Pergunta

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
Solução

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VitóryaMestre · Tutor por 5 anos
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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}
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}
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