The Bernoulli distribution is a distribution whose random variable can only take 0 or 1
- The value of E(x2) is p
- The value of V(x) is p(1 - p)
- The value of E(x79) is p
<h3>How to compute E(x2)</h3>
The distribution is given as:
p(0) = 1 - p
p(1) = p
The expected value of x2, E(x2) is calculated as:
![E(x^2) = \sum x^2 * P(x)](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Csum%20x%5E2%20%2A%20P%28x%29)
So, we have:
![E(x^2) = 0^2 * (1- p) + 1^2 * p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%5E2%20%2A%20%281-%20p%29%20%2B%201%5E2%20%2A%20p)
Evaluate the exponents
![E(x^2) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
Multiply
![E(x^2) = 0 +p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%20%2Bp)
Add
![E(x^2) = p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20p)
Hence, the value of E(x2) is p
<h3>How to compute V(x)</h3>
This is calculated as:
![V(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
Start by calculating E(x) using:
![E(x) = \sum x * P(x)](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20x%20%2A%20P%28x%29)
So, we have:
![E(x) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
![E(x) = p](https://tex.z-dn.net/?f=E%28x%29%20%3D%20p)
Recall that:
![V(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
So, we have:
![V(x) = p - p^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20p%20-%20p%5E2)
Factor out p
![V(x) = p(1 - p)](https://tex.z-dn.net/?f=V%28x%29%20%3D%20p%281%20-%20p%29)
Hence, the value of V(x) is p(1 - p)
<h3>How to compute E(x79)</h3>
The expected value of x79, E(x79) is calculated as:
![E(x^{79}) = \sum x^{79} * P(x)](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%20%5Csum%20x%5E%7B79%7D%20%2A%20P%28x%29)
So, we have:
![E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%5E%7B79%7D%20%2A%20%281-%20p%29%20%2B%201%5E%7B79%7D%20%2A%20p)
Evaluate the exponents
![E(x^{79}) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
Multiply
![E(x^{79}) = 0 + p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%20%2B%20p)
Add
![E(x^{79}) = p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%20p)
Hence, the value of E(x79) is p
Read more about probability distribution at:
brainly.com/question/15246027
Lets calculate the taxes that would been payed at each city:
city1 = 12000(7%) = 12000(0.07) = 840
city2 = <span>12000(8.5%) = 12000(0.085) = 1020
So if you go to city1 instead of city2, you save 1020 - 840 = 180
So you would save $180 from $12000, that is:
180 out of 12000
= 180/12000
= 0.015
and that is the 1.5%, that is what you could save</span>
Given:
12cm to 2.5mm
To find:
The simplified for of given statement.
Solution:
We know that,
1 cm = 10 mm
12 cm =120mm
We have,
12cm to 2.5mm
It can be written as
![\dfrac{12\ cm}{2.5\ mm}=\dfrac{120\ mm}{2.5 mm}](https://tex.z-dn.net/?f=%5Cdfrac%7B12%5C%20cm%7D%7B2.5%5C%20mm%7D%3D%5Cdfrac%7B120%5C%20mm%7D%7B2.5%20mm%7D)
![\dfrac{12\ cm}{2.5\ mm}=\dfrac{120}{2.5}](https://tex.z-dn.net/?f=%5Cdfrac%7B12%5C%20cm%7D%7B2.5%5C%20mm%7D%3D%5Cdfrac%7B120%7D%7B2.5%7D)
![\dfrac{12\ cm}{2.5\ mm}=48](https://tex.z-dn.net/?f=%5Cdfrac%7B12%5C%20cm%7D%7B2.5%5C%20mm%7D%3D48)
Therefore, the value of 12 cm to 2.5 mm is 48 or 48:1.
Answer:
D. All of the above
Step-by-step explanation:
A is right, 3(g) = 3g
I don't know about b, but c is right, so it is all of the above