Answer:
\(4,608(6- \pi)\ cm^{3}\)
Step-by-step explanation:
we know that
The volume of the empty space encircling the basketballs inside the box is equal to the volume of the box minus the volume of the two basketball inside the box
so
The volume of the box is equal to
\(V=LWH\)
in this problem we have
\(L=2*24=48\ cm\)
\(W=24\ cm\)
\(H=24\ cm\)
substitute
\(V=48*24*24=27,648\ cm^{3}\)
The volume of the two basketball inside the box is equal to
\(V=2*\frac{4}{3} \pi r^{3}\)
n this problem we have
\(r=24/2=12\ cm\)
substitute
\(V=\frac{8}{3} \pi 12^{3}\)
\(V=4,608 \pi\ cm^{3}\)
The expression is equal to
\((27,648-4,608 \pi)\ cm^{3}\)
\(4,608(6- \pi)\ cm^{3}\)
