## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 9 - Measurement - 9.2 Measuring Area and Volume - Exercise Set 9.2: 49

#### Answer

(a) The population density in 1900 was $25.6~people/mi^2$ The population density in 2010 was $87.4~people/mi^2$ (b) The percent increase in population density from 1900 to 2010 was $241.4\%$

#### Work Step by Step

(a) Let $P_1$ be the population in 1900. Let $A_1$ be the land area in 1900. We can find the population density in 1900. $\frac{P_1}{A_1} = \frac{75,994,575~people}{2,969,834~mi^2} = 25.6~people/mi^2$ The population density in 1900 was $25.6~people/mi^2$ 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 $87.4~people/mi^2$ (a) We can find the percent increase in population density from 1900 to 2010. $\frac{87.4~people/mi^2-25.6~people/mi^2}{25.6~people/mi^2}\times 100\% = 241.4\%$ The percent increase in population density from 1900 to 2010 was $241.4\%$

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