David invested $89,000 in an account paying an interest rate of 3.1% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 15 years?
The formula for calculating the amount of money in an account with continuous compounding is given by:
\[A = P \cdot e^{rt}\]
where:
- $A$ is the amount of money in the account after $t$ years
- $P$ is the principal amount (initial investment)
- $r$ is the annual interest rate (expressed as a decimal)
- $t$ is the time in years
- $e$ is Euler's number, approximately 2.71828
Plugging in the given values, we have:
\[A = 89000 \cdot e^{0.031 \cdot 15}\]
Calculating $e^{0.031 \cdot 15}$:
\[e^{0.031 \cdot 15} \approx 1.54532\]
Multiplying the principal by this value:
\[A \approx 89000 \cdot 1.54532\]
Rounding to the nearest ten dollars:
\[A \approx \$137,562\]
Therefore, to the nearest ten dollars, there would be approximately $137,562 in the account after 15 years. Answer: \boxed{137,560}.