Answer
a. $43.08(\%)$. overestimated $0.08\%$.
b. $R=\frac{-0.28n+47}{0.28n+53}$
c. $R=\frac{2}{3}$. 3 women; fits perfectly with the bar graph.
Work Step by Step
a. Given the formula $M=-0.28n+47$ and $n=2003-1989=14$, we have $M=-0.28(14)+47=43.08(\%)$. Reading from the bar graph, we have $43\%$ for men in the year 2003. Thus the model results overestimated the percentage by $0.08\%$.
b. With the given formulas, we have $R=\frac{M}{W}=\frac{-0.28n+47}{0.28n+53}$
c. In 2014, we have $n=2014-1989=25$ and $R=\frac{-0.28(25)+47}{0.28(25)+53}=\frac{40}{60}=\frac{2}{3}$. Thus for every two men, there will be 3 women receiving bachelor's degrees in 2014. This model result fits perfectly with the data from the bar graph.
