The population in the United States was roughly 161,000,000 in 1950. By 2000, it had grown to roughly 291,000,000. Assume that the population in the United States grew linearly during that period. Find a linear equation which models the population in the United States during the period 1950 to 2000

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The population in the United States was roughly 161,000,000 in 1950. By 2000, it had grown to roughly 291,000,000. Assume that the population in the United States grew linearly during that period. Find a linear equation which models the population in the United States during the period 1950 to 2000

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Jdoe0001id Jdoe0001id · Answer 1

to get the equation of any straight line, we simply need two points off of it, so let's use the ones provided in the table above.

[tex]\begin{array}{|cc|ll} \cline{1-2} \stackrel{millions}{population}&year\\ \cline{1-2} 161&1950\\ 291&2000\\ \cline{1-2} \end{array}\hspace{5em} (\stackrel{x_1}{161}~,~\stackrel{y_1}{1950})\qquad (\stackrel{x_2}{291}~,~\stackrel{y_2}{2000}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2000}-\stackrel{y1}{1950}}}{\underset{run} {\underset{x_2}{291}-\underset{x_1}{161}}} \implies \cfrac{ 50 }{ 130 }\implies \cfrac{5}{13}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1950}=\stackrel{m}{\cfrac{5}{13}}(x-\stackrel{x_1}{161}) \\\\\\ y-1950=\cfrac{5}{13}x-\cfrac{805}{13}\implies y=\cfrac{5}{13}x-\cfrac{805}{13}+1950\implies y=\cfrac{5}{13}x+\cfrac{24545}{13}[/tex]