First you need to find the area, the formula is A=LxW, multiply 14 by 11, that’s the area. If one back covers 22 ft we need to find how many 22s are in 154, to do that you divide. Make sense?
Answer: Third option is correct.
Step-by-step explanation:
Since we have given that
The number of freshmen entering college in a certain year = 621
It is considered as discrete data.
Since we know the definition of Discrete which refers to all types of data which take only numerical values.
It is based on counts.
According to question, we have given the exact number of freshmen instead of any range or interval.
Hence, Third option is correct.
Answer:
$0.11 per pound
Step-by-step explanation:
15 - 3 = 12
12/112 = 0.10714286
= about 0.11
Answer: Choice C

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Explanation:
The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve 
Think of the blue region as the floor of this weirdly shaped 3D room.
We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is
where 0 < x < 1
Let's compute the area of each general cross section.

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.
This is what we want to compute

Apply a u-substitution
u = -2x
du/dx = -2
du = -2dx
dx = du/(-2)
dx = -0.5du
Also, don't forget to change the limits of integration
- If x = 0, then u = -2x = -2(0) = 0
- If x = 1, then u = -2x = -2(1) = -2
This means,

I used the rule that
which says swapping the limits of integration will have us swap the sign out front.
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Furthermore,
![\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.5%5Cint_%7B-2%7D%5E%7B0%7De%5E%7Bu%7Ddu%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Be%5Eu%2BC%5Cright%5D_%7B-2%7D%5E%7B0%7D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%28e%5E0%2BC%29-%28e%5E%7B-2%7D%2BC%29%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
In short,
![\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B0%7D%5E%7B1%7De%5E%7B-2x%7Ddx%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
This points us to choice C as the final answer.
I think the answer is Sample 4.
Hope this helps!