Question

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Complete the table by finding the balance A when $18000 is invested at rate for t years, compounded continuously. (Round your answers to 2 decimal places.)
r = 1.5%

t A
10 $
20 $
30 $
40 $
50 $

Answer 1

`\begin{array}c\cline{1-2} \vphantom{\frac12} t& A\\\cline{1-2} \vphantom{\frac12} 10& \$20913.02 \\\cline{1-2} \vphantom{\frac12} 20& \$24297.46 \\\cline{1-2} \vphantom{\frac12} 30& \$28229.62 \\\cline{1-2} \vphantom{\frac12} 40& \$32798.14 \\\cline{1-2} \vphantom{\frac12} 50& \$38106.00 \\\cline{1-2} \end{array}`

Explanation:

`\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}}`

Given:

• P = $18,000
• r = 1.5% = 0.015

Substitute the values of P and r into the formula for continuous compounding interest to create an equation for A in terms of t:

`\implies A=18000e^(0.015t)`

To complete the table, substitute each given value of t into the equation.

`\begin{aligned}t=10 \implies A&=18000e^(0.015(10))\\&=18000e^(0.15)\\&=\$20913.02\; \; \sf (2\; d.p.)\end{aligned}`

`\begin{aligned}t=20 \implies A&=18000e^(0.015(20))\\&=18000e^(0.3)\\&=\$24297.46\; \; \sf (2\; d.p.)\end{aligned}`

`\begin{aligned}t=30 \implies A&=18000e^(0.015(30))\\&=18000e^(0.45)\\&=\$28229.62\; \; \sf (2\; d.p.)\end{aligned}`

`\begin{aligned}t=40 \implies A&=18000e^(0.015(40))\\&=18000e^(0.6)\\&=\$32798.14\; \; \sf (2\; d.p.)\end{aligned}`

`\begin{aligned}t=50 \implies A&=18000e^(0.015(50))\\&=18000e^(0.75)\\&=\$38106.00\; \; \sf (2\; d.p.)\end{aligned}`

Completed table:

`\begin{array}c\cline{1-2} \vphantom{\frac12} t& A\\\cline{1-2} \vphantom{\frac12} 10& \$20913.02 \\\cline{1-2} \vphantom{\frac12} 20& \$24297.46 \\\cline{1-2} \vphantom{\frac12} 30& \$28229.62 \\\cline{1-2} \vphantom{\frac12} 40& \$32798.14 \\\cline{1-2} \vphantom{\frac12} 50& \$38106.00 \\\cline{1-2} \end{array}`