Answer:
The volume is 
Step-by-step explanation:
The General Slicing Method is given by
<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is
The key observation is that the width is the distance between the upper bounding curve
and the lower bounding curve 
The width of each square is given by

This means that the area of the square cross section at the point x is

The intersection points of the two bounding curves satisfy
, which has solutions x = ±1.

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is
![V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}](https://tex.z-dn.net/?f=V%3D%5Cint%5Climits%5E%7B1%7D_%7B-1%7D%20%7B%282-2x%5E2%29%5E2%7D%20%5C%2C%20dx%5C%5C%5C%5CV%3D%5Cint%20_%7B-1%7D%5E14-8x%5E2%2B4x%5E4dx%5C%5C%5C%5CV%3D%5Cint%20_%7B-1%7D%5E14dx-%5Cint%20_%7B-1%7D%5E18x%5E2dx%2B%5Cint%20_%7B-1%7D%5E14x%5E4dx%5C%5C%5C%5CV%3D%5Cleft%5B4x%5Cright%5D%5E1_%7B-1%7D-8%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E1_%7B-1%7D%2B4%5Cleft%5B%5Cfrac%7Bx%5E5%7D%7B5%7D%5Cright%5D%5E1_%7B-1%7D%5C%5C%5C%5CV%3D8-%5Cfrac%7B16%7D%7B3%7D%2B%5Cfrac%7B8%7D%7B5%7D%5C%5C%5C%5CV%3D%5Cfrac%7B64%7D%7B15%7D)
Answer:
±square root 20
Step-by-step explanation:
x²= 20
x= square root 20
x= 2 square root 5
x≈ 4.4721
H. 100=10squared and 64=8squared 25=5squared
Answer:
It increases the mean value of the prizes
Step-by-step explanation:
If you order the numbers from the lowest to the highest, the median is the number placed in the middle of the ordered list. In this case, there are values repeated many times (there are 100 values, $1 is repeated 44 times and $5 is repeated 25 times, so $5 is the mean), then the presence of a high value with only one presence doesn't modify this median.
On the other hand, the mean is always affected but all data set, given that $100 is the highest value, the mean will increase.