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Unformatted text preview: we obtain = 2 . 5 + 2 . 33 = 2 . 73 . 3. a) No it takes negative values. b) This can be a pdf everywhere nonzero and R  f ( x ) dx = 1. c) This one is ne for the same reasons as b) d) No since R  f ( x ) dx = 1 . 1 6 = 1. 1 e) This one is ne. f) No it can take negative values (for large x ). g) No since R  f ( x ) dx = 1 2 6 = 1. 4. We need to choose C such that R  f ( x ) dx = 1. a) Z  f ( x ) dx = C Z exp(x ) dx = C 1 . So we have C 1 = 1, and, therefore, C = . b) Because of symmetry of the function we can write Z  f ( x ) dx = 2 C Z exp(x ) dx = 2 C 1 . So C = / 2. c) Z  f ( x ) dx = C Z 3 xdx = C 9 2 . So C = 2 9 . 2...
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This note was uploaded on 02/07/2011 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .
 Spring '08
 Perevalov

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