Let 
be the total amount of money paid by any given set of passengers. If there are 
passengers in a car, then the driver must pay a toll of 
.
Then has first moment (equal to the mean)
![E[Y]=E[0.5X+3]=0.5E[X]+3E[1]=0.5\mu_X+3=\boxed{4.35}](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5B0.5X%2B3%5D%3D0.5E%5BX%5D%2B3E%5B1%5D%3D0.5%5Cmu_X%2B3%3D%5Cboxed%7B4.35%7D)
and second moment
![E[Y^2]=E[0.25X^2+3X+9]=0.25E[X^2]+3E[X]+9E[1]=0.25E[X^2]+3\mu_X+9](https://tex.z-dn.net/?f=E%5BY%5E2%5D%3DE%5B0.25X%5E2%2B3X%2B9%5D%3D0.25E%5BX%5E2%5D%2B3E%5BX%5D%2B9E%5B1%5D%3D0.25E%5BX%5E2%5D%2B3%5Cmu_X%2B9)
Recall that the variance is the difference between the first two moments:
![\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]={\sigma^2}_X+{\mu_X}^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%5Cimplies%20E%5BX%5E2%5D%3D%7B%5Csigma%5E2%7D_X%2B%7B%5Cmu_X%7D%5E2)
![\implies E[Y^2]=0.25({\sigma^2}_X+{\mu_X}^2)+3\mu_X+9\approx19.22](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5E2%5D%3D0.25%28%7B%5Csigma%5E2%7D_X%2B%7B%5Cmu_X%7D%5E2%29%2B3%5Cmu_X%2B9%5Capprox19.22)
![\implies\mathrm{Var}[Y]=E[Y^2]-E[Y]^2=\boxed{0.3}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathrm%7BVar%7D%5BY%5D%3DE%5BY%5E2%5D-E%5BY%5D%5E2%3D%5Cboxed%7B0.3%7D)
Answer:
-7/40
Step-by-step explanation:
5/8 - 4/5
~Find common denominators
25/40 - 32/40
~Subtract
-7/40
Best of Luck!
Answer:
i think id say it is 10.5 i might be wrong though
Step-by-step explanation:
