Answer:
Cant see the picture can you zoom in?
\
Step-by-step explanation:
Answer:
Hello!
After reviewing the problem you have provided I have come up with the correct solution:
x= 9
Step-by-step explanation:
To come up with this solution you have to first realize that the smaller triangle is a proportionally scaled down version of the entire larger triangle! (I will show what I mean in a linked picture)
So after we have realized that the smaller triangle is a scaled down version of the larger one, we can then create a formula or ratio to calculate the value of the missing side of the larger triangle (being x+6=??).
To create the formula/ratio I divided 10inches by 4inches. Thus the larger triangle is 2.5 times larger than the smaller one.
I then use this ratio to figure out the missing length of the larger triangle by doing:
6inches x 2.5 = 15inches.
I then inputed the 15inches into the formula of the missing side:
x+6=15
Subtracted 6 from both sides to simplify, and came up with the solution!
x=9
Let me know if this helps!
2x + 4 = 3x - 10
-2x and +10 to both sides
14 = x
the answer is x = 14
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:

![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)