if the respective diameters are 9 and 4, then their respective radius is half that, or 4.5 and 2.
[tex]\bf \begin{array}{llll}
\textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}
\end{array}\qquad
\begin{array}{llll}
\stackrel{basketball}{\cfrac{4\pi (4.5^3)}{3}}\implies \cfrac{243\pi }{2}
\\\\\\
\stackrel{softball}{\cfrac{4\pi (2^3)}{3}}\implies \cfrac{32\pi }{3}
\end{array}\\\\
-------------------------------\\\\
\cfrac{\quad \frac{243\pi }{2}\quad }{\frac{32\pi }{3}}\implies \cfrac{243\pi }{2}\cdot \cfrac{3}{32\pi }\implies \cfrac{729}{64}\implies 11.390625[/tex]
