Answer:
Option a) [5.567, 5.633]
Step-by-step explanation:
We are given the following in the question:
Sample mean, 
= 5.6 feet
Sample size, n = 27
Alpha, α = 0.10
Sample standard deviation, s = 0 .1 feet
90% Confidence interval:
Putting the values, we get,
Option a) [5.567, 5.633]
Can you insert a photo lol
<span>4.5x10⁶ light bulbs
The actual answer is 4.26x10^6. multiply 4.56 and9.36 to get 42.6816 add the exponents of the 10^3 and 10^2 for 10^5 move the decimal 1 place to the left to get 4.26x10^6</span>
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.
</span>
Answer:
36,42,48
Step-by-step explanation:
