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anonymous

  • 4 years ago

Ok.\[f(x)=\left(\begin{matrix}ke ^{-2x} \\ 0\end{matrix}\right)\] \[x \ge 0\]otherwise, Find k

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  1. ash2326
    • 4 years ago
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    Is this a probability density function?

  2. anonymous
    • 4 years ago
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    Yes

  3. anonymous
    • 4 years ago
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    Is it piece wise functions?

  4. anonymous
    • 4 years ago
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    Just probability density.

  5. ash2326
    • 4 years ago
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    integrate it between 0 to infinity, it's value is 1, you'll be able to find k

  6. anonymous
    • 4 years ago
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    I'm not sure how to integrate it from infinity... that's the problem I'm having.

  7. ash2326
    • 4 years ago
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    we know for a probability density function \[\int_{-\infty}^{\infty} f(x)=1\] now f(x)=ke^-2x for x\(\ge\)0 \[\int_{0}^{\infty} f(x)=1\] so our integral becomes now f(x)=ke^-2x now \[\int_{0}^{\infty} ke^{-2x}=1\] integral of e^x=e^x so \[(\frac{-1}{2}*k*e^{-2x})=1\] now insert the limits e^(-2x) as x--->\(\infty\) is 0 so we get \[(0-(-1/2*k*e^0)=1\] or \[k/2=1\] or \[\large{k=2}\]

  8. ash2326
    • 4 years ago
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    did you get it order?

  9. anonymous
    • 4 years ago
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    Yes... Just one question, How did (0-(-1/2*k*e^0)=1 become 2?

  10. ash2326
    • 4 years ago
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    see we have, (0-(-1/2*k*e^0)=1 this is an equation now e^0=1 so (0+1/2k*1)=1 so k/2=1 or k=2

  11. anonymous
    • 4 years ago
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    Ah OK. I just figured that out :) Thanks!

  12. anonymous
    • 4 years ago
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    A computer ink cartridge has a life of X hours. The variable X is modelled by the probability density function \[f(x)=\left(\begin{matrix}kx^{-2} \\ 0\end{matrix}\right) \]\[x \ge 0 -otherwise -\] (a) Find K

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