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
