Question 831365
The bowling ball fits tightly into a box that is a cube.
 Sarah would like to fill the extra space in the box with foam packing material.
What percent of the volume (space) of the box that remains and needs to be filled with the packing material? 
:
Let s = the side of the box
then
s^3 = the volume of the box
:
Since the ball fits tightly we can say the length of the side = diameter of the ball
radius = .5s
{{{4/3}}}{{{pi*(.5s)^3}}} = the volume of the ball
{{{4/3}}}{{{pi*.125s^3}}} = the volume of the ball
We can write it
{{{pi*.167s^3}}}
:
the volume of the packing material = vol of the box - vol of the ball 
Then the percent packing material would be
{{{(s^3-(pi*.167s^3))/s^3}}} * 100 =
Cancel out s^3 mult by 100
{{{1 - (.3167*pi)}}} * 100 = 47.6% packing material