A person places $43900 in an investment account earning an annual rate of 9.2%, 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 19 years. Answer:

Question

A person places  $43900 in an investment account earning an annual rate of  9.2%, 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 19 years. Answer:

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Identify Given Values: Identify the given values from the problem. Principal (P) = $43900 Rate of interest (r) = 9.2% or 0.092 (as a decimal) Time (t) = 19 years We will use the formula for continuous compounding: V = Pe^(rt)Convert Rate to Decimal: Convert the rate of interest from a percentage to a decimal. r = 9.2% = 9.2/100 = 0.092Substitute Values in Formula: Substitute the given values into the formula. V = 43900 * e^(0.092 * 19)Calculate Exponent: Calculate the exponent part of the formula. Exponent = 0.092 * 19 = 1.748Calculate e Value: Calculate the value of e raised to the power of the exponent. Using a calculator, we find that e^1.748 ≈ 5.748Multiply Principal: Multiply the principal by the value of e raised to the power of the exponent. V = 43900 * 5.748Perform Multiplication: Perform the multiplication to find the final value. V ≈ 43900 * 5.748 ≈ 252,367.2Round Final Value: Round the final value to the nearest cent. V ≈ $252,367.20

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