A person places 7760 in an investment account earning an annual rate of 5% compounded continuously. Using the formula V=Pe^rt where V is the value of the account in t years, 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 12 years. Answer Attemptiontofs

To solve this problem, we will use the formula for continuous compounding interest, which is given as:

$$V = Pe^{rt}$$

where:
- $V$ is the value of 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 the money is invested for in years,
- $e$ is the base of the natural logarithm (approximately equal to 2.71828).

Given:
- $P = \$7760$
- $r = 5\% = 0.05$
- $t = 12$ years

Now, we can plug these values into the formula:

$$V = 7760 \times e^{0.05 \times 12}$$

First, calculate the exponent:

$$0.05 \times 12 = 0.6$$

Next, calculate $e^{0.6}$:

$$e^{0.6} \approx 1.82212$$

Now, multiply this by the principal amount:

$$V = 7760 \times 1.82212 \approx 14123.63$$

So, the amount of money in the account after 12 years, to the nearest cent, is approximately:

$$\boxed{14123.63}$$