A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx

... between 0 .65 7 and 0 .66 5. Eighth approximation: The eighth approximation to the root is given by 8 0 .65 7 0 .66 5 0 .66 1 2 x + == From last two approximations, i.e., 7 0 .6 65 x = and 8 0 .6 61 x = ... positive. Therefore f (0 .67 3) is positive and f(0 .65 7) is negative so the root lies between 0 .65 7 and 0 .67 3. ALGEBRAIC AND TRANSCENDENTAL EQUATION 41 Seventh approxi...
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A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx

... the argument half way between the arguments at q and r is + B A, 24 where A is the arithmetic mean of q and r and B is arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol. Given A is the arithmetic ... r and s corresponds to argument a, a + h, a + 2h and a + 3h respectively then the value of the argument lying half way between a + h and a + 2h will be a...
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A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx

... 18) 26 −− −− ×− + ×− = 235 – 24 .65 – 10. 766 25 – 1.0 766 25 = 235 – 36. 492875 = 198.507 { 198 ∴ Total no. of candidates who obtained fewer than 70 marks are 198. Example 6. The area A of a circle of ... of candidates who obtained marks between certain limits are as follows: Marks No of candidates −−−−−0192039405 960 798099 . 41 62 65 50 17 Find no. of candidates who ob...
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A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx

... dxa xdxa xdxa xdx 1111 3 5/2 2 4 012 0000 x dxa xdxa xdxa xdx=++ ∫∫∫∫ or Simplifying above equations, we get 12 0 012 0 12 2 233 2 2345 2 3457 aa a a aa aaa ++= ++ = ++= 284 COMPUTER BASED NUMERICAL ... NUMERICAL AND STATISTICAL TECHNIQUES Minimax polynomial approximation: Let f(x) be continuous on [a, b] and it is approximated by the polynomial P n (x) = a 0 + a 1 x +...
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A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx

... numerical solutions of differential equations. In researches, especially after the advent of computer, the numerical solutions of the differential equations have become easy for manipulations. Hence, ... (Proved) PROBLEM SET 6. 2 1. Use Trapezoidal rule to evaluate 1 3 0 xdx ∫ consisting five sub-intervals. [Ans. 0. 26] 2. Calculate an approximate value of integral /2 0 sin...
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A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx

... sensitive to small changes in A and B i.e., small change in A or B causes a large change in the solution of the system. On the other hand if small changes in A and B give small changes in the solution, the ... NUMERICAL AND STATISTICAL TECHNIQUES largest coefficient of y. We continue this process till last equation. This procedure is known as partial pivoting. In general...
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A textbook of Computer Based Numerical and Statiscal Techniques part 41 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 41 pptx

... taking suitable scale when the value of x are given at equally spaced intervals. Let h be the width of the interval at which the values of x are given and let the origin of x and y be taken ... a x b x=+ ∑∑∑ (5) The equation () 3 and () 4 are known as normal equations. On solving equations () 3 and () 4 , we get the value of a and b. Putting the value of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 46 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 46 pptx

... the type and nature of the variations in the data. 2. The segregation and study of the various components is of paramount importance to a businessman in the planning of future operations and in ... 1 = 64 .4 1978 2 58 + 3.4 × 2 = 64 .8 Example 8. Fit a straight-line trend equation by the method of least square and estimate the trend value. Year 1 961 1 962 1 963 1 964...
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A textbook of Computer Based Numerical and Statiscal Techniques part 50 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 50 pptx

... 0.147= 4 76 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. Mean Chart Mean of 10 sample mean 442 44.2 10 10 x X === ∑ Mean Range of 10 sample ranges 58 5.8 10 10 R R === ∑ As we have, ... removed. 478 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES whether there are any assignable causes. If assignable causes found, the process should be re-adjusted to...
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A textbook of Computer Based Numerical and Statiscal Techniques part 58 pptx

A textbook of Computer Based Numerical and Statiscal Techniques part 58 pptx

... i); break; } } } return 0; } Arrays The declaration int i; declares a single variable, named i, of type int. It is also possible to declare an array of several elements. 564 COMPUTER BASED NUMERICAL ... e where v is any vairable (or anything like a[ i]), op is any of the binary arithmetic operators, and e is any expression. 558 COMPUTER BASED NUMERICAL AND STATISTICA...
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