Question

7100 dollars is placed in an account with an annual interest rate of 7.75%. how much will be in the account after 30 years, to the nearest cent?

Explanation:

Step1: Identify the formula for compound interest

The formula for compound interest (assuming annual compounding, since it's not specified otherwise) is $A = P(1 + r)^t$, where:

• $A$ is the amount of money accumulated after $n$ years, including interest.
• $P$ is the principal amount (the initial amount of money).
• $r$ is the annual interest rate (decimal).
• $t$ is the time the money is invested for in years.

Step2: Convert the interest rate to decimal

The annual interest rate is $7.75\%$. To convert this to a decimal, we divide by 100: $r=\frac{7.75}{100}=0.0775$.

Step3: Identify the values of $P$, $r$, and $t$

• $P = 7100$ dollars (the principal amount).
• $r = 0.0775$ (the annual interest rate in decimal).
• $t = 30$ years (the time period).

Step4: Substitute the values into the formula

Substitute $P = 7100$, $r = 0.0775$, and $t = 30$ into the compound - interest formula $A = P(1 + r)^t$:

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Step5: Calculate $(1.0775)^{30}$

First, calculate $(1.0775)^{30}$. Using a calculator, $(1.0775)^{30}\approx 9.74702$.

Step6: Calculate the value of $A$

Multiply 7100 by 9.74702:

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Rounding to the nearest cent (two decimal places), we get $A\approx69203.84$ (if we consider more precise calculation of $(1.0775)^{30}$, we can get a more accurate result. Let's use a more precise value of $(1.0775)^{30}$. Using a calculator with higher precision, $(1.0775)^{30}\approx9.7470204$. Then $A = 7100\times9.7470204=7100\times9.7470204 = 69203.84484\approx69203.84$)

(Note: If we use continuous compounding, the formula is $A = Pe^{rt}$, where $e$ is the base of the natural logarithm ($e\approx2.71828$). Let's check the result for continuous compounding as well (in case the problem assumes continuous compounding, although it's not specified, but just for reference).

For continuous compounding:

$A = Pe^{rt}$

$P = 7100$, $r = 0.0775$, $t = 30$

$A=7100\times e^{0.0775\times30}$

First, calculate $0.0775\times30 = 2.325$

Then $e^{2.325}\approx10.237$

$A = 7100\times10.237=72682.7$

But since the problem doesn't specify continuous compounding, and usually, if not specified, we assume annual compounding. However, let's re - calculate $(1.0775)^{30}$ more accurately. Using a calculator, $(1.0775)^{30}=1.0775^{30}\approx9.7470204$

$7100\times9.7470204 = 7100\times9.7470204=69203.84484\approx69203.84$)

Answer:

The amount in the account after 30 years, to the nearest cent, is $\$69203.84$ (if we consider annual compounding). If we use more precise calculation or consider continuous compounding, the result may vary slightly, but with annual compounding as the standard assumption, the amount is approximately $\$69203.84$.