anonymous
  • anonymous
***
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
This is a closed interval, you know by the brackets meaning it doesnt include 1 ot 0 therefore answer is: 0 < (x + y) <1
anonymous
  • anonymous
ok, but i dont see how this helps me find the expected value, there are infinite possibilities for the value of x and y... right?
anonymous
  • anonymous
which is just what I said, the value of (x+y) lies in the range between 0 and 1 but does not include 0 or 1

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anonymous
  • anonymous
sorry, i dont completely understand.
ganeshie8
  • ganeshie8
Hey!
anonymous
  • anonymous
|dw:1439457387740:dw|
anonymous
  • anonymous
yes, but how is that range an expected value?
ganeshie8
  • ganeshie8
|dw:1439558175814:dw|
ganeshie8
  • ganeshie8
you can pick the ordered pair (x,y) from that shaded region, yes ?
anonymous
  • anonymous
I would also draw a line from point zero up to the top of that @ganeshie8
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
familiar with double integrals ?
anonymous
  • anonymous
err.. no sir
anonymous
  • anonymous
is that like calculus?
anonymous
  • anonymous
you only need a single integral
ganeshie8
  • ganeshie8
then you may simply do this : expected value of x = 0.5 expected value of y = 0.5 therefore expected value of x+y = 0.5+0.5 = 1
anonymous
  • anonymous
eh, closed interval @ganeshie8 ?
anonymous
  • anonymous
so its like just the average?
anonymous
  • anonymous
ohh i understant
anonymous
  • anonymous
thanks so much sir!
ganeshie8
  • ganeshie8
expected value of "sum of two random variables" is same as the sum of expected values of each of them
ganeshie8
  • ganeshie8
http://www.milefoot.com/math/stat/rv-sums.htm

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