Answer:
[See Below]
Step-by-step explanation:
✦ Turn fractions into decimals:
- (We will do something different for
and
.)
✦ Divide:
÷ 
- (Since the denominators are multiplies of each other we can divide them without having to change them into decimals.)
✦ Simplified Equation:
✦ Add:
✦ Subtract:
- (Subtraction rule is if there is 2 negatives next to each other in a problem it turns into a positive.)
✦ So your answer would be:
(Exact Form)
(Mixed Number Form)
(Decimal Form)
~<em>Hope this helps Mate. If you need anything feel free to message me. </em>
To get the average you need to add all the numbers together and divide by the match number. So, 98+25+105+62+65=355. 355/5=71, therefore the average is 71
Ms. Pacheco = P
Mr. Richards = R
Mr. Edwards = E
P = R + 3
E = 2R
R + E = 81
81 = R + 2R
81 = 3R
81 ÷ 3 = R
R = 27
P = R + 3
P = 27 + 3
P = 30
E = 2R
E = 2×27
E = 54
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)