A person places $$ \$ 1530 $$ in an investment account earning an annual rate of $$ 2.6 \% $$, compounded continuously. Using the formula $$ V=P e^{r t} $$, 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 11 years.

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Given:
- Initial investment: $$ P = \$1530 $$
- Annual rate of interest: $$ r = 2.6\% = 0.026 $$
- Time: $$ t = 11 $$ years

The formula to calculate the value of the account after $$ t $$ years is:
$$ V = P e^{rt} $$

Substitute the given values into the formula:
$$ V = 1530 e^{0.026 \times 11} $$

Now, we can calculate the value of the account after 11 years using the formula.
Calculate the value by following steps:
- step0: Calculate:
  $$1530e^{0.026\times 11}$$
- step1: Multiply the numbers:
  $$1530e^{0.286}$$
- step2: Convert the expressions:
  $$1530e^{\frac{143}{500}}$$
- step3: Rewrite the expression:
  $$1530\sqrt[500]{e^{143}}$$
The amount of money in the account after 11 years is approximately $2036.57 to the nearest cent.
The amount of money in the account after 11 years is approximately \$2036.57.

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