Chapter 2, Problem 10TP

Solution

To determine: By how many people per square mile of land did the population density increase from 1800 to 2000? Explain your reasoning.

The population density of the United States increased approximately by 74 people per square mile.

Given information:

The population of the United States is 1800 was $5.308\times {10}^6$ , and the land area was $8.647\times {10}^5$ square miles. By 2000, the population had increased to $2.814\times {10}^8$ , and the land area was $3.537\times {10}^6$ square miles.

Formula used:

  $\text{Population density}=\frac{\text{Area of the country}}{\text{Population of the country}}$

Calculation:

Find population density for years 1800 and 2000. Upon substituting the given values in the population density formula, the set-up would be:

  $\begin{array}{c}\text{Population density of US in 1800}=\frac{\text{5}\text{.308}\times {\text{10}}^6}{\text{8}\text{.647}\times {\text{10}}^5}\\ =\frac{\text{5}\text{.308}\times {\text{10}}^{6-5}}{\text{8}\text{.647}}\left(\text{Using exponent property }\frac{a^m}{a^n}=a^{m-n}\right)\\ =\frac{\text{5}\text{.308}\times \text{10}}{\text{8}\text{.647}}\\ =\frac{\text{53}\text{.08}}{\text{8}\text{.647}}\\ \approx 6\end{array}$

  $\begin{array}{c}\text{Population density of US in 2000}=\frac{\text{2}\text{.814}\times {\text{10}}^8}{\text{3}\text{.537}\times {\text{10}}^6}\\ =\frac{\text{2}\text{.814}\times {\text{10}}^{8-6}}{\text{3}\text{.537}}\left(\text{Using exponent property }\frac{a^m}{a^n}=a^{m-n}\right)\\ =\frac{\text{2}\text{.814}\times {\text{10}}^2}{\text{3}\text{.537}}\\ =\frac{\text{281}\text{.4}}{\text{3}\text{.537}}\\ =\text{79}\text{.5589}\\ \approx \text{80}\end{array}$

The difference between population density for both years would be $80-6=74$ .

Therefore, the population density of the United States increased approximately by 74 people per square mile.