Answer:
Step-by-step explanation:
Plug into distance formula:
Points: (5, -2), (3, -4)
sqrt( ((x2 - x1)^2) + ((y2 - y1)^2))
sqrt( ((3 - 5)^2) + ((-4 + 2)^2))
sqrt( 4 + 4)
sqrt8 ≈ 2.8 OR
sqrt8 = 2√2
Answer:
Blue cars, B = 63 cars
Step-by-step explanation:
Let the blue cars be B.
Let the red cars be R.
Given the following data;
Ratio of B:R = 9:7 = 9 + 7 = 16
Red cars, R = 49
To find the number of blue cars;
First of all, we would determine the total number of cars using the expression;
R = 7/16 * x = 49
7x = 49 * 16
7x = 784
x = 112 cars
Now, we can find the number of blue cars;
B = 9/16 * 112
B = 1008/16
Blue cars, B = 63 cars
Answer:
12
Step-by-step explanation:
I think you’re right because I put the same thing on my quiz
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:
![max(U_1, ....,U_n) \leq x](https://tex.z-dn.net/?f=max%28U_1%2C%20....%2CU_n%29%20%5Cleq%20x)
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:
![P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20P%28U_1%20%5Cleq%201%2C%20....%2C%20U_n%20%5Cleq%20x%29%20%5Cprod%20P%28U_i%20%5Cleq%20x%29%20%3D%5Cprod%20x%20%3D%20x%5En%20)
And then cumulative distribution would be expressed like this:
![0, x \leq 0](https://tex.z-dn.net/?f=0%2C%20x%20%5Cleq%200)
![x^n, x \in (0,1)](https://tex.z-dn.net/?f=x%5En%2C%20x%20%5Cin%20%280%2C1%29)
![1, x \geq 1](https://tex.z-dn.net/?f=1%2C%20x%20%5Cgeq%201)
For each value
we can find the dendity function like this:
![f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}](https://tex.z-dn.net/?f=f_X%20%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20F_X%20%28x%29%20%3D%20nx%5E%7Bn-1%7D)
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) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}](https://tex.z-dn.net/?f=E%28X%29%20%3D%5Cint_%7B0%7D%5E1%20s%20f_X%20%28x%29%20dx%20%3D%20%5Cint_%7B0%7D%5E1%20x%20n%20x%5E%7Bn-1%7D)
![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)