Answer:
![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Step-by-step explanation:
A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".
We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).
If we select a value 
we want this:
And we can express this like that:
for each possible i
We assume that the random variable 
are independent and 
from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:
And then cumulative distribution would be expressed like this:
For each value 
we can find the dendity function like this:
So then we have the pdf defined, and given by:
and 0 for other case
And now we can find the expected value for the random variable X like this:
Answer:
FV= $12,450
Step-by-step explanation:
Giving the following information:
David invests $10,000 in a savings account that pays 3.5% simple interest.
<u>To calculate the future value, we need to use the following formula.</u>
FV= PV*(1+i*n)
n= 7
i= 0.035
PV= 10,000
FV= 10,000*(1+0.035*7)
FV= $12,450
∠BEI ≅ ∠CEH because vertical angles are congruent; rotate ΔCEH 180° around point E, then translate point C to point B to confirm∠IBE ≅ HCE.
i got it right on the test
The correct answer should be A
