Answer
$36$
Work Step by Step
From the previous part, we have the equation for a learning curve: $N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$
Plug in $N=25$ to find $t$.
Therefore, $25= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$
Simplify: $\dfrac{5}{6}=1- e^{\frac{t}{20} \ln \frac{11}{30}}$
or, $\ln \dfrac{1}{6}=\ln [1- e^{\frac{t}{20} \ln \frac{11}{30}}]$
or, $\dfrac{1}{6}=\dfrac{t}{20} \ln \dfrac{11}{30}$
This gives: $t \approx 36$
