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)
Answer:
6, 8, and 10.
Step-by-step explanation:
You could work this out with the pythagorean theorem, by proving that 6^2, 36, plus 8^2, 64, equals 100. The fastest way, however, is to use pythagorean triples. These are predetermined sets of numbers that work as side lengths for right triangles. The first two are 3, 4, and 5, which form a right triangle, and 6, 8, and 10, shown here.
For question 1, the largest value would be 1 and the smallest value would be -1. A, B, and C are either all 1 or two is negative and one is positive, either way the number is 1 and it's opposite. 1^2 is 1, 1^3 is 1, 1^4 is 1. The product is 1. -1^2 is -1, -1^3 is -1, -1^4 is -1. The product is -1. Now the difference. 1 - (-1), which is 1 + 1, the answer is [D) 2]
This is an exponential function.
Without any transformations (up or down), the range is y > 0
So there is an identity we'll need to use to solve this:
cos(x+y) = cosxcosy - sinxsiny
replace the numerator with the right hand side of that identity and we get:
(cosxcosy - sinxsiny)/cosxsiny
Separate the numerator into 2 fractions and we get:
cosxcosycosxsiny- sinxsiny/cosxsiny
the cosx's cancel on the left fraction, the siny's cancel on the right fraction and we're left with:
cosy/siny - sinx/cosx
which simplifies to:
coty - tanx