Answer
a. $T=200e^{-0.11t}+20$
b. $42.$2 ${}^{o}F$.
Work Step by Step
a.
$\displaystyle \ln(\frac{T-20}{200})=-0.11t \qquad$ .. apply $e^{(..)}$ to both sides
$\displaystyle \frac{T-20}{200}=e^{-0.11t}\qquad$ .. /$\times 200$
$T-20=200e^{-0.11t} \qquad$ .. /$+20$
$T=200e^{-0.11t}+20$
b.
When $t=20,$
$T=200e^{-0.11(20)}+20$
$=200e^{-2.2}+20\approx 42.$2 ${}^{o}F$.