<span>Although two cylinders may have equal diameters, their volumes are not necessarily equal, as the volume of a cylinder is dependent also on the cylinder's height. The same holds true for cones. Spheres, however, are, by nature, proportional. Therefore, if two spheres have the same diameter, they also have the same volumes.</span>
Answer:
D) 50/3
Step-by-step explanation:
6/10 = 10/(6+x-6)
6/10 = 10/x
6x = 100
x = 100/6
x = 50/3
Answer:
![\boxed{4 \sqrt[8]{ {d}^{3} } }](https://tex.z-dn.net/?f=%20%5Cboxed%7B4%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20%7D%20)
Step-by-step explanation:
![= > 4 {d}^{ \frac{3}{8} } \\ \\ = > 4({d}^{3 \times \frac{1}{8} }) \\ \\ = > 4( {d}^{3} \times {d}^{ \frac{1}{8} } ) \\ \\ = > 4( {d}^{3} \times \sqrt[8]{d} ) \\ \\ = > 4 \sqrt[8]{ {d}^{3} }](https://tex.z-dn.net/?f=%20%3D%20%20%3E%204%20%7Bd%7D%5E%7B%20%5Cfrac%7B3%7D%7B8%7D%20%7D%20%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%20%3E%204%28%7Bd%7D%5E%7B3%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B8%7D%20%7D%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%20%7Bd%7D%5E%7B%20%5Cfrac%7B1%7D%7B8%7D%20%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%5Csqrt%5B8%5D%7Bd%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%20%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20)
Okay so a couple of things to know to best solve this problem
1) all angles in a triangle add to 180
2) a straight line the angles above it equal 180 and the angles below it add to 180
3) vertical angles are equal (vertical angles are angles that make an X the opposite angles are called vertical angles)
To compare the two classes, the Coefficient of Variation (COV) can be used.
The formula for COV is this:
C = s / x
where s is the standard deviation and x is the mean
For the first class:
C1 = 10.2 / 75.5
C1 = 0.1351 (13.51%)
For the second class:
C2 = 22.5 / 75.5
C2 = 0.2980 (29.80%)
The COV is a test of homogeneity. Looking at the values, the first class has more students having a grade closer to the average than the second class.