[tex]\begin{gathered} Circumference=2\pi r \\ \text{Here r is the radius,} \\ 28.5=2\times3.14\times r \\ r=4.54in \\ The\text{ volume of the basketball is,} \\ V=\frac{4}{3}\pi r^3 \\ V=\frac{4}{3}\times3.14\times(4.54)^3 \\ V=392in^3 \end{gathered}[/tex]
