## Intermediate Algebra (6th Edition)

$4.05 \text{ pounds}$
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 $4.05 \text{ pounds} .$