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Jacob invests $\$1049$ in a savings account with a fixed annual interest rate of $7\%$ compunded $3$ times per year. What will the account balance be after $10$ years?

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Exponential growth and decay: word problems

Full solution

Q. Jacob invests

\$1049

in a savings account with a fixed annual interest rate of

7\%

compunded

3

times per year. What will the account balance be after

10

years?

Identify Details: Identify the initial principal amount, interest rate, compounding frequency, and time period. $\newline$ - Principal (P) = $\$1049$ $\newline$ - Annual interest rate (r) = $7\%$ or $0.07$ $\newline$ - Compounding frequency per year (n) = $3$ $\newline$ - Time (t) = $10$ years
Calculate Compound Interest: Calculate the compound interest using the formula $A = P(1 + r/n)^{(nt)}$ . $\newline$ - Substitute the values: $A = 1049(1 + 0.07/3)^{(3*10)}$
Calculate Growth Factor: Perform the calculation for the growth factor $1 + \frac{r}{n}$ . $\newline$ - Calculate $1 + \frac{0.07}{3} = 1.02333$
Calculate Exponent: Calculate the exponent ( $n^t$ ). $\newline$ - Calculate $3\times10 = 30$
Calculate Total Growth: Calculate the total growth by raising the growth factor to the power of the exponent. $\newline$ - Calculate $(1.02333)^{30}$ $\newline$ - Using a calculator, $(1.02333)^{30} \approx 2.4273$
Calculate Final Balance: Calculate the final account balance. $\newline$ - Multiply the initial principal by the total growth: $1049 \times 2.4273 \approx 2547.04$

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