Answer:
1) 
2) 
3) 
And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
4) 
And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be:
Step-by-step explanation:
For this case we have the following distributions given:
Probability M J
0.3 14% 22%
0.4 10% 4%
0.3 19% 12%
Part 1
The expected value is given by this formula:

And replacing we got:

Part 2

Part 3
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
Part 4
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be:
Answer:
Step-by-step explanation:
Divide the figure into a rectangle and a triangle.
area of rectangle = 5×10 = 50 square units
area of triangle = 0.5×5×10 = 25 square units
total area = 50+25 = 75 square units
Answer:
IF THE question is After Fathi prints, what will be the balance in his printing account?
then it is: $1.9
There are 47 pages.
Printing on both sides would divide the number of pages into half.
47/2 = 23.5
2 pages on each side would mean 4 pages on one sheet. Therefore, the number of pages will be further divided by 2.
23.5/2 = 11.75
There cannot be 11.75 pages so we will round it up to 12 pages.
Each page costs $0.25 so 12 pages will cost:
12 x 0.25 = $3
Faithi has $1.1 so new account balance will be:
1.1 - 3 = $-1.9
Therefore, Fathi's balance in his printing account would be negative $1.9.
Your answer is: 2x^6-x^4+5x^3+2x^2+1\x^2
Convert to a decimal which would be 0.55 (11/20).
Then you want to multiply 0.55 by 100 (which is a whole number, all % is going to come from 100) to convert to a full % -> 0.55*100
Then you will simplify 0.55*100
You answer will be 55%.