A flat rectangular piece of aluminum has a perimeter of 60 inches. The
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Answer:
Step-by-step explanation:
Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:
And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:
Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.
So we are interested on the expected value of A+B, like this
Since the two random variables are assumed independent, then we have this
So we can find the individual expected values for each distribution and then we can add it.
For ths uniform distribution the expected value is given by where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:
The expected value for the exponential distirbution is given by :
If we use the substitution we have this:
Where X represent the random variable and the parameter. If we apply this formula to our case we got:
We can convert this into hours and we got E(B) =0.5 hours, and then we can find:
And in order to find the variance for the random variable A+B we can find the individual variances:
We have the following property:
Since we have independnet variable the Cov(A,B)=0, so then:
So is doubling, first off, meaning, if the current amount say P, then when it doubles is 2P, or double that, or P + P, so the rate of growth is 100%, since it's doubling.
and is doing it every 9 hour cycle, thus
![\bf \textit{Periodic/Cyclical Exponential Growth}\\\\ A=P(1 + r)^{\frac{t}{c}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\to &650\\ r=rate\to 100\%\to \frac{100}{100}\to &1.00\\ t=\textit{elapsed time}\to &10\\ c=period\to &9 \end{cases} \\\\\\ A=650(1 + 1)^{\frac{10}{9}}\implies A=650(2)^{\frac{10}{9}}\implies A=650\sqrt[9]{2^{10}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BPeriodic%2FCyclical%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DP%281%20%2B%20r%29%5E%7B%5Cfrac%7Bt%7D%7Bc%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%26650%5C%5C%0Ar%3Drate%5Cto%20100%5C%25%5Cto%20%5Cfrac%7B100%7D%7B100%7D%5Cto%20%261.00%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%2610%5C%5C%0Ac%3Dperiod%5Cto%20%269%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D650%281%20%2B%201%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%282%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%5Csqrt%5B9%5D%7B2%5E%7B10%7D%7D)
and is doing it every 9 hour cycle, thus