Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

Chapter 6 - Percent - Review Exercises - Page 469: 27

Answer

(a) 96.0% have a smoke alarm (b) 45.9% have a carbon monoxide detector

Work Step by Step

$\frac{part}{whole}=\frac{percent}{100}$ (a) 965 out of 1005 have a smoke alarm $\frac{965}{1005}=\frac{p}{100}$ cross multiply to solve for p $1005p=(965)(100)$ $1005p\div1005=96500\div1005$ $p=96.0199$ (b) 461 out of 1005 have a carbon monoxide detector $\frac{461}{1005}=\frac{p}{100}$ cross multiply to solve for p $1005p=(461)(100)$ $1005p\div1005=46100\div1005$ $p=45.8706$
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