Answer
$f(t)=-0.05t^{2}+0.5t+5.75$
In 2008, the model predicts $ 6.55\%$, which is $ 0.25\%$ higher than the actual value.
Work Step by Step
$ y_{1}\sim ax_{1}^{2}+bx+c$
Result:
$f(t)=-0.05t^{2}+0.5t+5.75$
The year 2008 is 8 years after $2000$,
$f(8)=6.55$ percent,
The chart in exercise $20$ has this value to be $ 6.3\%$
so the model is $ 0.25\%$ higher than the actual value.
