Abasketball is shipped in a cube-shaped box. the basketball just want touches the sides of the box. about what percent of the space in the cube is occupied by the basketball?

Volume
cube and sphere

the diameter of the sphere=height of box=legnth of box=width of box

volume of cube=side^3
volume of sphere=(4/3)pir^3

side=d=2r
Vcube=d^3

vsphere=(4/3)pi(d/2)^3
Vsphere=(pid^3)/6

percent
basketballV/boxV=( (pid^3)/6)/(d^3)=(pid^3)/(6d^3)=pi/3=aprox 0.523599=52.3599% (round if nececary)

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!