Answer
$4.05 \text{ pounds}$
Work Step by Step
The variation model described by the problem is $
w=kr^3
,$ where $w$ is the weight and $r$ is the radius.
Substituting the given values in the variation model above results to
\begin{array}{l}\require{cancel}
1.2=k(2)^3
\\\\
1.2=k(8)
\\\\
\dfrac{1.2}{8}=k
\\\\
k=0.15
.\end{array}
Therefore, the variation equation is $
w=0.15r^3
.$
Using the variation equation above, then
\begin{array}{l}\require{cancel}
w=0.15(3)^3
\\\\
w=0.15(27)
\\\\
w=4.05
.\end{array}
Hence, the weight of the ball is $
4.05 \text{ pounds}
.$
