a person places $1530 in an investment account earning an annual rate of 2.6%, compounded continuously…

a person places $1530 in an investment account earning an annual rate of 2.6%, 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 11 years.

a person places $1530 in an investment account earning an annual rate of 2.6%, 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 11 years.

Answer

Explanation:

Step1: Identify the values

$P = 1530$, $r=0.026$, $t = 11$

Step2: Substitute into the formula

$V=1530\times e^{0.026\times11}$

Step3: Calculate the exponent

$0.026\times11 = 0.286$

Step4: Calculate $e^{0.286}$

Using a calculator, $e^{0.286}\approx1.3317$

Step5: Calculate the final value

$V = 1530\times1.3317=2037.441\approx2037.44$

Answer:

$2037.44$