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anonymous

  • 4 years ago

The life, X, of the StayBright light bulb is modelled by the probability density function\[f(x)=\left(\begin{matrix}2x^{-2} \\ 0\end{matrix}\right)\]\[x \ge 0\]otherwise. Where X is measures in thousands of hours Find the probability that a StayBrite bulb lasts longer than 1000 hours.

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  1. anonymous
    • 4 years ago
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    f(x) is the function sine. Ok... How do you do that?

  2. anonymous
    • 4 years ago
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    sorry, it's 2e^-2

  3. anonymous
    • 4 years ago
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    so \[f(x)=2e ^{-2}\]

  4. anonymous
    • 4 years ago
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    and 0, yes

  5. anonymous
    • 4 years ago
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    But, it's basically just what you said.

  6. anonymous
    • 4 years ago
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    Oh no! Wait. It's e^-2x

  7. anonymous
    • 4 years ago
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    f(x)=2e^-2x

  8. anonymous
    • 4 years ago
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    Then integrate it from 1000 to infinty \[\int\limits_{1000}^{\infty} e ^{-2x}=-e ^{-2x}/2\]

  9. anonymous
    • 4 years ago
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    putting limits you get (e^-2000)/2.

  10. anonymous
    • 4 years ago
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    I thought it was -e?

  11. anonymous
    • 4 years ago
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    What exactly did you think was -e?

  12. anonymous
    • 4 years ago
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    No I mean, you wrote\[ -e^{-2x}\over2\]

  13. anonymous
    • 4 years ago
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    \[\int\limits_{}^{}e ^{ax} = e ^{ax}/a\]

  14. anonymous
    • 4 years ago
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    Yes, so, wouldn't it be -1/2 times e^-2x?

  15. anonymous
    • 4 years ago
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    thats exactly what i wrote.

  16. anonymous
    • 4 years ago
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    Yes, but when you wrote e^-2x putting in limits, would the e still carry the - sign?

  17. anonymous
    • 4 years ago
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    Substitue the limits with the minus sign you get the answer that i wrote only if a minus sign is there then we can reverse the limits and remove the minus sign(Property of definite integral.

  18. anonymous
    • 4 years ago
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    Ah Ok.

  19. anonymous
    • 4 years ago
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    So, how do you come to the answer?

  20. anonymous
    • 4 years ago
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    \[e ^{-\infty}=0\]

  21. anonymous
    • 4 years ago
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    I know the answer... Um, why do you do\[ e^{-\infty}=0?\]

  22. anonymous
    • 4 years ago
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    I just don't know how to get to the answer.

  23. anonymous
    • 4 years ago
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    After substitution, I got a strange answer...

  24. anonymous
    • 4 years ago
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    the limit is infinity right, so you get e^-2x= e power minus infinty which is 0.

  25. anonymous
    • 4 years ago
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    e^infinity= infinity so e^-infinity=1/infinity=0 (very vrey roughly). You can also observe from graph of e^-x|dw:1328079805509:dw|

  26. anonymous
    • 4 years ago
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    Um, that's true, but the answe in my book says the P = 0.454?

  27. anonymous
    • 4 years ago
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    Then you would need a scientific calculator

  28. anonymous
    • 4 years ago
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    It's not manditory here... It's statistics 2 A-levels...

  29. anonymous
    • 4 years ago
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    It's continuous random variables for the density functions.

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