Answer
a. $17.0\%$, underestimates $0.3\%$
b. $2016$
Work Step by Step
a. Given the model function $f (x) = 1.2 ln (x) + 15.7$, for year 2009, $x=2009-2006=3$, we have $f (3) = 1.2 ln (3) + 15.7\approx17.0$ percent. The bar graph gives a value of $17.3$ for year 2009; thus the model result underestimates the data by $0.3\%$
b. Letting $f(x)=18.5\%$, we have $ 1.2 ln (x) + 15.7=18.5$, which gives $ln(x)\approx2.333$ and $x\approx10\ yrs$, corresponding to year $2006+10=2016$
