Given that a<span>
gas station operates two pumps, each of which can pump up to 10,000
gallons of gas in a month and that the total of gas pumped at the station in a
month is a random variable y (measured in 10,000 gallons) with a
probability density function (p.d.f.) given by

Part A:
The value of c that makes f(y) a pdf is obtained as follows:
![F(\infty)= \int\limits^{\infty}_{-\infty} {f(y)} \, dy=1 \\ \\ \Rightarrow \int\limits^1_0 {cy} \, dy +\int\limits^2_1 {(2-y)} \, dy=1 \\ \\ \Rightarrow \left. \frac{cy^2}{2} \right]^1_0+\left[2y- \frac{y^2}{2} \right]^2_1=1 \\ \\ \Rightarrow \frac{c}{2} +4-2-2+ \frac{1}{2} =1 \\ \\ \Rightarrow \frac{c}{2} = \frac{1}{2} \\ \\ \Rightarrow \bold{c=1}](https://tex.z-dn.net/?f=F%28%5Cinfty%29%3D%20%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7D%20%7Bf%28y%29%7D%20%5C%2C%20dy%3D1%20%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cint%5Climits%5E1_0%20%7Bcy%7D%20%5C%2C%20dy%20%2B%5Cint%5Climits%5E2_1%20%7B%282-y%29%7D%20%5C%2C%20dy%3D1%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cleft.%20%5Cfrac%7Bcy%5E2%7D%7B2%7D%20%5Cright%5D%5E1_0%2B%5Cleft%5B2y-%20%5Cfrac%7By%5E2%7D%7B2%7D%20%5Cright%5D%5E2_1%3D1%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%20%5Cfrac%7Bc%7D%7B2%7D%20%2B4-2-2%2B%20%5Cfrac%7B1%7D%7B2%7D%20%3D1%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cfrac%7Bc%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cbold%7Bc%3D1%7D%20)
Part B:
We compute E(y) as follows:
![E(y)=\int\limits^{\infty}_{-\infty} {yf(y)} \, dy \\ \\ =\int\limits^1_0 {y^2} \, dy +\int\limits^2_1 {(2y-y^2)} \, dy \\ \\ =\left. \frac{y^3}{3} \right]^1_0+\left[y^2- \frac{y^3}{3} \right]^2_1 \\ \\ = \frac{1}{3} +4- \frac{8}{3} -1+ \frac{1}{3} \\ \\ =1](https://tex.z-dn.net/?f=E%28y%29%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7D%20%7Byf%28y%29%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7By%5E2%7D%20%5C%2C%20dy%20%2B%5Cint%5Climits%5E2_1%20%7B%282y-y%5E2%29%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%3D%5Cleft.%20%5Cfrac%7By%5E3%7D%7B3%7D%20%5Cright%5D%5E1_0%2B%5Cleft%5By%5E2-%20%5Cfrac%7By%5E3%7D%7B3%7D%20%5Cright%5D%5E2_1%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%2B4-%20%5Cfrac%7B8%7D%7B3%7D%20-1%2B%20%5Cfrac%7B1%7D%7B3%7D%20%20%5C%5C%20%20%5C%5C%20%3D1)
Therefore, E(y) = 1.
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Volume is measured in cubic units.
So the ratio would become 6^3: 5^3 = 216:125
The answer would be A.
If one lawn is 10.50, then to find 24 lawns you times it by the earnings of one lawn (i.e. 10.50)
The sum would be: 10.50 x 24
The answer to that would be 252.00.
(Remember to add in the currency sign in front of the answer (eg. £ $ etc)
Answer:

Step-by-step explanation:
h + 4 ≤ 20
Subtract 4 from both sides.
h ≤ 20 − 4
Subtract 4 from 20 to get 16.

Hope it helps and have a great day! =D
~sunshine~