"A gumball machine is in the shape of a sphere with a radius of 6 inches. A store manager wants to fill up the machine with miniature gumballs, which have a radius of 1/3in. How many gumballs will fit in the machine?"
So Volume (V) of a sphere is: [tex]V = \frac{4}{3} \pi {r}^{3} [/tex] so essentially we need to divide the machine's Volume (V) the gumball volume (v) [tex]V = \frac{4}{3} \pi {r}^{3} = \frac{4}{3} \pi {(6)}^{3} \\ V = \frac{4}{3}(216) \pi = 288\pi[/tex] [tex]v = \frac{4}{3} \pi {( \frac{1}{3} )}^{3} = \frac{4}{3} \times \frac{1}{27} \pi \\ v = \frac{4}{81} \pi[/tex] now we divide V by v: [tex]V/v = \frac{288\pi}{ \frac{4\pi}{81} } = \frac{288}{1} \times \frac{81}{4} = 5832[/tex] so b) 5,832 gumballs can fit in the machine
