Answer:
C. The height of each bar represents the category's frequency or relative frequency
Step-by-step explanation:
Answer:
Therefore,
![r=\sqrt[3]{\frac{3V}{4\pi }}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D)
is the required r
Step-by-step explanation:
Given:
Volume of inside of the sphere is given as

where r is the radius of the sphere
To Find:
r =?
Solution:
We have
......Given
![3\times V=4\pi r^{3} \\\\\therefore r^{3}=\frac{3V}{4\pi } \\\\\therefore r=\sqrt[3]{\frac{3V}{4\pi }} \textrm{which is the expression for r}](https://tex.z-dn.net/?f=3%5Ctimes%20V%3D4%5Cpi%20r%5E%7B3%7D%20%5C%5C%5C%5C%5Ctherefore%20r%5E%7B3%7D%3D%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%20%5C%5C%5C%5C%5Ctherefore%20r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D%20%5Ctextrm%7Bwhich%20is%20the%20expression%20for%20r%7D)
Therefore,
![r=\sqrt[3]{\frac{3V}{4\pi }}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D)
is the required r
Assuming x = dimes, y = quarters


Substitute that into,

Answer : He has 16 quarters.
Hope this helps. - M
Answer:
The answer is 103
Step-by-step explanation:
The given expression is:
300-7[4(3+5)]+3 to the 3rd power
3 to the 3rd power means 3^3
Therefore,
300-7[4(3+5)]+3^3
First we will solve the round bracket and find the cube
300-7[4(8)]+27
Now we will solve square bracket
300-7[32]+ 27
300-224+27
76+27
103
Thus the answer we get is 103....
To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².