Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 9 - Measurement - 9.2 Measuring Area and Volume - Exercise Set 9.2 - Page 594: 50

Answer

(a) The population density in 1800 was $6.1~people/mi^2$ The population density in 2010 was $87.4~people/mi^2$ (b) The percent increase in population density from 1800 to 2010 was $1332.8\%$

Work Step by Step

(a) Let $P_1$ be the population in 1800. Let $A_1$ be the land area in 1800. We can find the population density in 1800. $\frac{P_1}{A_1} = \frac{5,308,483~people}{864,746~mi^2} = 6.1~people/mi^2$ The population density in 1800 was $6.1~people/mi^2$ Let $P_2$ be the population in 2010. Let $A_2$ be the land area in 2010. We can find the population density in 2010. $\frac{P_2}{A_2} = \frac{308,745,538~people}{3,531,905~mi^2} = 87.4~people/mi^2$ The population density in 2010 was $87.4~people/mi^2$ (b) We can find the percent increase in population density from 1800 to 2010. $\frac{87.4~people/mi^2-6.1~people/mi^2}{6.1~people/mi^2}\times 100\% = 1332.8\%$ The percent increase in population density from 1800 to 2010 was $1332.8\%$
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