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$3000 are deposited in an account with 9% interest rate, compounded continuously.
What is the balance after 14 years?
F = $[?]
Round to the nearest cent.

Respuesta :

masifakrami

Answer:

the balance after 14 years with continuous compounding is approximately $10,575.00.

Step-by-step explanation:

To find the balance after 14 years with continuous compounding, we can use the formula for continuous compound interest:

[tex]\[F = P \cdot e^{rt}\][/tex]

Where:

[tex]- \(F\) is the future value (balance) of the investment[/tex]

[tex]- \(P\) is the principal amount (initial deposit)[/tex]

[tex]- \(e\) is the base of the natural logarithm (approximately equal to 2.71828)[/tex]

[tex]- \(r\) is the annual interest rate (expressed as a decimal)[/tex]

[tex]- \(t\) is the time the money is invested for (in years)[/tex]

Given:

[tex]- \(P = $3000\)[/tex]

[tex]- \(r = 9\% = 0.09\) (as a decimal)[/tex]

[tex]- \(t = 14\) years[/tex]

We'll plug these values into the formula to find the balance \(F\):

[tex]\[F = 3000 \cdot e^{0.09 \cdot 14}\][/tex]

[tex]\[F = 3000 \cdot e^{1.26}\][/tex]

Using a calculator, we can compute the value of [tex]\(e^{1.26}\)[/tex] and multiply it by $3000 to find F.

[tex]\[e^{1.26} \approx 3.525\][/tex]

[tex]\[F \approx 3000 \cdot 3.525\][/tex]

[tex]\[F \approx \$10575.26\][/tex]

So, the balance after 14 years with continuous compounding is approximately $10,575.26.

semsee45

Answer:

$10576.26

Step-by-step explanation:

To determine the balance of the account after 14 years, we can use the continuous compounding interest formula:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Continuous Compounding Interest Formula}}\\\\A=Pe^{rt}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{$A$ is the final amount.}\\\phantom{ww}\bullet\;\;\textsf{$P$ is the principal amount.}\\\phantom{ww}\bullet\;\;\textsf{$e$ is Euler's number (constant).}\\\phantom{ww}\bullet\;\;\textsf{$r$ is the interest rate (in decimal form).}\\\phantom{ww}\bullet\;\;\textsf{$t$ is the time (in years).}\end{array}}[/tex]

In this case:

  • P = $3000
  • r = 9% = 0.09
  • t = 14 years

Substitute the values into the formula:

[tex]A=3000\cdot e^{(0.09 \cdot 14)}[/tex]

Solve for A:

[tex]A=3000\cdot e^{1.26}\\\\A=3000\cdot 3.525421487...\\\\A=10576.264462...\\\\A=10576.26[/tex]

Therefore, the balance after 14 years, rounded to the nearest cent, is:

[tex]\Large\boxed{\boxed{\$10576.26}}[/tex]

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