A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
... 32. In a ∆ABC, b = 9.5 cm, c = 8.5 cm and A = 45 o, find allowable errors in b, c, and A such that the area of ∆ABC may be determined nearest to a square centimeter.Sol. Let area of the ∆ABC be ... standards:IEEE 7 54 is for binary arithmetic, and IEEE 8 54 covers decimal arithmetic as well. The only IEEE7 54, adopted almost universally by computer manufacturers. Unfortunately, not all manufacturersimplement ... semantics are also specified in detail by the IEEE standards, are notalways handled the same way. It turns out that many manufacturers believe (sometimes rightly and sometimes wrongly) that...
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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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A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx
... negative.Thus f(0.5) is negative and f(1) is positive. Then the root lies between 0.5 and 1. 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The second approximation to the root ... (approximation) a and b (where (a > b)) such thatf (a) . f(b) < 0.Step 2: Evaluate the mid point x1 of a and b given by x1 = 12 (a + b) and also evaluatef(x1).Step 3: If f (a) . ... f(0) is negative and f(1) is positive, therefore, a root lies between 0 and 1.ALGEBRAIC AND TRANSCENDENTAL EQUATION37Here ei and ei + 1 are the errors in ith and (i + 1)th iterations respectively....
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A textbook of Computer Based Numerical and Statiscal Techniques part 10 ppt
... x = 2 [Ans. 1.756](c) tan x = x [Ans. 4. 49 34] 84 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES x2= 1 – 10120−−−(–1) = 1 – 0. 047 62 = 0.95238Hence, f(x2) = 0.138 24 Now ... f(x5) = – 0.00002ALGEBRAIC AND TRANSCENDENTAL EQUATION77Since, imaginary roots occur in conjugate pairs roots are 1.915 ± 1.908i upto 3 places of decimal. Assuming other pair of roots to be α ... to four decimals.ALGEBRAIC AND TRANSCENDENTAL EQUATION796. Using N–R method, obtain formula for N and find 20 correct to two decimal places.[Ans. 4. 47]7. Find cube root of 3 correct...
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A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx
... marks obtained by 49 2 candidates in a certain examination: 040 4 045 4550505555606065. 21 043 547 43279MarksNo of Candidates−−−−−−Find out (a) No. of candidates, if they secure more than 48 but ... the argument half waybetween the arguments at q and r is +B A, 24 where A is the arithmetic mean of q and r and B is arithmeticmean of 3q – 2p – s and 3r – 2s – p.Sol. Given A is the arithmetic ... thousands[Ans. 54. 8528 thousands]2. If lx represents the number of persons living at age x in a life table, find as accurately asthe data will permit lx for values of x = 35, 42 and 47 ....
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A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx
... following table:11 44 5 5 622 172.2903 162.888 153.7 245 145 .3375 137. 648 3RateAnnuity Value178 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here, we have h = 20, a = 99∴ u = 70 991 .45 20−=−Now ... of candidates who obtained marks between certain limitsare as follows:MarksNo of candidates−−−−−019203 940 5960798099. 41 62 65 50 17Find no. of candidates who obtained fewer than 70 marks.Sol. ... (0.0000 04) 24 120−−++176 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. The difference table for the given date is as follows:xy y y y∇∇ ∇23 34. 83.6 4 8 .4 2.56.1 0.55 14. 5 39.1...
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A textbook of Computer Based Numerical and Statiscal Techniques part 27 ppt
... result.Example 7. The following are the mean temperatures (°F) on three days, 30 days apart round the pds. of summer and winter. Estimate the app. dates and values of max. and min. temperature.INTERPOLATION ... of 20th Jan. and its value can be obtained similarly.[f(x)]min = [f(x)]1.1 84 = 63. 647 °F approximately.Example 8. The mode of a certain frequency Curve y = f (x) is very near to x = 9 and ... differences of a polynomial of nth degree are constant.Let f(x) = A 0xn + A 1xn–1 + + A n–1 x + A n by a polynomial of degree n provided A 0 ≠ 0 and arguments be equally spaced so thatx1...
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A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx
... dxa xdxa xdxa xdx 111135/2 2 4 0120000x dxa xdxa xdxa xdx=++∫∫∫∫or Simplifying above equations, we get120012012223322 345 2 345 7aa a aaaaaa++=++ =++=2 84 COMPUTER BASED NUMERICAL ... bothdydx and the curvature22dydx are the same for the pair of cubics that join at eachpoint. The spline have possess the given properties.278 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example ... NUMERICAL AND STATISTICAL TECHNIQUES Minimax polynomial approximation: Let f(x) be continuous on [a, b] and it isapproximated by the polynomial Pn (x) = a 0 + a 1x + +a nxn , then the minimax...
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A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx
... x+−+Tabulating from 4x = to 4. 4x = we obtain (4. 3) 1.0 142 56y = and (4. 4) 1.018701y =. Ans.Example 5. Solve the equation 21,yxy′=+ given that 0,y = at 0,x = by the use of Taylor ... fields of Engineering and Science, we come across physical and natural phenomena which,when represented by mathematical models, happen to be differential equations. For example,simple harmonic ... −++∫326 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES []2(10)7(0) 32 (4) 12(7) 32(9) 7(12) 32(15) 12( 14) 32(8) 7(3) 45 = +++++ + ++ = 708Hence the required area of the cross-section of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx
... = 54. 8376 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES largest coefficient of y. We continue this process till last equation. This procedure is known aspartial pivoting. In general, ... 4. 4x = and 00z = for 1y and we obtain ()1151 8.8 0 4. 2210y=−−=Similarly, we obtain ()1161 4. 4 2 4. 22 4. 81610z=−−×=Now for second approximation, we obtain 2 4. 01 54 x ... 2.0137 741 0.9 141 70112=− − =3 84 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Third approximation:()()3120 3 2.0137 741 2 0.9 141 701 3.02662288x=+ − =()3133 4 3.0266228...
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