Answer:
No it is not a reasonable scale. She should use 1 inch= 6 inches.
Step-by-step explanation:
The width of the drawing is 48 inches, dividing it by 4 inches gives 48/4=12 inches.
This is exactly the width of her paper. Which means the drawing will get to the edge.
The height of the drawing is 120 inches,divide by 4inches you get: 120/4=30 inches. The height of paper is 24 inches. Which means her paper is too small to accommodate the height of the drawing. The preferred scale would be 1 inch=6inches.
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>
Solve: 2y(6 - 1)(6 - 1) = - 200
Simplify both sides of the equation:
50y = - 200
Divide both sides by 50
50y/50 = -200/50
y = - 4
Answer: y = - 4
Hope that helps!!!!!!! (Answer: y = - 4)
Answer:
Her total workout is
.
Step-by-step explanation:
Consider the provided information.
Let w be the amount of time Tika spends on weight training.
Tika spends twice as much time doing aerobics than weight training.
It means she spends
time doing aerobics.
She stretches for one-fifth of the time she spends on weight training.
It means she spends
time for stretches.
The total workout time is:

Hence, her total workout is
.
Answer:
$2,710.64
Step-by-step explanation:
Divide 43,720 by 100 so you know how many groups of 200s you have. Because it's 6.20 for each 100.
437.2
Do 6.20 times 437.2 to get your answer.
$2,710.64