Answer:
x-y=2
I don't get why they are asking you what is 2-5. That seems super basic compared to the other question it is with. Since 5-2=3, then 2-5 or -5+2=-3.
Step-by-step explanation:
We are going to use identity x^2-y^2=(x-y)(x+y)
x^2-y^2=10
(x-y)(x+y)=10
(x-y)(5)=10
Since (2)(5)=10, this implies x-y=2.
485, because the mean means you need to calculate the average. to find the average, you take the smallest number from the data set and the largest number. add the largest number by the smallest number and divide by the set of numbers you added (2). Example: (720+250=970 divided by 2 =485)
Answer: The height of the palm tree is 21 feet.
Step-by-step explanation:
We can use a ratio to solve this:
Actual height to shadow for both objects. The fraction equivalents must be equal.
6/8 = x/28 . Cross multiply
6(28) = 8x
168 = 8x Divide both sides by 8 (8's "cancel" on the right)
168/8 =8x/8
21 = x . This gives us the tree's height as 21 feet.
<em>Another way to solve this is to use the ratios, but simplify the first fraction</em>
<em>6/8 = 3/4</em>
<em>Then multiply the length of the shadow by 3/4</em>
<em>3/4 × 28 = height</em>
<em>28÷4 = 7 7 × 3 = 21</em>
<em>21 feet= the height of the palm tree.</em>
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)
