When estimating the number of spheres contained in a prism we should account for the empty space between the spheres. to do this we find 190% of the volume of the sphere and use that value in our estimation.
if gumballs have a 2 inch diameter, what should we use as the volume of each gumball in our calculation? (round your answer to the nearest tenth)
a. 4.2 cubic inches
b. 8.0 cubic inches
c. 33.5 cubic inches
d. 5280 cubic inches

  B.  8.0 cubic inches

Step-by-step explanation:

Sphere volume is ...

  V = (4/3)πr³ = (4/3)π(2 in/2)³ = 4π/3 in³ ≈ 4.19 in³

For estimation purposes, we should use a value that is 1.9 times this, or ...

  estimation volume = 1.9×4.19 in³ ≈ 7.96 in³ ≈ 8.0 in³

The factor 190% comes from the ratio 6/π ≈ 190.99%. Using this ratio means you're estimating volume as though the sphere were a cube with a side length equal to the diameter of the sphere.

A random packing of equal-size spheres (as, for example, gumballs) usually only requires a factor of about 160% to account for the empty space.

c) yes, because of asa or aas