anonymous
  • anonymous
Integration and probability
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I need help with question #3
amistre64
  • amistre64
|dw:1333402228973:dw| maybe?

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amistre64
  • amistre64
doh! -x+3 ... got me slope backwards
anonymous
  • anonymous
i need to use double integrals
anonymous
  • anonymous
or i guess we can do it visually
amistre64
  • amistre64
|dw:1333402352858:dw|
amistre64
  • amistre64
still amounts to the same thing if im reading it right
anonymous
  • anonymous
well like i used double integrals and got the answer 3 and the book got the answer 1
amistre64
  • amistre64
give me an overview of what it is we are looking for; see if my memory jogs
anonymous
  • anonymous
idk
anonymous
  • anonymous
it has to be the fraction of probabilty
anonymous
  • anonymous
so like obv it cant be 3
anonymous
  • anonymous
it can only be up to one but i dont how i am suppossed to figure that out
amistre64
  • amistre64
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amistre64
  • amistre64
\[\int_{0}^{2}\int_{0}^{1}\int_{0}^{xy}dz.dy.dx\] im assuming this would be the volume under the "cap"
anonymous
  • anonymous
ohh u use triple integrals???
anonymous
  • anonymous
oh that is weird i only used double
amistre64
  • amistre64
well, the inside simple turns into xy but it helps me to see where the parts are going to
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1333403335322:dw|
anonymous
  • anonymous
this was my setup. obv i did smth wrong
amistre64
  • amistre64
that 3-x is off; should be from 0 to 1 since all the points in the rectangle are included by the constraint
anonymous
  • anonymous
ohhhh i see
anonymous
  • anonymous
okk so i guess the the answer wld be 1
amistre64
  • amistre64
IF that line had cut through our rect; we would have had to divvy it up to determine what points we could assign to it
amistre64
  • amistre64
yep :) xy dy -> xy^2/2 , [0,1] = x/2 x/2 dx -> x^2/4, [0,2] = 4/4 = 1
anonymous
  • anonymous
yay
anonymous
  • anonymous
Thanks amistre
amistre64
  • amistre64
yep
anonymous
  • anonymous
that was cclear
anonymous
  • anonymous
thanks

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