Question

A person invests 9500 dollars in a bank. The bank pays 5.75% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 26100 dollars? A= P(1+! nt

Answer 1

1. Data input

P = 9500 dollars

r = 5.75% daily

A = 26100 dollars

n = 365

2. Equation

$A=P(1+\frac{r}{n})^{nt}$$\begin{gathered} 26100=9500(1+\frac{0.0575}{365})^{365t} \\ \frac{26100}{9500}=(1+0.00157)^{365t} \end{gathered}$

$t=17.6\text{ years}$