Answer: Hello mate!
In a fair die all the number have the same probability; 1/6
The probability of getting a 4 is the sides with the number 4 divided by the total amount of numbers in the dice, this is 1 divided by, the probability is 1/6
The probability o getting a number divisible by 3 is the number of numbers that are divisible by 3 divided by the total amount of numbers in the dice:
The numbers divisible by 3 are 3 and 6, and in the dice, there are 6 numbers in total, so the probability is 2/6 = 1/3
$11,000 will bet the cost in 7 years
Given:
Original cost: $25,000
Depreciation rate: 8%
Term: 7 years
Formula for Depreciation:
A = C ( 1 - ( r ) (t) )
A = Future Value
C = Original Cost
r = rate
t = term
Solution:
Substitute the given values to the formula for depreciation.
A = $25,000( 1 - ( 0.08)(7))
A = $25,000( 1 - .56 )
A = $25,000(0.44 )
A = $11,000
Firstly, we can distribute both sides, this gives us 5x + 2x - 6 = -2x + 2. Next, we can combine our like terms. 7x - 6 = -2x + 2. The second option, B, is our answer.
a. 
, 
, and 
each have mean 0, and by linearity of expectation we have
![E[R_1]=E[X+N_1]=E[X]+E[N_1]=0](https://tex.z-dn.net/?f=E%5BR_1%5D%3DE%5BX%2BN_1%5D%3DE%5BX%5D%2BE%5BN_1%5D%3D0)
![E[R_2]=E[X+N_2]=E[X]+E[N_2]=0](https://tex.z-dn.net/?f=E%5BR_2%5D%3DE%5BX%2BN_2%5D%3DE%5BX%5D%2BE%5BN_2%5D%3D0)
b. By definition of correlation, we have
![\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}](https://tex.z-dn.net/?f=%5Cmathrm%7BCorr%7D%5BR_1%2CR_2%5D%3D%5Cdfrac%7B%5Cmathrm%7BCov%7D%5BR_1%2CR_2%5D%7D%7B%7B%5Csigma_%7BR_1%7D%7D%7B%5Csigma_%7BR_2%7D%7D%7D)
where 
denotes the covariance,
![\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BR_1%2CR_2%5D%3DE%5B%28R_1-E%5BR_1%5D%29%28R_2-E%5BR_2%5D%29%5D)
![=E[R_1R_2]-E[R_1]E[R_2]](https://tex.z-dn.net/?f=%3DE%5BR_1R_2%5D-E%5BR_1%5DE%5BR_2%5D)
![=E[R_1R_2]](https://tex.z-dn.net/?f=%3DE%5BR_1R_2%5D)
![=E[(X+N_1)(X+N_2)]](https://tex.z-dn.net/?f=%3DE%5B%28X%2BN_1%29%28X%2BN_2%29%5D)
![=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2]](https://tex.z-dn.net/?f=%3DE%5BX%5E2%5D%2BE%5BN_1X%5D%2BE%5BXN_2%5D%2BE%5BN_1N_2%5D)
Because 
are mutually independent, the expectation of their products distributes over the factors:
![\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BR_1%2CR_2%5D%3DE%5BX%5E2%5D%2BE%5BN_1%5DE%5BX%5D%2BE%5BX%5DE%5BN_2%5D%2BE%5BN_1%5DE%5BN_2%5D)
![=E[X^2]](https://tex.z-dn.net/?f=%3DE%5BX%5E2%5D)
and recall that variance is given by
![\mathrm{Var}[X]=E[(X-E[X])^2]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D)
![=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
so that in this case, the second moment ![E[X^2]](https://tex.z-dn.net/?f=E%5BX%5E2%5D)
is exactly the variance of ,
![\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BR_1%2CR_2%5D%3DE%5BX%5E2%5D%3D%7B%5Csigma_X%7D%5E2)
We also have
![{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2](https://tex.z-dn.net/?f=%7B%5Csigma_%7BR_1%7D%7D%5E2%3D%5Cmathrm%7BVar%7D%5BR_1%5D%3D%5Cmathrm%7BVar%7D%5BX%2BN_1%5D%3D%5Cmathrm%7BVar%7D%5BX%5D%2B%5Cmathrm%7BVar%7D%5BN_1%5D%3D%7B%5Csigma_X%7D%5E2%2B%7B%5Csigma_%7BN_1%7D%7D%5E2)
and similarly,
So, the correlation is
![\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}](https://tex.z-dn.net/?f=%5Cmathrm%7BCorr%7D%5BR_1%2CR_2%5D%3D%5Cdfrac%7B%7B%5Csigma_X%7D%5E2%7D%7B%5Csqrt%7B%5Cleft%28%7B%5Csigma_X%7D%5E2%2B%7B%5Csigma_%7BN_1%7D%7D%5E2%5Cright%29%5Cleft%28%7B%5Csigma_X%7D%5E2%2B%7B%5Csigma_%7BN_2%7D%7D%5E2%5Cright%29%7D%7D)
c. The variance of 
is
![{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2]](https://tex.z-dn.net/?f=%7B%5Csigma_%7BR_1%2BR_2%7D%7D%5E2%3D%5Cmathrm%7BVar%7D%5BR_1%2BR_2%5D)
![=\mathrm{Var}[2X+N_1+N_2]](https://tex.z-dn.net/?f=%3D%5Cmathrm%7BVar%7D%5B2X%2BN_1%2BN_2%5D)
![=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2]](https://tex.z-dn.net/?f=%3D4%5Cmathrm%7BVar%7D%5BX%5D%2B%5Cmathrm%7BVar%7D%5BN_1%5D%2B%5Cmathrm%7BVar%7D%5BN_2%5D)
