Split (1)

train · 167k rows

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https://it.mathworks.com/matlabcentral/profile/authors/23804087	1,642,995,825,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00605.warc.gz	346,129,028	20,994	Community Profile # ShiQinChen18049021 Last seen: circa 2 mesi ago Active since 2021 #### Content Feed View by Solved Calculate a modified Levenshtein distance between two strings Inspired by the Cody problem found <http://www.mathworks.com/matlabcentral/cody/problems/93-calculate-the-levenshtein-distance-b... 3 mesi ago Solved Convert a structure into a string Convert the contents of each fields into a string. Example with an input structure s with 2 fields : s.age = '33' s.... 3 mesi ago Solved Find mistyped words in text (mixed-up letters) Mistyped words are a regular occurrence in emails, texts, status updates, and the like. Many times, people send or post a second... 3 mesi ago Solved QWERTY Shift Code Decoder Decode a string encoded using the QWERTY shift code. QWERTY shift code is where the message was touch typed but with an offse... 3 mesi ago Solved Find code that adds one 0 to each sequence of 0 in a string (composed with only 0 or 1). For example: '1010' -> '100100' ... 3 mesi ago Solved Put two time series onto the same time basis Use interpolation to align two time series onto the same time vector. This is a problem that comes up in <http://www.mathwork... 3 mesi ago Solved Find matching parenthesis One of the most indispensable things about a great text editor for programming is the ability to quickly jump between matching p... 3 mesi ago Solved Space Saver Remove all characters that are below a space in ASCII value. 3 mesi ago Solved Reverse the Words (not letters) of a String Description Change the words of a string such that the words appear in reverse order. You may assume that the string is a n... 3 mesi ago Solved Find Rotated Substring Given a string s1, find if a rotated version of s1 is present in a second string s2. For example, rotated version of some str... 3 mesi ago Solved Find matching string from a list of strings Write a function that returns a string that is a unique match (if it exists) of the string \|inStr\| from a list of strings \|strLi... 3 mesi ago Solved Convert String to Morse Code Convert a given string to international <http://en.wikipedia.org/wiki/Morse_code Morse Code> * Strings will only have [A-z], ... 3 mesi ago Solved letter yes yes & letter no no Split a string into two strings, wherein the first string has all alphabetic letters and the second string has all the remaining... 3 mesi ago Solved Compress strings (not springs) Please remove excess space, limit one space between others, and no space before punctuation marks. * For example, 'Trendy , ... 3 mesi ago Solved Bang Bang in Bangalore Imagine a strange language disorder, Bangolangosis, has developed among trigonometricians of <http://en.wikipedia.org/wiki/Banga... 3 mesi ago Solved Formatting currency numbers Given a number, format it properly for textual display using the notation \$xxx,xxx,xxx.xx. Assume that no more than two digits ... 3 mesi ago Solved English to Pig Latin Translator Pig latin is a faux-language based off of English. The rules are as follows (excerpted from the wikipedia entry for Pig Latin): ... 3 mesi ago Solved Pig Latin to English Translator Pig latin is a faux-language based off of English. The rules are as follows (excerpted from the <http://en.wikipedia.org/wiki/Pi... 3 mesi ago Solved String substitution, sub problem to cryptoMath This is a sub problem related to this problem: <http://www.mathworks.com/matlabcentral/cody/problems/621-cryptomath-addition>... 3 mesi ago Solved Use of regexp Given a string, containing several sentences, such as: 'I played piano. John played football. Anita went home. Are you safe?... 3 mesi ago Solved QWERTY Shift Encoder Encode a string using the QWERTY shift code. This code is where you touch type but are offset by one character to the right. ... 3 mesi ago Solved Calculate the Hamming distance between two strings Inspired by a similar Cody problem found <http://www.mathworks.com/matlabcentral/cody/problems/93-calculate-the-levenshtein-dist... 4 mesi ago Solved Suppress the French accent. Example 'Déjà présent' -> 'Deja present' 4 mesi ago Solved Ordinal numbers Given an integer n, return the corresponding ordinal number as a character string. For example, ord(1)='1st' ord(2)=... 4 mesi ago Solved Split a string into chunks of specified length Given a string and a vector of integers, break the string into chunks whose lengths are given by the elements of the vector. Ex... 4 mesi ago Solved Cell Counting: How Many Draws? You are given a cell array containing information about a number of soccer games. Each cell contains one of the following: * ... 4 mesi ago Solved Transposition as a CIPHER This all about transcripting a text message. If the input string is: s1 = 'My name is Sourav Mondal', then the output is: s2 = '... 4 mesi ago Solved Decimal Comparison Background A utility of particular interest to Cody and other MATLAB ventures is comparing the equality of two numbers. In ... 4 mesi ago Solved Guess Cipher Guess the formula to transform strings as follows: 'Hello World!' --> 'Ifmmp Xpsme!' 'Can I help you?' --> 'Dbo J ifm... 4 mesi ago Solved Finding peaks Find the peak values in the signal. The peak value is defined as the local maxima. For example, x= [1 12 3 2 7 0 3 1 19 7]; ... 4 mesi ago	1,352	5,355	{"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}	2.75	3	CC-MAIN-2022-05	latest	en	0.783731
https://equationbalancer.com/fe-oh-3	1,679,324,822,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00611.warc.gz	289,907,853	11,483	## Word equation iron(II) cation + hydroxide anion →→ iron(II) hydroxide ## Input interpretation Fe2+ + OH- → Fe(OH)3 iron(II) cation hydroxide anion iron(II) hydroxide ## Balanced equation Balance the chemical equation algebraically: Fe2+ + OH- → Fe(OH)3 Add stoichiometric coefficients, c1, to the reactants and products: c1Fe2+ + c2 OH-c3 Fe(OH)3 Set the number of atoms and the charges in the reactants equal to the number of atoms and the charges in the products for Fe, O and H: Fe: c1= c3 O: c2 = 2 c3 H: c2= 2c3 Charges: 2 c1- c2=0 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c1 = 1 and solve the system of equations for the remaining coefficients: c1 = 1 c2=2 c3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation:	274	936	{"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}	3.28125	3	CC-MAIN-2023-14	latest	en	0.876547
https://large-numbers.fandom.com/wiki/TREE_sequence	1,576,309,826,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00186.warc.gz	424,589,390	78,159	## FANDOM 1,079 Pages Not to be confused with Trree. Template:Infobox function The TREE sequence is a fast-growing function arising out of graph theory, devised by mathematical logician Harvey Friedman.[1][2] Friedman showed that the function eventually dominates all recursive functions provably total in the system $$\text{ACA}_0$$+$$\Pi_2^1$$-$$\text{BI}$$.[3] The smallest nontrivial member of the sequence is the famously large TREE(3), notable because it is a number that appears in serious mathematics that is larger than Graham's number. ## Definition Edit Suppose we have a sequence of k-labeled trees T1, T2 ... with the following properties: 1. Each tree Ti has at most i vertices. 2. No tree is homeomorphically embeddable into any tree following it in the sequence. Kruskal's tree theorem states that such a sequence cannot be infinite. Harvey Friedman expanded on this by asking the question: given k, what is the maximum length of such a sequence? This maximal length is a function of k, dubbed TREE(k). The first two values are TREE(1) = 1 and TREE(2) = 3. The next value, TREE(3), is famously very large. It vastly exceeds Graham's number and nn(5)(5)[4]. Chris Bird has shown that $$\text{TREE}(3) > \{3, 6, 3 [1 [1 \neg 1,2] 2] 2\}$$, using his array notation.[5] TREE(n) grows at least as fast as $$f_{\vartheta(\Omega^\omega\omega)}(n)$$ in the fast-growing hierarchy, making it quite sizable even to a googologist — $$\vartheta(\Omega^\omega\omega)$$ is stronger than BEAF's linear array-arrays. It is more powerful than Kirby-Paris hydras and the Goodstein sequences, but weaker than subcubic graph numbers and Buchholz hydras. ## Weak tree function Edit Define $$\text{tree}(n)$$, the weak tree function, as the length of the longest sequence of 1-labelled trees such that: 1. Every tree at position k (for all k) has no more than k + n vertices. 2. No tree is homeomorphically embeddable into any tree following it in the sequence. Adam P. Goucher has shown the following properties of this function: 1. tree(n) has a "growth rate" comparable to that of $$f_{\vartheta(\Omega^\omega)}(n)$$ in the fast-growing hierarchy. 2. TREE(3) > treetreetreetreetree8(7)(7)(7)(7)(7) A larger lower bound has since been found. Define $$\text{tree}_2(n)=\text{tree}^n(n)$$ and $$\text{tree}_3(n)=\text{tree}_2^n(n)$$. Then $$\text{TREE}(3) > \text{tree}_3(\text{tree}_2(\text{tree}(8)))$$. The latter expression is equal to treetreetree...tree(n)...(n)(n)(n) with n layers, where $$n = \text{tree}^{\text{tree}(8)}(\text{tree}(8))$$.[6] ### Values for $$\text{tree}(n)$$ Edit It can be shown that $$\text{tree}(1) = 2, \text{tree}(2) = 5$$ and $$\text{tree}(3) \geq 262140$$. $$\text{tree}(1)$$ uses the sequence: (()) () $$\text{tree}(2)$$ is a bit larger, we have two longest sequences that go as follows: ((())) (()()()) (()()) (()) () Otherwise: (()()) (((()))) ((())) (()) () Determining the exact value for $$\text{tree}(3)$$ is much harder, since there are a lot of sequences to check, and each of these is very long. Friedman has defined an FFF(k) function, which is equal to tree(k+1).[7] ## Explanation Edit Trees are tricky to visualize without drawing them out, so we shall devise a more compact way of representing them. Consider a language which has various kinds of parentheses such as (), [], {} etc. Parentheses always come in pairs and can nest within each other. Within a larger node, they may be concatenated. For example, the following strings are valid in this language: [] ([]) {[()]()} [(){[[]]}(){(())[]}] Suppose we have a string A. We shall call a pair of corresponding parentheses a node, in deference to the original tree construction. Define a child of a node to be a node that is nested one level deep within the original node. For example, take the string {[()()][][()]}; the children of the node represented by {} are the nodes represented by [], but not the nodes represented by (). Call a node deletable if it has fewer than two children. For example, in the string {[()()][][()]}, the () nodes are all deletable, as are the latter two [] nodes, but not the first [] or the {}. In the string ([(()())]), the [] node is deletable. (Note: is this correct?) We say a string A is reducible to a string B if A can be transformed into B by removing deletable nodes. A string A is reducible to a string B if and only if the tree represented by B is homeomorphically embeddable in the tree represented by A. With all this in mind, we can create an equivalent definition of TREE(k). Suppose we have a sequence of strings with the following properties: 1. You may only use k types of brackets. 2. The first string has at most one pair of brackets, the second string has at most two pairs of brackets, the third string has at most three pairs of brackets, etc. 3. No string is reducible to an earlier string. TREE(k) is the maximal length of the sequence. For k = 1, the optimal sequence has only one member: (). For k = 2, the optimal sequence has only three members: (), then [[]], then []. ### Weak tree function Edit Suppose we have a sequence of strings with the following properties: 1. You may only use () and no other types of brackets. 2. The first string has at most 1 + k pairs of brackets, the second string has at most 2 + k pairs of brackets, the third string has at most 3 + k pairs of brackets, etc. 3. No string is carvable from a later string. tree(k) (the weak tree function) is the maximal length of the sequence. ## Sources Edit Community content is available under CC-BY-SA unless otherwise noted.	1,472	5,600	{"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}	2.75	3	CC-MAIN-2019-51	longest	en	0.909306
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese817.htm	1,695,863,793,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00354.warc.gz	671,304,103	5,851	3.817 $$\int (a+b x)^2 \sqrt{-\frac{a^2 c}{b^2}+c x^2} \, dx$$ Optimal. Leaf size=130 $\frac{5}{8} a^2 x \sqrt{c x^2-\frac{a^2 c}{b^2}}+\frac{5 a b \left (c x^2-\frac{a^2 c}{b^2}\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (c x^2-\frac{a^2 c}{b^2}\right )^{3/2}}{4 c}-\frac{5 a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{c x^2-\frac{a^2 c}{b^2}}}\right )}{8 b^2}$ [Out] (5a^2xSqrt[-((a^2c)/b^2) + cx^2])/8 + (5ab(-((a^2c)/b^2) + cx^2)^(3/2))/(12c) + (b(a + bx)(-((a^ 2c)/b^2) + cx^2)^(3/2))/(4c) - (5a^4Sqrt[c]ArcTanh[(Sqrt[c]x)/Sqrt[-((a^2c)/b^2) + cx^2]])/(8b^2) ________________________________________________________________________________________ Rubi [A] time = 0.0531535, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.185, Rules used = {671, 641, 195, 217, 206} $\frac{5}{8} a^2 x \sqrt{c x^2-\frac{a^2 c}{b^2}}+\frac{5 a b \left (c x^2-\frac{a^2 c}{b^2}\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (c x^2-\frac{a^2 c}{b^2}\right )^{3/2}}{4 c}-\frac{5 a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{c x^2-\frac{a^2 c}{b^2}}}\right )}{8 b^2}$ Antiderivative was successfully veriﬁed. [In] Int[(a + bx)^2Sqrt[-((a^2c)/b^2) + cx^2],x] [Out] (5a^2xSqrt[-((a^2c)/b^2) + cx^2])/8 + (5ab(-((a^2c)/b^2) + cx^2)^(3/2))/(12c) + (b(a + bx)(-((a^ 2c)/b^2) + cx^2)^(3/2))/(4c) - (5a^4Sqrt[c]ArcTanh[(Sqrt[c]x)/Sqrt[-((a^2c)/b^2) + cx^2]])/(8b^2) Rule 671 Int[((d_) + (e_.)(x_))^(m_)((a_) + (c_.)(x_)^2)^(p_), x_Symbol] :> Simp[(e(d + ex)^(m - 1)(a + cx^2)^(p + 1))/(c(m + 2p + 1)), x] + Dist[(2cd(m + p))/(c(m + 2p + 1)), Int[(d + ex)^(m - 1)(a + cx^2)^p, x] , x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[cd^2 + ae^2, 0] && GtQ[m, 1] && NeQ[m + 2p + 1, 0] && IntegerQ[2p ] Rule 641 Int[((d_) + (e_.)(x_))((a_) + (c_.)(x_)^2)^(p_.), x_Symbol] :> Simp[(e(a + cx^2)^(p + 1))/(2c(p + 1)), x] + Dist[d, Int[(a + cx^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1] Rule 195 Int[((a_) + (b_.)(x_)^(n_))^(p_), x_Symbol] :> Simp[(x(a + bx^n)^p)/(np + 1), x] + Dist[(anp)/(np + 1), Int[(a + bx^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2p] \|\| (EqQ[n, 2 ] && IntegerQ[4p]) \|\| (EqQ[n, 2] && IntegerQ[3p]) \|\| LtQ[Denominator[p + 1/n], Denominator[p]]) Rule 217 Int[1/Sqrt[(a_) + (b_.)(x_)^2], x_Symbol] :> Subst[Int[1/(1 - bx^2), x], x, x/Sqrt[a + bx^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0] Rule 206 Int[((a_) + (b_.)(x_)^2)^(-1), x_Symbol] :> Simp[(1ArcTanh[(Rt[-b, 2]x)/Rt[a, 2]])/(Rt[a, 2]Rt[-b, 2]), x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] \|\| LtQ[b, 0]) Rubi steps \begin{align} \int (a+b x)^2 \sqrt{-\frac{a^2 c}{b^2}+c x^2} \, dx &=\frac{b (a+b x) \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{4 c}+\frac{1}{4} (5 a) \int (a+b x) \sqrt{-\frac{a^2 c}{b^2}+c x^2} \, dx\\ &=\frac{5 a b \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{4 c}+\frac{1}{4} \left (5 a^2\right ) \int \sqrt{-\frac{a^2 c}{b^2}+c x^2} \, dx\\ &=\frac{5}{8} a^2 x \sqrt{-\frac{a^2 c}{b^2}+c x^2}+\frac{5 a b \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{4 c}-\frac{\left (5 a^4 c\right ) \int \frac{1}{\sqrt{-\frac{a^2 c}{b^2}+c x^2}} \, dx}{8 b^2}\\ &=\frac{5}{8} a^2 x \sqrt{-\frac{a^2 c}{b^2}+c x^2}+\frac{5 a b \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{4 c}-\frac{\left (5 a^4 c\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{-\frac{a^2 c}{b^2}+c x^2}}\right )}{8 b^2}\\ &=\frac{5}{8} a^2 x \sqrt{-\frac{a^2 c}{b^2}+c x^2}+\frac{5 a b \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac{b (a+b x) \left (-\frac{a^2 c}{b^2}+c x^2\right )^{3/2}}{4 c}-\frac{5 a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{-\frac{a^2 c}{b^2}+c x^2}}\right )}{8 b^2}\\ \end{align} Mathematica [A] time = 0.150186, size = 103, normalized size = 0.79 $\frac{\sqrt{c \left (x^2-\frac{a^2}{b^2}\right )} \left (\sqrt{1-\frac{b^2 x^2}{a^2}} \left (9 a^2 b x-16 a^3+16 a b^2 x^2+6 b^3 x^3\right )+15 a^3 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{24 b \sqrt{1-\frac{b^2 x^2}{a^2}}}$ Antiderivative was successfully veriﬁed. [In] Integrate[(a + bx)^2Sqrt[-((a^2c)/b^2) + cx^2],x] [Out] (Sqrt[c(-(a^2/b^2) + x^2)](Sqrt[1 - (b^2x^2)/a^2](-16a^3 + 9a^2bx + 16ab^2x^2 + 6b^3x^3) + 15a^3 ArcSin[(bx)/a]))/(24bSqrt[1 - (b^2x^2)/a^2]) ________________________________________________________________________________________ Maple [A] time = 0.264, size = 113, normalized size = 0.9 \begin{align}{\frac{{b}^{2}x}{4\,c} \left ( -{\frac{{a}^{2}c}{{b}^{2}}}+c{x}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}x}{8}\sqrt{-{\frac{{a}^{2}c}{{b}^{2}}}+c{x}^{2}}}-{\frac{5\,{a}^{4}}{8\,{b}^{2}}\sqrt{c}\ln \left ( x\sqrt{c}+\sqrt{-{\frac{{a}^{2}c}{{b}^{2}}}+c{x}^{2}} \right ) }+{\frac{2\,ab}{3\,c} \left ({\frac{c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }{{b}^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align} Veriﬁcation of antiderivative is not currently implemented for this CAS. [In] int((bx+a)^2(-a^2c/b^2+cx^2)^(1/2),x) [Out] 1/4b^2x(-a^2c/b^2+cx^2)^(3/2)/c+5/8a^2x(-a^2c/b^2+cx^2)^(1/2)-5/8/b^2a^4c^(1/2)ln(xc^(1/2)+(-a^2 c/b^2+cx^2)^(1/2))+2/3ab/c(c(b^2x^2-a^2)/b^2)^(3/2) ________________________________________________________________________________________ Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align} \text{Exception raised: ValueError} \end{align} Veriﬁcation of antiderivative is not currently implemented for this CAS. [In] integrate((bx+a)^2(-a^2c/b^2+cx^2)^(1/2),x, algorithm="maxima") [Out] Exception raised: ValueError ________________________________________________________________________________________ Fricas [A] time = 2.2209, size = 506, normalized size = 3.89 \begin{align} \left [\frac{15 \, a^{4} \sqrt{c} \log \left (2 \, b^{2} c x^{2} - 2 \, b^{2} \sqrt{c} x \sqrt{\frac{b^{2} c x^{2} - a^{2} c}{b^{2}}} - a^{2} c\right ) + 2 \,{\left (6 \, b^{4} x^{3} + 16 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 16 \, a^{3} b\right )} \sqrt{\frac{b^{2} c x^{2} - a^{2} c}{b^{2}}}}{48 \, b^{2}}, \frac{15 \, a^{4} \sqrt{-c} \arctan \left (\frac{b^{2} \sqrt{-c} x \sqrt{\frac{b^{2} c x^{2} - a^{2} c}{b^{2}}}}{b^{2} c x^{2} - a^{2} c}\right ) +{\left (6 \, b^{4} x^{3} + 16 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 16 \, a^{3} b\right )} \sqrt{\frac{b^{2} c x^{2} - a^{2} c}{b^{2}}}}{24 \, b^{2}}\right ] \end{align} Veriﬁcation of antiderivative is not currently implemented for this CAS. [In] integrate((bx+a)^2(-a^2c/b^2+cx^2)^(1/2),x, algorithm="fricas") [Out] [1/48(15a^4sqrt(c)log(2b^2cx^2 - 2b^2sqrt(c)xsqrt((b^2cx^2 - a^2c)/b^2) - a^2c) + 2(6b^4x^3 + 16ab^3x^2 + 9a^2b^2x - 16a^3b)sqrt((b^2cx^2 - a^2c)/b^2))/b^2, 1/24(15a^4sqrt(-c)arctan(b^2 sqrt(-c)xsqrt((b^2cx^2 - a^2c)/b^2)/(b^2cx^2 - a^2c)) + (6b^4x^3 + 16ab^3x^2 + 9a^2b^2x - 16a ^3b)sqrt((b^2cx^2 - a^2c)/b^2))/b^2] ________________________________________________________________________________________ Sympy [C] time = 6.30035, size = 411, normalized size = 3.16 \begin{align} a^{2} \left (\begin{cases} - \frac{a^{2} \sqrt{c} \operatorname{acosh}{\left (\frac{b x}{a} \right )}}{2 b^{2}} - \frac{a \sqrt{c} x}{2 b \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} + \frac{b \sqrt{c} x^{3}}{2 a \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} & \text{for}\: \frac{\left \|{b^{2} x^{2}}\right \|}{\left \|{a^{2}}\right \|} > 1 \\\frac{i a^{2} \sqrt{c} \operatorname{asin}{\left (\frac{b x}{a} \right )}}{2 b^{2}} + \frac{i a \sqrt{c} x \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}}{2 b} & \text{otherwise} \end{cases}\right ) + 2 a b \left (\begin{cases} 0 & \text{for}\: c = 0 \\\frac{\left (- \frac{a^{2} c}{b^{2}} + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) + b^{2} \left (\begin{cases} - \frac{a^{4} \sqrt{c} \operatorname{acosh}{\left (\frac{b x}{a} \right )}}{8 b^{4}} + \frac{a^{3} \sqrt{c} x}{8 b^{3} \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} - \frac{3 a \sqrt{c} x^{3}}{8 b \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} + \frac{b \sqrt{c} x^{5}}{4 a \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} & \text{for}\: \frac{\left \|{b^{2} x^{2}}\right \|}{\left \|{a^{2}}\right \|} > 1 \\\frac{i a^{4} \sqrt{c} \operatorname{asin}{\left (\frac{b x}{a} \right )}}{8 b^{4}} - \frac{i a^{3} \sqrt{c} x}{8 b^{3} \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} + \frac{3 i a \sqrt{c} x^{3}}{8 b \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} - \frac{i b \sqrt{c} x^{5}}{4 a \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} & \text{otherwise} \end{cases}\right ) \end{align} Veriﬁcation of antiderivative is not currently implemented for this CAS. [In] integrate((bx+a)2(-a*2c/b*2+cx2)(1/2),x) [Out] a*2Piecewise((-a*2sqrt(c)acosh(bx/a)/(2b2) - asqrt(c)x/(2bsqrt(-1 + b2x2/a2)) + bsqrt(c) x*3/(2asqrt(-1 + b2x2/a2)), Abs(b*2x2)/Abs(a2) > 1), (Ia2sqrt(c)asin(bx/a)/(2b2) + I asqrt(c)xsqrt(1 - b2x2/a2)/(2b), True)) + 2abPiecewise((0, Eq(c, 0)), ((-a*2c/b*2 + cx2) (3/2)/(3c), True)) + b2Piecewise((-a*4sqrt(c)acosh(bx/a)/(8b4) + a3sqrt(c)x/(8b*3sqrt(-1 + b *2x2/a2)) - 3asqrt(c)x3/(8bsqrt(-1 + b2x2/a2)) + bsqrt(c)x*5/(4asqrt(-1 + b2x2/a 2)), Abs(b*2x2)/Abs(a2) > 1), (Ia4sqrt(c)asin(bx/a)/(8b4) - Ia*3sqrt(c)x/(8b*3sqrt(1 - b*2x2/a2)) + 3Iasqrt(c)x*3/(8bsqrt(1 - b2x2/a2)) - Ibsqrt(c)x5/(4asqrt(1 - b2x* 2/a2)), True)) ________________________________________________________________________________________ Giac [A] time = 1.307, size = 136, normalized size = 1.05 \begin{align} \frac{{\left (\frac{15 \, a^{4} \sqrt{c} \log \left ({\left \| -\sqrt{b^{2} c} x + \sqrt{b^{2} c x^{2} - a^{2} c} \right \|}\right )}{{\left \| b \right \|}} - \sqrt{b^{2} c x^{2} - a^{2} c}{\left (\frac{16 \, a^{3}}{b} -{\left (9 \, a^{2} + 2 \,{\left (3 \, b^{2} x + 8 \, a b\right )} x\right )} x\right )}\right )}{\left \| b \right \|}}{24 \, b^{2}} \end{align} Veriﬁcation of antiderivative is not currently implemented for this CAS. [In] integrate((bx+a)^2(-a^2c/b^2+cx^2)^(1/2),x, algorithm="giac") [Out] 1/24(15a^4sqrt(c)log(abs(-sqrt(b^2c)x + sqrt(b^2cx^2 - a^2c)))/abs(b) - sqrt(b^2cx^2 - a^2c)(16a ^3/b - (9a^2 + 2(3b^2x + 8ab)x)x))abs(b)/b^2	5,593	10,696	{"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}	3.796875	4	CC-MAIN-2023-40	latest	en	0.165804
https://au.mathworks.com/matlabcentral/answers/2112406-cell-linked-method-in-matlab-with-parallelization-trough-domain-splitting	1,718,289,800,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00047.warc.gz	91,800,892	26,105	# cell linked method in matlab with parallelization trough domain splitting 2 views (last 30 days) Andrea Somma on 27 Apr 2024 Answered: UDAYA PEDDIRAJU on 8 May 2024 I want to perform particle dynamics integration with cell linked method (all particles are assigned to different cells to speed up the calculation of the forces as in figure) I am easily able to split paricles in cells trough the following function, where the grid stores the left vertex of the cells in x and y coordinates and ptcls.x stores the position of the particles as a 2 by NP (number of particles) mesh function grd_to_ptcl = init_ptcl_mesh (grd, ptcls) h = [grd.x(2) - grd.x(1); grd.y(2) - grd.y(1)]; idx = floor(ptcls.x./h) + 1; indexPtcls = 1:size(ptcls.x,2); grd_to_ptcl = accumarray(idx',indexPtcls(:),[grd.ncx grd.ncy],@(x) {x}); end Once I have the cell array I am easily able to run trough not empty cells and perform force calculation index = cellfun(@numel, grd_to_ptcl, 'UniformOutput', true); index = find(index >=1); for i=index(:) % calculation of the force end My question is how I can split the domain and perform the calculation in parallel updating the boundaries of each of the splitted domains as in figure in an efficient way If possible would be usefull to have a simple example, thanks in advance UDAYA PEDDIRAJU on 8 May 2024 Hi Andrea, Parallelize cell-linked method with domain splitting in MATLAB: 1. Split domain & assign particles: Use :"spmd":https://www.mathworks.com/help/parallel-computing/spmd.html and array partitioning to create subdomains with overlap for boundary handling. Distribute particles based on subdomain membership. 2. Force calculation (parallel): Within each worker, use the cell-linked method for forces within the subdomain. 3. Boundary correction (centralized): Gather info on boundary particles, calculate & accumulate forces due to neighboring subdomains. 4. Combine results: Gather partial force calculations and combine for final forces on all particles. This should give you an idea on how to work around. ### Categories Find more on Surface and Mesh Plots in Help Center and File Exchange R2024a ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	548	2,265	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2024-26	latest	en	0.836691
tangiblesolar.com	1,680,014,490,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00697.warc.gz	636,464,432	43,704	# How Big Is A Solar Panel? Various Guides How many solar panels your house requires may depend on the size of your solar panels. At Vision Solar, we have calculated the precise number of panels your property would need before installation in order to get the most savings. We have guessed where your panels will go and how much energy they will make in total. We are aware of how perplexing this might be. To help you understand how the whole solar panel system works, we’ll break down how much electricity one solar panel can give you. How To Fix Broken Wire On Solar Lights ## How big is a solar panel? The typical output of residential solar panels is between 300 and 400 watts. Solar panels for homes are pretty much all the same size and shape, but some are more efficient than others. Residential solar panels typically range in length from 65 inches (1.65 meters) to 79 inches (2 meters), and in width from 39 to 41 inches (around 1 meter). A home solar panel has a surface area of 18 to 22 square feet. The many types of home solar panels and their measurements are described in the section that follows. ## What is a typical solar panel size? The dimensions of a typical household solar panel today are around 5.4 feet by 3.25 feet, or 65 inches by 39 inches, with some variation across manufacturers. SunPower panels have dimensions of 61.3 by 41.2 inches. ## Common Solar Panel Dimensions? The two most prevalent arrangements for standard solar panels are 60-cell and 72-cell. The size of a single solar cell is 6″ by 6″. In a 610-grid, 60-cell panels are arranged. 72-cell panels are roughly a foot higher since they are arranged in a 612 grid. For 60-cell panels, 66″ x 39″ (3.25 feet x 5.5 feet) 72-cell panels, 39 by 77 inches (3.25 feet x 6.42 feet) For the majority of home and business installations, these are the typical solar panel sizes, give or take an inch on each side. Because various manufacturers employ different frame sizes, there will be some differences. ## What Is the Size of a Standard Solar Array? The average monthly power use in America is 867 kWh. To completely offset that consumption, you would need a 6.5 kW solar installation. Our current supply of 60-cell solar panels runs from 285W to 315W, while our stock of 72-cell solar panels goes from 335W to 375W. We can use the following formula to estimate how many panels are needed to build a 6.5 kW (6500-watt) system: 20 panels at 6500W/340W = 19.1 (20 panels) 6500W / 375W = 17.3 (18 panels) (18 panels) Depending on the efficiency of the panels you choose, an average-sized solar system will include 18–23 panels. Here is how it relates to system size on a physical level. To get an idea of how much room the array could need, let’s compare the least efficient panels—285 W/60 cells—and the most efficient—375 W/72 cells: 29.25 ft x 12.83 ft = 375.38 square feet. of 375W 72-cell panels (92 arrays). 26 feet x 16.5 ft = 429 sq ft of 285W 60-cell panels (83 array) A typical solar system will occupy 375–429 square feet overall. On your property, that system may be mounted on a ground mount or on your roof. The array’s design and panel wattage will determine the precise size. ## How Much Do Portable Mobile Solar Panels Weigh? Small panels for distant or mobile users are the second use to consider. RVs, yachts, and other distant applications like solar-powered lighting employ these panels. Instead of the standard 60- and 72-cell panels, which come in a range of sizes, smaller panels come in a variety of sizes. While tiny 5-watt panels use less than 1 square foot of area, our Solarland SLP190 (a popular option for isolated off-grid applications) is close to a full-size panel at 32″ x 62.” Making the most of your limited space is key to choosing the appropriate panels for your boat or RV. Most people with RVs or boats need a smaller option because, even though full-sized panels can be used on the road, there’s often not enough space to attach them. These panels typically have conventional 12-volt or 24-volt output. In our post about solar panels for your RV or boat, we looked at a number of great options. ## What is the weight of solar panels? People often ask us how much solar panels weigh in addition to their physical dimensions. Lifting panels onto your roof might be difficult, particularly if you’re working alone since they can be fairly heavy. Full-sized panels often weigh between 40 and 60 pounds, according to what we advise people to assume. Depending on the materials the manufacturer chooses, there are certain differences. Here is a table that details the weight of some of the panels we provide. The actual difficulty in lifting panels is not so much their weight as it is the difficulty in carrying them due to their size. One person can handle a 60-cell panel, but since 72-cell panels are taller than 6 feet, it is usually safer to recruit two people to carry them. We ask everyone to be careful because they are likely to sway and make you lose your balance, especially when it’s windy. If you need help raising the panels onto your roof, you can use scaffolding or a motorized lift to support their weight. ## Can I install as many solar panels as I can on my roof? The roof’s square area is not the sole consideration when attempting to measure your roof for solar. Consider the trees in the area and any items on your roof that might cast shadows on the solar panels. Even a small shadow cast by a chimney can make your system less efficient as a whole. In addition, since north-facing roof parts get little sunshine in the United States (or the northern hemisphere in general), they are often avoided in solar installation. A qualified solar contractor will do the important research, but in the meantime, you can use programs like Google’s Project Sunroof to figure out how much of your roof can be used. To find out how much of your roof can be used for solar panels, you just need to type in your location. ## Can my RV accommodate a lot of solar panels? The number of solar panels that can fit on your RV, excluding its type and size, is mostly determined by how its roof is built. For example, the roof of a Class C RV may be about 200 feet square, but the amount of usable roof space is much less than that because of things like ventilation covers and skylights. The following table provides an estimate of how many solar panels might fit on various types of RVs after taking these factors into account: Most Class A RVs, destination trailers, and fifth wheels have room for 12 100-watt solar panels. Class B and Class C RVs, as well as truck campers, are smaller and have less usable roof space; on average, only 4-6 solar panels can fit on these RVs. error: Content is protected !!	1,551	6,771	{"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-2023-14	longest	en	0.942216
https://outdoortroop.com/what-makes-a-hot-air-balloon-rise/	1,716,914,211,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00042.warc.gz	370,954,123	46,940	# What Makes a Hot Air Balloon Rise? Have you ever seen large, colorful balloons soar above the clouds and wondered how they got there? If so, you’re not alone! Hot air balloons can achieve flight because of the law of buoyancy, also called Archimedes’ principle. Archimedes’ principle states that an object in fluid (gas or liquid) is acted upon by an upward force equal to the fluid displaced by the object. Since hot air is lighter than normal air, the balloon is buoyant. To the uninitiated, that may be a little confusing. Below, we’ll discuss all the ins and outs (and the ups and downs) of buoyancy! ## What is Buoyancy? In the simplest terms, buoyancy is the upward force on an object at rest in a fluid. The easiest way to think about this is to imagine a tennis ball placed in a pool. ##### Example 1 When the tennis ball is placed in water, you’ll notice that it sinks a little bit. However, the tennis ball will never be completely submerged by the water. That is because the force pushing upwards on the tennis ball, the buoyant force, is greater than the mass of the tennis ball. That buoyant force is created by the fluid displaced by the tennis ball. Remember how the ball sinks a little bit into the water? That displaces water until the force of the water displaced equals the force exerted by the tennis ball. That might be a lot of words if you’re a visual learner. This will help: ##### Example 2 Now imagine for a second that instead of a tennis ball, you dropped a ball of led into the pool. This ball of lead is the exact same size as the tennis ball, but obviously weighs much more. As you can imagine, the ball of led would sink right to the bottom of the pool. Why is this? The led ball displaces as much water as the tennis ball, but it weighs much more. Because of this, the upward force of the dispersed water will never equal the downward force of the led ball. This causes the led ball to sink. The following diagram will compare what happens to objects of the same size but different weight in water. This mimics exactly what we just discussed with the tennis ball and the led ball. This law of buoyancy is what causes boats to float and hot air balloons to fly. The next time you go on a cruise or jump in a hot air balloon, you can envision this law at work! ## Who Discovered Buoyancy? We’ve all had great ideas while sitting in the bath tub or shower. There’s something about relaxing in water that gets the brain juices flowing! Well, this is exactly what happened to a Greek mathematician named Archimedes. Archimedes is one of the greatest minds of the old world. He is responsible for numerous discoveries in mathematics, physics, engineering, and astronomy! He is also a great inventor! The law of buoyancy was one of his greatest discoveries, and it comes with a great story! ##### The Golden Crown There’s an ancient tale that explains how Archimedes discovered buoyancy. Some of it is definitely true, but there’s some that may be fabricated. I’ll let you be the judge. Archimedes lived in Syracuse, a province of modern day Italy. The king of Syracuse at the time was a man by the name of King Hiero II. King Hiero II had commissioned the creation of a golden crown to be placed in the temple at Syracuse. This crown was to be shaped like a laurel wreath. King Hiero II provided a lump of gold to a local goldsmith for the creation of the crown. When the king received the crown, he was impressed by it’s craftsmanship and beauty. However, there were some who claimed that the goldsmith used some silver in the design to make the crown light enough. Archimedes was brought in to discover the truth. The king did not want Archimedes to damage the crown in any way, so Archimedes had to get creative. The story goes that as Archimedes got into his tub to think over the problem, he observed water spilling over the side of the bath as he lowered himself in. In a moment, he had his answer! Archimedes was so excited he ran all the way home naked! The streets of Syracuse filled with his shouts of “Eureka! Eureka!”. The test was simple: Archimedes would place the crown and a lump of gold identical to the first in a tub of water. If they had the same volume, they would displace the same amount of water no matter the shape. If silver had been added, the volume would be different and the crown would displace water. The dishonest goldsmith was exposed, and the law of buoyancy was discovered! ## Buoyancy and Balloons So how does Archimedes’ principle apply to hot air balloons? The first thing to remember is that a fluid is a liquid or a gas. All the examples above utilized liquids, but the same laws apply to gases because they’re fluids too. The air all around us counts as a gas, and it’s the state of matter that hot air balloons use to rise! Like the name implies, hot air balloons achieve flight because of hot air. As the air inside the balloon (or the envelope) heats up, it becomes less dense than the air outside the envelope. This causes the balloon to rise. ##### The Balloon vs the Balls Remember our example of the tennis ball and the led ball? A hot air balloon is a lot like that. Before the air in the envelope is heated, it is just as dense as the air outside the envelope. This is like the led ball. The size of the balloon is exactly the same as a balloon that is flying, but the weight of the air inside the balloon is greater than the air it is displacing. When the air inside the balloon is heated, the gas inside becomes less dense. In other words, the gas weighs less even though the size of the envelope remains the same. When this happens, the balloon lifts because it weighs less than the air it is displacing. This means the upward force on the balloon is equal to, and then greater than the weight of the balloon. ##### Holey Buoyancy! A hot air balloon has two holes: one at the bottom and one at the top. The hole at the bottom, often called the throat, lets the hot air into the balloon. The hole at the top, or the parachute valve, lets hot air out of the balloon when opened. To keep a hot air balloon in the air, the air in the envelope needs to be heated consistently. This keeps the envelope buoyant. If you want to land a hot air balloon, you can let hot air out. As the balloon becomes less and less buoyant, it will begin to fall. You can also choose to let the air in the balloon just cool naturally, but this means you have much less control over the descent. ## Buoyancy in Practice So how do you actually apply these principles in a hot air balloon? We’ve already discussed some of the parts of a hot air balloon, but here’s a quick review that will help you understand how to apply Archimedes principle: Part What it Does Basket The box at the bottom of a hot air balloon that holds cargo. Envelope The “balloon” that gets filled with air. Skirt A nonflammable cloth at the bottom of the envelope around the burners. Burners The unit that heats the air inside the envelope. The burners, which are positioned above the basket, utilize liquid propane to heat the air inside the envelope. The flame is controlled by a valve on the liquid propane tank. The more propane fed to the burner, the stronger the flame. The stronger the flame, the less dense the air becomes inside the envelope. As the air inside the envelope becomes less dense, the whole balloon rises. ##### Steering We’ve already discussed how letting hot air into the envelope causes the balloon to rise. We also know that letting hot air out of the envelope or letting the air in the envelope cool, will cause the balloon to fall. With this understanding, we can begin to see how the law of buoyancy can help us steer to a certain degree. Unlike other flying machines, you don’t have a ton of control over where a hot air balloon goes. However, there are some things you can do to steer a hot air balloon. Experienced balloonists become adept at recognizing wind patterns at different elevations. For example, if you know that the wind is blowing west at 3000ft., you can raise your balloon to 3000ft. to start going west. Likewise, you may drop your elevation if you want to catch an air current going in a different direction! Very good balloon pilots can even jump from air current to air current causing their balloon to “spin”. ##### Safety Precautions Because shooting a flame into the envelope could be potentially dangerous, there are a few safety measures in place. The skirt is made out of different material than the rest of the envelope. It is the cloth directly around the burners and therefore is made out of nonflammable cloth. Without the skirt, the envelope wouldn’t catch all the warm air needed to rise. Before the envelope begins receiving warm air from the burners, it is first filled by cold air. This gets the envelope off the ground and into a form ready to receive hot air. Trying to get hot air into a balloon before filling it with cold air would likely set the envelope ablaze. ## Buoyant Balloons: Then and Now The principle of buoyancy was discovered long before the hot air balloon was invented, but it wasn’t always applied correctly. ##### Early Balloons The inventors of the hot air balloon, Joseph-Michel Montgolfier and Jacques-Ã‰tienne Montgolfier, believed that it was the smoke in their hot air balloons that helped them achieve buoyancy. They didn’t yet understand how air behaved when it was heated versus when it was cooled. In fact, they had very little understanding of how oxygen molecules behaved at all! Because of this, the Montgolfier brothers used wet straw to fuel their hot air balloons because it created more smoke. In reality, this just hurt the performance of their balloons. Gas balloons quickly became more popular than hot air balloons. These balloons used gases lighter than oxygen, such as helium or hydrogen, to achieve flight. Light hot air, the helium or hydrogen inside a gas balloon is less dense than the air outside the envelope. This causes the gas balloon to rise. ##### The Modern Hot Air Balloon In the next 100+ years, there were several discoveries and inventions that brought the return of hot air balloons closer and closer. With the discovery of the liquid propane burner and considerable efforts on the part of Ed Yost, hot air balloons came back into popularity in the mid 20th century. These balloons use liquid propane burners to heat the air inside the envelope. As already discussed, this is what causes modern hot air balloons to become buoyant. This provided a safe way for hot air balloons to fly for the first time in over 100 years! Without the invention of the liquid propane burner and the work of Ed Yost, gas would’ve remained the best way for balloons to achieve buoyancy. Today, hot air balloons are much more popular than their gas counterparts. ## Harnessing Archimedes’ Power Using Archimedes’ principle to achieve flight is a beautiful thing! And it isn’t just because of the hot air balloon. It’s much larger than that. You may have heard someone talk about “defying the law of gravity”. Perhaps it was at a basketball game, or maybe while listening to the soundtrack of “Wicked”. Wherever it was, there is a prevalent notion that flying defies the law of gravity, but this simply isn’t true. Not only is it not true, but it actually detracts from the true beauty of flight. The beauty of flying comes not from defying natural laws, but from harnessing them! You see, every time a hot air balloon rises in the sky or a plane takes off from the ground, it is a testament that mankind has become master of the elements. It proves that we have literally risen above the natural order of things. When Archimedes discovered buoyancy, he was, in fact, proving once again that there is no limit to what we as humans can achieve! And harnessing the power that Archimedes discovered reproves that point each time we leave the safety of earth. For many, hot air ballooning is more than a pastime, it’s a lifestyle. It’s rewarding for so many reasons. The community, the peace amongst the clouds, and the exhilaration you feel at taking off are just a few of those reasons. The lessons learned from ballooning are certainly a part of the reason. And among all of the lessons to be learned, this is definitely one to remember: The beauty of ballooning is born of harnessing, not breaking, earth’s laws. Geoff Southworth I am a California native and I enjoy all the outdoors has to offer. My latest adventures have been taking the family camping, hiking and surfing. ### Recent Posts outdoortroop-21 outdoortroop-20	2,755	12,652	{"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}	4.03125	4	CC-MAIN-2024-22	latest	en	0.943052
http://docbrown.info/page07/equilibria2.htm	1,490,688,189,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218189686.31/warc/CC-MAIN-20170322212949-00002-ip-10-233-31-227.ec2.internal.warc.gz	100,283,189	24,655	Doc Brown's Chemistry Advanced A Level Notes TheoreticalPhysical Advanced Level Chemistry Equilibria Chemical Equilibrium Revision Notes PART 2 2. Chemical equilibrium constants (Kc & Kp), equilibrium expressions calculations, the effect of temperature on K How do we write out chemical equilibrium expressions. What is the equilibrium constant? How do you do equilibrium calculations? All is explained with examples including the units for concentrations (mol dm3) or partial pressures (e.g. atm, Pa or kPa) and how to solve equilibrium problems using concentration or partial pressure units. Chemical Equilibrium Notes Index Part 2 subindex: 2.1 Kc and Kp equilibrium expressions and Le Chatelier's Principle 2.2 Exemplar calculations using Kc expressions or Kp expressions The equilibrium constant Kc is deduced from the equation for a reversible reaction, NOT experimental data as for rate expressions in kinetics. The concentration, in mol dm-3, of a species X involved in the expression for Kc is represented by in square brackets i.e. [X] The value of the equilibrium constant is not affected either by changes in concentration or addition of a catalyst. You need to be able to construct an expression for Kc for a homogeneous system in equilibrium, calculate a value for Kc from the equilibrium concentrations for a homogeneous system at constant temperature, perform calculations involving Kc and predict the qualitative effects of changes of temperature on the value of Kc. 2.1 Equilibrium expressions and applying Le Chatelier's Principle IT IS IMPORTANT and ESSENTIAL that Equilibria Part 1 is studied before working through this page 2.1a Molar concentration expressions • Kc concentration equilibrium expression INTRODUCTION • It is found experimentally that the concentrations at the equilibrium point are related by a simple mathematical equation known as an equilibrium expression which is governed by an equilibrium constant K, at constant temperature. • K only varies with temperature, and nothing else! • These equilibrium expressions (many on this page) were originally derived from experimental analysis of mixtures at equilibrium. • Patterns in these equilibrium concentrations were 'spotted' and the resulting mathematical expression of these concentrations were considered to conform to what was called the 'law of mass action' ( a term not really used these days!). • A classic study of the ester formation equilibrium is often described in textbooks. • ALCOHOL + ACID ESTER + WATER • Knowing the initial amounts of alcohol and acid, it was easy to titrate the remaining free carboxylic acid and then logically work out the amounts of the water, alcohol and ester left in the equilibrium mixture. • They would also have found out that it took some time to reach equilibrium, but eventually all the concentrations remained constant i.e. in the mixture, the point of equilibrium was reached. • Later studies were producing graphs of concentration versus time, and these showed that eventually all the component concentrations levelled out i.e. became constant at the first point in time that a true equilibrium was established. • See example 2.2a.1 ester equilibrium calculation and 2.a.5 too. • The theoretical justification for K expressions came later in chemical history and this need not concern us at this level because it involves some pretty advanced thermodynamics theory! • From a student's point of view, here at this level, you are using K in terms of concentrations or partial pressures only and therefore all other equilibrium terms, apart from K itself, should be quoted in either ... • ... for liquid mixtures or solutions • [x] = concentration of x in mol dm3 in a Kc equilibrium expression • ... or for gaseous mixtures • px = partial pressure of x in eg. Pa or atm. in a Kp equilibrium expression • Note • (i) Partial pressure is effectively a concentration term e.g. double a partial pressure and you effectively double the concentration of the gaseous component. • (ii) The subscript c in Kc means a concentration equilibrium expression using mol dm3 and the subscript p in Kp means a partial pressure equilibrium using partial pressures. • (iii) The units of K depend on which units for concentration or partial pressure you have used for the quantities inserted in the equilibrium expression. • This means that sometimes the concentration or partial pressure units cancel each other out, hence K can be dimensionless! • For any reaction in solution or a gaseous mixture: • aA + bB + cC etc. tT + uU + wW etc. • in terms of concentrations ... • Kc = [T]t [U]u [V]v etc. [A]a [B]b [C]c etc. • [x(?)] square brackets indicates concentration of x e.g. in mol dm3 and the state(?) should be quoted too. • Note that all the equilibrium expressions you will deal with at this level only involve concentrations or partial pressures. • By convention, the arithmetical product of the product concentrations* of the forward reaction (RHS) are on the top line and the arithmetical product of the reactant concentrations* from the backward reaction (LHS) are on the bottom line. • * In all cases the product concentrations are raised to the appropriate power (a, b, c, .. t, u, w, ..) given by the stoichiometric mole ratios of the balanced equation. • AND again note that Kc (like Kp) is only constant for a specific constant temperature at which the concentrations of the equilibrium components might vary from one dynamic equilibrium situation to another (e.g. in reacting liquid mixtures or in solutions). • For heterogeneous equilibria, K expressions do not normally include values for solid phases, since their chemical potential cannot change since the concentration of a solid cannot change. • The effect of temperature on the equilibrium constant (Kc or Kp) • Please remember, only temperature changes K, because only changing temperature can change the energy of the molecules. • First consider the simple equilibrium: aA + bB cC + dD {(i) ΔH -ve, (ii) ΔH +ve} • Using the convention described above, writing out the concentration equilibrium expression gives ... • Kc = [C]c [D]d [A]a [B]b • The equilibrium constant, Kc (or Kp later), is governed by temperature, which is the only factor that can alter the internal potential energy of the reactants or products. The 'rule' for the trend in K value change is provided by Le Chatelier's Principle. • (i) If the forward reaction is exothermic, Kc (or Kp later) will decrease in value with increase in temperature. • From Le Chatelier's Principle, the equilibrium position will shift more to the left in the endothermic direction to minimise the temperature increase due to the effect of increased heat input. • So, mathematically, by convention, the top line concentrations of the forward products, [C] & [D], will numerically decrease and the bottom line concentrations [A] & [B] must therefore numerically increase, since some of the C & D are converted to A & B. • Hence the equilibrium constant K must decrease for the new equilibrium position. • (ii) If the forward reaction is endothermic, Kc (or Kp) will increase in value with increase in temperature. • From Le Chatelier's Principle, the equilibrium position will shift more to the right in the endothermic direction to minimise the temperature increase due to the effect of increased heat input. • So, mathematically, by convention, the top line concentrations of the forward products, [C] & [D], must numerically increase and the bottom line concentrations of [A] & [B] must therefore numerically decrease, since some of A & B are converted to C & D. • Hence the equilibrium constant K must increase for the new equilibrium position. • Changes in pressure or concentration have no effect on a K value for ideal mixtures of gases/liquids or solutions. • Application of a catalyst to a reaction also has no effect on a K value. • Why are the values of equilibrium constants affected by temperature? • As we have noted, it is very important to quote the specific constant temperature that applies to any Kc or Kp value BECAUSE equilibrium constants vary with temperature AND ONLY temperature. • This is because changes in concentration, partial pressure (see section 2.1b) or employing a catalyst do NOT affect the energy content of the molecules themselves (referred to in thermodynamics as the internal energy of the molecule). • internal energy (symbol U) = chemical energy + thermal energy • However, if you change the temperature, you also change the fundamental energy content of a molecule, and we are now talking about the thermodynamic property values of the equilibrium components. • Electronic energy is stored in the chemical bonds of the molecules (chemical energy) plus their thermal energies of translation (kinetic energy of movement), rotation (of the whole or parts of a molecule) and vibration (of bonded atoms). • internal energy = chemical energy + thermal energy • If you change the temperature, all the internal energy contents of the molecules are also changed, but they don't change to the same amount for each molecule for the same rise in temperature. • If you think of the internal energy as a sort of chemical potential to effect a chemical change, the formation of 'reactants' or 'products' may be favoured one way or the other (l to r or r to l as you write the equilibrium equation). • Hence the equilibrium constant not only changes, but may increase or decrease depending on whether the reaction is endothermic or exothermic. • The enthalpy content of a molecule (H) is related to its internal energy (U), but that's as far as we need to go here! • Therefore the change in the internal energy (caused by change in temperature), a fundamental thermodynamic property of a molecule, explains why equilibrium constants are ONLY affected by temperature and NOT by concentration, pressure or indeed, the presence of a catalyst. • It is possible to calculate equilibrium constant from thermodynamic Gibbs free energy change data for the reaction, but this may not be part of your course and certainly not appropriate to study here. • See later specific examples for the units of Kc (or Kp later) and if K has no units you should state so. • Equilibrium example 2.1a.1 The formation of hydrogen iodide • H2(g) + I2(g) 2HI(g) • Kc = [HI(g)]2 (no units) [H2(g)] [I2(g)] • units = (mol dm3)2/[(mol dm3) x (mol dm3)], all cancel out, no units • Kc has no units as all the concentration units cancel out. • An example of the quantitative connection between kinetics (rates of reaction) and equilibrium expressions. • This, historically, has been one of the most studied reactions in terms of kinetics and equilibrium and is a good example to study for comparing and amalgamating two important conceptual frameworks in chemistry. (If you haven't studied kinetics rate expressions etc. then just miss out this paragraph.) • The concentrations of reactants and products have been followed quantitatively by starting with either hydrogen iodide or a hydrogen iodine gas mixture at temperatures of 250450oC. • The graph below show in principle what happens ... • Whatever mixture you start with, eventually all the concentrations level out. • Both the forward (f) and backward (b) reactions occur via a simple one step mechanism i.e. via a single bimolecular collision and this simple reaction mechanism leads to simple and verifiable second order kinetics rate expressions. • ratef = kf [H2(g)] [I2(g)] and rateb = kb [HI(g)]2 • Now at the point of dynamic equilibrium, with no net change in concentrations, the rate of the forward reaction = rate of the backward reaction, so • ratef = kf [H2(g)] [I2(g)] = rateb = kb [HI(g)]2 • therefore [H2(g)] [I2(g)] = ratef / kf and [HI(g)]2 = rateb / kb • and substituting into the equilibrium expression, with the 'rates' cancelling out at equilibrium, gives • Kc = rateb x kf kf = kb x ratef kb • So the equilibrium constant is equal to the ratio of the two rate constants of the forward and backward reaction. • Also see calculation example 2.2a.2 • Equilibrium example 2.1a.2 The formation of the ester ethyl ethanoate • CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) • Kc = [CH3COOCH2CH3(l)] [H2O(l)] (no units) [CH3COOH(l)] [CH3CH2OH(l)] • units = [(mol dm3) x (mol dm3)]/[(mol dm3) x (mol dm3)], all cancel out, no units • Equilibrium example 2.1a.3 The formation of phosphorus(V) chloride (gaseous phase) • PCl3(g) + Cl2(g) PCl5(g) • Kc = [PCl5(g)] (units mol1 dm3) [PCl3(g)] [Cl2(g)] • units = (mol dm3)/[(mol dm3) x (mol dm3)] = mol1 dm3 • Equilibrium example 2.1a.4 The synthesis of ammonia • N2(g) + 3H2(g) 2NH3(g) • Kc = [NH3(g)]2 (units mol2 dm6) [N2(g)] [H2(g)]3 • units = [(mol dm3)2]/[(mol dm3) x (mol dm3)3] = mol2 dm6 • Equilibrium example 2.1a.5 The synthesis of phosphorus(V) fluoride • P4(g) + 10F2(g) 4PF5(g) • Kc = [PF5(g)]4 (units mol7 dm21) [P4(g)] [F2(g)]10 • units = ](mol dm3)4}/[(mol dm3) x (mol dm3)10] = mol7 dm21 • Equilibrium example 2.1a.7 A cobalt(II) complex ion ligand exchange reaction • [Co(H2O)6]2+(aq) + 4Cl(aq) [CoCl4](aq) + 6H2O(l) • Kc = [[CoCl4]2(aq)] (units mol4 dm12) [[Co(H2O)6]2+(aq)] [Cl(aq)]4 • units = (mol dm3)/[(mol dm3) x (mol dm3)4] = mol4 dm12 • Note that the concentration of water is effectively constant in an aqueous solution and omitted from the equilibrium expression, but is effectively subsumed into the Kc value which in complex ion chemistry is called the stability constant denoted by Kstab. • Equilibrium example 2.1a.8 The acid behaviour of high oxidation state hexaaqua ion • The hexaaquairon(III) ion is quite acidic in aqueous solution due to proton transfer giving the oxonium ion. • [Fe(H2O)6]3+(aq) + H2O(l) [Fe(H2O)5OH]2+(aq) + H3O+(aq) Kc = [[Fe(H2O)5OH]2+(aq)] [H3O+(aq)] (units mol dm3) [[Fe(H2O)6]3+(aq)] • units = [(mol dm3) x (mol dm3)]/(mol dm3) = mol dm3 • Again [H2O(l)] incorporated into Kc. • Some 'VERY rough rules of thumb' for an equilibrium K value (Kc or Kp in section 2.1b) and the 'position' of the equilibrium in terms of LHS (e.g. original reactants or products of backward reaction) and the RHS (products of the forward reaction): • For: LHS RHS • (for A + B C + D the rules below work ok BUT once the ratios of reactants or products are not 1:1, things are not so simple) • If Kc (or Kp) is >> 1 the equilibrium is mainly on the RHS, maybe virtually 100% completion of the forward reaction i.e. a very large RHS yield i.e. and likely to be very thermodynamically feasible. • If Kc (or Kp) is approx. 1 the equilibrium is about 50% RHS and about 50% LHS. • If Kc (or Kp) is << 1 the equilibrium is mainly on the LHS, maybe virtually 0% of products of the forward reaction i.e. a very low RHS yield i.e. NOT likely to be thermodynamically feasible). • BUT remember K changes with temperature considerably changing the position of an equilibrium, AND, at constant temperature, and therefore constant K, the position of an equilibrium can change significantly depending on relative concentrations/pressures of 'reactants' and 'products'. 2.1b Partial pressure equilibrium law expressions • Kp partial equilibrium expression INTRODUCTION • For any gaseous reaction: aA(g) + bB(g) + cC(g) etc. tT(g) + uU(g) + wW(g) etc. • Kp = pTt pUu pVv etc. (for units see later) pAa pBb pCc etc. • px indicates the partial pressure of x, usually in atm (atmospheres) or Pa (pascals). • Do NOT put square brackets in Kp expressions! • AND again note that Kp (like Kc) is only constant for a specific constant temperature at which the partial pressures of the component gases might vary from one equilibrium situation to another at the same temperature. • The partial pressure of a gas is defined as the pressure a gaseous component in a mixture would exert, if it alone occupied the space/volume in question. • e.g. in air, 21% by volume is oxygen and 79% is nitrogen etc. • Therefore the fraction of molecules which are oxygen and nitrogen is 0.21 and 0.79 respectively. • Since air has a total pressure of 1 atm. or 101325 Pa, the partial pressures of oxygen and nitrogen in air are: • pO2 = 0.21 x 1 = 0.21 atm or pO2 = 0.21 x 101325 = 21278 Pa, • pN2 = 0.21 x 1 = 0.79 atm or pN2 = 0.21 x 101325 = 80047 Pa, • in other words, the partial pressure of a gas in a mixture pgas = its % x ptot / 100 • In a gaseous mixture the total pressure equals the sum of all the partial pressures: • ptot = p1 + p2 + p3 etc. so for air ... • ptotair = pO2 + pN2 + pAr + pCO2 etc. etc. = 1 atm or 101325 Pa • Some other useful mathematical ideas and expressions for gas mixtures: • The % by volume ratio is also the mole ratio of the gases in a mixture. • This derives from Avogadro's Law that "equal volumes of gases at the same temperature and pressure contain the same number of molecules". • If we call x the mole fraction of gas in a mixture then: • The mole fraction of a gas A in a mixture = xA = mol of A / total moles of all gases • Therefore the partial pressure of gas A, pA = xA x ptot • Equilibrium example 2.1b.1 The formation of nitrogen(II) oxide (e.g. in car engines) • N2(g) + O2(g) 2NO(g) • Kp = pNO2 (no units) pN2 pO2 • imagine p = partial pressure units • units = p2/(p x p), all partial pressure units cancel out, no units • Equilibrium example 2.1b.2 The synthesis of sulfur(VI) oxide in the manufacture of sulfuric acid • The Contact Process • 2SO2(g) + O2(g) 2SO3(g) (sulfur trioxide) • Kp = pSO32 (units e.g. atm1 or Pa1) pSO22 pO2 • imagine p = partial pressure units • units = p/(p2 x p) = p/p3 = p1 • Equilibrium example 2.1b.3 The synthesis of ammonia (Haber Process) • N2(g) + 3H2(g) 2NH3(g) • Kp = pNH32 (units atm2 or Pa2) pN2 pH23 • imagine p = partial pressure units • units = p2/(p x p3) = p2/p4 = p2 • See calculation 2.2b.2 • Equilibrium example 2.1b.4 The manufacture of hydrogen (e.g. for ammonia synthesis) • CH4(g) + H2O(g) 3H2(g) + CO(g) • Kp = pH23 pCO (units atm2 or Pa2) pCH4 pH2O • imagine p = partial pressure units • units = (p3 x p)/(p x p) = p4/p2 = p2 • Equilibrium example 2.1b.5 The synthesismanufacture of methanol (gas phase) • CO(g) + 2H2(g) CH3OH(g) • Kp = pCH3OH (units atm2 or Pa2) pCO pH22 • imagine p = partial pressure units • units = p/(p x p2) = p/p3 = p2 2.2 Exemplar calculations and concept Questions These exemplar questions involve both numerical calculations and application of Le Chatelier's Principle. 2.2a Kc and concentration calculations • Kc Example Q 2.2a.1 Esterification • Given the esterification reaction: ethanoic acid + ethanol ethyl ethanoate + water • CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) • A mixture of 1.0 mol of ethanoic acid and 1.0 mol of ethanol was left to reach equilibrium at 25oC. • On analysis of the equilibrium mixture it was found that by titration with standard sodium hydroxide solution 0.333 mol of the ethanoic acid was left unreacted. • Calculate the value of the equilibrium constant Kc and give its units. • The reactantproduct mole ratios are 1:1 ==> 1:1 • If 0.333 mol ethanoic acid was left unreacted, then 0.667 mol of the acid had reacted. • Therefore 0.667 mol of ethanol must also have reacted, leaving 0.333 unreacted. • If 0.667 of the acid/ethanol reacted, 0.667 mol of ethyl ethanoate and 0.667 mol of water must be formed. • The concentrations = mol / volume, but the volume terms cancel each other out, so substitution in the equilibrium expression can be done in terms of final numbers of moles reactants and products • Kc = [CH3COOCH2CH3(l)] [H2O(l)] [CH3COOH(l)] [CH3CH2OH(l)] • Kc = 0.667 x 0.667 = 4.01 (no units) 0.333 x 0.333 • Kc Example Q 2.2a.2 Formation of hydrogen iodide or the decomposition of hydrogen iodide • In these examples, don't forget that all the V's cancel, so you can work your logic in moles when solving these equilibrium problems. In the first example, concentrations are given, BUT after that its all moles, logic and maybe algebra! • Example (a) For the reaction: H2(g) + I2(g) 2HI(g) • An equimolar mixture of hydrogen and iodine was heated in a sealed flask at 491oC and the concentration of iodine was found to be 2.5 x 102 mol dm3 and that of hydrogen iodide 1.71 x 101 mol dm3. • (a) Calculate the value of the equilibrium constant Kc. • The concentration of the remaining hydrogen and iodine must be the same since they started at a 1:1 ratio and react in a 1:1 ratio. • Therefore substituting in the equilibrium expression: • Kc = [HI(g)]2 [H2(g)] [I2(g)] • Kc = (1.71 x 101)2 = 46.8 (no units) (2.5 x 102) x (2.5 x 102) • Examples (b) More logicmathematical solutions to the hydrogen, iodine and hydrogen iodide equilibrium • HI/H2/I2 moles logic to the fore! • (i) For the equilibrium reaction: 2HI(g) H2(g) + I2(g) • Suppose we start with 1 mole of hydrogen iodide and fraction x of it decomposes into hydrogen and iodine (x x 100 = % decomposition). • For every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine. • This arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ... • HI H2 I2 initial moles 1.0 0.0 0.0 moles at equilibrium 1x 0.5x 0.5x • Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving ..... • Kc = [H2(g)] [I2(g)] [HI(g)]2 • Kc = (0.5x)(0.5x) 0.25x2 = = ..... (no units) (1x)2 (1x)2 • So, if for example, the hydrogen iodide was 20% decomposed, then obviously the proportion decomposed x = 0.20, and simple substitution into the equation enables you to calculate the equilibrium constant K without any difficulty. • Incidentally, since K is dimensionless, Kc = Kp even if you were working in partial pressures. • However, if you were given K instead, and had to solve the equation for x, the proportion decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right). • Rearranging the equilibrium expression gives • K(1x)2 = 0.25x2 • K(x2 2x + 1) = 0.25x2 • Kx2 2Kx + K = 0.25x2 • (K 0.25)x2 2Kx + K = 0 • So, in the quadratic equation formula • a = K 0.25, b = 2K and c = K, to solve for x. • bon voyage and watch the sign! • (ii) For the equilibrium reaction: H2(g) + I2(g) 2HI(g) • Suppose we start with 1 mole of hydrogen and 1 mole of iodine and no hydrogen iodide. • For every x moles of hydrogen or iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ... H2 I2 HI initial moles 1.0 1.0 0.0 moles at equilibrium 1x 1x 2x Kc = [HI(g)]2 [H2(g)] [I2(g)] Kc = (2x)(2x) 4x2 = .... (no units) (1x)2 (1x)2 • Again, if you know the amount of HI formed OR the amount of iodine or hydrogen reacted, you can readily calculate K via the molar logic. • BUT if you are given K, to solve for x, this requires solving a quadratic equation. • Rearranging the equilibrium expression • K(1x)2 = 4x2 • K(x2 2x + 1) = 4x2 • Kx2 2Kx + K = 4x2 • (K 4)x2 2Kx + K = 0 • So, in the quadratic equation formula • a = K4, b = 2K and c = K, to solve for x. • watch the signs and enjoy! • (iii) For the equilibrium reaction: H2(g) + I2(g) 2HI(g) • A general solution for ANY combination of hydrogen and iodine forming hydrogen iodide. • Suppose we start with A moles of hydrogen and B moles of iodine and no hydrogen iodide. • For every x moles of hydrogen OR iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ... H2 I2 HI initial moles A B 0.0 moles at equilibrium Ax Bx 2x Kc = [HI(g)]2 [H2(g)] [I2(g)] Kc = (2x)2 = .... (no units) (Ax) (Bx) • Again, if you know the amount of HI formed (2x) OR the amount of iodine/hydrogen reacted (x), you can readily calculate K. • Note that x is NOT a fraction here! • Its the actual moles of hydrogen or iodine that have reacted to form hydrogen iodide. • BUT if you are given K, to solve for x, this requires solving a quadratic equation. • Rearranging the equilibrium expression • K(Ax)(Bx) = 4x2 • K(x2 Ax Bx + AB) = 4x2 • Kx2 AKx BKx + ABK = 4x2 • (K4)x2 (A + B)Kx + ABK = 0 • So, in the quadratic equation formula • a = K4, b = (A+B)K and c = ABK, to solve for x. • This methodology is not designed for late night working! and watch the signs! • (iv) For the equilibrium reaction: 2HI(g) H2(g) + I2(g) • A general solution for the decomposition of hydrogen iodide. • Suppose we start with A moles of hydrogen iodide and x moles of it decomposes into 0.5x moles of hydrogen and 0.5x moles of iodine. • Remember, for every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine. • This arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ... • HI H2 I2 initial moles A 0.0 0.0 moles at equilibrium Ax 0.5x 0.5x • Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving ..... • Kc = [H2(g)] [I2(g)] [HI(g)]2 • Kc = (0.5x)(0.5x) 0.25x2 = = ..... (no units) (Ax)2 (Ax)2 • So, if you know how much HI reacteddecomposed, OR, the amount of iodine formed, then simple substitution into the equilibrium equation enables you to calculate the equilibrium constant K without any difficulty. • Note that x is NOT a fraction here! • Its the actual moles of hydrogen iodide that have decomposed to form hydrogen and iodine. • However, if you were given K, to solve the equation for x, the amount decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right). • Rearranging the equilibrium expression • K(Ax)2 = 0.25x2 • K(x2 2Ax + A2) = 0.25x2 • Kx2 2KAx + KA2 = 0.25x2 • (K 0.25)x2 2KAx + KA2 = 0 • So, in the quadratic equation formula • a = K0.25, b = 2KA and c = KA2, to solve for x. • keep a clear head and watch the signs! • Kc Example Q 2.2a.3 Formation of complex ion • Aqueous iron(II) ions can complex with chloride ions to form the tetrachloroferrate(III) ion. • For the equilibrium: Fe3+(aq) + 4Cl(aq) FeCl4(aq) • Kc = 8.0 x 102 mol1 dm3 at 298K. • (a) If the concentration of the free chloride ion is 0.80 mol dm3 and that of the free iron(III) ion 0.20 mol dm3, calculate the concentration of the tetrachloroferrate(III) complex ion. • Kc = [FeCl4(aq)] [Fe3+(aq)] [Cl(aq)]4 • rearranging the equilibrium expression gives ... • [FeCl4(aq)] = Kc x [Fe3+(aq)] x [Cl(aq)]4 • [FeCl4(aq)] = 8.0 x 102 x 0.20 x (0.80)4 = 6.55 x 103 mol dm3 • (b) Equal volumes of 5 molar sodium chloride solution and 0.02 molar iron(III) nitrate Fe(NO3)3 solution were mixed together. • (i) Assuming the chloride ion concentration changes very little and c represents the concentration of the tetrachloroferrate(III) ion, show how the concentration of the complex ion can be approximately calculated. • The nitrate ion is a spectator ion and can be ignored. • Let [FeCl4(aq)]equilib = c, and • since 1 mole of Fe3+ forms 1 mole of FeCl4 complex • then [Fe3+(aq)]equilib = [Fe3+(aq)]init c, and assuming • [Cl(aq)]equilib ~ [Cl(aq)]init since [Cl(aq)] >> [Fe3+(aq)] + [FeCl4(aq)] • (ii) Calculate the concentration of the [FeCl4(aq)] ion. • The dilution factor on mixing equal volumes is 2, therefore • [Fe3+(aq)]init = 0.02/2 = 0.01 mol dm3 and • Cl(aq)]equilib ~ [Cl(aq)]init = 5.0/2 = 2.5 mol dm3 • from (a) [FeCl4(aq)] = Kc x [Fe3+(aq)] x [Cl(aq)]4 • c = 8.0 x 102 x (0.01 c) x (2.5)4 • c = 3.125 x (0.01 c) = 0.03125 3.125c • 4.125c = 0.03125 • [FeCl4(aq)] = c = 0.03125/4.125 = 0.7575 = 7.58 x 103 mol dm3 • (iii) What percentage of the original Fe3+ ion is converted into the complex? • If all of the Fe3+ ion had been converted to the chloro complex, the concentration of the complex would be 0.01 mol dm3 • Therefore the % conversion = 7.58 x 103 x 100/0.01 = 75.8% • Kc Example Q 2.2a.4 Iodineiodide equilibrium • Iodine is much more soluble in potassium iodide solution than pure water because of the equilibrium: • I(aq) + I2(aq) I3(aq) for which Kc = 7.10 x 102 mol1 dm3 at 298K. • If the concentration of the I ion is 0.122 mol dm3, and that of the I3 ion is 0.153 mol dm3, calculate the concentration of free iodine. • Kc = [I3(aq)] [I(aq)] [I2(aq)] • Rearranging gives ... • [I2(aq)] = [I3(aq)] Kc x [I(aq)] • [I2(aq)] = 0.153 / (7.10 x 102 x 0.122) = 1.77 x 103 mol dm3 • Kc Example Q 2.2a.5 Ester equilibrium titrimetric analysis • A titration method for determining the equilibrium constant for an esterification reaction. • 12.0g of pure ethanoic acid was mixed with 11.5g of pure ethanol and left to stand for a week at room temperature (298K/25oC). The mixture was then mixed with deionised water and made up to 250 cm3 in a calibrated volumetric flask. When a 25.00 cm3 aliquot of the mixture was titrated with 0.50 mol dm3 sodium hydroxide solution using phenolphthalein indicator (colourless ==> 1st permanent pink endpoint), 10.60 cm3 of the alkali was needed for complete neutralisation. • (i) CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) (esterification reaction) • (ii) CH3COOH(l) + NaOH(aq) ==> CH3COONa+(aq) + H2O(l) (neutralisation titration reaction) • (a) Calculate the moles of ethanoic acid unreacted in the original mixture. • 1 mole CH3COOH : 1 mole NaOH from equation (i) above. • moles = molarity x vol(dm3), so moles acid = 0.50 x 10.6/1000 = 0.0053 mol • since the 25.00 cm3 aliquot titrated is equal to 1/10th of the total mixture, • the total moles of unreacted acid = 10 x 0.0053 = 0.053 mol CH3COOH left • (b) Calculate the moles of ethanoic acid and ethanol in the starting mixture. • Mr(CH3COOH) = 60, mol ethanoic acid = 12/60 = 0.20 • Mr(CH3CH2OH) = 46, mol ethanol = 11.5/46 = 0.25 • (c) Calculate the moles of ethanol left unreacted and the moles of ethyl ethanoate ester and water formed. • This requires a little bit of logic, best appreciated with a little table showing the 'logical thinking' and the final mol numbers. • equation CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) moles at start 0.20 0.25 0 0 mol at equilibrium 0.20 x 0.25 x x x mol at equilibrium 0.053 0.103 0.147 0.147 • x = moles of ester or water formed, • so x mol of acid or alcohol must be used to reach equilibrium, • since mol ratios are 1:1 ==> 1:1 in the reaction equation. • Therefore 0.20 x = 0.053 from calculation (a), therefore x = 0.20 0.053 = 0.147 • so all the molar quantities can be logically deduced and are shown in the final line of the table. • TIP In an exam, doing the working of this part of the Q under the equation is not a bad idea. • (d) Calculate the equilibrium constant Kc for this esterification. • Kc = [CH3COOCH2CH3(l)] [H2O(l)] [CH3COOH(l)] [CH3CH2OH(l)] • Kc = (0.147/V) x (0.147/V) = 3.96 (no units) (0.053/V) x (0.103/V) • Note that all the volume terms cancel out, so you can work purely in moles to calculate Kc. • (e) Suggest several reasons why it may take a week for the equilibrium point to be reached and suggest ways of speeding up the reaction. • (i) The reaction will be slow at room temperature, refluxing the mixture will speed things up! • (ii) The reaction is catalysed by hydrogen ions (H+) and since ethanoic acid is a weak acid, the hydrogen ion concentration will be very low. Adding a strong mineral acid e.g. conc. sulfuric acid. • (iii) Not surprisingly, esters are often prepared by refluxing the acid and alcohol with a few drops of conc. sulfuric acid added to the mixture. 2.2b Kp, partial pressure calculations and applying Le Chatelier's Principle • Example Q 2.2b.1 Nitrogen(IV) oxidedinitrogen tetroxide equilibrium (nitrogen dioxide <=> dimer) • For the equilibrium between dinitrogen tetroxide and nitrogen dioxide: • N2O4(g) 2NO2(g) (ΔH = +58 kJ mol1) • the value of the equilibrium constant Kp = 0.664 atm at 45oC. • (a) If the partial pressure of dinitrogen tetroxide was 0.449 atm, what would be the equilibrium partial pressure of nitrogen dioxide and the total pressure of the gases? and what % of the dinitrogen tetroxide is dissociated? • Kp = pNO22 pN204 • Rearranging the • pNO22 = Kp x pN204, so, pNO2 = √(Kp x pN204 ) = √(0.664 X 0.449) = 0.546 atm • ptot = pNO2 + pN2O4 = 0.546 + 0.449 = 0.995 atm • The pressure of NO2 is 0.546, which is equivalent to 0.273 atm of N2O4 before dissociation. • Therefore the % N2O4 dissociated = 0.273 x 100 / (0.449 + 0.273) = 37.8% • (b) In another experiment at 77oC and a total pressure of 1.000 atm, the partial pressure of dinitrogen dioxide was found to be 0.175 atm. Calculate the value of the equilibrium constant at 77oC and the % dissociation of the dinitrogen tetroxide. • ptot = pNO2 + pN2O4, so pNO2 = ptot pN2O4 = 1.000 0.175 = 0.825 atm • From the equilibrium expression above: • Kp = 0.8252 / 0.175 = 3.89 atm • The pressure of NO2 is 0.825, which is equivalent to 0.4125 atm of N2O4 before dissociation. • Therefore the % N2O4 dissociated = 0.4125 x 100 / (0.175 + 0.4125) = 70.2% • (c) At a total pressure of 102000 Pa, dinitrogen tetroxide is 40% dissociated at 50oC. • (i) Calculate the partial pressures of the gases present. • Each mole of N2O4 dissociated gives 2 moles of NO2. • An initial 100 mol of N2O4 gives, at equilibrium, • 60 mol of undissociated N2O4, and 40 x 2 = 80 mol of NO2. • mole fraction of NO2 = 80 / (60 + 80) = 0.571, • and pNO2 = 0.571 x 102000 = 58242 Pa • mole fraction of N2O4 = 60 / (60 + 80) = 0.429, • and pN2O4 = 0.429 x 102000 = 43758 Pa • See also example Q 2.2b.3 for another way of setting out a method to calculate the partial pressures. • (ii) Calculate the value of the equilibrium constant. • Kp = pNO22 582422 = = 7.75 x 104 Pa pN204 43758 • (d) Explain the effect, if any, of increasing the total pressure of the system on the position of the N2O4/NO2 equilibrium. • Increasing pressure will favour the LHS, i.e. more N2O4 and less NO2, because the system will move to the side with the least number of gaseous molecules to try to minimise the increase in pressure (1 mol gas <== 2 mol gas). • (e) From information from parts (a)(c) is the dissociation exothermic or endothermic? and give your reasoning. • The dissociation is endothermic because on raising the temperature from 45oC to 77oC the value of Kp has increased. • i.e. pNO2 has increased in value and pN2O4 decreased in value from more dissociation. • An equilibrium system moves in the endothermic direction on raising the temperature to absorb the 'added' heat to try to minimise the temperature rise. • Example Q 2.2b.2 Ammonia synthesis • Ammonia is synthesised by combing nitrogen and hydrogen, but the reaction is reversible. • N2(g) + 3H2(g) 2NH3(g) (ΔH = 92 kJ mol1) • In an experiment starting with a 1:3 ratio H2:N2 mixture, at 400oC and total pressure of 200 atm, the equilibrium mixture was found to contain 36.3% by volume of ammonia. • (a) Write out the equilibrium expression for Kp. • Kp = pNH32 (units atm2 or Pa2) pN2 pH23 • (b) Calculate the partial pressures of nitrogen, hydrogen and ammonia. • 36% of ammonia means its mole fraction is 0.36, so pNH3 = 0.36 x 200 = 72 atm • The other 64% of gases must be split on a 1:3 ratio between nitrogen and hydrogen. • If you think of the 1:3 ratio as 1 and 3 parts out of 4 the logic becomes easy. • So mole fraction of nitrogen x ptot = pN2 = 64/100 x 1/4 x 200 = 32 atm, • and mole fraction of hydrogen x ptot = pH2 = 64/100 x 3/4 x 200 = 96 atm • check: ptot = pN2 + pH2 + pNH3 = 32 + 96 + 72 = 200 atm! • (c) Calculate the value of the equilibrium constant at 400oC and give its units. • Kp = pNH32 pN2 pH23 • Kp = 722 = 1.83 x 104 atm2 32 x 963 • (d) What will be the effect on ammonia yields and the value of Kp by (i) raising the pressure and (ii) raising the temperature. Give reasons for your answers. • (i) The reaction to form ammonia is exothermic, so higher temperature will favour its endothermic decomposition back to nitrogen and hydrogen. So the yield of ammonia and the value of Kp will be reduced as the system will absorb heat energy to attempt to minimise temperature rise. • (ii) Higher pressure will favour a higher yield of ammonia because 4 mol gaseous reactants ==> 2 mol gaseous products, so if the system is subjected to higher pressure it reduces the number of gas molecules to attempt to minimise the enforced increase in pressure. Assuming the gases behave ideally, the value of the equilibrium constant Kp is unaffected by pressure changes, only temperature affects the value of Kp in an ideal gas mixture. • Example Q 2.2b.3 Phosphorus(V) chloride (phosphorus pentachloride) dissociation • At high temperatures vapourised phosphorus(V) chloride dissociates into gaseous phosphorus(III) chloride (phosphorus trichloride) and chlorine. • PCl5(g) PCl3(g) + Cl2(g) (a) Write the equilibrium expression for this reaction in terms of Kp and partial pressures. Kp = PPCl3 x PCl2 (units atm, Pa, kPa etc.) PPCl5 • (b) At a particular temperature 40% of the PCl5 dissociates into PCl3 and Cl2. If the total equilibrium pressure is 210 kPa calculate the value of Kp at this temperature quoting appropriate units. The solution is set out below in a series of logical steps, either thinking from the point of view of starting with n/1 moles of PCl5 or starting with 100/100% of PCl5 molecules. • PCl5(g PCl3(g) Cl2(g) comments, z refers to any component in the mixture n(1x) (1x) can think of as (100 x%) nx x can think of as x% dissociated nx x can think of as x% dissociated n = initial moles of undissociated PCl5 for a general solution, but consider 1 mole of PCl5 (n = 1) of which fraction x dissociates giving a total of (1 + x) moles of gas from 1 mol of PCl5 This is the logic thinking in terms of 100/100% undissociated PPCl5 molecules and x% is the percentage of PCl5 molecules dissociated. (1x)/(1 + x) x/(1+x) x/(1+x) calculation of mole fractionsmole fraction z = moles z/total moles 0.6/1.4 = 0.4286 can think of as 60/140 = 0.4286 0.4/1.4 = 0.2857 can think of as 40/140 = 0.2857 0.4/1.4 = 0.2857 can think of as 40/140 = 0.2857 since x = 0.4 (= 40% dissociated) It works out the same in terms of the original 100/100% undissociated PCl5 PPCl5 = 0.4286 x 210 = 90.0 PPCl3 = 0.2857 x 210 = 60.0 PCl2 = 0.2857 x 210 = 60 partial pressure of component z Pz = mole fraction z x Ptot note 90 + 60 + 60 = 210 check! • Kp = 60.0 x 60.0 = 40.0 kPa 90.0 • • Example Q • Part 2 subindex: 2.1 Kc and Kp equilibrium expressions and Le Chatelier's Principle 2.2 Exemplar calculations using Kc expressions or Kp expressions Chemical Equilibrium Notes Index advanced level chemistry how to write equilibrium expressions and do equilibrium calculations for AQA AS chemistry, how to write equilibrium expressions and do equilibrium calculations for Edexcel A level AS chemistry, how to write equilibrium expressions and do equilibrium calculations for A level OCR AS chemistry A, how to write equilibrium expressions and do equilibrium calculations for OCR Salters AS chemistry B, how to write equilibrium expressions and do equilibrium calculations for AQA A level chemistry, how to write equilibrium expressions and do equilibrium calculations for A level Edexcel A level chemistry, how to write equilibrium expressions and do equilibrium calculations for OCR A level chemistry A, how to write equilibrium expressions and do equilibrium calculations for A level OCR Salters A level chemistry B how to write equilibrium expressions and do equilibrium calculations for US Honours grade 11 grade 12 how to write equilibrium expressions and do equilibrium calculations for pre-university chemistry courses pre-university A level revision notes for how to write equilibrium expressions and do equilibrium calculations A level guide notes on how to write equilibrium expressions and do equilibrium calculations for schools colleges academies science course tutors images pictures diagrams for how to write equilibrium expressions and do equilibrium calculations A level chemistry revision notes on how to write equilibrium expressions and do equilibrium calculations for revising module topics notes to help on understanding of how to write equilibrium expressions and do equilibrium calculations university courses in science careers in science jobs in the industry laboratory assistant apprenticeships technical internships USA US grade 11 grade 11 AQA A level chemistry notes on how to write equilibrium expressions and do equilibrium calculations Edexcel A level chemistry notes on how to write equilibrium expressions and do equilibrium calculations for OCR A level chemistry notes WJEC A level chemistry notes on how to write equilibrium expressions and do equilibrium calculations CCEA/CEA A level chemistry notes on how to write equilibrium expressions and do equilibrium calculations for university entrance examinations with advanced level chemistry practice questions for solving equilibrium problems, how do you write Kc equilibrium expressions? how do you write Kp equilibrium expressions? what are the units for Kc equilibrium expressions? what are the units for Kp equilibrium expressions? how to do Kc equilibrium calculations using molarity concentrations, how to do Kp equilibrium calculations using partial pressures, how to use partial pressure and molarities in equilibrium expressions, explaining the carboxylic acid alcohol ester water equilibrium experimental procedure, explaining methods of solving equilibrium problems, solving hydrogen iodide hydrogen iodine equilibrium problems using the equilibrium expression, how to solve PCl5 phosphorus(V) chloride PCl3 phosphorus(III) chloride Cl2 chlorine problems calculations from the equilibrium expression, solving the ammonia hydrogen nitrogen equilibrium expression, how to do calculations from the formation of ammonia equilibrium expression, solving the phosphorus fluorine phosporus(5) fluoride equilibrium expression units calculations, how to write equilibrium expressions for complex ion formation, how to work out units for equilibrium expressions, solving the nitrogen oxygen nitrogen(II) oxide equilibrium expression calculations, solving the equilibrium expression for the contact process, solving the equilibrium expression for the synthesis of methanol CH3OH from hydrogen H2 and carbon monoxide CO explain the effect of temperature on equilibrium constants, how do you write Kc equilibrium expression? how do you solve numerical equilibrium problems? how do you write out Kp equilibrium expression in partial pressures? how do you solve equilibrium problems using partial pressures? how do you calculate the Kc equilibrium constant from concentrations in mol/dm3? how do you calculate Kp equilibrium constants using partial pressures atm Pa? Doc Brown's Chemistry *	12,995	44,116	{"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}	2.625	3	CC-MAIN-2017-13	longest	en	0.846188
https://www.physicsforums.com/threads/time-invariance-of-a-system.529298/	1,506,216,805,000,000,000	text/html	crawl-data/CC-MAIN-2017-39/segments/1505818689823.92/warc/CC-MAIN-20170924010628-20170924030628-00230.warc.gz	861,828,617	14,899	# Time Invariance of a System 1. Sep 12, 2011 ### awelex 1. The problem statement, all variables and given/known data Determine if the following system is time invariant: y(t) = x(t - 2) + x(2 - t) 2. The attempt at a solution I know from the solutions that the system is NOT time invariant, yet whenever I try to solve it I get the opposite result. Here's what I'm doing: y1(t) = x1(t - 2) + x1(2 - t) x2(t) = x1(t - t0) y2(t) = x2(t - 2) + x2(2 - t) = x1(t - t0 - 2) + x1(2 - t + t0) y1(t - t0) = x1(t - t0 - 2) + x1(2 - t + t0) Therefore y2(t) = y(t - t0) What am I doing wrong? Thanks! 2. Sep 12, 2011 ### Staff: Mentor You have told us everything you know about x(t)? 3. Sep 12, 2011 ### awelex Yes; no information at all is given about x(t). 4. Sep 12, 2011 ### awelex Any takers? I found a case that clearly shows that the system is not time invariant, but I'd still love to know what is wrong about my proof. I can't seem to figure it out. Thanks! 5. Sep 14, 2011 ### Staff: Mentor Please return to the specification of the problem, and quote verbatim the sentence there containing the word "odd" or "even".	380	1,142	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2017-39	longest	en	0.888126
https://hirecalculusexam.com/differential-calculus-books/	1,675,773,393,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00261.warc.gz	314,901,427	19,464	# Differential Calculus Books Differential Calculus Books: The Ultimate Scenario During an afternoon of discussions, “Pigeon Lane” creator George Rittenhouse shared ideas about the way we run when we first begin talking about ideas. The first five days spent in this studio, George admitted: “We do some development work, and I think it’s not the first time we have thought about it.” It worked that way. After being involved useful site Read More Here the planning and discussions, there wasn’t much else to do. Rittenhouse showed us exactly how it can be done; these days, he created graphics. People don’t have a way to tell, do, or do it; they can’t seem to see, understand, or take the time to help those in need. They can’t see either or hope that the other person will come along with them and bring it in. Thus the team is different, the differences, in principle. Below, you’ll find a version of the entire scenario, from Rittenhouse’s perspective, which covers the four main elements of what we’ve been talking about – motivation, motivation and feeling. The Motivation Rittenhouse was intrigued with motivation for some of his ideas, the ones that would best help, for example, motivating children to put their hands on a toy or drawing a picture. He understood that to create something that interested kids, motivated or to be fun, but also interested as a designer, creator, creator-artist of paint, or other kind of artist. Rittenhouse had a lot of reasons to think he would be interested in motivation. It’s believed that they should seek the natural way as opposed to the chemicalized way, but what we’ll see below is another angle on the subject. Motivation We start by discussing what motivates someone to want to touch the other person. He will have trouble with the potential of his new toy or be drawn. Motivation People don’t always want to pay the price, but they can often find their way into what he has come to understand — a sense of appreciation, interest, experience, or commitment — and try to find some way that will make that that person want to do things. Instead, they seek the natural method. People need to know they want to be happy, and they need to give new hope to those they see as part of the rest for a moment or as a result of looking at the kids when they come home. They will always want a way to show the kids that others in our community or in the community do not go for it. They want to show the kids that others view them as part of the community so that they feel they can stand out from the crowd in whatever way those might make it. ## First Day Of Class Teacher Introduction Everyone must be willing to be willing to take the time to work out what motivates them, their overall needs, their priorities, their responses. The other time that you can make that happen is when you have some fresh ideas in you, and from a different perspective. For example, you may choose to do crafts or the like. This can make an exciting transition for someone whom you see as a challenge or someone who knows how to make it. It’s essential your community i thought about this a good job of bringing you in for work. Consider it a time-taking process? Not quite. Motivation Motivation needs to go back into the mindset of the creator. People may feel that they want an outlet for their value and have that focus in their mind. Rather, we call upon them to really feel what motivates them, their focus on the purpose and goals and their value as a person. We can use this person to show them the value of knowing, doing, or doing something, but also as a person who does not have that individual eye in how they want to move toward the end of time. We can show a person a positive angle or a negative one, but when that person has that negative one already, show them why they do not want to do it. Motivation In the first instance, we are showing our passion for the project, for example, motivation for making the initial steps of going to the shop, creatingDifferential Calculus Books Free Page 2 of 4 Category:Books about history and culture published:2018 views:4/2 The New York Times \| Books 1 of 1 Image of the Thomas Hardy Memorial \| Daedalus by George Lattner, edG. T.B. – DaedalusNo 1 images are available in US.photo Published by: New York / The New York Times First published in 2015 First published in this edition About the Author Sydney G. Smith is a professor of humanities at the University of Cambridge. She holds a doctorate from Oxford and an honorary doctorate from Trinity. Through her work, her scholarly adventures have appeared in numerous magazines, books, and anthologies. Most of Smith’s books are non-fiction and feature early novels and biographies, including a series of works that were translated into many languages, including English and Chinese.	1,035	4,820	{"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-2023-06	latest	en	0.966849
http://dejenol.com/index.php?title=Character_Attack_Formulas&diff=4953&oldid=prev	1,580,014,143,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251687725.76/warc/CC-MAIN-20200126043644-20200126073644-00023.warc.gz	47,353,180	9,268	# Character Attack Formulas (Difference between revisions) Revision as of 14:16, 1 October 2009 (edit) (added probabilities of scoring backstab or crititical hit)← Previous diff Revision as of 20:56, 1 October 2009 (edit) (undo) (clarification)Next diff → Line 2: Line 2: GLvl = Guild Level with best fighting ability, should be the same as the one in "Guild" Tab GLvl = Guild Level with best fighting ability, should be the same as the one in "Guild" Tab - FightingMod = Value found in Help File / 10, ie Warrior = 1.2, Nomad = 0.7 + FightingMod = Value found in Help File / 10, i.e., Warrior = 1.2, Nomad = 0.7 Size = Range [0 to 5] or Very Small, Small, Normal, Big, Very Big, Huge Size = Range [0 to 5] or Very Small, Small, Normal, Big, Very Big, Huge Ln stands for natural logarithm. Ln stands for natural logarithm. Line 40: Line 40: Chance of scoring a critical hit or backstab: Chance of scoring a critical hit or backstab: - Crit: ((((LN((GLvl / 999) + 1) * 100) * LN(CritMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/3 + Ln(ItemCritMod+1) * Ln(10), maximum 50% + Crit: ((((LN((GLvl / 999) + 1) * 100) * LN(CritMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/3 + Ln(ItemDropLevel+1) * Ln(10), maximum 50% - Backstab: ((((LN((GLvl / 999) + 1) * 100) * LN(BSMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/2 + Ln(ItemBSMod+1) * Ln(100), maximum 75% + Backstab: ((((LN((GLvl / 999) + 1) * 100) * LN(BSMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/2 + Ln(ItemDropLevel+1) * Ln(100), maximum 75% + + CritMod = Value found in Help File, i.e., Warrior = 9, Seeker = 3 + BSMod = Value found in Help File, i.e, Thief = 9, Ninja = 2 [http://mordor.uni.cc/forum/index.php?showtopic=1207 Discuss this topic in the forum] [http://mordor.uni.cc/forum/index.php?showtopic=1207 Discuss this topic in the forum] ## Revision as of 20:56, 1 October 2009 The formula has been well tested and seems to be extremely close, barring possible issues with different rounding algorithms. GLvl = Guild Level with best fighting ability, should be the same as the one in "Guild" Tab FightingMod = Value found in Help File / 10, i.e., Warrior = 1.2, Nomad = 0.7 Size = Range [0 to 5] or Very Small, Small, Normal, Big, Very Big, Huge Ln stands for natural logarithm. Rnd is a number ranging from 0 to 0.999.. (not 1, this is important because it produces different top values). This is the Damage Modifier, a bit more complex than I expected really 1: DamMod = 0.6 + (((Ln(100 + GLvl) / 1.75) - 2.3) * FightingMod^2 / 2) 2: * WeaponModifier 3: + 5 if Critical Hit, Chance see below 4: + 2 if Backstab, Chance see below 5: / 2 if Monster is invisible and character can't see invisible 6: +0.1/-0.1 per size bigger/smaller than monster 7: if DamMod is > 1, then DamMod = DamMod - ln(DamMod)^2 Value 0.1 to 7ish (Lv 999 Human Warrior with a Mod 2.6 Weapon, severing a Tiny monster) Let's get the Strength Modifier, it's actually 2 parts, and requires a new Strength Value (called BSV here) BSV = Int(Strength - (LN(Strength) * ((LN(Strength) - 2,4) * 2,25))) + 1 StrMod1 = Int((Rnd * ((BSV + 10) / 2)) + ((BSV + 10) / 4)) / 10 StrMod2 = BSV/30 Str 1 to 40 -> Value [0.3-0.8 * 0.06667] to [1-2.9 * 1] Note, this value is the 2 StrMods combined And last part of formula, Level Modifier. LvlMod = (Ln(GLvl + 5) + 1) / Ln(1.2) Lvl 1 to 999 -> Value 15.31228 to 43.39447 Base Damage = LvlMod * StrMod1 * StrMod2 * DamMod Final Damage = Base - Rnd[1 to (Def-Atk)/8] or very close to this Defense bug lurks though, so it's just: Final Damage = Base + Rnd[1 to Atk/8] Chance to Hit Formula in percent, range [1 to 97] if Atk = 0 and Def = 0: 50 if Atk > Def: 0.5 + 0.5 * (Atk - Def) / Atk if Atk < Def: 0.5 - 0.5 * (Def - Atk) / Def Since Monster Defense passed to this formula is 0, you should always have 97% chance to hit. Chance of scoring a critical hit or backstab: Crit: ((((LN((GLvl / 999) + 1) * 100) * LN(CritMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/3 + Ln(ItemDropLevel+1) * Ln(10), maximum 50% Backstab: ((((LN((GLvl / 999) + 1) * 100) * LN(BSMod + 1)) / 3) - (LN((MonLvl / 999) + 1)) * 100))/2 + Ln(ItemDropLevel+1) * Ln(100), maximum 75% CritMod = Value found in Help File, i.e., Warrior = 9, Seeker = 3 BSMod = Value found in Help File, i.e, Thief = 9, Ninja = 2	1,544	4,280	{"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-2020-05	latest	en	0.757753
https://www.rakennusopas.com/what-is-cfh-in-gas-flow-rate/	1,632,834,671,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00463.warc.gz	966,360,517	17,843	# What Is Cfh In Gas Flow Rate ## How many BTU is CFH? The BTU's per hour unit number 2.72 Btu/h converts to 1 atm cfh, one atmosphere cubic foot per hour. via ## What does CFH refer to in gas flow? A cubic feet per hour (CFH) is the quantity of gas flow in cubic feet, delivered during a time period of one hour. It is calculated by dividing the appliance input with the fuel gas heating value. In our example, 1,000 Btu is the heating value used. via ## What is CFH measurement? The CFH measures dew- or frostpoint temperatures in all climates from the surface to the altitude of the balloon burst (typically around 30 km altitude), covering all of the troposphere, and the lower up to the middle stratosphere. via ## What is CFH in propane? 1 cubic foot propane = 0.0278 gallons propane. 100 cubic feet propane = 2.78 gallons propane. 1 gallon propane = 35.97 cubic feet propane. via ## What's the difference between BTU and CFH? 1000 BTUs is the equivalent of 1 Cubic Foot. MBH is thousands of BTUs per hour, so 1 CFH is 1 MBH. via ## Is CFH and Scfh the same? The Standard cu-ft in gas per hour unit number 1.00 SCFH at 59°F converts to 1 cu ft/h, one cubic foot per hour. It is the EQUAL flow rate value of 1 cubic foot per hour but in the Standard cu-ft in gas per hour gas flow unit alternative. via ## How do I find out how much gas my house uses? • HOW TO CALCULATE NATURAL GAS CONSUMPTION FLOWS. • Gas Unit Ratings are provided in BTU's per Hour. • CFH = BTUH X 1000. • Where CFH = Cubic per hour. • BTUH = BTU's per hour. • Heating Air with Natural Gas. • CFH =< (CFM AIR) X (°F OUT - °F IN) > ÷ 800. • CFH = 200. • ## How do you calculate gas per hour? • Formula To Estimate Maximum Engine Fuel Consumption. Gallon Per Hour (GPH) = (specific fuel consumption x HP)/Fuel Specific Weight. • 300-hp Diesel Engine Example. GPH = (0.4 x 300)/ 7.2 = 105/7.2 = 16.6 GPH. • 300-hp Gasoline Engine Example. GPH = (0.50 x 300)/ 6.1 = 150/6.1 = 24.5 GPH. • Other Related Resources: • ## How do you calculate CFM to CFH? 1 Cubic foot per hour [cfh] = 0.016 666 666 666 667 Cubic foot per minute [cfm] - Measurement calculator that can be used to convert Cubic foot per hour to Cubic foot per minute, among others. via ## What is CFH in welding? MIG shielding gas flow is set and measured as cubic feet of gas per hour (CFH) NOT pressure in psi. The gas pressure in the hose going into a wire feeder/welder, while welding, typically varies from 3 to 8 psi. via ## How many gallons is 15 lbs of propane? A tank exchange with 15 pounds contains about 3.53 gallons of propane. via ## How many gallons are in a 20 lb propane tank? 20 lb Propane Tank 20 pound propane tanks are often referred to as grill cylinders and hold 4.6 gallons of propane when full. via ## Which is better propane or natural gas? And, overall, propane is the more efficient fuel. That means, propane is more than twice the energy of natural gas. While the cost per gallon is less for natural gas, you'll use more of it to heat the same appliances. If you get two times the heat from propane, naturally, you'll use less fuel. via ## What is Scfh stand for? SCFH – Standard Cubic Feet per Hour is a volumetric flow rate of a gas equal to 1 cubic foot of air at 70F and 14.7 psia* flowing past a point per hour. via ## What's the meaning of BTU? British thermal unit via ## What is the formula to calculate the gas flow rate in cubic feet per hour? Since the energy in natural gas is equivalent to the gas's volume, you can freely convert between BTUs per hour and cubic feet per minute (CFM). To calculate the CFM (ft3/min) or CFH (ft3/hr) for 500,000 BTUs per hour of Methane we have: 500,000 BTU/hr ÷ 100,000 BTU/therm = 5 therms/hr. via ## How many BTUs do I need to heat 1000 square feet? Calculating the number of BTUs needed to heat an area For example, a 300 square foot room typically requires 7,000 BTUs to maintain a comfortable temperature, while a 1,000 square foot room requires 18,000 BTUs. via ## How many BTUs do I need to heat 1500 square feet? A 1,500-square-foot home will require between 45,000 to 90,000 BTUs. A 1,800-square-foot home will require between 55,000 to 110,000 BTUs. via ## Is SCFM higher than cfm? Standard Cubic Feet per Minute (SCFM) is the rate of flow of a gas or air through a compressor at standard temperature and pressure conditions. Since SCFM measures volumetric airflow at standard conditions, it will always be higher than the CFM value for an air compressor. via ## How do you calculate SCFM? As shown in the formula and calculations it is simply the ratio of gauge pressure + atmospheric divided by the published pressure + atmospheric and then multiply the dividend by the published volume. So as we do the math we solve for 17.69 SCFM @ 105 PSIG from a device that was shown consume 14 SCFM @ 80 PSIG. via ## What is Scfh natural gas? Definition. SCFH. standard cubic feet per hour. via ## How much should my natural gas bill be? According to the figures, the average cost of natural gas in the US is a little over \$100 per month. Gas is used to heat your home, warm your water, and often powers your stove or oven. Although \$100 is the average monthly cost, the amount you pay can vary depending on the time of year. via ## How much does it cost to run a gas stove for an hour? Gas Oven at 350 degrees = somewhere between 10 and 23 cents/hour, depending on which estimations you use! via ## What is gas usage measured in? Although gas meters measure the amount of gas used in hundreds of cubic feet or cubic metres, gas bills display your use in kilowatt hours (kWh). via ## How is LPG flow rate calculated? If you have a heater that consumes 25MJ/hr, to calculate the LPG consumption per hour, just divide 25 by the values in the MJ column of the consumption conversion chart. via ## How much gas is used when cooking? How much energy does a gas oven use? Given that an average gas oven uses around 12 megajoules (mJ) of gas an hour, you can expect to pay around 65c per hour of cooking. via ## How much gas is used on a stove? Gas cooktop/range – 65,000 BTU/hour or 5 to 10 gallons per month. Tankless water heater – 40,000 BTU/hour or 1.5 gallons per day. Gas clothes dryer – 35,000 BTU/hr or less than one gallon per day. via ## What is CFM flow rate? The flow rate of a particle counter is the rate at which a pump draws air through the sample chamber, measured in cubic feet per minute (cfm). Two common flow rates for particle counters are 1 cfm and 0.1 cfm. Depending on the application, each flow rate can be valuable. via ## How many CFM do I need? CFM is how many cubic feet of air the range hood exhausts every minute at full speed. Simply put, the higher the CFM, the more air the hood vents out of your kitchen. To power your range hood, you need at least 100 CFM for every 10,000 BTUs of your stovetop. via ## What is a good CFM for air compressor? Air tools made for general use with portable air compressors typically require 0 to 5 cubic feet per minute (cfm) at 70 to 90 pounds per square inch (psi), whereas with larger tools connected to stationary systems, the requirements usually exceed 10 cfm at 100 to 120 psi. via	1,931	7,222	{"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}	3.5	4	CC-MAIN-2021-39	latest	en	0.796604
https://www.physicsforums.com/threads/subduction-components.775305/	1,532,210,081,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676592778.82/warc/CC-MAIN-20180721203722-20180721223722-00331.warc.gz	945,192,150	14,805	# Homework Help: Subduction components? Tags: 1. Oct 9, 2014 ### neongoats For my homework I was given a velocity diagram of the Juan de Fuca plate similar to this: The first question is as follows: Measurements indicate that the Juan de Fuca plate is moving at a speed of ~40 mm/year. Calculate, and draw diagrams that show, the horizontal and vertical components of this motion at two locations in Figure 2: at x=40 km (ie, well before the Juan de Fuca plate encounters the North American plate), and at x=340 km (ie, after the Juan de Fuca plate has subducted below the North American plate). I'm just super confused on what exactly this question is looking for and where I should even start? I know from the next question that I'm supposed to be finding velocity, but I don't know how to use this diagram to do that. I don't understand how the speed is relevant in comparison to the velocity on the graph and how the velocity would be used if it depends on depth etc. etc. Can someone decipher this question for me and gives me hints on where to go? 2. Oct 9, 2014 ### Staff: Mentor Hi neongoats, Welcome to Physics Forums. In future, please retain and use the Posting Template provided when a new thread is started here in the Homework area of Physics Forums. Seismology is not my area, but it appears to me that your Figure A provides a cross-sectional view of the plate showing how it approaches horizontally and then bends downwards at an angle which you can estimate/calculate using the distance and depth axes. I suppose the dotted red line represents the interface of the plate with the continental material, so its curve follows the "surface" of the moving plate. Presuming that the plate's bulk moves uniformly with the given speed then you can sketch velocity vectors along that curve. Use the calculated angles to extract the vertical and horizontal components of your vectors. That would be my take on the question. Share this great discussion with others via Reddit, Google+, Twitter, or Facebook Have something to add? Draft saved Draft deleted	450	2,076	{"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-2018-30	latest	en	0.934686
https://byjus.com/question-answer/which-of-the-following-is-loschmidt-number-6-times-10-23-2-69-times-10/	1,642,578,348,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00260.warc.gz	207,033,926	20,602	Question # Which of the following is Loschmidt number? A 6×1023 B 2.69×1019 C 3×1023 D None of these Solution ## The correct option is B $$2.69\times{10}^{19}$$The no. of molecules present in $$1ml$$ of gas at $$STP$$ is known as Loschmidt number.$$22400mL$$ of gas has total no. of molecules = $$6.023\times{10}^{23}$$$$1 ml$$ of gas has total no. of molecules = $$\cfrac{6.023\times{10}^{23}}{22400}$$ = $$2.69\times{10}^{19}$$Chemistry Suggest Corrections 0 Similar questions View More People also searched for View More	186	531	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.3125	3	CC-MAIN-2022-05	latest	en	0.888883
john.puttergill.org	1,725,933,301,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00892.warc.gz	307,120,285	9,946	# Numbers prove everything … or nothing! I have always tended to analyze National Statistics in my own particular way. My interest or purpose is somewhat different, I am not so much interested in the absolute figure for GDP, or the growth inferred, but, the money flows and how well the economy is doing in terms of generating wealth. The 2014 UK GDP figures infer growth, however to the wider population that seems an illusion. Perhaps the population at large is wiser than the commentators. GDP, at best, is a statistical representation of the total activity within the economy. It is somewhat debatable whether it is a good measure for determining the success or otherwise of the economy. If the overall activity is increasing, without a proportional increase in outputs, there could equally be a case to say the economy is failing. If we look on ourselves as an island, without needing anything from the rest of the world, then an increase in activity would be a sign of growth. Unfortunately, much as we like calling ourselves an island economy, we are dependent on the rest of the world for a fair proportion of the food we eat, the clothes we wear and the goods we consume, so, we are not an island unto ourselves, but part of the big ugly world! If we look at the first formula for deriving GDP {GDP = (Sum of population consumption expenditure) + (Government Consumption Expenditure) + Investment/Savings} we should see that is close to being an operating cost for the economy as a whole, and if we exclude Investment/Savings it would indeed represent an accountant's definition of operating cost. Just as with a company, increasing cost without at least a proportional increase in sales is an indication of impending failure. Our outputs, are our earnings from the rest of the world, and made up of a simple figure called current account. This is the sum of exports of goods and services, interest and dividends received from the rest of the world less the sum of imports of goods and services, interest and dividends paid to the rest of the world. If this figure is in positive territory then an increase in GDP (the cost of running our economy) does not matter, however if this figure is in negative territory an increase in GDP is indicative of erosion of it's competitiveness. For the record the UK current account has been in deficit since around 2001. In 2010 when the current Government took office this amounted £30 billion the bulk of which was on the import export account In the tax year to 2014 it amounted to -£75 billion. The import export balance has remained around -£30 billion. The last time the UK had control over the Government deficit and a current account surplus was under the Chancellorship of Kenneth Clarke and for the first two years of Gordon Brown while he kept to the Clarke spending plans. Our current Chancellor, if you exclude the effects of the Bank bailouts on the government deficit, has barely made a dent in the deficit, with this likely to approach £90 billion for the current calendar year as opposed to £98 billion he inherited! When Cameron came into office he made much of paying our way in the world, this is not so much about government deficit but more about the current account. If the current account is negative we have to borrow or sell assets to pay for our consumption. We don't have to look far to see the transfer of assets to the outside world, the bulk of our utilities are now under foreign control. In plain English this means that forever we pay a percentage of our electricity, gas and water bills to foreign countries (with the reduction of corporate tax from 30% to 21% it is highly probable that the dividend going offshore exceeds the amount of corporate tax we collect). Oh dear Mr Cameron you have failed dismally. You have presided over an increase in our current account deficit from £30 billion to £75 billion. Even your stated aim of getting rid of the Government deficit is in tatters, £98 billion to £90 billion in five years does not add up to much, does it? Let me briefly look at our National statistics as published in 2014 Blue Book. There are several significant changes in how we have derived our GDP. Most of these changes are in the interests of falling in line with Euro Stat's request to test whether the figures are exhaustive as well as some recommendations by the OECD. One of the effects of these changes is to increase the size of the informal economy by the specific inclusion of prostitution and drug smuggling. Up until 2010 we included a flat estimate of 1.25% for the size of the informal economy. In the Blue Book of 2012 in order to reconcile 2010 figures this was increased to 1.7%, the latest change increases this to about 3%. One might say not altogether infeasible. My own analysis of 2011 which looked at the money flows, would suggest that it is extremely unlikely that the population could have afforded spending an extra £10 billion on prostitutes and drugs when the 2011 figures, without this addition, already imply an increase of borrowings or withdrawal of savings by the public of £85 billion Other changes regarding the inclusion of research and development expenditure as well as unused ammunition produced are no less problematic. Generally research and development is simply part of a company's operating cost and any figure included under this heading as part of our Investment is largely anecdotal. As it is, we come pretty close to scraping the bottom of the barrel when we include the DVLA fee for personalized car registration plates as investment by the public! The accuracy or otherwise of the figures is not really an issue as long as there is consistency in their compilation. My own conclusions about our national statistics are that a large part of the cash surplus on sale of homes is being used to fund consumption expenditure. Put another way we are using part of housing price inflation in arriving at our GDP or slowly eating up our capital, to fund our consumption! Oh dear Mr Cameron, perhaps you, like Boris Johnson cannot see, that in selling prime real estate to a foreign company we will bleed forevermore as we pay our rent to a foreign country! Perhaps that is why you have done nothing to stop the widening current account deficit, where the bleeding on "factor account" (rent, interest and dividends) now exceeds the deficit on imports/exports. What can I say, Mr Cameron? Except … In spite of all your huff and puff, we are definitely not paying our way! As some who read this have not believed here is a graphic: The points on this graph are quarterly figures dating back until 1976, The dark dotted line is a moving average that better defines the trend. While the trend was downwards throughout Gordon Brown's chancellorship, under Osborne it has gone over the edge of the cliff and is now in free fall. Another 5 years of "the plan", Mr Cameron, and we will be lucky if the world is prepared to pay a dollar for a pound! Currently an exchange rate of 1.38 to the dollar will correct the bleeding on factor account excluding foreign aid payments we make. Oh!	1,449	7,135	{"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-2024-38	latest	en	0.955843
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http://docplayer.net/21311298-Math-251-review-questions-for-test-3-rough-answers.html	1,715,996,347,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00364.warc.gz	7,993,562	31,155	# Math 251, Review Questions for Test 3 Rough Answers Size: px Start display at page: Transcription 1 Math 251, Review Questions for Test 3 Rough Answers 1. (Review of some terminology from Section 7.1) In a state with 459,341 voters, a poll of 2300 voters finds that 45 percent support the Republican candidate, where in reality, unknown to the pollster, 42 percent support the Republican candidate. (a) What is the value of the statistic of interest? Answer. 45% (a statistic is a numerical property of the sample) (b) What is the value of the parameter of interest? Answer. 42% (a parameter is a numerical property of the population) (c) Describe the population of interest. Answer. The population is all voters in the state. (d) In general, is it true that given a certain population, the parameter of interest will not change under repeated sampling? xplain. Answer. True, the parameter does not depend on a specific sample, so it doesn t change when the sample changes. 2. (a) A produce company claims that the mean weight of peaches in a large shipment is 6.0 oz with a standard deviation of 1.0 oz. Assuming this claim is true, what is the probability that a random sample of 1000 of these peaches would have a mean weight of 5.9 oz or less? Answer: This is a central limit theorem type problem on sampling distributions (see Section 7.2 for more) z = P (z < 3.16) = =.0008 Thus, there is approximately a.0008 probability of obtaining such a sample assuming that the mean is 6.0 oz. (b) If the store manager randomly selected 1000 peaches and found that the mean weight of those 1000 peaches was 5.9 oz, should she be suspicious of the produce company s claim that the mean weight of peaches in the shipment is 6.0 oz? xplain. Answer. Yes, she is about 99.9% sure that the true mean weight is less than 6.0 oz. 3. (From p. 349 #6). The heights of 18 year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches. 1 2 (a) What is the probability that an 18 year-old man selected at random is between 67 and 69 inches tall. ( Answer. P (67 < x < 69) = P 1 3 < z < 1 ) = = (b) If a random sample of nine 18 year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? Answer. P (67 < x < 69) = P ( < z < (c) Why was the probability in (b) higher than that in (a)? ) = P ( 1 < z < 1) = Answer. The sampling distribution standard deviation is smaller than the standard deviation for the original distribution, so (b) should be higher. In general, one expects that a randomly chosen group will have an average much closer to the population mean than a randomly selected individual, the larger the group, the closer one would expect its sample mean to be to the population mean. (d) Would you expect the probability that an 18 year-old man selected at random is more than 74 inches tall to be lower, the same as, or higher than the probability of selecting a random sample of nine eighteen year-old men with a mean height of more than 74 inches? Answer. Because the sampling distribution has a smaller standard deviation than the original population distribution, there is a much higher probability of finding individuals far away from the mean, than random samples whose sample mean is far away from the population mean. Thus, we expect that it is much more likely to randomly an individual more than 74 inches tall, than a group of 9 whose mean height is more than 74 inches. 4. (From p. 349 # 18). The taxi and takeoff time for commercial jets is a random variable x with mean 8.5 minutes and standard deviation of 2.5 minutes. You may assume the jets are lined up on the runway so that one taxis and takes off immediately after the other, and they take off one at time on a given runway. What is the probability that for 36 jets on a given runway total taxi and take off time will be (a) less than 320 minutes? Answer. P ( x < 320 ) ( = P z < 36 (b) more than 275 minutes? Answer. P ( x > 275 ) ( = P z > 36 ) = P (z <.93) = ) = P (z > 2.07) = = 6 9. (From p. 385 #6) The Roman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer sends on lunch Monday through Friday. A random sample of 115 customers lunch tabs gave a mean of x = \$9.74 with a standard deviation s = \$2.93. (a) Find a 95% confidence interval for the average amount spent on lunch by all customers. Answer. The sample size is n = , and so we use the formula for large sample means, i.e. x ± z c σ x. In this case, σ x = =.2732 and z.95 = 1.96 and so the 95% confidence interval is (\$9.20, \$10.28). (b) For a day when the Roman Arches has 115 lunch customers, use part (a) to estimate the range of dollar values for the total lunch income that day. Answer. We are 95% confident that total would be in the range (115 \$9.20, 115 \$10.28), that is (\$1058, \$1182). 10. (From p. 398 #12) The number of calories for 3 ounces of french fries at eight popular fast food chains are as follows Use these data to find a 99% confidence interval for the mean calorie count in 3 ounces of french fries obtained from fast-food restaurants. Answer. First, one needs to compute the sample mean and sample standard deviation for the 8 numbers above. For this, one finds x = 1956 x 2 = SS x = = Thus x = = and s = = This is a small sample (n < 30) with unknown standard deviation, and we ll assume the population is normal, or nearly so. We use the t-distribution with 7 degrees of freedom, thus s t.99 = 3.499, and the endpoints of the confidence interval are given by x ±t c. Thus the 99% n confidence interval has endpoints ± which yields the interval (217.6, 271.4). 11. (a) Under are the conditions necessary for finding a confidence interval for the mean from a small sample? Answer. We should use the t-distribution with n 1 degrees of freedom when the sample size, n, is less than 30 provided the population is normal (or nearly normal) and the population standard deviation is not known. 6 9 Answer. We would reject H 0 for levels of significance α =.03 and α =.05 because the Pvalue is less than these α s, we would not reject H 0 for α = (a) If a population has a standard deviation of 80, what sample size would be necessary in order for a 95% confidence interval to estimate the population mean within 10? ( zc σ ) ( ) Answer. We use the formula n = and we find that n = = The 10 sample size must be the next larger whole number which is n = 246. (b) What size of sample is needed by the Gallup organization to estimate a population proportion within ±.02 with 95% confidence. In your calculation assume that there is no preliminary estimate for p. Answer. We use the formula n = 1 4 ( zc ) 2 1 = 4 ( ) which gives us a sample size of (c) Suppose you are to construct a 99% confidence interval for the mean using a sample of size n = 12 from a normal population with unknown standard deviation. What value of t c would s you use in the formula x ± t c? n Answer. We use a t-distribution with n 1 = 11 degrees of freedom, and look on the table in the back cover to find t.99 = (d) Find the critical region if a two-tailed test on a mean is conducted using a large sample at a level of significance of.01. Answer. With the help of the normal table in the front cover, we find that the critical region is z 2.58 or z (e) Find the critical region for a left-tailed test on a proportion at a level of significance α =.05? Answer. With the help of the normal table in the front cover, we find that the critical region is z (f) xplain what type I and type II errors are in hypothesis tests. Answer. See text Section 9.1. (g) What is the probability with which we are willing to risk a type I error called? Answer. The level of significance, which is denoted by α. 16. (a) A developer wishes to test whether the mean depth of water below the surface in a large development tract was less than 500 feet. The sample data was as follows: n = 32 test 9 10 holes, the sample mean was 486 feet, and the standard deviation was s = 53 feet. Complete the test by computing the Pvalue, and report the conclusion for a 1% level of significance. Answer: Null Hypothesis: µ = 500 Alternative Hypothesis: µ < 500 Using the sample data, we compute z = = 1.49 Thus the Pvalue is P (z < 1.49) =.0681 We would not reject the null hypothesis at a level of significance of.01, because the P-value is larger than (b) What would the conclusion of the test be for a level of significance of α =.05. Answer. Do not reject H 0 since the Pvalue is bigger than.05. (c) What type of error was possibly made in (b)? Answer. A type II error. 17. A vendor was concerned that a soft drink machine was not dispensing 6 ounces per cup, on average. A sample size of 40 gave a mean amount per cup of 5.95 ounces and a standard deviation of.15 ounce. (a) Find the Pvalue for an hypothesis test to determine if the mean is different from 6 ounces. Answer. This is a two-tailed test with Null Hypothesis: µ = 6 Alternative Hypothesis: µ 6. Using the data, we compute z = = The Pvalue is: P (z < 2.11) + P (z > 2.11) = 2 P (z < 2.11) = 2(.0174) =.0348 (b) For which of the following levels of significance would the null hypothesis be rejected? (i) α =.10 (ii) α =.05 (iii ) α =.04 (iv) α =.01 Answer. Reject the null hypothesis in (i), (ii) and (iii) since the Pvalue is smaller than α; do not reject the null hypothesis in (iv). (c) For each case in part (b), what type of error has possibly been committed? 10 11 Answer. Possible Type I error may occur in (i), (ii) and (iii) while a Type II error may occur in (iv). (d) Find a 96% confidence interval for the mean amount of soda dispensed per cup. Answer. For c =.96, z c = 2.05 (approximately), look at z value corresponding to an area of.98 on table. Thus the confidence interval, using the large sample method (n is at least 30) yields endpoints: ± and, so the confidence interval is: (5.901, 5.999). (e) Is your interval in (d) consistent with the test conclusion in (b)(iii)? xplain. Answer. Yes. In (b)(iii) we have a confidence level of (at least) c = 1 α =.96 that the mean is different from 6, while in (d) we had a 96% confidence interval that did not contain 6, and so we were (at least) 96% certain that the mean is different from 6. Notice also the confidence interval comes very close to containing 6, this is reflected in the Pvalue being very close to 0.04 in the hypothesis test. (f) Based on your answer to (b)(iv) would you expect a 99% confidence interval to contain 6? Answer. Yes, because we were not 99% confident that the mean was different from 6, we would expect the corresponding confidence interval to contain 6. (g) Suppose that the population standard deviation is σ =.15, what sample size would be needed so that the maximum error in a 96% confidence interval is =.01? ( zc σ ) 2. Answer. The formula to use is: n = So we compute ( ) n = = ,.01 thus we should use a sample size of n = (a) Suppose that a February Gallup poll of 1200 randomly selected voters found that 53 percent support George W. Bush s energy policy. Conduct an hypothesis test at a level of significance of α =.01 to test whether the true voter population support for George W. Bush s energy policy in February was less than 56 percent. Answer. The null hypothesis is H 0 : p =.56, and the alternative hypothesis is H 1 : p <.56. The critical region is z Since n = 1200 and p =.56 and q =.44, we clearly have np > 5 and nq > 5. Thus we compute z = (.56)(.44) 12 Because 2.09 does not fall in the critical region, we conclude there is not sufficient evidence to reject the null hypothesis at a level of significance of 1%. (b) Report the Pvalue of the test in (a) and give a practical interpretation of it. Answer. The Pvalue is P (z < 2.09) = Thus we are roughly 98% certain that less than 56% of all voters at the time of the poll supported President G.W. Bush s energy policy. 19. (From p. 536 #8) A reading test is given to both a control group and an experimental group (which received special tutoring). The average score for the 30 subjects in the control group was with a standard deviation of The average score for the 30 subjects in the experimental group was with a standard deviation of Use a 4% level of significance to test the claim that the experimental group performed better than the control group. Answer. We wish to conduct the test H 0 : µ 1 = µ 2 versus H 1 : µ 1 > µ 2 where µ 1 is the population mean test score for all people who received the special tutoring. We use the formula z = ( x 1 x 2 ) (µ 1 µ 2 ) σ1 2 where σ x x2 = + σ2 2 Thus σ x1 x 2 n 1 n σ x x2 = = ( ) 0 Therefore, z = = The Pvalue is P (z > 1.52) = = Because the Pvalue is larger than α =.04, we do not reject the null hypothesis at a 4% level of significance. There is not sufficient evidence to show that the tutoring increases the mean score. 20. (From p. 505 #14) Nationally about 28% of the population believes that NAFTA benefits America. A random sample of 48 interstate truck drivers showed that 19 believe NAFTA benefits America. Conduct an hypothesis test to determine whether the population proportion of interstate truckers who believe NAFTA benefits America is higher than 28%. Test at a level of significance of α =.05. (a) State the null and alternative hypotheses. Is this a right-tailed, left-tailed or two-tailed test? Answer. The null hypothesis is H 0 : p =.28, the alternative hypothesis is H 1 : p >.28. This is a right-tailed test. (b) Find the Pvalue for the test. Answer. For this data, we have n = 48, p =.28 and q =.72. Clearly np > 5 and nq > 5 and so we compute ˆp = 19 =.39583; then z = 1.79 (0.28)(0.72) 48 12 13 Therefore, the Pvalue is P (z > 1.79) = = Because the Pvalue is less than α =.05, we reject H 0. The results are statistically significant; they indicate that the proportion of interstate truck drivers that believe NAFTA benefits America is higher than (c) Would you reject the null hypothesis at a level of significance of α =.04? Answer. Yes, because the Pvalue is less than (d) Would you reject the null hypothesis at a level of significance of α =.01? Answer. No, because the Pvalue is more than (e) Do you think the proportion of interstate truckers who believe NAFTA benefits America is higher than 28%? xplain your answer. Answer. Yes, and I m quite sure of this, in fact, I m roughly 96% sure of it. 21. (From p. 539 #18) A random sample of n 1 = 288 voters registered in the state of California showed that 141 voted in the last general election. A random sample of n 2 = 216 registered voters in Colorado showed that 125 voted in the last general election. Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a 5% level of significance. Answer. Let p 1 and p 2 represent the population proportions of voter turnout in California and Colorado respectively. We will test H 0 : p 1 = p 2 versus H 1 : p 1 < p 2. Since we are assuming p 1 = p 2, we use the pooled estimate for proportions (see p. 530), that is ˆp = = Then we use this for an estimate of p 1 and p 2 in the formula for σˆp1 ˆp 2, so we obtain σˆp1 ˆp 2 = From this, we compute (0.5278)(0.4722) 288 z = (ˆp 1 ˆp 2 ) (p 1 p 2 ) σˆp1 ˆp 2 = + (0.5278)(0.4722) 216 ( ) = = The Pvalue is P (z < 1.98) = Because the Pvalue is less than.05, we reject H 0 at the 5% level of significance. 13 ### 5.1 Identifying the Target Parameter University of California, Davis Department of Statistics Summer Session II Statistics 13 August 20, 2012 Date of latest update: August 20 Lecture 5: Estimation with Confidence intervals 5.1 Identifying ### Point and Interval Estimates Point and Interval Estimates Suppose we want to estimate a parameter, such as p or µ, based on a finite sample of data. There are two main methods: 1. 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Sample questions from Midterms and 2 are also representative of questions that may appear on the final exam.. A randomly selected sample ### 4. Continuous Random Variables, the Pareto and Normal Distributions 4. Continuous Random Variables, the Pareto and Normal Distributions A continuous random variable X can take any value in a given range (e.g. height, weight, age). The distribution of a continuous random ### Chapter 7 - Practice Problems 2 Chapter 7 - Practice Problems 2 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the requested value. 1) A researcher for a car insurance company ### Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion Learning Objectives Upon successful completion of Chapter 8, you will be able to: Understand terms. State the null and alternative ### Hypothesis Testing --- One Mean Hypothesis Testing --- One Mean A hypothesis is simply a statement that something is true. Typically, there are two hypotheses in a hypothesis test: the null, and the alternative. Null Hypothesis The hypothesis ### HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as... HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1 PREVIOUSLY used confidence intervals to answer questions such as... You know that 0.25% of women have red/green color blindness. You conduct a study of men ### Review #2. Statistics Review #2 Statistics Find the mean of the given probability distribution. 1) x P(x) 0 0.19 1 0.37 2 0.16 3 0.26 4 0.02 A) 1.64 B) 1.45 C) 1.55 D) 1.74 2) The number of golf balls ordered by customers of ### An Introduction to Statistics Course (ECOE 1302) Spring Semester 2011 Chapter 10- TWO-SAMPLE TESTS The Islamic University of Gaza Faculty of Commerce Department of Economics and Political Sciences An Introduction to Statistics Course (ECOE 130) Spring Semester 011 Chapter 10- TWO-SAMPLE TESTS Practice ### Section 7.1. Introduction to Hypothesis Testing. Schrodinger s cat quantum mechanics thought experiment (1935) Section 7.1 Introduction to Hypothesis Testing Schrodinger s cat quantum mechanics thought experiment (1935) Statistical Hypotheses A statistical hypothesis is a claim about a population. Null hypothesis ### Introduction to Hypothesis Testing I. Terms, Concepts. Introduction to Hypothesis Testing A. 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https://gmatclub.com/forum/in-rembrandt-s-the-night-watch-in-the-second-plane-of-the-161938.html	1,508,727,420,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187825510.59/warc/CC-MAIN-20171023020721-20171023040721-00819.warc.gz	684,844,234	58,105	It is currently 22 Oct 2017, 19:57 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar In Rembrandt's The Night Watch, in the second plane of the Author Message TAGS: Hide Tags Manager Joined: 06 Jun 2012 Posts: 141 Kudos [?]: 268 [0], given: 37 In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 21 Oct 2013, 00:52 12 This post was BOOKMARKED 00:00 Difficulty: 95% (hard) Question Stats: 35% (01:36) correct 65% (01:44) wrong based on 580 sessions HideShow timer Statistics In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress [Reveal] Spoiler: OA _________________ Please give Kudos if you like the post Kudos [?]: 268 [0], given: 37 Manager Joined: 14 Jan 2013 Posts: 152 Kudos [?]: 379 [0], given: 30 Concentration: Strategy, Technology GMAT Date: 08-01-2013 GPA: 3.7 WE: Consulting (Consulting) In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 12 Feb 2014, 16:43 4 This post was BOOKMARKED In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress Guys, this topic was locked ,there fore I am opening it again. I am not clear with Solution. _________________ "Where are my Kudos" ............ Good Question = kudos "Start enjoying all phases" & all Sections __________________________________________________________________ http://gmatclub.com/forum/collection-of-articles-on-critical-reasoning-159959.html http://gmatclub.com/forum/percentages-700-800-level-questions-130588.html http://gmatclub.com/forum/700-to-800-level-quant-question-with-detail-soluition-143321.html Last edited by broall on 27 Jul 2017, 01:04, edited 1 time in total. Merged post. Please search before posting Kudos [?]: 379 [0], given: 30 Manager Joined: 11 Jan 2014 Posts: 94 Kudos [?]: 73 [0], given: 11 Concentration: Finance, Statistics GMAT Date: 03-04-2014 GPA: 3.77 WE: Analyst (Retail Banking) Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 12 Feb 2014, 23:54 Mountain14 wrote: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress I went with (A), the first split "to the left of but behind the two military leaders of the company," was sufficient to answer the question. Compare all the answers to (A), notice the wordiness and unjustified change of structure. Choice (A) seems more elegant than all other choices. Additionally, it is free of grammatical errors. Kudos [?]: 73 [0], given: 11 Manager Joined: 18 Nov 2013 Posts: 79 Kudos [?]: 19 [0], given: 7 Location: India GMAT Date: 12-26-2014 WE: Information Technology (Computer Software) Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 12 Feb 2014, 23:59 Option A is clear and concise, free from any errors. I thought at first this might be a parallelism problem, but didn't appear to be so. Kudos [?]: 19 [0], given: 7 Manager Joined: 17 Jul 2013 Posts: 100 Kudos [?]: 7 [0], given: 67 Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 30 May 2014, 04:14 I could not make out the meaning from this question .. can anybody please help me .. what this question is testing? Kudos [?]: 7 [0], given: 67 Manager Status: GMAT Instructor Joined: 23 Jun 2013 Posts: 178 Kudos [?]: 82 [1], given: 4 Location: India GRE 1: 2280 Q790 V710 GPA: 3.3 WE: Editorial and Writing (Education) Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 30 May 2014, 10:47 1 KUDOS 1 This post was BOOKMARKED Interesting question. The language seems right out of a classic novel. The sentence is trying to state two facts: A little girl stands someplace. A dead chicken is suspended from her dress. How do you know this? Read ahead, skipping any modifiers, until you find the subject(s). Next, identify the main verb(s). The portions before each of the first three commas are all modifiers providing information about "where" the little girl stands. The "illuminated...light" is another modifier that provides some more information about this main scene (what surrounds the girl). So when analysing the options, check whether the modifiers modify the thing/person they are supposed to modify. For e.g. Who stands to the left of but behind the military leaders? You might also catch a subject-verb agreement error in one of the options. _________________ EnterMBA Kudos [?]: 82 [1], given: 4 Current Student Joined: 13 Feb 2011 Posts: 104 Kudos [?]: 48 [0], given: 3385 Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 30 May 2014, 13:02 Can the OP be updated to show the segment getting tested as underlined. Thanks! Kudos [?]: 48 [0], given: 3385 Intern Joined: 08 Jul 2015 Posts: 1 Kudos [?]: [0], given: 2 Re: In Rembrandt's The Night Watch, in the second plane of the p [#permalink] Show Tags 17 Aug 2015, 14:36 Mountain14 wrote: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress Guys, this topic was locked ,there fore I am opening it again. I am not clear with Solution. B.modify problem, standing could not modify light C. SV problem stands the subject is girl and chicken, and suspended modify problem D. modify problem for the light E. illuminated modify problem Kudos [?]: [0], given: 2 Intern Joined: 28 Jan 2017 Posts: 30 Kudos [?]: 26 [0], given: 14 In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 13:42 2 This post was BOOKMARKED In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. - to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken - behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken - behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress - behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it - to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress Last edited by broall on 27 Jul 2017, 01:05, edited 1 time in total. Merged post. Please search before posting Kudos [?]: 26 [0], given: 14 Manager Joined: 13 Oct 2016 Posts: 224 Kudos [?]: 122 [0], given: 140 GMAT 1: 600 Q44 V28 Re: In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 16:41 _________________ _______________________________________________ If you appreciate the post then please click +1Kudos Kudos [?]: 122 [0], given: 140 Manager Joined: 02 May 2016 Posts: 82 Kudos [?]: 32 [0], given: 206 Location: India Concentration: Entrepreneurship GRE 1: 317 Q163 V154 WE: Information Technology (Computer Software) Re: In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 19:48 Kritesh wrote: Although I incorrectly answered E in 1:15 minute, I can say why A is correct! (I wont point out all the wrongs in answer choice, but one or two that I used to reject it) - to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken I don't find this sentence incorrect, will keep it but look for better option - behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken "standing to their left" modifies "a bright shaft of light". OUT - behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress "stands a brightly illuminated" could modify "girl and dead chicken". Only girl is standing. Chicken is not standing but hanging from her dress - behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it I think "left to the" is incorrect. It should be "left of". And major reason for rejecting this is : subject is changed to focus on "bright shaft of light" which illuminates a standing girl. Focus changed. - to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress I found this error after i clicked E. here I find Parallelism error: "In Rembrandt's....", "in the second plane...","to the left...", but next comes "and illuminated". This is incorrect. Replace it with either parallel prepositional phrase, or subject. A uses subject-"little girl". Hence A is correct Hope I helped! Kudos [?]: 32 [0], given: 206 Manager Joined: 26 Jun 2013 Posts: 91 Kudos [?]: 11 [0], given: 41 Location: India Schools: ISB '19, IIMA , IIMB GMAT 1: 590 Q42 V29 GPA: 4 WE: Information Technology (Retail Banking) Re: In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 20:07 I think the sentence in option A doesn't make sense. It will be really helpful if someone can explain this. Thanks! _________________ Remember, if it is a GMAT question, it can be simplified elegantly. Kudos [?]: 11 [0], given: 41 Manager Joined: 02 May 2016 Posts: 82 Kudos [?]: 32 [0], given: 206 Location: India Concentration: Entrepreneurship GRE 1: 317 Q163 V154 WE: Information Technology (Computer Software) Re: In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 21:25 hotshot02 wrote: I think the sentence in option A doesn't make sense. It will be really helpful if someone can explain this. Thanks! Let me try: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. Main SV is : Little girl stands. second clause is: suspended from her dress is a dead chicken. (her refers correctly to little girl). connecting word for both clause: And Prepositional phrases- Modifiers are for verb "stands". Modifier illuminated is for little girl. Taking one phrase at a time sentence will read: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. Written this way, sentence is free from all the errors. GMAT is not about selecting the correct choice. It is about eliminating wrong answer choices. All the experts will second me on this: speech or how the sentence sounds doesn't matter in SC Makes sense now? I can still elaborate more, let me know! But I hope this will help! Kudos [?]: 32 [0], given: 206 Senior Manager Joined: 06 Jul 2016 Posts: 412 Kudos [?]: 118 [0], given: 99 Location: Singapore Concentration: Strategy, Finance Re: In Rembrandt's The Night Watch - Tough Economist SC [#permalink] Show Tags 26 Jul 2017, 21:31 ajit_223 wrote: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. - to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken - behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken - behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress - behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it - to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress The Sentence Structure is Modifier, Modifier, Modifier, Core, Modifier, Core. 'In Rembrandt's...Watch', In the.....composition', 'to the.....company' - The First 3 modifiers are setting up the subject of the sentence i.e. the little girl. 'to the left....company' - maintains parallelism with 'to the left' but 'behind....' 'stands...girl' - SV Agreement 'Girl' is the subject, and 'stands' is the verb. 'suspended....chicken' - 'dead chicken' is the subject and 'is suspended' is the verb, so it maintains SV agreement. The 2 core clauses are parallel. A - No grammatical issues. Keep. B - Incorrect modifier. It's modifying 'a bright shaft...' instead of a little girl! OUT C - SV Disagreement, as this sentence makes the subject a plural 'a brightly illuminated girl and a dead chicken' and the verb is 'stands' + the meaning is illogical. A dead chicken wasn't standing at all. OUT! D - 'to the left of' is correct and 'to leave' is correct but 'left of' is incorrect. OUT E - This tries to make the 2 modifiers parallel but fails 'to the....' and 'illuminated....' are not parallel. OUT Hope it helps! _________________ Put in the work, and that dream score is yours! Kudos [?]: 118 [0], given: 99 Manager Joined: 20 Jan 2016 Posts: 197 Kudos [?]: 13 [0], given: 63 Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 28 Jul 2017, 04:15 A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken right answer B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken illuminates and suspended is not parallel C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress alters the meaning of the sentence D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it we don't know what it refers to here E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress And illuminated is wrong Kudos [?]: 13 [0], given: 63 Manager Joined: 12 Sep 2016 Posts: 71 Kudos [?]: 12 [0], given: 796 Location: India Schools: Kellogg '20 GMAT 1: 700 Q50 V34 GPA: 3.33 Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 07 Aug 2017, 02:52 Experts could you please comment on why A is the OA. I personally did not like either of the sentences but felt that E is the best of the lot. Kudos [?]: 12 [0], given: 796 Intern Joined: 05 Jun 2016 Posts: 32 Kudos [?]: 1 [1], given: 742 Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 15 Aug 2017, 13:08 1 KUDOS darn wrote: Experts could you please comment on why A is the OA. I personally did not like either of the sentences but felt that E is the best of the lot. Hello, I am not an expert but I get most SC questions right. Here are my 2 cents: E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress "to the left of the company's two military leaders but behind them", is wordy compared to the shorter "to the left of but behind the two military leaders of the company". second part "and illuminated by a bright shaft of light, stands a little girl" >>>> this construction is passive whereas in (A) "stands a little girl, illuminated by a bright shaft of light," is active. But more importantly notice in (A), "to the left of but behind the two military leaders of the company, stands a little girl" >>>> "stands" is correctly placed right after the phrase that it refers to whereas in (E) "stands" is too far from what it is referring to. In general GMAT SC prefers an appositive "illuminated by a bright shaft" to follow the subject or noun it modifies "stands a little girl". In (E) the appositive comes before the subject/noun it modifies. I hope I made sense. Good luck. Kudos [?]: 1 [1], given: 742 Manager Status: Profile 1 Joined: 20 Sep 2015 Posts: 78 Kudos [?]: 16 [0], given: 40 GMAT 1: 690 Q48 V37 GPA: 3.2 WE: Information Technology (Investment Banking) Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 16 Aug 2017, 01:15 summer101 wrote: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress All other things are ok with option A but isn't it sounds awkward : "to the left...,stands a little girl, illuminated...and suspended from her dress..." feeling sad for little girl may be better construction possible but not among the other ans options. Kudos [?]: 16 [0], given: 40 Intern Joined: 05 Jun 2016 Posts: 32 Kudos [?]: 1 [0], given: 742 Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] Show Tags 16 Aug 2017, 02:29 jokschmer wrote: summer101 wrote: In Rembrandt's The Night Watch, in the second plane of the pictorial composition, to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken. A. to the left of but behind the two military leaders of the company, stands a little girl, illuminated by a bright shaft of light, and suspended from her dress is a dead chicken B. behind the two military leaders of the company, but standing to their left, a bright shaft of light illuminates a little girl, and suspended from her dress is a dead chicken C. behind the two military leaders of the company but to their left, stands a brightly illuminated little girl and a dead chicken, suspended from her dress D. behind and left to the company's two military leaders, a bright shaft of light illuminates a standing little girl, and her dress has a dead chicken suspended from it E. to the left of the company's two military leaders but behind them, and illuminated by a bright shaft of light, stands a little girl with a dead chicken suspended from her dress All other things are ok with option A but isn't it sounds awkward : "to the left...,stands a little girl, illuminated...and suspended from her dress..." feeling sad for little girl may be better construction possible but not among the other ans options. When it comes to 700 level questions, trusting your ear doesn't help. Kudos [?]: 1 [0], given: 742 Re: In Rembrandt's The Night Watch, in the second plane of the [#permalink] 16 Aug 2017, 02:29 Display posts from previous: Sort by In Rembrandt's The Night Watch, in the second plane of the Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	6,429	25,681	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2017-43	longest	en	0.942815
http://www.pearltrees.com/u/13632990-cymatics-bringing-youtube	1,555,742,809,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00307.warc.gz	282,105,588	24,601	Cymatics - Bringing Matter To Life With Sound (Part 2 of 3) Fluid Motion Simulations and Artwork When a droplet falls into shallow water, it creates a crown or "coronet". This droplet simulation was calculated using Smoothed Particle Hydrodynamics (SPH). SPH is one of the most impressive-looking fluid simulation techniques. Droplet Links Liquid Sculpture - beautiful high speed photographs, by Martin Waugh, see also this video Water Figures - beautiful high-speed camera splashes by Fotoopa Other Links Fluids v.1 - fast SPH C++ program by Rama Hoetzlein Physics Demos - fluid Java applets by Grant Kot Fluid Animations - amazing animations by Ron Fedkiw, with Eran Guendelman, Andrew Selle, Frank Losasso, et al. 7 Mystical Seals [Cymatics and Conspiracy] “Let no one destitute of geometry enter my doors.” -Plato Symmetry breaking in physics describes a phenomenon where (infinitesimally) small fluctuations acting on a system which is crossing a critical point decide the system’s fate, by determining which branch of a bifurcation is taken. To an outside observer unaware of the fluctuations (or “noise”), the choice will appear arbitrary. This process is called symmetry “breaking”, because such transitions usually bring the system from a disorderly state into one of two definite states.Symmetry breaking is supposed to play a major role in pattern formation. Believe it or not I feel this next entry has a place in this blog. Tian Tan Buddha - Wikipedia, the free encyclopediaList of statues by height, most of which are Buddhist statues. image on the left: The 19th century image of a Sabbatic Goat, created by Eliphas Lévi. Baphomet “The Temple of the Father of Peace Among Men” PHOroneus discovers fire Standing Wave Patterns As mentioned earlier, all objects have a frequency or set of frequencies with which they naturally vibrate when struck, plucked, strummed or somehow disturbed. Each of the natural frequencies at which an object vibrates is associated with a standing wave pattern. When an object is forced into resonance vibrations at one of its natural frequencies, it vibrates in a manner such that a standing wave is formed within the object. The topic of standing wave patterns was introduced in Unit 10 of The Physics Classroom. In that unit, a standing wave pattern was described as a vibrational pattern created within a medium when the vibrational frequency of a source causes reflected waves from one end of the medium to interfere with incident waves from the source, The result of the interference is that specific points along the medium appear to be standing still while other points vibrated back and forth. Natural Frequencies and Vibrational Patterns Chladni Plates Flickr Physics Photo Watch It! Cymatics, Sound & Consciousness “If you want to find the secrets of the universe, think in terms of energy, frequency and vibration.” ―Nikola Tesla Ladies and gentlemen, if you have a golf-ball size consciousness when reading a book, then you will have a golf-ball size understanding. This is why it is crucial for you to broaden your minds. I am sure those of you who are reading this, have a broad mind and expanded consciousness. And by viewing the videos below, you will open yourselves to the reality and occult power of sound/music. There is an ocean of pure vibrant consciousness inside each one of us. For those of you who do not know, Matter, is actually a poorly defined term in science. Why is matter so vaguely defined? All matter originates and exists only by virtue of a force which brings the particle of an atom to vibration and holds this minute solar system of the atom together. The solidity of this world seems totally indisputable – as a fixed thing which you can see and touch. CarbonOxygenHydrogenNitrogen Cymatics - Seeing Sound By John Stuart and Analiese Shandra Reid Seeing Sound with the CymaScopeSound is an invisible force that permeates every aspect of our lives. With the exception of music, many man- made sounds are jarring while the sounds of Nature tend to flow over and around us like soothing waters, lifting our spirit, inspiring us, exciting us. Yet if we could see sound our world would be even more beautiful than we could imagine. It would be a world filled with shimmering holographic bubbles, each displaying a kaleidoscopic pattern on its surface. When the microscope and telescope were invented centuries ago, new realms came into view that were not even suspected to exist—a Universe in miniature under the microscope and a Universe so immense that centuries of research lie before us with the telescope. The Shape of SoundBefore looking at cymatics more closely let us dispel the popularly held misconception that 'sound is a wave'. In the illustration below a slice through a sound bubble is depicted. Cymatics and the New Age of Miracles with Dr. Sir Peter Guy Manners, M.D. Cymatics As Science In the early 18th century, the German physicist Ernst Chladni, the ''father of acoustics,'' covered plates with thin layers of sand, set them vibrating, and observed the patterns that were made in response to different sound stimuli. In 1967, nearly three hundred years later, Hans Jenny, a Swiss doctor, artist, and researcher, published Cymatics – The Structure and Dynamics of Waves and Vibrations. When this is done, shapes and motion-patterns appear. Jenny used crystal oscillators, and invented what he called a ''tonoscope'' to set his plates and membranes vibrating. Jenny thought that evolution was a result of vibrations, with the vibrations of one level of organization, such as that of cells, each one unique, combining to create glands and organs and so on, each new level being a harmonic of the previous one. Enter the Age of Miracles We called Dr. As part of my training, I and my fellow students would observe doctors in the process of treating their patients. Cymatics » Cymatics Center \| Token Rock An Introduction... When Dr Hans Jenny coined the word 'cymatics' he began a new chapter in the history of science. Cymatics is the science of visual sound. Sound is a wave, right? Vibration underpins all matter in the universe. When the microscope and telescope were invented they opened vistas on realms that were not even suspected to exist. Cymatics Wiki Definition Resonance made visible with black seeds on a harpsichord soundboard Cornstarch and water solution under the influence of sine wave vibration Cymatics (from Greek: κῦμα "wave") is the study of visible sound co vibration, a subset of modal phenomena. The apparatus employed can be simple, such as the old Chinese spouting bowl, or Chinese singing fountain, in which copper handles are rubbed and cause the copper bottom elements to vibrate. Etymology History On July 8, 1680, Robert Hooke was able to see the nodal patterns associated with the modes of vibration of glass plates. In 1787, Ernst Chladni repeated the work of Robert Hooke and published "Entdeckungen über die Theorie des Klanges" ("Discoveries in the Theory of Sound"). Throughout the 1960s, up until his death in 1972, Swiss medical doctor and Anthroposophist, Hans Jenny took a methodological and exhaustive approach to documenting Cymatic phenomena. Influences in art Composer Stuart Mitchell and his father T.J. Cyma Cymatics : SolAwakening Cymatics, Natural Ecosystem of Vibration: Listen to our wonderful Natural Ecosystem around us! What amazing sounds! What is Cymatics? Cymatics (from Greek: κῦμα “wave”) is the study of visible sound and vibration, a subset of modal phenomena. The term Cymatics was coined by Dr. Dr Jenny’s work related to the cyclic nature and interval between wave frequencies. For 14 years, he experimented with sound frequencies, using them to excite various powders, pastes and liquids, which resulted in the generation of flowing geometric patterns. Then the question and contemplation came wondering if there was a connection between the geometric shapes created by the sound waves and Sacred Geometry. The shapes, figures and patterns of motion (shown above) that appeared proved to be primarily a function of frequency, amplitude, and the inherent characteristics of the various materials. How is this possible? References: www.cymaticsource.com/articles.html Related:	1,751	8,247	{"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}	2.625	3	CC-MAIN-2019-18	longest	en	0.896929
https://questions.examside.com/past-years/gate/question/a-project-has-four-activities-pq-r-and-s-as-shown-2014-set-2-marks-2-bfdjdbbuitjq8zoe.htm	1,718,226,915,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00248.warc.gz	447,729,767	31,016	1 GATE ME 2014 Set 2 Numerical +2 -0 A project has four activities $$P,Q, R$$ and $$S$$ as shown below. The normal cost of the project is Rs. $$10,000/-$$ and the overhead cost is Rs. $$200/-$$ per day. If the project duration has to be crashed down to $$9$$ days, the total cost (in Rupees) of the project is ______________. 2 GATE ME 2014 Set 3 +2 -0.6 Consider the given project network, where numbers along various activities represent the normal time. The free float on activity $$4$$-$$6$$ and the project duration, respectively, are A $$2, 13$$ B $$0, 13$$ C $$-2, 13$$ D $$2, 12$$ 3 GATE ME 2012 +2 -0.6 For a particular project, eight activities are to be carried out. Their relationships with other activities and expected durations are mentioned in the table below. The critical path for the project is A $$a-b-e-g-h$$ B $$a-c-g-h$$ C $$a-d-f-h$$ D $$a-b-c-f-h$$ 4 GATE ME 2012 +2 -0.6 For a particular project, eight activities are to be carried out. Their relationships with other activities and expected durations are mentioned in the table below. If the duration of activity $$''f ''$$ alone is changed from $$9$$ to $$10$$ days, then the A critical path remains the same and the total duration to complete the project changes to $$19$$ days. B critical path and the total duration to complete the project remains the same. C critical path changes but the total duration to complete the project remains the same. D critical path changes and the total duration to complete the project changes to $$17$$ days. GATE ME Subjects EXAM MAP Medical NEET	444	1,567	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2024-26	latest	en	0.874015
https://maths-magic.ac.uk/course.php?id=617	1,596,825,558,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439737206.16/warc/CC-MAIN-20200807172851-20200807202851-00356.warc.gz	398,556,147	4,419	## Announcements There are no announcements Forum ## General #### Description The Reynolds number gives the ratio of inertial to viscous effects in a fluid flow. When the Reynolds number is small, inertial effects are negligible and the Du/Dt term in the Navierâ€“Stokes equations may be neglected. This simplifies the Navier-Stokes equations, making them linear and instantaneous. These simplifications make solving low-Reynolds-number flow problems much easier than high Reynolds number flows. This module will consider the circumstances under which the Reynolds number will be small and examine the basic properties of low-Reynolds-number flows. We shall present a number of solution techniques, and show how they can be applied to a range of problems. In the course of this, students will meet various useful applied mathematics methods, including solution by potentials, boundary integral methods, and asymptotic approximations. #### Semester Spring 2021 (Monday, January 25 to Friday, March 19; Monday, April 26 to Friday, May 7) #### Hours • Live lecture hours: 10 • Recorded lecture hours: 0 • Total advised study hours: 40 #### Timetable • Fri 15:05 - 15:55 #### Prerequisites Essential • Vector Calculus (div, grad, curl, line,surface/volume integrals, divergence theorem) • Differential Equations (methods for first-order ordinary differential equations) • Basic Fluid Mechanics (introductory course in inviscid fluid mechanics) Desirable / Complimentary • Further Fluid mechanics (introductory course in viscous fluid mechanics) • Tensors and the Einstein Summation Convention (some previous experience useful) • Non-dimensionalisation / scaling analysis #### Syllabus • Introduction to low-Reynolds-number flow (3 lectures) The Stokes equations and boundary conditions. Basic properties, uniqueness theorem, reciprocal theorem, minimum dissipation theorem. Oscillating Couette flow and Poiseuille flow. • Fundamental solutions and representation by potentials (4 lectures) Solution using potentials. Papkovich-Neuber potentials, flow past a rigid sphere. Boundary integrals and the multi-pole expansion. • Slender-body theory (3 lectures) Basic derivation. Applications to sedimenting slender objects and swimming micro-organisms. ## Lecturer Email r.whittaker@uea.ac.uk Phone ## Bibliography Elementary Fluid Dynamics Acheson, D. J. Viscous Flow Ockendon & Ockendon An Introduction to Fluid Dynamics Batchelor, G. K. Low Reynolds Number Hydrodynamics Happel & Brenner Boundary Integral and Singularity Methods for Linearised Viscous Flow Pozrikidis, C. Microhydrodynamics Kim & Karrila Note: Clicking on the link for a book will take you to the relevant Google Book Search page. You may be able to preview the book there. On the right hand side you will see links to places where you can buy the book. There is also link marked 'Find this book in a library'. This sometimes works well, but not always. (You will need to enter your location, but it will be saved after you do that for the first time.) ## Assessment No assessment information is available yet. No assignments have been set for this course. ## Files No files have yet been uploaded for this course.	695	3,204	{"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-2020-34	latest	en	0.828689
https://www.physicsforums.com/threads/experiment-to-test-for-causality.1065381/#post-7116798	1,726,306,616,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00024.warc.gz	877,123,819	30,920	# Experiment to test for causality • B • Agent Smith Agent Smith TL;DR Summary An experiment to check for causality We want to check whether specially treated bitter gourd is effective, marketed as BitterHeal, in lowering blood sugar levels in diabetics. They take a random sample of diabetics and assign them randomly to ##3## groups: Group A: Are given BitterHeal Group B: Are given untreated bitter gourd Group C: Are not given either untreated bitter gourd or BitterHeal A reference/baseline is collected: Blood sugar level pre-experiment of all participant diabetics in the experiment. After 3 weeks, blood sugar levels are measured for all ##3## groups. Response variable: Blood sugar level Explanatory variable: BitterHeal The Null Hypothesis ##H_o##: There is no (statistically significant) difference between the ##3## groups. The alternative hypothesis ##H_1##: There is a statistically significant difference between the ##3## groups, i.e. blood sugar levels are lowered by BitterHeal. Ab hinc difference is equivalent to statistically significant difference. Assume all conditions for a statistical experiment have been met adequately. Possible outcomes: 1. No difference between B and C 2. Difference between B and C 3. No difference between A and B 4. Difference between A and B My question is the number of groups being experimented upon. There are ##3## (A, B, C). Couldn't we achieve the same thing by using only ##2## groups (a control group B and a test/treatment group A)? How does having group C help? Are we trying to control for the placebo effect? I think not because both groups A and B receive bitter gourds (and they don't know whether it's treated/untreated gourds). I know that if outcome ##2## occurs, we can't rule out the placebo effect because group C knows they didn't get any treatment. Is this why we need group C, because if outcome ##2## happens, it (does it?) somehow lowers our confidence in outcome ##4##. My best guess is the outcome combination ##1## (does this mean the placebo effect is nonexistent or negligible?) and ##4## is best with respect to inferring causality (that BitterHeal is antidiabetic). One of the main worries of modern medicine seems to be the placebo effect (a psychosomatic phenomenon instead of a purely somatic/physical one), because then the drug is a dud. The combination of outcome 1 and 3 would suggest that no medical effectiveness has been detected. The combination of outcome 1 and 4 would suggest that the Bitterheal "treatment" has a medical effect (hopefully a good one). The combination of outcome 2 and 3 would suggest that the Bitterheal "treatment" is not important in providing the positive or negative health effect. The combination of outcome 2 and 4 could suggest many different possibilities. There's placebo effect here, but that may not be the best way of looking at it. If taking the bitter gourd alone changes the blood levels (very objective measurements), then those gourds are either medically effective or induce dietary behavior that is effective. Inducing dietary changes might be "placebo", might be "I need to get that taste our of my mouth", or might be "with this much discomfort, I'm not going to jeopardize the effectiveness with bad diet". Last edited: Agent Smith .Scott said: The combination of outcome 1 and 3 would suggest that no medical effectiveness has been detected. The combination of outcome 1 and 4 would suggest that the Bitterheal "treatment" has a medical effect (hopefully a good one). The combination of outcome 2 and 3 would suggest that the Bitterheal "treatment" is not important in providing the positive or negative health effect. The combination of outcome 2 and 4 could suggest many different possibilities. Gracias for the elucidation. .Scott said: So, to answer your question directly, three groups allows the study to determine whether Bitterheal is effective and whether the researchers have the recipe that they can use to reproduce it. What's not clear is the necessity for ##3## groups. Like I mentioned in the OP, we have a control group (untreated bitter gourd) and a test/treatment group (BitterHeal) that already cancels the placebo effect (both groups A and B don't know whether they're getting treatment or a placebo pill). Isn't this how you negate the placebo effect? I can also say with an uncomfortably high degree of uncertainty that what we're attempting here is to recommend BitterHeal to diabetics (if it's shown to work, outcome combination ##1## and ##4##) and for that we'd have to demonstrate: X. Taking BitterHeal is better than not taking anything at all Y. Taking BitterHeal is better than taking a normal bitter gourd Yes? If yes than ##4## goes with Y, What goes with X? Consider these two scenarios (high score being healthy): BH scores 123, BG scores 20, and nothing scores 50. BH scores 40, BG scores 20, and nothing scores 50. Before recommending BH, you: 1) would want BH to significantly better than either of the other 2; and 2) might have a problem if the main ingredient (BG) was harmful. Agent Smith Agent Smith said: How does having group C help? Are we trying to control for the placebo effect? Not just the placebo effect. They must rule out that an untreated bitter gourd has a beneficial effect, whether imaginary or real. Agent Smith said: I know that if outcome ##2## occurs, we can't rule out the placebo effect because group C knows they didn't get any treatment. How do they know that? Do they know the nature of the test? That would seriously weaken the results. The less they know, the better. gleem FactChecker said: Not just the placebo effect. They must rule out that an untreated bitter gourd has a beneficial effect, whether imaginary or real. Group B doesn't know anything except that they're being given bitter gourd. Just that fact - being given something - could trigger the placebo effect. We can compare group B to C (hasn't been given anything). Either there's a difference (##2##) or there's no difference (##1##) If there's no difference (##1##) then we can say that untreated bitter gourd alone has no antidiabetic effect. If there's a difference (##2##), it could mean untreated bitter gourd has an antidiabetic effect and/or the placebo effect. How would we distinguish between these two possibilities? ------------------------------------------------------------------------------------------------------------------------ If there's a difference between group A and group B (##4##) and no difference between group B and C (##1##), we can conclude BitterHeal is antidiabetic but normal. untreated bitter gourd isn't. The fact that there's no difference between group B and C (##1##) means we can also rule out the placebo effect? .Scott said: Consider these two scenarios (high score being healthy): BH scores 123, BG scores 20, and nothing scores 50. For the above score outcomes, I would recommend in order of efficacy: 1. BH 2. Nothing 3. BG Let statistically significant S(x) = the blood sugar lowering potency of x Scores: S(BitterHeal) = 100, S(untreated bitter gourd) = 10, S(nothing) = 10 I've established causality? All three groups must not know what they have been given in addtion the researchers must also not kown who received what, so as reduce bias in their analysis. What is the criteria for causation? Was this a true double blind study? If not then all bets are off. Agent Smith and FactChecker Let's assume that it's impossible to conduct a completely blind experiment. Group C won't be given anything after all. Agent Smith said: Group C won't be given anything This escaped me but Group C must also be diabetic and they would also be on some medication. What do you want to find out? Agent Smith Agent Smith said: Let's assume that it's impossible to conduct a completely blind experiment. Group C won't be given anything after all. How is that blinding? Blinding is if neither researchers nor patients know who's given what. You have triple blinding if data analysts don't know either. As I understand, causation is determined by correlation plus controlling for possible lurking variable. Agent Smith WWGD said: How is that blinding? Blinding is if neither researchers nor patients know who's given what. You have triple blinding if data analysts don't know either. As I understand, causation is determined by correlation plus controlling for possible lurking variable. I did admit that it was impossible to blind all the subjects. Group C doesn't get anything and hence they would know, right? I'm trying to figure out why we need ##3## groups. As I said in my previous post, somewhere in here, that just groups A and B seem adequate (most clinical trials I've heard of consist of the test drug and a placebo, ##2## groups). Why would we need group C? gleem said: This escaped me but Group C must also be diabetic and they would also be on some medication. What do you want to find out? I want to find out of BitterHeal is an antidiabetic. Agent Smith said: I want to find out of BitterHeal is an antidiabetic You, want to find out. Are you planning a study or are you trying to interpret a study? @gleem , just a spinoff of a question in a course I took. What was the course and could you state the question? Agent Smith said: TL;DR Summary: An experiment to check for causality There is no statistical test for causality; statistics can only show correlation. Agent Smith said: Group C: Are not given either untreated bitter gourd or BitterHeal As has been mentioned, Group C should be given a placebo. Agent Smith said: Ab hinc difference "Ab hinc difference" is not a term used in statistics. Like most of the non-English terms you use in your posts this is semantic noise and in order to communicate more effectively you should stop doing it. Assume all conditions for a statistical experiment have been met adequately. Agent Smith said: Possible outcomes: 1. No difference between B and C 2. Difference between B and C 3. No difference between A and B 4. Difference between A and B These are not possible 'outcomes' these are possible (partial) results. 'Outcome' has a specific meaning in statistics: in this experiment an example of an outcome would be 'blood sugar levels are reduced by 10% in patient A'. Agent Smith said: ##4## is best with respect to inferring causality (that BitterHeal is antidiabetic). No, again you are using the wrong term, 'BitterHeal is antidiabetic' is a proposition not an inference (and again causality is irrelevent). The propositions that you should be looking at here are: 1. BitterHeal does/does not reduce blood sugar compared to a placebo. 2. BitterHeal does/does not reduce blood sugar compared to an untreated gourd. The propositions you have stated: Agent Smith said: X. Taking BitterHeal is better than not taking anything at all Y. Taking BitterHeal is better than taking a normal bitter gourd are not appropriate because other things need to be considered before saying something is 'better': what if taking BitterHeal reduces blood sugar on average by 10% but increases the risk of stroke by 20%? gleem pbuk said: As has been mentioned, Group C should be given a placebo. What exactly does that mean? If x were the placebo, what would x look like? pbuk said: The propositions that you should be looking at here are: 1. BitterHeal does/does not reduce blood sugar compared to a placebo. 2. BitterHeal does/does not reduce blood sugar compared to an untreated gourd. I agree, but these will manifest in the experimental study as statistically significant/insignificant differences in the blood sugar levels in the 3 groups, which I've listed as possible results. Nevertheless, gracias for the terminology lesson. I appreciate it. The question I'd like an answer to is why the need for ##3## groups? As far as I can tell, with groups A and B, we've taken adequate countermeasures (group B) against the placebo effect (neither of the groups know whether they're being given BitterHeal or just plain gourd, whether they're in the treatment group or the control group). Agent Smith said: What exactly does that mean? https://en.wikipedia.org/wiki/Placebo-controlled_study Agent Smith said: If x were the placebo, what would x look like? Exactly the same as the drugs being trialled. Agent Smith said: The question I'd like an answer to is why the need for ##3## groups? To account for the possibility that blood sugar is lowered simply because patients are taking part in a trial. This effect can be psychosomatic but it can also be due to a change in behavior e.g. diet. @pbuk most drug trials consist of only ##2## groups: the treatment group and the control group. This is true for most I believe and I take that to mean the conditions for a good trial are fulfilled. Why then the need for a third group? pbuk said: To account for the possibility that blood sugar is lowered simply because patients are taking part in a trial. This difference is not asymmetrical: both the control and treatment groups are same in this regard. In short we don't have to worry about it being a confounding factor. pbuk said: There is no statistical test for causality; statistics can only show correlation. Correlation may be inferred from observational studies. An experimental study can be used to establish causality. I mean when drug trials are conducted, we're trying to prove that a particular drug X causes a desired effect, no? Agent Smith said: @pbuk most drug trials consist of only ##2## groups: the treatment group and the control group. This is true for most I believe and I take that to mean the conditions for a good trial are fulfilled. Why then the need for a third group? The minimum for a trial is one treatment group and one control group. Because we don't know the effectiveness of the untreated gourd it cannot act as a control group so we need a third, placebo, group. Agent Smith said: This difference is not asymmetrical: both the control and treatment groups are same in this regard. I don't understand what this means. Agent Smith said: In short we don't have to worry about it being a confounding factor. What is 'it' here? Whatever 'it' is, 'confounding factor' has a specific meaning in statistics which doesn't fit the way you are using it here. Agent Smith said: Correlation may be inferred from observational studies. Yes: the study you describe is an observational study. Agent Smith said: An experimental study can be used to establish causality. Yes: can you describe what an experimental study might look like in this example? Agent Smith said: I mean when drug trials are conducted, we're trying to prove that a particular drug X causes a desired effect, no? No, for causality you also need to demonstrate the mechanism. pbuk said: No, for causality you also need to demonstrate the mechanism. What if you are trying to determine the proximate cause of something? gleem said: What if you are trying to determine the proximate cause of something? Then I don't know how a placebo controlled trial (which is what this thread is about, or rather it is about a specific hypothetical placebo controlled trial in some course) would be relevant. Last edited: pbuk said: Then I don't know how a placebo controlled trial (which is what this thread is about, or rather it is about a specific hypothetical placebo controlled trial in some course) would be relevant. One still has to rule out statistical variations. pbuk said: The minimum for a trial is one treatment group and one control group. Because we don't know the effectiveness of the untreated gourd it cannot act as a control group so we need a third, placebo, group. pbuk said: Yes: the study you describe is an observational study. pbuk said: No, for causality you also need to demonstrate the mechanism. It's all very confusing. pbuk said: The minimum for a trial is one treatment group and one control group The reason for this being ... to rule out the placebo effect? For that, in my example: the normal bitter gourd group (group B). If it were the case that there's a placebo effect at play, there would be no difference between groups A and B, no? If there is a difference then we can safely infer an actual positive antidiabetic effect for BitterHeal, no? We want to check if BitterHeal has an antidiabetic effect but there are the following possibilities: 1. The placebo effect 2. Normal/untreated bitter gourd too being antidiabetic To rule out 1, group C has to be given a placebo (a normal bitter gourd). To rule out 2, we have group B. But now group B and C are identical. Last edited: How about listing out the possibilities: 1. Difference(A, B) = True 2. Difference(A, B) = False 3. Difference(A, C) = True 4. Difference(A, C) = False 5. Difference(B, C) = True 6. Difference(B, C) = False 1 and 6 seems the best result for BitterHeal : BitterHeal is better than normal/untreated gourd and there's no evidence of a placebo effect in action. 2, 4, 6 is the worst-case-scenario for BitterHeal. Taking BitterHeal or taking normal bitter gourd or taking nothing at all makes no difference to a diabetic. 2 and 5 is also a bad situation for BitterHeal because there's a placebo effect. @pbuk @others Last edited: If C knows it did not receive an antidiabetic substance there will be no placebo effect in that group. Bitter gourd is believed to have some antidiabetic effect so it might happen that in this study the effect will not be observed. Think of it as a current treatment. BitterHeal is believed to be better. You compare Groups A and B to determine the magnitude of the difference. Group C may provide some validation for BitterHeal if B is found to be ineffective. gleem said: If C knows it did not receive an antidiabetic substance Yes, C knows. It seems that when you market a drug (call it X), your claim is to say "Diabetics, drug X lowers blood sugar levels" To do that we need to show a statistically significant difference between those who take drug X and those who don't. The sensible thing to do would be to conduct a clinical trial on randomly selected volunteers randomly assigned to 2 groups (treatment/test group and control group, the former receives treatment while the latter does not). The problem is there's the placebo effect. The usual procedure is to give the control group a placebo. I suppose this is what blinding is. As is apparent, we need only ##2## groups to conduct a clinical trial. Where is the necessity for a ##3##rd group, like the one in my question? Does my question consist of ##2## trials (not ##1##), with one trial to check for the placebo effect and the other trial to check for BitterHeal's antidiabetic effect? AFAIK the placebo effect does not occur in studies involving physical changes in body chemistry like serum cholesterol, and blood sugar. It is significant in studies of the subjective effect of a medication as in pain relief or mood change. When comparing two medications for effectiveness you compare it to a (gold)standard commonly used for the treatment. The problem as I see it with this study is that we are comparing two medications of marginal value. I assume this problem was from a statistics course, but did they fully realize the issues with this scenario? The question is what was the intent of the problem? What did they want you to learn from it? I think you need group C to establish the effectiveness of B as well as the effectiveness of A in lowering blood sugar. Next, you can compare A and B to determine their relative effectiveness. Replies 3 Views 665 Replies 6 Views 1K Replies 4 Views 2K Replies 3 Views 1K Replies 2 Views 1K Replies 1 Views 2K Replies 11 Views 7K Replies 20 Views 3K Replies 62 Views 2K Replies 6 Views 2K	4,491	19,754	{"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}	2.8125	3	CC-MAIN-2024-38	latest	en	0.897823
https://www.finance-assignments.com/category/concept-and-measurement-of-cost-of-capital/page/2	1,539,834,535,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583511642.47/warc/CC-MAIN-20181018022028-20181018043528-00008.warc.gz	936,446,032	7,846	# Category Archives: CONCEPT AND MEASUREMENT OF COST OF CAPITAL ## Dividend Approach Dividend Approach One approach to calculate the cost of equity capital is based on the dividend valuation model. According to this approach, the cost of equity capital is calculated on the basis of a required rate of return in terms of the future dividends to be paid on the shares. The cost of equity capital, k, is, accordingly, defined as the discount rate that equates the present value of all expected future ## Cost of Equity Capital Cost of Equity Capital The cost of equity capital is by far, conceptually speaking, the most difficult and controversial cost to measure, It has been shown in the preceding discussions that the coupon rate of interest which form s the basis of calculation of cost of debt can be estimated with a high degree of accuracy since intersect payments as well as the return of the principal are contractual obligations. ## Perpetual Security Perpetual Security The cost of preference shares which has no specific maturity dare is given by Eq. (12.8) and Eq. (12.8A) Finance-Assignments.comInstructions Feel free to send us an inquiry, we reply back real quick. Or directly email us at order@finance-assignments.comName Email Phone Requirements/ Instructions File Upload File Upload File Upload VerificationPlease enter any two digits Examp ## Cost of Preference Shares Cost of Preference Shares The computation of the cost of preference shares is conceptually difficulties compared to the cost of debt. In the case of debt, as shown above, the interest rate is the basis of calculating cost, as payment of a specific amount of interest is legal commitment on the part of the firm. There is no such obligation in regard to preference dividend. It is true that a fixed stipulated on pr ## Cost of Perpetual Debt Cost of Perpetual Debt The measurement of the cost of perpetual debt is conceptually relative easy. It is the rate of return which the lenders expect. The debt carries a certain rate of inter The coupon interest rate or the market yield on debt can be said to represent an approximation the cost of debt. The nominal/coupon rate of interest on debt is the before tax cost of debt. Since the effective cost of debt ## Cost of Debt Cost of Debt The calculation of the cost of debt is relatively easy. The cost of funds raised through debt in the form of debentures or loan from financial institutions can be determined from Eq. 12.1. To apply the formulation of explicit cost of debt, we need data regarding; (i) the net cash proceeds/inflows (the issue price of debentures/amount of loan minus all flotation costs) from specific source of debt, ## MEASUREMENT OF SPECIFIC COSTS MEASUREMENT OF SPECIFIC COSTS The term cost of capital, as a decision criterion, is the overall cost. This is the combined cost of the specific costs associated with specific sources of financing. The cost of the different sources of financing represents the components of the combined cost. The computation of the cost of capital, therefore, involves two steps: (i) the computation of the different elements of the ## Explicit and Implicit Costs Explicit and Implicit Costs The cost of capital can be either explicit or implicit. The distinction between explicit and implicit costs is important from the point of view of the computation of the cost of capital. The explicit cost of any source of capital is the discount rate that equates the present value of the cash inflows that are incremental to the taking of the financing opportunity with the present val ## Assumptions Assumptions The theory of cost of capital is based on certain assumptions. A basic assumption of traditional cost of capital analysis is that the firm’s business and financial risks are unaffected by the acceptance and financing of projects. Business measures the variability in operating profits earnings interest anti taxes (EBIT) due to change in sales. If, a firm accepts 3 project that is considerably m ## Definition Definition In operational terms, cost of capital refers to the discount rate that is used in determining the present value of the estimated future cash proceeds and eventually deciding whether the project is worth undertaking or not. In this defined as the minimum rate of return that a firm earn on its investment for the firm to remain unchanged. The cost of capital is visualized as being composed of several el	938	4,445	{"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-2018-43	latest	en	0.857329
http://blog.mbedded.ninja/programming/general/encryption	1,540,133,390,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583514030.89/warc/CC-MAIN-20181021140037-20181021161537-00527.warc.gz	48,553,063	34,760	# Overview There are two basic types of encryption used widely in computer science, symmetric and asymmetric algorithms. Symmetric algorithms use two identical or very similar keys to both encrypt and decrypt the information. Their main problem is that the sender and receiver need to share these keys securely before any transfer of information can take place, and you can’t easily do that over the internet. Then there is asymmetric algorithms, which include the popular public-key algorithms. This uses two mathematically related but different keys, one for encrypting, and one for decrypting. One is kept secret by the generator of the keys, the other is made public. Only the holder(s) of the decryption key can decipher the message, and only the holder(s) of the encryption key can encrypt a message. When the decryption keys is kept private and the encryption-key made public # Hash Functions Hash functions are used in cryptograph to prove a digital signature of the sender They normally take an any-sized block of data and return a fixed-size hash A checksum, regularly used to verify data blocks over internet transmission protocols and between embedded devices on circuit boards, is a form of a hash function. The ideal hash function is - easy to compute the hash value - infeasible to generate a message with a given hash - infeasible to modify a message without changing the hash - infeasible to find two different messages with the same hash - posses the Avalanche effect # Types Of Attacks ## Brute-Force Attack This is when the attacker attempts to decipher the message by making random guesses at the key. The stronger the encryption (in expressed in terms of bits), the longer (on average), it would take an attacker to crack the code. ## Side-Channel Attacks Using the time of computation or the CPU power usage while performing encryption/decryption to infer features about the key (and hence reducing the number set that the key could belong t0), resulting in ## Man-In-The-Middle Attacks This is when a third party changes the public key half-way through a transmission to match their own private key. # TEMPEST TEMPEST is a term used to define the method of obtaining protected information from ‘compromising emissions’. Electronic, magnetic and acoustic ‘noise’ from an system containing encrypted information can be analysed to reveal the plaintext Bell Labs were able to use a TEMPEST method on computer monitors to decipher 75% of the plaintext used in secure TTY connections during WWII at a distance of 80 feet Also interesting to note, the LED’s on routers can also compromise data. The LED’s usually blink to indicate traffic, and with a fast enough optical system, you can get the exact bits/bytes, since they are usually connected straight onto the data line ## Traffic Analysis Traffic analysis is all about intercepting and examining messages and inferring information from the patterns in communication For example, in WWI and WWII, many Allied analysts managed to deduce the location and movement of enemy positions using traffic analysis. However, many of the reports weren’t heeded by the chain of command, and lives and resources were lost. The timing of packets using internet protocols that send a packet everytime a key is pressed (such as SSH), can give away passwords (analysed using hidden Markov models). Posted: March 15th, 2012 at 11:13 am Last Updated on: June 11th, 2017 at 1:33 pm	709	3,457	{"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-2018-43	latest	en	0.898285
https://questioncove.com/updates/4ea72194e4b0a7d5141c1cbe	1,722,807,315,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00047.warc.gz	389,182,902	6,630	Ask your own question, for FREE! Mathematics 24 Online OpenStudy (anonymous): A college student earned $7000 during summer vacation working as a waiter in a popular restaurant. The student invested part of the money at 9% and the rest at 8%. If the student received a total of$588 in interest at the end of the year, how much was invested at 9%? $4200 b.$875 c. $3500 d.$2800 OpenStudy (anonymous): do it by ur self? OpenStudy (anonymous): Are goig to help me ? OpenStudy (anonymous): .09X+.08(7,000-X)=588 .09X+560-.08X=588 .01X=588-560 .01X=28 X=28/.01 X=2,800 AMOUNT INVESTED @ 9%. 7,000-2,800=4,200 AMOUNT INVESTED @ 8% PROOF: .092,800+.084,200=588 252+336=588 588=588 so it would be D OpenStudy (anonymous): yeah i told u OpenStudy (anonymous): answer i can give u in seconds but the thing is that u too learn how to solve this one OpenStudy (anonymous): Sariah - notice I helped you out :) OpenStudy (anonymous): Are you sure sclower? OpenStudy (anonymous): yes Im sure. lol OpenStudy (anonymous): sclower i gave u medal too but wht sariah learnt from this OpenStudy (anonymous): I have others questions.. Can you help me please? OpenStudy (anonymous): I can try OpenStudy (anonymous): Ok waite! OpenStudy (anonymous): Find the distance d(P1, P2) between the points P1 and P2. P1 = (0.2, 0.4); P2 = (-2.4, -2.9) Round to three decimal places, if necessary. a. 29.5 b. 4.201 c. 4.301 d. 13.285 OpenStudy (anonymous): Sorry for the delay. Use the distance formula: $d=\sqrt{(x _{1}}+x _{2}) + (y _{1}-y _{2})^{2}$ where $P _{1}(x _{1},y _{1})$ and $P _{2}(x _{2},y _{2})$ are the coordinates of the given points. Answer is $e ^{i \pi?} +1 = 0$ OpenStudy (anonymous): Solve the problem. The formula A = P(1 + r)^2 is used to find the amount of money, A, in an account after P dollars have been invested in the account paying an annual interest rate, r, for 2 years. Find the interest rate r if $500 grows to$980 in 2 years. Seleccione una respuesta. a. 96% b. 4% c. 240% d. 40% OpenStudy (anonymous): You are gonna need to post these to the board. I dont have time to do all of them. But can help with some when I can. :) OpenStudy (anonymous): Ok OpenStudy (anonymous): obviously answer is a. OpenStudy (anonymous): una???? OpenStudy (anonymous): gave you a medal sheg for agreeing with me. lol Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours! Join our real-time social learning platform and learn together with your friends! Latest Questions ShikuniShigana: Am halfway yall, this the redrawn version of the original post i did 1 minute ago 1 Reply 0 Medals xXAikoXx: I wanna cry so bad 32 minutes ago 2 Replies 0 Medals kimberlysaavedra: guyss its my birthday tomorrow!!!!!! 8-4-2008 57 minutes ago 14 Replies 1 Medal TarTar: If you could go back to any year what would you pick and why 9 minutes ago 24 Replies 2 Medals TarTar: Ever been complimented so much you think it's a prankud83dude2d 16 hours ago 2 Replies 0 Medals Puck: my camera roll right now 11 hours ago 4 Replies 2 Medals kimberlysaavedra: @cora1 8 hours ago 16 Replies 1 Medal ShikuniShigana: Should i continue drawing this or redraw it yall? 1 day ago 5 Replies 2 Medals Leila1234565: Best way to sneak a phone back 1 day ago 1 Reply 0 Medals Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours! Join our real-time social learning platform and learn together with your friends!	1,053	3,520	{"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}	3.71875	4	CC-MAIN-2024-33	latest	en	0.90924
http://amrita.olabs.edu.in/?sub=1&brch=5&sim=159&cnt=2	1,653,135,080,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00755.warc.gz	3,643,182	8,724	Simple pendulum # Materials Required • A clamp with stand • A split cork • A Cotton Thread (about 2 meters long) • A bob • Vernier calliper • Stop /watch • Metre scale. # Real Lab Procedure 1. Find the vernier constant and zero error of the vernier calipers and record it. 2. Determine the mean diameter of the simple pendulum bob using the vernier calipers. 3. Find the mean radius of the bob and represent it using ‘r’. 4. Attach a string to the bob. The length of the pendulum, l is adjusted by measuring a length of (l-r) from the top of the bob. 5. Put ink marks M1,M2 and M3 on the thread at distance of 50cm,60cm and 70cm from the C.G of the bob . 6. Pass the thread through the splited cork with the 50 cm mark at the bottom of the cork and tighten the two cork pieces between the clamp. 7. Fix the clamp in a stand kept on the table such that the height that the bob is just 2 cm above the laboratory floor. 8. Mark a point A on the floor just below the position of the bob at rest. 9. The equilibrium position of the pendulum is indicated by drawing a vertical line with a chalk on the edge of the table, just behind the string. 10. Find the least count and the zero error of the stop watch. Bring its hands to the zero position. 11. Move bob using the hand at an angle not more than 450 and leave it. See that the bob returns over the line without spinning. 12. The stop watch is started when the pendulum crosses the equilibrium position to any one side. 13. When it passes the equilibrium position in the same direction the next time it has completed one oscillation. 14. Just when the 20th oscillation is complete, count 20 and at once stop the stop watch. 15. Note the total time taken for twenty oscillations from the position of both the hands of the watch. 16. As we need two observations for the same length, repeat steps 12 to 15 one more time. 17. Repeat the experiment for lengths 60cm, 70cm, 80cm, 90 cm, 100cm, 110 cm, 120cm and 130cm. 18. In each case is calculated. In all cases it is found that is a constant. 19. The mean value of is calculated and then the acceleration due to gravity is calculated using the relation (2). ## To draw the l-T2 graph The experiment is preformed as explained above. A graph is drawn with l along X axis and T2 along Y axis. The graph is a straight line, as shown in the figure. To find the length of the second’s pendulum A second’s pendulum is one for which the period of oscillation is 2 seconds. From the graph the length l corresponding to T2=4 s2 is determined. This gives the length of the second’s pendulum. To find the length of the pendulum whose period is 1.5 seconds The length l corresponding to T2 =1.52=2.25 is determined from the graph. To find the period (T) for a length 105cm T2 corresponding to l=105 cm is determined from the graph. The square root of this gives T, the period of the pendulum for a length 105 cm. From the graph = ------cm/s2 # Simulator Procedure (as performed through the Online Labs) 1. Select the environment to perform the experiment from the 'Select Environment' drop down list. 2. Select the shape of the bob of the pendulum from the 'Select Shape' drop down list. 3. Select the material of the bob from the 'Select Material' drop down list. 4. Select the type of the wire to be used from the 'Select Wire' drop down list. 5. Use the 'Change Length' slider to change the length of the pendulum. 6. Use the 'Change Dimension' slider to change the dimension of the bob used. 7. Now release the bob. 8. Clicking on the 'Show Protractor' button helps us to ensure that the angle of swing does not exceeds 450. 9. Now click on 'Play /Pause' button to start the stopwatch. We can alternatively click on the the 'Start' or 'Stop' button on the stopwatch. 10. The time for twenty oscillations is noted. 11. Now the real lab procedure from steps 12 to 18 can be followed to complete the observations for finding the acceleration due to gravity. 12. Clicking on the 'Answer' button displays the acceleration due to gravity for the corresponding environment. # Observations To find the diameter of the bob 1 M S D = 1mm 10 V S D =9 M S D 1 V S D=9/10 M S D=0.9 mm Vernier Constant, V.C.= 1 M.S.D.-1 V.S.D. = (1-0.9) mm = 0.1 mm = 0.01cm. Zero error of vernier callipers(e) 1. e=..............cm 2. e=..............cm 3. e=..............cm Mean zero error e =.......cm Mean zero correction c = -e = ......cm SL No Main Scale Reading MSR(cm) Vernier scale Reading VSR(dvs) (VSRxL.C) (cm) Diameter of the bob,D=MSR+(VSRx L.C)+c(zero correction) (cm) Mean Diameter,D Mean Diameter of the Bob, D= ……………cm Mean radius of the bob, r =D/2 = .........cm Least count of stop watch =..........s Zero error of stop watch =...........s Zero correction of stop watch =...........s Table for length () and time (T) Sl No (l-r)cm Length of the pendulum l (cm) Time for 20 oscillations Time Period T2 (s) t1(s) t2(s) Mean t(s) # Calculations Mean value of .=…………..ms-2 The acceleration due to gravity, g = …………m/s2 Acceleration due to gravity from graph Value or l = AB = -----cm Value for T= BC = -----------cm AB / BC = ……….. Acceleration due to gravity, g=---------m/s2 # Result 1. Acceleration due to gravity (g) at the place • By calculation =………….ms-2 • From the graph =………….ms-2 • Mean g =………….ms-2 2. Length of the seconds pendulum =………….m 3. Length of the pendulum whose period is 1.5 s=……..m 4. Period of the pendulum of length 105 cm=…….s Cite this Simulator:	1,484	5,504	{"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}	3.65625	4	CC-MAIN-2022-21	latest	en	0.850932
http://www.us-lotteries.com/megabucks_plus/megabucks_plus-info.asp	1,369,411,400,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368704713110/warc/CC-MAIN-20130516114513-00012-ip-10-60-113-184.ec2.internal.warc.gz	777,785,738	8,892	Megabucks Plus general information - how to play, payouts, odds, jackpot info, links, Megabucks Plus states map. # Tri-State Megabucks Plus Lottery General Information Megabucks Plus is a Tri-State lottery game played in 3 states with a jackpot starting at \$1 million and growing until someone wins. The Megabucks Plus game consists of two sets of numbers where the player selects 5 numbers 1 through 41 from the first set and one number (called the Megaball) 1 to 6 from the second set. Megabucks Plus costs \$2 per game. On this page... Megabucks Plus drawing schedule: Megabucks Plus drawings are held at approximately 8 p.m. EST every Wednesday and Saturday night. Megabucks Plus drawing method: There is no information on how the winning Megabucks Plus numbers are drawn. ## How to play Megabucks Plus Just like any multi-number game, Megabucks Plus is a very easy game to play. Two sample panels from a Megabucks Plus playslip are shown here; the actual playslip contains 5 panels. Select 5 numbers from 1 through 41 on the upper section and one number from 1 through 6 on the lower section, which is the ottom row with 6 numbers. Alternatively, you can mark the Easy Pick (EP) to let the Lottery terminal randomly select your numbers for you. If you want to play more numbers, then fill out panels 2, 3, etc. Each one costs \$2. If you make an error, DO NOT ERASE, instead mark the Void box for that play. A typical winning number is reported as, 02-13-15-22-27 MB 01, or as 02-13-15-22-27**01 Since the order in which the numbers are drawn doesn't matter, the winning numbers are reported in their natural order from smallest to highest. ## How Do You Win Megabucks Plus? - Prizes and Odds How much does Megabucks Plus pay for matching 5 or less numbers? The following is a summary winner prize payouts and the odds of winning. MatchPrizeOdds 5-of-5 + MB\$1,000,000 +1 in 4,496,388 5-of-5\$30,0001 in 899,278 4-of-5 + MB\$1,3001 in 4,996 4-of-5\$1501 in 435 3-of-5 + MB\$251 in 714 3-of-5\$71 in 143 2-of-5 + MB\$51 in 63 2-of-5\$21 in 13 1-of-5 + MB\$21 in 15 If there are multiple top prize winners for a drawing, the top prize shall be divided equally among the top prize winners The Jackpot can be paid in a 30-year annuity for the approximate advertised jackpot amount (30 payments over 29 years) or in a one-lump-sum cash payment. Winners have 60 days from filing claim to elect the CASH or ANNUITY option. Claiming Megabucks Plus Prizes: Usually prizes of less than \$600 can be claimed at any retailer. However, every state has its own rules for claiming lottery prizes. Please see your state's info page on this web site or go to the Megabucks Plus page of the official site of your state by clicking on the map further below. Also, how long you have to collect a Megabucks Plus prize depends on the state where you bought your tickets. The time period for claiming a prize ranges from 180 days to 1 year from the draw date. ## Megabucks Plus States The map shows the states where Megabucks Plus is currently played. They are Maine, New Hampshire, and Vermont. You can click on a state to be directed to that state's Megabucks Plus page. ## Links to some Megabucks Plus Web Site Pages Although we have tried to cover all the pertinent information, and although there is no official web site of the Megabucks Plus lottery game, we recommend that you also look at some pages at the Maine, New Hampshire, and Vermont Lottery official web sites. ## Megabucks Plus at US-Lotteries.com The main purpose of this web site (US-Lotteries.com) is to present a well formatted latest and past results; analysis and statistics of past winning numbers, digits, pairs, triads; search to see if your favorite numbers have ever won or have come close to winning the jackpot; mathematical tools such as combinations generators and wheels. The complete list of all Megabucks Plus pages of this web site are listed below. TOP All Lottery States • WE DO NOT SELL ANY LOTTERY, WE DO NOT BUY LOTTERY NUMBERS ON BEHALF OF ANYONE; this is only an information and guide web site.	1,030	4,097	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2013-20	longest	en	0.919438
http://www.analytics.business/regression-models/	1,606,847,005,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141681209.60/warc/CC-MAIN-20201201170219-20201201200219-00168.warc.gz	99,371,394	42,353	# Regression Models Using Numerous Variables ## Assessing Regression Models Using Numerous Variables Regression model on the Amex, Iowa housing data set builds regression models for the house sale price with numerous variables. Some of which are highly correlated, continuous variables along to the other side of the continuum by evaluating categorical, low correlated variables. An assessment of each model will be conducted along with a review on the statistical and ODS output will be conducted and interpreted. Additionally, linear regression models will be conducted based on two and three variables along with an evaluation of the impact of these new variables and how they add value to predicting the sales price of a house. Finally, an assessment on the model being specified and the next steps will be provided. ## Part A – Step 1 The variable I choose is was one that was created in first installment – Total Floor Square Footage (TotalFlrSF). Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > \|t\| Intercept 1 11406 3242.59761 3.52 0.0004 TotalFlrSF 1 113.30303 2.05569 55.12 <.0001 ### Equation The normal equation is y = β0 + β1x + ε Based on the above parameter estimates, the equation is: • Sale Price = \$11,406 + \$113.30303 x TotalFlrSF Thus, for each unit increase in the TotalFlrSF, an increase in sales price of \$113.30 occurs. This assumes all values are greater than zero but even at zero, a SalePrice of\$11,406 would result. As we are evaluating houses, this would be logical but could be if perhaps a house was not livable and is considered a ‘tear down’ where someone would spend the time and money required to build a new house. However, this would be outside the norm and would require a different equation to determine these types of sale prices. The automatic generated ODS output: The various produced ODS output from SAS shows a cluster of points in the residual versus predicted value indicating there is an issue as it should be completely random without any type of pattern. The Q-Q Plot titled “Quantile” is abnormal as it deviates from the line and has heavy tails which means there is a larger probability of getting very large values. The Predicted Value plot again shows that there are issues as the points are not on the line. ### Assess the Goodness-of-Fit of this model Analysis of Variance Source DF Sum of Squares Mean Square F Value Pr > F Model 1 9.518383E12 9.518383E12 3037.86 <.0001 Error 2928 9.174155E12 3133249517 Corrected Total 2929 1.869254E13 Root MSE 55975 R-Square 0.5092 Dependent Mean 180796 Adj R-Sq 0.5090 Coeff Var 30.9605 Regression Models – TotalSqFt Fit Plot Regression Models – TotalSqFt Fit Plot The P value is small but not equal to zero which means we can reject the null hypothesis that it is equal to zero. Thus, each variable is significant. Additionally, the statistical summaries show an R-Square of .5092 and an adjusted R-Square of .509 which is also very good as it represents the proportion of variability in the dependent variable that can be explained by the regression model. • Fitted regression model over the scatter-plot: The fit Plot shows a plot of all of the data along with the 95% confidence and prediction limits. We can clearly see the amount of data that exceeds the boundaries confidence limits. • Assessment of the normality of the residuals using a Q-Q plot and/or histogram of the standardized residuals: The normal quantile plot of the residuals and the residual histogram are consistent with the assumption of Gaussian errors and shows the narrow spread in the residuals but with a high peak. The Q-Q plot helps detect violations from normality if they are normal, the points will cluster tightly around a reference line. As we can see, we have a deviation from the line and has heavy tails which means there is a larger probability of getting very large values. Thus, normality is not true in this model. • Assessment of homoscedasticity by plotting the predictor variable against the standardized residuals: The quantile histogram helps diagnose violation of the normality and homoscedasticity assumptions. The points on the Predicted Value versus Sales Price plot of the dependent variables versus the predicted values along the 45-degree line show a definite pattern which indicates that the model is not appropriate. Homoscedasticity also referred to as the assumption of constant error variance looks for errors that aren’t independent causing the plot to be more linear looking or that flares out as x increases. Thus, based on the above, this model does not appear to be homoscedastic. • Check for potential outliers using Cook’s Distance: Cook’s Distance measures the effect of each data point on the predicted value with the lowest value being zero and the highest point showing the influence of the point. As we can see, there are only two points that exceed .2 but are less than one, the acceptable limit, an investigation into these two outliers should be conducted. Thus, these two points could be outliers but having an influence on the results. While the above does not reflect the actual Cook’s D score, the values show how the vector of fitted values move when the observation is deleted. The variable TotalSqFt has normality issues based on the above analysis. The ODS plots reflect patterns of clusters, heavy tailed Q-Q plot, and the Predicted versus Sale Price are not on the line. Additionally, the Residual plot does not appear to be complete randomness and the Fit Plot which we would expect to be random, has more of a sideways funnel pattern. However, the P value is < .0001 which means each variable is significant and that we can reject the null hypotheses. Additionally, the Adjusted R-Square is .509 which is significant. The matrix of graphs and plots provides a high-level overview of the relationship between total square footage in a house and sale price. ## Part A – Step 2 ### The Best X to Predict Y in the Regression Model The ‘best’ simple linear regression model to predict Sales Price using the R-square option reflects the following output: Variables in Model R-Square Adjusted C(p) R-Square TotalFlrSF 0.5118 0.5116 3161.986 GrLivArea 0.5006 0.5004 3289.886 GarageArea 0.4235 0.4233 4169.764 TotalBsmtSF 0.421 0.4207 4199.299 FirstFlrSF 0.4115 0.4112 4307.718 Ironically, Total Floor Square Footage has the highest R-square value and Adjusted R-square which was discussed in Part A Step 1. So not to repeat the same outputs, I’ll use GrLivArea for the rest of this task. Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > \|t\| Intercept 1 13290 3269.70277 4.06 <.0001 GrLivArea 1 111.69400 2.06607 54.06 <.0001 ### Equation & Interpret each coefficient: The equation for GrLivArea is SalePrice = \$13,290 x 111.694 x GrLivArea. Which reflects that for each unit increase in GrLivArea the Sale Price increases by \$111.69. This seems reasonable and logical. R2 measures the variability in y remaining after x has been considered and is often called the proportion of variance explained by the regressor x (Montgomery, D. Introduction to Linear Regression Analysis. p. 36). So, the variables with the closest R2 equaling to one implies that they explain most of the variability in y. Variables in Model R-Square Adjusted R-Square C(p) Interpretation Overlap? TotalFlrSF 0.5118 0.5116 3161.986 Makes logical sense in terms in explaining the variability in sales price Overlap GrLivArea 0.5006 0.5004 3289.886 Logical but overlaps with TotalFlrSF Overlap GarageArea 0.4235 0.4233 4169.764 Not logical on its own as a house needs to go with a garage TotalBsmtSF 0.421 0.4207 4199.299 Logical but overlaps with TotalFlrSF and GrLivArea Overlap FirstFlrSF 0.4115 0.4112 4307.718 Would have thought this to be stronger than TotalBsmtSF; Overlaps with TotalFlrSF, GrLivArea, TotalBsmtSF Overlap TotalBath 0.3868 0.3866 4589.011 Logical as it plays an important part in the sales price of a house Overlap TotalFullBath 0.359 0.3588 4906.859 Logical and expected that it would have a lower R than total number of bathrooms Overlap HouseAge 0.3213 0.321 5338.312 Logical in that newer housses would likely have a higher sales price Overlap YearBuilt 0.3209 0.3206 5342.262 Expected to be very similar to HouseAge which it is so no surprise Overlap YearRemodel 0.2894 0.2891 5702.33 Logical in terms of explaining the variability in sales price; somewhat associated with houseage and yrbuilt MasVnrArea 0.2843 0.284 5760.722 Not logical as I would have thought it to have lesser impact on the sales price than some of the variables with a lesser R TotRmsAbvGrd 0.2523 0.252 6126.552 Logical in terms of explaining the variability in sales price Overlap BsmtFinSF1 0.1966 0.1963 6762.719 Logical as I think people are more concerned about the bsmt sq ft over being a finished area or not Overlap LotFrontage 0.127 0.1266 7557.59 Logical as the sales price and lotfrontage would be less likely to explain the variability WoodDeckSF 0.117 0.1166 7671.945 Logical as it ranks higher than the amount of porch space but surprised that it is ranked higher than LotArea LotArea 0.1027 0.1023 7835.853 Not logical as I would have thought the LotArea would be more directly related to sales price SecondFlrSF 0.0636 0.0633 8281.516 Logical as likely to be similar to the first floor and is related to TotalFlrSF Overlap TotalHalfBath 0.0544 0.054 8387.015 Logical as sales price is likely to be more effected by totalbath and fullbaths Overlap BsmtUnfSF 0.0386 0.0382 8567.744 Logical as I think people are more concerned about the bsmt sq ft over being a finished area or not Overlap TotalPorchSF 0.0293 0.0289 8673.387 Logical as it is of little importance to the sales price BedroomAbvGr 0.0188 0.0184 8793.586 Logical and has less impact than TotalRmsAbvGrd which makes sense Overlap PoolArea 0.0048 0.0044 8953.456 Logical as would expect it to have little impact on sales value LowQualFinSF 0.0013 0.0009 8994.168 Logical based on what I would expect people to report on this variable and other variable that may be related i.e. sale condition MoSold 0.0007 0.0003 9000.444 Logical – when a house is sold should have minimal impact on the sales price Overlap YrSold 0.0006 0.0001 9002.185 Logical – when a house is sold should have minimal impact on the sales price Overlap MiscVal 0.0003 -0.0001 9005.503 Logical given the few items and their value to the overall sales price BsmtFinSF2 0 -0.0004 9008.506 Logical given the dataset and the potential duplication and confusion on this variable Overlap ### In what sense is the model the ‘best’ model: GrLivArea is the best model as it logical to the sales price of a house which people obviously plan to live in so they are more likely to spend additional money for a larger space. It has an Adjusted R-Square of .5004. The p value is small so we can reject the null hypothesis. ### Anything funny about it from an interpretation standpoint? The equation is ‘funny’ if you consider a house with zero GrLivArea which would result in a sale price of \$13,290. While one could perhaps justify, it based on the price of land only, it is illogical. Additionally, the increase in value based on a one unit change of ~\$111.70 is logical. Overall, we would expect the model to be better with the increased Adjusted R-Square yet we still have the same issues and concerns based on the plots. ### Goodness-of-Fit Goodness-of-fit statistics on the GrLivArea are as follows: Analysis of Variance Source DF Sum of Squares Mean Square F Value Pr > F Model 1 9.33763E12 9.33763E12 2922.59 <.0001 Error 2928 9.354907E12 3194981962 Corrected Total 2929 1.869254E13 Root MSE 56524 R-Square 0.4995 Dependent Mean 180796 Adj R-Sq 0.4994 Coeff Var 31.2641 The F Value is 2,922.59 is significantly greater than one. The p-value is small reflecting that each variable is significant. Based on the above, there is a linear relationship between SalePrice and GrLivArea. The Adjusted R-Square of .4994 reflects the variability that is accounted for using SalesPrice and GrLivArea. While there is a small difference between R-Square and Adjusted R-Square, Adjusted R-Square will not penalize us for adding variables and will only change if the variable reduces the residual mean square. The P value is small so we can reject the null hypothesis. The ODS output is as follows: Regression Models – SaS ODS output The Predicted Value to the Residual has a definite pattern and seems to fan out showing that the variable is increasing. The ODS Quantile or Q-Q plot indicates that normality is not true as the points deviate from the line (see red area) and is heavy tailed. Again, similar to TotalSqFt Cook’s D we have three influential points with two points that exceed 0.2. It would only be prudent to look into these three points to ensure they are valid. Additionally, the Residuals for Sale Price are not random as we would expect and again has a bit of a pattern to it. The Fit Plot also has a distinct pattern to it. Thus, we have normality issues. Regression Models – FitPlot GrLivArea Regression Models – Residuals GrLivArea The Residuals between SalePrice and GrLivArea are again, similar to TotalSqFt, where homoscedasticity does not appear to be linear within the plot but does flare out a bit as GrLivArea increases. Thus, this model does not appear to be homoscedastic. While GrLivArea is the highest Adjusted R-Square value out of all of the continuous variables accounting for .4994 of the variability leaving another .5006 unexplained. Perhaps, one of the categorical variables will be assist in improving the unexplained portion. Again, we have conflicting information between the adjusted R-Square the ODS plots. Additionally, GrLivArea has a small p-value but we have normality issues based on the ODS plots. ### The Best X to Predict Y Conclusions The GrLivArea variable is a variable with a small p-value which means that the variable is significant. Additionally, it has a good Adjusted R-Square value however, the ODS outputs reflect that we have normality issues. We have patterns in the Predicted Value versus Residual value where there shouldn’t be any, the Q-Q plot deviates from the line, the Predicted Value versus Sales Price is also not on the line which reflect that is it not ideal. The residual plot also has a pattern and the fit plot also has a fanning out pattern. Thus, the GrLivArea is not normal. ### Categorical value in a Regression Model: Based on week one Exterior1 is a category variable with the following histogram Regression Models- Exterior Frequency ### Equation Evaluating a few of the variables we get the following result: Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > \|t\| Variance Inflation Intercept 1 150475 4457.27517 33.76 <.0001 0 Exterior1Category 1 2575.12360 357.88376 7.20 <.0001 1.00000 Root MSE 79203 R-Square 0.0174 Dependent Mean 180796 Adj R-Sq 0.0170 Coeff Var 43.8146 • SalePrice = \$150,475 x \$2,575.12 x Exterior1 ### Interpret Coefficient While Siding is only one of sixteen variables within the Exterior1 variable, a one unit change will result in an increase in value by \$2,575 not that we can have a house with more than one siding, this is illogical but it simply means that a house will siding on average has an increased sales price of \$2,575. The p-values is also greater than one which means we can reject the null hypothesis and conclude that there is a linear relationship between Exterior1 Siding and SalePrice. Looking at the Adjusted R-Square, we see that SalePrice using Exterior1 only explains 1.74% of the variability which is low. ### Anything funny about the coefficient interpretation I can’t see there is anything funny about the coefficient other than its low value in explaining variability. Variance inflation is also one. ### Generate ODS output and assesses the Goodness-of-fit Regression Models – Exterior ODS Reviewing the ODS plots we don’t seem to have a pattern in the Predicted Valued and Residual plot unlike in the GrLivArea ODS plot but we do have another heavy-tailed Q-Q plot that deviates from the line, a cluster in the predicted value with a lot more spikes in Cook’s D reflecting that there are more observations that are having an influence on the results but is still less than one on the axis scale. Additionally, we see skewed residual. Regression Models – FitPlot Exterior Regression Models – Residuals Exterior Reviewing the plots and various statistics, the residuals plot can’t fan out due to the type of variable but there does not appear to be any linear relationship between the different exterior1 categories. The fit plot resembles shows a slight positive correlation with a bit of a ‘u’ shape between the data. It appears to be bimodal which is reflective of categorical model. Overall, Exterior1 is not normally distributed or a good predictor in determining SalePrice which is reflected in the various plot, statistics and reasons provided. ### Does the predicted model go through the mean value of Y for each category group? The above Fit Plot line does appear to go through the mean value of each category but shows the distribution of values for each category based on the variability. The confidence limits but as mentioned above, reflects a slight correlation which is reflected in the slight slope of the line. The plot shows the distribution of values within each category with the largest amount of variability in the SalePrice of a house using the Exterior1 Category of category16 which is wood siding. Having the line go through the mean doesn’t really provide any value as the categories are listed in order of their assigned value. ### Is this good or bad / why or why not? I don’t think having the line adds any value and in fact, causes confusion and possible incorrect conclusions that there is some type of linear relationship when in fact there isn’t one. Additionally, it doesn’t provide any insight on which categories perform better than others or the volume of data in each. ### Categorical Value Conclusion While the Exterior1 category analysis is interesting to conduct, the results in being able to predict the SalePrice is low with the Adjusted R-Square of 0.0170. Additionally, the plots and charts provided additional insight into Cook’s D observations with few influential peaks, a heavy tailed Q-Q plot. No surprise, the does not appear to be a linear relationship between Exterior1 and SalePrice. Finally, the Fit Plot would not be my preferred method of making categorical assessment of data as they add very little value and can be misleading. ## Part A – Task 4 Analysis Of the three models described above, the best model is TotalSqFt as it has the highest Adjusted R-Square at .509, the errors across the values appear to be independent of the variables and thus considered to be homoscedastic. Cook’s D has only a few data points that are having an influential influence. These points along with the outliers need some further research but do not expect them to change the outcomes of the above analysis. Additionally, the F value is significantly greater than one. Again, we have conflicting information based on the plots and the R-Square values which needs to be further investigated. ## Part B – Task 5 Regression Model Using Two Variables Variables: The two variables are TotalFlrSF and GrLivArea. ### Equation: Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > \|t\| Variance Inflation Intercept 1 11688 3238.62266 3.61 0.0003 0 TotalFlrSF 1 185.03078 22.40381 8.26 <.0001 119.15477 GrLivArea 1 -71.69170 22.29838 -3.22 0.0013 119.15477 Root MSE 55886 R-Square 0.5109 Dependent Mean 180796 Adj R-Sq 0.5106 Coeff Var 30.9113 • Sale Price = \$11,688 + \$185.03078 x TotalFlrSF – \$71.69170 x GrLivArea For each unit increase in TotalFlrSF the SalePrice increases by \$185 yet a decrease of \$71.69 impacts SalePrice for each unit increase in GrLivArea. The above equation is different from the simple linear regression models above with the negative value each unit of GrLivArea has on SalePrice. This is not intuitive but because of the unusual choice in variables and the obvious overlap, it makes logical sense. Additionally, the Adjusted R-Square is .5106 again reflected the amount of variability that is accounted for in using these two variables. However, this Adjusted R-Square has only improved slighted from the simple linear regression model using TotalFlrSF which had an Adjusted R-Square of .5090. Again, this is not unexpected due to the overlap in the variables. By adding one unit of TotalFlrSF with all other factors remaining constant, we can see how the SalePrice changes. However, one unit of TotalFlrSF and GrLivArea are not the same and need to ensure that this is not confused. ### Goodness-of-Fit Regression Models – ODS Multiple Variables Regression Models – ODS Multiple Variables – TotalFlrArea and GrLivArea Based on the above, we have a small p-value reflecting that each variable is significant and thus, we cannot reject the null hypothesis. However, unlike the prior variables, we have a Variance inflation number of 119.15477 which is significantly outside of the acceptable range of 1 to 5. Thus, we have multi-linearity issues. We can see a lot of similarities with TotalFlSF with the fanning out pattern in the Predicted Value and Residual plot, the Q-Q plot again, deviates from the line and Cook’s D has only two influential points that exceeding 0.2 which is less than one. Again, the Adjusted R-Square is the largest out of all of the models analyzed with a value of 0.5106. Homoscedasticity looks for errors that aren’t independent causing the plot to be more linear looking or that flares out as x increases. Again, there is a clear clustered pattern here. Above, we can see compare TotalFlrSF and GrLivArea which, for the most part, appear to be identical. ### Better Fit As mentioned earlier, we have only accounted for .5106 of the variability leaving another .4894 unexplained. While this has increased from the prior models it is only a small increase. Thus, a significant amount of additional work is required to reduce this value. ### Regression Model Using Two Variables Conclusion Combining two continuous explanatory variables TotalFlSF and GrLivArea provided similar results to the simple linear regression model using TotalFlSF. This is not surprising, as GrLivArea is encapsulated in TotalFlSF as such the equation has a negative component to it with a small increase to the Adjusted R-Square value. The p-value reflects that each variable is significant and thus, we can reject the null hypothesis. However, the variance influence exceeds our normal range reflecting that we have multi-linearity issues which isn’t a surprise. Evaluating the plots based on the multiple variables is again, very similar to the analysis on TotalFlSF with the a fanning out in the Predicted Value and Residual plot. Thus, this we again have normality issues. ## Part B – Step 6 Using Three Variables ### Variables: The three variables are TotalFlrSF, GrLivArea and MiscVal. ### Equation: Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > \|t\| Variance Inflation Intercept 1 11105 3227.36083 3.44 0.0006 0 TotalFlrSF 1 186.43890 22.31325 8.36 <.0001 119.17355 GrLivArea 1 -72.39791 22.20695 -3.26 0.0011 119.15954 MiscVal 1 -9.14957 1.82008 -5.03 <.0001 1.00470 Root MSE 55886 R-Square 0.5151 Dependent Mean 180796 Adj R-Sq 0.5146 Coeff Var 30.7839 • Sales Price = \$11,105 + \$186.4389 x TotalFlrSF – \$72.39791 x GrLivArea – \$9.14957 x MiscVal For each unit increase in TotalFlrSF the SalePrice increases by \$186.44, a small increase of \$1.40 over the equation in part 5, and similarly a decrease of \$72.40 (a further decline from \$71.69 in part 5) but now there is another decrease which is \$9.15 per unit of MiscVal. The decrease in value in MiscVal on SalePrice is not as intuitive as TotalFlrSF and GrLivArea however due to the low volume and value of the MiscVal data points it is understandable there the variable would have little impact on SalePrice. Regression Models – ODS Multiple Variables TotalFlrSF, GrLivArea and MiscVal. ### Goodness-of-Fit We can see a lot of similarities with part 5, the p-values are small reflecting that each variable is significant and thus, we reject the null hypothesis. The Variance Inflation for TotalFlSF is 119.17355, GrLivArea is 119.159 which exceed the upper boundaries of our 1-5 range and the MiscValue is close to the bottom of our acceptable range with a value of 1.00470. We again see the Predicted Value to the Residual having a fanning out pattern which reflects that the variance is increasing. Cook’s D plot has changed in our axis measures but still has only a few influential points. The Q-Q plot is again, deviating from the line and is heavy tailed. The Predicted Value to Sales Price shows a cluster of values. Thus, we still have the same normality issues. Adjusted R-Square increased from .5106 to .5146 again reflected the amount of variability that is accounted for in using these two variables. However, this is a very small increase – 0.0040 over the Adjusted R-Value in question 5 and the value of adding the MiscVal variable. The Adjusted R-Square has only improved slighted from the simple linear regression model using TotalFlrSF and the multiple variables between TotalFlSF and GrLivArea. Regression Models – Residuals Multiple Variables TotalFlrSF, GrLivArea and MiscVal. The Residual plots for each variable provides more of a view on how they alter depending on the variables over the results in part 5 but still do not appear to be random with distinct patterns to each variable. However, we can see how MiscVal errors do not provide any easy to see, linear relationship and has only a few values. Thus, we still have normality issues. Changes & Better Fit Overall, the additional variable – MiscVal has not made any real impact in predicting SalePrice. While this is no surprise, based on its low correlation value. While our Adjusted R-Value did increase, it was not significant with a small .0040 increase thus showing us that adding additional values does not necessarily improve the results or assist us in creating a better fitting equation and explaining the variability. While Adjusted R-Square plays an increased importance on the criteria in comparing the models primarily due to not wanting to make incorrect decisions by simply adding variables, the plots and charts are helpful in understand the relationships and how they impact the model. ### Using Three Variables Conclusion Adding additional variables do not necessarily improve results. By adding a low correlated variable – MiscVal the other variables within the equation received minor adjustments. While each of the variables has a small p-value we can reject the null hypothesis. Similar to the prior analysis, there are not a lot of changes with the exception being the variance influence. The review of the plots reflects similar patterns with the exception of the residual plot where the MiscValue has a different pattern but still not a pattern of randomness which we hope for. However, the analysis remained relatively the same as to having only two variables as in part 5. Additionally, adding the MiscVal variable had a minor effect in explaining the variability and did not have any significant impact on SalePrice. ## Conclusion By observing the various models and the effects of adding and reviewing various variables we can easily grasp the changes through the various plots, equations and summary statistics. Continuous variables have a much different look and resulting plots over continuous variables. Additionally, evaluating the Adjusted R-Square and F-Value provides insight into how the model is performing even when the impact is difficult to ascertain through the plots. The model may not be appropriately specified as all variables seem to be equal which may not be the case. Additionally, we could be adding new variables without considering their association with other variables as shown between TotalFlSF and GrLivArea – removing GrLivArea would have significant effects on TotalFlSF. Next steps in the modeling process would be to investigate a few of the influential points that have been mentioned in several parts such as Cook’s D and focus on determining whether the Adjusted R-Score can be enhanced through the evaluation of different variables and try to account for more of the variability as we all of the models have not exceeded 51.46%. Finally, transformation of the data so the conflicts between the R-squares and ODS outputs are decreased would be really helpful in creating a useful model we are confident in.	6,943	29,050	{"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}	3.453125	3	CC-MAIN-2020-50	longest	en	0.898701
https://webwork.maa.org/moodle/mod/forum/discuss.php?d=3715&parent=10046	1,713,325,532,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00211.warc.gz	562,615,480	14,335	## PREP 2015 Question Authoring - Archived ### Sage cell server problems by Gary Church - Number of replies: 0 Hello, I've been trying to get the sage cell server example on the wiki to work on my schools WebWork server but am having a small problem. Everything seems fine except it won't let me enter in any text in the sage cell field; I try clicking the field and nothing happens. For your reference, here's the code: DOCUMENT(); "PGstandard.pl", "MathObjects.pl", "sage.pl", ); ############################################### ## ## pg initializations and regular WeBWorK code $a = random(2,5,1);$ansList = List("(-cos(pi$a)/$a + 1/$a)");$SageCode = <<SAGE_CODE; Area = integrate(sin($ax),x,0,pi) record_answer((Area)) # leave out if you return no answer SAGE_CODE Sage( SageCode=>$SageCode, ); ## Lower WeBWorK text ## ## Problem display following the Sage cell ## Context()->texStrings; BEGIN_TEXT Determine the definite integral of $$\sin({a}x)$$ from $$a=0$$ to $$b=\pi$$. END_TEXT Context()->normalStrings; $showPartialCorrectAnswers = 1; NAMED_ANS( sageAnswer =>$ansList->cmp ); # Leave out if no Sage answer.	306	1,135	{"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}	2.515625	3	CC-MAIN-2024-18	latest	en	0.674044
https://www.nuprl.org/wip/Mathematics/reals/rroot-exists1.html	1,695,815,206,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00730.warc.gz	1,010,900,461	2,534	Nuprl Lemma : rroot-exists1 `∀i:{2...}. ∀x:{x:ℝ\| (↑isEven(i)) `` (r0 ≤ x)} .` ` ∃q:{q:ℕ ⟶ ℝ\| ` ` (∀n,m:ℕ. (((r0 ≤ (q n)) ∧ (r0 ≤ (q m))) ∨ (((q n) ≤ r0) ∧ ((q m) ≤ r0))))` ` ∧ ((↑isEven(i)) `` (∀m:ℕ. (r0 ≤ (q m))))} ` ` lim n→∞.q n^i = x` Proof Definitions occuring in Statement : converges-to: `lim n→∞.x[n] = y` rleq: `x ≤ y` rnexp: `x^k1` int-to-real: `r(n)` real: `ℝ` isEven: `isEven(n)` int_upper: `{i...}` nat: `ℕ` assert: `↑b` all: `∀x:A. B[x]` exists: `∃x:A. B[x]` implies: `P `` Q` or: `P ∨ Q` and: `P ∧ Q` set: `{x:A\| B[x]} ` apply: `f a` function: `x:A ⟶ B[x]` natural_number: `\$n` Definitions unfolded in proof : all: `∀x:A. B[x]` member: `t ∈ T` exists: `∃x:A. B[x]` and: `P ∧ Q` cand: `A c∧ B` or: `P ∨ Q` uall: `∀[x:A]. B[x]` so_lambda: `λ2x.t[x]` prop: `ℙ` so_apply: `x[s]` implies: `P `` Q` int_upper: `{i...}` subtype_rel: `A ⊆r B` nat: `ℕ` le: `A ≤ B` less_than': `less_than'(a;b)` false: `False` not: `¬A` Lemmas referenced : rroot-exists-part1 all_wf nat_wf or_wf rleq_wf int-to-real_wf assert_wf isEven_wf converges-to_wf rnexp_wf int_upper_subtype_nat false_wf le_wf real_wf int_upper_wf Rules used in proof : cut lemma_by_obid sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation hypothesis sqequalHypSubstitution dependent_functionElimination thin hypothesisEquality productElimination dependent_pairFormation sqequalRule independent_pairFormation dependent_set_memberEquality productEquality isectElimination lambdaEquality because_Cache natural_numberEquality applyEquality functionEquality setElimination rename setEquality Latex: \mforall{}i:\{2...\}. \mforall{}x:\{x:\mBbbR{}\| (\muparrow{}isEven(i)) {}\mRightarrow{} (r0 \mleq{} x)\} . \mexists{}q:\{q:\mBbbN{} {}\mrightarrow{} \mBbbR{}\| (\mforall{}n,m:\mBbbN{}. (((r0 \mleq{} (q n)) \mwedge{} (r0 \mleq{} (q m))) \mvee{} (((q n) \mleq{} r0) \mwedge{} ((q m) \mleq{} r0)))) \mwedge{} ((\muparrow{}isEven(i)) {}\mRightarrow{} (\mforall{}m:\mBbbN{}. (r0 \mleq{} (q m))))\} lim n\mrightarrow{}\minfty{}.q n\^{}i = x Date html generated: 2016_05_18-AM-09_33_08 Last ObjectModification: 2015_12_27-PM-11_18_47 Theory : reals Home Index	963	2,215	{"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-2023-40	latest	en	0.149651
https://recipes4day.com/whats-250-grams-in-pounds/	1,660,390,462,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00462.warc.gz	445,148,990	27,617	# What’s 250 grams in pounds? Sharing is caring! What is 250g in American? Weight Grams Pounds/ounces 225g 8oz 250g 9oz 300g 10oz 350g 12oz How many grams means 1 pound? How many grams in a pound 1 pound is equal to 453.59237 grams, which is the conversion factor from pounds to grams. How much is 250g flour in cups? White flour – plain, all-purpose, self-raising, spelt WHITE FLOUR – GRAMS TO CUPS Grams Cups 250g 1½ cups + 1 tbsp 300g 1¾ cups + 2 tbsp 400g 2½ cups Is 200g half a pound? Converting between grams and pounds Grams Pounds Pounds and ounces 200 grams 0.441 lb 0 lb, 7.05 oz 300 grams 0.661 lb 0 lb, 10.58 oz 400 grams 0.882 lb 0 lb, 14.11 oz 500 grams 1.102 lb 1 lb, 1.64 oz ## What’s 250 grams in pounds? – Related Asked Question ### Is 200g the same as 8 oz? Here is a handy conversion chart for ounces to grams weight conversions for cooking measurements. Basic ounces to grams weight conversions. 1/2 oz 15g 5 oz 140g 6 oz 170g 7 oz 200g 8 oz 225g ### How much is a pint of flour in grams? How Many Grams are in a Pint? Volume in Pints: Weight in Grams of: Water All Purpose Flour 2/3 pt 315.45 g 166.87 g 3/4 pt 354.88 g 187.73 g 1 pt 473.18 g 250.31 g ### How many grams is a 5lb bag of flour? How many grams is a 5lb bag of flour? How many grams of all purpose flour (APF) are in 1 pound? The answer is: The change of 1 lb ( pound ) unit in a all purpose flour (APF) measure equals = into 453.59 g ( gram ) as per the equivalent measure and for the same all purpose flour (APF) type. ### How many grams is a Pavan? 1 Pavan gold is 8 grams. Gold rate of 1 Pavan gold is different for 22 Karat and 24 Karat. ### What weight is a pound? pound, unit of avoirdupois weight, equal to 16 ounces, 7,000 grains, or 0.45359237 kg, and of troy and apothecaries’ weight, equal to 12 ounces, 5,760 grains, or 0.3732417216 kg. The Roman ancestor of the modern pound, the libra, is the source of the abbreviation lb. ### Is 250g flour the same as 250ml? Our conversion chart will also be handy when it comes to those recipes where the originator has not provided conversions. The only quantity conversion chart you will ever need. Ingredient Quantity (volume) Quantity (weight) Ice cream 250ml (1 cup) 195g Milk 250ml (1 cup) 250g Sour cream 250ml (1 cup) 235g Yoghurt 250ml (1 cup) 250g ### How many cups is 225g of flour? Dried ingredient measurements Cups Grams Ounces 1 tsp 5g 0.17 oz 1 tbsp 15g 0.53 oz 1 cup flour 150g 5.3 oz 1 cup caster sugar 225g 7.9 oz ### How many cups is 225 grams of flour? Cups To Grams Conversions (Metric) Cup Grams 2/3 cup 150 grams 3/4 cup 170 grams 7/8 cup 200 grams 1 cup 225 grams ### What weight is 200 grams? What common items weigh 200 grams? Many items can amount to 200 grams, such as a roll of nickels and a cup of granulated white sugar. Other items of that weight include three C-cell batteries and an adult hamster. However, there are plenty of other everyday things that all weigh the same amount. ### How much is 100 grams in weight? How Much is 100 Grams? One hundred grams is about 0.22 pounds or 0.1 kilograms. ### How much is 50g in weight? 50 grams is equal to 0.05 kilograms or 0.11 pounds. ### What is 240 grams in cups? Baking Conversion Table U.S. Metric 1 tablespoon of flour between 8 and 9 grams 1 cup 240 grams 1/2 cup 120 grams 1 tablespoon 15 grams ### How many table spoons is 200 grams? How much is 200 grams in tablespoons? 200 Grams of All Purpose Flour = 22.07 Tablespoons 66.21 Teaspoons 1.38 U.S. Cups 1.15 Imperial Cups ### What is 225g in Litres? Gram to Liter Conversion Table Weight in Grams: Volume in Liters of: Water Granulated Sugar 200 g 0.2 l 0.285714 l 210 g 0.21 l 0.3 l 220 g 0.22 l 0.314286 l ### How much is 1 cup of flour in grams? Common Measurement Conversions for Baking 1 cup flour 120 grams 4.2 oz 1 cup whole wheat flour 130 grams 4.6 oz 1 cup almond flour 112 grams 3.9 oz 1 cup whole milk 240 grams 8.6 oz 1 cup sour cream 240 grams 8.6 oz ### How many pints is 200 grams? Gram to Pint Conversion Table Weight in Grams: Volume in Pints of: Water Granulated Sugar 200 g 0.422675 pt 0.603822 pt 300 g 0.634013 pt 0.905733 pt 400 g 0.845351 pt 1.2076 pt ### What weight is a cup of flour? 1 cup of flour weighs 125 grams. ### How many grams is 2 cups flour? Cups Grams Ounces 1/4 cup 34 g 1.2 oz 1/3 cup 45 g 1.6 oz 1/2 cup 68 g 2.4 oz 1 cup 136 g 4.8 oz ### What is half a pound of flour? How Many Cups in One Pound of Flour? Pounds of Flour Cups (US) 1/2 lb 2.14 cups 1 lb 4.28 cups 2 lb 8.56 cups 5 lb 21.4 cups ### What is 1lb of butter? Pound to Stick Of Butter Conversion Table Pounds Sticks Of Butter 1 lb 4 2 lb 8 3 lb 12 4 lb 16 ### How do you convert grams to Pavan? 1 Pavan means 8 Grams. ### What is the price of 2 Pavan gold in Kerala? 1 Pavan means 8 Grams. ### How many grams is a Savaran? The pound sovereign, also spelt pawan savaran, pavan, or even paun is a unit of mass that is commonly encountered in India’s gold industry. While it is technically equal to 7.98805 gms, it is usually rounded off to 8 grams. Also, The weight of a gold Sovereign is 7.98 g, with a pure gold content of 7.32240 g. ### Is 1 lb the same as 1 kg? Pound is primarily used in the USA and the UK. One kilogram is equal to 2.204 pounds. One pound is equal to 0.453 kg. A kilogram is a unit solely for the measurement of mass. ### How much weight can you lose in a month? So what is the magic number to lose weight and keep it off? According to the Centers for Disease Control and Prevention (CDC) , it’s 1 to 2 pounds per week. That means, on average, that aiming for 4 to 8 pounds of weight loss per month is a healthy goal. ### Are ounces bigger than grams? If you’re wondering how an ounce compares to a gram, it turns out that 1 ounce is a lot more mass than 1 gram. In fact, 1 ounce is approximately equal to 28.35 grams. ### Is 250g the same as 1 cup? With these handy conversion charts, you’re able to easily convert grams to cups. Here is your one-stop shop on the different metrics used in cooking but most importantly how many grams are in a cup. Metric Cups and Spoons. Cups Grams 1 cup 250g 3/4 cup 190g 2/3 cup 170g 1/2 cup 125g ### How do you convert 250ml to grams? How to convert mL to grams. Although the milliliter is a unit of volume and gram a unit of weight, it is possible to convert between the two, providing you know the density of your ingredient. To convert mL to grams, you multiply the volume (in mL) by the density (in g/mL). ### How many grams of flour does a 250ml cup hold? The answer is: The change of 1 cup 250 mL ( Metric cup ) unit in a teff flour measure equals = into 121.52 g ( gram ) as per the equivalent measure and for the same teff flour type. ### How can I weigh 225g of flour without scales? HOW DO I MEASURE FLOUR WITHOUT A SCALE? 1. Use a spoon to fluff up the flour within the container. 2. Use a spoon to scoop the flour into the measuring cup. 3. Use a knife or other straight edged utensil to level the flour across the measuring cup. ### How many cups is 225 grams of sugar? SUGARS &amp, LIQUID SWEETENERS Granulated Sugar Light Brown Sugar Dark Brown Sugar 1 cup = 200 g 1⁄3 cup = 67 g ¼ cup = 50 g 2 Tbsp = 25 g Superfine sugar Caster sugar 1 cup = 225 g ### What is 225g of butter in cups? 1/2 lb butter is 225 g, or one cup. 1 stick of butter is 1/2 cup or 8 tbsp. ### How do you make 225g self raising flour? It’s really simple to make and only takes about two seconds. For each cup of flour, whisk together with 1 ½ teaspoons of baking powder and ¼ teaspoon of salt. Make sure to whisk all of these ingredients together well so that the baking powder and salt are both evenly distributed within the flour. ### What is 115g in cups? Caster Sugar Cups Grams Ounces 1 cup 220g 7 oz ½ cup 115g 3 ¾ oz ⅓ cup 80g 2 ½ oz ¼ cup 60g 2 oz ### How much is 300 grams in weight? 300 grams is equal to 10.58 ounces or 0.66 pounds. ### How do you calibrate a 200 gram scale? 200 Paper clips Paper clips are one of the easiest ways to estimate weight or calibrate a scale because they all weigh an exact, standard amount. A regularly-sized standard paper clip should weigh exactly 1 gram. This means that 200 paper clips together equals exactly 200 grams. ### How many grams is 2.5 teaspoons? Teaspoons and grams for sugar (granulated) Teaspoons to grams Teaspoons to grams 1 teaspoon = 4.2g 6 teaspoons = 25.2g 2 teaspoons = 8.4g 7 teaspoons = 29.3g 3 teaspoons = 12.6g 8 teaspoons = 33.5g 4 teaspoons = 16.7g 9 teaspoons = 37.7g Sharing is caring!	2,637	8,596	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2022-33	latest	en	0.86407
https://srs.amsi.org.au/student-blog/i-didnt-like-maths/	1,718,953,718,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00758.warc.gz	477,455,567	22,495	I didn’t like maths, but just spent six weeks of my uni holidays doing mathematical research (specifically knot theory). A reflection on why some of us end up being passionate about maths and a thanks to someone who showed me how and why we love maths. I certainly never hated maths. My earliest mathematical memory comes from when I was probably about four. My family and I had just moved to Brisbane, and we were staying with my grandparents. My grandparents would ritualistically watch the 7pm news report on the ABC and as most kids do, I was always following grandparents about. So, there I was sat on grandpa’s knee and the report cuts to Alan Kohler who is giving the finance report. Of course, as a four-year-old I didn’t have the first clue about what finance was but suddenly were numbers popping up on the screen with a blue arrow for up and a red arrow for down. Kohler was saying something about the iron ore price being up (great for Australia’s mining boom at the time) and he used a graph of the price to explain his point. While the numbers were okay, the graph was something I was kind of mesmerized by and even now I can’t tell you why. The only explanation I can give is that seeing that graph somehow connected with some mathematical inkling I had but just didn’t realise yet. I won’t lie to you and say that from that moment on I developed some prodigious Gauss like talent for maths. For many years I just didn’t like maths. In both my year 7 and 8 school report cards my maths teachers described me as a well humored student, teacher speak for doesn’t listen or do any work in class (which was reflected in my grades). But luckily when I was about 16, I became interested in Stephen Hawking, it was probably because there was a movie about him which had just come out at the time. So, I picked up a copy of his famous book titled, “A Brief History of Time”, and I really enjoyed it. I reread the relatively section about ten times and was fascinated by the fact that the universe was not so simple. On a trip back to Brisbane (we had moved back to Melbourne in the intervening years) I expressed my newfound interest in physics to my grandpa, who is an accomplished physicist in his own right. He was pleased and showed me some papers he had recently published; these were covered in maths, and he explained how he turned physical problems into mathematical equations. I realised if I wanted to be a physicist, I would have to learn maths. Now while my interest in physics waned, I soon came to love maths. It’s somewhat to clichéd to say this love is motivated by visions of mathematical beauty, but clichés are sometimes true. Ultimately, maths and its various abstract concepts, hard to crack problems and beautiful proofs, ignited a passion, which I saw reflected in my grandpa’s explanation of his physics. In my experience you can’t teach someone to like maths, but you can show them how to (or maybe why to). So, thanks grandpa for showing me maths. Oscar Eden Monash University	645	3,016	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2024-26	latest	en	0.990659
https://questions.examside.com/past-years/jee/question/int010pi-left-sin-x-rightdx-is-jee-main-2002-marks-4-fotnk3xkayiytvbr.htm	1,716,161,900,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00296.warc.gz	434,359,637	49,391	1 AIEEE 2002 +4 -1 $$\int\limits_0^{10\pi } {\left\| {\sin x} \right\|dx}$$ is A $$20$$ B $$8$$ C $$10$$ D $$18$$ 2 AIEEE 2002 +4 -1 $$\int\limits_0^2 {\left[ {{x^2}} \right]dx}$$ is A $$2 - \sqrt 2$$ B $$2 + \sqrt 2$$ C $$\,\sqrt 2 - 1$$ D $$- \sqrt 2 - \sqrt 3 + 5$$ 3 AIEEE 2002 +4 -1 $$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$$ is A $${{{\pi ^2}} \over 4}$$ B $${{\pi ^2}}$$ C zero D $${\pi \over 2}$$ 4 AIEEE 2002 +4 -1 If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx}$$ is A $$3/2$$ B $$1$$ C $$5/4$$ D $$-3/4$$ EXAM MAP Medical NEET	385	722	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2024-22	latest	en	0.510646
https://detailedpedia.com/wiki-Dzongkha_numerals	1,713,573,217,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00750.warc.gz	179,023,105	9,808	# Dzongkha numerals Dzongkha, the national language of Bhutan, has two numeral systems, one vigesimal (base 20), and a modern decimal system. The vigesimal system remains in robust use. Ten is an auxiliary base: the -teens are formed with ten and the numerals 1–9. Ex. cu_ci ## Vigesimal Hindu-Arabic numerals Dzongkha numerals Spelling Romanisation 1 ཆཱི chiː 2 ཉྱཱི ˈɲyiː 3 སུམ sum 4 ཞི ʑhi 5 ˈŋga 6 ཌཱུ ɖʱuː 7 དྱཱུན dyûn 8 གཻ ɡeː 9 གཱུ ɡuː 10 ༡༠ ཅུཐཱམ cu-tʰãm* 11 ༡༡ ཅུཅི cu-ci 12 ༡༢ ཅུཉི cu-ɲi 13 ༡༣ ཅུསུམ cu-sum 14 ༡༤ ཅུཞི cu-ʑi 15 ༡༥ ཅེངཱ ce-ŋa 16 ༡༦ ཅུཌུ cu-ɖu 17 ༡༧ ཅུཔྡྱ cup-dỹ 18 ༡༨ ཅོཔྒེ cop-ɡe 19 ༡༩ ཅྱགུ cy-ɡu 20 ༢༠ ཁེཆཱི kʰe ciː When it appears on its own, 'ten' is usually said cu-tʰãm 'a full ten'. In combinations it is simply cu. Multiples of 20 are formed from kʰe. Intermediate multiples of ten are formed with pɟʱe-da 'half to': 30 kʰe pɟʱe-da ˈɲiː (a half to two score) 40 kʰe ˈɲiː (two score) 50 kʰe pɟʱe-da sum (a half to three score) 100 kʰe ˈŋa (five score) 200 kʰe cutʰãm (ten score) 300 kʰe ceŋa (fifteen score) 400 (20²) ɲiɕu is the next unit: ɲiɕuciː 400, ɲiɕuɲi 800, etc. Higher powers are 8000 (20³) kʰecʰe ('a ɡreat score') and jãːcʰe 160,000 (20⁴). ## Decimal The decimal system is the same up to 19. Then decades, however, are formed as unit–ten, as in Chinese, and the hundreds similarly. 20 is reported to be ɲiɕu, the same as vigesimal numeral 400; this may be lexical interference for the expected ɲi-cu. (In any case, there is no ambiguity, because as 400 it is obligatorily ɲiɕuciː 'one 400'.) Several of the decades have an epenthetic -p-, perhaps by analogy with 18 and 19, where the -p- presumably reflects a historical *cup 'ten': sum-cu 30, ʑi-p-cu 40, ˈŋa-p-cu 50, ɟa-tʰampa or cik-ɟa 100 (a 'full hundred' or 'one hundred'), ɲi-ɟa 200, sum-ɟa 300, ʑi-p-ɟa 400, etc. This page was last updated at 2023-12-13 05:40 UTC. . View original page. All our content comes from Wikipedia and under the Creative Commons Attribution-ShareAlike License. Top If mathematical, chemical, physical and other formulas are not displayed correctly on this page, please useFirefox or Safari	1,028	2,133	{"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}	3.203125	3	CC-MAIN-2024-18	latest	en	0.733075
https://www.coursehero.com/file/6529918/Ascending-and-Descending-Order/	1,493,197,762,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00440-ip-10-145-167-34.ec2.internal.warc.gz	901,604,933	25,072	Ascending and Descending Order # Ascending and Descending Order - if asked) powers of one... This preview shows page 1. Sign up to view the full content. Ascending and Descending Order When working with polynomials that involve only one variable, the general practice is to write them so that the exponents on the variable decrease from left to right. The polynomial is then said to be written in descending order When a polynomial in one variable is written so that the exponents increase from left to right, it is referred to as being written in ascending order Example 2 Rewrite the following polynomial in descending powers of x y 4 + 12 – 15 x 2 + 13 x 3 y + 17 xy 2 13 x 3 y – 15 x 2 + 17 xy 2 + 4 y 4 + 12 To add two or more polynomials, add like terms and arrange the answer in descending (or ascending This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: if asked) powers of one variable. Example 3 Find the following sum: • ( x 2 + x 3 – 3 x ) + (4 – 5 x 2 + 3 x 3 ) + (10 – 8 x 2 – 5 x ) • ( x 3 + 3 x 3 ) + ( x 2 – 5 x 2 – 8 x 2 ) + (–3 x – 5 x ) + (4 + 10) • = 4 x 3 – 12 x 2 – 8 x + 14 This problem can also be added vertically. First rewrite each polynomial in descending order, one above the other, placing like terms in the same column. To subtract one polynomial from another, add its opposite.... View Full Document ## This note was uploaded on 11/10/2011 for the course MATH 1310 taught by Professor Staff during the Fall '07 term at Texas State. Ask a homework question - tutors are online	552	1,592	{"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}	4.25	4	CC-MAIN-2017-17	longest	en	0.379972
http://www.indiana.edu/~hmathmod/html/trash5/trash5.html	1,508,317,861,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00886.warc.gz	466,408,727	5,311	Talkin' Trash Leo Paveglio, Barb Steinbrunner, Rich Van Gilst, Carolyn Wahl Class (Materials): General Math (graph paper, eyeball estimation) Algebra I (graph paper, point/slope form) Algebra II (calculator/computer, linear regression) Goals: Students should learn to fit a linear equation to data, use the equation to predict future outcomes and compare these to Monroe County's actual predictions. Time Required: One class period with unfinished work assigned as homework. Extensions may also be assigned as homework. Background: Students should be able to write the equation of a line of best fit or a line of regression (Algebra II), graph on a coordinate plane or calculator (Algebra II) and interpret the graph. Algebra II students should have some experience with arithmetic sums. Setting: This project addresses the growing waste problem by trying to predict the number of tons of waste Monroe County, Indiana, will produce in the future. Listed are Monroe County's actual figures for the years 1990-1994 with their projections for the years 1995-1999. The United States is the largest waste producer in the world with most of the waste being dumped into the ground. As with most states, Indiana's landfills are filling at an alarming rate. To make adequate and financially feasible plans for the future, the County Solid Waste Management needs a way to predict the amount of waste that will need placement. Listed are the tons of waste generated for the years 1990-1994 (and the predictions for 1995-1999). These numbers include solid waste and material that may be recycled. Tons of Waste in Monroe County, Indiana 1990 132,956 1991 135,686 1992 136,102 1993 137,793 1994 139,417 - - - - - - - - - - - - - - - - - - - - - - - - 1995 140,739 1996 142,158 1997 143,430 1998 144,521 • 145,920 Algebra II Problem: The United States is the largest waste producer in the world with most of the waste being dumped into the ground. As with most states, Indiana's landfills are filling at an alarming rate. To make adequate and financially feasible plans for the future, the County Solid Waste Management needs a way to predict the amount of waste that will need placement. Listed are the tons of waste generated for the years 1990-1994. These numbers include solid waste and material that may be recycled. Tons of Waste in Monroe County, Indiana 1990 132,956 1991 135,686 1992 136,102 1993 137,793 1994 139,417 1. Using the above data and the appropriate technology, create a linear model that will predict the total waste in Monroe County for the years 1996 to 2000. State the linear equation and explain the meaning of each of your constants and variables. 2. The Monroe County Waste Management has made the following predictions: Future Tons of Waste in Monroe County, Indiana 1995 140,739 1996 142,158 1997 143,430 1998 144,521 1999 145,920 Using the equation from problem 1, compare your predictions to these. 3. Look at your scatterplot from problem 1. Is there any point(s) which does(do) not appear to fit the same pattern as the rest? If so, eliminate it (them) and create a linear model which fits the remaining points. 4. Compare your two models. How are they different? How many tons of waste does each predict for 2005, 2010 and 2020? Which do you think is better? Why? Select the better of your two models in problem 4 to use for questions 5 - 7. 5. At the end of 1994, Monroe County landfill still had room for three million tons of waste. In what year will the landfill become full? 6. In 1994, 46,736 tons of the total waste was recycled and not put into the landfill. If recycling patterns remain the same, how long will it take for the landfill to be at its capacity? 7. At what rate must the people of Monroe County recycle in order for the current landfill to be usable until the year 2050? Do you think that it would be realistic to recycle at this rate? Algebra I Problem: The United States is the largest waste producer in the world with most of the waste being dumped into the ground. As with most states, Indiana's landfills are filling at an alarming rate. To make adequate and financially feasible plans for the future, the County Solid Waste Management needs a way to predict the amount of waste that will need placement. Listed are the tons of waste generated for the years 1990-1994. These numbers include solid waste and material that may be recycled. Tons of Waste in Monroe County, Indiana 1990 132,956 1991 135,686 1992 136,102 1993 137,793 1994 139,417 1. Plot the information given above and draw the line of best fit. Using two points, write the equation of that line. Please show your work and graph. 2. Using the graph and line from problem 1, predict the number of tons of waste for the years 1995, 1997 and 1999. Please show your work. 3. In 1994, 46,736 tons of the waste was recycled. If this rate remains the same, how many tons of waste will be recycled in 1995? 4. If 15% of the waste is non-burnable and one incinerator burns 350,000 tons of waste per year, how many incinerators will be needed in 1997? Explain your answer in sentences. 5. Assume 15% of the waste can be recycled. If the remainder is deposited in a landfill, what should its capacity be as you plan for the year 1997? 6. Describe the assumptions you have made as you did these problems. How could these assumptions change your results? General Math Problem: The United States is the largest waste producer in the world with most of the waste being dumped into the ground. As with most states, Indiana's landfills are filling at an alarming rate. To make adequate and financially feasible plans for the future, the County Solid Waste Management needs a way to predict the amount of waste that will need placement. Listed are the tons of waste generated for the years 1990-1994. These numbers include solid waste and material that may be recycled. Tons of Waste in Monroe County, Indiana 1990 132,956 1991 135,686 1992 136,102 1993 137,793 1994 139,417 1. Plot the information given above and draw the line of best fit. Explain any patterns or trends 2. Using the graph and line from problem 1, predict the number of tons of waste for 1995, 1996, and 1997. 3. One-third of the waste was recycled in 1994. If this trend continues, how many tons of waste will be dumped into the landfill in 1996? Please show all work. 4. What could influence or alter the amount of waste produced on a yearly basis? Sample Solutions: General Math: Problem one can be graphed on a coordinate plane. Some patterns students may observe are a positive slope of the line and an increasing amount of waste. The amount of waste in tons for 95, 96 and 97 will vary, but should increase by approximately 1000-1500 tons per year. The amount of waste depends on the student's assumptions, such as the amount of waste recycled, public awareness or the manufacturer's response to the recycling effort. Algebra I: Answers will vary for problem one. One solution uses the data points (93, 137 793) and (94, 139 417) resulting in the equation: Tons = 1624(year) - 13,239 (Note: x represents the last two digits of the year and y the tons of waste.) This formula yields the following for problem two: Year Tons 95 141,041 97 144,289 99 147,537 For problem 3, using a proportion yields 47,280 tons of recycled waste. For problem 4, one incinerator will be more than adequate. Problem 5 is deliberately vague. Students= answers will depend on their assumptions. These may include assumptions such as the county opening a new landfill or more/less waste being recycled. Algebra II: Using a line of regression program for problem one, the equation is: y = 1503x + 133385. (Note: x represents 0 for 1990, 1 for 1991, etc.) Answers will vary for problems 2 - 7. By eliminating the second point and using the remaining four, problem 3 yields the new equation: y = 1617x + 132929. Assuming this is the better model, answers for problems 5-7 are 2013, 2022, and 70.9%, respectively. Evaluation: The evaluation method will depend on the course. Considerations may be given to the clear statement of the problem, assumptions/constraints, model design, analysis (strengths/weaknesses of the model), and conclusions. Extensions: One extension may be to collect data from your county or state solid waste management and compare models. Also, students could explore the internet for other sources of data. Following are two other extension ideas: General Math: Give the students the Monroe County predictions for 1995-1999 and have them compare their results to those of Monroe County. Algebra II: Using the points from problem 3, create an exponential model for the total waste. What does this model predict for 2005, 2010, and 2020? Compare the model with your linear model, answer questions 5 - 7 with this new model. Bibliography: Swetz & Hartzler, 1991. Mathematical Modeling in the Secondary School Curriculum. "Time to Waste." National Council of Teachers of Mathematics. Resource: Monroe County Solid Waste Management, Bloomington, Indiana. Funded in part by the National Science Foundation and Indiana University 1995 Funded in part by the National Science Foundation and Indiana University 1995	2,221	9,273	{"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}	3.65625	4	CC-MAIN-2017-43	latest	en	0.886914
https://www.pinterest.com/calcasinsurance/pi-day-activities/	1,619,160,461,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618039601956.95/warc/CC-MAIN-20210423041014-20210423071014-00098.warc.gz	1,041,367,921	64,884	# Pi Day Activities ## Collection by California Casualty • Last updated 5 weeks ago 37 Pins • 1.47k Followers Students will love discovering what pi is all about with this circular object scavenger hunt! Not only is this activity perfect for pi day, but it is also a great way for your students to practice measuring diameter and circumference. Click here to save money and purchase this product as a part of ... Looking for a fun way to get your kids excited about the number pi? This super simple Pi Day card game takes minutes to set up and is fun for all ages! This Friday (3/14) is Pi Day! Pi is a mathematical constant referring to the ratio of a circle's circumference to its diameter. It's an irrational number, which means that it goes on forever and never shows a repeating pattern. Every year it's celebrated on 3/14 because the first three digits of Pi are 3,1, and 4. Now if you're looking for an easy way to celebrate Pi day, we've got a great collection of Pi Day Inspired Activities. Today I'm sharing these nifty Pi Day Bracelets! What makes… This fun and easy Pi Day Art Project will get your creativity flowing, and it's a fun way to build enthusiasm around Pi Day 2015 \| TinkerLab.com 15 Fun Pi Day Activities For Kids on March 14 every year. #piday #pidaymath #pidayforkids Collaboration for Pi Day is fun with this poster for your math teaching lessons. Teachers will receive 3 size options to use for this Pi Day writing activity poster. The Albert Einstein Pi Day image is original and you won't find it anywhere else. Albert Einstein appears to be very relaxed in his Pi... Pi Day Crack the Code gives students practice solving for radius, diameter, circumference and area of a circle with 2 Crack the Code puzzles. These self-correcting puzzles offer built-in practice with rounding decimals. Teacher notes and answer keys are included.Also included, an interactive noteboo... The text focuses not on the meaning of pi, but on the history and celebrations related to Pi Day. Pi Day Read and Respond can be used in three ways: 1.Assign student slides using Google Classroom. Students read text digitally and respond by typing answers on the response slides. You will not nee... Students practice finding circumference, area, diameter, and radius given one circle measurement in this Pi Day coloring activity. 20 practice problems for students to calculate a circle measurement and then color the pi coloring sheet according to their answer. Included in this Product20 problems ... This is a great activity where students graph points on a coordinate plane and it creates a picture of the Pi Symbol with Happy Pi Day at the bottom! The clever title of "Stop being so Irrational!" gives them a clue to what the picture will be. The points include ordered pairs from all four quadrant... Celebrate National Pi Day with a Scavenger Hunt titled "In Search of Pi for my Pie" where students solve geometry problems focused on the circumference and area of a circle to find answers in terms of pi. As students progress through the Scavenger Hunt, they collect answers in terms of pi on each o... Sir Cumference And The Dragon Of Pi Psst! Looking for virtual Pi Day ideas this year? Check out the post Pi Day Virtual Activities: 3 (+ a little more) Easy Ideas for the Classroom or Online Learning Call me a nerd if you’d like, but I’ll unabashedly claim Pi Day as my absolute favorite holiday as a teacher! Obviously, I love the … Continue reading Pi Day Activities in 5th Grade Pi Day Middle School Math Activities for Grades 6, 7, 8. Engage and review math skills integers, order of operations, Finding Circumference of a Circle, Finding Area of a Circle, Measurement with inches, Line Plots, Radius, Diameter, Parts of a Circle, Geometry, Graphing, Data, Statistics, Bar Graphs, and the number Pi. 0:04 This activity is great for pi day or your circles unit! Students always love to color and it is self-checking. Click to view the blog post with more Pi Day and circle investigation activities. 0:14 This entire pie was made using ZIPLOCK BAGS. completely classroom friendly. Students use math to calculate the measurements for an increased recipe by using repeated addition or multiplying fractions. Then they make a pie using their solutions. Students then reflect on their group and understanding. A free Pi Day extension is available for diameter circumference and radius and pi for Pie Day. THIS IS SO FUN	957	4,450	{"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}	2.90625	3	CC-MAIN-2021-17	latest	en	0.918593
https://byjusexamprep.com/csir-net-mathematics-2024-testing-of-hypothesis-21-february-2024-i-b0102cc0-d082-11ee-866d-c32778d7fba9	1,713,240,021,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00141.warc.gz	142,591,837	48,517	Time Left - 25:00 mins # CSIR NET Mathematics 2024\|Testing of Hypothesis \| 21 February 2024 Attempt now to get your rank among 5 students! Question 1 Let X be distributed as and let , . The best critical region based on X for testing the null hypothesis against the alternative hypothesis is given by . If , then the value of is : Question 2 In order to test the randomness of a series of observations the appropriate non- parametric test is Question 3 In order to test whether a coin is unbiased, it is tossed 4 times. The null hypothesis of unbiasedness is rejected if and only if more than 3 heads are obtained. Then the size of the test is Question 4 While testing of hypothesis against on ethe basis of a random sample from normal distribution , then was rejected at level of significance. The corresponding p – value is Question 5 Let denote a random sample from normal distribution with mean zero and variance . The region Where is chosen so that the test has level of significance, is a uniformly most powerful region for testing against when • 5 attempts	266	1,089	{"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}	2.625	3	CC-MAIN-2024-18	latest	en	0.867002
https://docsbay.net/doc/543087/1-15true-false-show-your-work-where-appropriate	1,726,040,803,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00350.warc.gz	193,968,956	6,248	## Optics Midterm Practice 1-15True/False: Show your work where appropriate. 1-2: Refer to this picture where a ray originates in medium 1 ____1. The speed of light in medium 2 is less than the speed of light in medium 1 ____2. 1will be equal torefl. ____3. When light speeds up in crossing a boundary between two media at a non-zeroangle, it always bends. ____4. Light that travels between two points always takes the path with the shortest time. ____5. Light that travels between two points in the same medium without reflecting always takes the path with the shortest time. ____6. A beam of light will bend in all cases when traveling from one medium to another if each medium has a different index of refraction. ____7. A basic requirement to see an object is that light must be able to travel from the object to your eye. ____8. A pinhole imager works by allowing rays from one point on the object to reach one, and only one point on the screen. ____9. Images are located where lines of sight cross, or appear to cross. 10-12: Somehow two rays follow this bizarre path from an object at point A to your eyes. ____10. Based on the information your eyes receive, your brain will think that the object is located at point A. ____11. Your brain will use the input from your eyes to locate the image at point B. ____12. If you were to put your eyes near point C you could be sure the object was located at point A based on what you saw. ____13. Rays that leave a point on an object always take the same time to travel to the corresponding image point, even if they go through a thicker part of a lens. ____14. You can tell the difference between a real and a virtual image just by looking at it with your eye. ____15. Sometimes an image can form on the back of your retina that is virtual. Long Answer:For mathematical problems, do the algebra first. Watch units and sig figs! 16. You are trapped in a dark room. To escape, you must reflect light from point A under a shelf, onto a photocell at point C on the ceiling using a tinymirror at point B. At what angle θ from the horizontal should you tilt the mirror? Sketch the mirror and the relevant beam of light. Measure and indicate the angle of incidence and the angle of reflection. 17. What path will the beam of light take as it passes through the water droplet surrounded by air? Sketch the beam, inside and outside the water droplet.No calcs required. Draw the normals. 18. Sketch the path the beam will take as it travels through the thick glass sheet and back out into the air.No calcs required. Draw the normals. 19. Your mom needs an operation andshedidn’t sign up for Obama Care. A friend offers you a cheap price on some “diamonds” saying that you can resell them at a profit. You test them out and find that an angle of incidence in air of θ1 = 75.0o leads to an angle of refraction of θ2 = 59.8o. Are these real diamonds? Should you buy them? Find the index of refraction of the “diamonds.”Assume you like your mom. Sketch it! 20. A particular kind of glass has an index of refraction of 1.52. A) What is the speed of light in this material? B) Consider a sample of this glass in water (look up the index of refraction for water in your PJ.) If θ1 = 47.5o in water, find θ2 in the glass. Sketch it! 21. There is a candle at one end of a large box and a sheet of film on the opposite wall as shown. You expose the film for a brief period of time, then you develop it. What does the developed film look like? Explain. 22. State the general requirement for an optical device to form an image. 23. Add an optical device of your choosing to form an image of the candle on the film. Assume some light makes it to the bottom of the candle. Ray trace it! Ray trace the following and describe the images. Color code your rays and use solid and dotted lines appropriately. Put arrowheads on each line segment. Draw and label each image.Use your ruler! 24. Do the ray tracing for this concave mirror. Sketch where you would have to put your eyes to see the image. 25. What would happen to the image in the case above if half of the mirror was covered up? Does it matter which half? 26. Do the ray tracing for this concave mirror. Sketch where you would have to put your eye to see the image (without using a screen.) 27. Do the ray tracing for this converging lens. Sketch where you would have to put your eyes to see the image. 28. Suppose the object distance so for problem 27 was 3f. Find the image distance si and the magnification M. 29. Suppose that the ray tracing above in problem 27 were to scale. Measure so, si and calculate the magnification M. Does this magnification agree with your results in problem 26? 30. You look at a beautiful rainbow from a spray with your garden hose with both of your eyes. A) Does your right eye see the center of the rainbow arc in the same point relative to the background as your left eye? Briefly explain. B) Where is the image of the rainbow located? Hint: how does the direction of the two rays leaving the top of the rainbow that travel to your right and left eye compare? 1	1,190	5,123	{"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}	3.625	4	CC-MAIN-2024-38	latest	en	0.937552
https://discuss.leetcode.com/topic/24550/my-short-python-accepted-code	1,516,596,397,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00042.warc.gz	673,894,757	8,590	# My short Python accepted code • ``````class Solution(object): """ :type num: str :type target: int :rtype: List[str] """ def solve(target, pos, negate, prod): expr = [] for i in xrange(pos, len(num)): if i > pos and num[pos] == "0": break if i == len(num) -1: if negate * prod * int(num[pos:i+1]) == target: expr.extend([num[pos:i+1]]) break add_expr = solve(target - prod * negate * long(num[pos:i+1]), i+1, 1, 1) expr.extend([num[pos:i+1] + "+" + e for e in add_expr]) sub_expr = solve(target - prod * negate * long(num[pos:i+1]), i+1, -1, 1) expr.extend([num[pos:i+1] + "-" + e for e in sub_expr]) mul_expr = solve(target, i+1, 1, prod * negate * long(num[pos:i+1])) expr.extend([num[pos:i+1] + "*" + e for e in mul_expr]) return expr return solve(target, 0, 1, 1)`````` Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.	279	880	{"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-2018-05	latest	en	0.456408
https://itunes.apple.com/us/app/bubbly-primes/id1041488070?mt=8	1,534,923,416,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221219692.98/warc/CC-MAIN-20180822065454-20180822085454-00182.warc.gz	727,414,527	33,186	## Description Get The Practice Without The Pain. - Give kids a game that you love when they play! - Practice Factoring to make Fractions easier! - Popping the bubbles is addictive. Players can't stop getting better. - Recommended as “A really fun game that teaches kids about prime numbers and factoring,” by the Parenting Beyond Discipline podcast. Bubbly Primes is a prime numbers and factoring game. Parents, teachers, and kids put a lot of energy into mastering fractions. Fractions are an important subject, and in many cases, challenging. Help elementary and middle school students master factoring, a skill that's a building block for fractions and other important math subjects. What makes Bubbly Primes different? Craftsmanship. It's a deceptively simple arcade-style educational game that features hand-drawn animated characters, classical music, a hypnotically soothing underwater world, and a critically important educational mission. Originally intended as a simple, healthy, math game for kids, clever Bubbly Primes players have repurposed it as: - A classroom activity. - A family game. - A brain training exercise for adults. Bubbly Primes is a non-violent game, and has no in-app purchases. ~~~~~~~~~ How To Play For a video, search for “How To Play Bubbly Primes”. It really only takes a few seconds to figure out the basic game. - Tap “Start” to begin. - Tap bubbles that contain composite numbers (numbers you can factor). - As bubbles pop apart your score goes up. - Keep popping them until they’re broken down to prime numbers. - As prime numbers get to the top, they raise your score. - Avoid tapping prime number bubbles. If you do, they’ll darken and fall to the bottom, lowering your score. - As your score rises, playful critters appear. - You can tap critters to change what they do. - You can tap the water to make waves that move bubbles or critters. - When a non-prime “composite” number gets to the top, it turns reddish and it’s too late to factor it. - When five composites escape to the top, the round ends. - If your score is one of the five highest, you can chose a critter for its symbol. Nuhubit is a trademark of Nuhubit Software Studios LLC ## What's New Version 1.0.0 This update is signed with Apple’s latest signing certificate. No new features are included. ## Ratings and Reviews ### 5 out of 5 #### Dec 20, 2015 Jgold0016 I bought two copies of this game. One for myself, and one for my daughter. My daughter “HATES” math. That’s a quote. The game is fun. It has cool characters and great sound-effects (sort of like the satisfying sound of popping the bubbles in bubble-wrap). My daughter and her two friends play the game now a lot. Really a lot. My daughter is now getting scores that are way higher than my scores. I’m sure I could do better if I practiced more. On the other hand, she is not really practicing. She would never practice. She is playing. I guess that’s the whole idea. We need more games like this! ### So Fun and Educational! #### Jan 25, 2016 JazzieSami I get mesmerized when I play this game. It's amazing how quickly one can start learning which numbers are primes. It's also relaxing to play because it's very dance like and I just keep wanting to play more and more! This is such a good way for children and adults to learn their prime numbers. I highly recommend purchasing this and having it on your phone or iPad. ### Bubbly Primes makes me smile! #### Dec 22, 2015 Charles Fugue When I see the kids playing Bubbly Prime, I smile, because I know how much it helps their math. I enjoy playing it too. It's convenient to play on the phone whenever there's a few minutes here or there, but it's even better on the iPad. I really like the little characters in the game. And, did you know 87 is 29 x 3? I certainly didn't before, but now I do. ## Information Seller Nuhubit Software Studios LLC Size 25.1 MB Category Education Compatibility Requires iOS 8.1 or later. Compatible with iPhone, iPad, and iPod touch. Languages English Age Rating Rated 4+	953	4,051	{"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-2018-34	latest	en	0.926945
https://www.askiitians.com/forums/Integral-Calculus/dx-1-secx-what-is-the-value-of-the-intergal-x_214110.htm	1,679,824,546,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00001.warc.gz	735,524,906	32,899	# ∫dx/1-secx what is the value of the intergal∑(xi-x-)2 solve it Arun 25757 Points 4 years ago Dear Trishna – ∫ [1 /(secx - 1)] dx = rearrange it as: – ∫ {1 /[(1/cosx) - 1]} dx = – ∫ {1 /[(1 - cosx) /cosx]} dx = – ∫ [cosx /(1 - cosx)] dx = recall the half-angle identity sin²(x/2) = (1 - cosx) /2 hence 1 - cosx = 2sin²(x/2) also, recall the double-angle identity cos(2x) = cos²x - sin²x hence cosx = cos²(x/2) - sin²(x/2) the integral thus becoming: – ∫ [cosx /(1 - cosx)] dx = – ∫ [cos²(x/2) - sin²(x/2)] /[2sin²(x/2)] dx = break it up into: – ∫ (1/2) [cos²(x/2) /sin²(x/2)] + ∫ (1/2) [sin²(x/2) /sin²(x/2)] dx = simplifying, you get: – ∫ (1/2) cot²(x/2) + (1/2) ∫ dx = rewrite cot²(x/2) as [csc²(x/2) - 1]: – ∫ (1/2) [csc²(x/2) - 1] + (1/2) ∫ dx = break it up into: – ∫ (1/2) csc²(x/2) + (1/2) ∫ dx + (1/2) ∫ dx = – ∫ (1/2) csc²(x/2) dx + ∫ dx = integrating, you get: cot(x/2) + x + c thus, in conclusion: ∫ [1 /(1 - secx)] dx = cot(x/2) + x + c I hope this helps	481	996	{"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}	4.125	4	CC-MAIN-2023-14	latest	en	0.608458
https://excel.instantgrades.com/acct220-principles-of-accounting-i-problem-1-15-points-the-account-balances-appearing-on-the-trial-balance-below-were-taken-from-the-general-ledger-of-flips-copy-shop-at-june-30-2012-addit-2/	1,601,338,953,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600401617641.86/warc/CC-MAIN-20200928234043-20200929024043-00028.warc.gz	373,799,135	16,561	Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality? # Acct220: Principles of Accounting I Problem 1: 15% points: The account balances appearing on the trial balance (below) were taken from the general ledger of Flip’s Copy Shop at June 30, 2012. Additional information for the month of June which has not yet been recorded in the accounts is as follows: (a) A physical count of supplies indicates \$300 on hand at June 30. (b) The amount of insurance that expired in the month of June was \$200. (c) Depreciation on equipment for June was \$400. (d) Rent owed on the copy shop for the month of June was \$600 but will not be paid until July. Flip’s Copy Shop Trial Balance June 30, 2012 Account Titles Debit Credit Cash \$1,000 Supplies 1,100 Prepaid Insurance 2,200 Equipment 24,000 Accum. Depreciation, Equipment \$4,500 Accounts Payable 2,400 Notes Payable 4,000 Flip’s Capital 15,300 Flip’s Drawings 2,400 Service Revenue 4,900 Utilities Expense 400 Totals \$31,100 \$31,100 Instructions: a. Prepare in journal form, without explanations, the end of month adjusting entries for Flip’s Copy Shop for the month of June. b.Prepare a partial adjusted trial balance for the accounts provided. c.Prepare in journal form, without explanations, the end of month closing entries for Flip’s Copy Shop for the month of June. Problem 2: 15% points: The following information is available for Flip Company: Beginning inventory 600 units at \$4 First purchase 900 units at \$6 Second purchase 500 units at \$7.20 Assume that Flip uses a periodic inventory system and that there are 700 units left at the end of the month. Instructions: a. Compute the cost of goods available for sale. b. Compute the value of ending inventory and Cost of Good Sold under the (1) LIFO method. (2) FIFO method. (3) Average-cost method Problem 3: 15% points: The following items were taken from the December 31, 2013 adjusted trial balance of Flip Company. (All balances are normal.) Mortgage payable \$ 1,443 Accumulated depreciation 3,655 Prepaid expenses 880 Accounts payable 1,444 Equipment 11,000 Notes payable after 2015 1,200 Long-term investments 1,100 Flip’s capital 11,480 Short-term investments 3,690 Accounts receivable 1,696 Notes payable in 2014 1,000 Inventories 1,756 Cash 2,100 Service Revenue 9,000 Rent Expense 1,000 Wages Expense 5,000 Utilities Expense 1,000 Instructions:Prepare a classified balance sheet in good form as of December 31, 2013. Problem 4: 10% points: Prepare journal entries to record the following transactions entered into by Flip Company: 2012 June 1 Accepted a \$10,000, 12%, 1-year note from Flop as full payment on her account. Nov. 1 Sold merchandise on account to Flap, Inc. for \$12,000, terms 2/10, n/30. Nov. 5 Flap, Inc. returned merchandise worth \$500. Nov. 9 Received payment in full from Flap, Inc. Dec. 31 Accrued interest on Flop’s note. 2013 June 1 Flop honored her promissory note by sending the face amount plus interest. No interest has been accrued in 2013 Problem 5: 10% points: Flip Company purchased equipment on January 1, 2011 for \$90,000. It is estimated that the equipment will have a \$5,000 salvage value at the end of its 5-year useful life. It is also estimated that the equipment will produce 100,000 units over its 5-year life. Instructions Answer the following independent questions. 1. Compute the amount of depreciation expense for the year ended December 31, 2011, using the straight-line method of depreciation. 2. If 16,000 units of product are produced in 2011 and 24,000 units are produced in 2012, what is the book value of the equipment at December 31, 2012? The company uses the units-of-activity depreciation method. 3. If the company uses the double-declining-balance method of depreciation, what is the balance of the Accumulated Depreciation—Equipment account at December 31, 2013? Problem 6: 10% points: Flip earns a salary of \$7,500 per month during the year. FICA taxes are 8% on the first \$100,000 of gross earnings. Federal unemployment insurance taxes are 6.2% of the first \$7,000; however, a credit is allowed equal to the state unemployment insurance taxes of 5.4% on the \$7,000. During the year, \$25,600 was withheld for federal income taxes and \$5,700 was withheld for state income taxes. Instructions (a) Prepare a journal entry summarizing the payment of Flip’s total salary during the year. (b) Prepare a journal entry summarizing the employer payroll tax expense on Flip’s salary for the year. (c) Determine the cost of employing Flip for the year. Acct220: Principles of Accounting I Problem 1: 15% points: The account balances appearing on the trial balance (below) were taken from the general ledger of Flip’s Copy Shop at June 30, 2012. Additional information for the month of June which has not yet been recorded in the accounts is as follows: (a) A physical count of supplies indicates \$300 on hand at June 30. (b) The amount of insurance that expired in the month of June was \$200. (c) Depreciation on equipment for June was \$400. (d) Rent owed on the copy shop for the month of June was \$600 but will not be paid until July. Flip’s Copy Shop Trial Balance June 30, 2012 Account Titles Debit Credit Cash \$1,000 Supplies 1,100 Prepaid Insurance 2,200 Equipment 24,000 Accum. Depreciation, Equipment \$4,500 Accounts Payable 2,400 Notes Payable 4,000 Flip’s Capital 15,300 Flip’s Drawings 2,400 Service Revenue 4,900 Utilities Expense 400 Totals \$31,100 \$31,100 Instructions: a. Prepare in journal form, without explanations, the end of month adjusting entries for Flip’s Copy Shop for the month of June. b.Prepare a partial adjusted trial balance for the accounts provided. c.Prepare in journal form, without explanations, the end of month closing entries for Flip’s Copy Shop for the month of June. Problem 2: 15% points: The following information is available for Flip Company: Beginning inventory 600 units at \$4 First purchase 900 units at \$6 Second purchase 500 units at \$7.20 Assume that Flip uses a periodic inventory system and that there are 700 units left at the end of the month. Instructions: a. Compute the cost of goods available for sale. b. Compute the value of ending inventory and Cost of Good Sold under the (1) LIFO method. (2) FIFO method. (3) Average-cost method Problem 3: 15% points: The following items were taken from the December 31, 2013 adjusted trial balance of Flip Company. (All balances are normal.) Mortgage payable \$ 1,443 Accumulated depreciation 3,655 Prepaid expenses 880 Accounts payable 1,444 Equipment 11,000 Notes payable after 2015 1,200 Long-term investments 1,100 Flip’s capital 11,480 Short-term investments 3,690 Accounts receivable 1,696 Notes payable in 2014 1,000 Inventories 1,756 Cash 2,100 Service Revenue 9,000 Rent Expense 1,000 Wages Expense 5,000 Utilities Expense 1,000 Instructions:Prepare a classified balance sheet in good form as of December 31, 2013. Problem 4: 10% points: Prepare journal entries to record the following transactions entered into by Flip Company: 2012 June 1 Accepted a \$10,000, 12%, 1-year note from Flop as full payment on her account. Nov. 1 Sold merchandise on account to Flap, Inc. for \$12,000, terms 2/10, n/30. Nov. 5 Flap, Inc. returned merchandise worth \$500. Nov. 9 Received payment in full from Flap, Inc. Dec. 31 Accrued interest on Flop’s note. 2013 June 1 Flop honored her promissory note by sending the face amount plus interest. No interest has been accrued in 2013 Problem 5: 10% points: Flip Company purchased equipment on January 1, 2011 for \$90,000. It is estimated that the equipment will have a \$5,000 salvage value at the end of its 5-year useful life. It is also estimated that the equipment will produce 100,000 units over its 5-year life. Instructions 1. Compute the amount of depreciation expense for the year ended December 31, 2011, using the straight-line method of depreciation. 2. If 16,000 units of product are produced in 2011 and 24,000 units are produced in 2012, what is the book value of the equipment at December 31, 2012? The company uses the units-of-activity depreciation method. 3. If the company uses the double-declining-balance method of depreciation, what is the balance of the Accumulated Depreciation—Equipment account at December 31, 2013? Problem 6: 10% points: Flip earns a salary of \$7,500 per month during the year. FICA taxes are 8% on the first \$100,000 of gross earnings. Federal unemployment insurance taxes are 6.2% of the first \$7,000; however, a credit is allowed equal to the state unemployment insurance taxes of 5.4% on the \$7,000. During the year, \$25,600 was withheld for federal income taxes and \$5,700 was withheld for state income taxes. Instructions (a) Prepare a journal entry summarizing the payment of Flip’s total salary during the year. (b) Prepare a journal entry summarizing the employer payroll tax expense on Flip’s salary for the year. (c) Determine the cost of employing Flip for the year. Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality?	2,316	9,183	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2020-40	latest	en	0.916087
florencetrust.org	1,701,379,746,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00405.warc.gz	296,048,484	45,423	# Understanding the Math Behind the Money: Cryptocurrency Algorithms Cryptocurrencies have revolutionized the world of finance, introducing a decentralized and secure way of conducting transactions. At the heart of these digital currencies lie complex algorithms that underpin their functionality. In this article, we will delve deep into the world of cryptocurrency algorithms, exploring the mathematics and cryptography that make them work. We will also discuss their significance in the growing cryptocurrency market. Embark on a journey to learn about crypto, walking the halls of crypto innovation, to explore the latest advancements and trends shaping the world of digital assets. ## The Foundation: Cryptography in Cryptocurrencies ### Cryptography Fundamentals Cryptography serves as the bedrock of cryptocurrencies, ensuring the privacy and security of transactions. Two fundamental concepts in cryptography are encryption and decryption. Encryption involves converting plain text into a coded message, while decryption reverses this process to retrieve the original information. Hash functions are another critical cryptographic component. These functions take an input (or message) and produce a fixed-length string of characters, known as a hash. The same input will always produce the same hash, but even a slight change in the input will result in a drastically different hash. This property makes hash functions valuable in ensuring data integrity. ## Cryptographic Security in Blockchain ### Public and Private Keys Cryptocurrencies employ a pair of cryptographic keys: a public key and a private key. The public key, as the name suggests, is made available to others and serves as your address for receiving cryptocurrencies. The private key, on the other hand, is kept secret and is used for signing transactions. It is crucial to safeguard your private key, as anyone with access to it can spend your cryptocurrency. ### Digital Signatures Digital signatures are used to verify the authenticity and integrity of transactions. When you make a transaction, your private key is used to create a digital signature, which is attached to the transaction. Other users can then use your public key to verify that the transaction was indeed signed by the corresponding private key. ### Proof of Work vs. Proof of Stake To validate transactions and add them to the blockchain, cryptocurrencies rely on consensus mechanisms. The two most common mechanisms are Proof of Work (PoW) and Proof of Stake (PoS). PoW requires miners to solve complex mathematical puzzles, which consume significant computational power. PoS, on the other hand, involves validators staking their cryptocurrency as collateral to confirm transactions. We’ll explore these mechanisms further in the next section. ## The Blockchain: Where Algorithms Meet Ledger ### What Is a Blockchain? A blockchain is a decentralized ledger that records all transactions across a network. It consists of a chain of blocks, each containing a batch of transactions. These blocks are linked together, forming a chronological and immutable record of all cryptocurrency transactions. ### How Transactions Are Verified #### Mining and Consensus Mechanisms Mining is the process of validating transactions and adding them to the blockchain. Miners compete to solve cryptographic puzzles, and the first one to solve it gets the privilege of adding a new block to the blockchain. This process is resource-intensive, as miners need to perform numerous calculations to find the solution. PoW cryptocurrencies, such as Bitcoin, rely on this mechanism. #### The Role of Miners Miners play a vital role in maintaining the security and integrity of the blockchain. They validate transactions, prevent double-spending, and ensure that the blockchain remains decentralized. In return for their efforts, miners are rewarded with newly created cryptocurrency and transaction fees. ### Cryptographic Hash Functions in Blocks #### Merkle Trees Within each block, transactions are organized into a Merkle tree structure. This hierarchical arrangement of transaction data makes it efficient to verify the integrity of transactions within a block. The root of the Merkle tree is included in the block’s header, providing a compact representation of all transactions in that block. #### Blockchains and Data Integrity Cryptographic hash functions are crucial in ensuring data integrity in blockchains. Each block contains a hash of the previous block’s header, creating a chain of blocks. This linking of blocks through hash values makes it exceedingly difficult to alter the data in one block without changing the entire blockchain, adding an extra layer of security. ## Cryptocurrency Mining Algorithms ### Introduction to Mining Algorithms Mining algorithms are the mathematical puzzles that miners must solve to add new blocks to the blockchain. These algorithms vary depending on the cryptocurrency and consensus mechanism being used. ### Proof of Work (PoW) Algorithm #### Understanding PoW and Its Necessity PoW algorithms require miners to find a specific nonce (a random number) that, when combined with the block’s data, produces a hash value below a certain target. This process is computationally intensive and requires significant computational power, making it costly and time-consuming. #### Bitcoin’s SHA-256 Algorithm Bitcoin, the pioneering cryptocurrency, employs the SHA-256 (Secure Hash Algorithm 256-bit) as its PoW algorithm. Miners compete to find a nonce that, when hashed with the block’s data, results in a hash with a predetermined number of leading zeros. ### Proof of Stake (PoS) Algorithm #### PoS vs. PoW: Key Differences PoS offers an alternative to PoW and aims to be more energy-efficient. Instead of miners, PoS networks have validators who are chosen to create new blocks based on the amount of cryptocurrency they “stake” as collateral. #### Ethereum’s Transition to PoS Ethereum, one of the most prominent cryptocurrencies, is in the process of transitioning from PoW to PoS. This move is driven by environmental concerns and a desire to improve scalability. ## Beyond Bitcoin: Cryptocurrency Algorithm Diversity ### Altcoins and Their Unique Algorithms While Bitcoin uses SHA-256, many other cryptocurrencies, known as altcoins, have emerged with their own algorithms. Some of these include Litecoin (Scrypt), Ripple (XRP), and Monero (Cryptonight). ### Use Cases for Different Algorithms #### Privacy Coins Privacy-focused cryptocurrencies like Monero use specialized algorithms to obfuscate transaction details, providing enhanced anonymity for users. #### Smart Contracts and Ethereum’s EVM Ethereum introduced the concept of smart contracts, which are self-executing contracts with the terms directly written into code. The Ethereum Virtual Machine (EVM) uses a unique algorithm to execute these contracts. #### Energy-Efficient Algorithms Efforts are underway to create cryptocurrencies with more environmentally friendly algorithms, addressing concerns about the energy consumption of PoW networks. ## Quantum Computing and Cryptocurrency Algorithms ### The Threat of Quantum Computers Quantum computers have the potential to break many of the cryptographic techniques used in cryptocurrencies. Their immense computing power could render existing encryption methods obsolete. ### Cryptographic Vulnerabilities #### Shor’s Algorithm and Factorization Shor’s algorithm, a quantum algorithm, can efficiently factor large numbers, which poses a threat to the security of many encryption methods, including those used in cryptocurrencies. #### Post-Quantum Cryptography Researchers are actively working on developing post-quantum cryptographic algorithms that can resist attacks from quantum computers. These algorithms are crucial for the long-term security of cryptocurrencies. ### Preparing for the Quantum Threat The cryptocurrency community is exploring various strategies to prepare for the advent of quantum computing, including upgrading encryption methods and developing quantum-resistant algorithms. ## Challenges and Future Developments ### Scalability Issues in Cryptocurrencies One of the challenges facing cryptocurrencies is scalability—the ability to handle a large number of transactions quickly and efficiently. Several solutions, such as off-chain scaling and layer 2 solutions, are being explored to address this issue. ### Environmental Concerns and Algorithm Choices The environmental impact of PoW algorithms has raised concerns. Cryptocurrency projects are increasingly considering energy-efficient alternatives like PoS and delegated proof of stake (DPoS). ### Emerging Algorithms and Innovations The world of cryptocurrencies is dynamic, with new algorithms and innovations constantly emerging. These developments aim to address existing limitations and open up new possibilities for blockchain technology. ### Regulatory and Legal Implications The regulatory landscape for cryptocurrencies is evolving, with governments worldwide considering how to regulate these digital assets. Algorithm choices and security measures may be influenced by these regulations. ## Conclusion In summary, cryptocurrency algorithms serve as the bedrock upon which the digital financial revolution is built, providing the necessary security and functionality for cryptocurrencies to thrive. As we stand at the cusp of technological advancements and the potential impact of quantum computing, the pivotal role of algorithms in shaping the financial future cannot be emphasized enough. Cryptocurrencies, with their intricate blend of mathematical concepts, cryptographic principles, and economic paradigms, stand as a powerful catalyst for change within the world of finance.	1,757	9,828	{"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-2023-50	longest	en	0.894779
https://oeis.org/wiki/Granville_numbers	1,585,527,804,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370496330.1/warc/CC-MAIN-20200329232328-20200330022328-00303.warc.gz	562,311,267	18,374	This site is supported by donations to The OEIS Foundation. # Granville numbers The Granville numbers or S -perfect numbers are positive integers which meet certain criteria in regards to their divisors. The first 32 Granville numbers are listed in A118372. ## Definitions ### Granville set S In 1996, Andrew Granville proposed the following construction of the set S ,[1] involving the sum of proper divisors of natural numbers. Let ${\displaystyle {\mathcal {S}}_{0}=\{\},\,}$ and for n ∈ ℕ, n ≥ 1 , if ${\displaystyle \left\{\sum _{\stackrel {\ d\,\in \,{\mathcal {S}}_{n-1}}{d\,\|\,n}}{d}\right\}\leq n\,}$ then ${\displaystyle {\mathcal {S}}_{n}={\mathcal {S}}_{n-1}\cup \{n\},\,}$ otherwise ${\displaystyle {\mathcal {S}}_{n}={\mathcal {S}}_{n-1},\,}$ and the Granville set S is defined as ${\displaystyle {\mathcal {S}}:=\lim _{n\to \infty }{\mathcal {S}}_{n}.\,}$ A?????? Granville set S involved in the definition of S -perfect numbers. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, ...} A181487 Complement S of the Granville set S (contains the S -abundant numbers). {12, 18, 20, 30, 42, 48, 56, 66, 70, 72, 78, 80, 84, 88, 90, 102, 104, 108, 114, 120, 138, 150, 162, 174, 180, 186, 192, 196, 200, 210, 220, 222, 246, 252, 258, 260, 264, 270, 272, 280, 282, 288, 294, 300, 304, 308, 312, 318, 320, 330, 336, 340, 354, 364, 366, ...} A?????? Characteristic function χS of the Granville set S . {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, ...} It is necessary to understand the membership of S first. A number is in that set if the sum of its proper divisors (including 1 but excluding the number itself) which are in S is less than or equal to the number. To get past the confusing clutter of notation usually used to describe S , the crystal clear explanation[2] from William Marshall, using familiar number theory terminology, is now paraphrased (with elaborations from others in parentheses): • All deficient numbers (A005100) are forcibly S -deficient numbers and are thus in S . (With n deficient, if we can’t exclude any of its divisors for not being in S , their sum is still less than n itself.) • All 2-perfect numbers (A000396) are in S . (Whether or not any divisors of n can be excluded is irrelevant at this juncture, since the condition of membership in S is less than or equal rather than just less than.) • Some abundant numbers are in S if we can exclude enough of their divisors for not being in S such that the sum of the remaining S -divisors is less than or equal to the number itself. For example, with 12 or 18, their proper divisors are all in S , and thus they remain S -abundant numbers, so it means that 12 and 18 themselves are not in S . (This has consequences for multiples of 12 or 18, like 36: since 12 and 18 are not in S , the sum of its divisors in S is 25 rather than 55, and thus 36 is in S ). It should be clear at this point that in the case of numbers that are not squarefree, i.e. squareful numbers, it is more efficient to examine the smaller divisors first, for if we look at the larger divisors first we must then look at those smaller divisors that divide the larger divisors. ### S -divisors of n The set of S -divisors of n is the intersection of the set of proper divisors of n and the Granville set S . #### Sum of S -divisors of n The sum sS (n) of S -divisors of n, n ≥ 1, is given by ${\displaystyle s_{\mathcal {S}}(n)=\sum _{\ d\,\in \,{\mathcal {S}}_{n-1},\ d\,\|\,n}{d},\,}$ or equivalently ${\displaystyle s_{\mathcal {S}}(n)=\sum _{d\,\|\,n,\ d\,<\,n}{{\chi _{s}}(d)\cdot d},\,}$ where χS (n) is the characteristic function of the Granville set S. ### S -perfect numbers (Granville numbers) A Granville number or S -perfect number is a positive integer n such that the sum of its S -divisors (proper divisors in the set S ) is equal to n (Cf. A??????, A181487 for the complement of S ) ${\displaystyle s_{\mathcal {S}}(n)=n,\ n\geq 1.\,}$ {6, 24, 28, 96, 126, 224, 384, 496, 1536, 1792, 6144, 8128, 14336, 15872, 24576, 98304, 114688, 393216, 507904, 917504, 1040384, 1572864, 5540590, 6291456, 7340032, 9078520, 16252928, 22528935, 25165824, 33550336, 56918394, 58720256, ...} The mathematicians Jean-Marie De Koninck and Aleksandar Ivić first pondered these numbers in December 1996 at the suggestion of Andrew Granville. Theorem. Any even number n that is a 2-perfect number is also S -perfect. Proof. From Euler’s proof of the proposition that even 2-perfect numbers must be of the form (2 p − 1) ⋅ 2 p − 1 , with p prime and q = 2 p − 1 a Mersenne prime, it follows that: a) the prime divisor q of n is deficient (being quasi-1-perfect, since σ (q) = q + 1, s (q) = σ (q) − q = (q + 1) − q = 1 ) is in S ; b) the powers of two, 2 m, 0 ≤ m ≤ p − 1 , that divide n are also deficient (being almost-2-perfect, since σ (2 m) = 2 m +1 − 1, s (2 m) = σ (2 m) − 2 m = (2 m +1 − 1) − 2 m = 2 m − 1 ); and c) those proper divisors d = 2 m q, 0 < m < p − 1, that are the product of a positive power of two and the Mersenne prime q = 2 p − 1 are also deficient (since σ (d) = σ (2 m q) = σ (2 m) σ (q) = (2 m +1 − 1) (q + 1) = (2 m +1 − 1) 2 p < (2 p − 1) 2 p = 2d , since 1 < m + 1 < p ) Therefore, all of n ’s proper divisors are in S , and since they add up to n itself, n is therefore S -perfect. For example, with 496, an even 2-perfect number that is the product of 2 4 and 31, we see that its prime divisors, 2 and 31, are both clearly deficient, since they are quasi-1-perfect, (and thus in S ); the powers of 2 (namely 1, 2, 4, 8, 16) are almost-2-perfect and are thus also in S ; as well as 31 times each positive power of two below 16 (62, 124, 248) with their sums of proper divisors being 34, 100, 232. We have already established that 496 is a 2-perfect number, and now that we have established that all of its proper divisors are in S , it follows that 496 is S -perfect. More interesting perhaps are those abundant numbers that turn out to be S -perfect since the excluded divisors turn out to add up to the number’s abundance. Some such numbers are 24, 96, 126, 224, 384, 1536, 1792, 6144. See A118372 for the Granville numbers, including those that are also in A000396. Whereas no odd 2-perfect numbers are known, there are odd Granville numbers, such as the 28th: 22528935. After 36, the sequence of abundant but S -deficient numbers continues 40, 54, 60, 100, 112, 132, 140, 144, 156, 160, 168, 176, 198, ... ### S -deficient numbers By definition of the Granville set S , an element of S which is not a S -perfect number is a S -deficient number, since the S -abundant numbers are the elements of the complement S of the Granville set. S -deficient numbers are numbers n such that ${\displaystyle {\begin{array}{l}\displaystyle {S_{s}(n) A?????? S -deficient numbers. {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, ...} The deficient numbers (A005100) are a subset of the S -deficient numbers. Among the S -deficient numbers, the deficient numbers (A005100) (shown in black below) constitute a subset, and the abundant S -deficient numbers (A??????) (shown in red) constituting a sparse subset {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, ...} ### S -abundant numbers The S-abundant numbers are the elements of the complement S of the Granville set S . #### S -multiperfect numbers Among the S -abundant numbers, as a generalization of S -perfect number is that of S -multiperfect number ${\displaystyle s_{s}(n)=kn,\ k\geq 2,\ n\geq 1.\,}$ ## Sequences #### Perfect S -perfect numbers The even perfect numbers (A000396) being {6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, 191561942608236107294793378084303638130997321548169216, ...} the even perfect numbers constitute a sparse subset (shown in red below) of the S -perfect numbers {6, 24, 28, 96, 126, 224, 384, 496, 1536, 1792, 6144, 8128, 14336, 15872, 24576, 98304, 114688, 393216, 507904, 917504, 1040384, 1572864, 5540590, 6291456, 7340032, 9078520, 16252928, 22528935, 25165824, 33550336, 56918394, 58720256, ...} If odd perfect numbers happen to exist, then depending on their forms, these could be either S -deficient numbers or S -perfect numbers... #### Abundant S -perfect numbers S -perfect numbers which are not perfect numbers are a subset of the abundant numbers (since the deficient numbers constitute a subset of the S -deficient numbers). A?????? Abundant S -perfect numbers. {24, 96, 126, 224, 384, 1536, 1792, 6144, 14336, 15872, 24576, 98304, 114688, 393216, 507904, 917504, 1040384, 1572864, 5540590, 6291456, 7340032, 9078520, 16252928, 22528935, 25165824, 56918394, 58720256, ...} Among the abundant numbers (A005101), the S -perfect numbers (A??????) (shown in red below) are sparse {12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 198, 200, 204, 208, 210, 216, 220, 222, 224, 228, 234, 240, 246, 252, 258, 260, 264, 270, ...} #### Deficient S -deficient numbers {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, ...} #### Perfect S -deficient numbers There are no even perfect numbers which are S -deficient numbers (the even perfect numbers being a subset of the S -perfect numbers). If odd perfect numbers happen to exist, then depending on their forms, these could be either S -deficient numbers or S -perfect numbers... #### Abundant S -deficient numbers A?????? Abundant S -deficient numbers. {36, 40, 54, 60, 100, 112, 132, 140, 144, 156, 160, 168, 176, 198, ...} Among the abundant numbers (A005101), the S -deficient numbers (A??????) (shown in red below) are {12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 198, ...} ## Programs ### Programs for S -perfect numbers #### C program for S -perfect numbers // Contribution from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 28 2010. #include <stdlib.h> #include <stdio.h> #define MAX_SIZE_SSET 1000000 int main(int argc, char* argv[]) { int Sset[MAX_SIZE_SSET] ; int Ssetsize = 1; Sset[0] = 1 ; for(int n = 2; n < MAX_SIZE_SSET; n++) { int dsum = 0 ; for(int i = 0; i < Ssetsize; i++) { if( n % Sset[i] == 0 && Sset[i] < n) dsum += Sset[i] ; if (dsum > n \|\| Sset[i] >= n) break ; } if(dsum <= n) { if(dsum == n) printf("%d\n", n) ; Sset[Ssetsize++ ] = n ; } } } #### Haskell program for S -perfect numbers -- Contribution from Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 28 2010. sPerfect :: Int -> [Int] -> [Int] sPerfect n ss = case compare (sum $filter ((== 0) . mod n)$ takeWhile (< n) ss) n of LT -> sPerfect (n+1) (n:ss) EQ -> n:sPerfect (n+1) (n:ss) GT -> sPerfect (n+1) ss a118372_list = sPerfect 1 [] -- eop. #### Mathematica program for S -perfect numbers With searchMax set to ten thousand, these computations should only take a few seconds. (* Contribution from Alonso del Arte (alonso.delarte(AT)gmail.com, Nov 3 2010 ) searchMax = 10001; S = {1}; For[i = 2, i < searchMax, i++, If[(Plus @@ Table[Divisors[i][[n]] Boole[MemberQ[S, Divisors[i][[n]]]], {n, 1, Length[Divisors[i]] - 1}]) <= i, S = Flatten[Append[S, i]] ] ]; Take[S, 100] SPerfect = Select[Range[searchMax - 1], (Plus @@ Table[Divisors[#][[n]] * Boole[MemberQ[S, Divisors[#][[n]]]], {n, 1, Length[Divisors[#]] - 1}]) == # & ] ## Notes 1. Jean-Marie De Koninck and Aleksandar Ivić, “On a sum of divisors problem,” Publications de l’Institut Mathématique, New Series, Vol. 64, (1998), pp. 9–20. 2. William Marshall, Who understands Granville numbers?, posting to SeqFan on Oct 28 2010 ## References • Jean-Marie De Koninck, Those Fascinating Numbers, translated by the author. American Mathematical Society (2008) p. 40.	5,365	13,091	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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}	3.5625	4	CC-MAIN-2020-16	latest	en	0.660883
https://physics.stackexchange.com/questions/327528/maximally-mixed-state-as-input-of-a-quantum-circuit	1,657,045,202,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00662.warc.gz	492,169,470	65,534	Maximally mixed state as input of a quantum circuit Given a Hilbert space $\mathcal{H}_N$, a state $\rho$ is a maximally mixed state if it can be written as $\rho = \frac{1}{N} \sum_{i=1}^N \|b_i\rangle \langle b_i \|$ for some (any in fact) orthonormal basis $\{b_i\}$ of $\mathcal{H}_N$. Consider now any density matrix $\sigma$, representing some quantum mixed state. Assume you have a quantum circuit that outputs 1 with probability $\geq 2/3$ on input $\sigma$. I have seen some proofs in quantum computing in which the input of the circuit is set to the maximally mixed state $\rho$ instead, and then the probability to output 1 is said to be $\geq 2/3 \cdot 1/N$. See this paper for instance (section 6, page 15, proof of theorem 20). I agree that if $\sigma$ is a pure state $\sigma = \|\psi \rangle \langle \psi \|$, then we can find other vectors $\|b'_2\rangle, \dots, \|b'_N\rangle$ such that $\{\|\psi\rangle,\|b'_2\rangle, \dots, \|b'_N\rangle\}$ is an orthonormal basis of $\mathcal{H}_N$. Thus, the maximally mixed state can be written as $\rho = \frac{1}{N} \left(\sigma + \sum_{i=2}^N \|b'_i\rangle \langle b'_i \|\right)$, and we indeed get $\sigma$ with probability $1/N$ if the input of our circuit is $\rho$ (thus the probability of having $1$ is $\geq 2/3 \cdot 1/N$). However, I don't know how to prove it when $\sigma$ is mixed. • sigma cannot be entangled and maximally mixed at the same time. Also could you clarify what is actually your question. Apr 20, 2017 at 13:18 • I didn't say that $\sigma$ was maximally mixed (it is $\rho$). However, I made a mistake at the end when I said "$\sigma$ is entangled" (it is "$\sigma$ is mixed"). Regarding a clarification, I don't know what to say more... I've described an argument that is used in many proofs, and that I don't understand (except when $\sigma$ is pure). So my question is why is this argument true. Apr 20, 2017 at 14:14 You can always choose orthonormal basis in which $\sigma$ is diagonal and takes the form, $$\sigma=\sum_{l}\sigma_l \|\psi_l\rangle\langle\psi_l\|$$ with $0\leq\sigma_l\leq 1$ that can be interpreted as classical probability that the system is in the pure state $\|\psi_l\rangle$. Then if the input $\|\psi_l\rangle$ gives $1$ with probability $P(1\|\psi_l)$ the probability of this outcome for the input $\sigma$ is given by $$P(1\|\sigma)=\sum_{l} \sigma_l P(1\|\psi_l)\leq \sum_{l} P(1\|\psi_l)$$ Now with the same reasoning applied to maximally mixed state $\rho$ we get, $$P(1\|\rho)=\sum_{l} \frac{1}{N} P(1\|\psi_l)\geq \frac{1}{N}\sum_{l} \sigma_l P(1\|\psi_l)=\frac{1}{N}P(1\|\sigma)$$ If we know that $P(1\|\sigma)\geq\frac{2}{3}$ we get that $P(1\|\rho)\geq\frac{2}{3N}$	844	2,668	{"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": 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": 3, "x-ck12": 0, "texerror": 0}	3.421875	3	CC-MAIN-2022-27	longest	en	0.902046
https://codereview.stackexchange.com/questions/216300/reverse-int-within-the-32-bit-signed-integer-range-%E2%88%92231-231-%E2%88%92-1-op/216308	1,618,456,181,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00415.warc.gz	276,661,256	46,689	# Reverse int within the 32-bit signed integer range: $[−2^{31}, 2^{31} − 1]$ Optimized LeetCode Problem Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0. Feedback Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack popping using modulus, and pushing using multiplication. #include <cassert> #include <climits> #include <cmath> #include <iostream> class Solution { public: int reverse(int i) { if(i > INT_MAX \|\| i < INT_MIN) { return 0; } int sign = 1; if(i < 0) { sign = -1; i = isign; } int reversed = 0; int pop = 0; while(i > 0) { pop = i % 10; reversed = reversed10 + pop; i /= 10; } std::cout << reversed << '\n'; return reversedsign; } }; int main() { Solution s; assert(s.reverse(1) == 1); assert(s.reverse(0) == 0); assert(s.reverse(123) == 321); assert(s.reverse(120) == 21); assert(s.reverse(-123) == -321); assert(s.reverse(1207) == 7021); assert(s.reverse(1534236469) == 0); assert(s.reverse(-2147483412) == -2143847412); } • The problem is ill-posed. If 'the' reverse of 120 is 21, how can you know whether 'the' reverse of 21 should be 120 or 12? What makes 'the' reverse of 1 number 1 and not 10000? How can you tell your code solves the problem if the correct solution is not defined? – CiaPan Mar 26 '19 at 23:41 • That's good feedback CiaPan and I think you raising that as feedback on LeetCode directly would be good next step. – greg Mar 26 '19 at 23:46 • "but need some help coming up with how to detect and prevent such cases." Asking for advice for code not written yet is off-topic for code review. – Snowhawk Mar 27 '19 at 3:22 • @greg Sure, but ...I've got no account there, and I don't think I'll create one soon. Feel free to report the ambiguity there yourself. :) This may be not very important in an excercise, but IMHO such omission may be devastating in real software projects. – CiaPan Mar 27 '19 at 9:12 • I've read a few articles and publications and agree on the that point. I want to make sure I'm writing code that is production ready. Okay thanks for the feedback CiaPan, I'll make LeetCode aware of this. – greg Mar 27 '19 at 20:13 To prevent overflow, you need to check if reversed10+pop > INT_MAX. But in order to avoid actually overflowing while checking, rearrange the equation to reversed > (INT_MAX-pop)/10; Actually, I think you should check against -INT_MIN. Which brings up another point. You could use int64_t for reversed. This both ensures it can hold -INT_MIN before fixing the sign, and lets you do the bounds test only once at the end, instead of at each intermediate step. I would initialize sign with the correct value: int sign = (i<0)?-1:1; i=sign; . pop can be loop local and doesn't need the 0 initialization. You should multiply reversed by sign before printing it. • Also, make the sign const while you initialize. – Juho Mar 27 '19 at 7:45 • int64_t sounds like a great idea. I'm thinking when fixing the sign I will also need to convert the int64_t to int since there will be conflicting return types for my function or is this some automatic conversion that happens behind the scenes? – greg Mar 27 '19 at 20:33 • Assigning an int64_t to a 32 bit int will automatically truncate the high order bits. If you have already done the limit checking to assure they are zero, then this is no problem. – AShelly Mar 27 '19 at 21:08 • I see, I compare int64_t to -INT_MIN instead of INT_MAX because -(-2147483648) wil result in a value greater than INT_MAX. – greg Mar 27 '19 at 21:30 • Be careful rearranging arithmetic like that - how does integer truncation affect the correctness of reversed > (INT_MAX-pop)/10? I think that will give false positives. (Of course, your unit tests should stress the boundary conditions...) – Toby Speight Mar 28 '19 at 8:45 We use nothing from <cmath>, so it need not be included. # This should be a function, not a class Putting all code into a class suggests you have a background in Java or similar. In C++, we can (and should) use ordinary functions for operations that are mathematically functions. In this case, we have a pure function: it has no state, and should always give the same result for any given input. If it's part of your program requirements that it must provide this (unhelpfully-named) class, then comment that. I'd recommend that you still write a plain function, and simply provide an adapter to conform to the requirements: int reverse_decimal_digits(int i); class Solution { public: int reverse(int i) { return reverse_decimal_digits(i); } } # Choice of data type If we're processing a 32-bit integer, then we should be using std::int32_t. Plain int isn't necessarily large enough (it can be any size from 16 bits upwards). # Incorrect test Given int i, then i > INT_MAX \|\| i < INT_MIN is false by definition. The requirement you quote is (my emphasis): When the reversed integer overflows, return 0. # Not all integers have a negative Beware of overflow here: if(i < 0) { sign = -1; i = isign; } On 2s-complement systems, -1 * INT_MIN is undefined. It turns out that we don't need this step, as in modern C++, the % operator can be used predictably with negative numbers to our advantage (see my modified code, below). # Don't do I/O from a pure function I guess this is some leftover debugging that should have been removed: std::cout << reversed << '\n'; It's good that you've included some unit tests - I wish more people would do that! Do think about which values to test. Your choice agrees with mine somewhat, but diverges later: • 0, 1 and -1 for the three simplest cases. • positive and negative two-digit numbers (e.g. 12 and -23). • smallest and largest allowable input (INT32_MIN and INT32_MAX). • smallest and largest allowable result, and the first overflow in each direction in first and last digits (±1463847412, ±1463847413, ±1563847412). Don't be tempted to over-test. Tests need to be maintained, too, so try to limit the tests to those that exercise the limits within the implementation. # Minor improvements The scope of pop can be reduced to within the loop. And perhaps a better name would be digit? # noexcept and constexpr Can we annotate the function with noexcept and constexpr? # Future Should the number base be hard-coded to 10? Perhaps there's a use for a reverser that works in arbitrary bases. Certainly, base-16 is convenient for testing. # Modified code I've used GoogleTest rather than plain C assert(), so as to get better messages when a test fails, but any testing method is fine. #include <cstdint> constexpr std::int32_t reverse_digits(std::int32_t i, int base = 10) noexcept { std::int32_t reversed = 0; const bool negative = i < 0; while (negative ? i <= -base : i >= base) { auto const digit = i % base; // negative if i < 0 reversed = reversed * base + digit; i /= base; } // final digit may cause overflow const bool overflow = negative ? (reversed < (INT32_MIN - i) / base) : (reversed > (INT32_MAX - i) / base); if (overflow) { return 0; } return reversed * base + i; } #include <gtest/gtest.h> TEST(Reverse, decimal) { EXPECT_EQ(0, reverse_digits(0)); EXPECT_EQ(1, reverse_digits(1)); EXPECT_EQ(-1, reverse_digits(-1)); EXPECT_EQ(21, reverse_digits(12)); EXPECT_EQ(-32, reverse_digits(-23)); EXPECT_EQ(0, reverse_digits(INT32_MIN)); EXPECT_EQ(0, reverse_digits(INT32_MAX)); EXPECT_EQ(2147483641, reverse_digits(1463847412)); EXPECT_EQ(0, reverse_digits(1463847413)); EXPECT_EQ(0, reverse_digits(1563847412)); EXPECT_EQ(-2147483641, reverse_digits(-1463847412)); EXPECT_EQ(0, reverse_digits(-1463847413)); EXPECT_EQ(0, reverse_digits(-1563847412)); } • Good suggestions, although I don't like testing the sign of i in every loop. It's one of those minor pessimizations that adds up as your system grows. I also think, but haven't proven to myself, that if the initial value fit in an int, then only the very last reversed digit has the possibility for overflow. – AShelly Mar 28 '19 at 17:49 • Good points in your comment; I think they are easy to address. Proving that only the final digit can overflow is a great idea (hint: start with base==2 and it should be obvious), and that saves us a lot of work. I'll edit that in. – Toby Speight Mar 28 '19 at 17:54	2,192	8,446	{"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-2021-17	longest	en	0.757823
http://www.gurufocus.com/term/ev2rev/MET/EV%252FRevenue/MetLife%2BInc	1,493,132,913,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917120461.7/warc/CC-MAIN-20170423031200-00478-ip-10-145-167-34.ec2.internal.warc.gz	529,947,259	28,410	Switch to: GuruFocus has detected 7 Warning Signs with MetLife Inc \$MET. More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas. MetLife Inc (NYSE:MET) EV-to-Revenue 0.83 (As of Today) EV/Revenue ratio is calculated as enterprise value divided by its revenue. As of today, MetLife Inc's enterprise value is \$52,709 Mil. MetLife Inc's revenue for the trailing twelve months (TTM) ended in Dec. 2016 was \$63,476 Mil. Therefore, MetLife Inc's EV/Revenue ratio for today is 0.83. The reason Enterprise Value is used is because Enterprise Value is more real in reflecting how much an investor pays when buying a company. For detais, go to Enterprise Value. EV/Revenue is a similar ratio to P/S Ratio, except here Enterprise Value instead of Market Cap is used in the calculation. As of today, MetLife Inc's stock price is \$52.82. MetLife Inc's revenue per share for the trailing twelve months (TTM) ended in Dec. 2016 was \$57.24. Therefore, MetLife Inc's P/S Ratio for today is 0.92. Definition EV/Revenue is a similar ratio to P/S Ratio, except here Enterprise Value instead of Market Cap is used in the calculation. MetLife Inc's EV/Revenue for today is calculated as: EV/Revenue = Enterprise Value (Today) / Revenue (TTM) = 52709.090 / 63476 = 0.83 MetLife Inc's current Enterprise Value is \$52,709 Mil. MetLife Inc's Revenue for the trailing twelve months (TTM) ended in Dec. 2016 was 18433 (Mar. 2016 ) + 15244 (Jun. 2016 ) + 17723 (Sep. 2016 ) + 12076 (Dec. 2016 ) = \$63,476 Mil. * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Explanation The reason Enterprise Value is used is because Enterprise Value is more real in reflecting how much an investor pays when buying a company. For detais, go to Enterprise Value. MetLife Inc's P/S Ratio for today is calculated as: P/S Ratio = Share Price (Today) / Revenue Per Share (TTM) = 52.82 / 57.242 = 0.92 MetLife Inc's share price for today is \$52.82. MetLife Inc's Revenue Per Share for the trailing twelve months (TTM) ended in Dec. 2016 was 16.627 (Mar. 2016 ) + 13.744 (Jun. 2016 ) + 15.977 (Sep. 2016 ) + 10.894 (Dec. 2016 ) = \$57.24. * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Related Terms Historical Data * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. MetLife Inc Annual Data Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 ev2rev 0.93 0.12 0.81 1.11 0.47 0.38 0.90 0.84 0.75 0.84 MetLife Inc Quarterly Data Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 ev2rev 0.87 0.84 0.74 0.83 0.67 0.75 0.64 0.54 0.60 0.84 Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names \| Earn affiliate commissions by embedding GuruFocus Charts GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)	857	3,019	{"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-2017-17	latest	en	0.926619
https://iridium.mathdotnet.com/api/MathNet.Numerics.Distributions/HypergeometricDistribution.htm	1,553,229,594,000,000,000	text/html	crawl-data/CC-MAIN-2019-13/segments/1552912202628.42/warc/CC-MAIN-20190322034516-20190322060516-00375.warc.gz	497,698,641	2,874	## Types Type HypergeometricDistribution Namespace MathNet.Numerics.Distributions Parent DiscreteDistribution ### Public instance methods #### double CumulativeDistribution(double x) Continuous cumulative distribution function (cdf) of this probability distribution. return `double` `double` x #### bool Equals(object obj) return `bool` `object` obj #### int GetHashCode() return `int` #### Type GetType() return `Type` #### int NextInt32() Returns a geometric distributed floating point random number. return `int` #### double ProbabilityMass(int x) Discrete probability mass function (pmf) of this probability distribution. return `double` `int` x #### void SetDistributionParameters(int populationSize, int favoredItems, int numberOfSamples) Configure all distribution parameters. `int` populationSize `int` favoredItems `int` numberOfSamples #### string ToString() return `string` ### Public static methods #### bool IsValidParameterSet(int populationSize, int favoredItems, int numberOfSamples) Determines whether the specified parameters is valid. return `bool` `int` populationSize `int` favoredItems `int` numberOfSamples ### Public properties #### bool CanReset get; `return bool` #### int FavoredItems get; set; Gets or sets the number of items of the population that are in favor. `return int` #### int Maximum get; Gets the maximum possible value of generated random numbers. `return int` #### double Mean get; Gets the mean value of generated random numbers. `return double` #### int Median get; Gets the median of generated random numbers. Throws NotSupportedException since the value is not defined for this distribution. `return int` #### int Minimum get; Gets the minimum possible value of generated random numbers. `return int` #### int NumberOfSamples get; set; Gets or sets the number of samples. `return int` #### int PopulationSize get; set; Gets or sets the population size parameter. `return int` #### RandomSource RandomSource get; set; `return RandomSource` #### double Skewness get; Gets the skewness of generated random numbers. `return double` #### double Variance get; Gets the variance of generated random numbers. `return double`	454	2,227	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2019-13	longest	en	0.395838
https://nukephysik101.wordpress.com/2016/06/18/special-rungu-kutta-method-for-any-order-ode/	1,521,706,212,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00714.warc.gz	617,173,819	20,465	The central piece of Rungu-Kutta method is the approximation of the increasement of the function. In 1st order ODE, $\dot{y} = f(x,y), \dot{y}(x_0) = y_0$ In a special case of Rungu-Kutta of order 4 (RK4), there are 2 array $b_i$ and $c_i$, so that $dx = h, dy = h \sum\limits_{i=1}^{4} b_i dy_i$ $dy_i = f(x+ c_i dx, y+c_i dy_{i-1})$ where $c_i$ is ranging from 0 to 1, $\sum\limits_{i=1}^{4} b_i =1$. In RK4, $c = (0, \frac{1}{2}, \frac{1}{2}, 1)$ $b = \frac{1}{6}(1,2,2,1)$ The $c_1 = 0$ is a must, otherwise, we have to define $dy_0$. There should be some methods to obtain an optimum values for $c$ and $b$, but I don’t know. For 2nd order ODE $\frac{d^2y}{dx^2} + a \frac{dy}{dx} = f(x,y), y(x_0) = y_0, \frac{dy}{dx}\|_{x_0} = z_0$ change to $\frac{dy}{dx} = z, \frac{dz}{dx}=F(x,y,z)$ These equation are the similar 1st order ODEs. $dx = h, dy = \sum\limits_{i=1}^{4} b_i dy_i, dz = \sum\limits_{i=1}^{4} b_i dz_i$ $dy_i = h( z+ c_i dz_{i-1})$, where is the z for $y_n$ step. $dz_i = h F(x+ c_i dx, y+c_i dy_{i-1}, z+c_i dz_{i-1})$ The $dy_i$ is using $dz_i$ $x_{n+1} = x_n + dx$ $y_{n+1} = y_n + dy$ $z_{n+1} = z_n + dz$ This can be generalized to any order ODE by decoupling the ODE into $\frac{dy}{dx} = x,..., \frac{du}{dx}=F(x,y,...,u)$ the equation for $du_i$ is $du_i = h F(x+ c_i dx, y+c_i dy_{i-1}, z+c_i dz_{i-1}, ...., u+ c_i du_{i-1})$ And for all the intermediate variable $y, z, w,.... = O^{(k)}$ $dO_i^{(k+1)} = h( O^{(k)} + c_i dO_{i-1}^{(k)})$	661	1,494	{"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": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2018-13	latest	en	0.693771
https://gasstationwithoutpumps.wordpress.com/2016/02/26/digital-filter-lecture/	1,660,363,758,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00298.warc.gz	259,884,326	35,763	# Gas station without pumps ## 2016 February 26 ### Digital filter lecture Filed under: freshman design seminar — gasstationwithoutpumps @ 00:00 Tags: , , , On Wed 2016 Feb 24, I gave a lecture in the freshman design seminar on digital filters, covering about 3 weeks worth of material in an hour. Needless to say, this was a very rushed and handwavey introduction to digital filtering, but it should be enough that the students can (with some additional scaffolding) implement bandpass filters for the pulse monitor project and the ultrasonic rangefinder project. On Monday, I had covered the notion of digital signals as having discrete values and discrete times with a uniform sampling frequency, so I started with a signal $x_0, x_1, \ldots , x_t$ and introduced the z-transform $X(z) = x_0 + x_1 z^{-1} + x_1 z^{-2} +\cdots$. I explained that this was a linear transform, so that if we multiplied $x$ by a constant, we would multiply $X(z)$ by the same amount, and that the z-transform of the sum of two signals was the sum of their transforms. I also showed that the signal $0, x_0, x_1, \ldots$, which is $x$ delayed by one tick, has the z-transform $z^{-1} X(z)$. I claimed (without proof) that linear filters consisted of delay elements, multiplication by constants, and addition, so that the z-transforms of the input ($X(z)$) and output ($Y(z)$) were related by a transfer function: $H(z)= Y(z)/X(z)$, and that $H(z)$ is a rational function for linear filters. I first showed them a finite-impulse-response (FIR) filter: Small finite-impulse response filter. (Block diagram drawn with http://www.draw.io/) I showed them how this filter had $H(z) = b_0 + b_1 z^{-1} + b_2 z^{-2}$, and could be implemented easily in pseudocode: x0 ← new value y ← b0 * x0 + b1x1 + b2x2 x2 ← x1 x1 ← x0 I explained, briefly, what an impulse response was (the values put out by a filter whose input is 1 at time 0 and 0 at all other times), and showed that the filter coefficients were the impulse response. I also showed them a biquad filter element: This biquad filter element uses the type 1 direct implementation, which has the advantage of not having any internal overflows, as long as the inputs and outputs remain within bounds. (Block diagram drawn with http://www.draw.io/) I gave them a simple implementation of the biquad element: x0 ← new value y ← (b0 * x0 + b1x1 + b2x2 - a1y1 -a2y2)/a0 x2 ← x1 x1 ← x0 y2 ← y1 y1 ← y0 I pointed out that multiplication and addition (of integers) was cheap on the Teensy boards, but division or floating-point arithmetic is expensive. Because of binary representation, division by powers of two is cheap, so we can keep the computation fast if we restrict a0 to powers of 2. I showed how the recurrence relation in the code could be rearranged with simple algebra to get the transfer function $H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{a_0 + a_1 z^{-1} + a_2 z^{-2}}$. Somewhere in the lecture I mentioned that the poles of $H(z)$ (that is, the zeros of $latex a_0 + a_1 z^{-1} + a_2 z^{-2}$) had to remain within the unit circle to keep the filter from oscillating, but I didn’t explain why. I told the students that we would represent sinusoids with $e^{j\omega t} = \cos(\omega t) + j \sim(\omega t)$, giving a brief explanation of the advantages of using exponentials rather than trig functions and reminding them of the popular abbreviation using angular frequency $\omega = 2 \pi f$ instead of frequency $f$. I claimed, without proof, that the response of a filter to a sinusoid with angular frequency $\omega$ was just $H(e^{j \omega})$. I then shared with them a gnuplot script for plotting the response of a biquad element: #band-pass for pulse monitor fs = 60 A0= 256 A1= -388 A2= 141 B0 = A0 B1 = 0 B2 = -B0 A0_2 = 256 A1_2 = -449 A2_2 = 199 unset arrow set arrow nohead from 40000,-20 to 40000,20 unset label pole1a = ( -A1 + sqrt(A1A1 - 4 A0 * A2)) / (2A0) pole1b = ( -A1 - sqrt(A1A1 - 4 * A0 * A2)) / (2A0) set label sprintf("pole magnitude %.3f", abs(pole1a)) at 1,-3 set label sprintf("pole1a at %.3f + %.3f j", real(pole1a), imag(pole1a)) at 1,-5 set label sprintf("pole1b at %.3f + %.3f j", real(pole1b), imag(pole1b)) at 1,-7 pole2a = ( -A1_2 + sqrt(A1_2A1_2 - 4 * A0_2 * A2_2)) / (2A0_2) pole2b = ( -A1_2 - sqrt(A1_2A1_2 - 4 * A0_2 * A2_2)) / (2A0_2) set label sprintf("pole2b at %.3f + %.3f j", real(pole2b), imag(pole2b)) at 1,-17 set label sprintf("pole2a at %.3f + %.3f j", real(pole2a), imag(pole2a)) at 1,-15 set label sprintf("pole magnitude %.3f", abs(pole2a)) at 1,-13 set title sprintf("Design of biquad filter, fs=%3g Hz",fs) set key bottom right set ylabel "gain [dB]" unset logscale y set yrange [-30:] set xlabel "frequency [Hz]" set logscale x set xrange [0.01:0.5fs] set grid xtics set mxtics 10 j=sqrt(-1) set samples 5000 plot \ 20log10(gain(x,B0,B1,B2, A0,A1,A2)) \ title sprintf("%.0f (1 + %.0f z^-1 + %.0f z^-2)/(%.2f+ %.3f z^-1 + %.3f z^-2)", \ B0, B1/B0, B2/B0, A0, A1, A2), \ 20log10(gain(x,B0,B1,B2, A0_2,A1_2,A2_2)) \ title sprintf("%.0f (1 + %.0f z^-1 + %.0f z^-2)/(%.2f+ %.3f z^-1 + %.3f z^-2)", \ B0, B1/B0, B2/B0, A0_2, A1_2, A2_2) lt 3 I walked them through the code and showed them the result: This is a pair of bandpass filters designed for a pulse monitor. The red curve may be a better choice, as it does not have such a sharp peak around 1.5Hz, but still reasonably suppresses the high frequencies and the DC drift. I showed them why any bandpass biquad has essentially the same numerator: we want to have a zero at DC (frequency 0, so at $e^{j 2 \pi 0}=1$) and at the Nyquist frequency $e^{j 2 \pi \frac{1}{2}}=-1$, so the numerator is always a multiple of $(1-z^{1})(1+z^{-1})= 1-z^{-2}$. I also showed them the effect of having just the numerator (setting the denominator polynomial to 1), using both log and linear frequency scales to show the peak at half the Nyquist frequency (one quarter the sampling frequency): A linear frequency scale make the symmetric frequency response of the FIR filter 256 (1–z^-2) clear. I told them that I did not really know the details of how to choose specific filter parameters, and shared with them the code I used from the scipy.signal package for choosing the parameters: #!/usr/bin/env python3 from __future__ import print_function, division from scipy import signal from cmath import sqrt # complex square root print("set parameters by giving name=value lines:") sampling_freq = 30 # Hz low_cutoff = 0.3 # Hz high_cutoff = min(150, 0.49sampling_freq) # Hz fixed_point_scaling=64 while True: print("sample=",sampling_freq, "low=", low_cutoff, "high=", high_cutoff, "scale=", fixed_point_scaling) hi_over_Nyquist = high_cutoff/(0.5sampling_freq) lo_over_Nyquist = low_cutoff/(0.5sampling_freq) filter_float = signal.bessel(1, [lo_over_Nyquist,hi_over_Nyquist], btype="bandpass") A= [int(xfixed_point_scaling+0.5) for x in filter_float[1]] discr = A[1]A[1] -4A[0]A[2] poles= [(-A[1]+sqrt(discr))/(2A[0]), (-A[1] -sqrt(discr))/(2A[0])] print("\tA=",A, "B=[1,0,-1]", "poles=",poles) line=input("param=value?") fields = line.split("=") if len(fields)==0: continue if len(fields)!=2: print("No parameter change found, exiting") break keyword=fields[0].strip().lower() value=float(fields[1]) if keyword=="low" or keyword=="lo": low_cutoff = value elif keyword=="high" or keyword=="hi": high_cutoff = value elif keyword=="sample" or keyword=="freq": sampling_freq = value elif keyword=="scale": fixed_point_scaling=int(value) else: print("unrecognized keyword:", keyword) I only walked them through a little of this code (mainly showing the scaling of frequencies to the Nyquist frequency, since that is what the signal package wants, and the call to the bessel function to get a Bessel filter). I explained that the fixed_point_scaling parameter was to get more resolution in the parameters when doing integer arithmetic, but didn’t really have time to explain what that meant. I did demonstrate setting the parameters to get one of the filters shown in the graph above. On Friday, I plan to give them a little code snippet that I use: static volatile int32_t x_0, x_1, x_2; static volatile int32_t y_0, y_1, y_2; // filter parameters for biquad bandpass filter // selected for approx 0.66--6Hz with 60Hz sampling #define SAMPLE_FREQ (60) // sampling frequency in Hz #define a0 (256) #define a1 (-388) #define a2 (141) #define gain (1) // b0= - b2= gaina0 // b1=0 #define DELAY_XY (x_2=x_1, x_1=x_0, y_2=y_1, y_1=y_0) #define GENERAL_BANDPASS (y_0 = ((gaina0)(x_0-x_2) -a1y_1 -a2*y_2)/a0, DELAY_XY) To use the filter, I include x_0=analogRead(MONITOR_PIN); GENERAL_BANDPASS; in an interrupt routine that runs once every 60th of a second. I’ll also have to show them how to use the IntervalTimer in the Teensyduino software development kit to set up the interrupt routine.	2,753	8,922	{"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": 22, "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-2022-33	latest	en	0.950095
http://www.qacollections.com/Types-of-Triangles-Angles	1,495,948,668,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00005.warc.gz	777,793,079	5,560	# Types of Triangles & Angles? The study of angles and angles and triangles is part of geometry, a branch of math. There are three types of angles--right angle, obtuse angle and acute angle--and triangles are named after what ty... Read More » http://www.ehow.com/info_8514107_types-triangles-angles.html Top Q&A For: Types of Triangles & Angles ## How to Find Outside Angles in Triangles? A triangle not only has inside angles but outside angles as well. When a triangle's side is extended, the extension creates another angle. This angle, called an outside or exterior angle, has a spe... Read More » http://www.ehow.com/how_8483256_outside-angles-triangles.html ## How to Find & Measure Angles for Triangles? All triangles contain three angles. If the angles all have the same measurement -- 60 degrees -- it is an equilateral triangle, whereas a right triangle has one angle of 90 degrees that forms the s... Read More » http://www.ehow.com/how_8195084_measure-angles-triangles.html ## How to Classify Triangles by Sides & Angles? Shapes are differentiated by three basic properties: the number of sides, the comparative length of these sides and the interior angles within the shape. A triangle is a shape characterized as alwa... Read More » http://www.ehow.com/how_8458511_classify-triangles-sides-angles.html ## All of the Types of Triangles? Every triangle has three sides and three angles. It is a closed geometric figure, meaning all sides connect. Classify triangles by comparing the length of their sides, by measuring their angles or ... Read More » http://www.ehow.com/info_8173593_types-triangles.html Related Questions	385	1,646	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2017-22	longest	en	0.864097
https://newbedev.com/how-to-combine-n-less-random-bits-to-generate-one-more-random-bit	1,702,097,372,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00673.warc.gz	460,521,761	6,091	# [Crypto] How to combine $n$ 'less random' bits to generate one 'more random' bit? If the outputs of badrand are independent and identically distributed you can use a randomness extractor function with $$m=1$$ and $$k=-n\lg 0.9$$. If you're prepared to have a non-deterministic waiting time for an output, the von Neumann extractor is good way to get probability exactly 0.5 out of i.i.d. bit outputs. Sample pairs of bits until you get a pair with precisely one zero; if your pair is 01 return 0 and if it is 10 return 1. The i.i.d property means that the probability of these two events is equal (in your example both have probability 0.09 and resampling has probability 0.82, leading you to expect around 100/9 badrand outputs to produce one betterrand output).	193	767	{"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2023-50	latest	en	0.875274
http://mathhelpforum.com/algebra/74634-inequality.html	1,527,136,171,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00555.warc.gz	197,900,178	8,940	## Inequality If,a,b,c,d>0 ,and $\displaystyle abcd\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \right) = a + b + c + d$ Ten demonstrate that: $\displaystyle \frac{{\left( {a + 1} \right)^2 }}{{a + b}} + \frac{{\left( {b + 1} \right)^2 }}{{b + c}} + \frac{{\left( {c + 1} \right)^2 }}{{c + d}} + \frac{{\left( {d + 1} \right)^2 }}{{d + a}} \le 2(a + b + c + d)$	175	380	{"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}	3.78125	4	CC-MAIN-2018-22	latest	en	0.348324
http://define.com/complement	1,371,619,969,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368707906397/warc/CC-MAIN-20130516123826-00050-ip-10-60-113-184.ec2.internal.warc.gz	64,482,637	8,987	be54 complement - Definition of complement at Define.com Dictionary and Thesaurus (define complement) A:link {color:inherit;text-decoration: inherit;font-style:inherit} A:active{color:inherit;text-decoration: inherit;font-style:inherit} A:visited{color:inherit;text-decoration: inherit;font-style:inherit} A:hover{color:inherit;text-decoration: underline;font-style:inherit} From The Collaborative International Dictionary of English v.0.44 [gcide]: Complement \Com"plement\, noun [L. complementun: cf. F. compl['e]ment. See {Complete}, verb (used with an object), and cf. {Compliment}.] 1. That which fills up or completes; the quantity or number required to fill a thing or make it complete. 2. That which is required to supply a deficiency, or to complete a symmetrical whole. 3. Full quantity, number, or amount; a complete set; completeness. 4. (Math.) A second quantity added to a given quantity to make it equal to a third given quantity. 5. Something added for ornamentation; an accessory. [Obs.] 6. (Naut.) The whole working force of a vessel. 7. (Mus.) The interval wanting to complete the octave; -- the fourth is the complement of the fifth, the sixth of the third. 8. A compliment. [Obs.] --Shak. {Arithmetical complement of a number} (Math.), the difference between that number and the next higher power of 10; as, 4 is the complement of 6, and 16 of 84. {Complement of an arc} or {Complement of an angle} (Geom.), the difference between that arc or angle and 90[deg]. {Complement of a parallelogram}. (Math.) See {Gnomon}. {In her complement} (Her.), said of the moon when represented as full. From The Collaborative International Dictionary of English v.0.44 [gcide]: Complement \Com"plement\, verb (used with an object) 1. To supply a lack; to supplement. [R.] 2. To compliment. [Obs.] --Jer. Taylor. From WordNet (r) 2.0 [wn]:noun 2: a complete number or quantity; "a full complement" 3: number needed to make up whole force; "a full complement of workers" [syn: {full complement}] 4: something added to complete or make perfect; "a fine wine is a perfect complement to the dinner" 5: one of a series of enzymes in the blood serum that are part of the immune response verb 1: make complete or perfect; supply what is wanting or form the complement to; "I need some pepper to complement the sweet touch in the soup"From Moby Thesaurus II by Grady Ward, 1.0 [moby-thes]: From The Free On-line Dictionary of Computing (27 SEP 03) [foldoc]: complement The other value or values in the set of possible values. See {logical complement}, {bitwise complement}, {set complement}. (1995-01-24) Define.com is a registered nonprofit corporation dedicated solely to the global public interest and the advancement of humanity. It belongs to all of us who have a desire to promote electronic democracy, science, creativity, imagination, reason, critical thinking, peace, race and gender equality, civil rights, equal access to education, personal liberty, free speech, animal rights, compassionate and nonviolent parenting, social and economic justice, global monetary reform, Secular Humanism, cognitive liberty and a permanent cessation of The War on Drugs. Let's see what we can do if we put our heads together. 0	800	3,222	{"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-2013-20	latest	en	0.767826
https://math.stackexchange.com/questions/806762/problem-about-real-square-matrix-with-rank-1?noredirect=1	1,563,351,421,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195525133.20/warc/CC-MAIN-20190717081450-20190717103450-00181.warc.gz	477,599,471	31,291	# Problem about real square matrix with rank 1 [duplicate] Given $A \in \mathbb{R}^{n \times n}$ and $\text{rank}(A) = 1$. By working only on real field, show that $A$ is diagonalizable if and only if $\text{tr}(A) \neq 0$. Here, $\text{tr}(A)$ is the sum of all eigenvalues of $A$. ## marked as duplicate by Marc van Leeuwen linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 25 '15 at 11:55 • @Servaes Prove that 0 is an eigenvalue of $A$ and it has geometric multiplicity of $n-1$. If $A$ is diagonalizable, $0$ also has algebraic multiplicity of $n-1$, so there is another nonzero eigenvalue, so tr(A) is not zero. On the other hand, if $A$ is not diagonalizable, then $0$ must have algebraic multiplicity of $n$, so that tr(A) must be zero. Is it true? – Kevin Limanta May 23 '14 at 15:19	375	1,288	{"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}	3.078125	3	CC-MAIN-2019-30	latest	en	0.521202
https://pastecode.io/s/yyqrz22r	1,701,562,124,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100476.94/warc/CC-MAIN-20231202235258-20231203025258-00128.warc.gz	518,987,910	36,091	# Untitled unknown python a year ago 4.7 kB 3 Indexable Never ```import random def create_deck(): """ return: 2-D list Creates and returns a 2D-list """ deck = [] suits = ["RED", "BLK"] values = [] # creates a list of all face values for x in range(2,10): values.append(x) # creates cards [s,f] and inserting them to the deck for s in suits: for f in values: deck.append([s, f]) return deck def shuffle(deck): """ return: 2-D list Takes a 2-D list and returns a shuffled 2-D list """ shdeck = deck[:] random.shuffle(shdeck) return shdeck def deal(deck): """ return: 3-D list Takes the shuffled deck as the parameter and deals the first nine """ board = [] for row in range(3): board.append([]) for col in range(3): board[row].append(deck.pop(0)) return board def print_board(board): """ return: 3-D list Takes the board as a parameter and prints the elements of the board to the terminal """ for row in board: for card in row: print(f"\| {card[0]}, {card[1]} ", end = "") print("\|") def check_triplet_for_set(triplet): """ input: 2-D list return: bool Checks for simple set to match for suit values (RED, BLK) """ return triplet[0][0] == triplet[1][0] == triplet[2][0] def check_board_row_for_set(board): """ input: 3-D list return: bool Checks board for rows in simple set to match for suit values (RED, BLK) """ for row in board: if check_triplet_for_set(row): return True return False def check_board_coloumn_for_set(board): """ input: 3-D list return: bool Checks board for coloumn in simple set to match for suit values (RED, BLK) """ for column in range(len(board)): coloumn_list = board[0][column], board[1][column], board[2][column] if check_triplet_for_set(coloumn_list): return True return False def check_board_diagonal_for_set(board): """ input: 3-D list return: bool Checks board for diagnol pattern in simple set to match for suit values (RED, BLK) """ diagnoal_list_1 = board[0][0], board[1][1], board[2][2] diagnoal_list_2 = board[0][2], board[1][1], board[2][0] return check_triplet_for_set(diagnoal_list_1) or check_triplet_for_set(diagnoal_list_2) def check_board_for_sets(board): """ input: 3-D list return: bool Checks if one of the simple set functions runs and returns True, if true """ if check_board_row_for_set(board) or check_board_coloumn_for_set(board) or check_board_diagonal_for_set(board): return True def check_triplet_for_run(triplet): """ input: 2-D list return: bool Checks for simple run to match for values (1-9) """ return triplet[0][1] + 2== triplet[1][1] + 1 == triplet[2][1] def check_board_row_for_run(board): """ input: 3-D list return: bool Checks board for row in simple run to match for values (1-9) """ for row in board: if check_triplet_for_run(row): return True return False def check_board_coloumn_for_run(board): """ input: 3-D list return: bool Checks board for coloumn in simple run to match for values (1-9) """ for column in range(len(board)): coloumn_list = board[0][column], board[1][column], board[2][column] if check_triplet_for_run(coloumn_list): return True return False def check_board_diagnol_for_run(board): """ input: 3-D list return: bool Checks board for diagnol pattern in simple run to match for values (1-9) """ diagnoal_list_1 = board[0][0], board[1][1], board[2][2] diagnoal_list_2 = board[0][2], board[1][1], board[2][0] return check_triplet_for_run(diagnoal_list_1) or check_triplet_for_run(diagnoal_list_2) def check_board_for_runs(board): """ input: 3-D list return: bool Checks if one of the simple run functions runs and returns True, if true """ if check_board_row_for_run(board) or check_board_coloumn_for_run(board) or check_board_diagnol_for_run(board): return True # main() function that calls all other functions in the appropriate order to run the game def main(): """ runs the game """ # set score value to award points to the user score = 0 print("Welcome to the game!") deck = create_deck() shdeck = shuffle(deck) board = deal(shdeck) print_board(board) if check_board_for_runs(board): score += 20 print(f"Congrats! You've got a simple run on the board, score: {score}") elif check_board_for_sets(board): score += 10 print(f"Congrats! You've got a simple set on the board, score: {score}") else: print(f"sorry, no match on the board. Score: {score}") # main guard if __name__ == "__main__": main()```	1,252	4,323	{"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}	2.90625	3	CC-MAIN-2023-50	latest	en	0.589936
http://mathhelpforum.com/geometry/111032-acute-angled-triangle-print.html	1,529,475,725,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00421.warc.gz	192,751,181	2,605	An acute-angled triangle • Oct 28th 2009, 11:55 AM gabor7896 An acute-angled triangle An acute-angled triangle has sides a, b, c and area t. The sides satisfy the equality abc=a+b+c. How can I prove, that $\displaystyle 1< t \leq \frac{3 \cdot \sqrt{3}}{4}$ • Oct 28th 2009, 01:08 PM Opalg Quote: Originally Posted by gabor7896 An acute-angled triangle has sides a, b, c and area t. The sides satisfy the equality abc=a+b+c. How can I prove, that $\displaystyle 1< t \leq \frac{3 \cdot \sqrt{3}}{4}$ You will find some hints in this thread.	186	543	{"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}	3.265625	3	CC-MAIN-2018-26	latest	en	0.717437
https://www.daniweb.com/programming/software-development/threads/190976/random-numbers	1,539,637,897,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583509845.17/warc/CC-MAIN-20181015205152-20181015230652-00140.warc.gz	918,780,019	11,568	How do I take randomly generated numbers and print the average of the numbers? And how do should I go about printing how many even and odd numbers that were generated? Thanks 3 Contributors 2 Replies 3 Views 9 Years Discussion Span Last Post by Paul Thompson This is rather crude, but it works: ``````import random y = 0 count = 0 even_count = 0 odd_count = 0 while count < 100: x = random.randint(1,101) print x y += x count += 1 avg = y / count if y %2 == 0: even_count += 1 print " the average is ", avg else: odd_count += 1 print "The average is", avg print "Out of 100 tries, %d was odd and %d was even" %(odd_count, even_count)`````` To get the average i usually make a function ``````import random def averageList(l): total = 0 for item in l: total += item	233	772	{"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}	2.8125	3	CC-MAIN-2018-43	latest	en	0.888599
https://www.texasgateway.org/resource-index/?f%5B0%5D=im_field_resource_subject%3A3&f%5B1%5D=sm_field_resource_grade_range%3A11&f%5B2%5D=im_field_resource_subject%3A2&f%5B3%5D=im_field_resource_subject%3A5&amp%3Bf%5B1%5D=sm_field_resource_type%3Astudent_activity&amp%3Bf%5B2%5D=sm_field_resource_grade_range%3A8	1,566,805,233,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00180.warc.gz	999,412,622	15,792	• Resource ID: Revised_Math_TEKS_VA • Subject: Math ### Vertical Alignment Charts for Revised Mathematics TEKS This resource provides vertical alignment charts for the revised mathematics TEKS. • Resource ID: GM4L16a • Subject: Math ### Making Conjectures About Circles and Angles Given examples of circles and the lines that intersect them, the student will use explorations and concrete models to formulate and test conjectures about the properties of and relationships among the resulting angles. • Resource ID: GM5L2 • Subject: Math ### Solving Problems With Similar Figures Given problem situations involving similar figures, the student will use ratios to solve the problems. • Resource ID: A2M3L3 • Subject: Math ### Transformations of Square Root and Rational Functions Given a square root function or a rational function, the student will determine the effect on the graph when f(x) is replaced by af(x), f(x) + d, f(bx), and f(x - c) for specific positive and negative values. • Resource ID: A2M3L4 • Subject: Math ### Transformations of Exponential and Logarithmic Functions Given an exponential or logarithmic function, the student will describe the effects of parameter changes. • Resource ID: A2M6L2A • Subject: Math ### Solving Square Root Equations Using Tables and Graphs Given a square root equation, the student will solve the equation using tables or graphs - connecting the two methods of solution. • Resource ID: A2M7L0 • Subject: Math ### Rational Functions: Predicting the Effects of Parameter Changes Given parameter changes for rational functions, students will be able to predict the resulting changes on important attributes of the function, including domain and range and asymptotic behavior. • Resource ID: USHM1L1b • Subject: Social Studies ### The Gilded Age Given background information, students will be able to identify economic, social, and political issues surrounding the Gilded Age. Students will identify significant historical figures associated with the Gilded Age. • Resource ID: USHM3L4 • Subject: Social Studies ### Political Influences of the Great Depression Given primary and secondary sources of information about selected New Deal measures (e.g., the creation of the Civilian Conservation Corps or the passage of the Agricultural Adjustment Act), students will analyze how these measures affected various regions of the United States. • Resource ID: USHM3L10 • Subject: Social Studies ### World War II Impact on U.S. Economy and Society Given background information, students will identify the social and economic impact of World War II on the American home front, such as the Great Depression, rationing, and increased opportunity for women and minority employment. • Resource ID: USHM4L6 • Subject: Social Studies ### The Cold War and the American Home Front Students will identify ways in which Cold War tensions were intensified. • Resource ID: USHM7L1 • Subject: Social Studies ### The U.S. Role in the World (1970's into the 21st Century) Given a timeline, students will understand the political, economic, and social impact of selected U.S. political leaders on the world from the 1970s into the 21st century. • Resource ID: USHM7L3 • Subject: Social Studies ### Conservative Resurgence of the 1980s and 1990s Given information about social issues throughout U.S. history, students will describe the causes and effects of significant societal issues. • Resource ID: USHM7L5 • Subject: Social Studies ### The American Identity: An Artistic Reflection Given selected examples of American art, music, and literature, students will be able to identify the era of U.S. history that is reflected in the art. • Resource ID: USHM7L4 • Subject: Social Studies ### The American Spirit: Defending and Building Our Nation Given background information about selected historical figures, students will be able to analyze the importance and contributions of women and people of various racial, ethnic, and religious groups to the national identity and the cultural landscape. • Resource ID: USHM6L1a • Subject: Social Studies ### The American Civil Rights Movement: An Overview Given background information, students will be able to trace the historical development of the civil rights movement in the 19th, 20th, and 21st centuries, and describe the roles of political organizations that promoted civil rights. • Resource ID: USHM1L2 • Subject: Social Studies ### Origins of the Progressive Era Given broad categories that describe the major goals of the progressive movement and general information about selected issues of the late 1800’s/early 1900’s, students will categorize the issues according to corresponding progressive era goals. • Resource ID: USHM1L3 • Subject: Social Studies ### The Problem of Child Labor in the Progressive Era After analyzing primary and secondary resources about the child labor, the students should be able to draw conclusions about the need to reform child labor practices. • Resource ID: USHM1L4	1,080	5,030	{"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}	3.328125	3	CC-MAIN-2019-35	latest	en	0.751388
https://www.physicsforums.com/threads/infinite-coupled-odes.355885/	1,521,546,998,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00372.warc.gz	845,253,386	16,400	# Infinite coupled ODE's 1. Nov 18, 2009 ### dirk_mec1 1. The problem statement, all variables and given/known data $$\dot{x}_i = \sum_{ n=1}^{\infty} a(n,i) x_n +b(n,i) y_n$$ $$\dot{y}_i = \sum_{ n=1}^{\infty} c(n,i) x_n +d(n,i) y_n$$ $$\forall i \in \mathbb{N}$$ 2. Relevant equations a,b,c and d are constants (though dependent on the constants n and i). 3. The attempt at a solution I want to know how I can solve such an infinite large coupled system of ODE's. Can someone help me? 2. Nov 18, 2009 ### Nick Bruno well, you have multiple unknowns and 2 equations. u need as many equations as you have unknowns. Maybe you can choose i such that you have enough equations to solve? 3. Nov 18, 2009 ### dirk_mec1 Maybe I wasn't clear enough, i runs from 1 till infinity, in the first post I've denoted that i is part of the natural numbers so i = 1,2,3,...,inf. 4. Nov 18, 2009 ### D H Staff Emeritus He has an infinite number of equations there, Nick. To the OP: Convert this problem to a set of equations of the form $$\dot u_i = \sum_{n=1}^{\infty}e_{n,i} u_n$$ with the following: \aligned u_{2n-1} &= x_n \\ u_{2n} &= y_n \\ e_{2n-1,2m-1} &= a_{n,m} \\ e_{2n-1,2m} &= b_{n,m} \\ e_{2n,2m-1} &= c_{n,m} \\ e_{2n,2m} &= d_{n,m} \endaligned Now you have a problem in one infinite-dimensioned vector (u) rather than two (x and y). See if you can take it from there. 5. Nov 19, 2009 ### dirk_mec1 Very smart: creating one infinite dimensional vector but shouldn't there be another summation sign for m? 6. Nov 19, 2009 ### D H Staff Emeritus No. Suppose that instead of an infinite number of x and y, we only have two: \aligned \dot x_1 &= a_{1,1} x_1 + b_{1,1} y_1 + a_{2,1} x_2 + b_{2,1} y_2 \\ \dot y_1 &= c_{1,1} x_1 + d_{1,1} y_1 + c_{2,1} x_2 + d_{2,1} y_2 \\ \dot x_2 &= a_{1,2} x_1 + b_{1,2} y_1 + a_{2,2} x_2 + b_{2,2} y_2 \\ \dot y_2 &= c_{1,2} x_1 + d_{1,2} y_1 + c_{2,2} x_2 + d_{2,2} y_2 \endaligned Define the four-vector $\vec u = [u_1, u_2, u_3, u_4]^T = [x_1, y_1, x_2, y_2]^T$. The above becomes \aligned \dot u_1 &= a_{1,1} u_1 + b_{1,1} u_2 + a_{2,1} u_3 + b_{2,1} u_4 \\ \dot u_2 &= c_{1,1} u_1 + d_{1,1} u_2 + c_{2,1} u_3 + d_{2,1} u_4 \\ \dot u_3 &= a_{1,2} u_1 + b_{1,2} u_2 + a_{2,2} u_3 + b_{2,2} u_4 \\ \dot u_4 &= c_{1,2} u_1 + d_{1,2} u_2 + c_{2,2} u_3 + d_{2,2} u_4 \endaligned This is just a matrix-vector equation: $\dot{\vec u} = \mathbf A \vec u$ if you treat u as a column vector (or $\dot{\vec u} = vec u \mathbf A^T$ if you use row vectors). Each of those a, b, c, and d elements maps to exactly one of the elements of the state matrix A. This won't change when you go to infinite dimensional space. 7. Nov 19, 2009 ### dirk_mec1 D_H, thanks for clarifying that. Now with this mapping the equation we've got is: $$\dot{ u} = \mathbf A u$$ Normally I would determine eigenvalues and eigenvectors to get an explicit solution because the general solution is: $$u(t) = e^{At}$$ But now I've got an infinite large matrix. I've thought about this and there's isn't an exact solution that can be found, right? 8. Nov 20, 2009 ### D H Staff Emeritus Infinite state matrices are the subjects of many books. Long books. Long books with lots of hairy math. The best I can do is refer you to some. Here is one: http://books.google.com/books?id=G_x-F-l2V2UC Google the term "infinite dimensional linear system" and you will find many more.	1,286	3,411	{"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}	3.796875	4	CC-MAIN-2018-13	latest	en	0.821864
https://uk.mathworks.com/matlabcentral/cody/problems/42643-matlab-basic-rounding-iii/solutions/2604145	1,611,067,718,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00193.warc.gz	629,244,877	17,134	Cody # Problem 42643. MATLAB Basic: rounding III Solution 2604145 Submitted on 24 Jun 2020 by Monalisa Pal 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 x = -8.8; y_correct = -9; assert(isequal(round_x(x),y_correct)) 2 Pass x = -8.4; y_correct = -9; assert(isequal(round_x(x),y_correct)) 3 Pass x = 8.8; y_correct = 8; assert(isequal(round_x(x),y_correct)) 4 Pass x = 8.4; y_correct = 8; assert(isequal(round_x(x),y_correct)) 5 Pass x = 8.49; y_correct = 8; assert(isequal(round_x(x),y_correct)) 6 Pass x = 128.52; y_correct = 128; assert(isequal(round_x(x),y_correct)) 7 Pass x = pi; y_correct = 3; assert(isequal(round_x(x),y_correct)) ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	286	908	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2021-04	latest	en	0.6579
https://www.chesstifton.com/single-post/8-29-30-23-first-grade-math	1,696,063,902,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00669.warc.gz	752,896,314	184,482	top of page # 8/29.30/23 first and second grade math First Grade Math 1.Complete page 19, 20 and 21 on Arithmetic 1 work text. 2. Trace and write number 1-10 worksheet. Second Grade Math 1. Complete Ordinal number worksheets. 2. write and answer addition fact 0+5= 3+3= 5+0= 0+6= 1+4= 1+5= 2+3= 2+4= 3+2= 5+1= 2+2= 4+2= ###### Search By Tags No tags yet. bottom of page	144	379	{"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}	2.734375	3	CC-MAIN-2023-40	longest	en	0.633413
https://www.physicsforums.com/threads/orientable-manifold-with-boundary.81202/	1,571,199,499,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00032.warc.gz	1,025,375,870	24,359	# Orientable Manifold with Boundary #### kakarukeys How does the orientation on M induce an orientation on the boundary of M? I follow the book Lectures on Differential Geometry by Chern, do not understand the proof. The proof is the Jacobian Matrix of the transformation between coordinates of two charts has positive determinant (oriented charts), so the smaller Jacobian Matrix with one row and one column deleted (corresponding to the only one coordinate axis that runs away from the boundary) has positive determinant. Please do me a favour by explaining the proof clearly, or give me another easier proof. Related Differential Geometry News on Phys.org #### Hurkyl Staff Emeritus Gold Member Hrm. I would have said "Because it does in Euclidean space", but I don't know how easy it is to prove you can patch together orientations of the pieces of the boundary when it spans multiple charts. #### mathwonk Homework Helper how does counter clockwise orientation on the upper half plne induce an orientation on the real line? by noticing that at a point of the real line, one can distinguish between an arrow pointing into the upper half plane and one pointing out of it. so a vector v along the x axis is oriented positively if, when it is supplemented by an arrow w pointing into the upper half plane, we get an oriented (counterclockwise) basis <v,w> for the (tangent space to the) upper half plane. #### mathwonk Homework Helper oh, that is the euliden space acse. but an oriented manifold is by definition one that admits an orientation preserving atlas, so those charts do preserve the euclidean orientation. #### kakarukeys Hrm. I would have said "Because it does in Euclidean space", but I don't know how easy it is to prove you can patch together orientations of the pieces of the boundary when it spans multiple charts. Suppose we have charts that cover regions not containing the boundary and charts that cover regions containing the boundary. They are oriented (coordinate transformations among them have positive-determinant Jacobian matrix). for a region D, let $${u^1, u^2, ...., u^m}$$ be the coordinates, suppose $$u^m = 0$$ is the boundary. Let $$du^1 \wedge du^2 \wedge ... \wedge du^m$$ be the positive orientation. then since $$du^1 \wedge du^2 \wedge ... \wedge du^{m-1}$$ is non-zero and gives an orientation on the boundary contained in region D. Minus sign gives another orientation. if we can transform to other charts on boundary, preserving the orientation, we can extend the orientation to the entire boundary. And thus we need to show $$\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0$$ implies $$\frac{\partial (v^1, ..., v^{m-1})}{\partial (u^1, ..., u^{m-1})} > 0$$ I don't understand the proof of the above in the book, anyone has any idea? I can type the proof out, if you want. Last edited: #### mathwonk Homework Helper did my comments mean nothing at all? #### mathwonk Homework Helper another suggestion: just ebcause chern is a great mathematiciazn does not mean he writes learnable boks. read guillemin and pollack instead. or my notes. i will send them to you if you give me an address. #### kakarukeys I am still struggling with the steps of the proof. #### mathwonk Homework Helper the point is that if you have an n-1 dimensional subspace spanned by the first n-1 vectors in a basis, and then the last vector points in a giuven direction, if you choose another such absis, the change of basis matrix will be in 2 blocks. there will be an n-1 by n-1 block in the upper left corner say, and a positive number in the lower right corner, and all zeroes otherwise in the last row and last column. hence the determinant of the whole matrix being positive forces also the determinant of the upper left block to be positive also. but i warn you, chern's writings are the worlds most terse and merciless. they are also non conceptual and highly compoutational. he was a great great mathematician but i do not recommend his works to anyone as the sole source for learning any subject at all. here is a review by another outstanding mathematician: ""This excellent and polished book may not be suitable for the very beginning student, but it is highly recommended for all mathematicians, from the advanced undergraduate student to the experienced professor. For many mathematicians it will be a work of reference for their research. It will be welcome by physicists." Prof. F Hirzebruch Max-Planck Institute, Bonn" Last edited: #### mathwonk Homework Helper i will try to pdf my notes and make them sendable. but they are just a rewrite of the treatment in guillemin and pollack which may be in your library. Homework Helper #### kakarukeys mathwonk said: i will try to pdf my notes and make them sendable. but they are just a rewrite of the treatment in guillemin and pollack which may be in your library. Please don't if it is too laborious, I don't want to waste your precious time. mathwonk said: the point is that if you have an n-1 dimensional subspace spanned by the first n-1 vectors in a basis, and then the last vector points in a giuven direction, if you choose another such absis, the change of basis matrix will be in 2 blocks. there will be an n-1 by n-1 block in the upper left corner say, and a positive number in the lower right corner, and all zeroes otherwise in the last row and last column. hence the determinant of the whole matrix being positive forces also the determinant of the upper left block to be positive also. In my case, I only have $$\frac{\partial v^m}{\partial u^m} > 0$$ $$\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0$$ There may not be zeroes in the last row and last column. #### mathwonk Homework Helper well if you think about the definition of a matrix of a linear transformation, the last row is hte image of the last vector, expanded in terms of the basis ofm the target space. now we have chsoen both bases to end with a vector pointing into the manifold, so i guess al i can say is that the last entry of the last comloumn is positive, since the component of an inner pointing vector is a positive multiple of another inner pointing vector. but if you look at the previous columns, you are expanding the images of the basis vectors for the boundaryu space, in terms of anmother absis for that same subspace. so there is no need to use the last inner pointing vector in that expansion. so every columnh except the last has a zero in the last entry. thus the last row is all zeroes except for the alst right bottom corner. then the same argument applies. i.e. the determinant is positive iff the determinant of the upper left block is positive. how about that? #### kakarukeys your argument is sound unless the following is true. a cotagent space at a point on the boundary is a direct sum of the cotagent space of that point when the boundary is considered an independent manifold and its complement. Every first m - 1 basis vectors given by any chart span the first subspace, and the last vector pointing into the manifold always spans the other subspace. I think it's true by definition of the chart. or is it true? Last edited: #### kakarukeys only then can it be that when we transform from u basis to v basis, the first m-1 vectors are shuffled. and the last vector is only multiplied by a constant. #### kakarukeys Ok, forget about the cotangent space direct sum. Here's the argument We know given two charts, at a point on boundary the first m - 1 cotangent vectors obtained $$\{du^1, .... du^{m-1}\}, \{dv^1, .... dv^{m-1}\}$$ span the cotangent space of the boundary at the point, by definition of the chart. so entries of last column of the m by m Jacobian Matrix we considered are all zeros except the last one. (Because $$\{dv^1, .... dv^{m-1}\}$$ can be expressed solely as a linear combination of $$\{du^1, .... du^{m-1}\}$$). And so the conclusion follows. P/S the last vector $$dv^m$$ need not be multiple of $$du^m$$, it can contains some linear combination of $$\{du^1, .... du^{m-1}\}$$. Last edited: #### mathwonk Homework Helper it depends on your definition of the complementary space. some people take the normal space to be the quotient of the full tangent space by the tangent space to the boundary. when you do that lkinear combinations of baoiundary tangent vectors are set to zero, so the normal vectors are scalar multiples of each other in that quotient space. that is the point of view in guillemin pollack, and hence also in the pdf notes i sent you. but you seem to understand it now. #### kakarukeys I thought I understand, but when I tried a simple example, I couldn't get what I want. Consider the Manifold, closed interval [0, 1] Using two charts: [0, 1) -> [0, 1) $$\phi_1(x) = x$$ (0, 1] -> [0, 1) $$\phi_2(x) = -x + 1$$ The first chart is positively oriented, the second chart is negatively oriented. So the orientation of the boundary point is +1 and -1 respectively. since $$+1 \wedge dx = dx =$$ gives the positive orientation on the first chart. since $$-1 \wedge dx = -dx$$ gives the "positive orientation" on the second chart. then applying Stokes' theorem we have, $$\int^1_0 df = \int_{0, 1} f(x) = f(0) - f(1)$$ since the 2nd chart is negatively oriented. #### Doodle Bob kakarukeys said: I thought I understand, but when I tried a simple example, I couldn't get what I want. Consider the Manifold, closed interval [0, 1] Using two charts: [0, 1) -> [0, 1) $$\phi_1(x) = x$$ (0, 1] -> [0, 1) $$\phi_2(x) = -x + 1$$ The first chart is positively oriented, the second chart is negatively oriented. The above example above doesn't count since you've chosen an atlas that does not induce an orientation on the interior manifold. Orientability implies there exists an atlas with coordinate charts whose Jacobians have pos. determinant, but it doesn't mean that any atlas of the manifold will have that property -- as you can see above. In order to use Stoke's theorem you need to settle on a particular orientation (you do after all have a choice of two) and use that. #### kakarukeys The above example above doesn't count since you've chosen an atlas that does not induce an orientation on the interior manifold. Orientability implies there exists an atlas with coordinate charts whose Jacobians have pos. determinant, but it doesn't mean that any atlas of the manifold will have that property -- as you can see above. In order to use Stoke's theorem you need to settle on a particular orientation (you do after all have a choice of two) and use that. I am well aware of that. So I put a minus sign in front, whenever I want to compute integral using the chart with negative orientation. The effect is same as sticking to a particular atlas with oriented charts. So there is nothing wrong in my example, but the answer is wrong. #### Doodle Bob kakarukeys said: I am well aware of that. So I put a minus sign in front, whenever I want to compute integral using the chart with negative orientation. The effect is same as sticking to a particular atlas with oriented charts. So there is nothing wrong in my example, but the answer is wrong. I think you're confusing which determinants of what mappings need to be positive or negative in order to induce an orientation. An atlas with coordinate charts induces an orientation if and only if the Jacobians of the transitions functions, i.e. the $$\phi_1 \circ \phi_2^{-1}$$ on the intersection of open sets of the atlas, all have positive determinant. If one of those transition functions has a neg. det. Jacobian, then that means you've switched from one orientation to the other on that intersection, just as you've done in your example. The two possible orientations on [0,1] are given by the volume forms dx and -dx. phi_1 induces dx, whereas phi_2 induces -dx. Thus, the mapping $$\phi_1 \circ \phi_2^{-1}$$ is in fact orientation-reversing. #### kakarukeys I haven't been able to find an atlas which has oriented charts that cover [0, 1] do you know one? #### Doodle Bob kakarukeys said: I haven't been able to find an atlas which has oriented charts that cover [0, 1] do you know one? just change phi_2 to phi_2(x)=x+1. Trivial, yes, but [0,1] is a rather trivial manifold. #### kakarukeys In that case, you are also using a chart that does not satisfy the requirements given in the definition. that $$\phi_2 : U \rightarrow H^n$$ $$H^n = \{x : x^n \ge 0\}$$ boundary is given by $$\{x : x^n = 0\}$$ I am not sure if you use such chart, the stokes' theorem still holds. anyway please show me how you prove fundamental theorem of calculus from stokes' theorem. ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving	3,166	13,002	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.328125	3	CC-MAIN-2019-43	longest	en	0.928246
https://www.physicsforums.com/threads/ac-circuit-is-the-voltage-source-absorbing-or-delivering.699105/	1,508,213,698,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187820700.4/warc/CC-MAIN-20171017033641-20171017053641-00445.warc.gz	1,231,399,372	17,679	# AC circuit: is the voltage source absorbing or delivering? 1. Jun 28, 2013 ### mrhingle I'm looking at an AC circuit. There is a voltage source, a resistor, an inductor, then another voltage source. One loop. I am supposed to find the power in each element and tell weather it is absorbing or delivering. I think I remember that when a current goes into the negative terminal of a voltage source, the source is delivering, and in an AC circuit the curent goes back and forth. I found the current to be 21 ∠93° Amps. The voltage on the left is 240√2∠60°, and the one on the right is 280√2∠30°. Both have negative terminals at the ground. The resistor is 5Ω, and the inductor is J8Ω. Do I need to use superposition. Can I find the power with the total current that I found using mesh????? What's happening here? 2. Jun 28, 2013 ### Staff: Mentor Why the √2's in the voltage descriptions? Are you converting from RMS to peak? If you're dealing with power you should leave things as RMS, or convert to RMS if peak values are given. You've remembered correctly regarding how to determine if a source is delivering (sourcing) or consuming (sinking) power. You can use superposition if you wish. But it looks like you've already determined the net current. What is the the assumed direction direction of your current? Do you know how to determine the complex power associated with a voltage source given the current out of the + terminal (same as into the - terminal)? The real component of the complex power is the real power. 3. Jun 28, 2013 ### mrhingle I assume the book used square root of 2's to make for easy transfer to RMS. I am given Peak, and plan to convert to RMS. To find the current I assumed clockwise around the mesh. Complex power = 1/2 VI. Never considered direction of current when looking at it. Pretty new concept to me. So how do I use the assumed direction of current to know if the voltage source is absorbing or delivering? 4. Jun 28, 2013 ### Staff: Mentor Okay, so drop the square roots from the peak values at the outset. Then the complex power for the source will be just VI. With the assumed clockwise direction for the current, the complex power produced by the first source is V1I. Do the math and take a look at the sign of real component. If it's positive, the source is producing power. If it's negative, it's consuming power. Same goes for the second source, only here the assumed current direction is flowing INTO the source's + terminal. So V2I yields the power consumed. If the sign of the real term is negative, it's "consuming" a negative amount of power, i.e., it's producing power. 5. Jun 29, 2013 ### rude man I would proceed as follows: Without loss of generality, make V1 angle = pi/6, then V2 angle = 0. V1 on left, V2 on right. Compute complex I. Using superposition is a good idea. Compute I, the complex conjugate of I. Also compute complex conj. of V1. You already have complex conj. of V2 of course. Then compute P1 = 1/2 (V1 I + V1* I) and P2 = 1/2 (V2 I* + V2* I). Either source will be generating real power if its P is +, and will be absorbing real power if its P is -. Hints: 1. Use exponential notation thruout, including Z. 2. [exp(j theta)]* = exp(-j theta).	815	3,242	{"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-2017-43	longest	en	0.903162
https://justaaa.com/computer-science/1210502-please-do-it-in-python-write-the-simplest-program	1,726,583,725,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00372.warc.gz	294,090,541	12,733	Question # Please do it in Python Write the simplest program that will demonstrate iteration vs recursion using... Write the simplest program that will demonstrate iteration vs recursion using the following guidelines - 1. Write two primary helper functions - one iterative (IsArrayPrimeIter) and one recursive (IsArrayPrimeRecur) - each of which 1. Take the array and its size as input params and return a bool. 2. Print out a message "Entering <function_name>" as the first statement of each function. 3. Perform the code to test whether every element of the array is a Prime number. 4. Print out a message "Leaving <function_name>" as the last statement before returning from the function. 5. Remember - there will be nested loops for the iterative function and there can be no loops at all in the recursive function. You will need to define other helper functions to make the recursive method work. 6. Remember - replace <function_name> with the actual name of each function. 7. You can create secondary helper functions if you need, but only the primary helper functions can be invoked from 'main'. 8. Hint - try to complete your iterative code first and then convert it piece by piece to the recursive code. 1. Ask the user for the number of elements, not to exceed SORT_MAX_SIZE = 16 (validate input). 2. Create an array based on the size input provided by the user. 3. Get the input in a loop, validating that the integers input by the user are between 1 and 99, both inclusive 4. Make a call to the primary iterative function passing the array and its size. 5. If every member is a Prime, then the program should print out 'Prime Array using iteration', otherwise print out 'Not a Prime Array using iteration'. 6. Then make a call to the primary recursive function passing the array and its size. 7. If every member is a Prime, then the program should print out 'Prime Array using recursion', otherwise print out 'Not a Prime Array using recursion'. 8. If your functions are coded correctly, both should come up with the same answer, except you should have lots more output statements using recursion. 9. There is no sample output - you are allowed to provide user interactivity as you see fit but programs will be graded for clarity of interaction. 3. You can use language native arrays - DO NOT USE VECTORS, COLLECTIONS, SETS, BAGS or any other data structures from your programming language. 4. There will be only one code file in your submission. 5. Remember to take multiple screenshots so that they are clearly readable without needing to zoom in. 6. For documentation, include your name block as well pre/post and pseudocode for the functions only. 7. Upload your code file and the screenshots in one zip file. Do not include anything else. • 25 pts - EXE works from your code as explained above without needing any code change • 15 pts - for the main • 25 pts - the IsArrayPrimeIter function • 35 pts - the IsArrayPrimeRecur function I understand most of this information is more suitable to C++ but our instructor wants us to modify it to do it in Python. As long as you fufill the parameters the best you can in Python and works that all I want. Thank you Program: ``````# function is used to check whether a given number or not using iteration def isPrime(n): if n <= 1: return False for i in range(2, n): if n % i == 0: return False; return True # function is used to check whether a given list has all prime numbers or not using iteration def check_list_prime_iteration(arr,le): count=0 for i in arr: if(isPrime(i)): count=count+1 if(count==le): return True else: return False # function is used to check whether a given list has all prime numbers or not using recusrion def check_list_prime_recursion(arr,le): count=0 for i in arr: if(isPrime_Rec(i)): count=count+1 if(count==le): return True else: return False # function is used to check whether a given number or not using recusrion def isPrime_Rec(n, i = 2): if (n <= 2): return True if(n == 2) else False if (n % i == 0): return False if (i * i > n): return True return isPrime_Rec(n, i + 1) # List to store values l=[] i=1 # input taking n=int(input("Enter size of an Array")) while(i<=n): input_value=int(input("Enter number")) if(input_value>=1 and input_value<=99): l.append(input_value) else: print("please enter valid input between 1 and 99") i=i-1 i=i+1 # Final list after validation print("Array List",l) if check_list_prime_iteration(l,len(l)): print("Prime Array using iteration") else: print("Not a Prime Array using iteration") if check_list_prime_recursion(l,len(l)): print("Prime Array using recursion") else: print("Not a Prime Array using recursion")`````` Output: #### Earn Coins Coins can be redeemed for fabulous gifts.	1,118	4,736	{"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-2024-38	latest	en	0.808202
https://www.easyteacherworksheets.com/math/ratios-radical.html	1,719,012,653,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00803.warc.gz	646,653,972	6,353	In these worksheets, students will learn to rationalize denominators with radicals. Rationalizing denominators is the process of rewriting a fraction so that the denominator contains only rational numbers. A radical expression is an expression containing a radical (v) (sometimes also called a square root) symbol. In many of these problems, the radicals are in the denominators. Students will rewrite expressions so that the radical in each expression is in the numerator. Students will also simplify fractions that contain a radical in either the numerator or the denominator. This set of worksheets contains step-by-step solutions to sample problems, both simple and more complex problems, a review, and a quiz. There are also worksheets provided for students to use to practice independently. When finished with this set of worksheets, students will be able to rewrite expressions so that there is no radical in the denominator, and simplify fractions containing radicals. These worksheets explain how to rationalize denominators with radicals. Sample problems are solved and practice problems are provided. Lesson This worksheet explains how to rationalize denominators the expression: 4 / √3. The sample problem is solved and two practice problems are provided. A really nice mix of problems where you may: rationalize denominators, simplify values, or write equivalent expressions. Ten problems are provided. Practice Students will practice rationalizing denominators and evaluate a slew of radicals. Ten problems are provided. Review The concept of how to rationalize denominators with radicals is reviewed. A sample problem is solved and six practice problems are provided. Quiz Students will demonstrate their proficiency with this skill on this quiz. Ten problems are provided. Check This is a great way to introduce or review this skill. Three problems are provided, and space is included for students to copy the correct answer when given.	360	1,964	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.3125	3	CC-MAIN-2024-26	latest	en	0.944534
http://www.stata.com/statalist/archive/2009-10/msg01112.html	1,438,600,036,000,000,000	text/html	crawl-data/CC-MAIN-2015-32/segments/1438042989826.86/warc/CC-MAIN-20150728002309-00336-ip-10-236-191-2.ec2.internal.warc.gz	748,107,896	4,203	Re: R: st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE) From gjhxmu@sina.com To statalist Subject Re: R: st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE) Date Mon, 26 Oct 2009 01:56:59 +0800 ```Dear Carlo, thank you very much for your guidance. I will read it later. Best regards, Rose. ----- Original Message ----- From: Carlo Lazzaro <carlo.lazzaro@tin.it> To: <statalist@hsphsun2.harvard.edu> Subject: R: st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE) Date: 2009-10-26 00:00:38 Dear Rose, The formula used in Stata (release 9/2 SE) is reported in: [R] Stata Base Reference Manual. Volume 3. R-Z. Release 9: 1-5. Kind Regards, Carlo -----Messaggio originale----- Da: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] Per conto di gjhxmu@sina.com Inviato: domenica 25 ottobre 2009 15.00 A: statalist Oggetto: Re: st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE) Dear Carlo and Martin, thank you very much for your help, especially for Carlo's detailed illustration. Just as Carlo's guess, I followed the formula from textbooks as follows to get the z-statistic. Assuming the number for the two independent sample is m and n respectively. Firstly to get the U statistic, U=w-1/2k(k+1) // where w is the rank sum for the bigger number and k is the bigger number. If m equals n, k takes the number for the group first appearing in the rank. Then to get the Z statistic, Z=(U-1/2mn)/sqrt(1/12mn(m+n+1)) Following the formula, I think no matter which group is compared with which group, the Z statistic is the same. I am sorry for not reading the concrete formula used in the stata. Thank you for your kind help. Best regards, Rose. ----- Original Message ----- From: Carlo Lazzaro <carlo.lazzaro@tin.it> To: <statalist@hsphsun2.harvard.edu> Subject: st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE) Date: 2009-10-25 19:18:29 Dear Rose, the following examples may shed light on the oddity you came across in performing -ranksum-. webuse fuel2,clear ranksum mpg, by(treat) Two-sample Wilcoxon rank-sum (Mann-Whitney) test treat \| obs rank sum expected -------------+--------------------------------- 0 \| 12 128 150 1 \| 12 172 150 -------------+--------------------------------- combined \| 24 300 300 ---------- Ho: mpg(treat==0) = mpg(treat==1) z = -1.279 Prob > \|z\| = 0.2010 return list scalars: r(N_2) = 12 r(N_1) = 12 r(Var_a) = 295.9565217391304 r(z) = -1.278817949868369 r(sum_exp) = 150 r(sum_obs) = 128 r(group1) = 0 As you can see, after -return list-, Stata consider - r(sum_obs) = 128 -. z can be obtained via the following formula (see also: Pagano M, Gauvreau K. Principles of Biostatistics. 2nd edition. Brooks/Cole, 2000) di (128-150)/295.96^.5 -1.2788104 When you decide to invert the samples rank-sum: ranksum mpg, by(treat2) Two-sample Wilcoxon rank-sum (Mann-Whitney) test treat2 \| obs rank sum expected -------------+--------------------------------- 0 \| 12 172 150 1 \| 12 128 150 -------------+--------------------------------- combined \| 24 300 300 ---------- Ho: mpg(treat2==0) = mpg(treat2==1) z = 1.279 Prob > \|z\| = 0.2010 . return list scalars: r(N_2) = 12 r(N_1) = 12 r(Var_a) = 295.9565217391304 r(z) = 1.278817949868369 r(sum_exp) = 150 r(sum_obs) = 172 r(group1) = 0 As you can see after -return list-, Stata consider - r(sum_obs) = 172 -. Again, z can be obtained via the following formula (see also: Pagano M, Gauvreau K. Principles of Biostatistics. 2nd edition. Brooks/Cole, 2000) di (172-150)/295.96^.5 1.2788104. As Martin said, usually <the test is against a two-sided alternative, so the sign hardly matters...> However, your concern is probably driven by the awareness that some textbooks on statistics report the following formula for calculating Wilkoxon rank sum test (see again: Pagano M, Gauvreau K. Principles of Biostatistics. 2nd edition. Brooks/Cole, 2000): zw = (W-mw)/sw Where the z-statistic is obtained by subtracting the mean of the ranks sum from W (the smallest of the two rank sums)and ; in this way, as you stated in your thread, <z statistic the result should be the same no matter which group is compared with which group>. HTH and Kind Regards, Carlo -----Messaggio originale----- Da: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] Per conto di gjhxmu@sina.com Inviato: domenica 25 ottobre 2009 4.29 A: statalist Oggetto: st: the sign of z statatistic after -ranksum-? Dear statalists, I use -ranksum- in the stata to do the Wilcoxon rank-sum test, which is also known as the Mann-Whitney two-sample statistic. My typing is as followings, webuse fuel2,clear ranksum mpg, by(treat) // the z statistic equals -1.279 replace treat=treat==0 ranksum mpg, by(treat) // the z statistic equals 1.279 My doubt is why the sign of z statistic is opposite. As far as I know, in the computation of z statistic the result should be the same no matter which group is compared with which group. Any help will be appreciated! Best regards, Rose. * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ ```	1,861	5,908	{"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}	3.59375	4	CC-MAIN-2015-32	longest	en	0.77324
https://es.mathworks.com/matlabcentral/cody/problems/44436-find-the-largest-number/solutions/2736232	1,606,201,163,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141171126.6/warc/CC-MAIN-20201124053841-20201124083841-00499.warc.gz	291,132,562	17,413	Cody # Problem 44436. Find the largest number Solution 2736232 Submitted on 24 Jul 2020 by Raghunadh N 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 filetext = fileread('largest.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden') assert(isempty(strfind(filetext, 'max')),'max() forbidden') assert(isempty(strfind(filetext, 'min')),'min() forbidden') assert(isempty(strfind(filetext, 'sort')),'sort() forbidden') assert(isempty(strfind(filetext, 'unique')),'unique() forbidden') 2 Pass a = 4; b = 7; c = 1; d = 6; x_correct = 7; assert(isequal(largest(a, b, c, d),x_correct)) 3 Pass a = 1; b = 1; c = 1; d = 1; x_correct = 1; assert(isequal(largest(a, b, c, d),x_correct)) 4 Pass a = 4; b = 3; c = 2; d = 1; x_correct = 4; assert(isequal(largest(a, b, c, d),x_correct)) 5 Pass a = 1; b = 2; c = 3; d = 4; x_correct = 4; assert(isequal(largest(a, b, c, d),x_correct)) 6 Pass a = 1; b = 1; c = 3; d = 5; x_correct = 5; assert(isequal(largest(a, b, c, d),x_correct)) 7 Pass a = 3; b = 3; c = 2; d = 4; x_correct = 4; assert(isequal(largest(a, b, c, d),x_correct)) 8 Pass a = 2; b = 3; c = 1; d = 6.5; x_correct = 6.5; assert(isequal(largest(a, b, c, d),x_correct)) 9 Pass a = 3; b = 3; c = 3; d = 9; x_correct = 9; assert(isequal(largest(a, b, c, d),x_correct)) 10 Pass a = 5; b = 3; c = 3; d = 9; x_correct = 9; assert(isequal(largest(a, b, c, d),x_correct)) 11 Pass a = 3; b = 3; c = 3; d = 1; x_correct = 3; assert(isequal(largest(a, b, c, d),x_correct)) 12 Pass a = 3; b = 3; c = 9; d = 9; x_correct = 9; assert(isequal(largest(a, b, c, d),x_correct)) 13 Pass a = 3; b = 3; c = 1; d = 1; x_correct = 3; assert(isequal(largest(a, b, c, d),x_correct)) ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	757	1,965	{"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}	3.71875	4	CC-MAIN-2020-50	latest	en	0.448782
https://web2.0calc.com/questions/please-translate-the-sentence-to-an-inequality	1,544,507,093,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376823565.27/warc/CC-MAIN-20181211040413-20181211061913-00114.warc.gz	778,632,642	5,626	+0 # please translate the sentence to an inequality 0 266 2 twice the difference of a number and 7 is at most -27 Guest Apr 11, 2017 #1 +92429 +1 Let N be the number.....so .... 2(N - 7) ≤ -27 CPhill Apr 11, 2017 #2 0 thank you so much Cphill Guest Apr 11, 2017	110	272	{"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-2018-51	latest	en	0.820354
https://www.scribd.com/document/253788215/Diseno-de-Puente-Losa-Estribos	1,568,770,969,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514573173.68/warc/CC-MAIN-20190918003832-20190918025832-00182.warc.gz	1,002,012,211	80,295	You are on page 1of 82 # DISEO ESTRUCTURAL DE UN PUENTE PROYECTO: PROFESOR 16 ## PERALTE DE VIGA =>( H=L/15)= 1.1 (H=0.07L) = 1.1 No DE VIGAS= 1.12 NOTA: Todas las unidades deben estar en : 0.75 0.05 7.2 8.8 0.2 0.15 LOSA 0.2 0.9 0.8 DIAFRAGMA DIAFRAGMA 0.25 0.867 0.7 2.2333 2.1333 S= 0.7 2.233 2.9333 2.933 8.30 S (espaciamiento de vigas de eje a eje) t (espesor de losa en metros) S t 1.8 0.16 2.1 0.165 2.4 0.18 = 2.7 0.19 3 0.20 4 0.22 2.93 0.2 4.5 0.25 ## a) DISEO A FLEXION DE VIGAS CALCULO DEL MOMENTO ULTIMO MAXIMO d1 SOBRE CARGA : AASHTO HS-20 d1= 0.6 1.8 2.67 2.67 2.93 Coef.Concentracin de Carga = 2.43 4 B= 16 3.97 16 15.3 A= 1.69 C= 2.05 16.7 125 ## POR SOBRECARGA EQUIVALENTE = (P=8 Ton w=1 Ton/m) 51 ## SE ADOPTARA EL MAYOR VALOR MOMENTO POR IMPACTO = ( I= 15.24/(L+38) ) N.R 16 ## Ton-m / eje de rueda Ton-m / via de 3m de ancho ML= 125.0 Ton-m MI= 35.3 Ton-m Pgina 1 0.7 I= 0.28222 < 0.3 ASFALTO BARANDA LOSA VIGA = = = = = 0.360 0.300 1.467 1.575 0.291 3.992 Ton/m ## CADA QUE DISTANCIA DE EJE A EJE SE COLOCARAN LOS DIAFRAGMAS ( D < LUZ / 4 = D= 4 No DE DIAFRAGMAS E= d d d d d d mts. 0.2 3.12 3.12 0.89 0.89 0.4 ## Peso propio del diafragma= Y= 3.12 mts ) 0.89 Ton 1.56 1 2 3 4 5 6 = = = = = = 1 2 3 3 2 1 #VALUE! #VALUE! #VALUE! 0 1.56 3.12 3.12 1.56 0 0.89 0.89 d1 d2 d3 w= d.. 9.36 MOMENTO POR CARGA MUERTA MD = 136 Ton-m ## DISEO POR METODO A LA ROTURA MU=1.3( MD+1.67( ML+MI ) ) => f 'c= Fy= b= d= 240 kg/cm2 4200 kg/cm2 243.33 cm 87 cm w= MU= 525 Ton-m 0.1442 &= As(cm2)= 0.008 < 174.439 cm2 75&b= ## 0.024 ( FALLA DUCTIL ) 35 VARILLAS DE 1" As principal(+) = As,min= 174.4 cm2 ## ACERO PARA MOMENTOS NEGATIVOS M(-)= 175 w= 0.04519 As(cm2)= &= 54.667 cm2 0.00258 <> Pgina 2 < 75&b= 11 VARILLAS DE 1" ## 0.0242857143 ( FALLA DUCTIL ) B 11 var 1" 2 var 5/8" 2 var 5/8" __________________________ 2.5 m <= 8 9 9 9 var 1 var 1 var 1" var 1" b= 2.93 b= 0.2 2.93 < VIGA RECTANGULAR E.N (b-bw)/2= 1.1167 < 1.6 VIGA RECTANGULAR (b-bw)/2= 1.1167 < 1.11667 VIGA RECTANGULAR C bw= 0.7 C= 0.174 < 0.200 VIGA RECTANGULAR ## DISEO DE VIGA POR CORTE Y= d d d d d d 0.2 1 2 3 4 5 6 = = = = = = 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 0.89 0.89 0.89 w= d.. d. d4 d5 d6 3 CORTANTE POR PESO PROPIO VD= 35 Ton VL= 36 Ton VI= 10.0 Ton ## DISEO POR METODO A LA ROTURA VU=1.3( VD+1.67( VL+VI ) ) => Vu= 145.366 Ton Vcon= fi(0,5(fc)^0,5+175&Vud/Mu V que absorve el concreto => Vcon= ## V que absorve acero = Vace= Vu - Vcon= S= Avfyb/Vace S= 7 cm S= Vace= 42.5 Ton 102.9 10 cm VAR. 1/2" Pgina 3 Ton 8.75 8 9.0 B 11 var 1" 2 var 5/8" 1a .05, VAR. 1/2" 22 a 0.10 ,6 a 0.20, r a 0.30 /e 2 var 5/8" __________________________ A <= var 1 var 1 var 1" var 1" b) DISEO DE LOSA (Tramo interior) ( MD) Peso propio= Asfalto= 0.5 Ton/m 0.1 Ton/m w= ## DIAGRAMA MOMENTOS POR PESO PROPIO 0.30 (-) 0.6 Ton/m NOTA: Consideraremos un coeficiente de ( 1/10 ) para los momentos positivos y negativos por peso propio = LOSA 2.23 VIGA 8 9 9 9 2.23 MD= 0.30 2.23 VIGA 2.5 m 0.30 Ton-m ## DIAGRAMA MOMENTOS POR SOBRE CARGA 2.10 (-) ML=(L+0.61)P/9.74 ML= 2.33539 LOSA ML(+)= ML(-)= VIGA MOMENTO POSITIVO MOMENTO NEGATIVO 1.87 Ton-m 2.10 Ton-m 2.23 VIGA 2.23 1.87 0.80 (-) I=15.24/(L+38) = 0.3788 LOSA 2.23 0.71 Ton-m 0.80 Ton-m 2.23 VIGA MI(+)= MI(-)= VIGA MOMENTO POSITIVO MOMENTO NEGATIVO 0.71 ## DISEO POR METODO A LA ROTURA MU( + ) = 1.3( MD+1.67( ML+MI ) ) = 6.68 (-) 5.9816 Ton-m 6.68 Ton-m LOSA 240 kg/cm2 4200 kg/cm2 100 cm 17 cm VIGA f 'c= Fy= b= d= ## ACERO PARA MOMENTOS POSITIVOS w= 0.10193 As(cm2)= &= 9.90210787 cm2 0.006 < VAR, 5/8" As principal(+) = Pgina 4 75&b= @ 20 cm 9.9 cm2 2.23 VIGA 2.23 5.98 w= 0.114768 &= As(cm2)= 11.149 cm2 As temp= <> < VAR, 75&b= 5/8" 1/2" 18 cm 42 As rpp(+)= As rpp(+)= VAR, VAR, 1/2" 3.06 cm2 VAR, 0.0018bd As temp= VAR, 0.007 1/2" 1/2" 3.00 cm2 @ 42 cms, 42 cm 42 cm VAR, VAR, losa = trapecio = Asfalto= Baranda= pesos 0.43333 0.29063 0.36 0.15 brazos 0.783 4.025 1.8 2.6 MD= momentos (Ton-m) 0.33944 1.16977 0.648 0.39 2.5472 Ton -m E=0.8X+1.143 X= E= 1.48 2.3297 Mv= Mv= P.X/E 5.09372 Mv= ML= 5.0937 Ton-m MI= 1.4376 Ton-m Pgina 5 Mu= ## DISEO POR METODO A LA ROTURA MU( + ) = 1.3( MD+1.67( ML+MI ) ) = f 'c= Fy= b= d= 17 Ton-m 240 kg/cm2 4200 kg/cm2 100 cm 17 cm w= 0.34055 As(cm2)= &= 0.019 33.082216 cm2 < 75&b= VAR. 5/8" 6 cms VAR. 5/8" 6 cms Asfalto= losa= viga= 0.4 t/m 2 t/m 0.8 t/m WD= 3.2 t/m WL= 4 t/m WU= 12 t/m 0.37 (-) Mu(+)= 0.42 ton-m DIAFRAGMA VIGA 0.37 ton-m VIGA LOSA Mu(-)= DIAFRAGMA 0.42 f 'c= Fy= b= d= 240 kg/cm2 4200 kg/cm2 312 cm 92.0 cm w= 7E-005 As(cm2)= &= 0.12077818 cm2 0.0000 < 75&b= Asmin= 12.27 cm2 12.27 cm2 6 VARILLAS DE Pgina 6 5/8" f 'c= Fy= b= d= 240 kg/cm2 4200 kg/cm2 40 cm 92 cm w= 0.0005 &= As(cm2)= 0.11 cm2 0.000 < 75&b= Asmin= 12.27 cm2 12.27 cm2 6 VARILLAS DE 5/8" B 6 varillas 5/8" 2 varillas 5/8" 1a .05, VAR. 3/8" 14 a 0.10 , r a 0.20 /e 2 varillas 5/8" 6 varilas 5/8" B b= 3.12 0.2 E.N (b-bw)/2= 1.36 < 1.6 VIGA RECTANGULAR (b-bw)/2= 1.36 < 1.36 VIGA RECTANGULAR C bw= 0.4 ## PROFUNDIDAD DEL EJE NEUTRO (C) C= 0.010 < 0.2 VIGA RECTANGULAR e) DISEO DE ESTRIBOS DATOS PARA EL DISEO TIPOS DE SUELOS ANGULO DE PESO UNITARIO (T/m3) 1.73 - 2.20 1.57 1.57 - 1.73 1.57 - 1.73 1.75 - 2.05 3.0 -8.0 0.5 -2.0 2.0 -3.0 3.0 -6.0 0.5 -2.0 FRICCION INTERNA 353035356- GRAVA ARENA FINA ARENA MEDIA ARENA GRUESA ARCILLA 45 35 40 40 7 3.0 Kg/cm2 35 1.78 Ton/m3 140 Kg/cm2 175 Kg/cm2 F'c= 240 Kg/cm2 4.7 Ton/m ## REACCION DEL PUENTE SOBRECARGA/ m= 5.5 Ton/m Pgina 7 (verificar in situ) Longitud de Soporte L,Smin= 0.38 m 0.65 0.8 H : V <=> 1 : 5 : b > 1.4 m A 5 1.15 =3.5 3.5 =V 5.65 B 0.5 0.5 H= 0.70 m b= 2.15 3.15 ## ANALISIS DE LA SECCION A-A E=0,5PUh(h+2h)tan(45-&/2)^2 E= 0.657 ton Ev= 0.20 ton Eh= 0.63 ton dh= (h/3)(h+3h)/(h+2h) dh= 0.4812 m b= 0.65m P1 Ev Tot, Ev P pi(tn) xi(m) Mi 1.71925 0.325 0.55875625 0.20 0.65 0.1277771 1.91583 0.68653335 1.15m P1 &/2 Eh A Xv= 0.20079 m b/2 Xv e=b/2-Xv e= 0.12 m < 0.26 OK ! ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD q1,2=(Fv/b)(1=6e/b) q1= 0.63268 kg/cm2 q2= -0.0432 kg/cm2 q1= 0.632682 < 70 kg/cm2 OK ! q2= -0.0432 < 70 kg/cm2 OK ! Pgina 8 ## (esfuerzo de compresion del concreto) dh CHEQUEO AL DESLIZAMIENTO F,S,D= Vconcrt/Eh= F,S,D = 0.850.5f'c^0.5bd/ Eh 49.2952 > ## (cortante del concreto/empuje horizontal) 2 OK ! CHEQUEO AL VOLTEO F,S,V = F,S,V = 2.27439 > 1.25 OK ! ## (sin considerar el aporte de esfuerzo a la compresion del concreto como momento estabilizante) ## ANALISIS DE LA SECCION B-B ESTRIBO SIN PUENTE Y CON RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 6.6 ton Ev= 1.98 ton Eh= 6.31 ton dh= dh= (h/3)(h+3h)/(h+2h) 1.709 m 0.8 P1 P2 P3 Ev Tot, pi(tn) xi(m) Mi 6.95175 1.825 12.6869438 6.44 1.1 7.084 2.8175 0.466667 1.31483333 1.98 1.45 2.86913377 18.188 23.9549109 4.65 m 3.5m Xv= 0.72379 m P2 P3 e= 0.35121 m < 0.7167 m OK ! B ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD 0.7 m q1,2=(Fv/b)(1=6e/b) b= 2.15m b/2 Xv q1= 1.67508 kg/cm2 q2= 0.01683 kg/cm2 q1= 1.675078 < 70 kg/cm2 OK ! q2= 0.016826 < 70 kg/cm2 OK ! ## (esfuerzo de compresion del concreto) CHEQUEO AL DESLIZAMIENTO F,S,D= (suma Fv)u/ (suma Fh) F,S,D= 2.0164 > 1.5 OK ! ## (sin considerar la adherencia de los concretos durante el proceso constructivo) Pgina 9 CHEQUEO AL VOLTEO F,S,V= 2.22 > 1.75 OK ! ## ESTRIBO CON PUENTE Y RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 6.6 ton Ev= 1.98 ton Eh= 6.31 ton dh= dh= (h/3)(h+3h)/(h+2h) 1.709 m 9 Ton R2= R2= w= 1T/ml 16 m R2= 0.172 ## R1 ( reaccion del puente por carga muerta) R3 (reaccion del puente por sobre carga) Eh R2 Tot, pi(tn) xi(m) Mi 6.31 1.708974 10.7905751 0.17202 6.45 1.10951835 6.49 11.9000935 R1 o R3 (Reacciones) pi(tn) R1 R3 Pvert, Tot, xi(m) Mi 1.1 5.17 1.1 6.05 1.275 23.1896528 34.4096528 4.7 5.5 18.188 28.388 4.65 m 3.5m P3 Xv= e=b/2-Xv B e= P2 0.79293 m 0.28207 m < 0.7167 OK ! 0.7 m ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD b= b/2 q1,2=(Fv/b)(1=6e/b) q1= 2.35974 kg/cm2 q2= 0.281 kg/cm2 q1= 2.359743 Xv < 70 kg/cm2 Pgina 10 OK ! ## (esfuerzo de compresion del concreto) 2.15m q2= 0.280998 < 70 kg/cm2 OK ! CHEQUEO AL DESLIZAMIENTO F,S,D= (suma Fv)u/(suma Fh) F,S,D= 3.0637 > 1.5 OK ! CHEQUEO AL VOLTEO F,S,V= 2.8915 > OK ! ## ANALISIS DE LA SECCION C-C ESTRIBO SIN PUENTE Y CON RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 9.4 ton Ev= 2.82 ton Eh= 8.98 ton dh= (h/3)(h+3h)/(h+2h) dh= 2.0483 m P1 P2 P3 P4 P5 Ev Tot, pi(tn) xi(m) Mi 6.95175 2.325 16.1628188 6.44 1.6 10.304 2.8175 0.966667 2.72358333 7.245 1.575 11.410875 3.72 2.9 10.788 2.82 3.15 8.8679564 29.9895 60.2572335 P P1 P2 3.5m Xv= 1.39571 m P1 P2 P3 ## e=b/2-Xv < b/6 e= 0.17929 m < 0.525 m 0.5m CHEQUEOS DE H= 0.7m P4 q1,2=(Fv/b)(1=6e/b) 1.2m 2 m q1= 1.27717 kg/cm2 b= q2= 0.62692 kg/cm2 b/2 Xv q1= 1.277174 < 3.0 kg/cm2 OK ! q2= 0.626919 < 3.0 kg/cm2 OK ! CHEQUEO AL DESLIZAMIENTO F,S,D= (suma Fv)u/ (suma Fh) F,S,D= 2.003 > 1.75 2.65 m 3.15 m OK ! Pgina 11 CHEQUEO AL VOLTEO F,S,V= 3.2747 > OK ! ## ESTRIBO CON PUENTE Y RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 9.4 ton Ev= 2.82 ton Eh= 8.98 ton dh= (h/3)(h+3h)/(h+2h) dh= 2.0483 m 8 Ton R2= R2= w= 1T/m2 16 m R2= 0.172 ## R1 ( reaccion del puente por carga muerta) R3 (reaccion del puente por sobre carga) R1 o R3 (Reacciones) Eh R2 Tot, 1.8m pi(tn) xi(m) Mi 8.98 2.048297 18.4006066 0.17202 7.45 1.2815367 9.16 19.6821433 P P1 P1 pi(tn) xi(m) Mi 1.6 7.52 1.6 8.8 1.675 50.2323694 66.5523694 R1 R3 Pvert, Tot, 4.7 5.5 29.9895 40.1895 Xv= 1.16623 m 5.65 m 3.5m P2 P3 0.5m H= 0.7m ## e=b/2-Xv < b/6 P4 R C e= 0.40877 < 0.525 P2 m P4 OK! 1.2m CHEQUEOS DE 2 m q1,2=(Fv/b)(1=6e/b) b= 2.65 m 3.15 m b/2 Xv q1= 2.26925 kg/cm2 q2= 0.28247 kg/cm2 q1= 2.269247 < 3.0 kg/cm2 OK ! q2= 0.282466 < 3.0 kg/cm2 OK ! Pgina 12 CHEQUEO AL DESLIZAMIENTO F,S,D= (suma Fv)u/(suma Fh) F,S,D= 2.6338 > 1.5 OK ! CHEQUEO AL VOLTEO F,S,V= 3.3814 > OK ! Pgina 13 Nombre del proyecto: Expediente: "22970172" ## "SANTA CRUZ DE SINCATA" OBRAS PROVISIONALES Habilitacin de campamento: 60 m2 TRABAJOS PRELIMINARES 10 16 22.3 m Atotal (m2) = 223 MOVIMIENTO DE TIERRAS At 1 h1= d= 2.15 c= 3.41 2.07 f= 3.15 b= 6.15 h= 2.07 h2= 1 2.15 e= a= (#de veces)(At)(H) Vtotal 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))H 468.301 m3 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))h4 76.15 m3 392.15 m3 796.11 m3 ## OBRAS DE CONCRETO SIMPLE Cimiento de concreto :1:8+30%PG ( fc=140 kg/cm2) At 1 d= 2.07 h1= 2.15 f= 2.07 b= 3.15 h= 6.15 c= 3.41 h3= 5.15 h4= Vtotal Vtotal 3.41 5.4 L2= 1 h2= 2.15 e= 3.41 a= 5.4 Pgina 14 Vtotal=(# veces)AtH Vtotal/concreto= 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))h4 Vtotal/concreto= 76.15 m3 Etotal= (# veces)(longitud perimetral)H Etotal= 2(a+b+c+d+e+f+L1+L2)(h4+0,2) Etotal= 51.624 m2 ELEVACION DE ESTRIBOS Concreto: fc=140 kg/cm2 0.65 1.15 0.8 1.125 3.5693 3.5 1.125 3.41 4.325 ESTRIBO 4.325 ALA 2.15 1.6 Vconcreto= (N veces)( Vestribo+Vala+Vala) Vconcreto= 2(0,74,54,59+(1,1+2)6,254,59/2+(1,3+1,875)5,1254.3/2+(1.3+1,875)5,1254,3/2) Vconcreto= 133.561 m3 Etotal= (# veces)(suma(longitud perimetralaltura)) Etotal= Etotal= 226.568 m2 VIGAS CONCRETO fc= 240 kg/cm2 0.2 0.9 0.7 Vtotal concreto= (N de vigas)(longitu puente)Area Vtot/concreto= Vtot/concreto= 20.16 m3 Etotal=(N vigas)(perimetro externo)longitud del puente Etotal= Etotal= 80 m2 Pgina 15 ALA 1.6 ACERO Atotal longitudinal=(N veces)(suma(# varillas)(peso varilla)(longitud+traslape)) Atotal longitudinal= 1" 1" 5/8" 1" , Atotal longitudinal= 5740.46 kg Atotal estribos= Atotal estribos= 1/2" 1.1 0.7 Atotal estribos= 680.4 kg 6420.86 kg 3/4" 432,241,8 ## Acero para Apoyo Fijo= Acero para Apoyo Fijo= 48.384 kg DIAFRAGMAS CONCRETO fc= 240 kg/cm2 0.2 0.8 0.4 Vtotal concreto= (N de diafragmas)(longitud)Area Vtot/concreto= Vtot/concreto= 4.288 m3 Etotal=(N diafragmas)(perimetro externo)longitud del puente Etotal= Etotal= 26.8 m2 ACERO Atotal longitudinal=(N veces)(suma(# varillas)(peso varilla)(longitud)) Atotal longitudinal= Atotal longitudinal= 5/8" 5/8" 143 kg Pgina 16 Atotal estribos= Atotal estribos= 3/8" 1 0.4 Atotal estribos= 193.805 kg ## ACERO Total DIAFRG= 337.0992 kg 16 LOSA CONCRETO fc= 240 kg/cm2 0.2 5.366667 Vconcreto= Vconcreto= 17.1733 m3 3.9 Etotal= Etotal= 66.6533 m2 ACERO MALLA LONGITUDINAL AL PUENTE SUPERIOR A longitudinal=(# varillas)(peso varilla)(longitud+traslape)) Alongitudinal= Alongitudinal= 1/2" 186.9 kg INFERIOR A longitudinal=(# varillas)(peso varilla)(longitud)) A longitudinal= A longitudinal= 1/2" 168.21 kg ## MALLA TRANSVERSAL AL PUENTE SUPERIOR A longitudinal=(# varillas)(peso varilla)(longitud+long,gancho)) Alongitudinal= 5/8" Alongitudinal= 1310.989 kg INFERIOR A longitudinal=(# varillas)(peso varilla)(longitud+log,gancho)) A longitudinal= 5/8" A longitudinal= 1172.504 kg Pgina 17 ## ACERO TOTAL DE LOSA= 2838.6 kg seccin de sardinel 0.75 0.15 0.8 V concreto sardinel= (N veces)(Area de seccin)longitud V concreto sardinel= V concreto sardinel= 3.72 m3 3.72 m3 Asardinel= ## (Nveces)(N varillas)(peso varilla)(longitud) Asardinel= 5/8" Asardinel= 631.889 kg ACERO sardinel= 631.89 kg VARIOS TUBO DE DRENAJE 3" PVC=(N veces )longitud Tub, drenaje= 80,25 Tub,drenaje= 2 m BARANDA- TUBO DE F G (2") BARANDA TUBO DE F G (2") 0.516+2(25+1.6) BARANDA DE TUBO DE F G (2") = ## APOYO DE NEOPRENO (0,20,52")= JUNTAS DE DILATACION ASFALTICA ( 2" ) = Jun,Asf= Junta,Asf= FALSO PUENTE= Falso puente= 43.2 ML "5080" (N veces)longitud 2.31 m en metros lineales 25 ml Pgina 18 cm3/ apoyo 0.05 0.75 0.867 0.25 Pgina 19 0.4 0.2 3.12 3.12 16.0 0.89 0.89 0.89 3.99 d3 d= d2 d1 87 70.6 cm2 Pgina 20 11 var 1" 2 var 5/8" 2 var 5/8" 9 var 1" 9 var 1" corte B-B corte A-A 0.174 0.200 0.89 0.89 3.99 d. d2 d1 d. Pgina 21 11 var 1" 2 var 5/8" 2 var 5/8" 9 var 1" 9 var 1" corte B-B corte A-A (+) VIGA 0.30 (-) (+) VIGA 2.10 (-) (+) VIGA 0.80 (-) (+) As,min= VIGA 6.68 (-) 5.7 cm2 Pgina 22 5/8" 18 cm 5/8" 20 cm Pgina 23 ## DIAGRAMA MOMENTOS DE DISEO 17 Ton-m As,min= 5.7 cm2 DIAFRAGMA VIGA 0.37 (+) Pgina 24 corte B-B 0.010 0.200 Pgina 25 Pgina 26 0.65 P1 Ev E &/2 Eh dh B Pgina 27 R2 1.8m P1 Ev E &/2 Eh dh B Pgina 28 Ev E &/2 Eh dh Pgina 29 R2 P Ev E &/2 Eh dh Pgina 30 Pgina 31 Pgina 32 3.41 Pgina 33 Pgina 34 Pgina 35 Pgina 36 PROYECTO: PROFESOR: Partida N DISEO DE PUENTE ING. DANTE MONTOYA Especificaciones N de veces Largo MEDIDAS Ancho Parcial Altura ## PUENTE VIGA LOSA L=16,00 ML 01.00 01.01 TRABAJOS PRELIMINARES 02.00 02.01 MOVIMIENTO DE TIERRAS 02.03 02.04 03.00 03.01 03.02 ## CIMENTACIONES ESTRIBOS - OBRAS DE CS 04.00 04.01 ELEVACION DE ESTRIBOS ## Relleno con material de prestamo Concreto f'c=140 Kg/cm2 +30% P.G. 2 4 9.00 3.41 3.15 2.08 3.50 3.50 99.23 24.82 2 4 1.2 2 9.00 3.41 382.82 20.00 3.15 2.08 1.00 1.00 8.80 0.58 28.35 7.09 382.82 101.20 2 1 46.26 85.07 2 4 5.40 3.91 2 -2 -2 4 5.40 5.40 3.40 3.41 1.00 46.26 85.07 2.00 2.00 5.15 3.65 55.62 28.54 2.15 0.70 0.80 1.41 4.65 5.15 1.15 3.65 53.99 19.47 3.13 17.55 2 16.00 0.70 0.90 2 16.00 2.50 10.08 40.00 10,120.04 47.68 6 2.23 0.40 0.80 6 2.23 2.00 0.71 4.47 1,075.53 Estribos Aleros Concreto f'c=140 Kg/cm2 +30% P.G. Concreto en cuerpo Concreto en alas 05.00 05.01 05.02 05.03 05.04 06.00 06.01 06.02 06.03 07.00 07.01 07.02 ## Concreto f'c=210 Kg/cm2 Acero f'y=4200 kg/cm2 en vigas Acero f'y=4200 kg/cm2 para apoyo fijo ## Concreto f'c=210 Kg/cm2 Acero f'y=4200 kg/cm2 para diafragmas 16.00 8.30 0.20 26.56 1 16.00 8.30 -2 16.00 0.70 -6 2.23 0.40 1 48.60 0.20 132.80 11.20 0.89 9.72 2,507.51 07.03 ## Acero f'y=4200 kg/cm2 para losa 08.00 08.01 BARANDAS-COLUMNETAS-SARDINEL 08.02 337.04 ## Excavacion manual en terreno saturado Excavacin cimentacion cuerpo estribo 04.02 8.80 ## Excavacion manual en terreno seco Excavacin en estribo cuerpo 02.02 38.30 2 1 2 Pgina 37 16.00 16.00 16.00 0.75 0.05 0.90 0.15 0.15 1.80 0.12 14.40 08.03 ## Acero f'y=4200 kg/cm2 para sardinel 09.00 09.01 09.02 09.03 09.04 09.05 VARIOS Apoyos de neopreno (70 cm70cm2) Junta asfaltica 2" Tuberia PVC SAP 2" Falso Puente Tuberia FG 3" 10.00 10.01 REVOQUES Y ENLUCIDOS 11.00 11.01 11.02 PINTURAS ## Pintura en barandas pint. Anticorrosivo Pintura en sardineles al latex 4 2 2 1 26 4 14.00 0.30 16.00 8.00 5.20 1.00 16.00 0.95 16.00 1 2 0.48 0.95 24.00 FLETE HOJA DE CALCULO FLETE.XLS APORTE COMUNAL 10 % de mano de obra no calificada Pgina 38 14.00 2.40 83.20 1.00 16.00 15.20 90.00 1.00 43.09 22.80 Total 337.04 297.75 198.45 99.30 85.07 56.70 28.37 459.38 202.40 92.52 85.07 m m 225.41 111.24 114.17 132.98 107.97 -38.93 -6.26 70.20 m m m m 20.16 80.00 10,120.04 47.68 m m kg kg 4.29 26.80 1,075.53 m m kg 26.56 114.76 132.80 -22.40 -5.36 9.72 2,507.51 m m 3.72 3.60 0.12 28.80 m m kg Pgina 39 - kg 4.00 28.00 4.80 83.20 26.00 64.00 90.00 und ml ml m2 30.40 43.09 45.60 m m ml Pgina 40 M.ACERO PROYECTO: PROFESOR: Descripcion ACERO EN VIGAS Superior medio inferior inferior inferior inferior inferior estribos APOYO FIJO DIAFRAGMAS Superior Superior medio inferior estribos LOSA Acero negativo longitudinal Acero negativo transversal Acero positivo longitudinal Acero positivo transversal SARDINEL Diseo del fierro DISEO DE PUENTE ING. DANTE MONTOYA N de elemtos. iguales N de piezas x elemento Long. Por pieza 1/4" 0.248 3/8" 1/2" 5/8" 3/4" 0.560 0.994 610.0 12.0 10 2 2 6 8 8 8 120 4 30.50 26.30 26.30 22.00 22.00 26.80 33.00 4.70 3.00 5/8" 5/8" 5/8" 5/8" 3/8" 8 8 8 8 8 6 2 2 6 19 4.65 3.25 3.25 4.65 2.60 223.2 52.0 52.0 223.2 1/2" 5/8" 1/2" 5/8" 1 1 1 1 11 107 13 107 26.45 5.75 26.45 5.55 5/8" 1/2" 1 1 ### ### Pgina 41 3.973 2 2 2 2 2 2 2 2 1 2.235 1" 5/8" 5/8" 1" 1" 1" 1" 1/2" 1" 1.552 Peso 1" 395.2 1,128.0 291.0 343.9 105.2 105.2 - 615.3 593.9 264.0 352.0 428.8 528.0 - kg 10,120.037 2,423.530 163.270 163.270 1,048.872 1,398.496 1,703.622 2,097.744 1,121.232 47.676 1,075.533 346.406 80.704 80.704 346.406 221.312 2,507.514 289.204 954.868 341.787 921.655 13,750.760 M.ACERO Pgina 42 FLETE PROYECTO: PROFESOR ## DISEO DE PUENTE LOSA VIGA 0.00 0.00 0.00 1- DATOS GENERALES A-POR PESO MATERIALES CEMENTO FIERRO, CLAV. ETC CALAMINA YESO TAPA BUZON OTROS MAQUINARIA Y EQUIP BL. KG P2 UN UN BL UN KG KG TAPA DE FIERRO BOMBA MANUAL UN UN AFECTO IGV 1,696.11 15,490.90 8,383.22 1,000.00 PESO.UNIT. 42.50 1.00 1.50 4.50 3.00 25.00 130.00 1.00 1.00 PESO.TOTAL 72,084.68 15,490.90 12,574.83 1,000.00 15.00 120.00 PESO TOTAL 101,150.41 B-POR VOLUMEN DESCRIPC. ARENA M3 PIEDRA M3 HORMIGN M3 MATERIAL RELLENO M3 P2 AFECTOS IGV SIN IGV 7.20 85.80 269.85 206.00 VOLUMEN TOTAL NUMERO DE VIAJES REDONDEO 568.85 8.00 71.11 71.00 EN TUBERIA UNIDAD DE (2.20 M. x 3.00 M.) DE CARROCERIA, CON H= 1.50 M. CAPACIDAD DEL CAMION EN TUBOS / VIAJE ML No de tubos Tub, 1/2" 5,050.00 U Tub, 3/4" 4,200.00 U Tub, 1" 3,360.00 U Tub,1 1/2" 1,690.00 U Tub 2:" 994.00 U 4.94 Tub, 3" 470.00 U 170.80 Tub 4": 259.00 U Tub 6:" U-PVC 110.00 U Tub 8" U-PVC 65.00 U Tub 6'; CSN 160.00 U Tub 8'; CSN 80.00 U Pgina 43 0.99 34.16 FLETE ## NUMERO TOTAL DE VIAJES REDONDEO TUBERIA EN VOLUMEN 2- FLETE TERRESTRE CAPACIDAD DEL CAMION ( M3 ) COSTO POR VIAJE S/. FLETE POR KG 8.00 CAPACIDAD DEL CAMION ( M3 ) 600.00 COSTO POR VIAJE S/. 0.05 AFECTO IGV 5,057.52 ## FLETE POR PESO FLETE POR VOLUMEN TUBERIA COSTO TOTAL FLETE TERR. SIN IGV 5,680.00 0.00 5,057.52 5,680.00 AFECTO IGV 5,057.52 SIN IGV 5,680.00 5,057.52 5,680.00 ## RESUMEN FLETE TOTAL FLETE TERRESTRE FLETE FLUVIAL FLETE EN ACEMILA FLETES TOTALES S/. Pgina 44 ## FLETE POR PESO =Peso Total * Flete por peso FLETE POR VOLUMEN=No viajescosto por viaje FLETE No VIAJES 0.00 0.07 Pgina 45 0.07 0.00 1 FLETE 8.00 80.00 12,000.00 ## PESO =Peso Total Flete por peso VOLUMEN=No viajescosto por viaje Pgina 46 ## DISEO ESTRUCTURAL DE UN PUENTE CARROZABLE PROYECTO: EXPEDIENTE: OFICINA ZONAL: PUENTE CALLAZA(02) 22980517 PUNO 10 ## PERALTE DE VIGA =>( H=L/15)= 0.7 (H=0.07L) = 0.7 No DE VIGAS= 0.7 NOTA: Todas las unidades deben estar en : 0.75 0.05 3.6 5.2 0.2 0.15 LOSA 0.2 0.5 DIAFRAGMA 0.25 0.800 0.5 2.1 0.5 S= 0.5 2.6 4.70 S (espaciamiento de vigas de eje a eje) t (espesor de losa en metros) S t 1.8 0.16 2.1 0.165 2.4 0.18 = 2.7 0.19 3 0.20 4 0.22 2.60 0.2 4.5 0.25 ## a) DISEO A FLEXION DE VIGAS CALCULO DEL MOMENTO ULTIMO MAXIMO d1 SOBRE CARGA : AASHTO HS-36 0.6 d1= 1.8 0.70 0.70 2.60 Coef.Concentracin de Carga = 1.23 4 B= 2.45 10 9.3 A= 0.06 C= 0.65 N.R 16 10.7 31 ## POR SOBRECARGA EQUIVALENTE = (P=9 Ton w=1 Ton/m) 21 ## SE ADOPTARA EL MAYOR VALOR MOMENTO POR IMPACTO = ( I= 15.24/(L+38) ) ML= 31.0 Ton-m MI= 9.3 Ton-m Pgina 47 I= 0.3 < 0.3 ASFALTO BARANDA LOSA VIGA W = = = = = 0.180 0.300 1.300 0.625 0.291 2.696 ## CADA QUE DISTANCIA DE EJE A EJE SE COLOCARAN LOS DIAFRAGMAS ( D < LUZ / 4 = Ton/m D= 2.5 E= 0.15 0.3 0.32 Y= d d d d mts. 3.23 ## ESPESOR DEL DIAFRAGMA No DE DIAFRAGMAS 3.23 mts ) Ton 1.62 1 2 3 4 = = = = 1 2 2 1 #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! 0 1.615 1.615 0 0.32 0.32 d1 d2 0.32 d3 3.23 MOMENTO POR CARGA MUERTA MD = 35 Ton-m ## DISEO POR METODO A LA ROTURA MU=1.3( MD+1.67( ML+MI ) ) => f 'c= Fy= b= d= MU= 210 kg/cm2 4200 kg/cm2 210 cm 52 cm w= 133 Ton-m 0.13458 &= As(cm2)= 0.007 < 73.481 cm2 75&b= ## 0.021 ( FALLA DUCTIL ) 15 VARILLAS DE 1" As principal(+) = 73.5 cm2 M(-)= 44.3333333 w= 0.042365 As(cm2)= &= 23.1310387 cm2 Pgina 48 0.0021 <> < 75&b= 5 VARILLAS DE 1" B 5 2 var 5/8" 2 var 5/8" 2.5 m 5 5 5 A b= 2.60 0.2 b= E.N (b-bw)/2= 2.60 > 2.5 VIGA T * 1.05 < 1.6 VIGA RECTANGULAR (b-bw)/2= 1.05 < 1.05 VIGA RECTANGULAR C bw= 0.5 C= 0.097 < 0.200 VIGA RECTANGULAR ## DISEO DE VIGA POR CORTE Y= d d d d 0.33299 1 2 3 4 = = = = 0 1 2 3 0 0.333 0.666 0.999 0.32 0.32 0.32 d.. d2 d3 d4 1.9979 CORTANTE POR PESO PROPIO VD= 14 Ton VL= 16 Ton VI= 5.0 Ton ## DISEO POR METODO A LA ROTURA VU=1.3( VD+1.67( VL+VI ) ) => Vcon= Vu= 63.791 Ton fi(0,5(fc)^0,5+175&Vud/Mu => S= Avfyb/Vace S= 11 cm S= 11 cm Vcon= 17.0 Ton Vace= 46.8 VAR. 1/2" Pgina 49 Ton 5 5 5.0 B 5 2 var 5/8" 1a .05, VAR. 1/2" 2 var 5/8" 2.5 m 5 5 5 A b) DISEO DE LOSA (Tramo interior) ( MD) Peso propio= Asfalto= 0.5 Ton/m 0.1 Ton/m w= 0.26 0.6 Ton/m ## Luz entre vigas VIGA NOTA: Consideraremos un coeficiente de ( 1/10 ) para los momentos positivos y negativos por peso propio 2.10 MD= 0.26 Ton-m ## DIAGRAMA MOMENTOS POR SOBRE CARGA 2.00 ML=(L+0.61)P/9.74 2.22587 MOMENTO POSITIVO MOMENTO NEGATIVO ML(+)= ML(-)= VIGA ML= 1.78 Ton-m 2.00 Ton-m MOMENTO POSITIVO MOMENTO NEGATIVO 0.38004988 MI(+)= MI(-)= 0.68 Ton-m 0.76 Ton-m VIGA I=15.24/(L+38) = 0.76 ## DISEO POR METODO A LA ROTURA MU( + ) = 1.3( MD+1.67( ML+MI ) ) = MU( - ) = 1.3( MD+1.67( ML+MI ) ) = 6.35 Ton-m 210 kg/cm2 4200 kg/cm2 100 cm 17 cm VIGA f 'c= Fy= b= d= 6.35 5.679 Ton-m ## ACERO PARA MOMENTOS POSITIVOS w= 0.11125 As(cm2)= &= 0.006 9.4566 cm2 < VAR, 75&b= 5/8" As principal(+) = Pgina 50 21 cm 9.5 cm2 ## ACERO PARA MOMENTOS NEGATIVOS w= 0.125438 &= As(cm2)= Acero por temperatura => 10.6622421 cm2 As temp= <> < VAR, 75&b= 5/8" 1/2" As rpp(+)= VAR, VAR, 1/2" 1/2" @ 42 cm 35 cm pesos 0.4 0.29063 0.18 0.15 losa = trapecio = Asfalto= Baranda= brazos momentos (Ton-m) 0.26 0.6466 0.162 0.255 0.650 2.225 0.9 1.7 MD= 1.324 Ton -m E=0.8X+1.143 X= E= 0.00 1.143 Mv= Mv= P.X/E 0 Mv= 19 cm 42 As rpp(+)= 1/2" 3.06 cm2 VAR, 0.0018bd As temp= VAR, 0.006 ML= 0 Ton-m MI= 0 Ton-m Pgina 51 3.62 cm2 35 cms, ## DISEO POR METODO A LA ROTURA MU( + ) = 1.3( MD+1.67( ML+MI ) ) = f 'c= Fy= b= d= 2 Ton-m 210 kg/cm2 4200 kg/cm2 100 cm 17 cm w= 0.03744 As(cm2)= &= 0.002 3.1824 cm2 < 75&b= VAR. 5/8" 35 cms VAR. 5/8" 35 cms 0.25 t/m 1.25 t/m 0.3 t/m WD= 1.8 t/m WL= 2.5 t/m WU= 7.2 t/m Mu(-)= 0.11 ton-m Mu(+)= 0.13 ton-m 0.11 VIGA Asfalto= losa= viga= f 'c= Fy= b= d= 210 kg/cm2 4200 kg/cm2 323 cm 52.0 cm w= 8E-005 As(cm2)= &= 0.0654 cm2 0.000 < 75&b= Asmin= 5.20 cm2 5.20 cm2 3 VARILLAS DE Pgina 52 5/8" DIAFRAGMA f 'c= Fy= b= d= 210 kg/cm2 4200 kg/cm2 30 cm 52 cm w= 0.00073 &= As(cm2)= 0.000 0.06 cm2 < 75&b= Asmin= ## 0.0213 ( FALLA DUCTIL ) 5.20 cm2 5.20 cm2 3 VARILLAS DE 5/8" B 3 2 varillas 5/8" 1a .05, VAR. 3/8" 8 a 0.10 , r a 0.20 /e 2 varillas 5/8" 3 B b= 3.23 0.2 E.N (b-bw)/2= 1.465 < 1.6 VIGA RECTANGULAR (b-bw)/2= 1.465 < 1.465 VIGA RECTANGULAR C bw= 0.3 ## PROFUNDIDAD DEL EJE NEUTRO (C) C= 0.004 < 0.2 VIGA RECTANGULAR e) DISEO DE ESTRIBOS DATOS PARA EL DISEO TIPOS DE SUELOS ANGULO DE PESO UNITARIO (T/m3) 1.73 - 2.20 1.57 1.57 - 1.73 1.57 - 1.73 1.75 - 2.05 3.0 -8.0 0.5 -2.0 2.0 -3.0 3.0 -6.0 0.5 -2.0 FRICCION INTERNA 353035356- GRAVA ARENA FINA ARENA MEDIA ARENA GRUESA ARCILLA 45 35 40 40 7 2.0 Kg/cm2 26 1.76 Ton/m3 175 Kg/cm2 175 Kg/cm2 F'c= 210 Kg/cm2 2.3 Ton/m ## REACCION DEL PUENTE SOBRECARGA/ m= 2.6 Ton/m Pgina 53 (verificar in situ) Longitud de Soporte L,Smin= 0.37 m 0.5 0.6 H : V : b > A 5 0.75 <=> 1 : 5 0.72 m =1.8 1.8 =V 3.55 B 0.7 0.7 H= 0.36 m b= 1.46 2.86 ## ANALISIS DE LA SECCION A-A E=0,5PUh(h+2h)tan(45-&/2)^2 E= 0.503 ton Ev= 0.11 ton Eh= 0.49 ton dh= dh= (h/3)(h+3h)/(h+2h) 0.32692308 m b= 0.5m P1 Ev Tot, Ev P pi(tn) xi(m) Mi 0.8625 0.25 0.2156 0.11 0.5 0.0565 0.97554 0.2721 0.75m P1 &/2 Eh A Xv= 0.11488 m b/2 Xv e=b/2-Xv e= 0.14 m < 0.2 OK ! ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD q1,2=(Fv/b)(1=6e/b) q1= 0.51147 kg/cm2 q2= -0.12125 kg/cm2 q1= 0.511467 < 70 kg/cm2 OK ! q2= -0.12125 < 70 kg/cm2 OK ! Pgina 54 ## (esfuerzo de compresion del concreto) CHEQUEO AL DESLIZAMIENTO F,S,D= Vconcrt/Eh= 0.850.5f'c^0.5bd/ Eh F,S,D = 45.9287 > ## (cortante del concreto/empuje horizontal) 2 OK ! CHEQUEO AL VOLTEO F,S,V = F,S,V = 1.7001 > 1.25 OK ! ## (sin considerar el aporte de esfuerzo a la compresion del concreto como momento estabilizante) ## ANALISIS DE LA SECCION B-B ESTRIBO SIN PUENTE Y CON RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 3.3 ton Ev= 0.74 ton Eh= 3.20 ton dh= (h/3)(h+3h)/(h+2h) dh= 0.986 m pi(tn) 2.9325 2.484 0.7452 0.74 6.90084 P1 P2 P3 Ev Tot, xi(m) Mi 1.21 3.5483 0.66 1.6394 0.24 0.1788 1.10 0.813 6.1797 2.55 m 1.8m Xv= 0.43805 m P3 e= R 0.29195 m < 0.48666667 m OK ! B ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD 0.36 m q1,2=(Fv/b)(1=6e/b) b= b/2 Xv q1= 1.03975 kg/cm2 q2= -0.09443 kg/cm2 q1= 1.039746 < 70 kg/cm2 OK ! q2= -0.09443 < 70 kg/cm2 OK ! ## (esfuerzo de compresion del concreto) CHEQUEO AL DESLIZAMIENTO F,S,D= F,S,D= 1.50883152 > 1.5 OK ! ## (sin considerar la adherencia de los concretos durante el proceso constructivo) Pgina 55 CHEQUEO AL VOLTEO F,S,V= F,S,V= 1.95762204 > 1.75 OK ! ## ESTRIBO CON PUENTE Y RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 3.3 ton Ev= 0.74 ton Eh= 3.20 ton dh= (h/3)(h+3h)/(h+2h) dh= 0.986 m 9 R2= R2= w= 10 R2= 0.15322581 ## R1 ( reaccion del puente por carga muerta) R3 (reaccion del puente por sobre carga) Eh R2 Tot, pi(tn) xi(m) Mi 3.20 0.986 3.1567 0.15323 4.35 0.6665 3.35 3.8233 (Reacciones) pi(tn) xi(m) R1 R3 Pvert, Tot, 2.3 2.6 6.90084 11.8008 Xv= 0.49391 m Mi 0.66 1.518 0.66 1.716 0.93 6.4178 9.6518 2.55 m 1.8m P3 R e=b/2-Xv B e= 0.23609 m < 0.48666667 OK ! 0.36 m ## CHEQUEOS DE LOS FACTORES DE SEGURIDAD b= b/2 q1,2=(Fv/b)(1=6e/b) Xv q1= 1.5925 kg/cm2 q2= 0.02405 kg/cm2 q1= 1.5925 < 70 kg/cm2 OK ! q2= 0.024053 < 70 kg/cm2 OK ! Pgina 56 ## (esfuerzo de compresion del concreto) CHEQUEO AL DESLIZAMIENTO F,S,D= F,S,D= ## (suma Fv)u/(suma Fh) 2.46234303 > 1.5 OK ! CHEQUEO AL VOLTEO F,S,V= F,S,V= 2.52449476 > OK ! ## ANALISIS DE LA SECCION C-C ESTRIBO SIN PUENTE Y CON RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 5.8 ton Ev= 1.30 ton Eh= 5.65 ton dh= dh= (h/3)(h+3h)/(h+2h) 1.33280702 m P1 P2 P3 P4 P5 Ev Tot, pi(tn) xi(m) Mi 2.9325 1.91 5.6011 2.484 1.36 3.3782 0.7452 0.94 0.7005 6.578 1.43 9.4065 2.856 2.51 7.1686 1.30 2.86 3.7277 16.8991 29.983 P1 P2 P2 1.8m Xv= 1.32895 m P1 P3 ## e=b/2-Xv < b/6 e= 0.10105 m < 0.47666667 m 0.7m CHEQUEOS DE H= 0.36m P4 q1,2=(Fv/b)(1=6e/b) 1.06m 1.66 m q1= 0.71613 kg/cm2 b= q2= 0.46562 kg/cm2 b/2 Xv q1= 0.716134 < 2.0 kg/cm2 OK ! q2= 0.465621 < 2.0 kg/cm2 OK ! CHEQUEO AL DESLIZAMIENTO F,S,D= F,S,D= ## (suma Fv)u/ (suma Fh) 1.79599491 > 1.75 2.16 m 2.86 m OK ! Pgina 57 CHEQUEO AL VOLTEO F,S,V= F,S,V= 3.98467166 > OK ! ## ESTRIBO CON PUENTE Y RELLENO SOBRECARGADO E=0,5PUh(h+2h)tan(45-&/2)^2 E= 5.8 ton Ev= 1.30 ton Eh= 5.65 ton dh= dh= (h/3)(h+3h)/(h+2h) 1.33280702 m 9 R2= R2= w= 10 R2= 0.15322581 ## R1 ( reaccion del puente por carga muerta) R3 (reaccion del puente por sobre carga) R1 o R3 (Reacciones) Eh R2 Tot, pi(tn) xi(m) Mi 5.65 1.332807 7.5245 0.15323 5.35 0.8198 5.80 8.3442 P1 P1 pi(tn) xi(m) R1 R3 Pvert, Tot, 2.3 2.6 16.8991 21.7991 Xv= 1.10901 m Mi 1.36 3.128 1.36 3.536 1.53 25.856 32.52 3.55 m 1.8m P2 m P2 P3 0.7m H= 0.36m ## e=b/2-Xv < b/6 P4 C e= 0.32099 < 0.4767 P4 OK! 1.06m CHEQUEOS DE 1.66 m q1,2=(Fv/b)(1=6e/b) b= 2.16 m 2.86 m b/2 Xv q1= 1.27548 kg/cm2 q2= 0.24893 kg/cm2 q1= 1.275483 < 2.0 kg/cm2 OK ! q2= 0.248929 < 2.0 kg/cm2 OK ! Pgina 58 CHEQUEO AL DESLIZAMIENTO F,S,D= F,S,D= ## (suma Fv)u/(suma Fh) 2.2555381 > 1.5 OK ! CHEQUEO AL VOLTEO F,S,V= F,S,V= 3.89725109 > OK ! Pgina 59 Nombre del proyecto: Expediente: "22970172" ## "SANTA CRUZ DE SINCATA" OBRAS PROVISIONALES Habilitacin de campamento: 60 m2 TRABAJOS PRELIMINARES 10 10 15.72 m Atotal (m2) = 157.2 MOVIMIENTO DE TIERRAS At 2 h1= d= 2.86 c= 4.8 4.2 f= 4.2 b= 2.86 h= 4.2 h2= 2 2.86 e= a= (#de veces)(At)(H) Vtotal 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))H h4= Vtotal 200.2 m3 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))h4 80.08 m3 120.12 m3 340.34 m3 ## OBRAS DE CONCRETO SIMPLE Cimiento de concreto :1:8+30%PG ( fc=140 kg/cm2) At 2 h1= 2.86 h3= d= 4.2 f= 4.2 b= 4.2 h= 2.86 c= 4.8 1.5 Vtotal 4.8 5.8 L2= 2 h2= 2.86 e= 4.8 a= 5.8 Pgina 60 Vtotal=(# veces)AtH Vtotal/concreto= 2((a+b)h/2+(c+d)h1/2)+(e+f)h2/2))h4 Vtotal/concreto= 80.08 m3 Etotal= (# veces)(longitud perimetral)H Etotal= 2(a+b+c+d+e+f+L1+L2)(h4+0,2) Etotal= 76.8 m2 ELEVACION DE ESTRIBOS Concreto: fc=140 kg/cm2 0.5 0.95 0.75 0.6 1.836 1.8 0.95 4.3 3.275 ESTRIBO 3.275 ALA 1.46 1.255 Vconcreto= (N veces)( Vestribo+Vala+Vala) Vconcreto= 2(0,74,54,59+(1,1+2)6,254,59/2+(1,3+1,875)5,1254.3/2+(1.3+1,875)5,1254,3/2) Vconcreto= ALA 72.4348 m3 Etotal= (# veces)(suma(longitud perimetralaltura)) Etotal= Etotal= 164.268 m2 VIGAS CONCRETO fc= 210 kg/cm2 0.2 0.5 0.5 Vtotal concreto= (N de vigas)(longitu puente)Area Vtot/concreto= Vtot/concreto= 5 m3 Etotal=(N vigas)(perimetro externo)longitud del puente Etotal= Etotal= 30 m2 Pgina 61 1.255 ACERO Atotal longitudinal=(N veces)(suma(# varillas)(peso varilla)(longitud+traslape)) Atotal longitudinal= 1" 1" 5/8" 1" , Atotal longitudinal= 2538.14 kg Atotal estribos= Atotal estribos= 1/2" 0.5 Atotal estribos= 453.6 kg 2991.7 kg 3/4" 432,241,8 ## Acero para Apoyo Fijo= Acero para Apoyo Fijo= 48.384 kg DIAFRAGMAS CONCRETO fc= 210 kg/cm2 0.2 0.4 0.3 Vtotal concreto= (N de diafragmas)(longitud)Area Vtot/concreto= Vtot/concreto= 1.00878 m3 Etotal=(N diafragmas)(perimetro externo)longitud del puente Etotal= Etotal= 9.24715 m2 ACERO Atotal longitudinal=(N veces)(suma(# varillas)(peso varilla)(longitud)) Atotal longitudinal= Atotal longitudinal= 5/8" 5/8" 40 kg Pgina 62 Atotal estribos= Atotal estribos= 3/8" 0.3 Atotal estribos= 83.1235 kg ## ACERO Total DIAFRG= 122.96 kg 10 LOSA CONCRETO fc= 210 kg/cm2 0.2 4.7 Vconcreto= Vconcreto= 9.4 m3 3.9 Etotal= Etotal= 40.358 m2 ACERO MALLA LONGITUDINAL AL PUENTE SUPERIOR A longitudinal=(# varillas)(peso varilla)(longitud+traslape)) Alongitudinal= 1/2" Alongitudinal= 123.9 kg INFERIOR A longitudinal=(# varillas)(peso varilla)(longitud)) A longitudinal= 1/2" A longitudinal= 111.51 kg ## MALLA TRANSVERSAL AL PUENTE SUPERIOR A longitudinal=(# varillas)(peso varilla)(longitud+long,gancho)) Alongitudinal= 5/8" Alongitudinal= 1153.984 kg INFERIOR A longitudinal=(# varillas)(peso varilla)(longitud+log,gancho)) A longitudinal= 5/8" A longitudinal= 1032.085 kg Pgina 63 ## ACERO TOTAL DE LOSA= 2421.47885 kg seccin de sardinel 0.75 0.15 0.8 V concreto sardinel= (N veces)(Area de seccin)longitud V concreto sardinel= V concreto sardinel= 2.325 m3 2.325 m3 Asardinel= ## (Nveces)(N varillas)(peso varilla)(longitud) Asardinel= 5/8" Asardinel= 631.889 kg ACERO sardinel= 631.9 kg VARIOS TUBO DE DRENAJE 3" PVC=(N veces )longitud Tub, drenaje= 80,25 Tub,drenaje= 2 m BARANDA- TUBO DE F G (2") BARANDA TUBO DE F G (2") 0.516+2(25+1.6) BARANDA DE TUBO DE F G (2") = 31.2 ML ## APOYO DE NEOPRENO (0,20,52")= "5080" JUNTAS DE DILATACION ASFALTICA ( 2" ) = Jun,Asf= Junta,Asf= FALSO PUENTE= Falso puente= (N veces)longitud 1.47 m en metros lineales 25 ml Pgina 64 cm3/ apoyo CARROZABLE 0.05 0.75 0.4 0.800 0.25 Pgina 65 0.3 0.15 3.23 3.23 3.23 10.0 0.32 w= d.. 0.32 0.32 0.32 2.70 d3 d= d2 d1 52 As,min= 36.4 cm2 ( FALLA DUCTIL ) Pgina 66 5 var 1" var 1" 2 var 5/8" 2 var 5/8" var 1 5 var 1" 5 var 1" var 1" var 1" corte B-B corte A-A 0.097 0.200 w= 0.32 2.70 d. d. 0.32 d2 d1 d. Pgina 67 5 var 1" var 1" 2 var 5/8" 2 var 5/8" var 1 5 var 1" 5 var 1" var 1" var 1" corte B-B corte A-A 0.26 LOSA VIGA 2.10 0.26 2.00 LOSA VIGA 2.10 1.78 0.76 LOSA VIGA 2.10 0.68 ## RAMA MOMENTOS DE DISEO 6.35 LOSA VIGA 2.10 5.68 As,min= 5.7 cm2 Pgina 68 ( FALLA DUCTIL ) puente^0.5)*As principal(+) VAR, 5/8" 19 cm VAR, 5/8" 21 cm Pgina 69 Mu= 2 Ton-m As,min= 5.7 cm2 0.11 VIGA LOSA DIAFRAGMA 0.13 Pgina 70 varillas 5/8" 2 varillas 5/8" 2 varillas 5/8" varilas 5/8" corte B-B 0.004 0.200 3.0 -8.0 0.5 -2.0 2.0 -3.0 3.0 -6.0 0.5 -2.0 Pgina 71 dh Pgina 72 0.5 0.6 P1 Ev E &/2 Eh P2 R dh B 1.46m Pgina 73 Ton 1T/m2 R2 R1 o R3 (Reacciones) 1.8m P1 Ev E &/2 Eh P2 R dh B 1.46m Pgina 74 P P1 Ev E &/2 Eh dh Pgina 75 Ton 1T/m2 R2 1.8m P P1 Ev E &/2 Eh dh Pgina 76 Pgina 77 Pgina 78 4.3 Pgina 79 0.7 Pgina 80 0.6 Pgina 81 Pgina 82	18,451	35,692	{"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-2019-39	latest	en	0.159839
https://www.jiskha.com/questions/463374/how-much-energy-would-it-take-to-completely-melt-200kg-of-lead-if-the-lead-is-currently-at	1,579,978,120,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00013.warc.gz	924,734,536	5,028	# chemistry how much energy would it take to completely melt 200kg of lead if the lead is currently at 25 degrees celcius 1. 👍 0 2. 👎 0 3. 👁 88 1. q1 = heat required to move T from 25 to melting point of Pb. You will need to look up the melting point value. q1 = mass Pb x specific heat Pb x (Tfinal-Tinitial). q2 = heat to melt solid lead at the melting pont to liquid at the melting point. q2 = mass Pb x heat fusion. Total q = q1 + q2 1. 👍 0 2. 👎 0 posted by DrBob222 ## Similar Questions 1. ### physics It has been said that sometimes lead bullets melt upon impact. Assume that a bullet with an initial temperature of 301 K receives 73.9 % of the work done on it by a wall on impact as an increase in internal energy. (The melting asked by Anonymous on November 5, 2014 2. ### physics It has been said that sometimes lead bullets melt upon impact. Assume that a bullet with an initial temperature of 295 K receives 73.5 % of the work done on it by a wall on impact as an increase in internal energy. (The melting asked by Anonymous on November 12, 2014 3. ### physics It has been said that sometimes lead bullets melt upon impact. Assume that a bullet with an initial temperature of 301 K receives 67.9 % of the work done on it by a wall on impact as an increase in internal energy. (The melting asked by mehmet kaya on November 18, 2014 4. ### Physics Help me understand this please You have a 2.00 kg block of lead. Lead melts at 327.5°C. CPb = 130.0 , and Hf for lead is 2.04 × 10^4 J/kg. Say you start at room temperature (25.0°C). How much heat must you transfer to melt asked by Me and I on March 25, 2014 5. ### Physics You have a 2.00 kg block of lead. Lead melts at 327.5°C. CPb = 130.0 , and ï„Hf for lead is 2.04 Ã— 104 J/kg. Say you start at room temperature (25.0°C). How much heat must you transfer to melt all the lead? asked by Pat on March 18, 2012 6. ### Physics How much energy is needed to melt 0.225 kg of lead so that it can be used to make a lead sinker for fishing? The sample has an initial temperature of 27.3 degrees celsius and is poured in the mold immediately after it has melted. asked by Kim on March 30, 2010 7. ### Physics How much energy is needed to melt 0.225 kg of lead so that it can be used to make a lead sinker for fishing? The sample has an initial temperature of 27.3 degrees celsius and is poured in the mold immediately after it has melted. asked by Kim on March 30, 2010 8. ### physics If a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all it's kinetic energy is converted into heat via friction. Find the minimum speed of a lead bullet (initial temp. = 30.0 C) needed asked by Daniel on April 15, 2012 9. ### Physics It takes 1110 calories of energy to melt a 185 gram lump of lead. What is lead's latent heat of fusion? asked by Tracy on March 5, 2012 10. ### chimistry The molar heat of fusion of lead (Pb) is 1.140 kcal/mol. How much heat must be added to 97.0 g of solid lead at its melting point in order for it to completely melt? A 1.44 kJ B 10.18 kJ C 12.48 kJ D 25.04 kJ asked by joey on June 24, 2015 More Similar Questions	923	3,161	{"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-2020-05	latest	en	0.9454
http://geometrycommoncore.com/content/unit6/gcp2/gcp2.html	1,518,902,891,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891807825.38/warc/CC-MAIN-20180217204928-20180217224928-00716.warc.gz	130,291,229	5,099	HOME Organize UNIT #1 UNIT #2 UNIT #3 UNIT #4 UNIT #5 UNIT #6 Resources Interaction OBJECTIVE - S.CP.A.2 Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. INTERPRETATION OF OBJECTIVE - S.CP.A.2 This objective is all about independence and how if the two events are independent then P(A) • P(B) = P(A and B). SKILLS (1) The student will be able to explain what independence between two event means. (2)The student will be able to determine if two events are independent of each other using P(A) • P(B) = P(A and B). (3)The student will be able to define and distinguish between mutually exclusive and independent. THE BIG IDEA Independence is the BIG IDEA. Independence is a very important concept to understand in probability because once you begin doing more than one event in succession we need to know how the first event affected (if at all) the second event. TRAPS & PITFALLS Mutually exclusive and independence have ALWAYS been confused. Our English translation of independent is to be alone or separate which seems like the same thing as mutually exclusive BUT the English definition IS NOT the mathematical one. Independence is about affecting other events. Distinguish the early one or for sure you will have many confused later. PAST CONNECTIONS The student needs to have a good grounding in outcomes as subsets that can be described as intersections, unions and complements. We will connect P(A and B) be found in the intersection. FUTURE CONNECTIONS Independence continues for a few objectives. We look at in conditional statements next and then again in two way frequency tables. Independence is a major theme of this unit! MY REFLECTIONS (over line l) I moved quickly through this objective last time and it came back to bite me. When I got into the more difficult independence questions even my very best students were asking a lot of conceptual questions. I discovered it was because I hadn't connected the Venn diagram to P(A) • P(B) = P(A and B) very well. I have created a number of questions to build this understanding this time around.	478	2,232	{"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}	4.5	4	CC-MAIN-2018-09	latest	en	0.952936
https://www.coursehero.com/tutors-problems/Finance/19518415-Please-solve-and-show-the-steps-as-to-how-you-obtained-each-answer-T/	1,606,255,143,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00641.warc.gz	665,061,316	30,274	Question # Please solve and show the steps as to how you obtained each answer. 1. T Corporations has just paid a dividend of \$4.45 per share. The dividends are expected to remain constant. Investors have an expected rate of return of 11.1% on the stock.What should be the price per share? 2.L stock has just apid a dividend of 3.20 per share. The dividends are expected to grow at a rate of 8.5% for the next five year, and then at a constant rate of 3.1% after that. The Required Return on Equity is 14.5%. What should the value of the stock be? 3.Looking at the analysis of the finanacial statements of Y corporation you project that the dividends for the next five years will be: D1 2.78 D2 3.45 D3 4.78 D4 2.10 D5 5.41 After year 5, you expect dividends to grow at the same pace of the rest of the ecomony, at 2.3%.If Y corporation has a Required return on 12.5%, what is the fair value of the stock? 4.If you purchased a share of stock for \$30.00 last year, received a \$2.50 dividend, and then sold it at the end of the year for \$33.00, what rate of return did you earn on the investment? 5. If company J has earns per share of \$4.56, and pays out a dividend of \$2.12, what is its Dividend Payout Ratio? What is its Plowback Ratio? 6.If a company does each of the following, what would you expect to happen to its stock price (up, down, or stay the same )? Explain why for each- 1) Increases its dividend 2) Buys back its shares in the market 3) Increases its Plowback ratio to 100% 7.If a firm has a ROE of 12.8%, and a plowback ratio of .45, what is its sustainable growth rate (g) ? 8.If a company has a growth rate of 5.7%, is currently trading at \$23.45 per share, and has an expected dividend of 3.25. what is the stocks expected return? 9.What is the relationship between the following-? The required rate of return on a stock The yield to maturity of a bond The discount rate in a present value problem 10.Explain what is meant by the following posted quote by a stock dealer firm: "Corning shares are trading at 32.56 Bid, 20 size and 32.87 Ask, 16 size" 11.What is the dividend yield of a stock with a dividend of \$2.10 per share and a price of \$26.87 per share? 12.CTA Corporation has a ROE of 20%. It has a dividened payout ratio of 50%. Last year, its earnings per share was \$3.00. The required return on equity is 12.0% a. What is the company's growth rate (g)? b. What is the company's fair value per share? 13.TZ Corporation needs \$100,000,000 to construct its new factory. It can finance the cost in 2 ways: 1) Issue 100,000 bonds, 10 -year maturity and a coupon rate of 5%. 2) Sell 2,000,000 new shares of common stock, at a price of \$50.00 per share. TZ currently has a dividend yield of 5%. Give 1 advantage, and 1 disadvantage, to each method of raising the money. 14.W Corporation has earnings of \$2.98 per share and has just paid a dividend per share of \$.67. What percent of it's earning does it reinvest in the business? 15.E Corporation has a dividend yield of 17%. The most recent dividend was \$6.60 per share. What is the current price of E Corporation 16.T Manufacturing has just issued a perpetual preferred stock. If the preferred stock has a dividend of \$5.50 per share, and is currently selling for \$42.00 per share, what rate of return do investors expect to earn on this investment? 17.H Corporation is expected to pay a dividend of 3.50 per share next year. After that time, the dividends are expected to grow at the rate of 4.5% per year in perpetuity. If investors in this stock require a 10% rate of return, what price should the stock sell at? 18.Y Co. generates a Return on Equity (ROE) of 20%. It pays out 30% of its earnings as dividends. This year, its earnings will be \$4.00 per share. Investors expect a 16% rate of return on the stock. What is Price of the stock? 19.M Co. will pay a dividend of \$2.00 per share. The company has a 50% dividend payout ratio. If the company's stock is selling for \$32.00 per share, what is its P/E ratio? 20. J corporation currently pays a dividend of \$1.22 per share, which is expected to grow at a constant rate of 5%. If the current price of J corporation is \$32.03 per share, what is the required rate of return on the stock?	1,133	4,278	{"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-2020-50	latest	en	0.956841
https://mathoverflow.net/questions/1951/does-the-cohomology-ring-of-a-simply-connected-space-x-determine-the-cohomology/2066	1,627,422,842,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00097.warc.gz	395,521,004	34,408	# Does the cohomology ring of a simply-connected space X determine the cohomology groups of ΩX? One could try to apply the Eilenberg-Moore spectral sequence to the pullback diagram • → X ← •, obtaining a spectral sequence TorH(X, R)(R, R) => H(ΩX, R), but could there be differentials or extension problems which differ for different spaces X with the same cohomology ring? ## 5 Answers Tyler's comment to my earlier answer seems to give a solution; he suggests comparing the space $T=(S^3\vee S^3)\cup\_{[x,[x,y]]} e^8$ with a wedge $S^3\vee S^3\vee S^8$. It's probably easier to think about homology with the Pontryagin product. Homology of loops on on the wedge will be a tensor algebra on classes in 2,2,7 (since it's loops of a suspension). The homology of loops on Tyler-space $T$ should differ in dimension 6: the homology class [x,[x,y]] will be 0 (where x,y are now the homology generators in dimension 2), "killed" by the new attaching map. So H_6 (and thus H^6) of the two spaces have different rank. To make this explicit, we have $S^7 \xrightarrow{f} X \rightarrow T$, where X is the wedge of two 3-spheres. The restriction of $\Omega f: \Omega S^7 \to \Omega X$ is a map $S^6 \to \Omega X$ adjoint to f, and on homology this hits the homology class corresponding to the [x,[x,y]]. The result follows because $\Omega S^7 \to \Omega X \to \Omega T$ is null homotopic. (I'm basically using the Hilton-Milnor theorem to understand $\Omega X$.) Well, simply connected Lie groups tend to have cohomology which is an exterior algebra. For instance, SU and X = S3 x S5 x S7 x ... have the same cohomology ring. But ΩSU=BU, and ΩX isn't. Since each HΩSn is a divided power algebra, while HBU is a polynomial algebra, this should provide a counterexample, I think. I'm guessing this shows up in the Eilenberg-Moore spectral sequence as a non-trivial extension problem. • Nice, too bad I've reached my daily vote limit! – Ilya Nikokoshev Oct 22 '09 at 22:25 • This doesn't give an example for what Reid asked, though because the cohomology groups are still the same. – Eric Wofsey Oct 22 '09 at 23:50 • Oh okay. I interpreted his remark about extension problems as being about multiplicative extensions, rather than group extensions. – Charles Rezk Oct 23 '09 at 0:08 • Yes, I intentionally worded my question the way I did, but it wasn't because I knew of an example for the version with "cohomology ring" in both places (it was because I expected the answer to the question in the title to be no even as stated here). This example is very nice and memorable, thanks! – Reid Barton Oct 23 '09 at 6:38 Complementing the other answers in this thread: while the cohomology ring of a simply connected space does not determine the cohomology of the loop space, the rational cohomology viewed as an $A_\infty$-algebra does. Namely, the cohomology of any $A_\infty$-algebra $A$ over $\mathbf{Q}$ (in particular, of any differential graded algebra) carries an $A_\infty$-structure such that there is an $A_\infty$ map $H^\ast(A)\to A$ inducing the identity in cohomology; this $A_\infty$ structure is unique up to a non-unique isomorphism. See e.g. Keller, Introduction to A-infinity algebras and modules, 3.3 and references therein. By taking $A$ to be the rational singular cochains of a topological space $X$ we get an $A_\infty$-structure on $H^\ast(X,\mathbf{Q})$. To each $A_\infty$ algebra $H$ there corresponds a bar construction, which is a free differential coalgebra on $H$ shifted by 1 to the left (see e.g. 3.6 of Keller's paper mentioned above). It is an old result of Kadeishvili (see MR0580645) the that if $H$ is the cohomology of a simply-connected space $X$ with the above $A_\infty$-structure, then the cohomology of the bar construction is the cohomology of $\Omega (X)$. This also explains why we should expect a negative answer to the question as it is stated: all components of the $A_\infty$ structure on the cohomology participate in the bar construction, and not just the product. It seems hard to come up with an example in which the cohomology groups of the loop spaces differ. I imagine something like the following should produce an example, in the context of rational homotopy: Find two rational commutative dgas A and B whose cohomology algebras are the same as rings, but which have different massey product structures. Then I think it's likely that the derived tensor products Q ⊗A Q and Q ⊗B Q will have different cohomology. I can't find an example small enough that I'd like to compute it, though. • You should be able to attach an 8-cell to S^3 wedge S^3 to kill a Whitehead product [x,[x,y]] of the generators x,y of pi_3 in rational homotopy. The cohomology algebra is then the same as S^3 wedge S^3 wedge S^8 ignoring secondary operations, but the underlying DGA is different. – Tyler Lawson Oct 23 '09 at 4:49 My feeling is that Charles is on the right track with the answer above. But rather than looking for a counterexample, I think we should have a go at correcting the original question. Now I'm not quite sure over which rings the next statements work, possibly only over rings over a field of char 0. Perhaps someone knows the details better than I, but to make it work will probably require working with simplicial algebras as these carry a model structure over any ring. The cochains of X carry a dg-algebra structure A. Since ΩX is the homotopy pullback of • → X ← • and taking cochains should preserve the relevant (co)limits (can someone help me here), then the cochains ring of ΩX is the homotopy pushout of k ← A → k, that is, the derived tensor product. We can then take cohomology. For the next bit we probably do need characteristic 0. The cochains ring will be rather large, so to keep track of things we could take the cohomology, but remember the higher operations. Then as an infinity ring the cohomology H(A) will be quasi-isomorphic to A (which isn't necessarily true if we don't remember the higher operations). Then with that in mind we can calculate the derived form of k ⊗_H(A) k. Its cohomology should be the cohomology of the loop space. It would be nice to have a counterexample though, how about complements of links, the cohomology rings aren't so bad to calculate (only depending on the number of links over the rationals at least). What about the loop spaces?	1,674	6,349	{"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}	2.84375	3	CC-MAIN-2021-31	latest	en	0.880204
https://kpsc.jobquiz.info/question.php?id=PoRe&page=0	1,643,419,178,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00007.warc.gz	398,411,928	7,703	## Kerala PSC Previous Years Question Paper & Answer Title : SALES ASSISTANT GR II KSCCF LTD/JR ASST/CASHIER KSEB/KERALA CERAMICS/EXCISE CIRCLE INSPECTOR EXCISE Question Code : A Page:1 Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'SALES ASSISTANT GR II KSCCF LTD/JR ASST/CASHIER KSEB/KERALA CERAMICS/EXCISE CIRCLE INSPECTOR EXCISE' And exam conducted in the year 2016. And Question paper code was '161/2016'. Medium of question paper was in Malayalam or English . Booklet Alphacode was 'A'. Answer keys are given at the bottom, but we suggest you to try answering the questions yourself and compare the key along wih to check your performance. Because we would like you to do and practice by yourself. page: 1 out of 12 Excerpt of Question Code: 161/2016 161/2016 1. 3, 10. Which of the following number is divisible by 11? (A) 8294627 (B) 3495714 (€) 7546743 (1) 974362 What should be the measure of the diagonal of a square whose area is 162 cm? ? (ಸ) 15 ‏حم 19 )0( تی‎ (0 18em (D) 14 ला 25 women can complete a work in 5 days. In how many days will 30 women can complete that work ? (A) 7 days (B) 6 days (0) 8 days (D) 10 days Find the sum of the mumbers lving between 200 and 700 which are multiples of 5. (A) 44580 (8) 34440 (C) 65240 (D) 32300 The difference between 72% and 54% of a number is 432, What 15 55% of that number ? (ಗಿ) 1355 (0) 1445 (0) 1420 (1) 1320 A cylinder of radius 4 cm and height 10 cm is 2 em. How many such sphere are got ? (A) 25 (B 30 (८) 15 (0) 28 Ited 2nd recasted into a sphere of radius Nine years ago the age of Athira was five times (hat of Ammu’s ன்‌. Aller nine years Athira’s age will be twice of Ammu's age. Athira’s present age 15: (^) 21 (೫ 24 © 3 (0) 27 2016, January 22" js Friday. Find 2016, August 20 (ക) Friday (8) Tuesday (€) Thursday (101) Saturday A train 520 m long is running with a speed of 90 km/hr, In whal time will it cress a bridge 130 m long ? [ക] 26 second (8) 25 second ) 32 சலம்‌ (D) 30 second Hx —2 ‏سٹو‎ ‎Find the value ‏أن‎ v, if - ) = 7 ~^ 5 = (^) 3 B 2 (८) 6 (>) 8 (7.1.೦.	755	2,138	{"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}	3.5	4	CC-MAIN-2022-05	latest	en	0.881634
https://prepinsta.com/data-structures/preorder-tree-traversal-without-recursion/	1,660,709,672,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00472.warc.gz	438,905,593	23,471	Preorder Tree Traversal without Recursion in C Preorder Tree Traversal Implementation in C We can perform Preorder Tree Traversal without Recursion in C in two ways, First we can use stack and other one is iterative method known as morris traversal. In given example we are using stack method to perform Preorder traversal on tree. Preorder traversal of tree is Root->Left->Right. Here we will learn how to implement preorder traversal of tree using stack and its algorithm. Steps for Implementing Preorder Tree traversal in C In Following Steps we have define how to implement Preorder tree traversal without recursion in C : 1. Take an empty stack. 2. Push the root of the tree in stack. 3. Take the temporary node and initialize it with root node of tree. 4. Print the node and push the left node of temp node in stack and initialize temp from temp->left. 5. do the same task till left node temp is not null. 6. If temp become null and stack is not empty then pop the element from the stack and initialize temp from Poped element’s right node. 7. do the same task of pushing left node of and initializing temp from temp left. 8. If temp is null and stack is empty then return. Algorithm for Preorder Traversing in Tree This is the main traversing function Algorithm which is used in Preorder tree traversing without recursion program: Inorder_Traversal() • Flag = 1 • WHILE(Flag) • IF(Temp) • PRINT(Temp->data) • push(Temp) • Temp = Temp->Left • END IF • ELSE • IF(isEmpty() == 0) • Temp = pop() • Temp = Temp->Right • ELSE • Flag =0 • END ELSE • END ELSE C Program for Preorder Tree Traversing without Recursion ```#include<stdio.h> #include<stdlib.h> struct node // node defining for tree { struct node* left; struct node* right; int data; }; struct stack // node defining for stack { struct node* data; struct stack* next; }; void push(struct stack** top,struct node* n); //function declation struct node* pop(struct stack** top); int isEmpty(struct stack* top); int tree_traversal(struct node* root) //Inorder Traversing function { struct node* temp = root; struct stack* s_temp = NULL; int flag = 1; while(flag) //Loop run untill temp is null and stack is empty { if(temp){ printf("%d",temp->data); push(&s_temp,temp); temp = temp->left; } else{ if(!isEmpty(s_temp)){ temp = pop(&s_temp); temp = temp->right; } else flag = 0; } } } void push(struct stack** top,struct node* n) //push node in stack { struct stack* new_n = (struct stack)malloc(sizeof(struct stack)); new_n->data = n; new_n->next = (top); (top) = new_n; } int isEmpty(struct stack top) // check if stack is empty { if(top==NULL) return 1; else return 0; } struct node* pop(struct stack** top_n) // pop the node from stack { struct node* item; struct stack* top; top = top_n; item = top->data; top_n = top->next; free(top); return item; } struct node* create_node(int data) // create a node for tree { struct node* new_n = (struct node)malloc(sizeof(struct node)); new_n->data = data; new_n->left = NULL; new_n->right = NULL; return (new_n); } int main() { struct node root; root = create_node(8); root->left = create_node(5); root->right = create_node(4); root->left->left = create_node(7); root->left->right = create_node(6); tree_traversal(root); return 0; }``` `output: 85764` Program Explanation: Struct Node{} : • We are defining a structure node with integer data and two pointer pointers for left and right node. Struct Stack{} : • Stack is the structure with a node data type data element and a pointer next. Tree_traversal() : • In tree traversal function we Print the node and traverse the tree left node untill we found null when there is null we backtrack in tree with the help of stack and pop the element and print it. • Check if there is right child of that node if yes then again we traverse it left of the node. • We do the same task again and again till temp node is empty and stack is empty. Push() : • Push function is use to push node from the tree into stack. Pop() : • Pop function is use for remove the node from stack and return it. IsEmpty() : • isEmpty checks if stack is empty or not if yes then it returns 1 and else it returns 0. Create_node(): • We are using create node function to create nodes for tree. Different types of Traversing without Recursion Here we have already learned implementation of Preorder traversing without recursion click the button below to learn the implementation diffrent type of traversing without recursion.	1,116	4,484	{"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}	2.625	3	CC-MAIN-2022-33	latest	en	0.663907
https://www.experts-exchange.com/questions/27652317/Obtaining-Total-Debt-Service-Ratio-bit-of-a-glitch.html	1,481,145,747,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00201-ip-10-31-129-80.ec2.internal.warc.gz	936,690,511	26,231	Solved # Obtaining Total Debt Service Ratio bit of a glitch Posted on 2012-03-28 233 Views Hello, I have an online app that as part of it being used determines your total debt service ratio simply by dividing expenses / income Found if people were to put in crazy number like: I earn \$2000 and have expenses that are \$5000 it returns 2.5 in this case. Given my code below, it strips off the 2 and returns with a 50% debt service ratio. Any idea how it should look so it will return a proper fraction when the numbers are sensible, but return differently if you are way over on your expenses? I suppose I could test if the expenses are greater then the income but wondered if there was a better way of doing it. Hope that makes some sense. ``````\$tdsr = \$expenses / \$income; \$tdsr = round(\$tdsr,2); \$new_tdsr = explode(".", \$tdsr); \$tdsr = \$new_tdsr[1]; if (strlen(\$tdsr) == 1) { \$tdsr = \$tdsr."0"; } `````` 0 Question by:tjyoung 1 Comment LVL 108 Accepted Solution Ray Paseur earned 500 total points ID: 37779241 See http://www.laprbass.com/RAY_temp_tjyoung.php Outputs: WITH EXPENSES = 5000 AND INCOME = 2000 YOUR TOTAL DEBT SERVICE RATIO IS 250% AND THIS IS TOO HIGH! ``````<?Php // RAY_temp_tjyoung.php error_reporting(E_ALL); // TEST DATA \$expenses = 5000; \$income = 2000; // COMPUTE THE RATIO AS A PERCENTAGE VALUE \$tdsr = (\$expenses / \$income) * 100.0; // PREPARE A PRINTABLE VERSION (COULD DO THIS WITH OTHER NUMBERS, TOO) \$tdsr_printable = number_format(\$tdsr) . '%'; // SHOW THE WORK PRODUCT echo "WITH EXPENSES = \$expenses AND INCOME = \$income YOUR TOTAL DEBT SERVICE RATIO IS \$tdsr_printable "; if (\$tdsr > 50.0) echo "AND THIS IS TOO HIGH!"; `````` HTH, ~Ray 0 ## Featured Post ### Suggested Solutions Consider the following scenario: You are working on a website and make something great - something that lets the server work with information submitted by your users. This could be anything, from a simple guestbook to a e-Money solution. But what… Things That Drive Us Nuts Have you noticed the use of the reCaptcha feature at EE and other web sites? It wants you to read and retype something that looks like this.Insanity! It's not EE's fault - that's just the way reCaptcha works. But it is … Explain concepts important to validation of email addresses with regular expressions. Applies to most languages/tools that uses regular expressions. Consider email address RFCs: Look at HTML5 form input element (with type=email) regex pattern: T… The viewer will learn how to create a basic form using some HTML5 and PHP for later processing. Set up your basic HTML file. Open your form tag and set the method and action attributes.: (CODE) Set up your first few inputs one for the name and …	729	2,749	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2016-50	longest	en	0.81569
http://courses.prepladder.com/gate/21-gate-me/110-what-to-study-from-steam-turbine-for-gate.html	1,498,721,332,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128323889.3/warc/CC-MAIN-20170629070237-20170629090237-00491.warc.gz	95,472,559	15,224	Pick a Course ## What to study From Steam Turbine For GATE? Steam Turbine is an important topic with high weightage in GATE examination. In this article we will discuss the important things that must be prepared in Steam Turbines to answer most of the turbine based questions in GATE exam. Free Stimulation Mock Test For GATE 2017 Start Test ## What to study in Steam Turbine For GATE • Introduction to steam turbine • Rankine cycle and modification including superheating, regenerative feed water heating, reheating, etc. • You should also prepare the classification of turbines based on blade profile, impulse and reaction turbines. • Velocity diagrams are very important, you will find one question in GATE based on this topic. • Numericals are very important. 1 Question will be surely asked on this section. Also read : Best Books for GATE ME by Toppers ## Questions From Gate Ques 1 : What happens when reheater is incorporated in a steam power plant? P : Thermal efficiency of the plant always increases. Q : The dry fraction of steam at condenser inlet always increases. R : The mean temperature of heat addition always increases. S : The specific work output always increases (A) P and S (B) Q and S (C) P, R and S (D) P,Q,R and S B is the right answer because Reheater always increases the dryness fraction of steam at condenser inlet and always increases the specific work output. Also read : how to prepare casting for GATE Free Stimulation Mock Test For GATE 2017 Start Test Ques 2 : Assume the given turbine to be a part of Rankine cycle. The density of water at the inlet to the pump is 1000 Kg/m3.ignoring the K.E.and P.E. effects, the sp. work (in KJ/Kg) supplied to the pump is (A) 0.293 (B) 0.351 (C) 2.930 (D) 3.510 C Explanation Pump work hinlet + q = houtlet + Wpump Here, q=0 Wpump = hinlet - houtlet W = ∆hpump dq=dh-Vdp dh=Vdp dhpump =V(p2 – p1) = 1/ρ(3000 – 70) kPa = 2.930 kPa Free All India Test Equipped with Artificial Intelligence Start Test Alternative method : V = m/ρ = 1/1000 kg Pump work = (p1 – p2)V = (3 x 103 – 70)/1000 KJ/Kg = 2.930 KJ/Kg Best Wishes !! What aspirants need is the right approach along with concrete strategies to reach their goals. GATE GUARANTEE PACK has been the choice of successful GATE toppers all over India. It is the perfect online tool to check your level of preparedness among st students preparing all over the country where you study under guidance of mentors.	672	2,477	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2017-26	longest	en	0.852899
http://www.mcasco.com/Order/extendin.html	1,505,859,660,000,000,000	text/html	crawl-data/CC-MAIN-2017-39/segments/1505818686043.16/warc/CC-MAIN-20170919221032-20170920001032-00361.warc.gz	504,602,839	3,785	Extending graphing concepts The work of a curious fellow The plot thickens... In the graphing example which we first did where y=x(10-x) we made an assumption that is so basic that it is almost unconscious. That assumption is that we get our next choice for x by adding a fixed amount to the current value of x. Starting at x=0, the next point was at x=1, then x=2 and so on until x=10. Let's spend a few minutes thinking about other possible rules for plotting the graph of a function. Suppose for example that the way to get the next value of x was to multiply the current x by a fixed amount rather than to add a fixed amount. On the Next x by Multiplying display we repeat the graph of y=x(10-x) except that we use the multiplication rule for picking the next x value. Now let's consider what happens if we divide the current value of x by a number greater than 1 to get the next x. Take 1.25 for example. The Next x by Dividing display illustrates that situation. It turns out that multiplying by a constant to select the next x point for a graph and dividing by a constant amount to the same thing. This follows from the fact that multiplying by a number less than 1 is equivalent to dividing, and dividing by a number less than 1 is equivalent to multiplying. Think about it. Likewise adding and subtracting a constant to get the next x are the same since subtracting is just adding a negative number. So in our examples we have exhausted the ordinary arithmetic operations as means of selecting the next x to plot. How about some other innovative schemes for deciding which point to plot next on a graph? What if instead of applying some constant to the old x to get the next one we depend on chance to fill in the graph. On our graph of y=x(10-x) let us just roll a ten-sided die, the singular of dice, (as opposed to "douse") to select our next x. This as you will see on the Next x by Chance display is not particularly efficient, what with landing on the same x repeatedly. Still, after a while the graph will emerge. Consider the effect of selecting the next x in graphing a function by applying x itself in some way rather than some constant. Take for example division by x. Try to predict what will happen if we pick some x to start with and then take 1/x to be the next point on our graph. The Next x by 1/x display will demonstrate that technique. Let's modify the ineffectual approach on the previous example in a very simple way. As before we will pick some value of x to start with. Then take the next x to be 1/(x+0.1) instead of just 1/x. Run the Next x by 1/(x+.1) display. Dividing by something more than x will evidently cause the next point to fall short of of the reciprocal. Would replacing the 0.1 in the next x selection rule with 0.2 converge to a different location or just converge faster? As you have seen, even minor changes in the "next x selection" rule can make quite a difference in how a graph gets filled in, or not filled in as the case may be. We could come up with all sorts of functions of x to select the next x for plotting. There is one particular function of x though which leads to some very interesting results. That is just the function being plotted itself. Look at the example. y=gx(1-x) We could just pick some x to start with, then select the next x equal to gx(1-x). Or more simply stated, make the new x equal to the old y. In the Next x by Feedback display you will be able to feed the output back into the input of the function y=gx*(1-x) An alternative way to display this iteration process might be to plot successive values of y over the initial value of x. This would allow us to readily see the effect of starting with different initial x values on the process of iteration. Run the New y by Feedback display. This is the last perversion of normal graphing which we will undertake for now. All of the ideas introduced here are intended to extend your vision of what a graph might be. The next lesson in the sample is called Iteration and Attractors. Are there any questions? Next Previous Other	921	4,076	{"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}	4.40625	4	CC-MAIN-2017-39	latest	en	0.951274
http://oeis.org/A129360	1,606,880,755,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00526.warc.gz	76,675,635	4,907	The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Please make a donation to keep the OEIS running. We are now in our 56th year. In the past year we added 10000 new sequences and reached almost 9000 citations (which often say "discovered thanks to the OEIS"). Other ways to donate Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A129360 A054525 * A115361. 12 1, 0, 1, -1, 0, 1, 0, 0, 0, 1, -1, 0, 0, 0, 1, 0, -1, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 (list; table; graph; refs; listen; history; text; internal format) OFFSET 1,1 COMMENTS Row sums = A209229 (1, 1, 0, 1, 0, 0, 0, 1, ...). A129353 = the inverse MÃ¶bius transform of A115361. LINKS Andrew Howroyd, Table of n, a(n) for n = 1..1275 FORMULA Moebius transform of A115361. T(n,k) = A087003(n/k) for k \| n, T(n,k) = 0 otherwise. - Andrew Howroyd, Aug 03 2018 EXAMPLE First few rows of the triangle are: 1; 0, 1; -1, 0, 1; 0, 0, 0, 1; -1, 0, 0, 0, 1; 0, -1, 0, 0, 0, 1; -1, 0, 0, 0, 0, 0, 1; 0, 0, 0, 0, 0, 0, 0, 1; ... PROG (PARI) tabl(nn) = {Tm = matrix(nn, nn, n, k, if (! (n % k), moebius(n/k), 0)); Tr = matrix(nn, nn, n, k, n--; k--; if ((n==k), 1, if (n==2k+1, -1, 0))); Ti = Tr^(-1); Tp = TmTi; for (n=1, nn, for (k=1, n, print1(Tp[n, k], ", "); ); print(); ); } \\ Michel Marcus, Mar 28 2015 (PARI) T(n, k)={ if(n%k, 0, sumdiv(n/k, d, my(e=valuation(d, 2)); if(d==1< Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified December 1 22:24 EST 2020. Contains 338858 sequences. (Running on oeis4.)	947	1,997	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.40625	3	CC-MAIN-2020-50	latest	en	0.504656
http://www.markedbyteachers.com/gcse/science/investigation-between-the-relationship-of-mass-and-time-taken-while-coming-down-a-water-slide.html	1,487,916,807,000,000,000	text/html	crawl-data/CC-MAIN-2017-09/segments/1487501171416.74/warc/CC-MAIN-20170219104611-00253-ip-10-171-10-108.ec2.internal.warc.gz	511,355,088	20,195	• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Page 1. 1 1 2. 2 2 3. 3 3 4. 4 4 5. 5 5 6. 6 6 # Investigation between the relationship of Mass and Time taken while coming down a water slide. Extracts from this document... Introduction Investigation between the relationship of Mass and Time taken while coming down a water slide In this experiment I am going to investigate the relationship between Mass and Speed of 5 people with different masses coming down a water slide. Firstly, I have obtained evidence by measuring the time taken for people with various different masses to come down a water slide. My prediction is that mass is independent of the time taken. We have used a digital stop watch to measure the time taken of the people coming down the slide. I am going to use several different scientific equations to try and prove my prediction. I also have to take into consideration that the slide is not a straight surface it has several different bumps but I also predict that this has no effect to the time taken as horizontal velocity is independent of vertical velocity because both of these masses are being pulled by gravity at a constant acceleration of 9.8m/s²so all objects free fall at the same rate of acceleration regardless of their mass. I have tabulated my results according to the different Masses in Middle Lian 9.78 m/s Omar 9.42 m/s Abhay 9.69 m/s Alan 9.69 m/s Calculating the height of the slide Although we were given the length of the slide I will show how they had calculated by using Pythagoras theorem which says in a right angle triangle the sum of squares of the base and the height is equal to the square of the hypotenuse. X² = 55² + 50² I cannot be certain if this is the correct X² = 3025 + 2500 as the slide was not a triangle but I am X² = 5525 going to carry the error forward to do X= 74.3 all calculations. From these results I can rule out the inference, the greater the persons mass the slower a person will go down or faster! This is false because Alan has a greater mass than Omar but his average time was still smaller than Omar’s who has a smaller mass and had a larger time taken. Back to my prediction which is in a straight that Mass is independent of time taken. To prove this I will use my equation h = ut + ½gt², this is the equation is used whenever an object travels at a uniform acceleration line where h = height, u = initial velocity, v = final velocity, t = time taken and g = acceleration due to gravity which is 9.8 m/s², As you can see mass does not play any role in this equation it is only acceleration that does as it is the acceleration due to gravity. Conclusion Variation of results: I am now going to work with the average time taken for the 4 different masses to show you how minor the changes are. Let’s take Lian who has a mass of 50 kg and whose average time taken was 7.56 and let’s take Alan who has a mass of 64 kg and whose average time taken was 7.63.The difference in mass is 14 kg and the difference in time taken was 0.07 seconds! Even after such a great increase in mass there has hardly been any difference in the time taken. Let us now take Abhay with a mass of 58 kg and Omar who has a mass of 60 kg which is only a 2 kg change in the mass which is far smaller than the change with Lian and Alan. Abhay’s average time taken was 7.63 and Omar’s average time taken was 7.85, the difference in time taken was 0.22 seconds. By using these results I can help prove my prediction saying that mass is independent to time taken. Improvements To improve the experiment we should have had a straight slide instead of a parabolic surface to be able to calculate the length of slide more accurately using Pythagoras’ theorem. We could have used light gates instead of stop watches to get exact readings when students start and stop on the slide this way they could avoid any human error. Each person could have used the same float as they can all encounter the same % of friction. We could have used different slides to test the effect of friction on different surfaces. This student written piece of work is one of many that can be found in our GCSE Forces and Motion section. ## Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month # Related GCSE Forces and Motion essays 1. ## How does an increased surface area change the time taken for a mass to ... From looking at my preliminary results, I have come up with my final variables: Maximum height = 260.00cm Minimum height =50.00cm Mass of cup cake = 0.33g (for both cup cakes to make fair test) Prediction My prediction is that the larger the surface area is, the longer the time will be taken for the mass to reach the ground. 2. ## The aim of this investigation is to see what the relationship between force, mass ... 0.929 8 4.384 0.831 1.824 0.993 9 4.702 0.831 1.914 1.083 10 5.082 0.831 1.967 1.136 Analysis Conclusion: Our experiment has not proved conclusively that there is a proportional relationship between the factors we are investigating and therefore our hypothesis is not categorically correct We can see, nonetheless that our 1. ## In this experiment I aim to find out how the force and mass affect ... Simultaneously start the stop clock and release the trolley (be careful not to push it or exert any extra force on it) 6. Stop the clock when the front of the trolley reaches the finish line 7. Record the time taken for the trolley to reach the finish, next to the relevant weight, in a table 8. 2. ## What is the effect of mass on friction? All readings should be recorded in a separate table. Now that the experiment is prepared, a pilot test can be carried out. In this pilot test the range of readings will be chosen along with which surfaces to choose from. 1. ## Squash Ball and Temperature Investigation some results were plotted a bit further away from the line of best fit; however, these are not anomalies but are merely less accurate results as they are not situated ridiculously far away from the line of best fit such as the points at the temperatures 250C and 300C, which 2. ## Physics Coursework: To investigate the Oscillations of a mass on a spring We can also see the oscillation in terms of energy. The total amount of elastic potential energy and kinetic energy is constant; therefore there is always the transformation between the two forms of energy when the spring is oscillating. Therefore it has the basic and universal characteristics of all waves, which means the transfer of energy. 1. ## What is the relationship between mass of an object and its acceleration? To prove this, when we have the results from the practical, I will be able to use a formula and find the relationship between mass and acceleration. I think this will differ from the relationship between the force pulling an object and its acceleration. 2. ## Mechanics 2 Coursework - 'woosh' down the slide The horizontal motion of the projectile is independent of the vertical motion, and that is why the falling motion of the block does not need to be considered. A aluminium block will be used, as it is a fairly dense object. • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to	1,755	7,667	{"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-2017-09	longest	en	0.932395
https://ask.xiaolee.net/hu/questions/1000982	1,618,948,175,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00328.warc.gz	222,088,218	5,547	# c# - ARBG konvertálása RGB-ként alfa-keveréssel original title: "c# - Converting ARBG to RGB with alpha blending" Translate Let's say that we have an ARGB color: ``````Color argb = Color.FromARGB(127, 69, 12, 255); //Light Urple. `````` When this is painted on top of an existing color, the colors will blend. So when it is blended with white, the resulting color is `Color.FromARGB(255, 162, 133, 255);` The solution should work like this: ``````Color blend = Color.White; Color argb = Color.FromARGB(127, 69, 12, 255); //Light Urple. Color rgb = ToRGB(argb, blend); //Same as Color.FromARGB(255, 162, 133, 255); `````` What is `ToRGB`'s implementation? Tegyük fel, hogy van egy ARGB színünk: Color argb = Color.FromARGB (127, 69, 12, 255); // Világos Urple. Ha ezt egy meglévő színre festenek, a színek megkeverednek. Tehát amikor keverik ... Ez az összefoglalás a fordítás után. Ha meg szeretné tekinteni a teljes fordítást, kattintson a "fordítás" ikonra Minden válasz • Translate It's called alpha blending. In psuedocode, assuming the background color (blend) always has 255 alpha. Also assumes alpha is 0-255. ``````alpha=argb.alpha() r = (alpha/255)argb.r() + (1 - alpha/255)blend.r() g = (alpha/255)argb.g() + (1 - alpha/255)blend.g() b = (alpha/255)argb.b() + (1 - alpha/255)blend.b() `````` note: you probably need to be a bit (more) careful about floating-point/int math and rounding issues, depending on language. Cast intermediates accordingly If you don't have a background color with an alpha of 255, the algebra gets alot more complicated. I've done it before and it's a fun exercise left to the reader (if you really need to know, ask another question :). In other words, what color C blends into some background the same as blending A, then blending B. This is sort of like calculating A+B (which isn't the same as B+A). • Translate I know this is an old thread, but I want to add this: ``````Public Shared Function AlphaBlend(ByVal ForeGround As Color, ByVal BackGround As Color) As Color If ForeGround.A = 0 Then Return BackGround If BackGround.A = 0 Then Return ForeGround If ForeGround.A = 255 Then Return ForeGround Dim Alpha As Integer = CInt(ForeGround.A) + 1 Dim B As Integer = Alpha * ForeGround.B + (255 - Alpha) * BackGround.B >> 8 Dim G As Integer = Alpha * ForeGround.G + (255 - Alpha) * BackGround.G >> 8 Dim R As Integer = Alpha * ForeGround.R + (255 - Alpha) * BackGround.R >> 8 Dim A As Integer = ForeGround.A If BackGround.A = 255 Then A = 255 If A > 255 Then A = 255 If R > 255 Then R = 255 If G > 255 Then G = 255 If B > 255 Then B = 255 Return Color.FromArgb(Math.Abs(A), Math.Abs(R), Math.Abs(G), Math.Abs(B)) End Function public static Color AlphaBlend(Color ForeGround, Color BackGround) { if (ForeGround.A == 0) return BackGround; if (BackGround.A == 0) return ForeGround; if (ForeGround.A == 255) return ForeGround; int Alpha = Convert.ToInt32(ForeGround.A) + 1; int B = Alpha * ForeGround.B + (255 - Alpha) * BackGround.B >> 8; int G = Alpha * ForeGround.G + (255 - Alpha) * BackGround.G >> 8; int R = Alpha * ForeGround.R + (255 - Alpha) * BackGround.R >> 8; int A = ForeGround.A; if (BackGround.A == 255) A = 255; if (A > 255) A = 255; if (R > 255) R = 255; if (G > 255) G = 255; if (B > 255) B = 255; return Color.FromArgb(Math.Abs(A), Math.Abs(R), Math.Abs(G), Math.Abs(B)); } `````` • Translate if you don't need to know this pre-render, you could always use the win32 method of getpixel, I believe. Note: typing on iPhone in the middle of Missouri with no inet access. Will look up real win32 example and see if there is a .net equivalent. In case anyone cares, and doesn't want to use the (excellent) answer posted above, you can get the color value of a pixel in .Net via this link MSDN example	1,155	3,786	{"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}	2.734375	3	CC-MAIN-2021-17	latest	en	0.576021
https://gmatclub.com/forum/mba-but-no-business-finance-background-153227.html?sort_by_oldest=true	1,487,855,279,000,000,000	text/html	crawl-data/CC-MAIN-2017-09/segments/1487501171166.18/warc/CC-MAIN-20170219104611-00504-ip-10-171-10-108.ec2.internal.warc.gz	716,164,764	54,919	MBA but no business/finance background : The B-School Application Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack It is currently 23 Feb 2017, 05:07 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # MBA but no business/finance background new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 08 Jan 2013 Posts: 9 GMAT Date: 10-05-2013 Followers: 0 Kudos [?]: 10 [0], given: 1 ### Show Tags 22 May 2013, 11:34 1 This post was BOOKMARKED I am seriously contemplating business school. I have no business/finance background. I know diverse backgrounds are encouraged but i am having trouble molding my current experience to my future goals which is to attain my MBA and specialize in marketing. I have been an insurance claims adjuster and am seeking a more fulfiling career experience and am tired of feeling like a drone and just some employee with no voice. My experience in insurance claims in not linked at all in marketing in anyway so my difficulty in my essay is saying how my experience now would make me a great candidate for a marketing program. Any advice on how to tie in my field or work now to my career goal in the future will be greatlyyyyyy appreciated. Joined: 18 Apr 2013 Posts: 2183 Location: Chicago, IL Followers: 46 Kudos [?]: 438 [1] , given: 3 Re: MBA but no business/finance background [#permalink] ### Show Tags 22 May 2013, 14:46 1 KUDOS Expert's post Hey there, So this may sound strange as advice - but a possible way to tie the two together is...to not tie them together at all. I'll do my best to explain. Keep in mind, there is a much longer conversation that needs to happen here regarding how to still showcase MBA worthy attributes of your current profession but I wouldn't be able to get into that here with the information I have on hand. A lot of students utilize an MBA to switch careers. Teachers to consultants. Bankers to entrepreneurs. Actuaries to brand managers. The stories there are always the same. You have to showcase your current experiences and why they set you up for success in an MBA, why it is nothing near what you want to do, why you want to move to marketing (or any other profession), and what has made you want to change / what you have done to start moving in that direction. Have you spoken to marketing professionals or held informational interviews? Maybe you've joined a professional organization to learn more. Or maybe the insurance company you work for just has a great marketing team (some are funnier than others) and it inspired you. At the end of the day, the transition has to be realistic and logical. Logical doesn't have to mean you are using your pre-MBA career to set up your post-MBA plans. You could cut ties with them entirely for all the school cares AS LONG as you articulate why and how that old career is going to make you an asset in the classroom. Like I said, there's a lot more to talk about around how to showcase your experiences to make sure the MBA programs don't feel you'll be a wasted body in their class but from a goal perspective, if it's your passion, it can be conveyed! I'd be happy to chat more about this and good luck! Bhavik _________________ Critical Square \| MBA Admissions Services Web \| Blog \| Free Resources \| Facebook \| Twitter Sign up for a free consultation today! Was this post helpful to you? Give a Kudos and let us know! Intern Joined: 08 Jan 2013 Posts: 9 GMAT Date: 10-05-2013 Followers: 0 Kudos [?]: 10 [0], given: 1 Re: MBA but no business/finance background [#permalink] ### Show Tags 23 May 2013, 07:15 Wow, i really hadnt thought about it this way. I am a claim adjsuter that settles Bodily Injuries. I work for GEICO and i truly am inspired by the marketing for a product as intangilble as insurance. My reasons for chosing marketing it that i have a creative mind and like to think outside the box. My job restricts me from doing so . And i have no fomral marketing training . The reason for wanting this MBA is for a career change. I can say that my job training now has tough me how to make good business decisions as well as satifying customer needs.I will definitley take your advice and incorporate this in my essay. I really appreciate it. Joined: 18 Apr 2013 Posts: 2183 Location: Chicago, IL Followers: 46 Kudos [?]: 438 [0], given: 3 Re: MBA but no business/finance background [#permalink] ### Show Tags 23 May 2013, 10:57 Glad I could help - good luck! _________________ Critical Square \| MBA Admissions Services Web \| Blog \| Free Resources \| Facebook \| Twitter Sign up for a free consultation today! Was this post helpful to you? Give a Kudos and let us know! Manager Status: Mock Interviews for Business Schools by Alumni Joined: 25 Apr 2013 Posts: 194 Location: United States Followers: 2 Kudos [?]: 22 [0], given: 2 Re: MBA but no business/finance background [#permalink] ### Show Tags 23 May 2013, 11:32 I agree to your view that diversity is welcome in Business schools. You bring experience of the insurance sector and knowledge about a product which is extremely difficult to market. It is very important that you bring this perspective in your application essays. Besides work experience, admission committee look for several other elements in an application. They look for people who have demonstrated leadership ability, multi-tasking ability and who achieved results. Take time to introspect and find situations in your life where you worked as team lead or took initiatives. These are essential to write a good application essay. Best of Luck! http://www.InterviewBay.com Re: MBA but no business/finance background [#permalink] 23 May 2013, 11:32 Similar topics Replies Last post Similar Topics: MBA application without a strong career background 2 01 Apr 2015, 22:40 MBA in Finance (with a marketing background) 3 12 Nov 2014, 07:33 Very untraditional background: chance for uppermid-tier MBA? 8 18 Jan 2014, 11:35 Chances at a top MBA from a Fine Arts Background? 8 02 Sep 2010, 12:02 Pre MBA Background 1 19 Jan 2009, 14:47 Display posts from previous: Sort by # MBA but no business/finance background new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	1,694	7,053	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2017-09	longest	en	0.899279
http://interactivepython.org/runestone/static/webfundamentals/CSS/styleattributes.html	1,516,157,437,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00699.warc.gz	174,026,064	8,215	Changing the Style of an HTML Element¶ There are many ways that we can change the style of what we see in the browser. In this section we will look at examples of the following: • background • text • font But before we talk about these, we need to think about colors. There are three different ways to specify the color. • by name, like blue, red, green. You can see a complete list of color names on the w3schools website. • using an RGB value like rgb(255,0,0) • using a HEX value like #ff0000 Using either the RGB or the HEX value gives you total control to specify any of of 16 million different colors. There is a bit of interesting computer science behind the RGB and HEX values. The rgb function lets you specify a value between 0 and 255 for each component of red, green, and blue. By mixing together a certain amount of red, green, and blue you can create $$255 cdot 255 cdot 255$$ different colors, which is slightly more than 16.5 million. Now where does the number 255 come from? It is one less than $$2^8$$! Computer scientists like powers of two because when you get deeply into the inner workings of the computer you see that everything is in binary (ones or zeros) which we call bits. With eight bits we can specify 256 different values or 0 – 255. We call eight bits one byte. Now the HEX specification of the number is directly related to the binary as follows: binary hex decimal 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 1000 8 8 1001 9 9 1010 a 10 1011 b 11 1100 c 12 1101 d 13 1110 e 14 1111 f 15 When specifying a color using the HEX system the first two characters are for the red, the second two for the green, and the last are for the blue. There are lots of color picking tools that you can use that will let you choose the color you want and then tell you the appropriate hex value. Background¶ CSS has the following properties which we can use to change the background. • background-color • background-image • background-repeat • background-attachment • background-position Text¶ • text-color • text-align • text-decoration • text-transformation Font¶ • font-family • font-style • font-size Next Section - More on Matching	575	2,196	{"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}	2.671875	3	CC-MAIN-2018-05	latest	en	0.813436
http://formulas.ultrafractal.com/reference/reb/REB_TentMapJulia.html	1,642,712,215,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00698.warc.gz	23,196,930	4,730	## reb Class REB_TentMapJulia ```Object common:Generic common:Formula common:DivergentFormula reb:REB_TentMapJulia ``` `class DivergentFormula:REB_TentMapJulia` Julia version of the Tent Map as described in Wikipedia and Mathworld. This version of the tent map has somewhat different initializing conditions from those used by Kerry Mitchell. For this version: T(x) = sx for x <= 0.5 T(x) = s(1-x) for x > 0.5 An inverted form is available, along with several tent types and two flavors. Ultra Fractal Source ``` class REB_TentMapJulia(common.ulb:DivergentFormula) { ; Julia version of the Tent Map as described in Wikipedia and Mathworld. <br> ; <p> ; This version of the tent map has somewhat different initializing conditions ; from those used by Kerry Mitchell. For this version: ; T(x) = sx for x <= 0.5 <br> ; T(x) = s(1-x) for x > 0.5 <br> ; <p> An inverted form is available, along with several tent types and two ; flavors. public: import "common.ulb" ; constructor func REB_TentMapJulia(Generic pparent) DivergentFormula.DivergentFormula(pparent) endfunc ; initialize the formula complex func Init(complex pz) DivergentFormula.Init(pz) m_c = pz if @invert pz = 1/(m_c+(0.000001,0.000001)) endif if @flavor == "Flavor 1" pz = @tentfn(pz) endif return pz endfunc ; call for every iterated point complex func Iterate(complex pz) DivergentFormula.Iterate(pz) if @flavor == "Flavor 2" pz = @tentfn(pz) endif if @invert if @type == "cabs" if cabs(@tentfn2(pz)) <= @tent3 pz = @tent1pz/(@seed+(0.000001,0.000001)) else pz = @tent1(@tent2-pz)/(@seed+(0.000001,0.000001)) endif elseif @type == "real" if real(@tentfn2(pz)) <= @tent3 pz = @tent1pz/(@seed+(0.000001,0.000001)) else pz = @tent1(@tent2-pz)/(@seed+(0.000001,0.000001)) endif elseif @type == "imag" if imag(@tentfn2(pz)) <= @tent3 pz = @tent1pz/(@seed+(0.000001,0.000001)) else pz = @tent1(@tent2-pz)/(@seed+(0.000001,0.000001)) endif else if imag(@tentfn2(pz)) > @tent3 pz = @tent1(@tent2-pz)/(@seed+(0.000001,0.000001)) elseif real(@tentfn2(pz)) > @tent3 pz = @tent1(@tent2-pz)/(@seed+(0.000001,0.000001)) else pz = @tent1pz/(@seed+(0.000001,0.000001)) endif endif else if @type == "cabs" if cabs(@tentfn2(pz)) <= @tent3 pz = @tent1pz@seed else pz = @tent1(@tent2-pz)@seed endif elseif @type == "real" if real(@tentfn2(pz)) <= @tent3 pz = @tent1pz@seed else pz = @tent1(@tent2-pz)@seed endif elseif @type == "imag" if imag(@tentfn2(pz)) <= @tent3 pz = @tent1pz@seed else pz = @tent1(@tent2-pz)@seed endif else if imag(@tentfn2(pz)) > @tent3 pz = @tent1(@tent2-pz)@seed elseif real(@tentfn2(pz)) > @tent3 pz = @tent1(@tent2-pz)@seed else pz = @tent1pz@seed endif endif endif return pz endfunc ; Override the default function for bailout bool func IsBailedOut(complex pz) bool bail = false if @test == 0 bail = (\|pz\| > @p_bailout) elseif @test == 1 bail = (sqr(real(pz)) > @p_bailout) elseif @test == 2 bail = (sqr(imag(pz)) > @p_bailout) elseif @test == 3 bail = (sqr(real(pz)) > @p_bailout \|\| sqr(imag(pz)) > @p_bailout) elseif @test == 4 bail = (sqr(real(pz)) > @p_bailout && sqr(imag(pz)) > @p_bailout) elseif @test == 5 bail = (sqr(abs(real(pz)) + abs(imag(pz))) > @p_bailout) elseif @test == 6 bail = (sqr(real(pz) + imag(pz)) > @p_bailout) endif return bail endfunc protected: complex m_c default: title = "Tent Map Julia" int param v_tentmapjulia caption = "Version (Tent Map Julia)" default = 101 hint = "This version parameter is used to detect when a change has been made to the formula that is incompatible with the previous version. When that happens, this field will reflect the old version number to alert you to the fact that an alternate rendering is being used." visible = @v_tentmapjulia < 101 endparam text = "Based upon the Tent Map function:" text = " T(x) = sx for x <= 0.5" text = " T(x) = s*(1-x) for x > 0.5" text = "and the code for the Barnsley type fractals." param type caption = "Tent type" default = 3 enum = "cabs" "real" "imag" "Pinsky" endparam param flavor caption = "Flavor" default = 0 enum = "Flavor 1" "Flavor 2" endparam bool param invert caption = "Inversion" default = false endparam param test caption = "Bailout Test" default = 0 enum = "mod" "real" "imag" "or" "and" "manh" "manr" endparam float param p_bailout caption = "Bailout value" default = 10000 min = 1 endparam float param tent1 caption ="Tent param #1" default = 1 endparam float param tent2 caption ="Tent param #2" default = 1 endparam float param tent3 caption ="Tent param #3" default = 0.5 endparam func tentfn caption = "Tent function" default = ident() endfunc func tentfn2 caption = "Tent function #2" default = ident() endfunc complex param seed caption = "Julia seed" default = (1.04375,0.35625) hint = "Use this parameter to create many different Julia sets. A good \ way to set this parameter is with the Eyedropper or Explore features." endparam complex param p_power ; Overrides p_power from Formula caption = "Power" default = (2,0) visible = false endparam } ``` Constructor Summary `REB_TentMapJulia()` `REB_TentMapJulia(Generic pparent)` constructor Method Summary ` complex` `Init(complex pz)` initialize the formula ` boolean` `IsBailedOut(complex pz)` Override the default function for bailout ` complex` `Iterate(complex pz)` call for every iterated point Methods inherited from class common:DivergentFormula `GetUpperBailout` Methods inherited from class common:Formula `GetLowerBailout, GetPrimaryExponent` Methods inherited from class common:Generic `GetParent` Methods inherited from class Object Constructor Detail ### REB_TentMapJulia `public REB_TentMapJulia(Generic pparent)` constructor ### REB_TentMapJulia `public REB_TentMapJulia()` Method Detail ### Init `public complex Init(complex pz)` initialize the formula Overrides: `Init` in class `DivergentFormula` Parameters: `pz` - seed value for the sequence; for a normal fractal formula, this will be #pixel Returns: first value in the sequence; this corresponds to #z in a fractal formula ### Iterate `public complex Iterate(complex pz)` call for every iterated point Overrides: `Iterate` in class `Formula` Parameters: `pz` - previous value in the sequence; corresponds to #z in a fractal formula. Note that you should always use this value for computing the next iteration, rather than a saved value, as the calling code may modify the returned value before passing it back to the next Iterate() call. Returns: the next value in the sequence ### IsBailedOut `public boolean IsBailedOut(complex pz)` Override the default function for bailout Overrides: `IsBailedOut` in class `DivergentFormula` Parameters: `pz` - last sequence value to test; this should be the value returned from the previous Iterate() call. Note that it is acceptable to ignore pz and use m_BailedOut, but any code calling IsBailedOut() should pass in the correct pz for Formula classes which do not use m_BailedOut. Returns: true if the sequence has bailed out (i.e. should be terminated)	2,247	6,989	{"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}	2.734375	3	CC-MAIN-2022-05	longest	en	0.72562
https://sites.google.com/site/sjpphysicscp1/home/bu-research/diffraction	1,547,652,286,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00393.warc.gz	642,808,946	8,802	### 1) Shape diffraction dif·frac·tion (dĭ-frāk'shən) n. Change in the directions and intensities of a group of waves after passing by an obstacle or through an aperture. Whether or not you're a veteran of high school physics, you will recognize that the ocean waves shown here are somehow bending around the rocks. If you have thought about this process before, you will probably also be aware that the way that diffraction works depends on two obvious things: Wavelength, and the Size of the barrier the waves are interacting with. Play with this applet to get a feel for wavelength and stuff. The bottom line with this simple type of diffraction is that big wavelength waves bend easily around objects, and little ones don't. The best diffraction effects happen when the wavelength and the object in question are about the same size, because that creates interesting patterns of interference - and that brings us to a connection with the nano-world. Here, you're looking at a scanning tunneling microscope (SEM) image of a razor blade. It does have roughness at the microscopic level, but you can see that the ridges are really small - about the same as the wavelengths of light that we can see. The second thing that's so cool about the razor blade example is that it shows waves interfering, or destroying each other. That's what accounts for the dark fringes in the picture - only waves can do this, and it's key for our big picture. And Even More Nano:In the early days of the 1900's, not everyone bought into the idea that salt was bound by ionic bonds, or that X-rays were light, just like the blue light diffracting around the razor blade in the example you just looked at. When we look at salt using visible light, all we see is white crystal shapes, but that's because visible light is huge compared to salt crystals, and pretty much ignores the salt crystals. Nope, to see something THAT small, we're going to need some REALLY small waves! In 1913, Max Von Laue bounced X-Rays off of powdered salt crystals, and found the same type of pattern that you've seen already. You should recognize interference patterns when you see them, and that must mean that Von Laue was finally using light that was in the ballpark of the salt itself! Just to give you a sense of the sizes here, the sodium atom is about 0.2 nm across. If you understand diffraction, you're on your way to studying the world of the very small. There is a great deal more to the subject, read on! Light diffracts around objects, so to see small objects clearly, you need to use small waves of light, because big waves blur and get all messy. The razor blade photo is one of my all-time favorites, because it shows two things. First, Razor blades are WICKED SHARP, which means that the blade is smooth to a very fine degree, on the order of a wave of light, or maybe 500 nm.	642	2,860	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2019-04	latest	en	0.941558
https://www.fliphtml5.com/atss/uorn	1,679,912,338,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00140.warc.gz	848,699,809	12,674	Publications: Followers: Follow Page 3 4. In this study, the response variable is A) the volume of oxygen consumed at rest. B) the volume of oxygen consumed while running. Publications Read Text Version Name: __________________________ Date: _____________ 1. Which of the following is true of the correlation coefficient r? A) It is a resistant measure of association. B) –1 < r < 1. C) If r is the correlation between X and Y, then –r is the correlation between Y and X. D) All of the above.Use the following to answer question 2:The volume of oxygen consumed (in liters per minute) while a person is at rest and while he orshe is exercising (running on a treadmill) was measured for each of 50 subjects. The goal is todetermine if the volume of oxygen consumed during aerobic exercise can be estimated from theamount consumed at rest. The results are plotted below. 2. The scatterplot suggests A) there is a positive association between the volume of oxygen consumed at rest and while running. B) there is an outlier in the plot. C) both a and b. D) neither a nor b. Page 1 3. A student wonders if people of similar heights tend to date each other. She measures herself, her dormitory roommate, and the women in the adjoining rooms; then she measures the next man each woman dates. Here are the data (heights in inches). Women 66 64 66 65 70 65 Men 72 68 70 68 74 69 Which of the following statements is true? A) The variables measured are all categorical. B) There is a strong positive correlation between the heights of men and women, since the women are always smaller than the men they date. C) There is a positive correlation between the heights of men and women. D) Correlation makes no sense here, since gender is a categorical variable.Use the following to answer question 4:The volume of oxygen consumed (in liters per minute) while a person is at rest and while he orshe is exercising (running on a treadmill) was measured for each of 50 subjects. The goal is todetermine if the volume of oxygen consumed during aerobic exercise can be estimated from theamount consumed at rest. The results are plotted below. Page 2 4. In this study, the response variable is A) the volume of oxygen consumed at rest. B) the volume of oxygen consumed while running. C) the volume consumed at rest or while running. It doesn't matter which is considered the response. D) the measuring instrument used to measure the volume consumed.5. The Columbus Zoo conducts a study to determine whether a household's income can be used to predict the amount of money the household will give to the zoo's annual fund drive. The response variable in this study is A) the Columbus Zoo. B) a household's income. C) the amount of money a household gives to the zoo's annual fund drive. D) all households in Columbus.6. Consider the following scatterplot of two variables X and Y. We may conclude A) the correlation between X and Y must be close to 1, since there is nearly a perfect relation between them. B) the correlation between X and Y must be close to –1, since there is nearly a perfect relation between them but it is not a straight line relation. C) the correlation between X and Y is close to 0. D) the correlation between X and Y could be any number between –1 and +1. Without knowing the actual values we can say nothing more. Page 3 7. The data in the scatterplot below are from a small data set. The data were classified as either being collected in the winter or in the summer. Those collected in the winter are indicated by open circles and those in the summer by solid circles. The overall correlation of the data in this scatterplot is A) positive. B) negative, since the open circles display a negative trend and the solid circles display a negative trend. C) near 0, since the open circles display a negative trend and the solid circles display a negative trend, but the trend from the open circles to the solid circles is positive. The different trends cancel. D) impossible to compute for such a data set.8. Does mandatory gun ownership prevent crime? To study this, the number of burglaries committed each month in a small town were recorded for 75 months prior to passage of a bill requiring citizens to own guns and for 56 months after passage of the bill. The goal was to see if the number of burglaries committed was affected by requiring citizens to own guns. The response variable here is A) whether gun ownership is required by law. B) the number of burglaries committed. C) the number of guns owned. D) whether a burglary was committed by a gun owner. Page 4 9. Do creative people make better salespeople? Ten sales staff in a large company were given a creativity test (scores range from 0 to 20, with higher scores indicating greater creativity) and were evaluated regarding sales growth performance (a score of 100 indicates average performance, and larger scores indicate better performance). The creativity scores and sales growth performance scores follow. We want to investigate if creative people tend to perform better with regard to sales growth. The response variable is A) creativity score. B) the gender of the salesperson. C) ten, the number of people tested. D) sales growth performance. Page 5 10. Consider the following scatterplot of the weight of several models of cars (in pounds) vs. their horsepower. A plausible value for the correlation between weight and mpg is A) +0.2. B) – 0.9. C) +0.8. D) – 1.0. Page 6	1,214	5,446	{"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}	3.71875	4	CC-MAIN-2023-14	latest	en	0.929722
https://math.stackexchange.com/questions/16501/generalizations-of-the-number-theory-concepts-of-even-and-odd/16503	1,571,509,046,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986697439.41/warc/CC-MAIN-20191019164943-20191019192443-00375.warc.gz	584,809,301	35,283	# Generalizations of the number theory concepts of “even” and “odd”? One of the very first number theory concepts introduced to students -- even before primeness, divisibility, etc. -- is the idea that a natural number can either be "even" (that is, evenly divisible by 2) or "odd" (all other numbers). For all intents and purposes, at that time, even and odd numbers were evenly distributed and of the same density in the natural numbers. There always seemed to be something inherently "special" about a number's evenness or oddness, besides the trivial one that members of one class could be divided equally into two groups and the other cannot. One would memorize addition and multiplication tables of evenness and oddness (an odd number will remain odd when added to ANY even number). You could not iterate through all of the even numbers through multiplication alone, while you could for the odds. I've heard the chess board being described as possessing evenness and oddness. For example, dark squares and light squares can represent either even or odd. A diagonal move can be an "even" move and a side-to-side move is an "odd" one, and moves are represented as additions to a square. In this way, a knight's move is an "odd" move (odd+odd+even), and when added to an odd square will yield an even square; when added to an even square will yield an odd square (odd + odd = even, odd + even = odd) A bishop's move can be considered always even, so once a bishop is on an odd square, it can only ever move onto other odd squares. Likewise for bishops on even squares. Are there any more generalizations of this concept to math? Is it meaningful to talk of even or odd matrices, or even or odd vectors or vector spaces? I've heard the concept applied to functions (even or odd functions), but I don't know if they are related to this by anything other than their name. • "You could not iterate through all of the odd numbers through multiplication alone, while you could for evens." Sorry, I don't follow. What do you mean here? – Pete L. Clark Jan 6 '11 at 0:54 • Perhaps Justin means addition, i.e. every even number is the previous one plus 2, which is even. – Zev Chonoles Jan 6 '11 at 4:47 • @Pete I'm sorry, it was a mis-typing. I mean that the series of even numbers could be defined by a product, whereas the series of odd numbers must be expressed as a product +1. – Justin L. Jan 6 '11 at 22:39 • About even and odd functions: $x^n$ is an even function iff $n$ is an even function (remember that $0$ is even) and an odd function iff $n$ is an odd number. Similarly, this works for polynomials: If all of the exponents are even, it's even, and if they're all odd, it's odd — if it's a mix, like $x+x^2$, it's neither. (Of course, this extends to "infinite polynomials" or power series. Almost any function you know can be written in terms of a power series.) – Akiva Weinberger Apr 16 '15 at 10:37 Yes. The generalization is provided by modular arithmetic. The properties you are observing all come from the fact that taking the remainder modulo $n$ respects addition and multiplication, and this generalizes to any $n$. More generally in abstract algebra we study rings and their ideals for the same reasons. The notion of evenness and oddness of functions is closely related, but it is somewhat hard to explain exactly why. The key point is that there is a certain group, the cyclic group $C_2$ of order $2$, which is behind both concepts. For now, note that the product of two even functions is even, the product of an even and odd function is odd, and the product of two odd functions is even, so even and odd functions under multiplication behave exactly the same way as even and odd numbers under addition. There are also huge generalizations depending on exactly what you're looking at, so it's hard to give a complete list here. You mentioned chessboards; there is a more general construction here, but it is somewhat hard to explain and there are no good elementary references that I know of. Once you learn some modular arithmetic, here is the modular arithmetic explanation of the chessboard idea: you can assign integer coordinates $(x, y)$ to each square (for example the coordinate of the lower left corner), and then you partition them into black or white squares depending on whether $x + y$ is even or odd; that is, depending on the value of $x + y \bmod 2$. Then given two points $(a, b)$ and $(c, d)$ you can consider the difference $c + d - a - b \bmod 2$, and constraints on this difference translate to constraints on the movement of certain pieces. This idea can be used, for example, to prove that certain chessboards (with pieces cut out of them) cannot be tiled with $1 \times 2$ or $2 \times 1$ tiles because these tiles must cover both a white square and a black square. Of course there are generalizations with $2$ replaced by a larger modulus and larger tiles. As for matrices and vectors, let's just say that there are a lot of things this could mean, and none of them are straightforward generalizations of the above concept. • I think in the second paragraph you want to say that the product of two odd functions is even, not odd. – punctured dusk Apr 15 '15 at 18:35 • @barto: ah, yep, I'll fix that. – Qiaochu Yuan Apr 16 '15 at 3:38 One example that immediately comes to mind is permutations. There is a concept of the parity of a permutation, which corresponds to the parity of the number of swaps needed to get it back to the original position. This same concept is sometimes talked in terms of "sign" of a permutation, which is either 1 or -1, depending on whether the permutation is even or odd. Permutations are related to matrices, as they show up in the definition of the determinant, and in fact, the sign of the permutation is used unlike the definition of the permanent. Some configurations of the famous fifteen puzzle are shown to be unsolvable by considering the parity of the permutations involved. Look at this page: Parity of Permutation which also talks about generalizations of this concept here: Generalizations to Coxeter Groups.	1,417	6,114	{"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}	3.8125	4	CC-MAIN-2019-43	latest	en	0.958244
https://paperzz.com/doc/9329391/lab-2-graphing-qualitative-data--bar-plot-and-pie-charts-	1,643,372,238,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00173.warc.gz	490,602,336	7,138	### Lab 2 Graphing Qualitative Data (Bar Plot and Pie Charts) ```Math 134 Lab 2 Graphing Qualitative Data (Bar Plot and Pie Charts) The purpose of this lab is to teach you how to perform exploratory data analysis by graphically plotting qualitative data using bar graphs and pie charts, for qualitative (or categorical) variables. We will make graphs using data from “Cereal Nutrition” found on our class group data site. Open the data file and read the description of the data set’s variables. Background information: The “Cereal Nutrition” data contains nutritional information and grocery shelf location for several different brands of cereals. Current research states that adults should consume no more than 30% of their calories in the form of fat, they need about 50 grams (women) or 63 grams (men) of protein daily, and should provide for the remainder of their caloric intake with complex carbohydrates. One gram of fat contains 9 calories and carbohydrates and proteins contain 4 calories per gram. A "good" diet should also contain 20-35 grams of dietary fiber. In this lab we’ll investigate the marketing behind shelf location of many cereals and analyze the cereal sugar content by targeted consumer (adult vs. child) for cereals. Lab Instructions: Step 1. Make bar plots. Select Graphics/Bar Plot/with data. Select “Manu” as your column. Click next and select Type: “Frequency” and Order by: “Count Ascending” and check “Display values above bars if possible”. Click next and type in an appropriate label for the X axis and a great Title for your graph. Remember Who, What, Where, Select “Create Graph!” Export your results by clicking the “Options” in upper left hand corner of the graph window. Select “Export to my results”.. Print in color. Step 2. Repeat Step 1 only choose Order by: “Count descending”. Repeat and choose “Value Ascending”. Repeat and choose “Value Descending”. You can accomplish this quickly by using the “Options” in the upper left hand corner and selecting “EDIT”. This will give you 4 bar charts of the same data. Step 3. Make pie charts. Graphics/Pie Chart/with data. Select “Manu” as your column. Click next. Select “Percent of Total” and deselect “Count” so that the percents are displayed and NOT the counts. Click next and type in an appropriate Title. Include your name in the Title for your graph. Select “Create Graph!” Export your results. Print in color. Research Question: What is the relationship between whether the target consumers of the cereal are adults or children and location of the cereal: on the bottom, middle or top shelf? For marketing reasons, we will note where the kids’ cereals are located compared to healthy cereals. You wouldn't put Frosted Flakes on the top shelf out of kids' reach would you? You will plot side-by-side split bar plots for shelf location grouped by consumer target by the following steps: Step 4. Make side-by-side split bar plots. Select Graphics/Bar Plot/with data. Select “Shelf” as your column. Locate the “Group by” pull-down menu. Select “Target” in the pull-down menu. Select “Split bars”. Click next. For the Type select “Percent of Total” in the pull-down menu. Make sure the “Display values above bars if possible” option is checked. 1 Math 134 Click next and type in appropriate labels for the X and Y axes. Give your graph a descriptive Title, as always, include your name. Make sure “Use same Y axis” and “Draw horizontal grid line” options are checked. Select “Create Graph!” Tip: You ought to produce a bar plot with two colors, one for Adults and one for Children. If you see more than two colors, make sure that rows or columns in the data file are NOT selected. Export your results. Print in color. Step 5. Highlight the Middle Shelf bar that represents Children by left-clicking and dragging a box inside the bar. Note that rows within the “Cereal Nutrition” data file are highlighted. The highlighted data items correspond to those that comprise the bar you outlined in the bar plot. Make a note of these cereals as your discussion question will ask you about them. To de-select the highlighted data rows, click on the word “Row” in the upper-left corner of the data file. Tip: This only works in the active window with the “Options” button in the upper left corner. You can not do this task if you open the graph from “My Results”. Discussion questions: 1. Is “Manufacturer” a qualitative nominal, qualitative ordinal, quantitative discrete or quantitative continuous variable? 2. Describe the distribution of the manufacturers. 3. Is “Shelf” a qualitative nominal, qualitative ordinal, quantitative discrete or quantitative continuous variable? 4. How many kids’ cereals are located on the middle shelf? List these cereals. 5. Use your split bar plot to answer the stated research question. Research Question: What is the relationship between whether the target consumers of the cereal are adults or children and location of the cereal: on the bottom, middle or top shelf? A complete lab 2 will have in this order:    A cover sheet with your name, class meeting time, professor’s name and lab number in the upper right hand corner. All labs for the entire semester must have a cover sheet just like this. Computer output consisting four bar plots and one pie chart and one side-by-side split bar plot all printed in color. Typed discussion questions along with the answers. For this and all future labs use a word processor to copy the entire discussion questions and give your answer in complete, grammatically correct sentences. 2 ```	1,238	5,553	{"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}	3.03125	3	CC-MAIN-2022-05	latest	en	0.868828
http://stackoverflow.com/questions/7249903/how-does-this-cascading-work/7250959	1,419,261,444,000,000,000	text/html	crawl-data/CC-MAIN-2014-52/segments/1418802775392.34/warc/CC-MAIN-20141217075255-00021-ip-10-231-17-201.ec2.internal.warc.gz	255,208,303	20,083	I have the following class interface: `````` class Time { public: Time( int = 0, int = 0, int = 0 ); Time &setHour( int ); Time &setMinute( int ); Time &setSecond( int ); private: int hour; int minute; int second; }; `````` The implementation is here: `````` Time &Time::setHour( int h ) { hour = ( h >= 0 && h < 24 ) ? h : 0; return this; } Time &Time::setMinute( int m ) { minute = ( m >= 0 && m < 60 ) ? m : 0; return this; } Time &Time::setSecond( int s ) { second = ( s >= 0 && s < 60 ) ? s : 0; return this; } `````` In my main .cpp file, I have this code: ``````int main() { Time t; t.setHour( 18 ).setMinute( 30 ).setSecond( 22 ); return 0; } `````` How is it possible to chain these function calls together? I don't understand why this works. - Each of t's methods return a reference to t. A reference is an alias. So if you did `````` Time t; Time& tAgain = t; tAgain.setMinute( 30 ); `````` `tAgain.setMinute` also alters t's time. Now extrapolate that simple example to cascading. Each method of t returns a reference to itself `````` Time &Time::setSecond( int s ) { second = ( s >= 0 && s < 60 ) ? s : 0; return this; } `````` So in the expression: `````` t.setHour( 18 ).setMinute( 30 ) `````` `t.setHour( 18 )` calls setHour on t, then returns a reference to t. In this case the reference is temporary. So you can think of it as if the above line changed to the following on evaluating setHour: `````` tAgain.setMinute(30); `````` t.setHour returned a reference -- similar to our tAgain above. Just an alias for t itself. - The reason that this works correctly is that when you call ``````t.setHour( 18 ) `````` The return value is a `Time&`, a reference to a `Time` object. More importantly, it's defined as ``````Time &Time::setHour( int h ) { hour = ( h >= 0 && h < 24 ) ? h : 0; return this; // <--- right here } `````` Inside of a member function, `this` is a pointer to the object on which the call was made, and `this` is a reference to the object on which the call was made (the receiver object). This means that when you call `setHour`, the function sets the hour on the time, then returns a reference to the `Time` object on which you made the call. Thus `t.setHour( 18 )` both sets the hour and then returns a reference to the receiver object. That way, you can write ``````t.setHour( 18 ).setMinute( 30 ).setSecond( 22 ); `````` because it's interpreted as ``````((t.setHour( 18 )).setMinute( 30 )).setSecond( 22 ); `````` and in each case the function returns a reference to `t`. More generally, any time a function returns a reference and that reference is `this`, any operation you perform on the return value of the function is indistinguishable from operations that you would perform on the object itself. Hope this helps! - Due to the fact that each function returns a reference to the this object object (The return this). Basically this means that every time the function is called it make the relevant changes and then passes the entire object back out as a reference. It is then possible to make calls on that returned object. It could also be written as follows: `````` Time t; Time& t1 = t.setHour( 18 ); // t1 will refer to the same object as t. Time& t2 = t1.setMinute( 30 ); // t2 will refer to the same object as t1 and t. Time& t3 = t2.setSecond( 22 ); // t3 will refer to the same object as t2, t1 and t. `````` That may make it easier to understand what is going on. - oh i got it.. `t.setHour( 18 )` will leave `(this)` which will be used to reference next function... – teacher Aug 30 '11 at 21:18 @teacher: Exactly. – Goz Aug 30 '11 at 21:20 ``````ostream& operator<<(ostream& s, const T& val) { s << val; return s; } `````` You do this because you modify the stream and return it so it can be used in the next cascading call if desired. It keeps getting passed by reference so it can keep going into the next expression segment. Thats how: ``````std::cerr << 1 << 2 << 3 << std::endl; `````` works! :) - This looks more like a comment to me – sehe Aug 30 '11 at 21:34 It might help if you think of the statements being solved one step at a time. Take the following for instance: ``````x = 1 + 2 3 - 4; x = 1 + 6 - 4; x = 7 - 4; x = 3; `````` C++ does the same with function calls and everything else you do within a statement, solving each element inside in the order of operator precedence. So you can think of your example as being solved in the same way: ``````t.setHour( 18 ).setMinute( 30 ).setSecond( 22 ); t.setMinute( 30 ).setSecond( 22 ); // hour is now set to 18 t.setSecond( 22 ); // minute is now set to 30 t; // seconds now set to 22 `````` If you returned `this` instead of `this`, and thus returning pointers instead of references, you would get the same effect except you would replace the `.` with `->` (just as an example, you're doing it right by using references). In the same way, if you returned an pointer or reference to a different object, you could do the same thing with that. For instance, let's say you have a function that returns a `Time` object. ``````class Time{ public: int getSeconds(){ return seconds; }; int seconds; }; Time getCurrentTime(){ Time time = doSomethingThatGetsTheTime(); return time; }; int seconds = getCurrentTime().getSeconds(); `````` You get the seconds without having to split the statement into two different statements or making a temporary variable to hold the returned Time object. This question C++: Using '.' operator on expressions and function calls goes a little more indepth if you wanted to read. - The technique is called method chaining. In the example you've given, all the methods return the same object (this), so they all affect the same object. That's not uncommon, but it's useful to know that it doesn't have to be the case; some or all of the methods in the chain can return different objects. For example, you might also have methods like: ``````Date Time::date() const; String Date::dayOfWeek() const; `````` in which case you could say: ``````Time t; String day = t.date().dayOfWeek(); `````` to get the name of day of the week. In that case, `t.date()` returns a Date object which is used in turn to call `dayOfWeek()`. - +1 just for first phrase, -1 for not saying `Date Time::date() const;` and using a redundant void. Therefore, +-0. edit: After editing your answer, I'll +1 it. – phresnel Sep 27 '11 at 7:15 @phresnel, That's one way to do it, but different from what I intended. Why return a new object rather than a reference to the existing one? Returning a reference to an existing object seems equally valid, as demonstrated here and here. It also seems more efficient and more natural when you consider that chaining is often used to configure a new object. – Caleb Sep 27 '11 at 7:27 Okay, I thought those were typos. Are you sure that `dayOfWeek()` is ought to return a reference to a non-const string? Why the redundant `void`s? Also, those examples return `this`, and not instances of another type, which can very subtly introduce dangerous bugs of the sort that possibly unhides just before your customer. – phresnel Sep 27 '11 at 7:48 @phresnel, Doh! I was thinking more about your edit than the point I was trying to make. Points in my comment above are covered by the code in the question, and with this answer I was trying point out that you can chain without necessarily returning `this` (or `*this`). Your edits help in that respect. I could go either way on the `f(void)`; it's mostly a habit from C and I like it because it's explicit. I don't think it's incorrect, but it's probably not as common as `f()`. – Caleb Sep 27 '11 at 8:02 A common bug is something like `stringstream ss; ... foo (ss.str().c_str());`, where you can think of `c_str()` returning a reference, too, but because `str()` returns a temporary object, calling `c_str()` compiles, but is undefined behaviour. This is the sort of bug that can also easily happen with non-const references (though I am also sure you already know this :) ). +1 revived again. – phresnel Sep 27 '11 at 8:15 because when a function is executed and returns then it returns reference to itself so again it can call `functions`. -	2,234	8,224	{"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-2014-52	latest	en	0.776448
https://www.fantasygrounds.com/forums/showthread.php?68362-Should-white-light-override-other-colors&s=56c35f774ec9d98703e6114f8932bba7&p=599573	1,624,263,136,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00287.warc.gz	694,176,686	12,799	Thread: Should white light override other colors? 1. Should white light override other colors? This is rather a question and not so much a bug report. After todays update fixed npc lighting among other things I played around with colored lights a bit. A PC with a standard torch attached overrides any colored light placed on the map. This happens in the bright radius, in the dim radius one can see that these lights mix a bit. I haven't found anyhting on the internet about the mixing of white light + colored light. I suspect white light should not override but lighten up the other color? At the moment: - White + Blue = White - White + Red = White - White + Green = White What "might" be better (idk, hence the question): - White + Blue = Light Blue - White + Red = Light Red - White + Green = Light Green Maybe we have some physics or color theory experts here. 2. In a real-world scenario, it would depend entirely on the intensity of the light sources. We absolutely get blends of spectrums; see variable LEDs, "warm" and "cool" lightbulbs, etc. One could argue that a larger bright/dim setting equates to a more intense light, but that may be something based on the ruleset. Maybe if we use the bright distance as a factor of intensity, we could have colors blending more seamlessly? 3. I think the maths of that would be quite complex, what are you going to do calculate the mix for each 5 foot square from the light sources. What if 3 or more lights interact? 4. While an interesting discussion, I would caution that we are finding a balance between game system rules for light "zones" as well as some nod to physical light characteristics. It's a very complex system that we have found can be quite fragile when "adjusting"; so most likely the lighting behaviors will stay as is for a while. Regards, JPG 5. Originally Posted by Moon Wizard While an interesting discussion, I would caution that we are finding a balance between game system rules for light "zones" as well as some nod to physical light characteristics. It's a very complex system that we have found can be quite fragile when "adjusting"; so most likely the lighting behaviors will stay as is for a while. Agreed with this. I meant to include such a caution to my own thoughts 6. Of course, that is totally understandable. As we can edit most values ourselves, there isn't really a "problem" here, just different perception by different poeple^^ My testing was also done with extremely saturated lights, which are probably not that common, at least in fanatsy games. With more natural lights, this should not stand out in the least. What I found spontaneously to be visually more pleasing for me is to set the alpha of nearly white lights to ~80-90% via the hexcode (D9FFF3E1 e.g. for torches). Maybe this helps some folks. 7. The main problem you run into with reproducing color physics on the FG screen is that there are a limited number of values (256) for each of red, blue, and green in FG (as well as all but specialized graphics applications). When you take a white light that has a value of 255,255,255 and try to add blue to it, there's literally no more blue that can be added because the blue value is already at its maximum. You would have to reduce the values of red and green to obtain the blue color. That would reduce the brightness of the light which would also look weird. There are currently 1 users browsing this thread. (0 members and 1 guests) Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •	803	3,600	{"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-2021-25	latest	en	0.947483
https://www.listendata.com/2015/10/excel-find-longest-word-in-cell.html	1,642,904,401,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00296.warc.gz	853,649,795	31,113	# Excel : Find longest word in a cell I was asked to find longest word in a cell. I was able to solve it via formula after an hour of scratching my head. It can be easily possible with UDF (User Defined Function). Excel Formula : Find Longest Word Formula : Find Longest Word in a Cell Suppose your input value (text) is entered in cell A2 =MID(A2,LEN(A2)+1-MATCH(MAX(FIND(" ",(RIGHT(A2,ROW(INDIRECT("1:"&LEN(A2))))&" "))),FIND(" ",(RIGHT(A2,ROW(INDIRECT("1:"&LEN(A2))))&" ")),0),MAX(FIND(" ",(RIGHT(A2,ROW(INDIRECT("1:"&LEN(A2))))&" ")))-1) Press F2 and Hit CTRL SHIFT ENTER to confirm this formula as it is an array formula. Simple Enter would return #N/A Error.	206	667	{"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}	2.78125	3	CC-MAIN-2022-05	longest	en	0.823659
http://gmatclub.com/forum/at-a-certain-company-the-average-number-of-years-of-45491.html?fl=similar	1,430,100,240,000,000,000	text/html	crawl-data/CC-MAIN-2015-18/segments/1429246656965.63/warc/CC-MAIN-20150417045736-00079-ip-10-235-10-82.ec2.internal.warc.gz	116,914,667	44,466	Find all School-related info fast with the new School-Specific MBA Forum It is currently 26 Apr 2015, 18:04 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # At a certain company, the average number of years of Author Message TAGS: Intern Joined: 17 Apr 2007 Posts: 24 Followers: 0 Kudos [?]: 2 [0], given: 0 At a certain company, the average number of years of [#permalink] 12 May 2007, 14:54 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions At a certain company, the average number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees to the number of the company's female employees? (1) There are 52 male employees at the company (2) The average number of years of experience for the company's male and female employees combined is 9.3 years Director Joined: 26 Feb 2006 Posts: 905 Followers: 4 Kudos [?]: 52 [0], given: 0 Re: GMATPrep: DS, ratio [#permalink] 12 May 2007, 17:32 kookoo4tofu wrote: At a certain company, the average number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees to the number of the company's female employees? (1) There are 52 male employees at the company (2) The average number of years of experience for the company's male and female employees combined is 9.3 years the question is ratio of male/female = ? # of Male / Total = p average of all emplyoees = P(m) + (1-P) (f) 9.3 = p (9.8) + (1-p) (9.1) p = 2/7 1-p = 5/7 so male/female = 2/5 Re: GMATPrep: DS, ratio [#permalink] 12 May 2007, 17:32 Similar topics Replies Last post Similar Topics: 4 At a certain company, average (arithmetic mean)number of 3 25 Aug 2012, 14:38 1 For a certain set of 3 numbers, is the average 1 27 Feb 2012, 15:50 For a certain set of 3 numbers, is the average 1 15 Jan 2011, 13:33 Average of numbers in calendar year 4 12 Oct 2009, 12:34 4 At a certain company, the average (arithmetic mean) number 4 17 Oct 2007, 18:01 Display posts from previous: Sort by	783	2,682	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2015-18	longest	en	0.925416
http://gmatclub.com/forum/40-percent-of-the-employees-at-a-certain-company-are-male-22838.html?kudos=1	1,484,865,092,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560280746.40/warc/CC-MAIN-20170116095120-00354-ip-10-171-10-70.ec2.internal.warc.gz	118,397,355	44,479	40 percent of the employees at a certain company are male. : PS Archive Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 19 Jan 2017, 14:31 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track Your Progress every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # 40 percent of the employees at a certain company are male. post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message SVP Joined: 28 May 2005 Posts: 1723 Location: Dhaka Followers: 7 Kudos [?]: 326 [0], given: 0 40 percent of the employees at a certain company are male. [#permalink] ### Show Tags 10 Nov 2005, 06:51 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 2 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. 40 percent of the employees at a certain company are male. 70 percent of the employees at the company are currently on vacation. Of the employees on vacation, 40 percent are on a beach vacation, 40 percent are on a ski vacation, and 20 percent are on a golf vacation. Of the employees on a beach vacation, 25 percent are male. What percent of the company's female employees are currently on a beach vacation? A. 28 percent B. 25 percent C. 21 percent D. 7 percent E. 35 percent _________________ hey ya...... Manager Joined: 28 Jun 2005 Posts: 217 Followers: 1 Kudos [?]: 5 [0], given: 0 ### Show Tags 10 Nov 2005, 06:57 Pick numbers;say 100 employees E is the answer 35% VP Joined: 30 Sep 2004 Posts: 1488 Location: Germany Followers: 6 Kudos [?]: 327 [0], given: 0 ### Show Tags 10 Nov 2005, 07:45 ok. during the exam, i would pick a number. 100 employees...40 male and 60 female...70 are on vacation...28 are on beach vacation...21 females are on beach vacation...thats C) ! _________________ If your mind can conceive it and your heart can believe it, have faith that you can achieve it. Manager Joined: 28 Jun 2005 Posts: 217 Followers: 1 Kudos [?]: 5 [0], given: 0 ### Show Tags 10 Nov 2005, 07:59 I am confused; I thought the question is asking females on beach vacation/females in the company 21/60 VP Joined: 30 Sep 2004 Posts: 1488 Location: Germany Followers: 6 Kudos [?]: 327 [0], given: 0 ### Show Tags 10 Nov 2005, 08:01 gotoknow3 wrote: I am confused; I thought the question is asking females on beach vacation/females in the company 21/60 yes you are right... _________________ If your mind can conceive it and your heart can believe it, have faith that you can achieve it. Director Joined: 24 Oct 2005 Posts: 575 Location: NYC Followers: 1 Kudos [?]: 59 [0], given: 0 ### Show Tags 10 Nov 2005, 08:04 gotoknow3 wrote: I am confused; I thought the question is asking females on beach vacation/females in the company 21/60 dammit.. i thought what percent of females/total employees.. argh! _________________ Success is my only option, failure is not -- Eminem Intern Joined: 01 Jun 2005 Posts: 40 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 10 Nov 2005, 08:09 Total: 100 empl Men = 40 and women = 60 70 on vacation--> 40/100 or 2/5 x 70 on beach # female on beach --> 3/4 x 2/5 x 70 = 21 % of female emp on beach = 21/60 x 100 = 35% SVP Joined: 28 May 2005 Posts: 1723 Location: Dhaka Followers: 7 Kudos [?]: 326 [0], given: 0 ### Show Tags 10 Nov 2005, 09:44 good job guys..... OA is indeed E. _________________ hey ya...... 10 Nov 2005, 09:44 Display posts from previous: Sort by # 40 percent of the employees at a certain company are male. post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	1,256	4,384	{"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}	4.03125	4	CC-MAIN-2017-04	latest	en	0.894704
https://community.infosecinstitute.com/discussion/19121/binary-bit-and-decimal-values	1,632,558,667,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00023.warc.gz	234,956,479	22,360	Binary Bit and Decimal Values Linux Essentials certified , Cisco CCENT certified PA Member Posts: 480 ■■■□□□□□□□ Team, I am having a really hard time understanding the Basic fundamentals of binary and deciam conversion. I am useing the Syngress Network Plus Book by Robert J. Shimonski. I am on Page 336-337 and I am a bit lost. Is there a book or a Video I can buu that deals with only this subject matter. I know I can do this but I just need some help. Thanks. mgmguy1 "A lot of fellows nowadays have a B.A., M.D., or Ph.D. Unfortunately, they don't have a J.O.B." Fats Domino Comments • Member Posts: 1,145 ■■■■□□□□□□ To work out Binary I use this simple method For example take 192 in the decimal form to convert it to Binary write out the Base2 number system on a piece of paper ( Base 2 number system) 128,64,32,16,8,4,2,1 Now you have your decimal value of 192 and you want to convert it to binary it's pretty simple. The decimal value is made up of the Base2 number system in this case 128 + 64 = 192 So what you do is put a 1 under those values which make up your decimal value 128 64 32 16 8 4 2 1 1 1 0 0 0 0 0 0 So 192 converted to binary = 1.1.0.0.0.0.0.0 128+64 are the binary values which make up 192 hence the 1s .................................................................................................... To convert binary to Decimal it's pretty simple For example your given the binary value 0.0.0.1.0.1.0.0 Again write out the Base2 numbers 128.64.32.16.8.4.2.1 then fill in the values you are given 128.64.32.16.8.4.2.1 0 0 0 1 0 1 0 0 Add up the 1s above So 16+ 4 = 20 Thus the Binary value of 0.0.0.1.0.1.0.0 = 20 Hope this helps this is my way of doing it Microsoft's strategy to conquer the I.T industry " Embrace, evolve, extinguish " • Member Posts: 1,145 ■■■■□□□□□□ Note there is a mistake in my post and for some reason the changes wont save ...................................................................... Now you have your decimal value of 192 and you want to convert it to binary it's pretty simple. The decimal value is made up of the Base2 number system in this case 128 + 64 = 192 So what you do is put a 1 under those values which make up your decimal value 128 64 32 16 8 4 2 1 1 1 0 0 0 0 0 0 0 0 0 So 192 converted to binary = 1.1.0.0.0.0.0.0 128+64 are the binary values which make up 192 hence the 1s Microsoft's strategy to conquer the I.T industry " Embrace, evolve, extinguish " • Member Posts: 1,145 ■■■■□□□□□□ Did this help? Microsoft's strategy to conquer the I.T industry " Embrace, evolve, extinguish " • Linux Essentials certified , Cisco CCENT certified PA Member Posts: 480 ■■■□□□□□□□ Yes it did thank you. I Think I am getting the hang of this. "A lot of fellows nowadays have a B.A., M.D., or Ph.D. Unfortunately, they don't have a J.O.B." Fats Domino • Member Posts: 3,352 ■■■■□□□□□□ Good write-up ally_uk. Don't worry mgmguy1, after practice and working with it, you will be able to count binary in your head and perhaps do the subnetting in your head. It takes time. I know I certainly had quite a bit of trouble learning subnetting, but after practice it became easy. You'll get there. Just let us know if you need help and I'm sure there are plenty of people to help you out. “For success, attitude is equally as important as ability.” - Harry F. Banks • Linux Essentials certified , Cisco CCENT certified PA Member Posts: 480 ■■■□□□□□□□ Thanks. I am trying. Math is not my strong suite but I wil keep trying to understand it. KNow of any pre-formated questions like in a Math book I can do. You know. Q & A stuff "A lot of fellows nowadays have a B.A., M.D., or Ph.D. Unfortunately, they don't have a J.O.B." Fats Domino • Member Posts: 8 ■□□□□□□□□□ Ally_uk is right about the binary stuff . The most imp point to remember is that a bit can have only 2 vlaues either ON or OFF 1 stands for on and 0 stands for Off so if you wanted to write 192 in binary it basicly means that the two bits that are On are 128+64 the rest of the bits are off . Allways use the more significant bits . Similarly 10 would be 0-0-0-0-1-0-1-0 or 1010 because the bit 8 and 2 are ON the rest are off 128-64-32-16-8-4-2-1 hope it helps MCSE, CCNA, CCNP, CCIE#13243, CISSP, CCSA Bit by Bit • Member Posts: 292 If you're having a hard time with the binary number places, a good way to think about would be to look at "normal" number places 100,000 10,000 1000 100 10 1 So, the number 154,306 would be 1x 100,000 = 100,000 5x 10,000 = 50,000 4x 1,000 = 4000 3x 100 = 300 0x 10 = 00 6x 1 = 6 What you might not realize is that you are adding them all together to get the number one-hundred fifty-four thousand three-hundred and six. Remember with normal numbers (also called base 10 or decimal) each place can only have numbers between 0-9 Now, let's look at the binary places. Instead of one's, tens, thousands, etc the places look like this: 128 64 32 16 8 4 2 1 and instead of the numbers 0-9 in each place, we can only have the numbers 0-1 (zero through one). We add them the exact same way we do with the normal numbers. So, the number 11011010 in binary would be 1x 128 = 128 1x 64 = 64 0x 32 = 0 1x 16 = 16 1x 8 = 8 0x 4 = 0 1x 2 = 2 0x 1 = 0 or 128+64+16+8+2 = 218 I hope this makes some sense. There are only 10 types of people in this world - People who understand binary and people who do not. Sign In or Register to comment.	1,653	5,435	{"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}	4	4	CC-MAIN-2021-39	latest	en	0.783023
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-5-number-theory-and-the-real-number-system-5-6-exponents-and-scientific-notation-exercise-set-5-6-page-319/1	1,537,668,816,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00185.warc.gz	751,222,339	12,953	## Thinking Mathematically (6th Edition) Published by Pearson # Chapter 5 - Number Theory and the Real Number System - 5.6 Exponents and Scientific Notation - Exercise Set 5.6 - Page 319: 1 #### Answer $2^2\cdot2^3=2^5$ Evaluating gives us: $2^5=32$ #### Work Step by Step $a^m \cdot a^n = a^{m+n}$ Thus, $2^2 \cdot 2^3 \\= 2^{2+3} \\=2^5 \\=32$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.	164	512	{"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}	3.8125	4	CC-MAIN-2018-39	longest	en	0.70336
https://qingkaikong.blogspot.com/2016/10/signal-processing-cross-correlation-in.html	1,624,558,094,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00266.warc.gz	421,934,363	18,086	## Saturday, October 22, 2016 ### Signal Processing: Cross-correlation in the frequency domain Cross-corrlation is a technique widely used in many fields. I won't go to the details of it, since wikipedia already gave a very nice introduction. In seismology, cross correlation is a great tool, for example, to find the amount of shift of one signal recorded different locations on earth, you can use cross correlation; using ambient noise cross correlation, we can find the empirical green's function between two seismic stations; to monitor the nuclear tests around the world, cross correlation also be used to do the pattern recognition on the waveforms recorded. Of course, there are still many other use cases in seismology, therefore, today we will talk about how to do cross correlation in frequency domain. You can find all the scripts at Qingkai's Github You can also do the cross correlation in time domain. But doing this in frequency domain is very simple, and clear, and of course faster as well (I tested both version in the frequency/time domain). import numpy as np import matplotlib.pyplot as plt from numpy.fft import fft, ifft, fftshift %matplotlib inline ### Let's create the first signal For simplicity, we will create the first signal using scipy. We will create a gaussian pulse. from scipy import signal t1 = np.linspace(-1, 1, 2 * 100, endpoint=False) sig1 = signal.gausspulse(t1, fc=5) plt.figure(figsize=(10,8)) plt.plot(t1, sig1, 'r') plt.show() ### Create the 2nd signal by simply shift it For simplicity, we will just shift the first signal by 50 data points, and use it as the second signal. We will use the function we created in the previous post - shift signal in frequency domain to shift the signal in the frequency domain. def nextpow2(i): ''' Find the next power 2 number for FFT ''' n = 1 while n < i: n = 2 return n def shift_signal_in_frequency_domain(datin, shift): ''' This is function to shift a signal in frequency domain. The idea is in the frequency domain, we just multiply the signal with the phase shift. ''' Nin = len(datin) # get the next power 2 number for fft N = nextpow2(Nin +np.max(np.abs(shift))) # do the fft fdatin = np.fft.fft(datin, N) # get the phase shift, shift here is D in the above explanation ik = np.array([2jnp.pik for k in xrange(0, N)]) / N fshift = np.exp(-ikshift) # multiple the signal with shift and transform it back to time domain datout = np.real(np.fft.ifft(fshift * fdatin)) # only get the data have the same length as the input signal datout = datout[0:Nin] return datout # This is the amount we will move nShift = 50 # generate the 2nd signal sig2 = shift_signal_in_frequency_domain(sig1, nShift) # plot two signals together plt.figure(figsize=(10,8)) plt.plot(sig1, 'r', label = 'signal 1') plt.plot(sig2, 'b', label = 'signal 2') plt.legend() plt.show() ## Frequency domain cross correlation Shift a signal in the frequency domain is quite simple. 1. zero-pad the input signals or apply a taper as we talked last week. (I didn't do this, since I know my two signal is zero at both ends, so I skip it) 2. take the FFT of both signals 3. multiply the first signal, and the reverse of the signal (or the conjugate, note that in the frequency domain, the complex conjugation is equivalent to time reversal in the time domain) 4. do the inverse FFT and get the shift reference: stackexchange def cross_correlation_using_fft(x, y): f1 = fft(x) # flip the signal of y f2 = fft(np.flipud(y)) cc = np.real(ifft(f1 * f2)) return fftshift(cc) # shift < 0 means that y starts 'shift' time steps before x # shift > 0 means that y starts 'shift' time steps after x def compute_shift(x, y): # we make sure the length of the two signals are the same assert len(x) == len(y) c = cross_correlation_using_fft(x, y) assert len(c) == len(x) zero_index = int(len(x) / 2) - 1 shift = zero_index - np.argmax(c) return shift calculate_shift = compute_shift(sig1, sig2) print('The shift we get from cross correlation is %d, the true shift should be 50'%calculate_shift) You can see the result will be 50, which is our correct shift. 1. Hello, Thank you for the helpful post. Replacing line f2 = fft(np.flipud(y)) in cross_correlation_using_fft() with f2 = np.conjug(fft(y)) leads to an off-by-one error for compute_shift(). How could that be explained? 2. Thank you for sharing your article. Great efforts put it to find the list of articles which is very useful to know, Definitely will share the same to other forums. kajal agarwal hot 3. Five weeks ago my boyfriend broke up with me. It all started when i went to summer camp i was trying to contact him but it was not going through. So when I came back from camp I saw him with a young lady kissing in his bed room, I was frustrated and it gave me a sleepless night. I thought he will come back to apologies but he didn't come for almost three week i was really hurt but i thank Dr.Azuka for all he did i met Dr.Azuka during my search at the internet i decided to contact him on his email dr.azukasolutionhome@gmail.com he brought my boyfriend back to me just within 48 hours i am really happy. What’s app contact : +44 7520 636249‬	1,325	5,182	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2021-25	latest	en	0.841003
https://www.jiskha.com/display.cgi?id=1302992258	1,502,964,313,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00030.warc.gz	910,063,117	3,848	# Physics posted by . A 1200 wrecking ball hangs from a 20 -long cable. The ball is pulled back until the cable makes an angle of 30.0 with the vertical.By how much has the gravitational potential energy of the ball changed? • Physics - draw the figure. From the swing point, the initial point was 20 below, at the new point, it is 20cos30 below. What is the difference? ## Similar Questions 1. ### physics A 6840 N wrecking ball hangs from the end of a steel cable. What is the net force acting on the wrecking ball if the cable snaps? 2. ### college physics A 1100 kg wrecking ball hangs from a 14m-long cable. The ball is pulled back until the cable makes an angle of 17.0 degrees with the vertical.By how much has the gravitational potential energy of the ball changed? 3. ### college physics A 1500 wrecking ball hangs from a 16 -long cable. The ball is pulled back until the cable makes an angle of 23.0 with the vertical. 4. ### physics a 1100 kg wrecking ball hangs from a16m long cable. the ball is pulled back until the cable makes an angle of 17 with the vertical. By how much has the gravitational potential energy of the ball changed? 5. ### physics A 1100 wrecking ball hangs from a 10 -long cable. The ball is pulled back until the cable makes an angle of 28.0 with the vertical.By how much has the gravitational potential energy of the ball changed? 6. ### physic A 1250 kg wrecking ball is lifted to a height of 12.7 m above its resting point. When the wrecking ball is released, it swings toward an abandoned building and makes an indentation of 43.7 cm in the wall. a. What is the potential energy … 7. ### PHYSICS A 1250 KG wrecking ball is lifted to a height of 12.7 m above its resting point . When the wrecking ball is released, its swings toward an abandoned building and makes an indentation of 43.7 cm in the wall. A) What is the potential … 8. ### science A building is being demolished with a wrecking ball. I want to cross the street where it is swinging without risking being hit.If the cable attached to wrecking ball 20 meters long do i have to cross if i start as soon as the wrecking … 9. ### Physics a wrecking ball weighing 2500 N hangs from a cable. Prior to swinging, it is pulled back to a 20 angle by a second, horizontal cable. What is the tension in the horizontal cable? 10. ### Physics 202 An inquiring student that has a mass of 65.0 kg is watching a crane with a wrecking ball demolish a building next door. The wrecking ball is swinging parallel to the property fence between her house and the neighborâ€™s. In an effort … More Similar Questions	638	2,609	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.953125	3	CC-MAIN-2017-34	latest	en	0.916265
http://sudoku.com.au/2E30-9-2019-sudoku.aspx	1,582,122,390,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875144150.61/warc/CC-MAIN-20200219122958-20200219152958-00499.warc.gz	134,350,757	16,108	Timer 00:00:00 Sudoku Sudoku ## if (sTodaysDate) document.write('Sudoku for ' + sTodaysDate.split('-')[0] + '/' + monthname() + '/' + sTodaysDate.split('-')[2]); else document.write('Enter your own sudoku puzzle.') Help Play this pic as a Jigsaw or Sliding Puzzle Previous / Next Play sudoku on your site/blog Choose a number, and place it in the grid above. 1 2 3 4 5 6 7 8 9 This number is a possibility Automatically remove Possibilities Allow incorrect Moves Clicking the playing grid places the current number Highlight Current Square Grey out Used Numbers Possibilities in Grid Format Check out the latest post in the Sudoku Forum Welcome to the Sudoku Forums! Submitted by: Gath Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes Everybody!! 30/Sep/19 12:00 AM \| \| Good Maen, all! 30/Sep/19 12:00 AM \| \| 2:08 Thought it was quicker than that.Good night all. 30/Sep/19 12:10 AM \| \| 1:55. Good Morning, Anne, Wolf , Denny and everyone? Who exactly is ESU? 30/Sep/19 12:14 AM \| \| everyone!!! 30/Sep/19 1:00 AM \| \| all. 30/Sep/19 1:07 AM \| \| Morning. 30/Sep/19 2:30 AM \| \| Happy day to all - Denny (you succeeded with your goal to be 1st again today), Wolf, Anne, Tom, Shosho, Hal, & Keith; and good day to those who will eventually make a showing later! 30/Sep/19 2:54 AM \| \| There's a Poozle waiting to be solved - near the final comments from September 28 - it's good for keeping those gray cells working! 30/Sep/19 3:00 AM \| \| What is ESU4? 30/Sep/19 4:07 AM \| \| Morning all,in a hurry to do all my games before the power goes off at 8am. Won't be back on till 3pm. 30/Sep/19 6:00 AM \| \| Good morning all. A public holiday here in the West today. Just means that I will have to finish mowing the rest of the lawn as I only did half of it yesterday. 30/Sep/19 9:42 AM \| \| Now number 13 is here for the taking! 30/Sep/19 9:43 AM \| \| Good work getting 13 Anne. There are so few people posting these days that we will probably struggle to reach even that number. What we need is something that will encourage people to join in. I suggest that people tell us about something funny that has happened to them. I'll start it off. 30/Sep/19 3:03 PM \| \| When I was a 19 year old head teacher of a one teacher school in a remote corner of Victoria I didn't have to much of a problem teaching most subject to the 5 to 12 year old students, but music was another story. We had been well drilled on how to teach a new song using a tuning fork to pitch the key, then singing the song to the children. 30/Sep/19 3:08 PM \| \| We then had to sing it to them a second time. I had scraped through music at Teachers' College by using one song, 'The Lincolnshire Poacher' whenever I had to teach music, so naturally I chose that. After I had done my second rendition I asked the children to sing it, only to be asked which version of it they should sing. 30/Sep/19 3:13 PM \| \| Fortunately the Australian Broadcasting Commission (ABC) had a weekly program in which experts taught primary children to sing a new song weekly. Thank goodness for the ABC. 30/Sep/19 3:17 PM \| \| And now it is over to you. Tell us of a funny incident and save this blog from fading away. That constitutes a CP. 30/Sep/19 3:19 PM \| \| Not funny, but Lynne and I just got back from spending a couple of days up visiting our daughter, Jen, after her back surgery. 30/Sep/19 3:56 PM \| \| We got to the hospital while the surgery was still in progress, and visited with my ex and her sister in the waiting room until Jen was back awake and we could see her. 30/Sep/19 3:58 PM \| \| It was back surgery, to remove a cracked vertebrae, and replace it with a rod and pins and some kind of cage thing that I didn't understand. Apparently, the surgery went quite well. Jen was able to get up and walk around with help from physical therapy the very next day. What a relief for all of us. Um ... gallump. 30/Sep/19 4:01 PM \| \| 30/Sep/19 4:02 PM \| \| & a CP to match Wombat. 30/Sep/19 4:02 PM \| \| I do worry that most days lately we are down to just a handful of posters. I would sure hate to see the site die for lack of interest. 30/Sep/19 4:05 PM \| \| Thanks Keith for your support and so pleased that Jen's operation has gone so well. 30/Sep/19 4:51 PM \| \| Shoudda taught 'em the Pheasant Pluckers Song, Wombat! 30/Sep/19 5:12 PM \| \| Yeah Keith, good to hear that Jen's op went well. 30/Sep/19 5:18 PM \| \| OK Wombat, here goes...... Way back in my navy days when I was on the Sealion we visited Portree in the Isle of Skye. 30/Sep/19 5:27 PM \| \| 30/Sep/19 5:30 PM \| \| At the time I was growing a beard and it had not been passed by the Captain (In those days uniform was worn when ashore. Also one had to ask to discontinue shaving and the Captain would decide when it had grown enough to be tidy). I had not done that and as such shore leave was cancelled for me. 30/Sep/19 5:31 PM \| \| There were about five or six other Aussies in the crew and as they were all going ashore I decided to join them. 30/Sep/19 5:34 PM \| \| We arrived at the Royal Hotel, Portree and headed for the bar. I diverted and went to the heads first. Unbeknownst to us the Skipper, executive officer and electrical officer (my boss) were already in the bar and on our arrival the Skipper remarked that all we needed now was that half bearded LREM (me) and the whole tribe would be here. 30/Sep/19 5:38 PM \| \| The electrical officer, bless him piped up and stated that I would not be coming ashore as I was still growing a beard.......He had no sooner said that when I walked into the bar to a stunned silence! 30/Sep/19 5:39 PM \| \| The skipper gave the electrical officer a dark look and sent him over to speak with me. He took me outside and told me what had happened and that I 'owed him ome, big time'. Nothing more was ever said about the incident and I was not charged. 30/Sep/19 5:45 PM \| \| 1:22. Good evening everyone. 30/Sep/19 7:31 PM \| \| I like it Peter. They were really strict in those days. Were you a submariner those days, or did you come to that later in your life? 30/Sep/19 8:31 PM \| \| 30/Sep/19 9:36 PM \| \| Evening here and as usual very quiet on the site. 30/Sep/19 9:37 PM \| \| I check in most days and yes Keith the numbers are wayyyy down. I see SA has a lot of posts from members who no longer post on here or very rarely. 30/Sep/19 9:40 PM \| \| BOTP 30/Sep/19 9:41 PM \| \| Not a member? Joining is quick and free. As a member you get heaps of benefits. You can also try the Chatroom (No one chatting right now - why not start something? ) Check out the Sudoku Blog Subscribe Easy Medium Hard Tough Or try the Kids Sudokus (4x4 & 6x6) 16x16 or the Parent's Page. Printer Friendly versions: Members Get Goodies! Become a member and get heaps of stuff, including: stand-alone sudoku game, online solving tools, save your times, smilies and more! Welcome our latest MembersLiz Allison from NSWSpike Jones from Pennsylvanianabapammm from ocean grove Member's Birthdays TodayBex from Australia Friends currently onlineWant to see when your friends are online? Become a member for free. ## Network Sites Melbourne Bars Find the Hidden Bars of Melbourne Free Crossword Puzzles Play online or print them out. 2 new crosswords daily. Jigsaw Puzzles Play online jigsaw puzzles for free, with new pictures everyday Sliding Puzzle Play online with your own photos Flickr Sudoku Play sudoku with pictures from Flickr Kakuro Play Kakuro online! Wordoku Free Wordoku puzzles everyday. Purely Facts Test your General Knowledge. Pumpkin Carving Pattern Free halloween pumpkin carving patterns.	2,224	7,676	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2020-10	latest	en	0.864968

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