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anonymous

  • 5 years ago

Double integration... with x+y dy dx where y goes from x to infinity and x goes from 0 to 1/2... anybody up for this???

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  1. anonymous
    • 5 years ago
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    \[P(X \le1/2)=(I think)\int\limits_{0}^{1/2} \int\limits_{x}^{\infty} (x+y)dy dx\]

  2. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{1/2}\int\limits_{0}^{x}(x+y)dydx\] is your integral. Since all they gave you was the probability x is at least 1/2 or smaller, the probability has to be between 0 and 1, so your dy limit will start at zero and go as far as x. That is just like takin the limit as x goes to infinity on the upper limit. But it does have to be at least zero by order of the definition of a probability. So you just solve that and you've got it. I got 1/64 +1/24, which is some small decimal and consistant with the definition. I'm sure you can do the integration just fine.

  3. anonymous
    • 5 years ago
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    Thank You!!! the infinities were making me dizzy....

  4. anonymous
    • 5 years ago
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    No problem, I had to spend some time looking it up, and I almost confused myself.

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