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:
The graph should be stretched rather than become narrower.
Step-by-step explanation:
To figure this out, just create some example points.
At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.
If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph
Answer:
2x+59.
Step-by-step explanation:
Let <em>J </em>represent Jessica's weight and <em>R</em> represent Ronda's weight.
Jessica weighs <em>x+34</em> pounds. Thus:
Ronda weighs 12 pounds less than Jessica. In other words:
The sum of their weights, therefore, is:
Now, if Jessica gains 5 pounds and Ronda loses 2 pounds, the net gain of the total weight would be 3 pounds. Thus, we only need to add 3 to the original total to find the sum of their new weights:
The sum of the new [weights] is represented by 2x+59.
Bror cancelled terms which should NEVER be done.
x² + 3x + 2 / (x + 2) =
(x+1) * (x+2) / (x+2)
Cancelling the (x+2) factor we get
x + 1
