## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Section 1.2 - Fractions in Algebra - Exercise Set - Page 31: 130

#### Answer

$a.\displaystyle \qquad\frac{23}{25}$ $b.\qquad$The model overestimates the data for 2014 by $\displaystyle \frac{1}{50}$ $c.\displaystyle \qquad\frac{19}{20}$

#### Work Step by Step

$a.$ Substitute $14$ for x ($2014$ was 10 years after 2000): $I=\displaystyle \frac{3}{100}\cdot 14+\frac{1}{2}=$... reduce the product by $2$ $=\displaystyle \frac{3\cdot 7}{50}+\frac{1}{2}\qquad$... LCD is $50$ $=\displaystyle \frac{21}{50}+\frac{1\times 25}{2\times 25}$ $=\displaystyle \frac{21+25}{50}$ $=\displaystyle \frac{46}{50}\qquad$... reduce by 2 $=\displaystyle \frac{23}{25}$ $b.$ The graph shows the fraction to be $\displaystyle \frac{9}{10}$. The LCD for 10 and 50 is $50$. Model: $\displaystyle \frac{23\times 2}{25\times 2}=\frac{46}{50}$ Graph: $\displaystyle \frac{9\times 5}{10\times 5}=\frac{45}{50}$ (The model overestimates the data for 2014 by $\displaystyle \frac{1}{50}$) $c.$ Substitute $15$ for x ($2015$ was $15$ years after 2000): $I=\displaystyle \frac{3}{100}\cdot 15+\frac{1}{2}=$... reduce the product by $5$ $=\displaystyle \frac{3\cdot 3}{20} +\frac{1}{2}$ $=\displaystyle \frac{9}{20} +\frac{1}{2}\qquad$... LCD is $20$ $=\displaystyle \frac{9}{20}+\frac{1\times 10}{2\times 10}$ $=\displaystyle \frac{9+10}{20}$ $=\displaystyle \frac{19}{20}$

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