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求統計學高手打救
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Question No. 1 ......... Given a set of simultaneous equation : y = x² + 3x - 4 -----------------(1) y = 2x + 4 -----------------------(2) Question No. 2 .................. Given a G.P. series 2,4,8,16,32,........ for n nos. : Calculation the value of T(100) - T(50) 更新: 有Ｄ不明白nelsonywm2000 ( 博士級 2 級 )的 Ans. 可否detail D 更新 2: 點解 x = [-1 +/- √(1 + 32)] / 2 x = (-1 +/- √33) / 2 when x = (-1 + √33)/2, y = 3 + √33 when x = (-1 - √33)/2, y = 3 - √33 Question 2 點解= 2 * 2^99 - 2 * 2^49 = 2^100 - 2^50 if u = 2^x log u = x log 2..........
最佳解答:
(1) y = x^2 + 3x - 4 ... (1) y = 2x + 4 ... (2) Sub (2) into (1), 2x + 4 = x^2 + 3x - 4 x^2 + x - 8 = 0 x = [-1 +/- √(1 + 32)] / 2 x = (-1 +/- √33) / 2 when x = (-1 + √33)/2, y = 3 + √33 when x = (-1 - √33)/2, y = 3 - √33 (2) GP: 2, 4, 8, 16, 32... a = 2 r = 4/3 = 2 T(100) - T(50) = 2 * 2^99 - 2 * 2^49 = 2^100 - 2^50 if u = 2^x log u = x log 2 u = 10^(x log 2) 2^100 = 10^(100 log 2) = 1.26765 x 10^30 2^50 = 10^(50 log 2) = 1.1259 x 01^15 2^100 - 2^50 = 1.268 x 10^30 (4 sig fig) 2009-10-31 19:52:35 補充： which part you do not understand? I have been more detailed that I usually do. So you have to tell me how I can help your further. 2009-11-01 00:45:43 補充： (1) ax^2 + bx + c = 0 Quadratic equation formula, x = [-b +/- sqrt(b^2 - 4ac)] / 2a a = 1, b = 1, c = -8 x = [-1 +/- sqrt(1 + 32)]/2 = [-1 +/- sqrt(33)]/2 2009-11-01 00:49:39 補充： (2) Geometric sequence a, ar , ar^2 ... , ar^(n-1) a = 2, r = 4/2 = 2 Geometric sequence becomes 2, 2 * 2 , 2 * 2^2, ... 2 * 2^(n-1) T(100) = 2 * 2^(100- 1) = 2 * 2^99 = 2^100 T(50) = 2 * 2^(50 -1) = 2 * 2^49 = 2^50 Let u = 2^100 log u = 100 log(2) = 30.103 u = anti-log(30.103) = 1.268 x 10^30 ...
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