To determine how many oil changes you will need the first year, you can calculate the number of miles driven the first year and then use that to determine how many groups of 3000 miles you will drive after the first 5000 miles.
1. 52 x 250 = 13000 miles
2. 13000 miles - 5000 miles (1 oil change) = 8000 miles
3. 8000 miles - 3000 miles (1 oil change) - 3000 miles (1 oil change) = 2000 miles
The first year, you will need 3 oil changes.
Answer:
7x+39>=53 and 16x+ 15>31 x>=2 (no solutions exist)
Step-by-step explanation:
Answer:
44
because each that's ¼ has 11 so 11x4 =44
Answer:
a) 
b) 
And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
c) 
d) 
e) 
f) 
And replacing we got:

And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:

Step-by-step explanation:
We have the following distribution
x 0 1 2 3 4
P(x) 0.2 0.3 0.1 0.1 0.3
Part a
For this case:

Part b
We want this probability:

And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
Part c
For this case we want this probability:

Part d

Part e
We can find the mean with this formula:

And replacing we got:

Part f
We can find the second moment with this formula

And replacing we got:

And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:
