Problems 3 - Tài liệu

501 Challenging Logic And Reasoning Problems - 3

... QUESTIONS – 33 233 . 234 . 235 . 236 . a. b. c. d. a. b. c. d. a. b. c. d. a. b. c. d. – QUESTIONS – 34 237 . 238 . 239 . 240. a. b. c. d. a. b. c. d. a. b. c. d. a. b. c. d. – QUESTIONS – 35 241. a. ... related. 222. 2 23. 224. a. b. c. d. a. b. c. d. a. b. c. d. – QUESTIONS – 31 225. 226. 227. 228. a. b. c. d. a. b. c. d. a. b. c. d. a. b. c. d. – QUESTIONS – 32 229. 230 . 231 . 232 ....

Ngày tải lên: 02/11/2013, 16:20

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Problems 3

... specific message x 1 x 3 x 2 x 4 when it emanates from the following statistically independent symbol source: M = 4, p(x 1 ) = 1/2, p(x 2 ) = 1/4, p(x 3 ) = p(x 4 ) = 1/8. 3. 3 Calculate the loss ... source. b) Determine the redundancy of the source. 3. 2 a) Consider a source having an M = 3 symbol alphabet where p(x 1 ) = ½, p(x 2 ) = p(x 3 ) = ¼ and symbols are statistically indep...

Ngày tải lên: 09/11/2013, 05:15

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solutions to exercises and problems of Introduction of algorithm 3 (Bài giải sách Introduction of algorithm )

... 4.2-4 If you can multiply 3  3 matrices using k multiplications, then you can multiply n  n matrices by recursively multiplying n =3  n =3 matrices, in time T .n/ D kT .n =3/ C ‚.n 2 /. Using the ... is ‚.n log 3 k / D ‚.n log 3 21 / D O.n 2:80 / (since log 3 21  2:77). Solution to Exercise 4.4-6 The shortest path from the root to a leaf in the recursion tree is n ! .1 =3/ n !...

Ngày tải lên: 02/11/2013, 08:12

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Tài liệu Two Point Boundary Value Problems part 3 docx

... 760 Chapter 17. Two Point Boundary Value Problems Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521- 431 08-5) Copyright (C) 1988-1992 by Cambridge University ... integrating from one side and from the other, y i (x f ; V (1) )=y i (x f ;V (2) ) i =1, ,N (17.2 .3) In some problems, the N matching conditions can be better described (physically, mathematic...

Ngày tải lên: 24/12/2013, 12:16

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coulson & richardson - solutions to the problems in chemical engineering volume 2 & 3

... sphere: (R  0 /ρu 2 0 )Re  2 0 = (2d 3 /3 2 )ρ(ρ s − ρ)g (equation 3. 34) = (2 × 0.002 3 /3 × 0.05 2 )900(7870 −900)9.81 = 131 .3 log 10 131 .3 = 2.118 From Table 3. 4: log 10 Re  0 = 0. 833 or: Re  0 = 6.80 Thus: ... d 1 (mm) n 1 nd 1 nd 2 1 nd 3 1 nd 4 1 6.00 5.00 0.26 1 .3 6.5 32 .5 162.5 4.00 3. 00 0.18 0.54 1.62 4.86 14.58 2.00 1 .37 5 0. 23 0 .31 6 0. 435 0.598 0.8...

Ngày tải lên: 01/04/2014, 11:15

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