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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.
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0f(x) is the function sine. Ok... How do you do that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But, it's basically just what you said.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh no! Wait. It's e^2x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then integrate it from 1000 to infinty \[\int\limits_{1000}^{\infty} e ^{2x}=e ^{2x}/2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0putting limits you get (e^2000)/2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What exactly did you think was e?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No I mean, you wrote\[ e^{2x}\over2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}e ^{ax} = e ^{ax}/a\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, so, wouldn't it be 1/2 times e^2x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats exactly what i wrote.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but when you wrote e^2x putting in limits, would the e still carry the  sign?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Substitue 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, how do you come to the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know the answer... Um, why do you do\[ e^{\infty}=0?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just don't know how to get to the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After substitution, I got a strange answer...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the limit is infinity right, so you get e^2x= e power minus infinty which is 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0e^infinity= infinity so e^infinity=1/infinity=0 (very vrey roughly). You can also observe from graph of e^xdw:1328079805509:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Um, that's true, but the answe in my book says the P = 0.454?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then you would need a scientific calculator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's not manditory here... It's statistics 2 Alevels...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's continuous random variables for the density functions.
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