Answer
See below
Work Step by Step
From part a, we found: $V=\frac{S\sqrt S}{6\sqrt \pi}$
We have the volume of a bowling ball whose surface area is 232 inches: $V=\frac{232\sqrt 232}{6\sqrt \pi}\approx 332.28$ inches.
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 6 Rational Exponents and Radical Functions - Mixed Review of Problem Solving - Lessons 6.1-6.3 - Page 436: 2c
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