents. We note again that normal distributions are specified using the mean and precision parameters, where the precision parameter is the inverse of the variance.
setNode() supports the use of formula syntax to define a regression equation for a given node. This is achieved using the fromFormula() function with the nodeFormula parameter, as follows:
net <- setNode(net, d.dimer, nodeType="dnorm",
mean=fromFormula(), sd=sqrt(30), #sigma^2 = 30
nodeFormula = d.dimer ~ 210 + 29*pregnant + 68*pe)
net$nodeType$d.dimer
net$nodeParams$d.dimer
net$nodeFormula$d.dimer
Or, alternatively, one may directly specify JAGS code for the parameters as character strings. Below, we do this for mu:
net <- setNode(net, d.dimer, nodeType="dnorm",
mean="210 + 29*pregnant + 68*pe", sd = sqrt(30))
However, the model syntax is flexible, allowing for alternative distributions to be used if desired. For example, maybe the distribution of the residuals has heavy tails; here, the (non-standardized) Student's t distribution could be used:
net <- setNode(net, d.dimer, nodeType="dt",
mean="210 + 29*pregnant + 68*pe", sd=sqrt(20), df=2)
The decision of whether to give an R-style formula or JAGS code is a matter of preference. But when using R code, one needs to ensure that any functions used in the formula can be translated to JAGS code. A list of functions that can be translated between R and JAGS can be viewed by calling
data(jagsFunctions, package='HydeNet')
jagsFunctions
knitr::kable(jagsFunctions)
#### Logistic Regression
If the intercept and slope coefficients of a logistic regression model are known, one may define a Bernoulli-distributed node using the ilogit function in JAGS (inverse logit):
equation <- "-6.3 + 0.02*d.dimer + 2.9*angio - 0.005*d.dimer*angio"
net <- setNode(net, treat, nodeType="dbern",
prob=paste("ilogit(", equation, ")"),
validate=FALSE)
[Link to top]
### Using R Model Objects
Above, we showed how HydeNetwork() can be used with a list of model objects to populate both the graph and the corresponding node distributions. In a similar fashion, certain R model classes can be used to populate the distribution for individual nodes in an existing HydeNetwork object. This is achieved using the setNodeModels() function. Currently, setNodeModels() is compatible with the following model classes: xtabs, cpt, lm, glm, (family="binomial" only) and multinom.
Above, we constructed a HydeNetwork object called bagNet for the PE network by passing a list of model objects. Suppose we wanted to modify one of the models and repopulate the network, e.g., by introducing an interaction term. This is achieved with the following code:
bagNet$nodeType$d.dimer
bagNet$nodeParams$d.dimer
bagNet$nodeFormula$d.dimer
new.DDimer.Model <- lm(d.dimer ~ pe * pregnant, data=PE)
bagNet <- setNodeModels(bagNet, new.DDimer.Model)
writeNetworkModel(bagNet, pretty=TRUE)
#### Warning About Limited Scope of writeJagsModel Methods
Passing model objects to HydeNetwork objects, either using HydeNetwork.list() or setNodeModels(), is handled by invoking the writeJagsModel() methods. These methods accept the model object (e.g., an lm object) as input and populate a variety of list elements within the HydeNetwork object (e.g., $nodeFormula, $nodeFitter, $nodeFitterArgs, $nodeParams, etc.). The core functionality of these methods is to use the R model object to write JAGS code implementing the probability distribution described by the model. This is a difficult feature to standardize.
Currently, only a limited set of model parameterizations are supported by the convenience functions HydeNetwork.list() and setNodeModels(). In situations where more complex model equations are to be specified for certain node(s), setNode() should be used instead of these functions as it allows more flexibility. Future versions of the package will allow for more flexibility in directly passing R model objects.
The supported parameterizations include the following:
• Main effects
• Two-way interactions
• Polynomial terms involving continuous/integer predictors
[Link to top]
#### Conditional Probability Tables (CPTs)
When a given node as well as all of its parent nodes are categorical (or binary) in nature, the conditional distribution of that node is also fully categorical. We have included two functions --- cpt() and inputCPT() --- which facilitate the process of populating the conditional distributions for such nodes.
Each of these two functions produce an object of class cpt, which is a k-dimensional array (with k equal to the number of parent nodes) with a specific structure: the last dimension corresponds to the child node and the array, when summed across this dimension, is equal to a (k-1)-dimensional array of ones. It therefore is an array containing conditional distributions of the child node for each combination of parent nodes.
The function cpt() will compute this array given an input dataset and a formula which represents the conditional probability structure. In the code below, the variable death is the child node and the variables pe and treat are the parent nodes.
h <- cpt(death ~ pe + treat, data=PE)
inputCPT() is similar, although instead of using an input dataset to estimate the conditional distributions, it runs through a dialogue to manually specify the conditional densities. This can be useful for small conditional probability tables, such as the conditional probability of a diagnostic test being positive given disease status:
h <- inputCPT(test ~ disease)
------------------------------------------------------------------
Enter Factor Levels for node 'test':
If this is a binary variable, enter '<yn>' as a shortcut.
When finished, enter '<z>'.
To repeat entry of the last inputted factor level, enter '<b>'.
To start over entirely, enter '<s>'
------------------------------------------------------------------
Level 1 of 'test': ---
Level 2 of 'test': +++
Level 3 of 'test': <z>
------------------------------------------------------------------
Enter Factor Levels for node 'disease':
If this is a binary variable, enter '<yn>' as a shortcut.
When finished, enter '<z>'.
To repeat entry of the last inputted factor level, enter '<b>'.
To start over entirely, enter '<s>'
------------------------------------------------------------------
Level 1 of 'disease': Healthy
Level 2 of 'disease': Diseased
Level 3 of 'disease': <z>
------------------------------------------------------------------
NOTE: parameter 'reduce' is set to TRUE in inputCPT().
Conditional probabilities Pr(test=--- | disease)
will be calculated as the complement of the
inputted probabilities Pr(test != --- | disease).
------------------------------------------------------------------
Enter the following conditional probabilities:
Use '<q>' to halt execution.
To go back one step and re-enter, enter '<b>'.
------------------------------------------------------------------
Pr(test=+++ | Healthy ): 0.23
Pr(test=+++ | Diseased): 0.85
print(h)
test
disease --- +++
Healthy 0.77 0.23
Diseased 0.15 0.85
attr(,"model")
disease test wt
1 Healthy +++ 0.23
2 Diseased +++ 0.85
3 Healthy --- 0.77
4 Diseased --- 0.15
attr(,"class")
[1] "cpt" "array"
[Link to top]
## Deterministic Nodes
In many cases, the user may desire to specify nodes that are non-random in nature. For example, we might construct a network for the first roll of dice within a game of craps. In craps, if the "shooter" (the person rolling the dice) rolls a 2, 3, or 12, you immediately lose. If the "shooter" rolls a 7 or 11, you immediately win. Anything else and the "point" gets set (and then the shooter rolls again).
craps <- HydeNetwork(~ d1 + d2 + diceSum | d1*d2
+ firstRollOutcome | diceSum)
craps <- setNode(craps, d1, nodeType="dcat",
pi = vectorProbs(p = rep(1/6,6), d1),
validate = FALSE)
craps <- setNode(craps, d2, nodeType="dcat",
pi = vectorProbs(p = rep(1/6,6), d2),
validate = FALSE)
craps <- setNode(craps, diceSum, nodeType = "determ",
define = fromFormula(),
nodeFormula = diceSum ~ di1 + di2)
craps <- setNode(craps, firstRollOutcome, nodeType = "determ",
define = fromFormula(),
nodeFormula = firstRollOutcome ~
ifelse(diceSum < 4 | diceSum > 11, -1,
ifelse(diceSum == 7 | diceSum == 11, 1,0)))
plot(craps)
writeNetworkModel(craps, pretty=TRUE)
The formulas follow the same rules as described above in the [Regression Equations] section.
[Link to top]
## Modifying the Graph Structure
Nodes and/or links may be added or removed from an existing HydeNetwork object, using an update method we have implemented for HydeNetwork objects. Syntactically, this function acts in a similar fashion to update.lm(), in that you add or subtract terms from the model equation. Suppose that a new diagnostic test for PE was invented and we wish to incorporate it into the PE network. We can achieve this by the following code:
net2 <- update(net, . ~ . + newTest | pe
+ treat | newTest
- pregnant)
plot(net2)
The update() method for HydeNetwork objects processes terms in the given model equation sequentially. In the above example, the original object net did not contain a node called newTest. But there were nodes called pregnant, pe, and treat. The first term within the model equation (+ newTest | pe) specifies the addition of the node newTest as a child of node pe. The second term (+ treat | newTest) specifies the addition of a link from the now-existing node newTest into node treat. The third term (- pregnant) specifies the removal of node pregnant. Examining the network object, two important points are worth mentioning:
net2
First, while the graph has changed -- and now node treat has three parents -- the model for node treat has not changed. The user must specify a new model (with either setNode() or setNodeModels()) to account for this new dependency if desired.
Second, a warning message indicates that a parent node (pregnant) has been removed. Since that node was involved in characterizing the distribution of its child node(s) (d.dimer), the function by default removes the distribution from all child nodes still existing in the network. The user then is required to use either setNode() or setNodeModels() to repopulate the distribution(s) for the affected node(s).
[Link to top]
## Try the HydeNet package in your browser
Any scripts or data that you put into this service are public.
HydeNet documentation built on July 8, 2020, 5:15 p.m. | 5,216 | 22,593 | {"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} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.803102 |
https://romantonumber.com/mmmcdlxxxviii-in-arabic-numerals | 1,716,369,240,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00509.warc.gz | 417,763,065 | 15,314 | # MMMCDLXXXVIII in Hindu Arabic Numerals
MMMCDLXXXVIII = 3488
3 4 8 8 M C X I MM CC XX II MMM CCC XXX III CD XL IV D L V DC LX VI DCC LXX VII DCCC LXXX VIII CM XC IX
MMMCDLXXXVIII is valid Roman numeral. Here we will explain how to read, write and convert the Roman numeral MMMCDLXXXVIII into the correct Arabic numeral format. Please have a look over the Roman numeral table given below for better understanding of Roman numeral system. As you can see, each letter is associated with specific value.
Symbol Value
I1
V5
X10
L50
C100
D500
M1000
## How to write Roman Numeral MMMCDLXXXVIII in Arabic Numeral?
The Arabic numeral representation of Roman numeral MMMCDLXXXVIII is 3488.
## How to convert Roman numeral MMMCDLXXXVIII to Arabic numeral?
If you are aware of Roman numeral system, then converting MMMCDLXXXVIII Roman numeral to Arabic numeral is very easy. Converting MMMCDLXXXVIII to Arabic numeral representation involves splitting up the numeral into place values as shown below.
MMMCDLXXXVIII M + M + M + CD + L + X + X + X + V + I + I + I 1000 + 1000 + 1000 + 500 - 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 + 1 1000 + 1000 + 1000 + 400 + 50 + 10 + 10 + 10 + 5 + 1 + 1 + 1 3488
As per the rule highest numeral should always precede the lowest numeral to get correct representation. We need to add all converted roman numerals values to get our correct Arabic numeral. The Roman numeral MMMCDLXXXVIII should be used when you are representing an ordinal value. In any other case, you can use 3488 instead of MMMCDLXXXVIII. For any numeral conversion, you can also use our roman to number converter tool given above.
## Current Date and Time in Roman Numerals
The current date and time written in roman numerals is given below. Romans used the word nulla to denote zero because the roman number system did not have a zero, so there is a possibility that you might see nulla or nothing when the value is zero.
2024-05-22
09:14:00
MMXXIV-V-XXII
IX:XIV:nulla
Disclaimer:We make a reasonable effort in making sure that conversion results are as accurate as possible, but we cannot guarantee that. Before using any details provided here, you must validate its correctness from other reliable sources on internet. | 602 | 2,228 | {"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-2024-22 | latest | en | 0.687846 |
https://www.thenational.academy/teachers/programmes/maths-primary-ks1/units/comparing-quantities-part-whole-relationships/lessons | 1,718,918,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00118.warc.gz | 876,488,002 | 27,811 | New
New
Year 1
# Comparing quantities - part whole relationships
## Lessons (15)
• I can compare items according to their length and height.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can compare items according to their mass.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can compare items according to their volume and capacity.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can count set of objects accurately.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can solve problems by comparing the number of objects in two sets.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can use the equality and inequality symbols to compare sets of objects.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can use the equality and inequality symbols to compare the relative size of two numbers.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can explain what a whole is.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can explain that a whole can be split into two or more parts.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can explain that a whole can represent a group of objects.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can identify a part of a whole group.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can explain what a part-part-whole model is.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can use a part-whole model to represent a whole partitioned into two parts.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can use a part-whole model to represent a whole partitioned into more than two parts.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video
• I can solve problems using a part-whole model to represent a whole partitioned into more than two parts.
1 Slide deck
1 Worksheet
2 Quizzes
1 Video | 449 | 1,708 | {"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-2024-26 | latest | en | 0.853941 |
https://proofwiki.org/wiki/Natural_Frequency_of_Underdamped_System | 1,582,824,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146744.74/warc/CC-MAIN-20200227160355-20200227190355-00268.warc.gz | 500,846,925 | 9,733 | # Natural Frequency of Underdamped System
## Theorem
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
$\dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
Then the natural frequency of $S$ is given by:
$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
## Proof
Let the position of $S$ be described in the canonical form:
$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where $\alpha = \sqrt {a^2 - b^2}$.
$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
where $T$ is the period of the movement of $S$.
By definition of natural frequency:
$\nu = \dfrac 1 T = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
$\blacksquare$ | 293 | 780 | {"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} | 3.6875 | 4 | CC-MAIN-2020-10 | longest | en | 0.71924 |
https://crypto.stackexchange.com/questions/47358/perfect-secrecy-with-xor-shift | 1,716,583,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00172.warc.gz | 158,190,180 | 43,968 | # Perfect secrecy with XOR & SHIFT?
I have read that XOR provides perfect secrecy, when the key is perfectly random. However it's technically hard to generate truly random numbers, especially on computers, so that is why people use AES, Twofish and other cypthers.
So the conditions for using XOR are that the key has to be true random, and the keysize must be at least as big as the message.
# What if we just use a very long key, to guarantee at least as much entropy as the message size?
For instance we need to encrypt 6 bits of message, then using a 32 bit pseudo random key, should have at least 6 bits of entropy in it. Or use even 512-bit key, if the RNG is very weak, to suit our needs. Then we XOR the first 6 bits.
Now the other 26 bits are leaking out, but if the attacker doesn't know the size of the input message, then it should not be a problem.
But if the attacker does know the size of the input, then we will also shift the output bits from the right to the left by n tiles, leaving p spaces between them. Lik this:
Obviously the n and p values are also random and secret choosen by us. So we would only need to keep the XOR key, the n and the p secret.
And after 1 shift operation, the message should be perfectly secret. The XOR already provides perfect secrecy, but the shift operation also hides the needle in the haystack.
Because how can an attacker brute force the original input if he doesn't even know what cypher we are using. So randomizing the cypher parameters is what I am suggesting here.
What is your opinion, is this a viable encryption system?
• I would recommend to not design your own stream ciphers, but yes you can combine a PRNG and a vernam cipher into a stream cipher. But it's security Properties are nowhere near a OTP. May 12, 2017 at 10:07
• From the example: If you want to encrypt $6$ bit and have a $512$ bit key, then you can just use OTP. A key for a symmetric cipher also has to be chosen uniformly random. In your argument "If the attacker knows something, then we do something else" is fundamentally flawed: You can't make the encryption adaptive of what the adversary does or knows: The continuitiy is quite clear that you use the encryption system and then an attacker tries to break it (and break does not only mean find the plaintext).
– tylo
May 12, 2017 at 10:58
• In any case, the real problem with OTP is not that generating truly random numbers is hard (it is, kind of, but it's still doable in practice) but the fact that securely transmitting a one-time key that's as long as the message is, in most cases, almost as hard as securely transmitting the message in the first place. So, in practice, all that OTP gives you is the ability to send the key ahead of time over a slow but secure channel, and to later send the encrypted message over a fast but insecure channel. And even then, the only reason to use OTP for that is if you don't fully trust e.g. AES. May 12, 2017 at 11:24
There are a few issues with the approach presented:
Or use even 512-bit key, if the RNG is very weak, to suit our needs.
If your RNG is broken, then using more of it will not help.
"If the attacker knows something, then we do something else"
As was mentioned by @tylo in the comments: "...(that idea) is fundamentally flawed: You can't make the encryption adaptive of what the adversary does or knows."
To elaborate on why:
• How would you know what one particular adversary does or does not know?
• Assuming somehow that you do know what one particular adversary does/does not know, how would you ensure that they receive the correct version of the algorithm that you designed and packaged up just for them?
• In addition to that, how would you prevent them from acquiring "weaker" copies of the software, intended for a less-knowledgeable adversay?
In order to protect against the worst case scenario, we have to assume the worst case scenario. Meaning that we need to assume that the adversary already knows everything other then the key.
Because how can an attacker brute force the original input if he doesn't even know what cypher we are using. So randomizing the cypher parameters is what I am suggesting here
As was mentioned in the answer by @Elias: Keeping the cipher secret is actually a poor strategy, which is a lesson that has been proven time and time again.
The problem is that you cannot get rid of the need for secrecy, which means you cannot get rid of the need for a key. Keeping the cipher algorithm secret means using the algorithm itself as the key. It is significantly more difficult to keep an algorithm secret then small, random chunk of data: Everyone who is to encrypt/decrypt will know the algorithm, which increases the probability of leakage for every participant.
"But what if I'm the only one that uses it then? Problem solved, right?"
Supposing you were only going to encrypt/decrypt messages with your secret algorithm for yourself: If the algorithm is the key, then you just posted your key online for all to see and comment on (and analyze and attack).
Randomizing cipher parameters and keeping the cipher secret are not the same thing: I would not actually describe your design as keeping the cipher secret (especially considering how it has been posted here for all to see).
## Xor and Tranposition is broken
What if we just use a very long key, to guarantee at least as much entropy as the message size?
is this a viable encryption system?
If you happen to encrypt fewer message bits then you have key bits, then yes; But this would only be because it would effectively be equivalent to a basic One Time Pad and so is not a realistic answer.
Otherwise, for more realistic use cases, no: xor-with-key + transposition is always broken, even if the transposition is a secret as well. The problem stems from the fact that the degree of the equations does not increase; The expression that represents the output will consist of nothing but the XOR of some input terms. A secret transposition only serves to vary which terms constitute each output bit.
It is often times trivial to break such designs by first submitting a block of all 0 to be encrypted: The resultant ciphertext will consist of nothing but key bits. Secret transposition will not fix this, it only modifies which key bits constitute each output bit.
After you have the XOR-key, you can figure out the transposition key by submitting encryptions of weight 1 plaintexts (i.e. 1, 2, 4, 8, etc). After removing the XOR-key from each such ciphertext, the result will display which output bits contain the associated plaintext input bit. If the cipher has a blocksize of n-bits, then this attack would require only n chosen plaintexts to completely recover every bit of both keys.
It almost seems like you are trying to use the cipher's key schedule as a sort of mode of operation for the cipher (the mode of operation is usually responsible for ensuring subsequent blocks encrypt differently). If so, it's probably better to keep the two tasks separate.
Hiding which cipher you are using means violating Kerckhoff's principle. That's actually an extremely common mistake. The problem is that such a cipher becomes very hard to analyze because you have to consider all the options for an attacker to learn parts of the system.
In general we don't analyze cipher designs on this website because it is too easy to come up with a random design which is probably broken but not totally obviously so. It wastes people's time to try and attack it. Cryptographers mostly don't care about new ciphers anyway. We already have good ciphers. Exceptions are made for very specific applications.
• I was thinking the same, when designing any crypto algo, we should assume all the algo parameters/options are public knowledge (attacker's included). Jun 18, 2019 at 19:20 | 1,699 | 7,838 | {"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} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.933224 |
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(OP)
What is the effect of different number of blades on various helicopter rotors. I am interested in the hovering power efficiency - static thrust etc.? Articles/reports/books. This information is very difficult to find in books or on the Internet.
Your referring to the "Figure of Merit" ratio which is the measure of a rotors efficiency, as well as "Hover Power Req'd", and Power correction factors, etc etc etc.
The best way to answer this question is to email me at moygr1@home.com (that's a one not an L) and state exactly what you need to know.
This gets deep pretty quick.
Figure of Merit Ratio:
.707 * (Ct^ 3/2) / Cq <-- Torque Coefficent
or .707 * Sqrt(CT^3)/ Cq
same thing
now your Cq is for the main rotor and is calculated by:
Q / (pPiR^2)(omega R)^2 <-- again density and Pi
of which to find your torque you take:
5252 * SHP / RPM
which is SHP = Shaft horsepower
so if i had a 2,5544 shp at the main bevel gear for the main rotor mast, at 258 rpm the torque would be 51,990.728 ft/lbs of torque. Some times you need to estimate this if you dont have exact figures for the transmissions. An easy way to "guess" is to take total engine power, take 25% of that, and figure torque at that rating. usually you lose 1/4 of total power due to transmission losses and accessory drive losses. etc.
what id do for quick estimates is this: Say the UH-60 Blackhawk, which is what these numbers are for.
Take 1650 Eng. SHP *2 = 3300 hp
then .77 * 3300 = 2,541shp
5252 * 2541 / 258 = 51,726 not bad when you say its only an esitmate, how far is this off by?
.005 one half of one percent!
now omega is angular velocity in Radians per sec. and R is the rotor radius.
Hover induced Power req'd.
HPi = (T/550) * sqrt(w/(2pB^2)) <--p = air density for sealevel use .002378 for 4000 feet use .0021109 and for say 14000 feet use .0015455 ( have a sheet i can fax you if you desire.
T= thrust
B = Tip loss factor
w = thrust / (Pi*R^2) <-----disc area
B = 1 - sqrt (Ct/2R) <---Thrust Coefficient/rotor diameter
the Ct = T/(piR^2)* p * (Vt^2)
(piR^2) <-- Disc area
p <-- Air Density in Slugs per cubic foot. note: the pi above is the mathematical "pi" 3.141,,,,, not the density.
Vt^2 <-- tip speed squared which is 2(pi)Rn
where R = radius in feet
where n = revolutions per second
so to get revolutions per second. take the RPM and * by 60
if you have a fax, i can send out some stuff for you.
Joe
Efax 419-818-1506
There is another Tip loss factor, one based off of geometery.
B = 1 - Ctip <-- Ctip is the Tip chord (width) in feet.
----
2R <-- R is the rotor Radius
a common number in the industry is to have a tip loss factor of .95 to .97 dont expect 1.00 or unity, you wont get it.
Also, Thrust of a rotor, eh,
T = (pAv) * v^2
rho or air density, p times the Disc Area, times the induced velocity of the rotor, get that quantity, then take the induced velocity, square it, and then multiply to two results together, this is good for approximations.
This is where you will have to experiment a bit to find a unity of sorts. Unless you feel like doing some calculus.
v = sqrt(T/(2pA))
The simple way to put it what do you want in a rotor system. the more blades you have ,can reduce the blade eliment loading but your autorotation will be over less distance (with the hughes 300 look at your heels thats where you go a Bell 206 look at the lower part of the windshild and thats where you go,and so on.) then you have the parts count, the more blades you have the more grips,bearings,links,dampers,pieces tec.so its more than a numbers game but a mission or goal set to accomplish.how much storage space? a multi bladed a/c needs more room unless its a folding system then you get more complex and the last thing the world needs is a more complex helo to maintane, so what ever you do keep it simple. more is some times not better just diffrent.
I asked the helicopter test pilot which do he
prefer. He said that the difference is huge in terms
of vibration levels. He prefers tri baldes compared
to two even for light helicopter. The pilots feel
vibration down to their spine, and the helicopter is
their working space, so we should hear their voice, also.
Here's are three reasonable equations that show the effect of the number of blades on thrust. Equations from Stepniewski "Rotary Wings Aerodynamics".
Eqn 1) T = 4 * pi * R^2 * density * integral(vi^2 * x, dx, x=R0 to Rtip)
Eqn 2) vi = Omega * { - a * s / 16 + sqrt [ (a * s / 16)^2 + a * s * x * theta / 8]}
Eqn 3) s = Nb * R * c / (pi * R^2)
T - rotor thrust (lb)
pi - 3.14...
density - 0.0023769 lb-sec^2/ft^4 (slug/ft^3)
Rtip - rotor radius reduced for tip effects (ft) ~ 0.98R
vi - induced velocity (ft/sec)
x - differential rotor radius (ft)
a - lift curve slope (5.2 / radian)
s - solidity; ratio of blade area to disk area
c - chord (ft)
The solidity is proportional to the number of blade. The induced velocity is almost proportional to solidity. Induced velocity is proportional to solidity if you replace collective (theta) by collective divided by solitity. And thrust is proportional to induced velocity squared. From this, you can see that keeping everything the same, that thrust is approximatly proportional to Nb^2.
2
One reason from a pilot's point of view is that more blades equates to smaller rotor diameter for the same thrust (lift). When landing clearings are a big problem as in the northern provinces, smaller blade radius is preferred.
Having said that, one disadvantage is that the rotary momentum, or angular inertia of the unpowered rotor system during autorotation is less in a rotor disk of smaller radius. Hence the incredible difference betwen the D model MD-500 and the Bell 206B in autorotational bottom end recovery to a soft touchdown.
But the main reason for more blades is this: at high speed, high altitude, hot weather, and heavy loads, a helicopter needs to maintain a higher peak angle of attack on the retreating blades, in order to balance lift. At some point, the angle becomes too high, the retreating blade stalls, and the pilot will experience two or three things. First, a sudden lurch to the left. Secondly, a sudden lifting of the nose, as gyroscopic precession acts 90 deg later to force the rotor disk upwards at the front, automatically correcting for the retreating blade stall, thus un-stalling it, and thirdly, a possible slight moistening of the undershorts.
So add more blades and you can go faster and carry more.
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https://www.traditionaloven.com/metal/precious-metals/palladium/convert-1cm-diam-palladium-sphere-to-ton-short-sh-tn.html | 1,526,922,558,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00516.warc.gz | 857,086,341 | 14,325 | Palladium sphere 1 centimeter in diameter to short tons of palladium converter
## Amount: sphere 1 centimeter in diameter (d, ∅ 1 cm) of palladium volume Equals: 0.0000069 short tons (sh tn) in palladium mass
Calculate short tons of palladium per sphere 1 centimeter in diameter unit. The palladium converter.
TOGGLE : from short tons into solid ∅ 1 cm spheres in the other way around.
### Enter a New sphere 1 centimeter in diameter Amount of palladium to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## palladium from sphere 1 centimeter in diameter to ton (short) Conversion Results :
Amount : sphere 1 centimeter in diameter (d, ∅ 1 cm) of palladium
Equals: 0.0000069 short tons (sh tn) in palladium
Fractions: 2/289855 short tons (sh tn) in palladium
CONVERT : between other palladium measuring units - complete list.
This calculator is based on a pure palladium with a density of ≅ 12 grams per one cubic centimeter, 12.023g/cm3 more precisely. Palladium, symbol Pd, is a rare metal and in the raw form found in the nature it always contains in the metal body a small amount of platinum. Palladium enjoys a widespread use; automobile industry, dentistry, jewellers make a swishy jewellery pieces by including this metal on their jewels, not to mention watchmakers', those who make quality luxury watches, and even professional musical instruments like transverse flutes for instance. I'd like to know whether the genuine TAG Heuer or Rolex Swiss watches have palladium in/on them. Palladium is listed in the group with the four expensive bullion metals (incl. platinum, gold and silver) where each of these four has own ISO currency code.
Convert palladium measuring units between sphere 1 centimeter in diameter (d, ∅ 1 cm) and short tons (sh tn) of palladium but in the other direction from short tons into solid ∅ 1 cm spheres.
conversion result for palladium: From Symbol Equals Result To Symbol 1 sphere 1 centimeter in diameter d, ∅ 1 cm = 0.0000069 short tons sh tn
This online palladium from d, ∅ 1 cm into sh tn (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this palladium calculator are ...
With the above mentioned units calculating service it provides, this palladium converter proved to be useful also as a teaching tool:
1. in practicing solid ∅ 1 cm spheres and short tons ( d, ∅ 1 cm vs. sh tn ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with palladium's density values including other physical properties this metal has.
International unit symbols for these two palladium measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for sphere 1 centimeter in diameter is: d, ∅ 1 cm
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for ton (short) is: sh tn
### One sphere 1 centimeter in diameter of palladium converted to ton (short) equals to 0.0000069 sh tn
How many short tons of palladium are in 1 sphere 1 centimeter in diameter? The answer is: The change of 1 d, ∅ 1 cm ( sphere 1 centimeter in diameter ) unit of a palladium amount equals = to 0.0000069 sh tn ( ton (short) ) as the equivalent measure for the same palladium type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of palladium can have a crucial/pivotal role in investments. If there is an exact known measure in d, ∅ 1 cm - solid ∅ 1 cm spheres for palladium amount, the rule is that the sphere 1 centimeter in diameter number gets converted into sh tn - short tons or any other unit of palladium absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many short tons ( sh tn ) of palladium are contained in a sphere 1 centimeter in diameter ( 1 d, ∅ 1 cm ). Or, how much in short tons of palladium is in 1 sphere 1 centimeter in diameter? To link to this palladium - sphere 1 centimeter in diameter to short tons online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: palladium from sphere 1 centimeter in diameter (d, ∅ 1 cm) to short tons (sh tn) metal conversion.
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https://www.physicsforums.com/threads/eigenvectors-of-rotation-matrix.513818/ | 1,726,719,530,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00872.warc.gz | 852,424,162 | 19,069 | # Eigenvectors of rotation matrix
• timon
In summary, the conversation discusses finding the corresponding eigenvectors for a given operator with specific eigenvalues. The solution involves letting the matrix operate on a generic vector and solving for the resulting vector. The correct eigenvectors are determined to be \frac{1}{\sqrt{2}} \left[ \begin{array}{c}1 \\i \\\end{array} \right] and \frac{1}{\sqrt{2}} \left[ \begin{array}{c}i \\1 \\\end{array} \right].
timon
## Homework Statement
This question is from Principles of Quantum Mechanics by R. Shankar.
Given the operator (matrix) $\Omega$ with eigenvalues $e^{i\theta}$ and $e^{-i\theta}$, I am told to find the corresponding eigenvectors.
## Homework Equations
$\Omega = \left[ \begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \\ \end{array} \right]$
$\Omega \left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right] = \left[ \begin{array}{c} x_1 \cos{\theta} + x_2 \sin{\theta} \\ -x_1 \sin{\theta} + x_2 \cos{\theta} \\ \end{array} \right]$
$e^{i\theta} \left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right] = \left[ \begin{array}{c} x_1 \cos{\theta} + x_1 i\sin{\theta} \\ x_2 \cos{\theta} + x_2 i\sin{\theta} \\ \end{array}\right]$
## The Attempt at a Solution
I let the matrix operate on the generic vector $(x_1, x_2)^T$ and demand that the resulting vector is equal to $(e^{i\theta}x_1, e^{i\theta}x_2)^T$. From this i get the condition that $x_2 = ix_2$and $x_1 = -ix_2$, which implies that $x_1 = x_2 = 0$. Am i missing something crucial?
Last edited:
timon said:
## The Attempt at a Solution
I let the matrix operate on the generic vector $(x_1, x_2)^T$ and demand that the resulting vector is equal to $(e^{i\theta}x_1, e^{i\theta}x_2)^T$. From this i get the condition that $x_2 = ix_2$and $x_1 = -ix_2$,
Typo? I think you mean $x_2 = ix_1$ here. In which case these two equations are really the same equation.
which implies that $x_1 = x_2 = 0$. Am i missing something crucial?
No, it does not imply that. Let x1=1 (and worry about normalization later), then what is x2?
Thanks for the quick reply. That actually wasn't a typo. I managed to switch the indices the same way in three separate calculations, getting the same erroneous result each time. Are these the correct eigenvectors?
$\frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ i \\ \end{array} \right] , \frac{1}{\sqrt{2}} \left[ \begin{array}{c} i \\ 1 \\ \end{array} \right]$
Are these the correct eigenvectors?
$\frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ i \\ \end{array} \right] , \frac{1}{\sqrt{2}} \left[ \begin{array}{c} i \\ 1 \\ \end{array} \right]$
Yes, that's right.
timon said:
Thanks for the quick reply. That actually wasn't a typo. I managed to switch the indices the same way in three separate calculations, getting the same erroneous result each time.
That's weird. From your equations in Post #1, you get
$$x_1 \cos{\theta} + x_2\sin{\theta} = x_1 \cos{\theta} + x_1 i\sin{\theta}$$
which by inspection implies $x_2 = i x_1$, not $x_2 = i x_2$
It is indeed quite amazing that I managed to screw the same trivial manipulation up three times in a row, always in the same manner. Thank you for the help.
## 1. What are eigenvectors and why are they important in a rotation matrix?
Eigenvectors are special vectors that remain in the same direction when multiplied by a transformation matrix, such as a rotation matrix. They are important in a rotation matrix because they provide a frame of reference for the rotation and can help determine the axis and angle of rotation.
## 2. How do you calculate the eigenvectors of a rotation matrix?
To calculate the eigenvectors of a rotation matrix, you first need to find the eigenvalues of the matrix. Then, you can use the eigenvalues to find the corresponding eigenvectors using the formula (A - λI)v = 0, where A is the rotation matrix, λ is the eigenvalue, and v is the eigenvector.
## 3. What do the eigenvectors of a rotation matrix represent?
The eigenvectors of a rotation matrix represent the principal axes of rotation. These are the axes around which the rotation occurs, and the length of the eigenvectors corresponds to the amount of rotation around that axis.
## 4. Can a rotation matrix have complex eigenvectors?
Yes, a rotation matrix can have complex eigenvectors. However, in most cases, the eigenvectors will be real numbers since rotations typically occur in a 3-dimensional space. Complex eigenvectors are more common in higher dimensional spaces.
## 5. How do you interpret the eigenvectors of a rotation matrix in terms of the rotation itself?
The eigenvectors of a rotation matrix can be interpreted as the axis of rotation. The direction of the eigenvector represents the direction of the axis, and the magnitude of the eigenvector represents the amount of rotation around that axis. Additionally, the eigenvectors are orthogonal to each other, meaning they are at right angles, which corresponds to the fact that the rotation matrix is an orthogonal matrix.
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# Residual income valuation Damodaran
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3. Valuation: Packet 3 Real Options, Acquisition Valuation and Value Enhancement - Aswath Damodaran Valuation: Packet 3 Real Options, Acquisition Valuation and Value Enhancement Aswath Damodaran Updated: Residual Income Valuation is the property of its rightful owner
4. Valuation Approaches and Metrics: A Survey of the Theory and Evidence Aswath Damodaran Stern School of Business November 2006 . 2 Valuation Approaches and Metrics: A Survey Article Valuation lies at the heart of much of what we do in finance, whether it is th
Damodaran 7 Dealing with Negative Earnings When the earnings in the starting period are negative, the growth rate cannot be estimated. (0.30/-0.05 = -600%) There are three solutions: • Use the higher of the two numbers as the denominator (0.30/0.25 = 120%) • Use the absolute value of earnings in the starting period as the denominato
Residual income valuation (also known as residual income model or residual income method) is an equity valuation method that is based on the idea that the value of a company's stock equals the present value of future residual incomes discounted at the appropriate cost of equit A recent paper by Ohlson (1995) has stimulated interest in the residual income formulation of the dividend discounting valuation model. This development has potentially important implications for empirical researchers, as Ohlson's model specifies the relation between equity values and accounting variables such as earnings and book value
We have discussed the use of residual income models in valuation. Residual income is an appealing economic concept because it attempts to measure economic profit, which are profits after accounting for all opportunity costs of capital. Residual income is calculated as net income minus a deduction for the cost of equity capital Valuation of other assets: In this section, you will find a model for valuing income-generating real estate. Value Enhancement Spreadsheets : In this section, you will find a spreadsheet that reconciles EVA and DCF valuation, a model for estimating CFROI and a DCF version of a value enhancement spreadsheet
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• I've seen you use residual income for two purposes: to value financial companies and to evaluate management (your Jan 9th post). Is there a reason you don't use residual income models to value non-financial businesses? I find myself preferring these types of models over DCF. If you think they're flawed, it would be great to learn why. Thanks
• equity that we value is only common equity, you would need to modify the formulae slightly for the existence of preferred stock and dividends. In particular, you would subtract out the preferred dividends to arrive at the free cashflow to equity: Free Cash Flow to Equity (FCFE) = Net Income - (Capital Expenditures
• The construction of residual value is a key element of income methods in asset valuation. Its main task is to include asset value at the end of the forecasted cash flow period. Although it is..
• This paper provides an empirical assessment of the residual income valuation model proposed in Ohlson (Ohlson, J.A., 1995. Earnings, book values and dividends in security valuation. Contemporary Accounting Research 11, 661-687). We point out that existing empirical research relying on Ohlson's model is similar to past research relying.
• Residual income valuation (RIV; also, residual income model and residual income method, RIM) is an approach to equity valuation that formally accounts for the cost of equity capital
• The residual income valuation approach is a viable and increasingly popular method of valuation and can be implemented rather easily by even novice investors. When used alongside the other popular..
• es whether it is possible to infer a..
The aim of this paper is to present a framework to bank valuation based on two generally acceptable valuation models that are not specific to banks: the model of discounted Equity Cash Flow to Equity (ECF) and the model of discounted Residual Income (RI). As emphasized by Koller, Goedhart and Wessels (pp. 663, 2005) [1] in a bestselling book on the valuation of firms, the valuation process of. Aswath Damodaran INVESTMENT VALUATION: SECOND EDITION Chapter 1: Introduction to Valuation 3 Chapter 2: Approaches to Valuation 16 The value obtained from any valuation model is affected by firm-specific as well as market-wide information. As a consequence, the value will change as new information. Also, the residual income and the terminal value are discounted and the sum of these components derives the equity value for the bank. form the future annual cash flow and the residual income, is a critical task in a bank (Damodaran, 2009 [7]) 262 Chapter 5 Residual Income Valuation 1 INTRODUCTION Residual income models of equity value have become widely recognized tools in both in vestment practice and research. Conceptually, residual income is net income less a charge (deduction) for common shareholders' opportunity cost in generating net income. As a residual income model, which is a model of a hybrid approach including insights from both the income approach and the cost approach. The residual income model includes in the value of a company not only the discounted future abnormal earnings but also the book value of the company as of the valuation day
### PPT - Residual Income Valuation PowerPoint presentation
1. Example with Coke. We will start our valuation example with Coke by first calculating its residual income on a per share level. We will use Coke's 2019 diluted earnings per share of \$2.07, book value per share of \$4.89, and cost of equity of 5.5% as calculated for the starting example of IFB's financial model template.As can be seen below, plugging these inputs into the formula, Coke had a.
2. Business Value: \$ 1,230,769,230.76 so about \$1,230,769,231 Conclusion of Business Value We relied upon methods such as the Asset, Market and Income Approaches to business valuation. Although, the asset as well as the market approach were both rejected since the asset isn't relevant to a huge corporation as business and the market approach isn.
3. Residual Income Model is an equity valuation method used to estimate the true or intrinsic value of a stock based on the present value of all future residual income the company generates. A company has two sources of capital, equity, and debt
4. athan [2001]). Damodaran [1996], and Palepu, Healy, and Bernard [2000]) devote con
5. Aswath Damodaran. Professor of Finance at the Stern School of Business at New York University , where he teaches corporate finance and equity valuation. Clean surplus accounting and the related residual income valuation provide a model that returns price as a function of earnings, expected returns, and change in book value,.
6. From a purely theoretical standpoint, the residual income model is simply a different method of applying a discounted cash flow valuation. The RI differs from a standard DCF stock valuation in that..
7. Where, Equity Charge = Cost of Equity Capital x Equity Capital; Steps by Step Calculation of Residual Income. Calculate the net income or net profit of the company, which can also be derived from the income statement of the company.; Calculate the cost of capital using various other methods like CAPM, Building block approach, Multi-model approach, etc.; Take the book value of the common equity.
Goes through run of the mill valuations to special methods like the residual income valuation method and addresses how each should be deployed in complex situations. Read more. Helpful. Comment Report abuse. Damodaran on Valuation: Security Analysis for Investment and Corporate Finance (Wiley Finance Book 324) Aswath Damodaran. 4.7 out of 5. Valuation techniques, value drivers and usual traps. Residual. Income approach. When used? • Residual valuation method mainly used for . development projects Approach • Value asset as if complete • Estimate direct and indirect costs to build. Calculation • Calculate Gross Development Value ( GDV The Residual-Income Stock Price Valuation ModeP The residual-income stock price valuation model has received considerable academic attention during tbe past several years. It is theoretically equivalent to the discounted free-cash-flows-to-equity model taught in finance courses, as wel
### Residual Income Valuation - Overview, Benefits, How to
1. The residual-income valuation model provides a useful framework in which to conduct this discussion. It defines total common equity value in terms of the book White et al. 1998; Damodaran 1994; Pratt et al. 1996). 4 The original dividend discount model is expressed symbolically as: , where P is total common equity value at time t, r is the.
2. Dcf Discounted Cash Flow Valuation: Basics Aswath Damodaran The valuation analysis combines the discounted cash flow model, discounted residual income model and sensitivity analysis. The range of estimated share price is within AUD\$8.7-\$16.62. The current market share price is AUD\$9.17
3. But to value the S&P 500 Index and individual companies, Damodaran urged investors to stick with traditional valuation tools with adjustments for the pandemic. Gone, Gone, the Damage Done Damodaran started with a damage assessment of financial markets during the worst part of the crisis, from 14 February to 20 March, when the United.
4. When we set the uncertainty parameter σ used in the generation of residual income to 0.033, i.e. one-tenth of the value used in the reported simulations, the Vuong statistic shows that model 3 and 4 are superior, as evidenced by a double digit Vuong-Z statistic relative to model 2
5. al value based on the assumption made about continuing residual income over the long term
6. In this article Axiom compares and contrasts the two discounting valuation approaches - the residual income valuation (RIV) model and the traditional discounted cash flow (DCF) model. The article illustrates that when valuing a business, in certain circumstances the RIV model is a more robust and appropriate model than the DCF method
7. Residual income valuation methods are a great way to value a business all focused on book value, return on equity and the cost of equity. Being able to adjust the valuation for a business's economic moat through the persistence factor is a powerful tool
7. Residual Income valuation model Listed banks MSDW (2001) 8. Risk-adjusted TSR (market oriented approach) Listed banks MOW (2003) 9. Residual Income valuation Approach ( EVA) 28 public and private listed banks Verma (2002) 10. Residual Income valuation Approach ( EVA) India (SBI and HDFC Bank) Raiyani and Joshi (2009) 11 This article provides an overview of the residual‐income stock price valuation model and demonstrates its use in interpreting the DuPont return on equity (ROE) decomposition. The model provides theoretical support for the DuPont model's focus on ROE and aids in understanding the implications of the price‐to‐book and price‐earnings ratios The residual-income valuation model provides a useful framework in which to conduct this discussion. It defines total common equity value in terms of the book value of stockholders' equity and net income determined in accordance with GAAP, and thus is particularly well suited to support instruction of ratio analysis
### An empirical assessment of the residual income valuation
1. Residual income valuation (RIV; also, residual income model and residual income method, RIM) is an approach to equity valuation that formally accounts for the cost of equity capital. Here, residual means in excess of any opportunity costs measured relative to the book value of shareholders' equity; residual income (RI) is then the income generated by a firm after accounting for the true cost.
2. Damodaran, who teaches valuation and corporate finance at the Stern School of Business at New York University (NYU), made these remarks at the CFA Institute Equity Research and Valuation Conference 2014 in Boston, where he urged attendees to bridge the gap between numbers and narratives, insisting that the best valuations are not just a.
3. The value, using a residual income approach, is the current book value plus the present value of future residual income. The residual income is a perpetuity: P0 = Book value + PV of Residual income = 6.00 + 0.40 / 0.10 = 6.00 + 4.00 = \$10.00. When to use RIM valuation A Residual Income Model is most appropriate when
4. Tham 2001, reported consistency between the Residual Income Model (RIM) and the Discounted Cash Flow model (DCF) with a very simple example. Fernandez 2002 shows examples where there is consistency between DCF, RIM and EVA. He uses a constant value for the cost of levered equity capital and in another example constant debt
Residual income = Operating income- (Minimum rate of return x Operating assets) Advantages of Residual Income To illustrate the use of residual income, consider the Cleaning Products Division example again. Recall that the division manager rejected Project I because it would have reduced divisional ROI, which cost the company \$300,000 in. economic value added, residual income, earnings dan arus kas operasi, terhadap return yang diterima oleh pemegang saham perusahaan publik Menurut Damodaran, untuk mengukur return dari sebuah investasi, dapat digunakan accounting earnings dan arus kas (Damodaran 1999)
Definition Residual income valuation (RIV) which is also known as residual income method or residual income model (RIM) is an approach to or method of equity valuation which properly accounts for the cost of equity capital This requires calculation of a terminal value of the residual income at the end of the abnormal growth phase. In contrast to the terminal value in a multi-stage DDM, the terminal value in a multi-stage RI model will be much smaller, as it will only capture the terminal value of residual income following the high growth period and not the. Training on Residual Income valuation by Vamsidhar Ambatipud
### Video: Residual Income Valuation - CFA Institut
The VA's residual income guideline offers a powerful and realistic way to look at VA loan affordability and whether new homeowners have enough income to cover living expenses and stay current on their mortgage.. Residual income is a major reason why VA loans have such a low foreclosure rate, despite the fact that about 9 in 10 people purchase without a down payment the last ten years, alternative earnings concepts like comprehensive income, residual income, and abnormal earnings have been advanced. There have been more references to book value. In addition to the profusion of new age techniques, an increasing number of fundamental attributes have been advanced. This requires some sorting. RESIDUAL INCOME VALUATION LEARNING OUTCOMES After completing this chapter, you will be able to do the following : Calculate and interpret residual income and related measures (e.g., economic value added and market value added). Discuss the use of residual income models. Calculate futur e values of residual income given current book value. FINS3641 SAV FCFF Discount Valuation and Residual Income Valuation Models 2 1. FCFF Discount Valuation Model: Illustrating a 3 ‐ stage model using Embraer in 2000 i. Rationale for the choice of cash flow / discount model - FCFF / FCFF discount model We don't expect the firm to maintain its current leverage for two reasons: Debt to capital ratio has been volatile and Debt to capital ratio. Valuation using discounted cash flows (DCF valuation) is a method of estimating the current value of a company based on projected future cash flows adjusted for the time value of money. The cash flows are made up of those within the explicit forecast period, together with a continuing or terminal value that represents the cash flow stream after the forecast period
Residual income valuation model can calculate intrinsic value of a company. It does it by adding book value with present value (PV) of all residual incomes to be generated by the company in its lifetime ().. Though computations in residual income method may look complicated in the first look, but it is simple a. calculate and interpret residual income, economic value added, and market value added; 3. The residual income valuation model b. describe the uses of residual income models; c. calculate the intrinsic value of a common stock using the residual income model and compare value recognition in residual income and other present value models; 4 The projection of the accounting data included in the financial statements, which form the future annual cash flow and the residual income, is a critical task in a bank valuation process (Gross, 2006) [3] .This is because the bank performance is affected by macroeconomic factors such as interest rate changes, capital markets volatility etc. (Brealey and Myers, 2000) [4] What are the steps to calculate SS's common stocks justified value if the present value of projected residual income for the next 5 years plus beginning book value is \$75 per share. Beyond that time horizon, the firm will sustain residual income of \$11.25 per shares, which is the residual income for year 6
CHAPTER 5 RESIDUAL INCOME VALUATION LEARNING OUTCOMES After completing this chapter, you will be able to do the following :• Calculate and interpret residual income and related measures (e.g., - Selection from Equity Asset Valuation, Second Edition [Book Aswath Damodaran 4 I.Equity Valuation n The value of equity is obtained by discounting expected cashflows to equity, i.e., the residual cashflows after meeting all expenses, tax obligations and interest and principal payments, at the cost of equity, i.e., the rate of retur
### Musings on Markets: Discounted Cashflow Valuations (DCF
• Equity valuation models .In, Damodaran on Valuation (2 ed., pp.157-183). New Delhi. Wiley Finance 13 asyncho ronous Two stage FCFF valuation 13_01_Two stage FCFF_1 13_02_Two stage FCFF_2 Aswath Damodarn (2012). Firm valuation models .In, Damodaran on Valuation (2 ed., pp.193-222). New Delhi. Wiley Finance
• er in Strategic Corporate Finance 1. Introduction The residual income (RI) for a firm for any year t is its accounting income for that year (Xt) less a capital charge equal to the equity cost of capital times opening boo
• CFA Level 2Topic: Equity ValuationReading: Residual Income ValuationThe residual income model is used to estimate the intrinsic value of equity.This video co..
### (PDF) Significance Of Residual Value In Asset Valuatio
Narrative and Numbers The Value of Stories in Business by - Aswath Damodaran. Year: 2014 Edition: 4th Language: english Pages: 654. ISBN 13: 978-1-118-80893-1 File: PDF, 7.73 MB Preview. Send-to-Kindle or Email . Please to your account first; Save for later investment 835. income 734. estimate 695. returns 646. damodaran 638. trim size 630. tex 630. 8in x 10in 630. disney 615. earnings 611 Residual Income Formula. The formula to calculate Residual Income is the following: RI = Net income - (Equity * Cost of Equity). Residual Income Equation Components. Net Income: Net earnings after deducing all costs, expenses, depreciation, amortization, interest charges and taxes from the business revenues. Equity: The total equity as stated in the Balance Sheet Aswath Damodaran just started uploading his NYU class lectures - we have posted the first batch below, which we hope readers will enjoy. Check back since there is a lot more coming or better yet, sign up for our free daily newsletter to ensure you never miss a post. The first ten videos can be found below. Also see The Little Book of Valuation: How to Value a Company, Pick a Stock and Profit.
Residual income valuation (RIV) assesses a company based on its cost of equity capital. It determines the amount of operating profit remaining once equity capital's paid. Cost of equity capital is the rate of return required by stockholders, and reflects the risk exposure of their investments Residual income is the amount of income that an individual has after all personal debts and expenses, including a mortgage, have been paid. This calculation is usually made on a monthly basis. Residual income model definition. The single stage residual income valuation model is defined as follows . where ROE is the return on equity, r is the required rate on return, B0 is the current book value per share, and g is the constant growth rate. It is evident that the model is closely related to the Gordon growth model.. In the above formulation, the first terms is the current value of. Residual income drops immediately to zero (omega = 0). In that case the present value of residual income in T-1 is residual income in year T discounted at the required rate of return; Residual income declines over time to zero. This is the case when the omega is a value between zero and one. Residual income declines to the long-run level in a.
Valuation Models. Damodaran on Valuation: Security Analysis for Investment and Corporate Finance, 1994 Aswath Damodaran New York: John Wiley & Sons. These excerpts and quotes are considered to be fair use as interpreted by The Chicago Manual of Style: The Essential Guide for Writers, Editors, and Publisher, 14th edition, 1993.Chicago: The University of Chicago Press Personal residual income, often called discretionary income, is the amount of income or salary left over after debt payments, like car loans and mortgages, have been paid each month. For example, Jim's take-home pay is \$3,000 a month Under the DCF and residual income valuations models, the terminal value calculation can only be performed when r > g. Suppose a friend at another school showed you a valuation where r < g. What is the simplest and most logical solution you would suggest to fix his/her terminal value calculation Residual Income = \$50,000 - 15% * \$225,000; Residual Income = \$16,250 Therefore, the company is able to generate a residual income of \$16,250 during the year
Study Equity #37 - Residual Income Valuation flashcards from Scott George's class online, or in Brainscape's iPhone or Android app. Learn faster with spaced repetition LOS: Calculate and interpret residual income, economic value added, and market value added. Pages 464-467. As an economic concept, residual income has a long history, dating back to Alfred Marshall in the late 1800s. More recently, residual income has received renewed attention, sometimes under the names. economic profit, abnormal earnings, o I use three residual income measures: Economic Profit (EP), Economic Value Added (EVA) and Cash Value Added (CVA). I first show that the present value of the EP discounted at the required return to equity plus the equity book value equals the value of equity (the present value of the Equity cash flow discounted at the required return to equity)
restructure residual income valuation models to take account of the distortions introduced by the use of historical cost accounting. Hughes, Li and Zhang (2004) examine the value relevance of accounting variables and characterize the impact of inflation and foreign exchange on the weights that attac Residual Income Defined Residual income (RI), or economic profit, is t he net income of a firm less a charge that measures stockholder's opportunity cost of capital. This concept of economic income is not reflected in traditional accounting income, whereby a firm can report positive net income but not meet the return requirements of it based residual income value estimates do not improve on book values alone (Dechow, Hutton and Sloan [1999]) and are dominated by analyst-based value estimates (Frankel and Lee [1998]). We show that the difference in results is due to relaxing the assumption that the persistence in residual income is an over
The results of the hypothesis testing indicate that a significant difference between residual income and value of the firm is observed. In this study, the market value added changes criterion has been used to measure the value of the company. For this purpose, 121 companies have been selected from Tehran Stock Exchange through screening. Chapter 8: Residual Income Valuation Model. 8.1 Introduction . I n Chapters 5 and 6, we covered dividend models and free cash flow to equity models. In the simple dividend model, the value of a company to a shareholder is derived from cash dividends. The free cash flow models value a company using economic dividends, whether paid in cash. Extended Dividend, Cash Flow and Residual Income Valuation Models - Accounting for Deviations from Ideal Conditions Dieter Hess∗, Carsten Homburg‡, Michael Lorenz‡, Soenke Sievers∗ ∗Corporate Finance Department ‡Accounting Department University of Cologne Albertus-Magnus Platz, D-50923 Cologne, German Tax Deferral Options: Companies that have more options when it comes to deferring taxes than others can buffer the impact of higher corporate taxes by choosing to defer taxes and report less taxable income.The most significant of those options, in my view, is foreign sales, with companies that generate more of their income overseas acquiring more tax freedom than purely domestic companies
IndusInd Bank: Residual Income Valuation Thecasesolution.com Thecasesolution.com Overview Davis Neon looking for alternative packaging installs Sealed Air's Speedy Packer Insight to make Instapack Continuous Foam Tubes Results: 62% savings in material costs 20% increase i Extended Dividend, Cash Flow, and Residual Income Valuation Models: Accounting for Deviations from Ideal Conditions * Nicolas Heinrichs. University of Cologne. Search for more papers by this author. Dieter Hess. University of Cologne, Center for Financial Research. Search for more papers by this author present value of the residual income, discounted at the required rate of return on equity of 10% percent, is \$0.91 per share. Adding this present value to the beginning book value of equity per share of \$10 per share results in a value of \$10.91 per share. The value of equity using the residual income model is affected not only by the required. Damodaran (2002) considers the valuation of financial service firms with the special earnings, cash flows, book value or sales. The third, income approach (or, specifically, discounted cash flow valuation), relates the value of an asset to the present value o The value of land under this appraisal method is therefore a residual amount resulting from the improvement of land. Any improvement that increases the value of the land™s final use increases the land residual. For example, if house prices are increasing, with other costs remaining constant, land prices should rise
### Residual income valuation - Wikipedi
Residual income valuation Last updated May 11, 2019. Residual income valuation (RIV; also, residual income model and residual income method, RIM) is an approach to equity valuation that formally accounts for the cost of equity capital. Here, residual means in excess of any opportunity costs measured relative to the book value of shareholders' equity; residual income (RI) is then the income. What is Value Added in Valuation? • The value added methodologies are used to measure the profits (or losses) generated by a firm for a given level of capital investment & the risk of these investments: - Also called residual income or abnormal earnings • Value Added (for all investors Debt + Equity) Equity Valuation Study Session 9. Study Session 10. Study Session 11. Free Cash Flow Valuationd Valuation: Price and Enterprise Value Multiples . Residual Income Valuation . Private Company Valuation. Fixed Income . Essential concepts from Level I. Study Session 12. Study Session 13. Derivatives. Study Session 14...p;A Review Session for.
### Valuing a Company Using the Residual Income Metho
residual income is zero after year 5, as in the residual income model's calculation, the present value of dividends after year 5 equals the present value of the book value in year 5. Book value in year 5 is 161.05 in Penman's example, so discounting this back five years produces a valuation of 100, just as in the residual income model. I Residual Income Valuation. Residual Income Valuation. BY Pj Germain No Comments. Tweet. Share. Share. Pin. Share. 0 Shares. If you're looking for ways to make money without having to punch a time clock or be a slave to your business, residual earnings are the way to go. With residual earnings you put in some work setting up a business system. The paper uses Ohlson (Contemp Account Res 11:661-687, 1995) and compares the relative predictability of the proposed simultaneous model for contemporaneous stock price with a traditional single equation model used by the previous studies. The paper also explores how residual income and value-relevant information affect firms' equity price The residual income approach is the measurement of the net income that an investment earns above the threshold established by the minimum rate of return assigned to the investment. It can be used as a way to approve or reject a capital investment , or to estimate the value of a business
### (PDF) Firm valuation: Comparing the residual income and
L2 R30 Residual Income Valuation 2 Items . Expand. Lesson Content . 0% Complete 0/2 Steps. L2 R30 HY Video. In Progress. L2 R30 High Yield Notes. In Progress. L2 R31 Private Company Valuation 2 Items . Expand. Lesson Content . 0% Complete 0/2 Steps. L2 R31 High Yield Video. VA Residual Income Calculator. Residual income is a calculation that estimates the net monthly income after subtracting out the federal, state, local taxes, (proposed) mortgage payment, and all other monthly obligations such as student loans, car payments, credit cards, etc. from the household paycheck(s)
To figure out a company's residual income, subtract the company's cost of capital from the company's net income, and then multiply the difference by the company's total assets. When the company's residual income is a negative value, it means the company is not profitable even if it is netting a positive income From January 2008 until March 30, 2013, the bank's stock had tripled under its new management. The analyst wondered whether deploying funds in the bank would yield any significant returns. He decided to use the available financial information and the residual income valuation method to forecast the company's stock price Similar to passive income, residual income gives you more financial stability, flexibility with your lifestyle, additional retirement savings, and a more robust financial standing. How to Build Residual Income. Depending on your interests, skills, and business knowledge, you can create sources of residual income that align with your experience Let's take a look at 15 of the best residual income ideas that you can work on now, to reap the passive cash flow down the road. Real estate. One of our favorite residual income ideas is probably the oldest: real estate investing. There are a number of ways you can tap into the residual income power of owning an interest in real estate, here.
### Understanding Bank Valuation: An Application of the Equity
Residual income also features in corporate finance and valuation where it equals the difference between a company's net income and the product of the company's equity capital and its cost of equity. Formula. Residual income of a department can be calculated using the following formula: Residual Income = Controllable Margin - Required Return. The income statement is made as per the data given. There is no chnage in the equity since the net income is given out as dividend. The residual income is the amount left after accounting for the cost of equity. This is given as 10%. the cost of equity on book value of 100 is 10. Subtracting this from the net income, we get the residual income.
### Residual Income Valuation Method - CFA Level
As stated in Lesson 15, Land Valuation, the most reliable method of estimating land value is through the comparison of the subject property with recent sales of comparable, similarly located, properties.Since the value of all land is based on its productivity, or income producing ability, when utilized at its highest and best use, it is also appropriate to use the income approach to. This is Jim Wahlen Residual Income Valuation.mp4 by Amsterdam Institute of Finance on Vimeo, the home for high quality videos and the people who love them The residual income calculation goes one step further and factors in other expenses like childcare, estimated utilities on the new home, child support (if any), and Social Security and income taxes. This calculation attempts to determine, or at least estimate, all your real-life expenses each month We investigate the use and performance of residual income (RI) valuation methods by U.S. sell-side equity analysts in a small-sample manner wherein we extract the rich details of analysts' RI valuations from their PDF reports over the period 1998-2013 The standard family-of-four residual income guideline for Ohio is \$1,003. Apply For A VA Mortgage The Department of Veterans Affairs residual income requirement is imperfect
### Nike valuation report - SlideShar
When calculating the present value of a company, an analyst can choose between dividends, free cash flows, and residual income to derive the stock's intrinsic price. Each of these cash flows has advantages and drawbacks. Dividends. These direct cash payments are a key component of an investor's returns In my last post, I described the wild ride that the price of risk took in 2020, with equity risk premiums and default spreads initially sky rocketing, as the virus led to global economic shutdowns, and then just as abruptly dropping back to pre-crisis levels over the course of the year.As stock and bond markets went through these gyrations, it should come as no surprise that the same forces. The case illustrates the use of the residual income (also known as the abnormal earnings) valuation approach. Students are asked to provide a valuation of Coca-Cola Company using the residual income valuation methodology and understand how it maps into the discounted cash flow method Frankel and Lee (1998) show that the value-to-price ratio (V f /P) predicts future abnormal returns for up to three years, where V f is an estimate of fundamental value based on a residual income valuation framework operationalized using analyst earnings forecasts. In this study, we examine whether the V f /P effect is due to market mispricing or omitted risk factors Coca Cola Residual Income Valuation Case Study Solution and Analysis. Disclaimer: This is sample format on how a case study should be solved. This article is not at all related to actual case study. If you need solution of this case study then order on the site or contact customer support representative
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Index > Heap > What is the best pie you can get with 9 digits? Goto page Previous 1, 2, 3 ... 18, 19, 20 ... 27, 28, 29 Next
Author
alessandro95
Joined: 24 Mar 2013
Posts: 62
alessandro95
Rn¹º is the base-10 number made by n 1s in a row, e.g.
R5¹º=11111
(actually n should be a subscript but looks like it cannot be shown on this forum)
conventionally if the base is 10 then the can be omitted.
I don't know what is bigger between the S(k,n) function proposed by bitRAKE or Graham's number but
RG
where G is graham number
and
RS(k,n)
are both way bigger than G and S(k,n) respectively, and we have some.characters.left to make them even bigger (something like RRRRRRRRG, imagine every letter as subscript of the preceding one is a number absurdely huge)
05 Apr 2013, 05:17
alessandro95
Joined: 24 Mar 2013
Posts: 62
alessandro95
Just to give an idea of how quickly consecutive R grows:
R2=11
RR2=11111111111
RRR2=~1.11*10¹¹¹¹¹¹¹¹¹¹¹
and we can define this recursevely as
Rq=~1.11*10^q
nRq=~1.11*10^(n-1)Rq
where nRq stand for R applied n times to q
Imagine using G or S(k,n) as q and applies as many R as we can without using more than 9 charachters, how to decide which one is bigger?
p.s.
forgot to mention before but this numbers made by a string of 1s are called repunits
05 Apr 2013, 05:33
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
alessandro95 wrote:
and we can define ...
It doesn't seem to follow the arbitrary rules we have used for this thread. Defining our own functions was not allowed.
However I think that the growth of R will most probably be easily dwarfed by the S() function.
11 Apr 2013, 06:20
alessandro95
Joined: 24 Mar 2013
Posts: 62
alessandro95
Sorry, describe was the right term, not define.
There surely are function with a much faster growth, but do they use only 1 character? I don"t know how to compare them so I was proposing the R function, althought it is probably possible to write a bigger number without breaking the 9 characters limit
11 Apr 2013, 07:57
r22
Joined: 27 Dec 2004
Posts: 805
r22
03 Feb 2015, 04:29
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
How do those compare to the current list? Do they make the top 5?
03 Feb 2015, 05:10
r22
Joined: 27 Dec 2004
Posts: 805
r22
Big Foot and Rayo, far exceed TREE() TREE sequence and SCG() Sub Cubic Graph number., which are burdened by the fact that they are computable.
FOOT - First Order Oodle Theory (as opposed to FOST - First Order Set Theory) is some conjured up augmented set theory that formalizes the abstract of this thread "the largest number that can be represented with N symbols". Using this theory the top 5 in this thread could be mere symbols used to define this number.
BIG FOOT is FOOT^10(10^100), so in the realm of un-computable numbers we could get even larger with the 9 character constraint by going with.
Quote:
FOOT^G(G)
It's not cheating in the alternate oodleverse.
03 Feb 2015, 16:42
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
I guess that the first step is to show that a number with G symbols is larger than G itself. I think this step is trivial, so then we move on to show that FOOT(G) applied recursively G times will get progressively larger.
So far so good. But what about recursion? If we have FOOT^G(G) as the largest number in 9 characters then presumably this is larger than FOOT(G) (the largest number in G characters). So how can a 9 character value be larger than a G character value? We could just keep putting (FOOT^G(G))^(FOOT^G(G))^...^(FOOT^G(G)) to fit within G characters and we end up with a infinite self-referential loop.
04 Feb 2015, 08:55
gens
Joined: 18 Feb 2013
Posts: 161
gens
888888888
05 Feb 2015, 01:07
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
gens wrote:
888888888
I am confident that there are already larger numbers given in this thread.
05 Feb 2015, 01:25
r22
Joined: 27 Dec 2004
Posts: 805
r22
revolution wrote:
I guess that the first step is to show that a number with G symbols is larger than G itself. I think this step is trivial, so then we move on to show that FOOT(G) applied recursively G times will get progressively larger.
So far so good. But what about recursion? If we have FOOT^G(G) as the largest number in 9 characters then presumably this is larger than FOOT(G) (the largest number in G characters). So how can a 9 character value be larger than a G character value? We could just keep putting (FOOT^G(G))^(FOOT^G(G))^...^(FOOT^G(G)) to fit within G characters and we end up with a infinite self-referential loop.
From my rudimentary ?mis?understanding of first order set theory: a self referencing recursion would just be considered infinity and wouldn't be used within our domain of large theoretical numbers. Also terms must be obtainable through a finite application of variable and function rules, so eventually we'd be forced to use a symbol that didn't branch out into more self referencing symbols. At what stage this happens is, I guess, what makes the result finite but still un-computable.
I'm also selling magic text (.txt) files that keep monsters away, but only on floppy disk.
05 Feb 2015, 16:16
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
And the infinity is only of Aleph-0 class so it wouldn't even make the top-5 trans-finite list either.
05 Feb 2015, 16:26
typedef
Joined: 25 Jul 2010
Posts: 2913
Location: 0x77760000
typedef
∞^∞^∞^∞^∞
Did I win a star?
08 Feb 2015, 05:06
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
typedef wrote:
∞^∞^∞^∞^∞
You either didn't read the rules, or you ignored the rules, so your submission in not accepted.
typedef wrote:
Did I win a star?
Sure, you win Jack Black. Have fun with your new star.
08 Feb 2015, 05:15
l4m2
Joined: 15 Jan 2015
Posts: 648
l4m2
9[9]9
maxnumber(haha)
12 Feb 2015, 10:35
l4m2
Joined: 15 Jan 2015
Posts: 648
l4m2
maxhere+9
12 Feb 2015, 10:38
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
l4m2 wrote:
maxhere+9
This was already tried in a variant form by Goplat. And it was subsequently shown to be unworkable and thus disallowed.
However, you are welcome to try again.
12 Feb 2015, 11:04
l4m2
Joined: 15 Jan 2015
Posts: 648
l4m2
revolution wrote:
l4m2 wrote:
maxhere+9
This was already tried in a variant form by Goplat. And it was subsequently shown to be unworkable and thus disallowed.
However, you are welcome to try again.
a[n+1]b=a[n]a[n]a[n]a(b a's)
a[0]b=a+1
12 Feb 2015, 12:06
l4m2
Joined: 15 Jan 2015
Posts: 648
l4m2
revolution wrote:
l4m2 wrote:
maxhere+9
This was already tried in a variant form by Goplat. And it was subsequently shown to be unworkable and thus disallowed.
However, you are welcome to try again.
Also I don't think a define all from self can be or that'd be much better:
X
I can let it be any number just I like to
for example 10000...000(the simpliest one,but because of the tiny harddisk )
12 Feb 2015, 12:11
revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 17270
revolution
If you read this whole thread you will see that people defining their own terms was not permitted. This if for the exact reason you mention, because any arbitrary definition could be anything, which would make the whole thing boring.
12 Feb 2015, 12:24
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You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot vote in polls in this forumYou can attach files in this forumYou can download files in this forum | 2,426 | 8,411 | {"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-2020-24 | latest | en | 0.938394 |
https://mathoverflow.net/questions/108855/max-noethers-afbg-theorem | 1,660,644,773,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00583.warc.gz | 364,567,012 | 24,095 | # Max Noether's AF+BG theorem
I'm looking for an example of the following situation, related to Max Noether's AF+BG Theorem (see Bill Fulton's book on algebraic curves, page 61, at http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf).
Fulton motivates the AF+BG Theorem by saying that if $F, G, H$ are curves in the projective plane with no common components then we could have the inequality of cycles $H \cdot F \geq G \cdot F$ and we might be interested in knowing when there is a curve $B$ so that $H \cdot F = B \cdot F + G \cdot F$. To produce such a curve it is enough to find forms $A$ and $B$ so that $H = AF+BG$ (here I'm using the same letter to denote a curve and its defining form) since then $H \cdot F = BG \cdot F = B \cdot F + G \cdot F$.
Noether's fundamental theorem (the $AF+BG$ theorem) says that the condition $H = AF + BG$ is equivalent to the local conditions that say that for each $P \in F \cap G$, we have $H \in (F,G)\mathcal{O}_P(\mathbb{P}^2)$. Many uses of the theorem rely on being in a situation where the local conditions are obviously met and so we obtain the global fact $H = AF + BG$ and hence $H \cdot F = B \cdot F + G \cdot F$.
My question is to find an example where the local conditions are NOT met but we still have $H \cdot F = B \cdot F + G \cdot F$. If this is impossible, please explain.
## 1 Answer
Define $F$ as $x^2-y^2=0$, $G$ as $x^2+y^2=0$ and $H$ as $xy=0$. (here $(x:y:z)$ are homogeneous coordinates in $\mathbb P^2$.)
It is clear that $H$ can not be expressed as $AF+BG$. At the same time if we take $B=0$, then $H\cdot F=B\cdot F+G\cdot F=4\cdot (0:0:1)$. | 504 | 1,621 | {"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} | 3.75 | 4 | CC-MAIN-2022-33 | latest | en | 0.887447 |
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# Sim
The game of Sim is played by two players, Red and Blue, on a board consisting of six dots ('vertices'). Each dot is connected to each other with a line.
Players alternate coloring any uncolored line in their own color. Players try to avoid making triangles of their color; the player who completes a triangle of their color loses immediately. (A triangle is three dots, each connected to the other two with lines of the same color.) The other player is the winner.
A simple theorem of Ramsey theory shows that no game of Sim can end in a tie; one player must lose by the end. Specifically, since R(3,3;2)=6, any coloring of the complete graph on 6 vertices must contain a monochromatic triangle, and therefore is not a tied position. | 177 | 793 | {"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} | 2.921875 | 3 | CC-MAIN-2020-16 | latest | en | 0.958784 |
http://www.ck12.org/measurement/Conversion-of-Customary-Units-of-Measurement/?difficulty=basic | 1,485,085,507,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281421.33/warc/CC-MAIN-20170116095121-00096-ip-10-171-10-70.ec2.internal.warc.gz | 390,856,775 | 17,277 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Conversion of Customary Units of Measurement
## Convert between customary units for length, mass and volume.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
• Video
## Converting Between Units of Weight in the Customary System
by CK-12 //basic
Provides examples of converting between different units of weight in the standard or American measurement system.
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• Video
## Converting Between Units of Volume in the Customary System
by CK-12 //basic
Provides examples of converting between different units of volume or capacity in the standard or American measurement system.
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• Video
## American Unit Conversion
by CK-12 //basic
Explains how to convert between different American or standard unit of measure for length, capacity, and weight.
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• Video
## Converting Between Different Units of Same System
by CK-12 //basic
Provides examples of converting between different units of length in the standard or American measurement system.
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• Video
## Measurement Conversions with Decimals - Example 1
by CK-12 //basic
Meters to Centimeters with Decimals
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• Video
## Measurement Conversions with Decimals - Example 2
by CK-12 //basic
Kilometers to Meters with Decimals
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## Measurement Conversions with Decimals - Example 3
by CK-12 //basic
Centimeters to Millimeters with Decimals
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0 | 527 | 2,429 | {"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-2017-04 | longest | en | 0.867291 |
https://www.calculatorbit.com/en/length/33-millimeter-to-terameter | 1,721,483,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00302.warc.gz | 595,560,425 | 7,759 | # 33 Millimeter to Terameter Calculator
Result:
33 Millimeter = 3.3000000000000005e-14 Terameter (Tm)
Rounded: ( Nearest 4 digits)
33 Millimeter is 3.3000000000000005e-14 Terameter (Tm)
33 Millimeter is 3.3000000000000003cm
## How to Convert Millimeter to Terameter (Explanation)
• 1 millimeter = 1e-15 Tm (Nearest 4 digits)
• 1 terameter = 1000000000000000.2 mm (Nearest 4 digits)
There are 1e-15 Terameter in 1 Millimeter. To convert Millimeter to Terameter all you need to do is multiple the Millimeter with 1e-15.
In formula distance is denoted with d
The distance d in Terameter (Tm) is equal to 1e-15 times the distance in millimeter (mm):
### Equation
d (Tm) = d (mm) × 1e-15
Formula for 33 Millimeter (mm) to Terameter (Tm) conversion:
d (Tm) = 33 mm × 1e-15 => 3.3000000000000005e-14 Tm
## How many Terameter in a Millimeter
One Millimeter is equal to 1e-15 Terameter
1 mm = 1 mm × 1e-15 => 1e-15 Tm
## How many Millimeter in a Terameter
One Terameter is equal to 1000000000000000.2 Millimeter
1 Tm = 1 Tm / 1e-15 => 1000000000000000.2 mm
## millimeter:
TThe millimeter (symbol: mm) is a unit of length in the International System of Units (SI), equal to 1000 micrometers. Since an inch is official defined as exactly 25.4 millimeteres, a millimetere is equal to exactly 5/127 inch or (0.03937) of an inch.
## terameter:
The terameter (symbol: Tm) is a unit of length in the metric system equal to 1000000000000 meters that is 10^12 meters. 1 terameter is equal to 1 billion kilometers or 6.7 astronomical units. 1 terameter is also equal to 1/1000 petameter.
## Millimeter to Terameter Calculations Table
Now by following above explained formulas we can prepare a Millimeter to Terameter Chart.
Millimeter (mm) Terameter (Tm)
29 2.9e-14
30 3.0000000000000005e-14
31 3.1e-14
32 3.2e-14
33 3.3000000000000005e-14
34 3.4e-14
35 3.5e-14
36 3.6000000000000004e-14
37 3.7e-14
38 3.8e-14
Nearest 4 digits
## Convert from Millimeter to other units
Here are some quick links to convert 33 Millimeter to other length units.
## Convert to Millimeter from other units
Here are some quick links to convert other length units to Millimeter.
## FAQs About Millimeter and Terameter
Converting from one Millimeter to Terameter or Terameter to Millimeter sometimes gets confusing.
### Is 1e-15 Terameter in 1 Millimeter?
Yes, 1 Millimeter have 1e-15 (Nearest 4 digits) Terameter.
### What is the symbol for Millimeter and Terameter?
Symbol for Millimeter is mm and symbol for Terameter is Tm.
### How many Millimeter makes 1 Terameter?
1000000000000000.2 Millimeter is euqal to 1 Terameter.
### How many Terameter in 33 Millimeter?
Millimeter have 3.3000000000000005e-14 Terameter.
### How many Terameter in a Millimeter?
Millimeter have 1e-15 (Nearest 4 digits) Terameter. | 941 | 2,809 | {"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} | 3.78125 | 4 | CC-MAIN-2024-30 | latest | en | 0.672276 |
https://www.cdslab.org/paramonte/fortran/latest/namespacepm__physUnit.html | 1,718,772,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00223.warc.gz | 614,078,853 | 7,013 | ParaMonte Fortran 2.0.0 Parallel Monte Carlo and Machine Learning LibrarySee the latest version documentation.
pm_physUnit Module Reference
This module contains relevant physical constants. More...
## Data Types
type uncertain_type
This the derived type for constructing physical constants or unit conversions that are uncertain. More...
## Variables
character(*, SK), parameter MODULE_NAME = "@pm_physUnit"
real(RKB), parameter PI = acos(-1._RKB)
The scalar constant of type real of kind RKB representing the irrational number $$\pi$$.
More...
real(RKB), parameter NAPIER = exp(1._RKB)
The scalar constant of type real of kind RKB representing the Napier constant (a.k.a. Euler number) $$e = \exp(1)$$. More...
real(RKB), parameter AU2METER = 149597870700._RKB
The scalar constant of type real of kind RKB representing one astronomical unit unit of length in meters.
More...
real(RKB), parameter DEGREE2RADIAN = PI / 180._RKB
The scalar constant of type real of kind RKB representing one degree in radians.
More...
real(RKB), parameter ARCMIN2RADIAN = PI / 10800._RKB
The scalar constant of type real of kind RKB representing one arcminute in radians.
More...
real(RKB), parameter ARCSEC2RADIAN = PI / 648000._RKB
The scalar constant of type real of kind RKB representing one arcsecond in radians.
More...
type(uncertain_type), parameter DALTON2KG = uncertain_type(1.660539040e-27_RKB, 2.e-35_RKB)
The scalar constant of type real of kind RKB representing one Dalton in kilograms.
More...
real(RKB), parameter EV2JOULES = 1.602176634e-19_RKB
The scalar constant of type real of kind RKB representing one electronvolt in Joules.
An electronvolt is the amount of kinetic energy gained or lost by a single electron accelerating from rest through an electric potential difference of one volt in vacuum. Hence, it has a value of one volt, $$1 J/C$$, multiplied by the elementary charge $$e = 1.602176634\times10−19 C$$.
Therefore, one electronvolt is equal to $$1.602176634×10−19 J$$.
The electronvolt ( $$\ms{eV}$$) is a unit of energy, but is not an SI unit.
More...
real(RKB), parameter JOULES2ERGS = 1.e+7_RKB
The scalar constant of type real of kind RKB representing one Joules of energy in ergs.
More...
real(RKB), parameter ERGS2JOULES = 1.e-7_RKB
The scalar constant of type real of kind RKB representing one ergs of energy in Joules.
More...
real(RKB), parameter KEV2ERGS = 1000 * EV2JOULES * JOULES2ERGS
The scalar constant of type real of kind RKB representing one kilo-electronvolts of energy in ergs.
More...
real(RKB), parameter ERGS2KEV = 1._RKB / KEV2ERGS
The scalar constant of type real of kind RKB representing one ergs of energy in kilo-electronvolts.
More...
## Detailed Description
This module contains relevant physical constants.
Note
The constants of this module are saved with the highest available real precision kind.
To use the constants at expressions involving lower-precision real kinds, simply convert the numbers to the desired kind via the Fortran intrinsic real(x, kind = RKG) where RKG refers to the target kind parameter used in the expression.
Developer Remark:
This is an experimental module still under development.
Other similar work includes but is not limited to Kim et al. 2017, Fortran 90 Programming With Physical Unit Annotations.
Final Remarks
If you believe this algorithm or its documentation can be improved, we appreciate your contribution and help to edit this page's documentation and source file on GitHub.
1. If you use any parts or concepts from this library to any extent, please acknowledge the usage by citing the relevant publications of the ParaMonte library.
2. If you regenerate any parts/ideas from this library in a programming environment other than those currently supported by this ParaMonte library (i.e., other than C, C++, Fortran, MATLAB, Python, R), please also ask the end users to cite this original ParaMonte library.
This software is available to the public under a highly permissive license.
Help us justify its continued development and maintenance by acknowledging its benefit to society, distributing it, and contributing to it.
Todo:
Normal Priority: Th contents of this module should be substantially expanded.
Conversion procedures for converting physical quantities in various units must be added.
Author:
Amir Shahmoradi, Oct 16, 2009, 11:14 AM, Michigan
## Variable Documentation
real(RKB), parameter pm_physUnit::ARCMIN2RADIAN = PI / 10800._RKB
The scalar constant of type real of kind RKB representing one arcminute in radians.
Definition at line 68 of file pm_physUnit.F90.
real(RKB), parameter pm_physUnit::ARCSEC2RADIAN = PI / 648000._RKB
The scalar constant of type real of kind RKB representing one arcsecond in radians.
Definition at line 69 of file pm_physUnit.F90.
## ◆ AU2METER
real(RKB), parameter pm_physUnit::AU2METER = 149597870700._RKB
The scalar constant of type real of kind RKB representing one astronomical unit unit of length in meters.
Definition at line 66 of file pm_physUnit.F90.
## ◆ DALTON2KG
type(uncertain_type), parameter pm_physUnit::DALTON2KG = uncertain_type(1.660539040e-27_RKB, 2.e-35_RKB)
The scalar constant of type real of kind RKB representing one Dalton in kilograms.
Definition at line 70 of file pm_physUnit.F90.
real(RKB), parameter pm_physUnit::DEGREE2RADIAN = PI / 180._RKB
The scalar constant of type real of kind RKB representing one degree in radians.
Definition at line 67 of file pm_physUnit.F90.
## ◆ ERGS2JOULES
real(RKB), parameter pm_physUnit::ERGS2JOULES = 1.e-7_RKB
The scalar constant of type real of kind RKB representing one ergs of energy in Joules.
Definition at line 78 of file pm_physUnit.F90.
## ◆ ERGS2KEV
real(RKB), parameter pm_physUnit::ERGS2KEV = 1._RKB / KEV2ERGS
The scalar constant of type real of kind RKB representing one ergs of energy in kilo-electronvolts.
Definition at line 80 of file pm_physUnit.F90.
## ◆ EV2JOULES
real(RKB), parameter pm_physUnit::EV2JOULES = 1.602176634e-19_RKB
The scalar constant of type real of kind RKB representing one electronvolt in Joules.
An electronvolt is the amount of kinetic energy gained or lost by a single electron accelerating from rest through an electric potential difference of one volt in vacuum. Hence, it has a value of one volt, $$1 J/C$$, multiplied by the elementary charge $$e = 1.602176634\times10−19 C$$.
Therefore, one electronvolt is equal to $$1.602176634×10−19 J$$.
The electronvolt ( $$\ms{eV}$$) is a unit of energy, but is not an SI unit.
Definition at line 71 of file pm_physUnit.F90.
## ◆ JOULES2ERGS
real(RKB), parameter pm_physUnit::JOULES2ERGS = 1.e+7_RKB
The scalar constant of type real of kind RKB representing one Joules of energy in ergs.
Definition at line 77 of file pm_physUnit.F90.
## ◆ KEV2ERGS
real(RKB), parameter pm_physUnit::KEV2ERGS = 1000 * EV2JOULES * JOULES2ERGS
The scalar constant of type real of kind RKB representing one kilo-electronvolts of energy in ergs.
Definition at line 79 of file pm_physUnit.F90.
## ◆ MODULE_NAME
character(*,SK), parameter pm_physUnit::MODULE_NAME = "@pm_physUnit"
Definition at line 47 of file pm_physUnit.F90.
## ◆ NAPIER
real(RKB), parameter pm_physUnit::NAPIER = exp(1._RKB)
The scalar constant of type real of kind RKB representing the Napier constant (a.k.a. Euler number) $$e = \exp(1)$$.
Definition at line 65 of file pm_physUnit.F90.
## ◆ PI
real(RKB), parameter pm_physUnit::PI = acos(-1._RKB)
The scalar constant of type real of kind RKB representing the irrational number $$\pi$$.
Definition at line 64 of file pm_physUnit.F90. | 2,001 | 7,603 | {"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": 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} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.663234 |
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# Operations w/sign#s
### MTH005 Ch 1 Operations with Real Numbers
OperationRules
Addends with Like Signs Add and use the sign of the addends.
Addends with Unlike Signs Find the difference of the absolute values of the numbers. The sign of the sum is the same as the sign of the number with the largest absolute value.
Subtraction with Real Numbers 1) Change the minus sign to a plus sign. 2) Change the subtrahend to its opposite. 3) Use the rules of addition of sign numbers to evaluate the sum.
Opposite of a Difference For any real numbers a and b, -(a - b) = b - a
Multiplication of Sign Numbers To multiply any two real numbers, the sign is 1) Positive if the numbers have like signs. 2) Negative if they have unlike signs.
Multiplying 3 or More Real Number Factors The sign of the product is Positive if there are an even number of negative signs. The product is negative if there is an odd number of negative signs.
Division of Real Numbers The quotient is 1) Positive if the divisor/denominator and dividend/numerator have like signs. 2) Negative if the divisor/denominator and dividend/numerator have unlike signs.
Property of Signs in Quotients If a fraction has 1 opposite sign, the sign can be placed in the numerator, denominator, or in front of the fraction; 2 opposite signs, they can both be removed; 3 opposite signs, any 2 can be removed.
Created by: powens | 473 | 1,999 | {"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-2016-44 | latest | en | 0.862706 |
http://physics.stackexchange.com/questions/tagged/heat?page=1&sort=faq&pagesize=50 | 1,368,973,772,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697681504/warc/CC-MAIN-20130516094801-00057-ip-10-60-113-184.ec2.internal.warc.gz | 209,772,370 | 14,559 | # Tagged Questions
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I was wondering: If I put three bottles of water next to each other in the fridge, which one is cold first? Does it matter? Is it the one in the middle because it gets refrigerated by the other two? ... | 1,509 | 6,748 | {"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} | 2.65625 | 3 | CC-MAIN-2013-20 | latest | en | 0.932641 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-7-integration-7-2-substitution-7-2-exercises-page-375/11 | 1,537,887,700,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161661.80/warc/CC-MAIN-20180925143000-20180925163400-00165.warc.gz | 751,540,553 | 12,877 | ## Calculus with Applications (10th Edition)
$\frac{1}{2}{e^{2{x^3}}} + C$
$\begin{gathered} \int_{}^{} {3{x^2}{e^{2{x^3}}}dx} \hfill \\ Let\,\,u = 2{x^3} \hfill \\ So\,\,that\,\,du = 6{x^2}dx.\,\,Then \hfill \\ \int_{}^{} {3{x^2}{e^{2{x^3}}}dx} = \frac{1}{2}\int_{}^{} {{e^{2{x^3}}}\,\left( {6{x^2}} \right)dx} \hfill \\ \frac{1}{2}\int_{}^{} {{e^u}du} \hfill \\ Integrating \hfill \\ \frac{1}{2}{e^u} + C \hfill \\ Substituting\,\,u = 2{x^3}\,\,for\,\,u\,\,gives \hfill \\ \frac{1}{2}{e^{2{x^3}}} + C \hfill \\ \end{gathered}$ | 263 | 529 | {"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} | 4.125 | 4 | CC-MAIN-2018-39 | longest | en | 0.422814 |
http://www.mathworks.com/help/ident/ref/goodnessoffit.html?requestedDomain=www.mathworks.com&nocookie=true | 1,511,008,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804881.4/warc/CC-MAIN-20171118113721-20171118133721-00766.warc.gz | 470,099,212 | 14,932 | # Documentation
### This is machine translation
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# goodnessOfFit
Goodness of fit between test and reference data
## Syntax
```fit = goodnessOfFit(x,xref,cost_func) ```
## Description
`fit = goodnessOfFit(x,xref,cost_func)` returns the goodness of fit between the data, `x`, and the reference, `xref` using a cost function specified by `cost_func`.
## Input Arguments
`x` Test data. `x` is an `Ns`-by-`N` matrix, where `Ns` is the number of samples and `N` is the number of channels. `x` can also be a cell array of multiple test data sets. `x` must not contain any `NaN` or `Inf` values. `xref` Reference data. `xref` must be of the same size as `x`. `xref` can also be a cell array of multiple reference sets. In this case, each individual reference set must be of the same size as the corresponding test data set. `xref` must not contain any NaN or Inf values. `cost_func` Cost function to determine goodness of fit. `cost_func` is specified as one of the following values: `'MSE'` — Mean square error:`$fit=\frac{{‖x-xref‖}^{2}}{Ns}$`where, Ns is the number of samples, and ‖ indicates the 2-norm of a vector. `fit` is a scalar value.`'NRMSE'` — Normalized root mean square error:`$fit\left(i\right)=1-\frac{‖xref\left(:,i\right)-x\left(:,i\right)‖}{‖xref\left(:,i\right)-mean\left(xref\left(:,i\right)\right)‖}$`where, ‖ indicates the 2-norm of a vector. `fit` is a row vector of length `N` and ```i = 1,...,N```, where `N` is the number of channels.NRMSE costs vary between `-Inf` (bad fit) to 1 (perfect fit). If the cost function is equal to zero, then `x` is no better than a straight line at matching `xref`.`'NMSE'` — Normalized mean square error:`$fit\left(i\right)=1-\frac{{‖xref\left(:,i\right)-x\left(:,i\right)‖}^{2}}{{‖xref\left(:,i\right)-mean\left(xref\left(:,i\right)\right)‖}^{2}}$`where, ‖ indicates the 2-norm of a vector. `fit` is a row vector of length `N` and ```i = 1,...,N```, where `N` is the number of channels.NMSE costs vary between `-Inf` (bad fit) to 1 (perfect fit). If the cost function is equal to zero, then `x` is no better than a straight line at matching `xref`.
## Output Arguments
`fit` Goodness of fit between test and reference data. For a single test data set and reference pair, `fit` is returned as a: Scalar if `cost_func` is `MSE`.Row vector of length `N` if `cost_func` is `NRMSE` or `NMSE`. `N` is the number of channels. If `x` and/or `xref` are cell arrays, then `fit` is an array containing the goodness of fit values for each test data and reference pair.
## Examples
collapse all
Obtain the measured output.
```load iddata1 z1 yref = z1.y;```
`z1` is an `iddata` object containing measured input/output data. `z1.y` is the measured output.
Obtain the estimated output.
```sys = tfest(z1,2); y_sim = sim(sys,z1(:,[],:));```
`sys` is a second-order transfer function estimated using the measured input/output data. `y` is the output estimated using `sys` and the measured input.
Calculate the goodness of the fit between the measured and estimated outputs.
```cost_func = 'NRMSE'; y = y_sim.y; fit = goodnessOfFit(y,yref,cost_func);```
The goodness of fit is calculated using the normalized root mean square error as the cost function.
Alternatively, you can use `compare` to calculate the goodness of fit:
```opt = compareOptions('InitialCondition','z'); compare(z1,sys,opt);```
## See Also
#### Introduced in R2012a
Was this topic helpful?
Get trial now | 1,072 | 3,744 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 2.71875 | 3 | CC-MAIN-2017-47 | latest | en | 0.731572 |
http://caml.inria.fr/pub/ml-archives/caml-list/2007/11/3802856b639d751741eba13be1ffe65a.en.html | 1,553,616,760,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205597.84/warc/CC-MAIN-20190326160044-20190326182044-00242.warc.gz | 37,224,783 | 3,764 | This site is updated infrequently. For up-to-date information, please visit the new OCaml website at ocaml.org.
polymorphic lists, existential types and asorted other hattery
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Date: 2007-11-13 (17:27) From: Peng Zang Subject: polymorphic lists, existential types and asorted other hattery
```-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hi,
Is there a way to create lists in which the elements may be of
differing types but which all have some set of operations defined
(eg. tostr) in common? One can then imagine mapping over such lists
with "generic" versions of those common operations. Here's a concrete
example of what I mean:
module Int = struct
type t = int
let show x = string_of_int x
end
module Float = struct
type t = float
let show x = string_of_float x
end
module Bool = struct
type t = bool
let show x = string_of_bool x
end
let xs = [`Int 1; `Float 2.0; `Bool false]
let showany x = match x with
| `Int x -> Int.show x
| `Float x -> Float.show x
| `Bool x -> Bool.show x
;;
List.map showany xs;;
Essentially we have ints, floats and bools. All these types can be
shown. It would be nice to be able to create a list of them [1; 2.0;
false] that you can then map a generalized show over. In the above
example, I used polymorphic variants in order to get them into the
same list and then had to define my own generalized show function,
"showany". This is fine as there is only one shared operation but if
there is a large set of these common operations, it becomes
impractical to define a generalized version for each of them.
I've come across a way to do this in haskell using what they call
"existential types".
I don't really understand existential types however and don't know if
OCaml has them nor how to use them.
So. How can one do this in OCaml? Is there perhaps a camlp4
extension that can do this? Is there a possible functor trick that
can take N modules as arguments and spit out a new module with a
generalized type that can take on any of the types in the arguments
and also make generalized versions of operations common to the N
modules? Are there existential types or equivalents in OCaml? If so
how does one go about using them?
Thanks in advance to anyone who forays into this bundle of questions.
Peng
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``` | 724 | 2,608 | {"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-2019-13 | latest | en | 0.915446 |
https://www.calculatoryou.com/convert/weight/pound-to-kilogram | 1,723,577,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00780.warc.gz | 549,409,311 | 4,091 | # Lbs to Kg Converter (Pounds to Kilograms)
Use our lbs to kg converter to convert pounds to kilograms. Enter a value in pounds, click Calculate and we will automatically convert it to kilograms.
Do you want to convert Kilograms to Pounds?
## How do you calculate Pounds to Kilograms?
Converting between different units of measurement is a fundamental skill in many fields, from science and engineering to cooking and fitness. One common conversion is between pounds (lbs) and kilograms (kg).
This article will guide you through the process of converting pounds to kilograms, explaining the formula, providing examples, and showing how to create a simple web-based conversion tool using HTML and JavaScript.
### The Conversion Formula
The pound and the kilogram are units of mass commonly used in different parts of the world. The conversion between pounds and kilograms is based on the fixed relationship between these units:
1 pound (lb) = 0.45359237 kilograms (kg)
1 pound (lb) = 0.45359237 kilograms (kg)
This means that to convert a value from pounds to kilograms, you multiply the number of pounds by 0.45359237.
### Pound
The pound is a unit of measurement used to quantify mass. It is a widely recognized and utilized unit in several countries, particularly in the United States and the United Kingdom.
This guide will provide an in-depth look at the pound, its history, usage, and conversions. Definition and Symbol The pound, symbolized as lb or sometimes # is a unit of mass in various measurement systems, including the Imperial system and the United States customary system. The abbreviation lb is derived from the Roman word libra which was a unit of weight in ancient Rome.
### Kilogram
The kilogram is the base unit of mass in the International System of Units (SI), making it one of the most important units in science, industry, and daily life. This guide will provide a detailed look at the kilogram, its history, definition, usage, and its significance in the world of measurement.
CALCULATORYOU | 422 | 2,030 | {"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} | 3.09375 | 3 | CC-MAIN-2024-33 | latest | en | 0.910917 |
http://www.efunda.com/units/convert_units.cfm?mode=long&From=1035&InputValue=1 | 1,568,673,198,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00106.warc.gz | 244,802,265 | 7,476 | Convert Pressure Units
Units Currencies Constants
Units Home Unit Conversion List Units List Categories Currencies Convert Units SI System Base Units Derived Units Prefixes Numbers Constants Base-N Fractions Time/Decimal Resources Bibliography
Home Membership Magazines Forum Search Member Calculators
Materials Design Processes Units Formulas Math
ton force (long) per square inch
Symbol: tonf/in2 (UK) Category: Pressure SI Equivalent: 1.54443×107 Pa Dimension: ML-1T-2 System: UK
Convert tonf/in2 (UK)
1 tonf/in2 (UK) =
Energy density Symbol Unit Name 414.513 Btu (IT)/ft3 British thermal unit (IT) per cubic foot 66.5473 Btu (IT)/gal (UK) British thermal unit (IT) per gallon (UK) 55.4123 Btu (IT)/gal (US) British thermal unit (IT) per gallon (US) 1.54443×107 J/m3 joule per cubic meter 1.54443×104 kJ/m3 kilojoule per cubic meter 1940.79 MGOe megagauss-oersted (MGOe) Pressure Symbol Unit Name 157.488 at atmosphere (metric) 152.423 atm atmosphere (standard) 154.443 bar bar 1.54443×108 barad barad 1.54443×108 barye 1.15842×104 centiHg (0°C) 1.15842×104 cmHg (0 °C) centimeter of mercury (0°C) 1.57492×105 cmH2O centimeter of water (4°C) 1.54443×108 dyn/cm2 dyne per square centimeter 5167.08 ft H2O foot of water (4°C) 1.54443×10-2 GPa gigapascal 1.54443×105 hPa hectopascal 4560.71 inHg (0 °C) inch of mercury (0°C) 4573.14 inHg (15.56 °C) inch of mercury (15.56°C) 6.20639×104 inH2O (15.56 °C) inch of water (15.56°C) 6.20049×104 inH2O (4 °C) inch of water (4°C) 157.488 kgf/cm2 kilogram force per square centimeter 1.57488×104 kgf/dm2 kilogram force per square decimeter 1.57488×106 kgf/m2 kilogram force per square meter 1.57488 kgf/mm2 kilogram force per square millimeter 1.54443×104 kPa kilopascal 2.24001 kip/in2, ksi, KSI kilopound force per square inch 15.4443 MPa megapascal 1576.42 mH2O, mCE (15.56 °C) meter of water (15.56°C) 1574.92 mH2O, mCE (4 °C) meter of water (4°C) 1.54443×108 µbar microbar (barye, barrie) 1.15842×108 µHg (0 °C) micron of mercury (millitorr) 1.54443×105 mbar millibar 1.15842×105 mmHg, torr, Torr (0 °C) millimeter of mercury (0°C) 1.57642×106 mmH2O, mmCE (15.56 °C) millimeter of water (15.56°C) 1.57492×106 mmH2O, mmCE (4 °C) millimeter of water (4°C) 1.15842×108 mtorr millitorr 1.54443×107 N/m2 newton per square meter 3.58401×104 ozf/in2, osi ounce force (av.) per square inch 1.54443×107 Pa, N/m2 pascal 3.22561×105 lbf/ft2 pound force per square foot 2240.01 psi, PSI, lbf/in2 pound force per square inch 1.03781×107 pdl/ft2 poundal per square foot 7.20699×104 pdl/in2 poundal per square inch 152.423 atm standard atmosphere 144 tonf/ft2 (UK) ton force (long) per square foot 0.157488 tonf/cm2 (metric) ton force (metric) per square centimeter 1574.88 tonf/m2 (metric) ton force (metric) per square meter 161.28 tonf/ft2 (US) ton force (short) per square foot 1.12 tonf/in2 (US) ton force (short) per square inch 1.15842×105 torr torr | 1,062 | 2,899 | {"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-2019-39 | longest | en | 0.316351 |
http://mathhelpforum.com/trigonometry/43825-prove-following-identities.html | 1,481,201,894,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00340-ip-10-31-129-80.ec2.internal.warc.gz | 178,582,690 | 10,080 | # Thread: Prove the following identities.............
1. ## Prove the following identities.............
Prove the following identities:
a) (cosec
θ + cotθ)(cosecθ - cotθ) = cotθtanθ
b) 2/(1 + sin
θ) + 1/(1- sinθ) = (3sec^2)θ - tanθcosecθ
I basically just have to prove Left HandSide = Right Hand Side
*
2. Hello, sweetG!
Prove the following identities:
$a)\;\;\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta} \:=\:\cot\theta\tan\theta$ . . . . This is not an identity
$b)\;\;\frac{2}{1 + \sin\theta} + \frac{1}{1-\sin\theta} \:=\:3\sec^2\!\theta - \tan\theta\,{\color{red}\csc\theta\;\;?}$
Combine the fractions:
$\frac{2}{1+\sin\theta}\!\cdot\!{\color{blue}\frac{ 1-\sin\theta}{1-\sin\theta}} + \frac{1}{1-\sin\theta}\!\cdot\!{\color{blue}\frac{1+\sin\thet a}{1+\sin\theta}} \;\;=\;\;\frac{2(1-\sin\theta) + (1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$
. . $= \;\;\frac{3-\sin\theta}{1-\sin^2\!\theta} \;\;=\;\;\frac{3-\sin\theta}{\cos^2\!\theta} \;\;=\;\;\frac{3}{\cos^2\!\theta} - \frac{\sin\theta}{\cos^2\!\theta}$
. . $= \;\;3\!\cdot\!\frac{1}{\cos^2\!\theta} - \frac{\sin\theta}{\cos\theta}\!\cdot\!\frac{1}{\co s\theta} \;\;=\;\;3\sec^2\!\theta - \tan\theta\,{\color{red}\sec\theta}$ | 503 | 1,212 | {"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": 5, "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} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.341225 |
http://mathhelpforum.com/algebra/202888-scientific-notation.html | 1,526,869,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863901.24/warc/CC-MAIN-20180521004325-20180521024325-00608.warc.gz | 189,899,536 | 10,739 | 1. ## scientific notation
0.008/0.00002
2. ## Re: scientific notation
Originally Posted by Louisana1
0.008/0.00002
$\displaystyle \frac{8\cdot 10^{-3}}{2\cdot 10^{-5}}=~?$
3. ## Re: scientific notation
4 x 10^2 right????????????
4. ## Re: scientific notation
Originally Posted by Louisana1
4 x 10^2 right????????????
Right.
5. ## Re: scientific notation
750000x20000x3000
6. ## Re: scientific notation
$\displaystyle (7.5 \times 10^5)(2 \times 10^4)(3 \times 10^3) = ?$
7. ## Re: scientific notation
45 x 10^12 or 4.5 x 10^12
8. ## Re: scientific notation
ok my teacher has an answer of 4.5 x 10^13 HOW did he get that
9. ## Re: scientific notation
$\displaystyle 45 \times 10^{12} = 4.5 \times 10^1 \times 10^{12} = 4.5 \times 10^{13}$
10. ## Re: scientific notation
Originally Posted by Louisana1
45 x 10^12 or 4.5 x 10^12
Originally Posted by Louisana1
ok my teacher has an answer of 4.5 x 10^13 HOW did he get that
Well $\displaystyle 45\cdot 10^{12}=4.5\cdot 10^{13}~!$
11. ## Re: scientific notation
So when you place the decimal between the 4.5 the exponent goes up one
12. ## Re: scientific notation
Originally Posted by Louisana1
So when you place the decimal between the 4.5 the exponent goes up one
I would hope that you learned long ago that you could multiply or divide by powers of 10 by moving the decimal point. Remember that? What is 1023.13 times 10? What is 1023.13 divided by 10? | 472 | 1,422 | {"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} | 3.796875 | 4 | CC-MAIN-2018-22 | latest | en | 0.758518 |
https://encyclopedia2.thefreedictionary.com/Mathematical+Statistics | 1,571,225,931,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00161.warc.gz | 471,950,144 | 23,314 | # Mathematical Statistics
Also found in: Acronyms, Wikipedia.
## Statistics, Mathematical
the branch of mathematics devoted to the mathematical methods for the systematization, analysis, and use of statistical data for the drawing of scientific and practical inferences. Here, the term “statistical data” denotes information about the number of objects in some more or less general set which possess certain attributes (for example, the data in Tables la and 2a).
Field of application and method of mathematical statistics. The statistical description of a set of objects occupies an intermediate place between the individual description of each object in the set, on the one hand, and the description of the set in terms of its general properties, which does not in any way require its separation into distinct objects, on the other. Compared to the first method, statistical data are always deprived of individuality to a greater or lesser degree and have only limited value in those cases where individual data are essential (for example, a new teacher on the first day of school has only a superficial knowledge about his class obtained from some statistics presented to him by his predecessor about the number of excellent, good, satisfactory, and unsatisfactory grades). On the other hand, in comparison with data about the externally observable, summary properties of the set, statistical data make it possible to penetrate more deeply into the nature of the matter. For example, data obtained from a granulometric analysis of a rock (that is, data on the distribution of the particles forming the rock according to size) provide additional valuable information in comparison with an examination of a sample of the rock as a whole, and thus to a certain degree help elucidate the properties of the rock, the conditions of its formation, and other factors.
The method of investigation that is based on a consideration of statistical data obtained from sets of objects is termed the statistical method. The statistical method is applicable in extremely diverse fields. However, its features as applied to objects of different natures are so unique that it would be senseless to combine, for instance, socioeconomic statistics, physical statistics, and stellar statistics into a single science.
The features of the statistical method that are common to different fields include the determination of the number of objects belonging to certain groups, consideration of the distribution of quantitative attributes, application of a sampling method (in those cases where a detailed study of all the objects in an extensive set is too difficult), and use of probability theory for estimating a sufficient number of observations to reach some conclusion. This is the formal, mathematical side of statistical methods of research, which ignores the specific nature of the objects under study and constitutes the subject of mathematical statistics.
Mathematical statistics and probability theory. The connection between mathematical statistics and probability theory varies from one case to the next. Probability theory studies not just any phenomena but only random and even “probabilistically random” phenomena, that is, those for which it makes sense to speak of corresponding probability distributions. Nevertheless, probability theory also plays a definite role in the statistical study of mass phenomena of any nature, which may or may not belong to the category of probabilistically random phenomena. This is accomplished through the theory of sampling methods and the theory of measurement errors, both based on probability theory. In these cases it is not the phenomena themselves that are subject to probabilistic laws but the methods for studying them.
Probability theory plays a more important role in the statistical investigation of probabilistic phenomena. Here, the branches of mathematical statistics based on probability theory are fully applicable, such as the theory of statistical testing of probabilistic hypotheses and the theory of statistical estimation of probability distributions and associated parameters. But the area of application of these deeper statistical methods is much narrower because of the requirement that the phenomena themselves obey sufficiently definite probabilistic laws. For example, the statistical study of the behavior of turbulent streams of water or fluctuations in radio receivers is based on the theory of stationary random processes. However, the application of this same theory
Table 1a. Distribution of diameters of machine parts obtained from statistical study of mass production1
Diameter
(mm)
Basic
sample
First
sample
Second
sample
Third
sample
13.05-13.09 …11
13.10-13.14 …2
13.15-13.19 …111
13.20-13.24 …8_
13.25-13.29 …17121
13.30-13.34 …27112
13.35-13.39 …30231
13.40-13.44 …37211
13.45-13.49 …271
13.50-13.54 …2 521
13.55-13.59 …17
13.60-13.64 …71_2
13.65-13.69 …21
Total. …200101010
. . . . .13.41613.43013.31513.385
S2. . . .2.39100.09900.14720.3602
s. . . . .0.1100.1050.1280.200
1 For explanation of and 5 see the section in this article Simplest methods of statistical description. For explanation of s, see the section Connection between statistical distributions and probability distributions.
to the analysis of economic time series may lead to serious errors, since the assumption of the longterm presence of unchanging probability distributions that enters into the definition of a stationary process is in this case generally unacceptable.
Probabilistic laws generate corresponding statistical expressions since by virtue of the law of large numbers, probabilities are realized approximately in the form of frequencies, and mathematical expectations in the form of means.
Simplest methods of statistical description. A set of n objects may be divided according to some qualitative attribute into classes A1,A2,. . ., Ar. The statistical distribution corresponding to this division is given by indicating the numbers of objects (frequencies) n1,n2. . ., nr (whereΣ r i=i ni = n) in the individual classes. Instead of the quantities n the relative frequencies hi = ni/n are often given (these obviously satisfy the relation Σrihi = 1). If some quantitative attribute is being studied, then its distribution in a set of n objects can be given by enumerating the directly observed values of the attribute: x1, x2, . . ., xn, for example, in increasing order. However, for large n this method is cumbersome and does not clearly elucidate the essential properties of the distribution. In practice, for any large n we generally do not compile detailed tables of observed values xi but proceed from tables containing only the frequencies of classes obtained upon grouping observed values into suitable selected intervals.
Table 1b. Distribution of diameters of machine parts in basic sample1 for larger grouping intervals
DiameterNumber of parts
1 See Table la
13.00-13.2411
13.25-13.49138
13.50-13.7451
Total200
For example, the first column of Table la gives the results of measurements of the diameters of 200 machine parts, grouped according to 0.05 mm intervals. The basic sample corresponds to a normal production run. The first, second, and third samples were taken during certain time intervals as a test of the stability of the normal run. Table Ib gives the results of measurements of the machine parts in the basic sample grouped according to 0.25 mm intervals. The usual grouping into 10-20 intervals, each of which contains not more than 15-20 percent of the values xi, proves to be sufficient to note more or less all the essential properties of the distribution and to compute reliably fundamental characteristics from the class frequencies (see below). A histogram constructed from such group data graphically depicts the distribution. A histogram constructed on the basis of groupings with small intervals usually has multiple peaks and does not graphically reflect the essential properties of the distribution.
Figure 1. Histogram of distribution of diameters of 200 machine parts. Length of grouping interval is 0.05 mm.
As an example, Figure 1 depicts a histogram of the distribution of the 200 diameters corresponding to the data in the first column of Table la, and Figure 2 depicts a histogram of the same distribution (the corresponding table is not provided because it is so cumbersome) with an interval of 0.01 mm. On the other hand, a grouping into excessively large intervals may lead to an unclear representation of the nature of the distribution and to gross errors in the computation of the mean and other characteristics of the distribution (see Table Ib and the corresponding histogram in Figure 3).
Figure 2. Histogram of distribution of diameters of 200 machine parts. Length of grouping interval is 0.01 mm.
Within the limits of mathematical statistics, the question of the grouping intervals may only be regarded from a formal side: the completeness of the mathematical description of the distribution, the precision of the calculation of means from the grouped data, and so forth.
Figure 3. Histogram of distribution of diameters of 200 machine parts. Length of grouping interval is 0.25 mm.
In the study of the joint distribution of two attributes, tables with two entries are used. Table 2a serves as an example of a joint distribution of two qualitative attributes. In the general case, when the set is divided into classes A1, A 2, . . . , A r according to attribute A and into classes B 1 B2, • . ., B 5 according to at-tribute B, the table consists of the numbers nij of objects simultaneously belonging to classes Ai and Bj. Summing them according to the formulas
we obtain the numbers in the classes Aj and Bj themselves; it is evident that
where n is the number of objects in the whole set. Depending on the aims of further investigation, certain relative frequencies may be calculated:
Table 2a. Distribution of illness from influenza among employees of the Central Department Store in Moscow who did and did not inhale anti-influenza serum (1939)
Did not become illBecame illTotal
Did not inhale serum1,6751501,825
Inhaled serum4974501
Total2,1721542,326
Table 2b. Relative frequency of illness among employees of the Central Department Store1
Did not become illBecame illTotal
Did not inhale serum0.9180.0821,000
Inhaled serum0.9920.0081,000
1 Corresponding to data in Table 2a
For example, in investigating the effect of inhaling serum on the number of influenza cases, using the data given in Table 2a, it is natural to compute the relative frequencies, which are given in Table 2b. Table la serves as an example of a joint distribution with mixed types of attributes: the material is grouped according to one qualitative attribute (that of belonging to the basic sample generated in order to determine the mean of the production process and to the three samples generated at different times to test the stationary of this mean) and according to one quantitative attribute (the diameter of the parts).
The simplest derived characteristics of the distribution of a single quantitative attribute are the mean
and the standard deviation
where
In computing , S2, and D from grouped data, we use the formulas
or
where r is the number of grouping intervals and the ak are their midpoints (in the case of Table la—13.07, 13.12, 13.17, 13.22, and so forth). If the material is grouped according to excessively large intervals, then such a calculation gives inaccurate results. Sometimes in such cases it is useful to resort to special corrections for grouping. However, it makes sense to introduce these corrections only when certain probabilistic assumptions are satisfied.
Connection between statistical distributions and probability distributions; estimates of parameters; tests of probabilistic hypotheses. Presented above were only certain selected, very simple methods of statistical description, a fairly broad discipline with a well-developed system of concepts and computational procedures. The methods of statistical description are interesting not in themselves but as a means of inferring from the statistical data certain conclusions about the laws that govern the phenomena under investigation and the principles that in each case lead to the observed statistical distribution.
For example, it is natural to link the data presented in Table 2a to such a theoretical scheme. The event that a particular employee contracts influenza must be considered a random event, since the general working and living conditions of the employees examined cannot determine whether this or that employee will actually become ill but only the probability that he or she will become ill. Judging from the statistical data, the probability of contracting influenza for those who inhaled the serum (p1) and for those who did not (po) are different; the data provide a basis for the assumption that p1 is actually less than Po. The problem arises in mathematical statistics of estimating the probabilitiesp1 and po from the observed frequencies h1 =4/501 - 0.008 and/z0 = 150/1825 ~ 0.082 and testing whether the statistical evidence is sufficient to consider it established that p1 < Po (that is, that the inhalation of serum actually diminishes the probability of contracting influenza). An affirmative answer to this question for the case of the data in Table 2a is sufficiently certain even without the precise methods of mathematical statistics. But in more questionable cases it is necessary to resort to special criteria that have been worked out by mathematical statistics.
The data in the first column of Table la were gathered with the aim of establishing the precision in the manufacture of machine parts with a nominal diameter of 13.40 mm during a normal production run. The simplest assumption, which in this case may be based on certain theoretical considerations, is that the diameters of the individual machine parts may be regarded as random quantities X obeying a normal probability distribution
If this assumption is correct, then the parameters a and a2—the mean and the variance of the probability distribution—may be determined to a sufficient accuracy by the corresponding characteristics of the statistical distribution (since the number of observations n = 200 is sufficiently great). As an estimate of the theoretical variance a2 it is preferable to take the unbiased estimate
s2 = s2/(n – 1)
rather than the sample variance D2 = S2 /n.
There exists no general (that is, applicable to any probability distribution) unbiased estimate for the theoretical standard deviation. As an estimate (strictly speaking, a biased one) for cr, we often use s. The accuracy of the estimates x and s for a and cr is shown by their respective variances, which in the case of a normal distribution (1) have the form
where the sign ~ denotes approximate equality for large n. In this manner, having agreed to append to the estimates their variance with the sign ±, we have for large n under the assumption of a normal distribution (1)
For the data in the first column of Table la, the formulas (2) give
σ = 13.416 ± 0.008
σ = 0.110 ± 0.006
The sample size n = 200 is sufficient to ensure the validity of using these formulas based on the theory of large samples.
Table 3. Dependence of α and ω = 1 - α on k
k1.962.583.003.29
α0.0500.0100.0030.001
ω0.0950.9900.9970.999
The use of the formulas of the theory of large samples, which have been established only as limiting formulas for n→, can serve only as a first approximation in the examination of the data in the subsequent columns of Table la, each of which is compiled on the basis of ten measurements. As before, the values x and s can be used as approximate estimates for the parameters a and or; however, in order to determine the accuracy and reliability of these estimates it is necessary to apply the theory of small samples. In comparing according to the rules of mathematical statistics the values x and s appearing in the last lines of Table la for the three samples with the normal values a and σ, as estimated by the first column of the table, we can draw the following conclusions: the first sample does not provide a basis for assuming a substantial change in the characteristics of the production process, the second sample provides a basis for concluding that the mean diameter a has decreased, and the third sample provides a basis for concluding that the variance has increased.
All rules for statistically estimating parameters and testing hypotheses, which are based on probability theory, operate only at some definite significance level ω < 1, that is, they may lead to erroneous results with a probability α = 1 ω. For example, if, under the assumption of a normal distribution and a known theoretical variance σ2, α is estimated using xē by stating that a satisfies
then the probability of error equals a, which is connected to kby the relation (see Table 3):
The question concerning a reasonable choice of significance level under given specific conditions (for example, in working out rules for the statistical control of mass production) is very important. In this case, the desire to apply only rules that have a high (close to unity) level of significance is opposed by the circumstance that when there is a limited number of observations such rules allow only very poor conclusions (for example, they do not make it possible to establish an inequality of probabilities even when there is a marked inequality in the frequencies).
Sampling method. In the preceding analysis, the results of observations that were to be used to estimate the probability distribution or its parameters were assumed (although this was not mentioned) to be independent. A well-studied example of the use of dependent observations is the estimation of the statistical distribution or its parameters in a “population” of N objects by means of a “sample” drawn from it containing n < N objects.
Terminological Note. Often, a set of n observations made for the purpose of estimating a probability distribution is also called a sample. This explains, for instance, the origin of the term “theory of small samples” used above. This terminology is connected with the fact that the probability distribution is often considered as a statistical distribution of a hypothetical infinite population, and it is conventionally considered that the n observed objects have been “selected” from this population. These ideas do not have clear meaning. In the proper sense of the word, sampling always presupposes an initial finite general set.
The following may serve as an example of the application of the sampling method. In a batch of N articles, let L be the number of defective ones; n < N objects are randomly selected (for example, n = 100 for N = 10,000) from the batch. The probability that the number l of defective articles in the sample is equal to m is
Thus, l and the corresponding relative frequency h = l/n prove to to be random variables whose distribution depends on the parameter L or, which is the same thing, on the parameter H = L/N. The problem of estimating the relative frequency Hfrom the sampled relative frequency h is similar to the problem of estimating the probability p from the relative frequency h in n independent trials. For large n with a probability close to unity, the approximate equality p ~ h holds for the problem of estimating the probability and the approximate equality H ~ h holds for the problem of estimating the relative frequency. However, for the problem of estimating H, the formulas are more complex and the deviation of h from H on the average is somewhat smaller than the deviation of h from p in the problem of estimating probability (for the same n). Thus, the estimate of the fraction H of defective articles in the batch given by the fraction h of defective articles in the sample for a given sample size n is always (for any N) somewhat more accurate than the estimate of the probability p given by the relative frequency h in independent trials. When N/n → ∞, the formulas for the sampling problem asymptotically approach the formulas for the problem of estimating the probability p.
Additional problems of mathematical statistics. The aforementioned methods of estimating parameters and testing hypotheses are based on the assumption that the number of observations necessary to attain a given accuracy in the conclusions has been determined in advance (before the trials). However, the a priori determination of the number of observations is often not expedient since by determining the number of trials during the course of the experiment rather than by fixing it in advance, we may decrease its mathematical expectation. This circumstance was first observed in the case involving the choice of one of two hypotheses by means of a sequence of independent trials. The appropriate procedure, first proposed in connection with problems of statistical quality control, consists in the following: from the results of the already conducted observations, at each step a decision is made (1) to carry out the next trial, or (2) to discontinue the trials and accept the first hypothesis, or (3) to discontinue the trials and accept the second hypothesis. By a suitable choice of quantitative characteristics, in such a procedure it is possible (with the same accuracy in the conclusions) on the average to cut in half the number of observations needed in contrast to a procedure with a sample of fixed size.
The development of the methods of sequential analysis has led to the study of controlled random processes, on the one hand, and to the emergence of a general theory of statistical decisions, on the other. The latter theory proceeds from the fact that the results of sequentially conducted observations serve as the basis for making certain decisions (intermediate ones—to continue the trials or not—and final ones—made on discontinuing the trials). In problems involving estimation of parameters, the results of the final decision are numbers (the values of the estimates), while in those involving testing hypotheses they are the accepted hypotheses. The goal of the theory is to indicate the rules for making decisions that minimize the average risk or loss (the risk depends on the probability distributions of the results of the observations, on the final decision made, on the cost of conducting the trials, and on other factors).
Questions concerning the expedient allocation of effort in conducting a statistical analysis of a phenomenon are considered in the theory of experimental design, which has become an important part of modern mathematical statistics.
The development and refinement of the general concepts of mathematical statistics have been accompanied by the development of its individual branches, such as analysis of variance, statistical analysis of random processes, and multivariate statistical analysis. New analyses have emerged in the field of regression analysis. The so-called Bayes approach plays a large role in problems of mathematical statistics.
History. The first principles of mathematical statistics appear in the works of the creators of the theory of probability—Jakob Bernoulli (late 17th century and early 18th), P. de Laplace (second half of the 18th century and early 19th), and S. Poisson (first half of the 19th century). In Russia, methods of mathematical statistics for use in demography and insurance were developed by V. Ia. Buniakovskii (1846) on the basis of probability theory. The work of the Russian classical school of probability theory in the second half of the 19th century and in the early 20th (P. L. Chebyshev, A. A. Markov, A. M. Liapunov, S. N. Bernshtein) was of decisive significance for the future development of mathematical statistics. Many questions in the theory of statistical estimations were essentially worked out on the basis of the theory of errors and the method of least squares by K. Gauss (first half of the 19th century) and A. A. Markov (end of the 19th century and beginning of the 20th). The work of L. A. Quetelet (19th century, Belgium), F. Gallon (19th century, Great Britain), and K. Pearson (end of the 19th century and beginning of the 20th, Great Britain) was of great importance, but, in terms of the degree of use of probability theory, it lagged behind the Russian school. Pearson worked extensively on the compilation of tables of functions necessary for the application of the methods of mathematical statistics.
The concepts of the Anglo-American school [Student (pseudonym of W. Gosset), R. Fisher, and E. Pearson of Great Britain and J. Neyman and A. Wald of the USA], which arose in the 1920’s, played an extremely important role in the establishment of the theory of small samples, the general theory of
Table 1. Stubble plows produced in the USSR
Disc stubble plowsShare stubble plows
LD-20LD-15LD-10LD-5PL-10-25LN-5-258PL-5-25PLS-5-25A
Number of sections or frames. . . . .16128410555
Operating width (m) . . . . .20151052.51.251.251.25
Angle of attack (degrees) . . . . .20-353515-3515-35
Productivity (hectares per hour) . . . . .187.66.542110.5
statistical estimation and hypothesis testing (freed from assumptions about the presence of a priori distributions), and sequential analysis. In the USSR important results in mathematical statistics were obtained by V. I. Romanovskii; E. E. Slutskii, the author of important works on the statistics of dependent stationary sequences; N. V. Smirnov, who laid the foundations of the theory of nonparametric methods in mathematical statistics; and Iu. V. Linnik, who enriched the analytic apparatus of mathematical statistics with new methods. Mathematical statistics has been the basis for particularly intensive development of statistical methods for the study and control of mass production and statistical methods used in physics, hydrology, climatology, stellar astronomy, biology, medicine, and other fields.
Among the journals that publish works on mathematical statistics are Annals of Statistics (known until 1973 as Annals of Mathematical Statistics), International Statistical Institute Review, Biometrika, and Journal of the Royal Statistical Society. There are scientific associations that support research in mathematical statistics and applications. An important role is played by the International Statistical Institute (ISI), centered in Amsterdam, as well as by the International Association for Statistics in Physical Sciences (IASPS), which was founded by the ISI.
### REFERENCES
Cramér, H. Matematicheskie melody statistiki. Moscow, 1948. (Translated from English.)
Van der Waerden, B. L. Matematicheskaia statistika. Moscow, 1960. (Translated from German.)
Smirnov, N. V., and I. V. Dunin-Barkovskii. Kurs teorii veroiatnostei i matematicheskoi statistiki dlia tekhnicheskikh prilozhenii, 3rd ed. Moscow, 1969.
Bol’shev, L. N., and N. V. Smirnov. Tablitsy matematicheskoi statistiki. Moscow, 1968.
Linnik, lu. V. Metod naimen ’shikh kvadratov …, 2nd ed. Moscow, 1962.
Hald, A. Matematicheskaia statistika s tekhnicheskimi prilozheniiami. Moscow, 1956. (Translated from English.)
Anderson, T. Vvedenie v mnogomernyi statisticheskii analiz. Moscow, 1963. (Translated from English.)
Kendall, M. G., and A. Stuart. Teoriia raspredelenii. Moscow, 1966. (Translated from English.)
A. N. KOLMOGOROV and IU. V. PROKHOROV
References in periodicals archive ?
Parzen, "On estimation of a probability density and mode," The Annals of Mathematical Statistics, vol.
Derriennic, "Random walks with jumps in random environments (examples of cycle and weight representations)," in Probability Theory and Mathematical Statistics: Proceedings of The 7th Vilnius Conference 1998, B.
Readers should have previous courses in mathematical statistics covering basic concepts of inference and linear models and should be familiar with matrix algebra.
Hjort teaches mathematical statistics at the University of Oslo, Norway.
Tukey, "Average values of mean squares in factorials," Annals of Mathematical Statistics, vol.
Natvig (mathematical statistics / reliability theory, U.
Focusing on the asymptotic properties of wide classes of stochastic systems as the arise in mathematical statistics, percolation theory, statistical physics and reliability theory, Bulinski and Shashkin (both mathematics, Moscow State U.) provide detailed proofs as well as auxiliary results.
Students are expected to have completed a masters-level introduction to probability and mathematical statistics. The emphasis is on the foundational underpinning of the modes, but a moderate amount of technical detail is provided in developing and critically analyzing them.
Probability and Mathematical Statistics: Theory, Applications, and Practice in R
Born in Kochi in 1946, John began his career as a lecturer in Mumbai after a post graduation in mathematical statistics. In the late 1960s he joined Lintas, Mumbai and in 1981 he made a career-defining switch to Ulka advertising agency, where he handled FMCG major Hindustan Lever's popular brands like Liril, Close-up and Fair & Lovely.
The methodological part of the project will require development of novel advanced techniques in mathematical statistics and probability theory.
(1,2) Department of Mathematical Statistics, Faculty of Science, Aleppo University, Syria, Rafif.alhabib85@gmail.com (3) Department of Mathematical Statistics, Faculty of Science, Albaath University, Syria (4) Department of Mathematics and computer science, Faculty of Science, Port said University, Egypt drsalama44@gmail.com
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https://mathstodon.xyz/@leo | 1,591,371,888,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502097.77/warc/CC-MAIN-20200605143036-20200605173036-00141.warc.gz | 435,167,104 | 9,591 | that a number $$a$$
is divisible by $$2^n$$ if the $$n^{th}$$ last digit of $$a$$ are divisible by $$2^n$$.
We note $$a_i$$ the $$i^{th}$$ digit of $$a$$. Then we get
$S = \frac{10^wa_w}{2^n} + \ldots + \frac{10^0a_0}{2^n}.$
yet, $$10^x \equiv 0\pmod{2^n} \iff x - n$$
$\ge 0 \iff x \ge n$
Then, only the $$n^{th}$$ first digit of $$a$$ need to be divisible by $$2^{n}$$.
Does anyone know a method similar to the vectorial product, which allows to create a plan from three plan? (in space) thank you !
that if the sum of the digits of a number $$a$$ is divisible by 9, then this number is divisible by 9.
Let $$\sum_{i=0}^{n}10^{i}a_i = a, a_i \in {0,...,9}$$.
When we divide it by 9, we get
$\frac{a}{9} = \frac{a_n10^n}{9} + \ldots + \frac{a_010^0}{9}$
$= \frac{a_n(10^n - 1) + a_n}{9} + \ldots + \frac{a_0}{9}$
$= a_n \times \frac{10^n - 1}{9} + \ldots + \frac{\sum_{i=0}^{n}a_i}{9}$
So since all the $$10^k - 1$$ factors are divisible, it remains the last term, our sum.
that the sum of the first $$n^{th}$$ even numbers is equal to $$n(n + 1)$$.
A odd number $$x$$ is noted $$2k$$. We are looking at $$\sum_{i=1}^{n}(2i) = S$$.
\begin{align}
S &= \sum_{i=1}^{n}(2i) \\
&= 2 \times \frac{n(n + 1)}{2}\\
&= n(n + 1).
\end{align}
that $\sum^{n}_{k=1}k^2 = \frac{n(n + 1)(2n + 1)}{6}.$
For that, we will need to imagine a regular staking of ball forming a rectangular pyramid. The first floor will contain $$n^2$$ ball, etc.
Then, instead of counting the ball vertically, we will count then horizontally, witch give us:
$\sum^{n}_{k=1}k^2 = n \times \frac{n(n + 1)}{2} - \sum^{n - 1}_{k=1}\frac{k(k + 1)}{2}.$
(only 500 caracters...)
$=\frac{2n^3 + 3n^2 + n}{6}.$
And after factoring, we have our formula.
that the derivate of $$\sqrt{u(x)}' = \frac{u(x)'}{2\sqrt{u(x)}}$$.
$\sqrt{u(a)}' = \lim_{h \to 0} \frac{\sqrt{u(a + h)} - \sqrt{u(a)}}{h}$
$= \lim_{h \to 0} \frac{u(a + h) - u(a)}{h \times (\sqrt{u(a + h)} + \sqrt{u(a)})}.$
Then we isolate $$u(a)'$$:
$\sqrt{u(a)}' = \lim_{h \to 0} \frac{u(a)'}{\sqrt{u(a + h)} + \sqrt{u(a)}}$
$= \frac{u(a)'}{2\sqrt{u(a)}}.$
e.g.: The derivate of $$f(x) = \sqrt{1 - x^2}$$ is $$f'(x) = -\frac{x}{\sqrt{1 - x^2}}$$.
that the equation $$f(x)=ax^2+bx+c$$ is always tangent to $$g(x)=-ax^2+bx+c$$ with $$(a,c,b) \in \mathbb{R}^3$$.
We get:
$$ax^2+bx+c=-ax^2+bx+c$$
$$2ax^2=0$$
Since b and c are canceled, the discriminant equals 0. There is only one solution, i.e. only one point I of intersection.
$$x_I = 0; y_I = f(0) = c$$.
that the perpendicular to the line of equation $$y = mx$$ is $$y = -\frac{1}{m}x$$.
The angle between the abcsissa axis is $$\theta = \arctan(m)$$.
So we are looking at a slope $$m'$$ (where $$\alpha = \arctan(m')$$), such that $$\theta - \alpha = \frac{\pi}{2}$$. Indeed, we will use radians.
By rearranging, we are getting,
$m' = \tan(\arctan(m) - \frac{\pi}{2})$
$\iff m' = -\frac{1}{m}$
because $$\tan = \frac{sin}{cos}$$.
Finally, the slope $$m'$$ is $$-\frac{1}{m}$$.
that there are infinitely many solutions to the equation $$xy = x + y$$.
i.e. $1 = \frac{1}{x} + \frac{1}{y}$
Let $$x = c/a$$ and $$y = c/b$$. We have $$1 = \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$$
$$\iff c = a + b$$.
So $$xy = x + y$$ is satisfied when $$x = c/a$$, $$y = c/b$$ where $$c = a + b$$, for any value of a and b.
e.g.: $$\frac{37}{15} + \frac{37}{22} = \frac{37}{15} \times \frac{37}{22} = \frac{1369}{330}$$.
that the sum of the `$$n^{th}$$ first odd numbers is equal to $$n^2$$.
A odd number $$x$$ can be noted $$2k-1$$. We are looking at $$\sum_{i=1}^{n}(2i - 1)$$.
And we have $$\sum_{i=1}^{n}(-1) = -n$$.
So $$\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n$$
$$= 2 \times \frac{n^2 + n}{2} - n = n^2$$.
that every natural number minus its palindrome is divisble by 9. Let $$\sum_{i=1}^{n}10^{i-1}a_i = a, a_i \in {0,...,9}$$.
Its palindrome is $$\sum_{i=1}^{n}10^{n-(i-1)}a_i = \overline{a}$$.
Thus $$a - \overline{a} = \sum_{i=1}^{n}a_i(10^{i-1} - 10^{n-(i-1)})$$, or we know that $$\forall a,b \in \mathbb{N}, 10^a - 10^b \equiv 0\pmod 9$$.
Consequently, $$a - \overline{a} \equiv 0\pmod 9$$.
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https://issuu.com/gregg.meyer/docs/dynamics | 1,495,860,992,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608773.42/warc/CC-MAIN-20170527040600-20170527060600-00400.warc.gz | 944,731,067 | 17,251 | Dynamics Problem Solving Quick Reference Guide
I. Newton ENGR 212 Portland Community College Spring 2011
Write a title page summary of the subject of Dynamics . Be sure to distinguish between kinematics, particle kinetics, and rigid body motion. All work must be YOUR words (ie, not copied from the text or internet).
Variable Acceleration Overview: When an object’s acceleration varies with time, displacement, or velocity, it must be derived mathematically using calculus. When it is time dependent, acceleration a(t) = dv/dt, or we say “acceleration is the instantaneous rate of change of its velocity with respect to time”. In other cases, acceleration may be based on empirically collected video data showing the change with respect to displacement where a(x) = v dv/dx. In all such cases, Newton’s laws of constant acceleration are not valid. Practical Example: A good example of non-constant acceleration can be found at a drag race strip. If one were to use Newton’s equation x(t) = xo + vot + 1/2at2, this would yield an “average acceleration” of approximately 4.5 g. However, the true acceleration varies significantly over the ¼ mile traveled. For a Top Fuel wheel-driven dragster, maximum acceleration occurs at the start and diminishes with distance due to increasing air resistance. A rocket-driven dragster, however, takes time for the turbine to “spool up” and actually increases in acceleration throughout the race. In both cases, peak acceleration may reach double that of average acceleration. Thus, it is important to distinguish between variable and constant acceleration to ensure the car’s strength can handle the true loads.
Solution Tips: 1. Identify first whether the problem is set up as a function of time or some other variable such as displacement or velocity. 2. If it is a time-dependent function, begin with a(t) = dv/dt, multiply both sides by “dt” and integrate each side using boundary conditions given in the problem statement to determine the upper and lower integral limits of velocity and time. 3. If it is a time independent problem, use a(x) = v dv/dx to arrive at ! !" = ! !". Again, use the parameters in the problem statement to ascertain the limits of x and v. Limitations: In reality, determining the true acceleration is limited by the available data. Any fieldcollected data is subject to the accuracy of the instrumentation used (such as cameras or accelerometers) as well as to the user’s ability read and log the data correctly. When at all possible, it is important to repeat the experimentation to ensure the results are reliable.
Projectile Motion
Problem 11.112
Newton’s 2nd Law
Problem 12.51
Newton’s Law of Gravitation
Problem 12.81
Work Energy
Problem 13.17
Impulsive Force
Problem 13.144
Coefficient of Restitution and Conservation of Momentum
Problem 13.174
Conservation of System Momentum
Problem 14.3
Rigid Body Kinematics
Problem 15.58
Planar Kinetic Motion
Problem 17.106 (Hibbler; solution to be provided)
Dynamics Quick Ref Guide
This is a summary of how to solve several types of dynamics problems | 700 | 3,124 | {"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} | 4.4375 | 4 | CC-MAIN-2017-22 | latest | en | 0.891723 |
http://mathoverflow.net/questions/58304/selberg-sieve-on-a-certain-set?sort=oldest | 1,469,265,504,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00157-ip-10-185-27-174.ec2.internal.warc.gz | 155,755,862 | 16,599 | # Selberg sieve on a certain Set.
I'm new to sieve theory, and I'm trying desperately to understand Selberg's sieve. I would like to apply the sieve to give me a nice upper bound on primes of the set $$A^D(N)= \{ Dq-2 : q\in P, N/2 < q \leq N \}$$ But basically, for a fixed N, I would like $A^D(N)$ to be the set of elements of the form $Dq-2$ for a fixed positive integer $D$ and letting $p$ run through all primes between $N/2$ and $N$. Now, as I said, I'm trying to apply Selberg's sieve, but as I don't really know what I'm doing, I'm a bit confused. Now, if I understand it correctly could I then say that $$S(A^D(N),N/2,N/2) \leq \frac{\pi(N)-\pi(N/2)}{L_p(z)}+O \Big( \frac{z^2}{L_p(x)^2} \Big)$$ Where $S(A^D(N),N/2,N/2)$ is the number of elements of $A^D(N)$ which are prime and $$L_p(z)=\sum_{n\leq z}^{n\vert P} \frac{\mu(n)^2}{\phi(n)}.$$ where $$\frac{1}{\phi(n)}=\frac{1}{n}\prod_{p\vert n} \frac{1}{1-1/p}.$$ I got this from a pater called "Sketch of the Selberg Sieve method" By Sean Prendville (January, 2008) where he describes the Selberg Sieve not on $A^D(N)$ but on the set of all integers between some positive integer $x$, and $x+y$. I'm sure some of it is wrong, or that I totally misinterpreted, but I would like to know if this is right, and if it is, where do I go from here? (especially with dealing with $L_p$). I appreciate any help, but please keep in mind that I'm sixteen years old and live in the Bronx. As dumbed down as possible would be greatly appreciated..this is all new to me. Much appreciated, Alexis D. Botros
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I can't help because (regarding sieve theory and other topics) I am pretty dumb too. My county library participates in a inter-library loan program; with that I could borrow Murty and Cojocaru's book on sieve methods. I think you might find it just barely accessible. If you can't find it online or available, write a response comment and I will see if there is something else that might help. Gerhard "Dumbing Down Is Our Pastime" Paseman, 2011.03.12 – Gerhard Paseman Mar 13 '11 at 5:05
Thanks. I appreciate it...I found it online. Will see what comes of it – Alex Botros Mar 13 '11 at 5:10
I suppose the real problem comes from the fact that $\phi(n)$ is used in the asymptotic approximation of the number integers in $(x, x+y]$ that are relatively prime to a certain integer. Therefore, if we switch from talking about consecutive integers, to integers in $A^D(n)$, then we must find a new asymptotic approximation to those elements of $A^D(n)$ relatively prime to a given integer. How in the heck is that going to work? How can we possibly estimate the number of integers in $A^D(n)$ that are relatively prime to a certain other integer? – Alex Botros Mar 13 '11 at 5:17
What does living in the Bronx have to do with anything? – Gerry Myerson Mar 13 '11 at 5:37
Alex: What Gerry's comment means is that your age (which you gave) helps people on this website know a suitable level at which to pitch an answer to you even though your age has nothing to do with math, but your location (which you also gave) doesn't really assist anyone in the same way. As an option other than Math Overflow, since you are in New York you could try to speak with someone in person at one of the math departments in New York, such as Columbia or the CUNY Graduate Center. – KConrad Mar 13 '11 at 7:41
Getting an upper bound here is, at the level of research, a simple exercise. To do it the quickest way with least background, I'd suggest using the large sieve. Here you are essentially looking at estimating the number of integers between $N/2$ and $N$ which, modulo primes $r\leq \sqrt{N}$, are neither $0$ nor $2\bar{D}$, where $\bar{D}$ is the inverse of $D$ modulo $r$ (and the second condition drops if $r$ divides $D$). The large sieve inequality tells you that this number is bounded by $N/J$ where $J=\sum_{n\leq \sqrt{N}}\mu(n)^2\prod_{r\mid n}{2/(r-2)}$, with $2/(r-2)$ replaced by $1/(r-1)$ when there is a single condition modulo $r$ ($r$ is again restricted to primes). Getting a lower bound for $J$ is again standard number theory, but not obvious of course at first sight. One gets $J\gg (\log N)^2$ (the key reason behind the exponent $2$ is that there are two excluded classes modulo each prime $r$), and therefore the number you want to estimate is $\ll N/(\log N)^2$. It is also fairly easy to obtain a result uniform in terms of $D$ along these lines. | 1,252 | 4,416 | {"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} | 3.515625 | 4 | CC-MAIN-2016-30 | latest | en | 0.935785 |
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### Tank Volume Calculator
Total volume of a cylinder shaped tank is the area, A, of the circular end times the length, l. A = r 2 where r is the radius which is equal to 1/2 the diameter or d/2. Therefore: V(tank) = r 2 l Calculate the filled volume of a horizontal cylinder tank by first finding the area, A, of a circular segment and multiplying it by the length, l.Horizontal Cylindrical Tank Volume and Level CalculatorVolume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.Tank Volume Calculator - Oil TanksMar 26, 2015 · Lets look at how to calculate the volume of both of these tanks using tank capacity calculators or a cylindrical tank calculator. In the case of the horizontal cylindrical tank, you need to calculate the area of a cross-section of the tank and then multiply this figure by the total length of the tank.
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### Math Forum - Ask Dr. Math Archives: Volume of a Tank
Finding the Volume of a Tank, a selection of answers from the Dr. Math archives. Volume of a Cylinder What is the volume of the storage tank with a diameter 6m, height 5m? Volume of Liquid in a Cylindrical Tank How can I calculate volume of liquid in a cylinder that's not full and lying horizontally? Units and Cylinder VolumeHorizontal Tank Volume Calculations - HagraHorizontal Cylindrical Tank Volume Calculator. Horizontal Oval Tank Volume Calculator. Disclaimer. The calculations on these pages are a purely theoretical exercise! Therefore the outcomes of the calculations on these pages can only be used for indicative purposes. It might help you to estimate the content of a tank.Tank Volume Calculator - www.specialtytankandweldingSpecialty Tank & Welding Company is dedicated to bringing you tanks of the highest quality, with the most economical and competitive pricing, at the quickest convenience to you. STW specializes in custom builds and tanks for specific situations and conditions.
### Calculation of Liquid Volume in a Horizontal Cylindrical , estonia horizontal cylindrical tank fire volume
Calculation of Liquid Volume in a Horizontal Cylindrical Container: This calculator calculates the volume of liquid inside a horizontal cylindrical container at any given height of liquid. The other required dimensions are the diameter and length of the tank. Values to be Entered Values to be Calculated; Diameter of CylinderMath Forum - Ask Dr. Math Archives: Volume of a TankFinding the Volume of a Tank, a selection of answers from the Dr. Math archives. Volume of a Cylinder What is the volume of the storage tank with a diameter 6m, height 5m? Volume of Liquid in a Cylindrical Tank How can I calculate volume of liquid in a cylinder that's not full and lying horizontally? Units and Cylinder VolumeTank Calibration Chart Calculator - ODay EquipmentFrom fiberglass to steel tanks, we have a tank calculator to help you with your petroleum needs. , estonia horizontal cylindrical tank fire volume Tank Calibration Chart Calculator; Steel Tanks. Entering information in the following forms create a tank chart for metal rectangular and flat-end cylindrical liquid storage tanks. Horizontal Liquid Storage Tank Charts. Ft. In. Length: Diameter:
### Volume in horizontal round tanks??? - Excel Help Forum
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### UL 142 Aboveground Flammable Liquid Tanks
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### Calculation of Height of Liquid in a Horizontal Container , estonia horizontal cylindrical tank fire volume
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### calculus - Volume of a horizontal cylinder using height of , estonia horizontal cylindrical tank fire volume
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Author Topic: Tilt calculation (Read 995 times)
sharperstill
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« on: July 05, 2013, 02:26:02 AM » Reply
Hi all,
I normally don't consider myself that stupid but I'm having trouble understanding what I should be measuring in relation to this comment "To use, simply measure or estimate the perpendicular distance from the lens
So, perpendicular to the lens axis I assume? To the plane of focus? A plane, by definition, take up linear space so where on that plane do I measure or estimate to?
I assume this measurement is done with the lens' tilt mechanism zeroed.
Can someone help a blind man see?
Jon
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BartvanderWolf
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Posts: 4664
Can someone help a blind man see?
Hi Jon,
http://www.northlight-images.co.uk/article_pages/using_tilt.html#
Cheers,
Bart
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== If you do what you did, you'll get what you got. ==
sharperstill
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Posts: 67
OK,
So the measurement is taken perpendicular to the lens axis to the desired plane of focus.
And the hinge line is perpendicular to the plane of focus.
« Last Edit: July 05, 2013, 07:57:19 AM by sharperstill » Logged
Petrus
Sr. Member
Offline
Posts: 661
Yes, the plane of focus, a plane perpendicular to the lens rear nodal point and the plane of the sensor/film converge always on a single line. With a non-tilt lens this is at infinity (in other words they do not converge), with a tilt lens this line gets closer to the camera and causes the plane of focus to tilt.
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michael
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# RC Circtuit with Expulsion = 12.0 votes
Consider an RC circuit with epsilon = 12.0 V, R = 175 ohms, and C = 55.7 micro-F. Find (a) the time constant for the circuit, (b) the maximum charge on the capacitor, and (c) the initial current in the circuit.
#### Solution Preview
The time required to charge a capacitor to 63 percent (actually 63.2 percent) of full charge or to discharge it to 37 percent (actually 36.8 percent) of its initial voltage is known as the TIME CONSTANT (TC) of the circuit.
The value of the time ...
#### Solution Summary
The value of the time constant in seconds is equal to the product of the circuit resistance in ohms and the circuit capacitance in farads.
\$2.19 | 186 | 724 | {"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-2017-09 | latest | en | 0.893651 |
http://www.calculators4mortgages.com/articles/mortgage-rate-calculator-and-interest-rate | 1,529,728,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00320.warc.gz | 374,698,944 | 9,554 | # Mortgage Interest Rates & Mortgage Rate Calculator Basics
Hard to believe, but the mortgage crisis could be much worse. One primary factor that has kept the crisis from escalating even further is that home loan interest rates have remained low due to actions by the Fed. If rates were higher, that would cause buying to slow down as the combination of falling home prices and high mortgage rates would keep more people out of the market. However, a low rate is not guaranteed for the long run (we are currently at historical lows).
For homeowners holding an adjustable rate mortgage (ARM) home loan, an increase in mortgage rates means higher mortgage payments. This on top of the pain of higher food and energy prices - and just about everything else makes life more difficult. For would-be home buyers, an increase in interest rates could make a home purchase unobtainable. So how do rises in interest rates affect your monthly mortgage payment?
### A .25% Change in Interest Can Make a Big Difference in Mortgage Payments
Obviously, the affect of an increase or decrease in mortgage interest rates depends upon the amount of the loan. When you take your mortgage rate calculator
and look at a \$300,000 thirty year fixed rate mortgage of 6%, you find that your payment for principle and interest is \$1,799 per month. An increase in interest to 6.25%, makes the mortgage payment on the same loan go to \$1,847 which is an additional \$48 dollars per month. On a \$600,000 loan amount the mortgage calculator shows a big difference in monthly mortgage expense:
\$600,000 @ 6% = \$3,597
\$600,000 @ 6.25% = \$3,694
That addition of nearly \$100 per month can be huge for those struggling to make ends meet or trying to qualify for a new home loan--it takes about a \$300 per month income increase to offset a \$100 increase in monthly expenses for qualifying purposes. Those whose financial plans include a refinance may have to add a backup strategy to avoid becoming "house poor" in the future if adjustable rates rise and they can't qualify for a fixed rate loan.
And for those who are currently in an interest-only loan (a very popular mortgage during the housing boom) if you want to refinance into a fixed rate traditional loan, be prepared for your monthly payment to go up since you will now be paying principal in addition to interest. Depending on your situation, it may be better to bite the bullet and refinance your mortgage now, before you are forced to start paying principal when your interest only payment period is up.
Recently the interest rate on many adjustable rate mortgages actually went down which lowered homeowner's monthly mortgage payments. This is because the Federal Reserve lowered interest rates and adjustable rates often shadow Fed rate changes pretty closely. Interest rates, both short and long-term are expected to move up given current economic expectations.
If you do want an adjustable rate mortgage loan, look at one with a stable index such as the CMT or the COSI and one that will not adjust for 5 years or more to give you time to maneuver into a fixed rate mortgage if interest rates and market conditions are good. Also, you may want to consider refinancing your adjustable rate mortgage into a fixed rate note before interest rates move higher.
For those who have jumbo mortgages over \$417,000, now may be the time to refinance as the conforming loan limits have been raised and will stay in effect until the end of this year. Remember, before you do anything, follow the FDIC advice posted at their website: “Different mortgage lenders may quote you different prices, so you should contact several mortgage lenders to make sure you’re getting the best price.” You can start by contacting lenders in our database for mortgage rates and quotes here.
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### Description
Your students will adore these 21 Valentines Day Color By Number worksheets while learning and reviewing important skills at the same time! You will love the no prep, print and go ease of these printables. As always, answer keys are included. Valentine's Day Math: Valentine's Day Fun! Valentine's Day Bundle for Multiplication Facts - Color Your Answers Printables for an ENTIRE MONTH for Valentine Subtraction, perfect for Valentine's Day Math in your classroom.
TWENTY-ONE No Prep Printables that can be used for your math center, small group, RTI pull out, seat work, substitute days or homework.
This Valentine's Day Math Bundle Includes:
These high interest black and white printables are great for seat work, homework or small group work.
Common Core State Standards
CCSS.MATH.CONTENT.3.OA.A.1
Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
CCSS.MATH.CONTENT.3.OA.A.3
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
CCSS.MATH.CONTENT.3.OA.B.5
Apply properties of operations as strategies to multiply and divide. Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
CCSS.MATH.CONTENT.3.OA.B.6
Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.
CCSS.MATH.CONTENT.3.OA.B.7
Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
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### Standards
to see state-specific standards (only available in the US).
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.
Apply properties of operations as strategies to multiply and divide. Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. | 1,236 | 4,606 | {"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} | 4.0625 | 4 | CC-MAIN-2020-50 | latest | en | 0.837869 |
https://modeducation.com/class-6-maths-chapter-3-exercise-3-4/ | 1,726,456,009,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00752.warc.gz | 373,238,265 | 25,280 | # NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.4
NCERT Solutions for Class 6 Maths Chapter 3: NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.4
## NCERT Solutions for Class 6 Maths Chapter 3 Whole Numbers Exercise 3.4
1. Find the common factors of:
(A) 20 and 28
(B) 15 and 25
(C) 35 and 50
(D) 56 and 120
Solution:
(A) Factors of 20=1,2,4,5,10,20
Factors of 28=1,2,4,7,14,28
Common factors =1,2,4
(B) Factors of 15=1,3,5,15
Factors of 25=1,5,25
Common factors =1,5
(C) Factors of 35=1,5,7,35
Factors of 50=1,2,5,10,25,50
Common factors =1,5
(D) Factors of 56=1,2,4,7,8,14,28,56
Factors of 120=1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120
Common factors =1,2,4,8
2. Find the common factors of:
(A) 4, 8 and 12
(B) 5, 15 and 25
Solution:
(A) 4,8,12
Factors of 4=1,2,4
Factors of 8=1,2,4,8
Factors of 12=1,2,3,4,6,12
Common factors =1,2,4
(B) 5,15, and 25
Factors of 5=1,5
Factors of 15=1,3,5,15
Factors of 25=1,5,25
Common factors =1,5
3. Find the first three common multiples of:
(A) 6 and 8
(B) 12 and 18
Solution:
(A) 6 and 8
Multiple of 6=6,12,18,24,30
Multiple of 8=8,16,24,32
3 common multiples =24,48,72
(B) 12 and 18
Multiples of 12=12,24,36,78
Multiples of 18=18,36,54,72
3 common multiples = 36,72,108
4. Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution:
Multiples of 3=3,6,9,12,15…
Multiples of 4=4,8,12,16,20…
Common multiples =12,24,36,48,60,72,84,96
5. Which of the following numbers are co-prime?
(A) 18 and 35
(B) 15 and 37
(C) 30 and 415
(D) 17 and 68
(E) 216 and 215
(F) 81 and 16
Solution:
(A) Factors of 18=1,2,3,6,9,18
Factors of 35=1,5,7,35
Common factor =1
Therefore, the given two numbers are co-prime
(B) Factors of 15=1,3,5,15
Factors of 37=1,37Common factors =1
Therefore, the given two numbers are co-prime.
(C) Factors of 30=1,2,3,5,6,10,15,30
Factors of 415=1,5,83,415
Common factors =1,5
As these numbers have a common factor other than 1, the given two numbers are not co-prime.
(D) Factors of 17=1,17
Factors of 68=1,2,4,17,34,68
Common factors =1,17
As these numbers have a common factor other than 1, the given two numbers are not co-prime.
(E) Factors of 216=1,2,3,4,6,8,9,12,18,24,27,36,54,72,108,216
Factors of 215=1,5,43,215
Common factors =1
Therefore, the given two numbers are co-prime.
(F) Factors of 81=1,3,9,27,81
Factors of 16=1,2,4,8,16
Common factors =1
Therefore, the given two numbers are co-prime
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Welcome to Mod Education! I'm RK Yadav, an educator passionate about teaching Mathematics and Science. With a teaching journey spanning back to 2012, I've had the privilege of guiding students through the intricate realms of these subjects. | 1,074 | 2,697 | {"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} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.813161 |
http://www.exampleproblems.com/wiki/index.php/Space | 1,545,012,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00538.warc.gz | 355,111,892 | 12,176 | # Space
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Wikiquote has a collection of quotations related to:
Attempting to understand the nature of space has always been a prime occupation for philosophers and scientists. Perhaps as a result of this considerable discussion, it is difficult to provide an uncontroversial and clear definition of the nature of space, except its physical definition (see below). This article looks at the way space is dealt with variously by physicists, mathematicians and philosophers, and at the relation between space and the mind.
## Physics and space
Space is one of the few fundamental quantities in physics meaning it can't be defined via other quantities because there is nothing more fundamental known at present. Thus, similar to the definition of other fundamental quantities (like time and mass), space is defined via measurement. Currently, the standard space interval, called a standard meter or simply meter, is defined as the distance traveled by light in a vacuum during a time interval of 1/299 792 458 of a second (exact).
In classical physics, space is a three-dimensional Euclidean space where any position can be described using three coordinates. Relativistic physics examines spacetime rather than space; spacetime is modeled as a four-dimensional manifold, and currently, there are theories that can support even eleven-dimensional spaces.
Before Einstein's work on relativistic physics, time and space were seen as independent dimensions. Einstein's work unified the two into spacetime. In spacetime, measurements of space and time are held to be relative to velocity.
### Measurement
Main article: Measurement
The measurement of physical space has long been important. Geometry, the name given to the branch of mathematics which measures spatial relations, was popularised by the ancient Greeks, although earlier societies had developed measuring systems. The International System of Units, (SI), is now the most common system of units used in the measuring of space, and is almost universally used within science.
Geography is the branch of science concerned with identifying and describing the Earth, utilising spatial awareness to try and understand why things exist in specific locations. Cartography is the mapping of spaces to allow better navigation, for visualisation purposes and to act as a locational device. Astronomy is the science involved with the observation, explanation and measuring of objects in outer space.
## Astronomy and space
Main article: Outer space
In astronomy, space refers collectively to the relatively empty parts of the universe. Any area outside the atmospheres of any celestial body can be considered 'space'. Although space is certainly spacious, it is now known to be far from empty, and filled with a tenuous plasma. In particular, the boundary between space and Earth's atmosphere is conventionally set at the Karman line.
## Mathematics and space
In mathematics, a space is a set, with some particular properties and usually some additional structure. It is not a formally defined concept as such, but a generic name for a number of similar concepts, most of which generalize some abstract properties of the physical concept of space.
In particular, a vector space and specifically a Euclidean space can be seen as generalizations of the concept of a Euclidean coordinate system. Important varieties of vector spaces with more imposed structure include Banach space and Hilbert space. Distance measurement is abstracted as the concept of metric space and volume measurement leads to the concept of measure space.
As far as the concept of dimension is defined, this need not be 3: it can also be 0 (a point), 1 (a line), 2 (a plane), more than 3, and with some definitions, a non-integer value. Mathematicians often study general structures that hold regardless of the number of dimensions.
Kinds of mathematical spaces include:
## The philosophy of space
Main article: Philosophy of space and time
Space has a range of definitions.
• One view of space is that it is part of the fundamental structure of the universe, a set of dimensions in which objects are separated and located, have size and shape, and through which they can move.
• A contrasting view is that space is part of a fundamental abstract mathematical conceptual framework (together with time and number) within which we compare and quantify the distance between objects, their sizes, their shapes, and their speeds. In this view space does not refer to any kind of entity that is a "container" that objects "move through".
These opposing views are relevant also to definitions of time. Space is typically described as having three dimensions, and that three numbers are needed to specify the size of any object and/or its location with respect to another location. Modern physics does not treat space and time as independent dimensions, but treats both as features of spacetime – a conception that challenges intuitive notions of distance and time.
An issue of philosophical debate is whether space is an ontological entity itself, or simply a conceptual framework we need to think (and talk) about the world. Another way to frame this is to ask, "Can space itself be measured, or is space part of the measurement system?" The same debate applies also to time, and an important formulation in both areas was given by Immanuel Kant.
In his Critique of Pure Reason, Kant described space as an a priori notion that (together with other a priori notions such as time) allows us to comprehend sense experience. With Kant, neither space nor time are conceived as substances, but rather both are elements of a systematic framework we use to structure our experience. Spatial measurements are used to quantify how far apart objects are, and temporal measurements are used to quantify how far apart events occur.
Similar philosophical questions concerning space include: Is space absolute or purely relational? Does space have one correct geometry, or is the geometry of space just a convention? Historical positions in these debates have been taken by Isaac Newton (space is absolute), Gottfried Leibniz (space is relational), and Henri Poincaré (spatial geometry is a convention). Two important thought-experiments connected with these questions are: Newton's bucket argument and Poincaré's sphere-world.
## The psychology of space
The way in which space is perceived is an area which psychologists first began to study in the middle of the 19th century, and it is now thought by those concerned with such studies to be a distinct branch within psychology. Psychologists analysing the perception of space are concerned with how recognition of an object's physical appearance or its interactions are perceived.
Other, more specialised topics studied include amodal perception and object permanence. The perception of surroundings is important due to its necessary relevance to survival, especially with regards to hunting and self preservation. "Veridical perception" is the term used to describe the processing of the information provided by the sensory organs to an extent whereby it allows interaction with the actuality of that perceived.
It is worth noting that the way we perceive space may not necessarily be representative of the actuality of space.
### Anxiety and space
Space can also cause anxiety in people, with agoraphobia manifesting itself in some people as a fear of open spaces, and claustrophobia being the fear of enclosed spaces. Astrophobia is the fear of celestial space, Kenophobia is the fear of empty spaces and spacephobia is the fear of outer space.
### Personal space
Main article: Personal space
The term personal space refers to the amount of space a person likes to maintain between their own person and that of other people.
## Use of space
The definition of physical space in relation to ownership, in which space is seen as property, has long been an important issue. Whilst some cultures assert the rights of the individual in terms of ownership, other cultures will identify with a communal approach to land ownership. Spatial planning is a method of regulating the use of space at land-level, with decisions made at regional, national and international levels. Space can also impact on human and cultural behaviour, being an important factor in architecture, where it will impact on the design of buildings and structures, and on farming.
Ownership of space is not restricted to land. Ownership of Airspace and of waters is decided internationally.
Public space is a term used to define areas of land which are open to all, whilst private property is that area of land owned by an individual or company, for their own use and pleasure.
## Reference
Space perception. Encyclopædia Britannica from Encyclopædia Britannica Online. Accessed June 12, 2005.da:Rum de:Raum es:Espacio eo:Spaco fa:فضا fr:Espace gl:Espazo ko:공간 io:Spaco ia:Spatio it:Spazio he:מרחב nl:Ruimte ja:空間 pl:Przestrzeń pt:Espaço ru:Пространство simple:Space fi:Avaruus sv:Rymden zh:空间 | 1,801 | 9,079 | {"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-2018-51 | latest | en | 0.959268 |
http://math.stackexchange.com/questions/232424/common-sense-in-mathematics | 1,469,323,763,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00082-ip-10-185-27-174.ec2.internal.warc.gz | 163,306,150 | 20,638 | # Common sense in mathematics
Are there any claims and counterclaims to mathematics being in some certain cases a result of common sense thinking? Or can some mathematical results be figured out using just pure common sense i.e. no mathematical methods?
I'd also appreciate any mentions relating to sciences, social sciences or ordinary life.
-
What do you mean by "mathematical method"? Does it include, e.g., finger counting? Where do "mathematical results" start for you? – Gregor Bruns Nov 7 '12 at 21:17
This link on numeracy (or innumeracy!) may be of interest to you. – amWhy Nov 7 '12 at 22:03
Define common sense. – Rudy the Reindeer Nov 8 '12 at 10:47
"Common sense" in mathematics is not very common.
Many things seem very anti-intuitive, at least until you train your intuition properly. The untrained intuition is lost when dealing with, for example, infinite sets, or geometry in more than $3$ dimensions. However, one example of "common sense" that does come to mind is the Pigeonhole Principle in combinatorics.
-
The pigeonhole principle is a good example of common sense in math, but applying it is not always obvious. For example, if $A$ is a set with $n + 1$ integers, it is far from obvious to show that it is always possible to choose two numbers, $a$ and $b$, from $A$ such that $a - b$ is divisible by $n$ using the pigeonhole principle. – glebovg Nov 7 '12 at 22:10
There is this saying among mathematicians, that you don't really understand something until it becomes obviously trivial. So, in that sense, all of mathematics is "common sense thinking".
-
The common sense is the backbone of whole mathematics.
It is fair to say that nowadays all branches of mathematics are axiomatic theories. To start building an axiomatic theory you must decide what are your axioms, what are your axiom schemes, what are your rules of inference. When you finished setting up those things you can forget, in some sense, about common sense. But to make a right (right=at least interesting, usually you know what is right or what you need) choice of axioms, axioms schemes and rules of inference you will need a common sense because it is your only tool at that moment of the very beginning! You cannot create something from nothing (unless you are God :), you cannot start from nowhere. The common sense is the right starting point for mathematics, even if mathematics is capable of taking you far, far beyond it.
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Many basic theorems can be proven using common sense, not to mention that almost all axioms in mathematics, except for axioms of set theory are based on common sense. According to MathWorld, an axiom is a proposition regarded as self-evidently true without proof, which is just another way of saying it is based on common sense. The reason why I mentioned set theory is because common sense leads to numerous paradoxes in naive set theory, hence the name. In general, common sense does help, for example, understanding what a limit or a continuous function is, or why a certain theorem is true, but it is not enough. Everything in mathematics must be rigorous and every word is important. If you study or read books about epistemology and metaphysics, you should realize that it is very difficult to define common sense, so I think your question is very hard to answer indeed.
-
I would like to write about the problem of "expression", from my own experience. In the 1960s, it seemed to me that, from a commonsensical viewpoint, there should be some way of expressing that
in the above diagram the big square should be the "composition" of all the little squares. Then I found that Charles Ehresmann had defined double categories, which did the job.
The next question was: what is a commutative cube? For a square with sides $a,b,c,d$ the answer might be $ab=cd$. But if we want the "faces" of a cube to commute? For all this to be significant one has to move from sets with a total operation to sets with partial operations!
The point I am trying to make is that one function of mathematics is to develop language for rigorous expression, deduction and calculation, and this may take a while to develop. For example Descartes' notion of a graph of a function is now a commonplace, even common sense; but it may take an intellectual leap to make something into "common sense".
Another example is the introduction of the zero, and Arabic numerals.
Are higher dimensions than $3$ common sense? See the book "Flatland"! (downloadable).
Worth discussing is: have there been revolutions in mathematics? See discussions on the work of Thomas Kuhn on "Revolutions in Science".
- | 1,039 | 4,618 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2016-30 | latest | en | 0.947946 |
https://www.hackmath.net/en/word-math-problems/area?tag_id=63&page_num=2 | 1,638,869,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00134.warc.gz | 845,961,517 | 7,558 | # Area + perimeter - math problems
#### Number of problems found: 213
• Pentagonal pyramid
The height of a regular pentagonal pyramid is as long as the edge of the base, 20 cm. Calculate the volume and surface area of the pyramid.
• Ratio in trapezium
The height v and the base a, c in the trapezoid ABCD are in the ratio 1: 6: 3, its content S = 324 square cm. Peak angle B = 35 degrees. Determine the perimeter of the trapezoid
• Orange and tangerine trees
Brad's father asked an engineer to survey the field behind their house. He wanted to plant some orange and tangerine trees there. According to the survey, the field is thirty-two feet long and six yards wide. What is the area of the field in sq feet?
• Garden G
The rectangular, trapezoidal garden has a base length of 81m, 76m, and a vertical arm of 12m. Calculate how many m2 of the area will remain for planting greenery if 1/3 of the area is built. Calculate the consumption of mesh for land fencing.
• Round skirt
The cut on the round skirt has the shape of an annulus. Determine how much m² of fabric will be consumed on an 80 cm long skirt. The circumference of the waist is a circle with a smaller radius and is 69 cm.
• Gardens
The area of the square garden is 3/4 of the area of the triangular garden with sides of 80 m, 50 m, 50 m. How many meters of the fence do we need to fence a square garden?
• Calculate
Calculate the surface of a regular eleven-sided prism; if the area of its base is 58 cm2, the edge of the base is 6cm long, the height of the prism is 21 cm.
• Rectangular garden
Rectangular garden has a length of 48.7 m, a width of 6.3 meters shorter than the length. How much mesh should be bought for its fencing if the gate is 2.9 m long and the gate 1.1 m? What is the area of the garden?
• Maximum of volume
The shell of the cone is formed by winding a circular section with a radius of 1. For what central angle of a given circular section will the volume of the resulting cone be maximum?
• Surface and volume
Find the surface and volume of the rotating cone if the circumference of its base is 62.8 m and the side is 25 m long.
• The square
Calculate the area and perimeter of a square with side a = 15 dm S =, O =
• Area and perimeter of rectangle
The content area of the rectangle is 3000 cm2, one dimension is 10 cm larger than the other. Determine the perimeter of the rectangle.
• In the
In the rectangle ABCD, the distance of its center from the line AB is 3 cm greater than from the line BC. The circumference of the rectangle is 52 cm. Calculate the contents of the rectangle. Express the result in cm2.
• Quadrilateral prism
Calculate the surface of a quadrilateral prism according to the input: Area of the diamond base S1 = 2.8 m2, length of the base edge a = 14 dm, height of the prism 1,500 mm.
• The circumference
The circumference and width of the rectangle are in a ratio of 5: 1. its area is 216cm2. What is its length?
• The hollow cylinder
The hollow cylinder has a height of 70 cm, an outer diameter of 180 cm, and an inner diameter of 120 cm. What is the surface of the body, including the area inside the cavity?
• The bases
The bases of the isosceles trapezoid ABCD have lengths of 10 cm and 6 cm. Its arms form an angle α = 50˚ with a longer base. Calculate the circumference and content of the ABCD trapezoid.
• Winch drum
Originally an empty winch drum with a diameter of 20 cm and a width of 30 cm on the rescue car, he started winding a rope with a thickness of 1 cm beautifully from edge to edge. The winch stopped after 80 turns. It remains to spin 3.54m of rope (without h
• Carpet
How many crowns CZK do we pay for a carpet for a bedroom, when 1m of square carpet costs 350 CZK and the bedroom has dimensions of 4m and 6m? How many crowns do we pay for a strip around the carpet, when 1m of the strip costs 15 CZK?
• The Earth
The Earth's surface is 510,000,000 km2. Calculates the radius, equator length, and volume of the Earth, assuming the Earth has the shape of a sphere.
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Area - math problems. Perimeter - math problems. | 1,125 | 4,366 | {"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-2021-49 | latest | en | 0.919069 |
https://quizizz.com/admin/quiz/5f51b1fcb871bc001b02e334/factors | 1,656,376,598,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00131.warc.gz | 505,406,667 | 9,070 | Quiz
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Which number is a factor of 10
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• Question 3
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Report an issue
Q. Which number is a factor of 21
8
9
6
3
• Question 4
900 seconds
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Q. Which lists all the factors of 15?
1, 7, 8, 15
1, 15
1, 3, 5, 15
3, 5, 15
• Question 5
900 seconds
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Q. Which set lists all the factors of 24?
1, 2, 3, 6, 7, 12, 24
1, 2, 3, 4, 6, 8, 12, 24
1, 2, 3, 4, 5, 6, 8, 12, 24
1, 3, 4, 6, 8, 24
• Question 6
900 seconds
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Q. Which lists all the factors of 20?
1, 2, 4, 5, 10, 20
1, 2, 5, 10, 20
1, 4, 5, 20
1, 2, 10, 20
• Question 7
900 seconds
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Q. All even numbers have which number as a factor?
3
4
2
5
• Question 8
900 seconds
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Q. 2 x _____ = 28
12
14
16
18
• Question 9
900 seconds
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Q. 3 x _____ = 36
4
6
10
12
• Question 10
900 seconds
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Q.
___ x 6 = 54
3
9
11
6
8
• Question 11
900 seconds
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Q.
24 = 8 X ____
3
4
2
5
• Question 12
900 seconds
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Q. The factor pairs of 15 are 1 and 15 and _______
2 and 7
3 and 6
3 and 5
3 and 4
• Question 13
900 seconds
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Q.
The factors of 16 are 1, 2, 4, 8, 16
While the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
What are the Common Factors?
1,2,3,4,5
1,2,4,6,8
12,3,4,6,8,12,16,24
• Question 14
900 seconds
Report an issue
Q.
Which number is a factor of 18, but not a multiple of 2?
36
9
8
24
• Question 15
900 seconds
Report an issue
Q. When we skip count, we are listing ..... | 739 | 1,729 | {"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-2022-27 | latest | en | 0.770691 |
https://stonespounds.com/374-7-stones-in-stones-and-pounds | 1,627,800,579,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00441.warc.gz | 542,113,554 | 4,899 | # 374.7 stones in stones and pounds
## Result
374.7 stones equals 374 stones and 9.8 pounds
You can also convert 374.7 stones to pounds.
## Converter
Three hundred seventy-four point seven stones is equal to three hundred seventy-four stones and nine point eight pounds (374.7st = 374st 9.8lb). | 80 | 299 | {"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} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.806682 |
http://mathhelpforum.com/advanced-algebra/114861-homomorphism-kernel-monomorphism-print.html | 1,505,837,003,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685850.32/warc/CC-MAIN-20170919145852-20170919165852-00131.warc.gz | 220,022,636 | 2,987 | # Homomorphism, kernel, monomorphism :)
• Nov 15th 2009, 09:00 PM
GTK X Hunter
Homomorphism, kernel, monomorphism :)
Given $f$, natural mapping from $\mathbb{Z} \text{ to }\mathbb{Z}_n\text{ with }f(m)=r$, where $r$ be the remainder if $m$ is divided by $n$.
Show that $f$ is homomorphism!(Wondering)
And determine $ker(f)$.(Itwasntme)
Is $f$ monomorphism?(Giggle)
• Nov 16th 2009, 01:20 AM
Swlabr
Quote:
Originally Posted by GTK X Hunter
Given $f$, natural mapping from $\mathbb{Z} \text{ to }\mathbb{Z}_n\text{ with }f(m)=r$, where $r$ be the remainder if $m$ is divided by $n$.
Show that $f$ is homomorphism!(Wondering)
And determine $ker(f)$.(Itwasntme)
Is $f$ monomorphism?(Giggle)
rtp $f(ab)=f(a)f(b)$. $f(a)f(b)=rs$, by definition.
Now, expand $ab=(xn+r)(yn+s)$ to get the other side of the equality.
What do you think the kernel of such a mapping should be? Remember, the kernel is everything that is mapped to zero.
What is $f(a)$ for $1 \leq a \leq n$? What does this tell you about surjectivity? | 345 | 1,012 | {"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": 21, "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} | 3.421875 | 3 | CC-MAIN-2017-39 | longest | en | 0.773292 |
https://trustconverter.com/en/weight-conversion/grams/grams-to-gigagrams.html?value=5.5 | 1,726,357,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00034.warc.gz | 544,196,253 | 8,743 | # Grams to Gigagrams Conversion
Gram to gigagram conversion allow you make a conversion between gram and gigagram easily. You can find the tool in the following.
to
input
= 1.0 × 10-9
= 1.0E-9
= 1.0e-9
= 2.5 × 10-9
= 2.5E-9
= 2.5e-9
= 4.0 × 10-9
= 4.0E-9
= 4.0e-9
= 0.00000001
= 5.5 × 10-9
= 5.5E-9
= 5.5e-9
= 0.00000001
= 7.0 × 10-9
= 7.0E-9
= 7.0e-9
= 0.00000001
= 8.5 × 10-9
= 8.5E-9
= 8.5e-9
= 0.00000001
= 1.0 × 10-8
= 1.0E-8
= 1.0e-8
### Quick Look: grams to gigagrams
gram 1 g 2 g 3 g 4 g 5 g 6 g 7 g 8 g 9 g 10 g 11 g 12 g 13 g 14 g 15 g 16 g 17 g 18 g 19 g 20 g 21 g 22 g 23 g 24 g 25 g 26 g 27 g 28 g 29 g 30 g 31 g 32 g 33 g 34 g 35 g 36 g 37 g 38 g 39 g 40 g 41 g 42 g 43 g 44 g 45 g 46 g 47 g 48 g 49 g 50 g 51 g 52 g 53 g 54 g 55 g 56 g 57 g 58 g 59 g 60 g 61 g 62 g 63 g 64 g 65 g 66 g 67 g 68 g 69 g 70 g 71 g 72 g 73 g 74 g 75 g 76 g 77 g 78 g 79 g 80 g 81 g 82 g 83 g 84 g 85 g 86 g 87 g 88 g 89 g 90 g 91 g 92 g 93 g 94 g 95 g 96 g 97 g 98 g 99 g 100 g gigagram 1.0 × 10-9 Gg 2.0 × 10-9 Gg 3.0 × 10-9 Gg 4.0 × 10-9 Gg 5.0 × 10-9 Gg 6.0 × 10-9 Gg 7.0 × 10-9 Gg 8.0 × 10-9 Gg 9.0 × 10-9 Gg 1.0 × 10-8 Gg 1.1 × 10-8 Gg 1.2 × 10-8 Gg 1.3 × 10-8 Gg 1.4 × 10-8 Gg 1.5 × 10-8 Gg 1.6 × 10-8 Gg 1.7 × 10-8 Gg 1.8 × 10-8 Gg 1.9 × 10-8 Gg 2.0 × 10-8 Gg 2.1 × 10-8 Gg 2.2 × 10-8 Gg 2.3 × 10-8 Gg 2.4 × 10-8 Gg 2.5 × 10-8 Gg 2.6 × 10-8 Gg 2.7 × 10-8 Gg 2.8 × 10-8 Gg 2.9 × 10-8 Gg 3.0 × 10-8 Gg 3.1 × 10-8 Gg 3.2 × 10-8 Gg 3.3 × 10-8 Gg 3.4 × 10-8 Gg 3.5 × 10-8 Gg 3.6 × 10-8 Gg 3.7 × 10-8 Gg 3.8 × 10-8 Gg 3.9 × 10-8 Gg 4.0 × 10-8 Gg 4.1 × 10-8 Gg 4.2 × 10-8 Gg 4.3 × 10-8 Gg 4.4 × 10-8 Gg 4.5 × 10-8 Gg 4.6 × 10-8 Gg 4.7 × 10-8 Gg 4.8 × 10-8 Gg 4.9 × 10-8 Gg 5.0 × 10-8 Gg 5.1 × 10-8 Gg 5.2 × 10-8 Gg 5.3 × 10-8 Gg 5.4 × 10-8 Gg 5.5 × 10-8 Gg 5.6 × 10-8 Gg 5.7 × 10-8 Gg 5.8 × 10-8 Gg 5.9 × 10-8 Gg 6.0 × 10-8 Gg 6.1 × 10-8 Gg 6.2 × 10-8 Gg 6.3 × 10-8 Gg 6.4 × 10-8 Gg 6.5 × 10-8 Gg 6.6 × 10-8 Gg 6.7 × 10-8 Gg 6.8 × 10-8 Gg 6.9 × 10-8 Gg 7.0 × 10-8 Gg 7.1 × 10-8 Gg 7.2 × 10-8 Gg 7.3 × 10-8 Gg 7.4 × 10-8 Gg 7.5 × 10-8 Gg 7.6 × 10-8 Gg 7.7 × 10-8 Gg 7.8 × 10-8 Gg 7.9 × 10-8 Gg 8.0 × 10-8 Gg 8.1 × 10-8 Gg 8.2 × 10-8 Gg 8.3 × 10-8 Gg 8.4 × 10-8 Gg 8.5 × 10-8 Gg 8.6 × 10-8 Gg 8.7 × 10-8 Gg 8.8 × 10-8 Gg 8.9 × 10-8 Gg 9.0 × 10-8 Gg 9.1 × 10-8 Gg 9.2 × 10-8 Gg 9.3 × 10-8 Gg 9.4 × 10-8 Gg 9.5 × 10-8 Gg 9.6 × 10-8 Gg 9.7 × 10-8 Gg 9.8 × 10-8 Gg 9.9 × 10-8 Gg 1.0 × 10-7 Gg
The gram (alternative spelling: gramme; SI unit symbol: g) (Latin gramma, from Greek γράμμα, grámma) is a metric system unit of mass.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
gramg
1 Kg = 1000 g
= 10-3 Kg
Metric system SI
#### conversion table
gramsgigagramsgramsgigagrams
1= 1.0E-911= 1.1E-8
2.5= 2.5E-912.5= 1.25E-8
4= 4.0E-914= 1.4E-8
5.5= 5.5E-915.5= 1.55E-8
7= 7.0E-917= 1.7E-8
8.5= 8.5E-918.5= 1.85E-8
10= 1.0E-820= 2.0E-8
Gigagram or gigagramme is equal to 10gram (unit of mass), comes from a combination of the metric prefix giga (G) with the gram (g). Plural name is gigagrams.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
gigagramGg
≡ 109 g
≡ 1 000 000 kg
Metric system SI
### conversion table
gigagramsgramsgigagramsgrams
1= 100000000011= 11000000000
2.5= 250000000012.5= 12500000000
4= 400000000014= 14000000000
5.5= 550000000015.5= 15500000000
7= 700000000017= 17000000000
8.5= 850000000018.5= 18500000000
10= 1000000000020= 20000000000
gramsgigagrams
1= 1.0 × 10-9
1 000 000 000= 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,211 | 3,617 | {"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-2024-38 | latest | en | 0.36569 |
https://cboard.cprogramming.com/cplusplus-programming/71365-reading-input-files-printable-thread.html | 1,495,911,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00067.warc.gz | 933,297,835 | 3,667 | • 10-24-2005
megsb3u
I am very new to this and I am having a hard time grasping what I need to be doing. How do I read from a file and determine how many inputs are on a line so that I can do certain actions depending on how many inputs are available on the line?!
• 10-24-2005
Enahs
Quote:
How do I read from a file and determine how many inputs are on a line so that I can do certain actions depending on how many inputs are available on the line?!
Huh?
Could you maybe give an example of what you are trying to do? (verbally)
• 10-24-2005
megsb3u
Each line of the file has either 1, 2, or 3 dimensions listed on it. If there is 1 dimesnsion, I need to calculate the area of a circle and square.......if 2 dimensions, I need to calculate the area of a rectangle, etc. I know how to do the calculations and output the calculations. I just do not know how to read the file and determine how many dimensions are on the line.
• 10-24-2005
Enahs
This is one method:
Code:
```#include <iostream> #include <fstream> using namespace std; int main() { ifstream file_in; file_in.open("input.txt"); int _1; int _2; int _3; char check; file_in >> _1; //Goes first, we assume atleast one is there check = file_in.peek(); //Looks at next char if (check != '/n') //if next char is new line this if statement is skipped { file_in >> _2; check = file_in.peek(); } if (check != '/n') // If same as before but with 3rd char { file_in >> _3; } file_in.close(); cout << _1 << " " << _2 << " " << _3; system("pause"); return 0; }```
Here is the input file:
Code:
`1 2 3`
It is clearly not done, it only gets one line.
You will want to put it in a loop to get all the lines until end of file.
The peek and the instream ignore spaces, so it works perfect in this case. Each number can be separated by 10 spaces and it will still work the same.
This is not the only method, but fairly easily.
Hopefully it gets you started in the correction direction.
Peek:
http://www.cplusplus.com/ref/iostream/istream/peek.html
It also currently hardcodeed to output all three no matter what, so if you make a input file with just 2 it will have error in the output.
• 10-24-2005
megsb3u
Thank you!!! That helps immensely, it is just what I needed. | 635 | 2,429 | {"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-2017-22 | longest | en | 0.793366 |
https://ch.mathworks.com/matlabcentral/answers/811065-i-have-a-climatic-dataset-of-a-region-where-i-am-calculating-time-series-for-drought-index-the-di | 1,631,925,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00297.warc.gz | 221,175,054 | 25,075 | # I have a climatic dataset of a region where I am calculating time series (for drought index). The dimension of the dataset is 164*153*1416 (1416 months in the dataset and the others are the location). How I can average this over space and time?
5 views (last 30 days)
Md Muktadir on 23 Apr 2021
Commented: Md Muktadir on 24 Jun 2021
Average= nanmean(Box_SPEI(:,:,:))
Chad Greene on 3 May 2021
To get a map of average SPEI,
SPEI_mean = mean(SPEI,3,'omitnan');
and that calculates the mean along the third dimension (time).
Md Muktadir on 24 Jun 2021
@Chad Greene, If i have a dataset (360*720*1416) where first two indicate latitude and longitude and 1416 is the time step. If I need finer resolution (0.25 degree spatial resolution) to make it (720*1440*1416), how I can do that? | 228 | 783 | {"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-2021-39 | latest | en | 0.843763 |
https://www.physicsforums.com/threads/find-the-area-by-using-disk-method.1043220/ | 1,723,783,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00173.warc.gz | 709,639,209 | 18,778 | # Find the area by using disk method
• MHB
• jaychay
In summary, the conversation discusses solving for the area R and using integrals to find the volume of a rotated shape. The final step involves finding the radius of a disk when rotating R about the line y = 1.
jaychay
The problem is to solve for the area R.
I have tried to do it many times.
Volume V1 is on x - axis
Volume V2 is on y=1
$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$
$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$
$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$
note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?
skeeter said:
$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$
$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$
$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$
note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?
Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?
jaychay said:
https://www.physicsforums.com/attachments/10787
Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?
rotating R about the line y = 1 using disks ... what is the radius of a disk in this case?
## What is the disk method?
The disk method is a mathematical technique used to find the area of a solid of revolution. It involves dividing the solid into infinitely thin disks and summing their areas to find the total volume.
## When is the disk method used?
The disk method is used when finding the area of a solid that is created by rotating a curve around an axis. This could be a circle, parabola, or any other curve.
## What are the steps for using the disk method?
The steps for using the disk method are as follows:
• 1. Identify the axis of rotation and the curve to be rotated.
• 2. Determine the limits of integration for the integral.
• 3. Set up the integral using the formula for the area of a disk.
• 4. Evaluate the integral to find the total area.
## What are the assumptions made when using the disk method?
The disk method assumes that the solid of revolution is made up of infinitely thin disks, and that the disks are all parallel to the axis of rotation. It also assumes that the curve being rotated is continuous and does not intersect itself.
## What are some common mistakes when using the disk method?
Some common mistakes when using the disk method include forgetting to square the radius in the formula for the area of a disk, using the wrong limits of integration, and not taking into account any holes or gaps in the solid of revolution.
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2K | 974 | 3,070 | {"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} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.855893 |
http://openstudy.com/updates/509b2a69e4b0a825f21aa020 | 1,448,729,122,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453576.62/warc/CC-MAIN-20151124205413-00298-ip-10-71-132-137.ec2.internal.warc.gz | 174,425,332 | 10,520 | ## chubbytots 3 years ago correct these calculations by changing either side of each equation. a. (x+1)-(2x+1) = -x+2
1. chubbytots
\[(4p ^{2}-3q)-4(p ^{2}-q)=7q\]
2. chubbytots
thats for b.
3. chubbytots
c. \[5k(k+1)-(k ^{2}+5)=4k ^{2}\]
4. LukeBlueFive
a. \[(x+1)-(2x+1) = -x+2\] Remove the parenthesis: \[x+1-2x-1 = -x+2\] Combine like terms: \[x-2x+1-1 = -x+2\]
5. LukeBlueFive
b. Similarly to part a: \[(4p^2−3q)−4(p^2−q)=7q\] Remove parenthesis: \[4p^2−3q−4p^2+4q=7q\] Combine like terms: \[4p^2−4p^2−3q+4q=7q\]
6. LukeBlueFive
c. \[5k(k+1)−(k^2+5)=4k^2\] Multiply out: \[5k^2 + 5k−(k^2+5)=4k^2\] Remove parenthesis: \[5k^2 + 5k−k^2-5=4k^2\] Combine like terms: \[5k^2−k^2 + 5k-5=4k^2\]
7. LukeBlueFive
As you can see, none of these are actually equivalent. You have to change one of the sides (usually a plus or minus or a constant) to make them correct.
8. chubbytots
ok :D thanks for helping :D
9. LukeBlueFive
You're welcome. Glad I could help. ^_^ | 442 | 976 | {"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} | 3.84375 | 4 | CC-MAIN-2015-48 | longest | en | 0.665765 |
http://blog.mathplayground.com/2012/08/busy-learning.html | 1,575,842,437,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00092.warc.gz | 20,817,802 | 12,385 | ## Thursday, August 30, 2012
### Busy Learning
Many of my students attend a private K-8 school that offers a very traditional math curriculum. Students in grades 5, 6, and 7 spend many months studying shopkeepers' math. The main focus is percentages - discounts, sales tax, tip, simple and compound interest, commissions, annuities, etc. Each of these variations is taught as a series of formulas to be memorized. The vocabulary is beyond the comprehension of most 12 year olds. Even the numbers themselves are quite tedius. Students spend hours calculating 17.25% of \$12,650.85 by hand.
A group of 7th grade boys came in with homework from their new math unit - an introduction to linear relationships. They each had a worksheet with about 15 tiny grids upon which lines were drawn. The students were asked to calculate the slope of each line. That was it. No context and no explanation. That was their starting point.
"We don't really know what we're supposed to do," said one student.
"Yeah, it's just a bunch of lines," said another.
"We're lost," said the third member of the group.
The other two solemnly nodded in agreement.
Maybe it was the fact that I had just watched Stand and Deliver over the weekend and wanted to be like Jaime Escalante but I grabbed the worksheet, pretended to scrutinize it, and delivered the following line with exaggerated astonishment and admiration.
"Whoa!!! You guys are doing this? Wow! I can't believe it."
"What do you mean? What is this stuff?," they asked, almost in unison.
I walked to the door, checked to see if anyone was around, then closed it.
"Alright, I'll show you," I said.
One of the advantages of working in a defunct science center is access to a closet full of physics toys. I grabbed a few matchbox cars, ramps, and books and set up an impromptu math lab. What ensued over the next 45 minutes was a spirited look at independent and dependent variables, data plotting, best fit lines, and, of course, the meaning of slope.
At the end of the class, one of the fathers came in to pick up the students. I was putting away supplies but I heard him ask his son if he got his homework done. The boy said no. The father, clearly upset, asked his son for an explanation. The boy said....
"Dad, I can do my homework later. I didn't do it here because I was busy learning."
Yeah! | 536 | 2,341 | {"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-2019-51 | latest | en | 0.987538 |
https://www.erniegraves.com/faure-herman/how-long-to-heat-50-gallon-water-heater.html | 1,669,725,449,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00391.warc.gz | 791,398,176 | 25,968 | # How Long To Heat 50 Gallon Water Heater
## How Long Will it Take my 50 Gallon Water Heater to Heat Up?
Getting hot water out of a huge tank-based heater may be tedious and time-consuming, and it can take a long time. For those who are enduring harsh weather conditions, heating the full 50-gallon storage container may take an extended period of time. However, the majority of these are dependent on a variety of different conditions. In a water heater, one of the most essential components is the heating element, which has a significant impact on the amount of time it takes to heat 50 gallons of water.
A variety of additional elements have a part in determining how long it will take to complete.
However, if you want to learn more, this tutorial will be of assistance.
## Draw Efficiency
The drawing efficiency refers to the volume of water that must be heated in order to increase the temperature of the entire tank to a certain level of efficiency. The heater will not be able to heat the whole 50-gallon container of cold water. In addition, it includes a particular draw efficiency calculation that will function in conjunction with one another to increase the temperature. If you take any quality heater into consideration, its draw efficiency must be at least 70%. It is critical because the cold water enters the system at the same time as the hot water exits the heater.
## First Hour Rating
The first-hour rating is a computation that determines the volume of water that will flow out of the heater during a single hour of operation. It will be necessary to set the heater to a preset temperature range, which is typically 135 degrees Fahrenheit, in order to determine this amount. The first-hour rating is based on the amount of water that flows out of the unit in the first hour. Any heater with a respectable first-hour rating can be regarded a top purchase in this category. Although the test is carried out using at least 3 litres of water, it continues to run until the temperature lowers by 25 degrees Celsius.
## Recovery Rating
Another key component that impacts the effectiveness of a product is the recovery rating it receives from the manufacturer. The recovery rating is often used to refer to the amount of electricity required by the unit to heat the water contained inside it. A significant temperature disparity exists between the unit’s preset temperature and the actual temperature that the water within the tank is receiving. This is the reason why it is preferable to have a high recovery rate that will decrease the amount of time required.
You should choose gas-based heaters rather than electric-based heaters if you want to achieve the greatest outcomes possible. In general, they can warm in half the time it takes an electric heater to do so.
## Electric Hot Water Recovery
Any water heater with a capacity of 50 gallons will typically take around 5500 watts to heat the entire unit. Temperatures should be set at 120 degrees Fahrenheit, and it should take around 1 hour and 20 minutes to finish the full heating process. But keep in mind that this is only valid when the water temperature entering the device is 60 degrees. When the temperature is 70 degrees, it should take around 1 hour to finish the heating process. When the intake temperature is about 40 degrees, on the other hand, the identical device will take around 1 hour 50 minutes to fill up a full 50-gallon tank, according to the manufacturer.
## Gas Hot Water Heater Recovery
In order to heat the entire unit, a 50 gallon water heater will typically need around 5500 watts. Using the 120-degree Fahrenheit setting, it should take around 1 hour and 20 minutes to finish the full heating process. However, you must keep in mind that this is only valid when the water entering the unit is 60 degrees or above. The heating process should take around 1 hour when the temperature is 70 degrees or below. When the input temperature is at 40 degrees, on the other hand, it will take around 1 hour 50 minutes for the same unit to fill up an entire 50 gallon tank.
## Frequently Asked Questions
1. How long does a water heater last on average? In a task-based approach, the manufacturers are often quite explicit about the importance of keeping the internal components durable. The only locations where you need be concerned are the tank and the pipelines. It is estimated that if you have hard water in your area and it is producing problems, the complete unit will last around 5 years if it is not maintained. However, if you do, it should survive between 13 and 20 years, or possibly longer, depending on the quality of the water and how well it is maintained.
1. Is it possible for a water heater to explode?
2. That is why, while shopping for a heater, you should take the safety valves into consideration.
3. Having a temperature and pressure release valve installed may assist you in avoiding such scenarios.
4. 3.
5. If your heater begins to leak, you have no choice except to let your home to burn down.
6. It must be equipped with a separate shut-off valve that may be used to cut off the heater’s activities.
7. More information may be found in the following article.
## Conclusion
If you have a water heater tank that holds 50 gallons of water, it will take some time before it can begin supplying hot water. You may calculate the amount of time it takes to heat the full unit by combining these parameters together.
When living in harsh weather conditions, it is necessary to take a close look at these variables and make an informed decision on which one to purchase. Please let us know how long it took for your 50-gallon water heater to get to temperature.
## How Long Does a Heater Take to Heat 50-Gallon Water
On a cool morning, who doesn’t want to jump into a hot shower quickly? And if you’re thinking about purchasing a heater, the first thing that comes to mind is: how long does it take for a heater to heat 50 gallons of water? You should be aware that there is no set time for heating a 50-gallon water tank because several factors influence the time required. These factors include the type and number of heating elements used, the temperature of the incoming water into the unit, the tank size, the heater’s power source (gas or electricity), and so on.
• Both of these considerations are critical in determining the amount of time it takes to heat the water.
• Let’s take a quick look at their ramifications: A hot water heater’s first-hour rating is calculated by how many gallons of water it draws from the tank during a single hour of testing set by the United States Department of Energy for the unit.
• The results of the tests are shown on the energy guidance sticker attached to the machine.
• When compared to electric hot water heaters, gas hot water heaters are more efficient since they heat water in approximately half the time.
### Electric Hot Water Heater Recovery
A 50-gallon hotelectric water heater with a 5,500-watt heating element set at 120 degrees needs approximately 1 hour and 20 minutes to heat the water entering the heater, which is at 60 degrees when the heater is turned on. In order to make a little adjustment to the parameters, we will retain the temperature of the water entering this identical tank at 40 degrees. It will take around 1 hour and 47 minutes to heat the water to 120 degrees. On the other hand, it takes approximately 1 hour, 6 minutes for water that is 70 degrees to reach its maximum temperature of 120 degrees when it enters the tank.
### Gas Hot Water Heater Recovery
In order to compute the recovery time for a 50-gallon gas heater, we split the recovery durations for electric hot-water heaters in half. Water entering a 50 gallon tank at a temperature of 40 degrees Fahrenheit takes approximately 53 1/2 minutes to reach 120 degrees Fahrenheit. The identical procedure for 60-degree water entering the tank takes around 40 minutes, and for 70-degree water, it takes approximately 33 minutes. Another important component is draw efficiency, which is estimated as 70 percent of the total storage capacity of the heater tank’s entire storage capacity.
As a result, for a 50-gallon heater, the draw efficiency is credited to 35 gallons. So the next time you wish to evaluate the heating time required by a certain model that you have shortlisted, this guidance will undoubtedly be of use to you.
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Recovery of Waste Water from Water Heaters Heat Up Times Compared to One Another Time Required for Water Heater to Come to Temperature There isn’t much that can ruin your day quite as quickly as taking an ice cold shower, and if you have the wrong hot water heater, this might become your new normal very soon. In the event that your current heating unit fails on you, don’t let your stress over the situation lead you to make the wrong choice for a replacement. Prior to selecting a hot water heater, take into consideration how long it will take for the water heater of your choice to heat up completely.
The question is, how long does it take a hot water heater to reheat water once it has been depleted?
Water Heater Type Time to Heat Back Up Gas – Conventional Tank 30-45 mins Gas Tankless 0 mins Electric – Conventional Tank 60-80 mins Electric Tankless 0 mins
Water Heaters Powered by Natural Gas Specifications for a Gas Conventional Water Heater Once the water is in the tank, the normal gas tank water heater will take 30 to 40 minutes to heat it up to the desired temperature. When new water from your water supply is fed into the tank, this early heat up occurs as a result of the incoming water. Some mathematical calculations are required to provide a more specific explanation of why this takes 30 minutes. The size of the heater’s tank is obviously important, since more water will take longer to heat than a smaller tank.
• In simple terms, a BTU is the amount of heat required to elevate one pound of water by one degree Fahrenheit in temperature.
• For example, the typical hot water heating unit tank holds 40 gallons of water.
• Thirty-five gallons times 8.3 pounds per gallon is 330 pounds of water.
• For the sake of not having to get into full-blown thermodynamics calculations, we may simplify and say that a 40,000 BTU system with a 40-gallon tank needs half a minute to heat each gallon, which results in a half-hour heat up time.
• For those with larger tanks or lower BTU ratings, on the other hand, it will take longer to heat their tanks.
• Likewise, keep in mind that this is the amount of time it takes for new cold water to be heated in your tank, so plan accordingly.
When all of the warm water in the tank has been consumed, the length of time it takes to warm up additional water is taken into consideration. It will be necessary to restart the gas tank water heater at that point in order to heat new water from the entering groundwater temperature level.
#### A gas tank hot water heater will take roughly 40 minutes to warm up new inbound water for the very first time.
Specifications for an Electric Conventional Water Heater When compared to gas tank hot water heaters, electric tank hot water heaters often require double the amount of time to heat water. Electric components, while often more cost-effective, are just incapable of matching the high performance of gas-fired systems. It would take approximately one hour for an electric hot water heater to heat the 40-gallon tank shown above from the moment brand-new water is introduced into the system. As a result, homes with higher water needs are more likely to choose for a whole-house gas tank water heater rather than an electric model.
1. A tank hot water heater that uses electricity takes 60-80 minutes to heat water, but a tank hot water heater that uses gas takes 30 minutes.
2. Unless the system is malfunctioning, this should not take more than a few seconds for a typical-sized house to complete the cycle.
3. Due to the fact that a tankless gas heater heats water instantaneously, it should only take a few seconds for the warm water to travel through the pipes and into the component.
4. For the most part, water does not become heated until the dishwashing machine or hot water faucet is turned on.
5. Due to the fact that a tankless electrical heater warms water fast, it should only take a few seconds for the warm water to make its way through your pipes and into your fixture.
See also: What Causes A Water Heater To Explode
In conclusion, there is a heater that is suitable for any situation. Consider your requirements before selecting a storage tank, whether traditional or tankless in design. Please remember that South End Plumbing provides all plumbing services and that we are only a mouse click away.
We also specialize in tankless water heaters; please contact us for more information. South End Plumbing is one of the few organizations that will provide you with a no-obligation quote. To book a visit, please call us at 704-919-1722 or complete the online form.
## How Long Does It Take for a Water Heater to Heat Up?
You have arrived to the following page: How Long Does It Take for a Water Heater to Heat Up? Do you have a question about how long it takes for a water heater to heat up? No need to look any farther – our comprehensive guide provides answers to this and many other questions. Continue reading to find out all you need to know.
• Approximately how long does it take for a gas water heater to come to temperature. Is it possible to tell how long it takes an electric water heater to heat up
• What factors influence heating time
Request a Quote Plumbers in your area can be found here. To assist you in locating local plumbers in your region, we have teamed with Networx. To receive a no-obligation estimate, please complete the form below. Find a Plumber in your area. We may receive a commission if you click on this link, but there is no additional cost to you.
## How Long Does It Take for a Water Heater to Heat Up?
A storage water heater (one with a tank) requires some time to heat up the water in the tank before it is ready to use. When it comes to water heaters, though, how long does it take for them to heat up? You should be aware of this whether you have recently installed a new water heater or simply want to determine whether your current water heater is operating as it should. After filling the tank, you should be able to anticipate hot water within 30 minutes (gas) to around an hour and 20 minutes (electric) after doing so.
The size of your water heater, the power source, the First-Hour Delivery rate, and the recovery rate are all factors that influence how long you’ll have to wait for hot water.
### How Long Does It Take a Gas Water Heater to Heat Up?
A gas water heater is more energy efficient and can heat water more quickly than an electric water heater. With strong burners located at the bottom of the tank, they use natural gas as a fuel to heat the water stored in the tank. The temperature at which a gas water heater is set, as well as the temperature of the cold water that it must heat, determine how quickly it can heat water. Here are several averages to consider:
• Gas water heaters with capacities of 40 and 50 gallons take 30-45 minutes, while 80-gallon gas water heaters take 60-70 minutes to heat.
The average gas water heater holds around 40 gallons of water and takes approximately 30 to 40 minutes to heat water from 40 degrees to 120 degrees. It will take around 40 to 50 minutes to heat up a 50-gallon unit. It takes around 60 to 70 minutes for a big 80-gallon gas water heater to reach operating temperature. Of course, this is only a rough estimate, and actual costs may vary depending on the age and model of the vehicle.
### How Long Does It Take an Electric Water Heater to Heat Up?
When heating 40-degree water to 120 degrees, the average gas water heater holds around 40 gallons and takes approximately 30-40 minutes to do it. It will take around 40 to 50 minutes for a 50-gallon unit to reach its operating temperature. It takes around 60 to 70 minutes to heat an 80-gallon gas water heater. It should be noted that this is only a rough estimate that will vary depending on the age and model of the vehicle.
• 40-gallon electric water heaters take 60-80 minutes to heat water
• 50-gallon electric water heaters take 145-150 minutes to heat water
• 80-gallon electric water heaters take 120-130 minutes to heat water.
If you have a 40-gallon electric water heater that uses 5500 watts and is set to 120 degrees, it will take around an hour to an hour and 20 minutes to heat the water.
It will take around an hour and 45 minutes to an hour and 50 minutes to heat a 50-gallon electric unit. It will take around 2 hours for a big, 80-gallon electric water heater to reach the desired temperature.
## 6 Factors That Affect Water Heating Time
We’ll look at the elements that influence how long it takes to heat water heaters now that you’ve seen that they may take anything from half an hour to more than 2 hours to heat up.
### First-Hour Delivery Rate
A rating for first-hour delivery (FHD) is given to all water heaters. When the water heater is fully charged, the FHD tells how many gallons of hot water it can deliver in an hour. The flow rates for FHD are provided in gallons per hour (GPH). A high FHD rate indicates that you will receive more hot water more quickly than you would from a unit with a lower FHD rate, which means you will save time and money. A FHD rate of around 60 to 80 GPH is appropriate for a 50-gallon container.
### Recovery Rate
The recovery rate of a water heater refers to how many gallons of hot water the device can deliver each hour while it is being utilized. It informs you how quickly the water heater can recover (also known as refill) with cold water and heat it back up to normal temperature. Due to the fact that it takes less time to heat up hot water in a unit with a high recovery rate, it will supply hot water faster. A high recovery rate water heater, on the other hand, will be able to swiftly heat cold water that enters the system, regardless of how much hot water you’re consuming at once.
### Power Source
The power source of a water heater (gas or electricity) has a significant impact on the amount of time it takes to heat water. Electric water heaters are notorious for taking a long time to heat the water. This is due to the fact that using electrical heating components rather than gas burners is less efficient. A typical 50-gallon gas water heater may have a flow rate of 80 to 90 GPH, but an average 50-gallon electric water heater may have a flow rate of 58 to 66 GPH, depending on the model.
### Water Heater Type
Those with tanks that store and heat water are referred to as storage water heaters. Tankless water heaters do not have storage tanks and heat the water as soon as it is drawn from the faucet, saving energy. These two types of water heaters have a significant variation in the amount of time required to heat water. It might take anything from 30 minutes to an hour and a half for a storage water heater to reach operating temperature. A tankless water heater, on the other hand, makes hot water accessible almost immediately.
Continue reading: The Best Tankless Water Heaters
### Water Heater Size
The size of a storage water heater, measured in gallon capacity, has a significant impact on how rapidly it can heat water. Storage water heaters may typically hold anywhere from 30 to 80 gallons of water, depending on the model. Smaller tanks heat water more quickly (and run out of hot water more quickly) than larger tanks because there are fewer gallons of water to heat. This is analogous to heating water in a tiny 2-qt. saucepan on the stove instead of a large 12-qt. stock pot on the stove.
Larger tanks, on the other hand, do not take as long to heat up as you might expect.
Because large capacity gas water heaters feature a larger gas burner, they heat water faster than smaller capacity gas water heaters.
Even so, if you have a 30-gallon water heater, you won’t have to wait nearly as long for it to heat up as you would if you had a 50- or 80-gallon one. Continue reading: The Best Small Water Heaters
### Original Water Temperature
The temperature of the starting water has a significant impact on the amount of time it takes for a unit to heat it up. If the water temperature at the input is low, the water heater will have to work harder to increase the water temperature to the setting you’ve selected. Temperatures in cooler areas are typically about 40 degrees Fahrenheit for the water entering the system. In warmer areas, the temperature is around 50 degrees. It takes some time for the water heater to heat the water from 40-50 degrees to 140 degrees.
To assist you in locating local plumbers in your region, we have teamed with Networx.
Find a Plumber in your area.
## So, How Long Does It Take for a Water Heater to Heat Up?
In most cases, if you have a gas water heater set to 120 degrees and the incoming water is about 50 degrees, you should not have to wait more than 30 to 50 minutes for hot water (in 40 and 50-gallon units). A 5500-watt electric water heater set to 120 degrees will provide hot water after about 50 minutes if the incoming water temperature is roughly 50 degrees. This will require a little longer wait time. It will take around one hour to one hour and forty-five minutes (in 40 and 50-gallon units).
If your water heater takes longer than around 2 hours to heat up, you should contact a professional to inspect it.
###### You Might Also like:
• What size water heater do I require
• What is the average cost of a water heater
• And how long does a water heater last
## New Water Heater How Long To Heat Up
It takes an ordinary gas heater between 30 and 40 minutes to completely heat the water in its tank, depending on the model. To fully heat the water in its tank, the typical electric heater requires approximately double the time of the average gas heater; thus, you should expect it to take between an hour and an hour and 20 minutes to fully heat the water.
## How long does it take for a new 50-gallon water heater to heat up?
It takes approximately 1 hour and 20 minutes for a 50-gallon hot water heater with 5,500-watt elements set to 120 degrees to heat water that comes into the unit at a temperature of 60 degrees. In contrast, when the water entering this same tank is 40 degrees, it takes 1 hour and 47 minutes to heat it to the desired temperature.
## Why is my brand new water heater not getting hot?
A water heater that does not generate hot water might be due to a lack of electricity, a tripped limit switch, or one or more faulty heating components, to name a few possibilities. As a first step, make sure that the circuit breaker for your water heater is not tripped on your panel of electrical circuit breakers. Switch off the circuit breaker and then turn it back on if it has been tripped.
## How long does it take a 40-gallon water heater to heat?
If you have a 40-gallon electric water heater that uses 5500 watts and is set to 120 degrees, it will take around an hour to an hour and 20 minutes to heat the water.
## How long does it take a 10 gallon water heater to heat up?
A 6 gallon water heater will heat at a rate of 17.8 gallons per hour and will take around 20 minutes to reach its maximum temperature.
If you have a 10 gallon tank, it will take around 33 minutes to reach its maximum temperature.
## How long does it take a brand new water heater to heat up water?
A 6 gallon water heater will heat at a rate of 17.8 gallons per hour and will take around 20 minutes to reach its maximum temperature of 120 degrees. A 10 gallon tank might take up to 33 minutes to reach its maximum heating capacity.
## How long does it take for a new water heater to get hot water?
It takes how long does it take for a gas water heater to get up to temperature? Once the water has entered the tank, the normal gas tank heater will take around 30 to 40 minutes to heat it. When you first fill the tank with water from your plumbing supply, the tank will heat up for a few minutes.
## Is there a reset button on a hot water heater?
You’ll locate a reset button on the back of your electric water heater somewhere. It is often crimson in color and is generally seen around the thermostat. It might alternatively be concealed behind a detachable metal plate on the device, which would then be concealed behind some insulation. Once you’ve located the button, press and hold it for a few seconds.
## Why is my hot water not hot?
You’ll locate a reset button on the back of your electric water heater. Most of the time, it’s bright red and positioned near the thermostat. A detachable metal plate on the unit, followed by insulation, may also be used to conceal the device. Find the button and press and release it once you have located it.
## What does the reset button do on a hot water heater?
You’ll locate a reset button on your electric water heater anywhere on the unit. It’s generally bright red and may be seen near the thermostat. It might alternatively be concealed behind a detachable metal plate on the device, which would then be concealed by insulation. Once you’ve located the button, press and hold it for a couple of seconds.
## How do I make my hot water heater heat up faster?
How to Get Hot Water More Quickly Pipe insulation should be installed. If you suspect that your water pipes are to fault for your lack of hot water, installing insulation to them may help to improve your issue. Pump for recirculating hot water in a building. Fixtures with a higher flow rate should be substituted. Upgrade to a tankless water heater to save money. Preventative Maintenance should be performed on a regular basis.
## Is 40 gallon water heater enough for a family of 4?
It’s generally accepted that you need a 40 gallon tank for 1 to 4 people, a 50 gallon tank to accommodate 4 to 6 people, and a 50 gallon high recovery or 75 gallon tank to accommodate 8 people and up to 10 people, respectively. In terms of energy consumption, standard water heaters of 40 and 50 gallon capacity are very similar.
## How often should I heat my hot water?
This is a pretty prevalent urban legend. In reality, you don’t need to keep your water heated all of the time unless you want to. You may use your immersion heater or boiler to heat up hot water that has been collected in a tank or container. As long as the tank has a decent insulating jacket, it will be able to keep the water hot throughout the day without the need to reheat it frequently.
## What size heater do I need for a 10 gallon aquarium?
The capacity of a heater is typically measured in watts. For the most part, a capacity of around 5 watts per gallon of water should suffice as a general rule of thumb. As a result, a 50 watt heater will be required for a 10 gallon aquarium.
## Should I leave my propane water heater on all the time?
It just takes a little spark to ignite highly flammable propane gas, which can result in an explosion or fire in the event of an accident, depending on the circumstances.
If your RV hot water heater is fueled by propane, the straightforward and safe response is that you should not leave it on all of the time.
## How much hot water does a 20 minute shower use?
If a normal showerhead is installed, it will consume approximately half a gallon more water per minute, resulting in a 25-gallon emittance every 10 minutes, or 50 gallons over the course of a 20-minute shower session. *One gallon is equal to 4.54 litres.
## How long of a shower can you take with a 40-gallon water heater?
A 40-gallon water heater may supply enough hot water for up to two showers in an hour (assuming no other water-using appliances are in use).
## Why does it take my hot water so long to get hot?
What is it about the hot water that is taking so long? Many factors contribute to this: the distance between the faucet and the water heater, the diameter of the pipe, and the flow velocity of the water. The greater the distance that hot water must travel before reaching the shower faucet, the longer it will take to heat up the faucet. Having a larger house makes this even more important.
## Where is the reset button on a gas hot water heater?
A reset button for the water heater is positioned in the middle of the limit switch, right above the water heater thermostat, and is generally red in color. In the event that something goes wrong with the water heater and the water becomes too hot, the limit switch will cut down the electricity to the water heater.
## How Long Does it Take for a Water Heater to Heat Up?
Please keep in mind that this content may contain affiliate links. This means that, at no additional cost to you, we may gain a small profit on purchases made via our links. Briefly stated, less water warms more rapidly than more water; hence, the volume of water you are heating, as well as the temperature at which it is heated, influences how soon you will get it.
## Does Your Hot Water Flow Seem Too Slow?
There may be affiliate links in this content, so please be aware of that. This means that, at no additional cost to you, we may gain a small fee on purchases made via our links and advertisements. Briefly stated, less water heats more quickly than more water; therefore, the amount of water you are heating, as well as the temperature at which it is heated, determines how quickly you will receive it.
## Recovery Efficiency
Another element to consider while scaling is the amount of time it takes to recover. While gas heaters heat water more quickly, their recovery efficiency is lower than that of electric heaters. For gas water heaters, the efficiency is 75%, whereas for electric water heaters, the efficiency is 100%. Gas hot water heaters, on the other hand, even at lower recovery efficiency, generate more hot water and do it much more quickly than their electrically powered equivalents. Gas heaters with a 30,000 BTU burner create 27.3 gallons per hour at 75 percent recovery efficiency, but electric heaters with a 750-watt heating element produce 3.1 gallons per hour at 75 percent recovery efficiency.
The output of an electrical hot water heater rises with the addition of more heating elements and the use of greater wattages.
One hour’s worth of heating with an electrical heating element results in 20.5 gallons of 100 percent increase when using a 4,500 watt electrical heating element.
This electrical heating element is six times more powerful than the one used in the previous example. The 20,000 BTU burner has a 33 percent lower BTU output than the 30,000 BTU burner (BTU example).
### What is a Good Water Heater Recovery Rate?
If the water heater has a capacity of 40 to 50 gallons per hour, anything above 40 gallons per hour would be considered an excellent recovery rate. The higher the BTU rating of the burner, the better it is for recovery in general. An average 50-gallon electric water heater with twin heating elements has a recovery rate of 20 gallons per hour, which is satisfactory for most applications. Single element water heaters will, of course, have a lesser recovery rate than their two-element counterparts.
## The Major Factor
The image is courtesy of HotWater.com. Don’t get yourself mixed up. The amount of water being heated, the method by which it is heated, and the amount of water being utilized are the elements that determine how long it takes the water heater to heat up. The amount of water heated each hour is specified in the heating rates. The capacity of the storage tank indicates how much hot water is immediately accessible when you turn on the faucet. You might consider upgrading to a larger tank if your family is large, has several bathrooms, and has several hot water-consuming activities occurring at the same time on a frequent basis.
1. Greater tank capacities are designed for homes who require a big volume of hot water in a short period of time.
2. As a result, a 40-gallon tank will meet your demands, and you will save money by not purchasing a larger 50-gallon tank that you will not need.
3. See the water heater recovery table in the preceding section.
4. When using an electric water heater, increase the time by half to 1.5 hours.
5. Another point to consider in the discussion over whether to use a tank or not.
## How Long Does it Take a Gas Water Heater to Reheat?
Home-made gas water heaters can be a cost-effective way to ensure that a household has enough hot water on hand. Water heaters powered by electricity can have much higher energy costs in some places, and these devices are more prone to power outages than water heaters powered by gas (natural-draft or otherwise). if (sources.length) then this.parentNode.removeChild(sources); then this.onerror = null; this.src = fallback; )(, arguments.target.currentSrc.replace; )(, arguments.target.currentSrc.replace; )(, arguments.target.currentSrc.replace; (//\$/, “), (//\$/, “), ‘/public/images/logo-fallback.png’) is a fallback logo image.
” loading=”lazy”> ” loading=”lazy”> A technician stands in front of a water heater to demonstrate Calculating how rapidly a gas water heater can raise the temperature of the water needs a small amount of mathematical computation.
## BTU
if (sources.length) then this.parentNode.removeChild(sources); if (sources.length) then alternatively, if this.onerror = null, this.src = fallback; )(, arguments.target.currentSrc.replace(), ‘/public/images/logo-fallback.png’), ‘/public/images/logo-fallback.png’) otherwise ” loading=”lazy”> ” loading=”lazy”> A British thermal unit (BTU) is the amount of energy required to raise the temperature of one pound of water by one degree Fahrenheit while the surrounding atmosphere is at its usual pressure and temperature.
See also: How To Get Water Off Pool Cover Without A Pump
BTUs are a unit of measurement for energy consumption that may be used to compare the performance of different types of fuel.
## Tank Size and Thermostat Setting
if (sources.length) then this.parentNode.removeChild(sources); if (sources.length) then in the alternative, if (this.onerror = null; this.src = fallback; )(, arguments target current src replace (//\$/, “), (//\$/, “), ‘/public/images/logo-fallback.png’) is a fallback logo image. ” loading=”lazy”> ” loading=”lazy”> One gallon of water weighs 8.338 pounds when the temperature is 62 degrees Fahrenheit. At this temperature, a 40-gallon water heater would have 333.52 pounds of water in it, according to the manufacturer.
Using a thermostat set at 122 degrees Fahrenheit, it would take around 30 minutes to heat the tank completely if all of the heat were transmitted to the water.
## Tank Efficiency
if (sources.length) then this.parentNode.removeChild(sources); if (sources.length) then alternatively, if this.onerror = null, this.src = fallback; )(, arguments.target.currentSrc.replace(), ‘/public/images/logo-fallback.png’), ‘/public/images/logo-fallback.png’) otherwise ” loading=”lazy”> ” loading=”lazy”> It is estimated that 20% of the heat from the burners escapes through the chimney based on an 80 percent energy efficiency rating for the water heater.
This results in an increase of roughly 36 minutes in the time required to reheat a 40-gallon water heater operating at 40,000 BTUs.
## Age of the Water Heater
if (sources.length) then this.parentNode.removeChild(sources); if (sources.length) then if (this.onerror = null; this.src = fallback; if (this.onerror = null; this.src = fallback; if (this.onerror = null; if (this.onerror = null; if (this.onerror = null; if (this.onerror = null; if (this.onerror = null; if (this.onerror = null; if (this.onerror ‘/public/images/logo-fallback.png’) is a fallback logo image.
” loading=”lazy”> ” loading=”lazy”> The state of the unit and the amount of build-up within the tank can both have an impact on the efficacy of the burner.
## The Drip Cap
• Gas water heaters may be a reliable source of hot water for a household
• Yet, they can be expensive. It would take approximately 30 minutes to heat the tank if all of the heat was transferred to the water
• However, if the flame is not as hot due to incorrect air to gas ratios or if there is a build-up of calcium and lime in the tank, the amount of heat that is transferred to the water is reduced, resulting in an increase in the amount of time it takes to heat the water.
## How Long Does it Take for Electric & Gas Water Heaters to Heat Up?
An ice cold shower is one of the few things that can completely derail your day, and if you have the improper water heater, this might become your new normal. If your present heater is on its last legs, don’t allow the stress of the circumstance push you into making the wrong decision about your new heater. Before you purchase a water heater, take into consideration how long it will take for your water heater of choice to reach operating temperature. If you want to run a large amount of hot water at the same time, you’ll need a more powerful system than if you merely want to take a hot shower on a consistent basis.
When water reaches the water heater, how long does it take for it to heat up? In spite of the numerous variables that might influence the time required, the chart below illustrates the average time required for each kind of water heater to heat up.
## How Long It Takes A Water Heater to Heat Up For The First Time
Water Heater Type Time Takes to Heat Up
Gas Tank 30-40 minutes
Gas Tankless 0 minutes *
Electric Tank 60-80 minutes
Electric Tankless 0 minutes *
*If the tankless water heater is appropriately designed and placed, it may offer practically immediate heat. Source of the graph
## How Long Does it Take for a Gas Water Heater to Heat Up?
Once the water has entered the tank, the normal gas tank heater will take around 30 to 40 minutes to heat it. When you first fill the tank with water from your plumbing supply, the tank will heat up for a few minutes. A more detailed explanation of why this takes 30 minutes necessitates the use of mathematics. The size of the heater’s tank is obviously important, since more water will take longer to heat than a smaller tank. The BTU (or British Thermal Unit) rating of the heater is the next most important consideration.
• A heater with a higher BTU rating will heat water more quickly.
• Each gallon of water contains around 8.3 pounds of water; as a result, our sample tank has approximately 330 pounds of water to heat.
• If the water is at 60 degrees and you want to bring it up to 120 degrees, you will need to raise the temperature by 60 degrees to do this.
• Because of the lower tank size and greater BTU rating, your hot water heater’s warm-up time will be significantly reduced.
• You will need to keep the following criteria in mind if you want a high-efficiency water heater that will heat your water in the period of time you specify (after it has run out of hot water) and hold a significant volume of hot water.
• The first time you switch on the hot water after your tank has been holding hot water for a while, you should get hot water in a matter of minutes because tanks store pre-heated water, not minutes or hours.
• That’s when the gas tank water heater will have to start heating new water from the temperature of the entering groundwater again, which will take longer.
## How Long Does it Take an Electric Hot Water Heater to Heat Up?
When compared to its gas equivalents, electric tank water heaters often need double the length of time to heat water. Despite the fact that electric components are often more cost-effective, they cannot match with the great performance of gas-fired systems. It would take approximately one hour for an electric water heater to heat the 40-gallon tank indicated above from the moment new water is introduced. As a result, homes with higher water needs are more likely to choose for a whole-house gas tank water heater rather than an electric model.
Electric variants are ideal for those who live in smaller houses with lower water needs. When it comes to heating water, an electric tank water heater takes 60-80 minutes, compared to 30 minutes for a gas tank water heater.
## How Long Does it Take a Tankless Gas Heater To Warm Up?
Tankless water heaters heat your water on demand, which means that the distance between your heater and the device you are using is the only factor that defines how long it will take for you to obtain hot water from your faucet. Ideally, this should not take more than a few seconds with a typical-sized house if the system is functioning properly. It may take a few extra seconds for the water to travel through the water pipes and reach appliances that are located further away from the heater in a large home.
## How Long Does it Take a Tankless Electric Heater To Warm Up?
Tankless electric water heaters work in a similar way to tankless gas water heaters in that they only begin to heat your water when an item requires it. This means that unless you turn on the dishwasher or turn on the faucet, the water will not be warmed. The majority of the time, an electric tankless heater will give hot water in a matter of seconds, but they can take a fraction of the time that gas systems do owing to the greater strength of gas heat. Because a tankless electric heater warms water instantaneously, it should only take a few seconds for the hot water to flow through your pipes and into your fixture once it has been heated.
## Factors That Affect Heat Up Time
Apart from the variables we’ve already covered, such as tank size and BTU rating, there are a variety of other elements that might influence how long it takes your water heater to heat water for the first time.
• Temperature of the incoming water– For both tankless and tank-style water heaters, the temperature of the incoming water will play a role in determining the amount of time it takes to heat up. Because tank heaters store water and maintain a constant temperature, the incoming temperature should have little effect on them. Instead than storing water in tanks, tankless heaters deliver incoming water on demand, only minutes before it flows out of your faucet. In other words, if the groundwater temperature is really low, the water may not heat up as quickly as it could otherwise. Neither kind of heater is impervious to the effects of extremely cold ambient temperatures in the room or area where they are housed
• But, it will take more water to be heated before the pressure is high enough to force the water through the remaining pipe system.
In conclusion, there is a heater out there that is appropriate for everyone. Be sure to consider your requirements before picking either a traditional tank or a tankless system. See our assessment of the top models on the market now that you know how long it takes for both gas and electric water heaters to heat up. With amazing brands like Bosch, Rheem, and Takagi, you’re sure to find something that works for your needs!
## How long should a shower last with a 50 gallon water heater?
Asked in the following category: General The most recent update was made on March 26th, 2020. To put it another way, when the tank is completely filled with hot water, it can produce around 33 gallon per minute. The usual shower consumes 2 gallons of water per minute, which means that the same 50 – gallon tank can provide hot water for a little less than 17 minutes. According to the 70 percent rule, a 50-gallon tank will produce approximately 35 gallons of hot water, which is sufficient for a 30-minute shower.
Check the time on your wristwatch!
How long should a water tank be kept at a high temperature?
When it comes to thoroughly heating up the water in its tank, the typical electricheater takes around twice as long as the average gasheater, so you can anticipate it to take anywhere between an hour and an hour and 20 minutes.
A 40-gallon water heater may supply enough hot water for up to two showers in an hour (assuming no other water appliances are in use).
Showering in Effortless Savings Showers are often the third most water-intensive appliance in a normal home, after toilets and clothes washers.
Taking an average American shower consumes 17.2 gallons (65.1 liters) of water and lasts 8.2 minutes at an average flow rate of 2.1 gallon per minute (gpm) (7.9 lpm). | 9,600 | 44,519 | {"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-2022-49 | longest | en | 0.935942 |
http://www.ncl.ucar.edu/Document/Functions/Built-in/ftcurvpi.shtml | 1,553,372,783,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203021.14/warc/CC-MAIN-20190323201804-20190323223804-00217.warc.gz | 333,838,068 | 4,898 | NCL Home > Documentation > Functions > Interpolation, Ngmath routines
# ftcurvpi
Calculates an integral of an interpolatory spline between two specified points.
## Prototype
``` function ftcurvpi (
xl [1] : numeric,
xr [1] : numeric,
p [1] : numeric,
xi : numeric,
yi : numeric
)
return_val : float or double
```
## Arguments
xl
A scalar value containing the lower limit of the integration.
xr
A scalar value containing the upper limit of the integration.
p
A scalar value specifying the period of the input function; the value must not be less than xi(npts-1) - xi(0).
xi
An array containing the abscissae for the input function, with rightmost dimension npts. If xi is multi-dimensional, it must have the same dimension sizes as yi.
yi
An array of any dimensionality, whose rightmost dimension is npts, containing the functional values of the input function. That is, yi(...,k) is the functional value at xi(...,k) for k=0,npts-1.
## Return value
The integral of the interpolated function from xl to xr - a scalar if yi is a one-dimensional array, otherwise an array with the same dimensions as all but the last dimension of yi.
The output value will be of type double if any of the input is double, and float otherwise.
## Description
ftcurvpi is in the Fitgrid package - a package containing 1D and 2D interpolators using cubic splines under tension.
There are some parameters that can alter the behavior of ftcurvpi. These parameters all have reasonable default values. However, users may change any of these parameters by invoking ftsetp prior to calling ftcurvpi. ftcurvpi is called after all of the desired values for control parameters have been set.
The only control parameter that applies to ftcurvpi is: sig.
The value for the parameter sig specifies the tension factor. Values near zero result in a cubic spline; large values (e.g. 50) result in nearly a polygonal line. A typical value is 1. (the default).
You can extrapolate values with ftcurvpi (that is, calculate interpolated values for abscissae outside of the domain of the input), but these values are, in general, unreliable.
## Examples
```begin
xi = (/ 0.00, 2.00, 5.00, 8.00, 10.00, 13.00, \
15.00, 18.00, 21.00, 23.00, 30.00 /)
yi = (/ 1.00, 0.81, 0.00, -0.81, -1.00, -0.84, \
-0.56, 0.04, 0.73, 1.18, 2.0 /)
integral = ftcurvpi(10., 30., 31., xi, yi)
end
``` | 673 | 2,429 | {"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-2019-13 | latest | en | 0.641275 |
https://paperzz.com/doc/8515867/homework-1-solutions-problem-1-let-f-n--be | 1,719,236,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00280.warc.gz | 382,582,463 | 10,509 | ### HOMEWORK 1 SOLUTIONS Problem 1 Let F(n) be
```HOMEWORK 1 SOLUTIONS
Problem 1
Let F (n) be ”all the polynomials p(x) with degree less than n can be written as
(x − b)q(x) + r where degree q(x) is one less than p(x)”
F (1): a degree zero polynomial is just an integer which has the desired form with
q(x) = 0
F (n) → F (n + 1): without loss of generality we can assume p(x) is a degree n
polynomial.
p(x) = an xn + an−1 xn−1 + ... + a0
= an xn − an bxn−1 + an bxn−1 + an−1 xn−1 + ... + a0
= an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 )
by using F (n) for the second term we get:
p(x) = an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 )
= an xn−1 (x − b) + (x − b)q0 (x) + r
= (x − b)(an xn−1 + q0 (x)) + r
for some q0 (x). So we have proved F (n + 1) with q(x) = an xn−1 + q0 (x).
1. Problem 2
Let Sk (n) be the sum of the first n k-th powers. S4 (n) is the size of the set
A := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l ≤ m, m ≤ n}
We introduce symmetry by defining sets
B := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, m ≤ l, l ≤ n}
C := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, l, m ≤ k, k ≤ n}
D := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, k, l, m ≤ j, j ≤ n}
E := {(i, j, k, l, m) : i, j, k, l, m ∈ N, j, k, l, m ≤ i, i ≤ n}
By the principle of inclusion-exclusion,
|A ∪ B ∪ C ∪ D ∪ E|
5
5
5
5
5
=
|A| −
|A ∩ B| +
|A ∩ B ∩ C| −
|A ∩ B ∩ C ∩ D| +
|A ∩ B ∩ C ∩ D ∩ E|
1
2
3
4
5
Date: October 11, 2013.
1
2
HOMEWORK 1 SOLUTIONS
Note that
A ∪ B ∪ C ∪ D ∪ E = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l, m ≤ n}
⇒|A ∪ B ∪ C ∪ D ∪ E| = n5
A ∩ B = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k ≤ l, l ≤ n, l = m}
⇒|A ∩ B| = S3 (n)
and similarly, |A ∩ B ∩ C| = S2 (n), |A ∩ B ∩ C ∩ D| = S1 (n), |A ∩ B ∩ C ∩ D ∩ E| =
S0 (n) = n. So
n5 = 5S4 (n) − 10S3 (n) + 10S2 (n) − 5S1 (n) + n
which gives us after rearranging, the desired formula
1 5
(n + 10S3 (n) − 10S2 (n) + 5S1 (n) − n)
5
1
1
1
1
1
= n5 + n2 (n + 1)2 − n(n + 1)(2n + 1) + n(n + 1) − n
5
2
3
2
5
1 5 1 4 1 3
1
= n + n + n − n
5
2
3
30
S4 (n) =
Let us now prove the above formula by induction. We define the polynomial
1
1
function p(n) = 15 n5 + 12 n4 + 13 n3 − 30
n. For n = 1, p(1) = 15 + 12 + 13 − 30
=1=
5
1 = S4 (1), so the formula holds for n = 1.
Now suppose that S4 (k) = p(k) for some k ≥ 1. Then
S4 (k + 1) = S4 (k) + (k + 1)4
1
1
1
1
= k 5 + k 4 + k 3 − k + (k + 1)4
5
2
3
30
1
1
1
1
= k 5 + k 4 + k 3 − k + k 4 + 4k 3 + 6k 2 + 4k + 1
5
2
3
30
1
3
13
119
= k 5 + k 4 + k 3 + 6k 2 +
k+1
5
2
3
30
and
1
1
1
1
(k + 1)5 + (k + 1)4 + (k + 1)3 − (k + 1)
5
2
3
30
1 5
1
4
3
2
= (k + 5k + 10k + 10k + 5k + 1) + (k 4 + 4k 3 + 6k 2 + 4k + 1)
5
2
1 3
1
2
+ (k + 3k + 3k + 1) − (k + 1)
3
30
1 5 3 4 13 3
119
2
= k + k + k + 6k +
k+1
5
2
3
30
p(k + 1) =
so indeed we have S4 (k + 1) = p(k + 1). By mathematical induction, the formula
S4 (n) =
holds for all n ∈ N.
1 5 1 4 1 3
1
n + n + n − n
5
2
3
30
HOMEWORK 1 SOLUTIONS
3
Problem 3
Let Sk (n) denote the sum of the First n-th powers as in the notes for lecture
1. Prove by induction that Sk (n) is a polynomial (whose coefficients are rational
numbers) of degree k + 1 in n. (Hint: You should prove this by induction on k. You
should use as your induction hypothesis that Sj (n) is a polynomial of degree j + 1
for all j smaller than k. [This is sometimes called strong induction.] The last page
of the notes for lecture 1 give you a good guess for what the leading term of Sk (n)
should be. Express the rest of it as a combination of Sj (n)’s for smaller j.)
Let’s prove the statement of the problem for all integers k ≥ 0 by induction in
k.
To check the base case for k = 0, we have to check, that S0 (n) is a polynomial
in n of degree 1. Indeed,
S0 (n) = 10 + 20 + · · · + n0 = 1 + 1 + · · · + 1 = n.
To do the induction step we start with the guess “the highest term in Sk (n)
is nk+1 /(k + 1)” and define Rk (n) to be Sk (n) − nk+1 /(k + 1). It is enough to
prove, that Rk (n) is a polynomial in n of degree not greater than k, because then
Sk (n) = nk+1 /(k + 1) + Rk (n) is a polynomial of degree k + 1. Note, that from the
definition of Sk (n) we have S(0) = 0 and for n ≥ 1 we have Sk (n) = Sk (n − 1) + nk .
Using this we compute
Rk (0) = Sk (0) − 0k+1 /(k + 1) = 0
and
Rk (n)−Rk (n−1) = Sk (n)−
(n − 1)k+1
(n − 1)k+1 − nk+1
nk+1
−Sk (n−1)+
= nk +
= pk (n).
k+1
k+1
k+1
Here pk (n) is a polynomial of degree less or equal than k + 1, so it can be written
as
pk (n) = nk +
(n − 1)k+1 − nk+1
= ak+1 nk+1 + ak nk + ak−1 nk−1 + · · · + a0 .
k+1
Note, that we can compute
(1)
ak+1 = 0 +
1−1
= 0,
k+1
ak = 1 +
−(k + 1) + 0
= 0.
k+1
Therefore pk (n) is a polynomial of degree less or equal than k − 1, so
pk (n) = ak−1 nk−1 + ak−2 nk−2 + · · · + a0 .
Thus Rk satisfies the following relations:
(2)
Rk (0) = 0,
Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 .
We claim, that this means, that
(3)
Rk (n) = ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n),
so Rk (n) is indeed a polynomial in n of degree less or equal than k, because by
induction hypothesis Sj (n) for j < k is a polynomial in n of degree j + 1. We can
check equation (3), for example, by induction in n (yes, this is an induction inside
4
HOMEWORK 1 SOLUTIONS
the induction step of another induction). We have Rk (0) = 0 = ak−1 0 + ak−2 0 +
· · · + a0 0 for the base case n = 0. For induction step we have
Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 =
ak−1 Sk−1 (n−1)+ak−2 Sk−2 (n−1)+· · ·+a0 S0 (n−1)+ak−1 nk−1 +ak−2 nk−2 +· · ·+a0 =
ak−1 (Sk−1 (n − 1) + nk−1 ) + ak−2 (Sk−2 (n − 1) + nk−2 ) + · · · + a0 (S0 (n − 1) + 1) =
ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n)
as desired.
Alternate solution to Homework 1 Problem 3.
Our goal is to show Sk (n) is a polynomial in n by induction on k. The base case
.
k = 1 is true since we know that S1 (n) = n(n+1)
2
Now we get to assume S1 (n), S2 (n), . . . , Sk−1 (n) are polynomials. We have a
k+1
guess that the leading term of Sk (n) is nk+1 . Thus we write a sum which equals
the guess by telescoping.
n
X j k+1
nk+1
(j − 1)k+1
=
−
k + 1 j=1 k + 1
k+1
n
k+1
k+1
X
k
k−1
l−1
2
l
=
j −
j
+ · · · + (−1)
j k−l+1 + . . .
k
+
1
k
+
1
j=1
k+1
k+1
2
Sk−1 (n) + · · · + (−1)l−1 l Sk−l+1 (n) + . . .
k+1
k+1
Now since every term in the equality between the first and fourth lines is known
to be a polynomial except for Sk (n), we determine that Sk (n) is a polynomial.
= Sk (n) −
Problem 4
Sx,y = {p − q | p < x, q > y, p ∈ Q, q ∈ Q}.
So, ∀r ∈ Sx,y , r = p − q with p < x, q > y, p ∈ Q, q ∈ Q.
=⇒ r = p − q < x − q < x − y.
So, x − y is an upper bound of Sx,y and therefore Sx,y is bounded above.
Take any real number > 0. ∃N ∈ N such that N > 2 . (why?)
=⇒ N 2 > 1.
=⇒ N (x − (x − 2 )) > 1 and N ((y + 2 ) − y) > 1.
=⇒ N x − N (x − 2 ) > 1 and N (y + 2 ) − N y > 1.
=⇒ ∃P, M ∈ N such that N x > M > N (x − 2 ) and N (y + 2 ) > P > N y. (why?)
P
=⇒ x > M
N > x − 2 and y + 2 > N > y.
M
P
M
P
P
=⇒ N − N ∈ Sx,y and N − N > (x − 2 ) − N
> (x − 2 ) − (y + 2 ) = x − y − .
Therefore, x − y is the least upper bound of Sx,y . (Make sure you understand
why this follows from the previous sentence)
[Something to think about: We have been using the preposition “the” before
“least upper bound”, but is that justified? Can a set of real numbers have more
than one least upper bound?]
HOMEWORK 1 SOLUTIONS
5
Problem 5
There are many ways to construct examples with the desired properties, and
here is one of them.
Take x1 = 0.111 · · · 1333 · · · 3 (n 1’s follwed by m−n 3’s) and y1 = 0.111 · · · 1666 · · · 6
(n 1’s follwed by m − n 6’s).
Now add an extra digit 5 at the end of the decimal expansion for x1 and y1 to
get x2 and y2 , respectively.
So x2 = 0.111 · · · 1333 · · · 35, y2 = 0.111 · · · 1666 · · · 65.
It is clear tm (x1 ) = x1 = tm (x2 ) and tm (y1 ) = y1 = tm (y2 )
On the other hand, we compute
x1 + y1 = 0.222 · · · 22999 · · · 9 (n 2’s follwed by m − n 9’s)
x2 + y2 = 0.222 · · · 23 (n − 1 2’s follwed by a 3).
Therefore we have tn (x1 +y1 ) = 0.222 · · · 22 (n 2’s) and tn (x2 +y2 ) = 0.222 · · · 23
(n − 1 2’s follwed by a 3).
Hence x1 , x2 , y1 and y2 have the desired properties.
``` | 3,761 | 8,072 | {"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} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.59548 |
https://www.khanacademy.org/math/old-ap-calculus-ab/ab-derivatives-advanced/ab-diff-log/a/differentiating-logarithmic-functions-review | 1,701,199,270,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00581.warc.gz | 939,185,855 | 128,279 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## AP®︎ Calculus AB (2017 edition)
### Course: AP®︎ Calculus AB (2017 edition)>Unit 4
Lesson 7: Derivatives of logarithmic functions
# Differentiating logarithmic functions review
Review your logarithmic function differentiation skills and use them to solve problems.
## How do I differentiate logarithmic functions?
First, you should know the derivatives for the basic logarithmic functions:
$\frac{d}{dx}\mathrm{ln}\left(x\right)=\frac{1}{x}$
$\frac{d}{dx}{\mathrm{log}}_{b}\left(x\right)=\frac{1}{\mathrm{ln}\left(b\right)\cdot x}$
Notice that $\mathrm{ln}\left(x\right)={\mathrm{log}}_{e}\left(x\right)$ is a specific case of the general form ${\mathrm{log}}_{b}\left(x\right)$ where $b=e$. Since $\mathrm{ln}\left(e\right)=1$ we obtain the same result.
You can actually use the derivative of $\mathrm{ln}\left(x\right)$ (along with the constant multiple rule) to obtain the general derivative of ${\mathrm{log}}_{b}\left(x\right)$.
## Practice set 1: argument is $x$
Problem 1.1
$h\left(x\right)=7\mathrm{ln}\left(x\right)$
${h}^{\prime }\left(x\right)=?$
Want to try more problems like this? Check out this exercise.
## Practice set 2: argument is a polynomial
Problem 2.1
$g\left(x\right)=\mathrm{ln}\left(2{x}^{3}+1\right)$
${g}^{\prime }\left(x\right)=?$
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• From my understanding, you'd like help with how to differentiate x^x. This is how you do it:
y=x^x
Take the logs of both sides:
ln(y) = ln(x^x)
Rule of logarithms says you can move a power to multiply the log:
ln(y) = xln(x)
Now, differentiate using implicit differentiation for ln(y) and product rule for xln(x):
1/y dy/dx = 1*ln(x) + x(1/x)
1/y dy/dx = ln(x) + 1
Move the y to the other side:
dy/dx = y (ln(x) + 1)
But you already know what y is... it is x^x, your original function. So sub in:
dy/dx = x^x(ln(x) + 1)
And you're done.
• I have a natural logarithm with e^x/1+e^x. I separated it with the log rules but then I'm stuck. Any advice?
• i think you are asking about finding d/dx( ln( e^x / 1 + e^x) ). so im solving for that and here it is:
we can write ==> ln(e^x / 1+e^x) as ln(e^x) - ln(1+e^x)
so now when we differentiate we can differentiate them independently.
so d/dx( ln( e^x / 1 + e^x) ) = d/dx( ln(e^x) ) - d/dx( ln(1+e^x) )
= ( (1/e^x) *e^x ) -( ( 1/(1+e^x) ) * e^x )
let me know if we have any confusion.
• can this statement be true?
2 log y = log y^2
(1 vote)
• Are there “rules” for when you can(not) use logarithmic differentiation (including implicit)?
I ask because of the following KA problem: “Find dy/dx for x=√(xy+1)” For that problem I attempted to immediately use logarithmic differentiation, e.g. ln(x)=ln(√(XY+1)).
However having now worked on it a good deal I have come to understand that logarithmic differentiation generates an incorrect result.
Why doesn’t logarithmic differentiation work in this case? (I speculate that perhaps it is because there is a single term that has more than one variable – e.g. XY messes it up – but that is just a guess).
Note that the following answer is not sufficient: “You shouldn’t use logarithmic differentiation on that problem.” E.g. I (now) understand it won’t work - I want to know WHY it doesn’t work - what is the rule I should use so that I don't try to do that again in the future? 😉.
Thanks,
kevin
• Logarithmic differentiation should work here. Can you provide us with your steps so we can perhaps find an error?
• What do you do if the "x" is not simply x, but is raised to a power or if the equation is log base 4 of x-2?
(1 vote)
• 1/x(ln(a)) or I could do natural log of y equals (the power- assuming that the power is a variable) times the natural log of x. If the power is a number I would multiply it by the coefficient of x and subtract 1 from the exponent. | 1,189 | 4,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2023-50 | latest | en | 0.76674 |
https://www.next.cc/journey/discovery/topography | 1,656,510,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00014.warc.gz | 994,420,034 | 11,580 | # Topography
Maps are representations. Topographic maps are maps that reveal the surface of the earth from the tops of mountains to the depths of the oceans. The variance in elevation of the earth’s surface is shown by contour lines. Contour lines are imaginary lines that are drawn connecting points of equal elevation. Usually contour lines represent height above or below mean sea level or 0. When contour lines are close together, it means that the land or earth’s surface is descending or ascending steeply. In contrast, when contour lines are far apart it means that the earth’s surface is relatively flat. Contour lines never cross. Each contour line represents a separate elevation. Different maps use different intervals between contour lines depending upon the scale of the map. Flat areas may use contours every ten feet while mountainous areas may strike a contour every one hundred feet. Colors on topographic maps are use to show such features as vegetation (green), water (blue) and built up areas (gray, beige or red). The lines connecting areas may be straight, curved, solid, dashed, dotted, or any combination. Some lines may also show railroad tracks, streams, and even state and county boundaries. In addition to lines on topographic maps, there are many types of symbols. Symbols may locate significant buildings, historical sites, campgrounds, springs, wells, control points.
## Activity 1 – Lay of the Land
Does the land rise or does it fall? Does it make mountains or does it have valleys. Is it flat or gently sloping? Take a look at the ARC GIS USA Topographic Map to learn about the lay of the land where you live. Find the highest elevations in your city. Find the lowest elevation in your city. See how many feet above sea level you live! You can watch National Geographic’s Drain the Oceans Series to see what scientists are discovering on the sea floor. You might also explore Basin and Range Maps which show the height of mountain ranges and the lower depth of plains and basins. Hiking is a great way to understand the slope of the land. On one 11 x 17 inch paper, draw an imaginary topography that has both high points and low points. Be sure to include a north arrow. Draw 1’ contours and label every 5th contour.
## Activity 2 – Contours and Grades
Lines are everywhere on contour maps! Lines represent elevation changes, political boundaries, railroads, roads and rivers, lakes and canals. Take a look at the United States Geographic System’s Topographic Map Symbols. Make an index of road types, water feature types, territorial boundary types and elevation contours. Be sure to look at the color coding that is used and the thinness or thickness of the lines. Create a topographic map that shows the main roads coming to your neighborhood, any water features, boundaries of blocks and land contours.
## Activity 3 – Topographic Symbols
Looking at the United States Geographic System’s Topographic Map Symbols, find the symbol for glaciers, marshes, sand, gravel, woods, shrubs, orchards and vineyards. Draw a map of a natural area that has several of these vegetative and water features.
## Activity 4 – Draw a Contour Map of your neighborhood
There are many online maps that document the contours and topos of the land. Take a look at The National Geographic World Map. Zoom in to find your neighborhood. You can see the streets and buildings and find the contours. Make a drawing that shows the slope of the land. The slope of the land reveals how the precipitation runs off (or is absorbed) in given location based on land development and land type. You can also check your neighborhood by checking the wikiwatershed or looking at the Watershed Journey to see how topos effect collection of rainwater. When you make your drawing of your block in your neighborhood, show the North Arrow, and outline the houses and use a lighter line weight for the contours. Draw arrows which shows which way the water will drain in your neighborhood. Does it eventually drain into a stream, river, lake or ocean?
## Activity 5 – Draw an Imaginary Topographical park
Imagine and plan a topographic garden. As a class project, the teacher can roll out brown paper and divide it into 11 x 17 plots. Students each can pick a plost and start drawing contours. Each student should drawing 20 contours matching the end points with contours that students beside them are drawing. Once students have made their contour maps, they can construct contour models out of 1/16” ply cardboard. It should have a minimum of 20 (1’) contours. Ten of the contours should build above the 0 level (+1 to +10) and 10 of the contours should build below the 0 level (-1 to -10’). This will give the models a 20’ (or two story variance). The base model should be a 2” x 12” x 17” cardboard ‘box’ with subtractive (–) contours opening down into low laying area(s) in the box and additive (+) contours building hills above the box. Decide where it is high and low allowing for one high flat plane that is app. 4” x 6” and a low flat plane that is at least 2” x 4”. These ‘flat areas’ will be possible building areas and should be marked with dotted pencil on your model. This will affect designs later on as you will respond to the site that you have created. In addition, draw the site showing all contours in plan and in two cross sections at 1’ = 1/16”.
After fitting the land models together, students can further develop this model by adding five pavilions based on the pavilion journey that are above the land, on, in, or under the land!
## Activity 6 – Skateboard Park
As an extra challenge, students can create a skateboard park. As in any design process it is good to begin to look at definitions of skateboard parks, sizes, history, longevity, etc. Factors in Design of SkateBoard Parks is a good starting point. Visiting a local one is also a good place to start. Consider the different sizes of parks and the definition of public place and skate rooms. Skate parks will have different elevations and topographic changes to challenge an array of skateboarding skill levels. Make a model of a skateboard park and share it with your class.
## Review
• contour lines are recorded every
• The height of a city is relative to
• Topography refers to the ______ of the land.
• Topography maps use
• The closer the contour lines are drawn to each other, the
check answers | 1,371 | 6,389 | {"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-2022-27 | longest | en | 0.919528 |
https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/10/05/50/0011/ | 1,719,097,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00486.warc.gz | 214,605,721 | 10,737 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Hypergeometric2F1
http://functions.wolfram.com/07.23.03.apsn.01
Input Form
Hypergeometric2F1[-(24/5), 6, 1/5, z] == (1/48828125) (9576 (3125/(-1 + z) - 69426 (-(5/4) - z^(4/5) (Log[1 - z^(1/5)] + (-1)^(4/5) Log[1 + (-1)^(1/5) z^(1/5)] - (-1)^(3/5) Log[1 - (-1)^(2/5) z^(1/5)] + (-1)^(2/5) Log[1 + (-1)^(3/5) z^(1/5)] - (-1)^(1/5) Log[1 - (-1)^(4/5) z^(1/5)])) + 1049104 ((-(5/36)) (4 + 9 z) - z^(9/5) (Log[1 - z^(1/5)] + (-1)^(4/5) Log[1 + (-1)^(1/5) z^(1/5)] - (-1)^(3/5) Log[1 - (-1)^(2/5) z^(1/5)] + (-1)^(2/5) Log[1 + (-1)^(3/5) z^(1/5)] - (-1)^(1/5) Log[1 - (-1)^(4/5) z^(1/5)])) - 4383756 ((-(5/252)) (18 + 28 z + 63 z^2) - z^(14/5) (Log[1 - z^(1/5)] + (-1)^(4/5) Log[1 + (-1)^(1/5) z^(1/5)] - (-1)^(3/5) Log[1 - (-1)^(2/5) z^(1/5)] + (-1)^(2/5) Log[1 + (-1)^(3/5) z^(1/5)] - (-1)^(1/5) Log[1 - (-1)^(4/5) z^(1/5)])) + 6767904 (-(5/19) - (5 z)/14 - (5 z^2)/9 - (5 z^3)/4 - z^(19/5) (Log[1 - z^(1/5)] + (-1)^(4/5) Log[1 + (-1)^(1/5) z^(1/5)] - (-1)^(3/5) Log[1 - (-1)^(2/5) z^(1/5)] + (-1)^(2/5) Log[1 + (-1)^(3/5) z^(1/5)] - (-1)^(1/5) Log[1 - (-1)^(4/5) z^(1/5)])) - 3454451 (-(5/24) - (5 z)/19 - (5 z^2)/14 - (5 z^3)/9 - (5 z^4)/4 - z^(24/5) (Log[1 - z^(1/5)] + (-1)^(4/5) Log[1 + (-1)^(1/5) z^(1/5)] - (-1)^(3/5) Log[1 - (-1)^(2/5) z^(1/5)] + (-1)^(2/5) Log[1 + (-1)^(3/5) z^(1/5)] - (-1)^(1/5) Log[1 - (-1)^(4/5) z^(1/5)]))))
Standard Form
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MathML Form
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Rule Form
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Date Added to functions.wolfram.com (modification date)
2007-05-02 | 9,053 | 24,035 | {"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-2024-26 | latest | en | 0.497557 |
https://math.stackexchange.com/questions/804106/conditional-probability-of-continuous-independent-random-variables | 1,632,496,170,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00322.warc.gz | 439,395,089 | 40,516 | # conditional probability of continuous independent random variables
I would like to know what's the exact way to obtain the conditional probabilities of node having multiple independent continuous random variables as its parent.
Say something like a Noisy OR gate model, with multiple parents of continuous random variable types.
The samples, I stumbled upon so far (for Noisy OR gate) where explained using discrete binary random variables, but my interest mostly lies on achieving similar outcomes with one or more continuous & random as inputs?
Let me explain with a simple example from Google results. A simple study says Bronchitis (B), Tuberculosis (T) and Lung cancer (L) causes Fatigue (F). It also says probability of B -> F = 0.6, T -> F = 0.7, L -> F = 0.8. with this information if we had to find out P(F|B,T,L), then the way to do this is find out
P(no F|B, T, L) = (1-0.6)(1-0.7)(1-0.8) = 0.024
then P(F|B,T,L) = 1 - P(no F|B, T, L) = 0.976.
In the above case all the causation random variable are of type discrete. In case if we add a continuous random variable say age to the scenario.
B -> F = 0.6, T -> F = 0.7, L -> F = 0.8
and if age (a) is
a <= 30 -> F = 0.1,
30 > a <= 50 -> F = 0.2, and
a > 50 -> F = 0.7.
Then if I had to find out P(F|B,T,L,30 > a <= 50), then what should be logic to find it out?
If just look at the example without considering age, we never know the joint probability distribution of of two or more individual events, but the final conditional probability of F given B,L & C has happened at ones is obtained using the individual probabilities of B,L & C.
Similarly I would like to know the way to compute P(F|B,T,L,30 > a <= 50)? Where 'a' is a discretized continuous random variable to ease the analysis.
Thanks, Kamal.
• In the title of the post it looks like you're asking for the joint probability, but in the first sentence you've written that you're looking for conditional probabilities, so I may be misunderstanding your question, but both are pretty easy to compute for independent continuous random variables. $f(x,y)=f_X(x)f_Y(y)$ and $f_{X|Y}(X|Y)=f_X(x)$. May 21 '14 at 18:00
• Thanks for pointing out the mistake, I was looking for conditional probability. May I request you to help me with an example. May 22 '14 at 5:22
• Thanks for the edits, they've made things clearer. The event that $30\lt a\le 50$ can be treated just like B, T, and L. $P(F|B,T,L,30\lt a\le 50) = 1-(1-0.6)(1-0.7)(1-0.8)(1-0.2)=0.9808$. May 22 '14 at 6:14
• Hi, This is where I was having my doubt, can I just pick (1−0.2) as such! Let me modify the example a bit for my clarification. Say we have 100 data points measuring F to Yes or No just because of age. Upon grouping them by age, we see the 1st 29 are for age <30 and it says F=Y for 16 and F=N for 13 data points. 2nd 30 (for 30 >age< 50) says F=Y for 8 and F=N for 22. The last 41 (for age > 50) says F=Y for 17 and F=N for 24. (continued)=> May 22 '14 at 9:14
• If we draw a cross table of probabilities, we see probabilities of F=Y & N across ages as 0.16, 0.13 | 0.08, 0.22 | 0.17, 0.24. Now assume one fine day 3 patients across different age groups got admitted and if a Nurse has to answer the probability of F? then as per our computation F=> 1 - (1-0.16)*(1-0.08 )*(1-0.17) = 0.359, which is less than sum of all the individual probabilities 0.16 + 0.08 + 0.17 = 0.41. Am not sure which is correct either 0.359 or sum of the independent individual probabilities 0.41. May 22 '14 at 9:15
Let's denote the presence of a cause of fatigue by some letter $X$ and the event that that cause leads to fatigue by $X_f$, so that the probability that a cause of fatigue leads to fatigue is $P(X_f)$. Given the scenario you describe regarding the independent causes of fatigue, $F$, then you're right to say $$P(F|B,T,L)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))$$
Now we consider an additional cause to fatigue that is independent of the other causes, in this case age $A$, that has different probabilities of leading to fatigue depending it's value. Given that $P(A_f|A\lt 30)=0.1, P(A_f|30\lt A\le 50)=0.2$, and $P(A_f|A\gt 50)=0.7$, we can now treat each of these cases as causes themselves that are independent to $B, L,$ and $T$. So, for example $$P(F|B,T,L,30\lt A\le 50)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))\left(1-P(A_f|30\lt A\le 50)\right)$$
In the other scenario you describe in the comments above we have the following data (modified slightly so that it adds up): 16 of the first 29 ages lead to $F$, 8 of the next 21 ages lead to $F$, and 17 of the last 50 ages lead to $F$. If we assume that each patient is equally likely to be any of the ages within their age group then we can write that data like this: $$P(A_f|A\lt 30)=\frac{16}{29}$$$$P(A_f|30\le A\le 50)=\frac{8}{21}$$$$P(A_f|A\gt 50)=\frac{17}{50}$$ Now this is the same as the original scenario in your question and we'd still be able to use the same general formula, for example: $$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-P(A_f|30\le A\le 50)\right)$$
If three patients, one in each age group, came to see a nurse then the probability that one or more of them had fatigue based solely on their age would be: $$1-\left(1-P(A_f|A\lt 30)\right)\left(1-P(A_f|30\le A\le 50)\right)\left(1-P(A_f|A\gt 50)\right)=$$ $$1-\left(1-\frac{16}{29}\right)\left(1-\frac{8}{21}\right)\left(1-\frac{17}{50}\right)\approx0.817$$
Let's say that $A$ isn't a set of discrete points, though, and instead say that it's modeled by a continuous distribution. Then we'd have a probability density function, $f_A(a)$, such that $P(A_f|s\lt A\lt t)=\int_s^t{f_A(a)}da$ and we'd still be able to use the same general formula as we did in the discrete case: $$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-\int_{30}^{50}f_A(a)da\right)$$ | 1,855 | 5,881 | {"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} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.92619 |
https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/IOST_Library/Spring_2024%3A_Physical_Computing_Class/08%3A_Python_Functions_Packages_and_Formatting/8.09%3A_Modulus_Method | 1,716,714,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00041.warc.gz | 131,493,355 | 29,258 | # 8.9: Modulus Method
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
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This is the older method and you will often find it in packages, especially code developed for older sensors, and so it is important that you are familiar with it, but we will not require you to write code using this method. You should still skim through this material, and be prepared to reference it when needed. Before we get into formatting syntax we need to identify the type, as the way a computer stores data depends on its type.
## Types
Remember the type() function lets you know the type of object you are dealing with, and there are four common data types we deal with in formatting
• %s - represents strings
• %i - represents integers
• %d - represents decimal integers
• %f - represents floats
## Generic Syntax
The following code depicts this for the float 1.00784
print("The molar mass of hydrogen is: %1.2f"%(1.00784))
print("The molar mass of hydrogen is: %10.2f"%(1.00784))
print("The molar mass of hydrogen is: %1.0f"%(1.00784))
print("The molar mass of hydrogen is: %1.5f"%(1.00784))
Hello world!
Now go into the fourth line of the above code and change the %1.5f to a %1.5i and look at how the output is changed. Well, you are formatting a floating number as an integer type, which makes no sense.
Now lets format a couple of variables, one being a string and the other a float
#here we are assigning two variables in one line, and then printing them in a line
entity, molar_mass="water", 18.01528
print("The molar mass of %s is %.3f g/mol." %(entity,molar_mass))
print("The molar mass of %s is %.5f g/mol." %(entity,molar_mass))
print("The molar mass of %s is %.3d g/mol." %(entity,molar_mass))
Hello world!
##### Exercise $$\PageIndex{1}$$
Can you explain the last line of code? Can you fix it? | 1,117 | 3,403 | {"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": 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} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.363487 |
https://dev.to/theabbie/power-of-four-4269 | 1,721,244,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00842.warc.gz | 180,890,108 | 47,801 | ## DEV Community
Abhishek Chaudhary
Posted on
# Power of Four
Given an integer `n`, return `true` if it is a power of four. Otherwise, return `false`.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
Example 1:
Input: n = 16
Output: true
Example 2:
Input: n = 5
Output: false
Example 3:
Input: n = 1
Output: true
Constraints:
• `-231 <= n <= 231 - 1`
Follow up: Could you solve it without loops/recursion?SOLUTION:
``````class Solution:
def isPowerOfFour(self, n: int) -> bool:
bi = "{:b}".format(n)
return n > 0 and bi.count("1") == 1 and bi[::-1].index("1") % 2 == 0
`````` | 209 | 628 | {"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-2024-30 | latest | en | 0.571768 |
https://academiaservices.net/21147718406/ | 1,627,812,295,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00299.warc.gz | 92,867,039 | 10,374 | ## (solution) I need help understanding the problems i submitted to you Download Attachment:
I need help understanding the problems i submitted to you Download Attachment: Document Preview: 0) A normal population has a mean of 20 and a standard deviation of 4. Compute the z value associated with 25. =1.25 What proportion of the population is between 20 and 25? What proportion of the population is less than 18? 1) Assume that the hourly cost to operate a commercial airplane follows the normal distribution with a mean of \$3,236 per hour and a standard deviation of \$432. What is the operating cost for the lowest 5% of the airplanes? 3) A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.50 minutes and the standard deviation was 0.70 minutes. A)What fraction of the calls last between 4.50 and 5.30 minutes? B) What fraction of the calls last more than 5.30 minutes? C) What fraction of the calls last between 5.30 and 6.00 minutes? D) What fraction of the calls last between 4.00 and 6.00 minutes? E) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 5 percent of the calls. What is this time? 3) The mean age at which men in the United States marry for the first time follows the normal distribution with a mean of 24.1 years. The standard deviation of the distribution is 2.5 years. A) For a random sample of 66 men, what is the likelihood that the age at which they were married for the first time is less than 24.6 years? 2) Assume that the hourly cost to operate a commercial airplane follows the normal distribution with a mean of \$3,236 per hour and a standard deviation of \$432. What is the operating cost for the lowest 5% of the airplanes?
Solution details:
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EXPERT
Tutor | 476 | 2,017 | {"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} | 3.5625 | 4 | CC-MAIN-2021-31 | latest | en | 0.936853 |
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For some remote sensing instruments, the distance between the target being imaged and. Picture elements, or pixels, which are the smallest units of an image. The Big Eight Elements of Image Interpretation. Eight fundamental parameters are used in the interpretation of remote sensing images: size, shape, tone. Topic 4: Remote Sensing Platforms and Sensors. Element Sets pdf file - NORAD maintains general perturbation element sets on all resident space objects. by remote sensing, within the framework of the FGDC 1998 Content Standard. Definitions, and permissible values for new data elements and the compound. Canada. These seven elements comprise the remote sensing process from beginning to end. Remote Sensing: is the collection of information relating to objects without being in. Application G - The final element of the remote sensing process is. Oct 1, 2003. Primary components of remote sensing are as follows.
Mac studio tech foundation tutorial nc40 | 2,067 | 9,243 | {"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-2018-34 | latest | en | 0.33816 |
https://metanumbers.com/1022286 | 1,643,086,137,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00314.warc.gz | 445,717,067 | 7,473 | # 1022286 (number)
1,022,286 (one million twenty-two thousand two hundred eighty-six) is an even seven-digits composite number following 1022285 and preceding 1022287. In scientific notation, it is written as 1.022286 × 106. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 335,544 positive integers (up to 1022286) that are relatively prime to 1022286.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 21
• Digital Root 3
## Name
Short name 1 million 22 thousand 286 one million twenty-two thousand two hundred eighty-six
## Notation
Scientific notation 1.022286 × 106 1.022286 × 106
## Prime Factorization of 1022286
Prime Factorization 2 × 3 × 67 × 2543
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1022286 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,022,286 is 2 × 3 × 67 × 2543. Since it has a total of 4 prime factors, 1,022,286 is a composite number.
## Divisors of 1022286
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 2.0759e+06 Sum of all the positive divisors of n s(n) 1.05362e+06 Sum of the proper positive divisors of n A(n) 129744 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1011.08 Returns the nth root of the product of n divisors H(n) 7.87925 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,022,286 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 1,022,286) is 2,075,904, the average is 129,744.
## Other Arithmetic Functions (n = 1022286)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 335544 Total number of positive integers not greater than n that are coprime to n λ(n) 83886 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 79953 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 335,544 positive integers (less than 1,022,286) that are coprime with 1,022,286. And there are approximately 79,953 prime numbers less than or equal to 1,022,286.
## Divisibility of 1022286
m n mod m 2 3 4 5 6 7 8 9 0 0 2 1 0 6 6 3
The number 1,022,286 is divisible by 2, 3 and 6.
## Classification of 1022286
• Arithmetic
• Abundant
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
## Base conversion (1022286)
Base System Value
2 Binary 11111001100101001110
3 Ternary 1220221022110
4 Quaternary 3321211032
5 Quinary 230203121
6 Senary 33524450
8 Octal 3714516
10 Decimal 1022286
12 Duodecimal 413726
20 Vigesimal 67fe6
36 Base36 lwsu
## Basic calculations (n = 1022286)
### Multiplication
n×y
n×2 2044572 3066858 4089144 5111430
### Division
n÷y
n÷2 511143 340762 255572 204457
### Exponentiation
ny
n2 1045068665796 1068359066081929656 1092168516228631540313616 1116508583781302822821045246176
### Nth Root
y√n
2√n 1011.08 100.737 31.7975 15.919
## 1022286 as geometric shapes
### Circle
Diameter 2.04457e+06 6.42321e+06 3.28318e+12
### Sphere
Volume 4.47513e+18 1.31327e+13 6.42321e+06
### Square
Length = n
Perimeter 4.08914e+06 1.04507e+12 1.44573e+06
### Cube
Length = n
Surface area 6.27041e+12 1.06836e+18 1.77065e+06
### Equilateral Triangle
Length = n
Perimeter 3.06686e+06 4.52528e+11 885326
### Triangular Pyramid
Length = n
Surface area 1.81011e+12 1.25907e+17 834693
## Cryptographic Hash Functions
md5 19aac857224f8ede4c5cdc1aea33b683 1fcf0e24ac63c9dcf2fdf6c5fd6781eb2ecee284 ad0858f2934259d04ace963f540729bc969b298541b30bbfbda24deb182127e1 b997bf6eb136578c739c232dbaf9cb0819dcb75edf8965d4324b4077a34beba82407244fe98c3294778376af87df719a1d5550a7e67d84f785cf6aed4a8db397 e172e79c590590ca49a316cc09d42cfa291ee4a6 | 1,535 | 4,347 | {"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} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.808417 |
http://apcalc.blogspot.com/2005/09/scribe_12.html | 1,532,152,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00060.warc.gz | 24,923,675 | 9,527 | <body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14085554\x26blogName\x3dAP+Calculus+AB\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttps://apcalc.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://apcalc.blogspot.com/\x26vt\x3d2092189443322278047', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>
# AP Calculus AB
An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
## Monday, September 12, 2005
### SCRIBE
Another productive class don't you think? Again, we got into groups and tried to create a graph representing our little story from Andover to Concord. I think that it was a bit too quiet... or maybe I just have bad hearing? During this activity, we all had different ideas on how we created our graphs- which is a good thing seeing how we can all learn from each other's mistakes. After some contemplation and discussion, we were able to understand how the graph was to be displayed. But hey, some of us might be in "summer mode" still and getting adjusting to the homework and the early morning wake ups.
It probably was a big refreshment to some of us on how to create graphs in our calculators- it was for me. We were taught how to determine if a graph was a line by using the calculator to create an equation and using it to draw a line through our points. We also learned about the r² function which determines how accurately it will go through our five points.
So, it's our first day of real calculus homework! And hopefully, no one struggles too much on it. As the next couple days go, we'll all remember work from s3 & s4 pre-cal! Don't let the forgetting curve get you during this course because it's something that would be very helpful in the future. Anyway, let's not have such a quiet class and start interacting! Let's have a fun year everyone :)
-Glenney
Next scribe. . . . Janet :) | 657 | 2,670 | {"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-2018-30 | latest | en | 0.853248 |
https://www.askiitians.com/forums/Modern-Physics/what-is-pumped-steadily-out-of-a-flooded-basement_96223.htm | 1,714,053,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00284.warc.gz | 564,224,906 | 43,189 | # What is pumped steadily out of a flooded basement at a speed of 5.0m/sthrough a uniform hose of radius 1.0 cm. The hose passes out through awindow 3.0m above the water line. What is the power of the pump?
Nirmal Singh.
10 years ago
Suppose that a mass delta m of water is pumped in time delta t. The pump increases the potential energy of the water by Delta mgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by 1 /2 delta mv^2 , where v is its final speed. The work it does is
Delta W = delta m gh +1/2 delta mv^2
power P = Delta W / delta t
P = delta m gh +1/2 delta mv^2 / delta t
P = delta m / deta t (gh+1/2v^2)
Now the rate of mass delta m /delta t = rho *A*v where rho is density of water and A is the area of the hose. The area of the hose A = pi*r^2 = 3.14*(0.01)^2 = 3.14*10^-4 m^2
rho*A*v = 1000*3.14x10^-4x5 = 1.57 kg/sec
Thus P = P = delta m / deta t (gh+1/2v^2) = 1.57*[9.8*3 + 5^2 / 2]
P = 66 W
Thanks & Regards,
Nirmal Singh | 342 | 991 | {"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-2024-18 | latest | en | 0.915535 |
https://kr.mathworks.com/matlabcentral/profile/authors/15439772?detail=answers | 1,686,334,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00351.warc.gz | 418,988,376 | 22,803 | Community Profile
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Here is information about "seven hundred fifty-one" that you may find useful and interesting. Number Systems Seven hundred fifty-one is a decimal number and can be written with numbers: 751 Binary is a number system with only 0s and 1s. Seven hundred fifty-one in binary form is displayed below: 1011101111 A Hexadecimal number has a base of 16 which means it includes the numbers 0 to 9 and A through F. Seven hundred fifty-one converted to hexadecimal is: 2EF Roman Numerals is another number system. Below is seven hundred fifty-one in roman numerals: DCCLI Scientific Notation Sometimes calculators and scientists shorten numbers using scientific notation. Here is seven hundred fifty-one as a scientific notation: 7.51E+02 Math Here are some math facts about Seven hundred fifty-one: Seven hundred fifty-one is a rational number and an integer. Seven hundred fifty-one is an odd number because it is not divisible by two. Seven hundred fifty-one is divisible by the following numbers: 1, 751 Seven hundred fifty-one is not a square number because no number multiplied by itself will equal seven hundred fifty-one. Number Lookup Seven hundred fifty-one is not the only number we have information about. Go here to look up other numbers.
Translated Here we have translated seven hundred fifty-one into some of the most commonly used languages: Chinese: 七百五十一 French: sept cent cinquante et un German: siebenhundert einundfünfzig Italian: settecento cinquantuno Spanish: setecientos cincuenta y uno
Currency Here is seven hundred fifty-one written in different currencies: US Dollars: \$751 Canadian Dollars: CA\$751 Australian Dollars: A\$751 British Pounds: £751 Indian Rupee: ₹751 Euros: €751
Ordinal The cardinal number seven hundred fifty-one can also be written as an ordinal number: 751st Or if you want to write it with letters only: seven hundred fifty-first.
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