Answer
(a) $A(t)=12000(1+\frac{0.056}{12})^{12t}$
(b) $14195.06$
(c) $9.12$ years
Work Step by Step
(a) Given $P_0=12000, r=0.056, n=12$, using the formula in this section, we have
$A(t)=P_0(1+\frac{r}{n})^{nt}=12000(1+\frac{0.056}{12})^{12t}$
(b) In this case $t=3,n=365$, we have
$A(3)=12000(1+0.056/365)^{365\times3}=14195.06$
(c) For interest compounded continuously, the formula is $A(t)=P_0e^{rt}$ so we have
$20000=12000e^{0.056t}$, solve for $t$ to get $t=ln(20/12)/0.056\approx9.12$ years
