A rumor $is$ spread $in$ a school. For $b>0tpt(p)={\displaystyle \underset{a}{\overset{p}{}}}\frac{b}{x(1-x)}dxt(p)ln{\displaystyle \underset{a}{\overset{p}{}}}\frac{b}{x(1-x)}dx=t=0p=0\mathrm{.}02aa=t=1p=0\mathrm{.}5bb=p=0\mathrm{.}95t=0$ and $b>0,$ the time $tat$ which a fraction $pof$ the school population has heard the rumor $is$ given $by$

$t(p)={\displaystyle \underset{a}{\overset{p}{}}}\frac{b}{x(1-x)}dx$

$(a)$ Evaluate the integral $to$ find $an$ explicit formula for $t(p).$ Write your answer $soit$ has only one $ln$ term.

${\displaystyle \underset{a}{\overset{p}{}}}\frac{b}{x(1-x)}dx=$

$(b)At$ time $t=0,$ two percent $of$ the school population $(p=0\mathrm{.}02)$ has heard the rumor. What $isa?$

$a=$

$(c)At$ time $t=1,$ fifty percent $of$ the school population $(p=0\mathrm{.}5)$ has heard the rumor. What $isb?$

$b=$

$(d)At$ what time has ninety$-$five percent $of$ the population $(p=0\mathrm{.}95)$ heard the rumor?

$t=$