Answer:
D. 3.35
Step-by-step explanation:
First we need to form an equation and solve it to find the number of weeks when the prices were the same. Because the prices were the same we can say that b = c, and therefore form the equation:
2.35 + 0.25x = 1.75 + 0.4x - Now we nee to solve it and find x.
2.35 - 1.75 = 0.4x - 0.25x
0.6 = 0.15x
x = 0.6 ÷ 0.15
x = 4 weeks
So now we substitute x into the equation for beef and find the price.
b = 2.35 + (0.25 × 4)
b = 2.35 + 1
b = $3.35 per pound
Find sketch of parallelogram attached
Answer and explanation:
A parallelogram is a quadrilateral which is wider than it is long. A parallelogram has two sides parallel to the other. From the above, we are told the parallelogram PQRS has an angle QPS 120, with sides 10cm and 8cm, we use this to sketch the parallelogram
We find that angle QPS is 120 degrees hence angle RSP is 60 degrees since angle on straight line is 60 degrees from 180-120 in angle QPS and alternate angles are equal. Also opposite angles of a parallelogram are equal
It is only 1 proportion, and it is 1 problem.
10 8
---- = ---
X 4
1). It is 8x=40
2). Divide both sides
by 8
3). X=5
Step-by-step explanation:
there are three parts first one is hypotenuse, base, height....
hope you understood
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)