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A textbook of Computer Based Numerical and Statiscal Techniques part 7 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 7 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 7 potx

... 1.40360.10203=Since, x3 and x4 are approximately the same upto four places of decimal, hence the requiredroot of the given equation is 1.4036.48 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Using ... 4 .79 77 – 4.8446 = – 0.0469Thus, the root lies between 1.6866 and 2.50 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES and f(3) = (3)3 – 2(3) – 5 = 16Therefore, a root lies between 2 and ... ei, xi+1 = α + ei+152 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Since f(2) and f(3) are of opposite signs, therefore the root lies between 2 and 3, so takingx0 = 2, x1 = 3,...
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A textbook of Computer Based Numerical and Statiscal Techniques part 28 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 28 potx

... 0.42 073 5 + 0.445605 + 0.00 675 375 – 0.005 67 sin (1.05) = 0.8 674 2. Ans.Example 4. A switching path between parallel railroad tracks is to be a cubic polynomial joiningpositions (0,0) and (4,2) and ... (x) and vi(x) are polynomials in x of degree≤ 2n + 1 and satisfy. (i) ui (xi) = 0,1,ijij≠= ( 3a) 260 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (ii) vi (xj) ... function as well as itsderivative at each of the points. Sometimes it is also called osculating interpolation formula.Let the set of data points (xi, yi, y’i), 0≤ i≤ n be given. A polynomial...
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A textbook of Computer Based Numerical and Statiscal Techniques part 38 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 38 potx

... the previous evaluation of y and ()1,yfxy= at a certain number of evenly spaced pivotal point (discrete points of 1x of x-axis) in the neighbourhood of 0.xIn general the Predictor-Corrector ... differential equation (),dyyfxydx′==by this method we first obtain the approximate value of 1ny+ by Predictor formula and thenimprove the value of 1ny+by means of a corrector formula. ... corrector formula, we get []53 3 4543hyy f ff=+ + + ()()0.20.6841 1.4681 4 2.05 97 3.36 67 3=+ + + 1.555 673 3 1.55 57= =.362 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒...
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A textbook of Computer Based Numerical and Statiscal Techniques part 39 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 39 potx

... error. NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATION 371 Inherent stability is determined by the mathematical formulations of the problem and isdependent on the eigen values of Jacobian Matrix ... complex ()aib=+, then it is stable if()[1 ] 1aibh+= <i.e., ()()2211ah bh++ < 374 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 11 12 13 1ax ay az b++=11122 23 2ay az b+= ... Matrix of the differential equation. Numerical stability is a function of the error propagation in the numerical method. Threetypes of errors occur in the application of numerical integration...
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A textbook of Computer Based Numerical and Statiscal Techniques part 42 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 42 potx

... the values of x∑, etc. and calculated by means of above table in the normalequations. We get,5 .75 36 = 5A + 34B and 40.8862 = 3 4A + 300BOn solving these equations we obtain, A = 0. 976 6; ... the values in the normal equations, we obtain3 .73 93 8 36AB=+ and 22 .73 85 36 204AB=+⇒ 0.1406B = and A = 1.8336⇒ log 1.38banti B== and log 0.68.aanti A= =Thus the required curve of best ... normal equations are given by20Synab x a ∂=⇒ = +∂∑∑ (1)2240Sxy a x b xb∂=⇒ = +∂∑∑∑ (2)404 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here 6n =Normal equations are, 17. 25 573 ...
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A textbook of Computer Based Numerical and Statiscal Techniques part 54 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 54 potx

... 0(Samples are large)522 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. 1,000 apples are taken from a large consignment and 100 are found to be bad. Estimatethe percentage of bad apples ... Rejected at both level 1% and 5%]16. A normal population has a mean of 0.1 and standard deviation of 2.1. Find the probabilitythat mean of a sample of size 900 will be negative? [Ans. 0. 076 4] 17. ... population with mean height 162.5 cms and standard deviation 4.5 cms? [Ans. H0: Accepted]13. A random sample of 200 measurements from a large population gave a mean value of 50 and a S.D. of 9. Determine...
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A textbook of Computer Based Numerical and Statiscal Techniques part 55 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 55 potx

... E(c)=accdN++bgbg and E(d)=bdcdN++bgbgababcdcdac bd N++++∴ χ2=aEaEabEbEbcEcEcdEdEd––––afafafafafafafaf2222+++ (2) a – E (a) =a – abacN++bgbg= a a b c d a ac ... 11. 07. Calculated value of χ2 is less than the tabulated value, it is not significant at 5% level of significance and hence the null hypothesis of equal probability for male and female births maybe accepted.Example ... expected frequencies of the accidents on each of the dayswould be:528 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Days Sun. Mon. Tues. Wed. Thus. Fri. Sat. TotalNo. of accidents 12 12 12...
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A textbook of Computer Based Numerical and Statiscal Techniques part 56 potx

A textbook of Computer Based Numerical and Statiscal Techniques part 56 potx

... store.After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153 .7 and showed a standard deviation of 17. 2. Was the advertising campaign successful?Sol. We are ... may take the product conforming to specifications.Example 17. A random sample of size 16 has 53 as mean. The sum of squares of the derivation frommean is 135. Can this sample be regarded as ... 544 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES =836 74 0 872 22af af++ – = 1659. 076 . ∴ s = 40 .73 17 The t-statistic t =XXsnn121211−+=1234 103640 73 17 181 7 –. +...
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A textbook of Computer Based Numerical and Statiscal Techniques part 62 docx

A textbook of Computer Based Numerical and Statiscal Techniques part 62 docx

... the RangeLower limit a = 0Upper limit b = 6Enter the number of subintervals = 6Value of the integral is: 1.3 571 598 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Enter the value of y2 ... ALGORITHM FOR TRAPEZOIDAL RULEStep 1. Start of the program for numerical integrationStep 2. Input the upper and lower limits a and bStep 3. Obtain the number of subinterval by h = (b a) /nStep 4. ... form of x–Enter the value of x1 – 5Enter the value of x2 – 7 Enter the value of x3 – 11Enter the value of x4 – 13Enter the value of x5 – 17 Enter the value in the form of y–Enter the value of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt

A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt

... This pageintentionally leftblankThis pageintentionally leftblank A TEXTBOOK OF COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Anju Khandelwal M.Sc., Ph.D.Department of MathematicsSRMS ... Based Numerical and Statistical Techniques is primarily writtenaccording to the unified syllabus of Mathematics for B. Tech. II year and M.C .A. I year students of all Engineering colleges affiliated ... affiliated to U.P. Technical University, Lucknow.The subject matter is presented in a very systematic and logical manner. In each chapter, allconcepts, definitions and large number of examples...
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