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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It is £21 but I have to be 20 characters long to post this comment but that’s the answer
Answer:
42
Step-by-step explanation:
5 more than 2 times the girls
89=5+2x
x=42
Method A: 7.6666.... = 7 + 0.6666... = 7 + 2/3 = 21/3 + 2/3 = 23/3
Method B: 10(7.666...) - 1(7.666...) = 76.666... - 7.666... = 69.000...
(10 - 1)(7.666...) = 69
7.666... = 69/9 = 23/3
okay well usually what I would tell you would be to use the ABC method. you would multiply A (2) by C (6) to get twelve. so you need two numbers that multiply to twelve and add to -4 (the B value). but there are no numbers that fit this description so you have to use the GCF method instead
for this method you must find the GCF for all three numbers which in this case is 2. 2(x^2-2x+3) all these numbers are now prime so this is the factored form
