Jose deposits $\$ 7,500$ every year into an account earning an annual interest rate of $5.9 \%$ compounded annually. How much would he have in the account after $10$ years, to the nearest dollar? Use the following formula to determine your answer. $\newline$ $A=d\left(\frac{(1+i)^{n}-1}{i}\right)$ $\newline$ $A=$ the future value of the account after $n$ periods $\newline$ $d=$ the amount invested at the end of each period $\newline$ $i=$ the interest rate per period $\newline$ $n=$ the number of periods $\newline$ Answer:

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Math Problems

Grade 7

Compound interest

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Q. Jose deposits

\$ 7,500

every year into an account earning an annual interest rate of

5.9 \%

compounded annually. How much would he have in the account after

10

years, to the nearest dollar? Use the following formula to determine your answer.

\newline

A=d\left(\frac{(1+i)^{n}-1}{i}\right)

\newline

A=

the future value of the account after

n

periods

\newline

d=

the amount invested at the end of each period

\newline

i=

the interest rate per period

\newline

n=

the number of periods

\newline

Answer:

Identify Given Values: Identify the given values from the problem. $\newline$ We are given: $\newline$ $d$ (the amount invested at the end of each period) = $\$7,500$ $\newline$ $i$ (the interest rate per period) = $5.9\%$ or $0.059$ when converted to decimal $\newline$ $n$ (the number of periods) = $10$ years $\newline$ We will use these values in the compound interest formula to find $A$ (the future value of the account).
Convert Interest Rate: Convert the annual interest rate from a percentage to a decimal. $\newline$ To convert $5.9\%$ to a decimal, divide by $100$ . $\newline$ $i = \frac{5.9\%}{100} = 0.059$
Substitute Values in Formula: Substitute the values into the compound interest formula. $\newline$ Using the formula $A = d \times \left(\left(1 + i\right)^{n} - 1\right) / i$ , we substitute the values we have: $\newline$ $A = \$(7,500) \times \left(\left(1 + 0.059\right)^{10} - 1\right) / 0.059$
Calculate Exponential Value: Calculate the value inside the parentheses $(1 + i)^{n}$ . $(1 + 0.059)^{10} = (1.059)^{10}$ Now, calculate the exponentiation. $(1.059)^{10} \approx 1.77729$ (rounded to five decimal places for precision in further calculations)
Subtract One: Continue with the formula by subtracting $1$ from the result obtained in Step $4$ . $\newline$ $1.77729 - 1 = 0.77729$
Divide by Interest Rate: Divide the result from Step $5$ by the interest rate $i$ . $\newline$ $0.77729 / 0.059 \approx 13.17559$
Multiply by Amount: Multiply the result from Step $6$ by the amount invested at the end of each period $d$ . $\$7,500 \times 13.17559 \approx \$98,817.42$
Round Final Result: Round the final result to the nearest dollar. $\newline$ The future value of the account, rounded to the nearest dollar, is approximately $\$98,817$ .