Given:
The radius of the sphere is 6 in.
To find:
The volume of the sphere.
Solution:
We know that, the volume of a sphere is
![V=\dfrac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
Where, r is the radius of the sphere.
Putting r=6, we get
![V=\dfrac{4}{3}\pi (6)^3](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20%286%29%5E3)
![V=\dfrac{4}{3}\pi (216)](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20%28216%29)
![V=288\pi](https://tex.z-dn.net/?f=V%3D288%5Cpi)
![V\approx 904.78](https://tex.z-dn.net/?f=V%5Capprox%20904.78)
The volume of the sphere is 904.78 inches cubed.
Therefore, the correct option is C.
Answer:
10m.
Step-by-step explanation:
Since there is no remainder we simply need to find the least common multiple of each of the lengths of Janice and Jasmin's strings until we find a multiple that matches for each. Like so...
2*1 = 2 5*1 = 5
2*2 = 4 5*2 = 10
2*3 = 6
2*4 = 8
2*5 = 10
Finally, we have found the first common multiple which is 10m. That means that the shortest equal length for both Janice's and Jasmin's ribbon is 10m.
The last answer is correct, "Dave is equally likely to select an even number as an odd number."
Numbers 1-10:
1
2
3
4
5
6
7
8
9
10
Out of those, 5 are even and 5 are odd.
So that means there's an equal chance that he selects an odd number as the chance of selecting an even number.
Answer:
x= 7, y=0 (7,0)
Step-by-step explanation:
Area: 12+30= 42
42/2=21
21 x 24=504
Area= 504
Perimeter = 12+30+24.7+26.8=93.5
Hope this helps :)