Question

A person places $3950 in an investment account earning an annual rate of 9.9%, compounded continuously. Using the formula V, equals, P, e, start superscript, r, t, end superscriptV=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 11 years.

Answer 1

After 11 years, an investment of $3950 at an annual interest rate of 9.9% compounded continuously will grow to approximately $11,732.88.

Step-by-step explanation:

The question involves calculating the future value of an investment with continuous compounding interest. We use the formula V = Pert, where V is the future value of the investment, P is the principal amount, e is the base of the natural logarithm (approximately equal to 2.71828), r is the annual interest rate (expressed as a decimal), and t is the time in years. Let's apply this formula to the provided example.

Here, the principal P is $3950, the annual interest rate r is 9.9% or 0.099, and the time t is 11 years. Replacing these values into the formula, we get:

V = 3950 × e(0.099 × 11)

Using a calculator with an exponent and e function, we see that:

V ≈ 3950 × e1.089

V ≈ 3950 × 2.97035 (rounded to five decimal places)

V ≈ $11,732.88 (rounded to the nearest cent)

Therefore, after 11 years, the investment will grow to approximately $11,732.88 when compounded continuously at an annual rate of 9.9%.