Answer
$7.6$ hours
Work Step by Step
$y(t)$ = $\frac{1}{1-\frac{e^{-kt}}{C}}$
the initial condition $y(0)$ = $0.08$ allow us to determine the value of C
$\frac{2}{25}$ = $\frac{1}{1-\frac{1}{C}}$
$C$ = $-\frac{2}{23}$
then
$y(t)$ = $\frac{2}{2+23{e^{-kt}}}$
the condition $y(4)$ = $0.5$ allow us to determine the value of $k$
$\frac{1}{2}$ = $\frac{2}{2+23{e^{-kt}}}$
$k$ = $\frac{1}{4}\ln(\frac{23}{2})$ $\approx$ $0.6106$ $hours^{-1}$
90% of the students have heard the rumor when y(t) = 0:9. Thus
$\frac{9}{10}$ = $\frac{2}{2+23{e^{-0.6106t}}}$
$t$ = $7.6$ hours
