A person places $\$ 38100$ in an investment account earning an annual rate of $4 \%$, compounded continuously. Using the formula $V=P e^{r t}$, where $V$ is the value of the account in t years, $\mathrm{P}$ is the principal initially invested, e is the base of a natural logarithm, and $r$ is the rate of interest, determine the amount of money, to the nearest cent, in the account after 9 years.
Answer
\(\boxed{54,609.85}\) is the amount of money in the account after 9 years, to the nearest cent.
Step 1 ：Given the formula: \(V = P e^{rt}\), where \(V\) is the value of the account, \(P\) is the principal, \(r\) is the interest rate, and \(t\) is the time in years.
Step 2 ：Plug in the values: \(P = 38100\), \(r = 0.04\), and \(t = 9\) into the formula.
Step 3 ：Calculate the value of the account: \(V = 38100 e^{(0.04)(9)}\)
Step 4 ：Evaluate the expression: \(V \approx 54609.85\)
Step 5 ：\(\boxed{54,609.85}\) is the amount of money in the account after 9 years, to the nearest cent.
