Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 3 - Section 3.3 - Slope - Exercise Set - Page 242: 45

Answer

$a.\quad -0.8$ $b.\quad $ $0.8, $ $ 0.8\%$ per year

Work Step by Step

Use: $m=\displaystyle \frac{\text{change in y}}{\text{change in x}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ $(a)$ The slope of the line passing through $(22,49)$ and $(62,29)$ is $m_{1}=\displaystyle \frac{29-49}{62-49}=\frac{-20}{25}\cdot\frac{4}{4}=-\frac{80}{100}=-0.8$ $(b)$ The slope written as $\displaystyle \frac{\text{change in y}}{\text{change in x}}=\frac{-0.8}{1}$ is interpreted as: For a 1 year change in x, stress (y) decreases by $ 0.8\%$. The unit for slope is (unit for y)/(unit for x). Here: "percent per year"
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