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Can anyone verify this? | Can anyone verify this? | ||
--[[User:Leedj|Leedj]] 21:46, 29 March 2009 (UTC) | --[[User:Leedj|Leedj]] 21:46, 29 March 2009 (UTC) | ||
+ | |||
+ | I thought that the part b is | ||
+ | |||
+ | <math> f_{x}(x)=\int_{0}^x 4 dy = 4x, 0\leq x \leq 0.5 </math> | ||
+ | <br /> | ||
+ | <math> f_{x}(x)=\int_{x}^1 4 dy = 4 - 4x, 0.5\leq x \leq 1.0 </math> | ||
+ | <br /> | ||
+ | <math> f_{y}(y)=\int_{0}^{0.5} 4 dx = 2, 0\leq y \leq x </math> | ||
+ | <br /> | ||
+ | <math> f_{y}(y)=\int_{0.5}^{1} 4 dx = 2, x\leq y \leq 1 </math> | ||
+ | <br /> | ||
+ | |||
+ | and part c, they are not independent because | ||
+ | |||
+ | <math>f_{xy}(xy)</math> is not equal to <math>f_{y}(y)*f_{x}(x).</math> | ||
+ | |||
+ | part d is confused for me. Could anyone tell me how to do for part d?--[[User:Kim415|Kim415]] 03:54, 30 March 2009 (UTC) |
Revision as of 22:54, 29 March 2009
For part b, I am getting
$ f_{x}(x)=\int_{0}^x 4 dy = 4x, 0\leq x \leq 0.5 $
$ f_{x}(x)=\int_{x}^1 4 dy = 4 - 4x, 0.5\leq x \leq 1.0 $
$ f_{y}(y)=\int_{y}^{0.5} 4 dx = 2 - 4y, 0\leq y \leq 0.5 $
$ f_{y}(y)=\int_{0.5}^y 4 dx = 4y - 2, 0.5\leq y \leq 1 $
For part c, they are not independent.
Can anyone verify this?
--Leedj 21:46, 29 March 2009 (UTC)
I thought that the part b is
$ f_{x}(x)=\int_{0}^x 4 dy = 4x, 0\leq x \leq 0.5 $
$ f_{x}(x)=\int_{x}^1 4 dy = 4 - 4x, 0.5\leq x \leq 1.0 $
$ f_{y}(y)=\int_{0}^{0.5} 4 dx = 2, 0\leq y \leq x $
$ f_{y}(y)=\int_{0.5}^{1} 4 dx = 2, x\leq y \leq 1 $
and part c, they are not independent because
$ f_{xy}(xy) $ is not equal to $ f_{y}(y)*f_{x}(x). $
part d is confused for me. Could anyone tell me how to do for part d?--Kim415 03:54, 30 March 2009 (UTC)