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

$a.\displaystyle \qquad\frac{4}{5}$ $b.\qquad$The graph also shows the fraction to be $\displaystyle \frac{4}{5}$. (The model coincides with the data for 2010) $c.\displaystyle \qquad\frac{49}{50}$
$a.$ Substitute 10 for x (2010 was 10 years after 2000): $I=\displaystyle \frac{3}{100}\cdot 10+\frac{1}{2}=$... reduce the product by 10 $=\displaystyle \frac{3}{10}+\frac{1}{2}\qquad$... LCD is 10 $=\displaystyle \frac{3}{10}+\frac{1\times 5}{2\times 5}$ $=\displaystyle \frac{3+5}{10}$ $=\displaystyle \frac{8}{10}\qquad$... reduce by 2 $=\displaystyle \frac{4}{5}$ $b.$ The graph also shows the fraction to be $\displaystyle \frac{4}{5}$. (The model coincides with the data for 2010) $c.$ Substitute $16$ for x ($2016$ was $16$ years after 2000): $I=\displaystyle \frac{3}{100}\cdot 16+\frac{1}{2}=$... reduce the product by $2$ $=\displaystyle \frac{3\cdot 8}{50} +\frac{1}{2}$ $=\displaystyle \frac{24}{50} +\frac{1}{2}\qquad$... LCD is $50$ $=\displaystyle \frac{24}{50}+\frac{1\times 25}{2\times 25}$ $=\displaystyle \frac{24+25}{50}$ $=\displaystyle \frac{49}{50}$