Since both of the equations equal Y can substitute one for the Y and say that
X-1=3x+3
-4=2x
-2=x
6/x= -2/18
Cross multiply. ( Multiply the numerators together ) . ( Multiply the denominators together (.
6*18= - 2*x
108= -2x
Divide by -2 for 108 and -2x
108/-2= -2x/2
x= -54
Answer: x= -54
Karmin will make $18.4 for working an hour after the increase.
Step-by-step explanation:
Step 1:
First, we need to determine how much Karmin made an hour.
Karmin's hourly rate 
So Karmen made $16 every hour that she worked. She worked for 30 hours and earned $480.
She will get a 15% increase in her hourly rate this month.
Step 2:
We need to determine how much 15% of $16 is.
15% of $16 
So her 15% increase amounts to $2.4.
So the hourly rate for this month 
So Karmin will make $18.4 for working 1 hour.
Answer:
a) ![\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)
b) ![\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Step-by-step explanation:
Given:
The lifetimes of the individual items are independent exponential random variables.
Mean = 200 hours.
Assume, Ti be the time between (
)st and the
failures. Then, the
are independent with
being exponential with rate
Therefore,
a) ![E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]](https://tex.z-dn.net/?f=E%5BT%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20E%5Cleft%5B%5Ctau_%7Bi%7D%5Cright%5D)

![\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)

The variance is given by, ![\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20%5Cmathrm%7BVar%7D%5BT%5D)
![\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)