Question

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Complete the table by finding the balance A when P dollars is invested at rate r for t years and compounded n times per year. (Round your answers to the nearest cent)

P = $2100, r= 8.5%, t = 9 years

n A
1 $
2 $
4 $
12 $
365 $
Continuous $

Answer 1

`\begin{array}\cline{1-2} \vphantom{\frac12} n&A \\\cline{1-2}\vphantom{\frac12} 1& \$4376.10\\\vphantom{\frac12} 2& \$4442.10\\\vphantom{\frac12} 4& \$4476.86\\\vphantom{\frac12} 12& \$4500.73\\\vphantom{\frac12} 365& \$4512.49\\\vphantom{\frac12} \sf Continuous& \$4512.89 \\\cline{1-2} \end{array}`

Explanation:

`\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+(r)/(n)\right)^(nt)$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}`

Given:

• P = $2100
• r = 8.5% = 0.085
• t = 9 years

Substitute the given values into the formula to create an equation for A in terms of n:

`\implies A=2100\left(1+(0.085)/(n)\right)^(9n)`

Substitute each value of n into the equation:

`\begin{aligned}n=1 \implies A&=2100\left(1+(0.085)/(1)\right)^(9 * 1)\\&=2100\left(1.085\right)^(9)\\&=\$4376.10\end{aligned}`

`\begin{aligned}n=2 \implies A&=2100\left(1+(0.085)/(2)\right)^(9 * 2)\\&=2100\left(1.0425\right)^(18)\\&=\$4442.10\end{aligned}`

`\begin{aligned}n=4 \implies A&=2100\left(1+(0.085)/(4)\right)^(9 * 4)\\&=2100\left(1.02125\right)^(36)\\&=\$4476.86\end{aligned}`

`\begin{aligned}n=12 \implies A&=2100\left(1+(0.085)/(12)\right)^(9 * 12)\\&=2100\left(1.00708333...\right)^(108)\\&=\$4500.73\end{aligned}`

`\begin{aligned}n=365 \implies A&=2100\left(1+(0.085)/(365)\right)^(9 * 365)\\&=2100\left(1.00023287...\right)^(3285)\\&=\$4512.49\end{aligned}`

`\boxed{\begin{minipage}{8.5 cm}\underline{Continuous Compounding Formula}\\\\$ A=Pe^(rt)$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\\phantom{ww}$\bullet$ $P =$ principal amount \\\phantom{ww}$\bullet$ $e =$ Euler's number (constant) \\\phantom{ww}$\bullet$ $r =$ annual interest rate (in decimal form) \\\phantom{ww}$\bullet$ $t =$ time (in years) \\\end{minipage}}`

`\implies A=2100e^(0.085 * 9)`

`\implies A=2100e^(0.765)`

`\implies A=2100(2.14899437...)`

`\implies A=\$4512.89`

Input the calculated values into the table:

`\begin{array}\cline{1-2} \vphantom{\frac12} n&A \\\cline{1-2}\vphantom{\frac12} 1& \$4376.10\\\vphantom{\frac12} 2& \$4442.10\\\vphantom{\frac12} 4& \$4476.86\\\vphantom{\frac12} 12& \$4500.73\\\vphantom{\frac12} 365& \$4512.49\\\vphantom{\frac12} \sf Continuous& \$4512.89 \\\cline{1-2} \end{array}`