Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 4 - Review - Page 320: 71

Answer

$\frac{91}{150}$

Work Step by Step

$\frac{4}{25}+\frac{23}{75}+\frac{7}{50}=\frac{4\times6}{25\times6}+\frac{23\times2}{75\times2}+\frac{7\times3}{50\times3}=\frac{24}{150}+\frac{46}{150}+\frac{21}{150}=\frac{24+46+21}{150}=\frac{91}{150}$
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