Answer:
[tex]\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} t& A\\\cline{1-2} \vphantom{\dfrac12} 10& \$20913.02 \\\cline{1-2} \vphantom{\dfrac12} 20& \$24297.46 \\\cline{1-2} \vphantom{\dfrac12} 30& \$28229.62 \\\cline{1-2} \vphantom{\dfrac12} 40& \$32798.14 \\\cline{1-2} \vphantom{\dfrac12} 50& \$38106.00 \\\cline{1-2} \end{array}[/tex]
Step-by-step explanation:
[tex]\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}}[/tex]
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:
[tex]\implies A=18000e^{0.015t}[/tex]
To complete the table, substitute each given value of t into the equation.
[tex]\begin{aligned}t=10 \implies A&=18000e^{0.015(10)}\\&=18000e^{0.15}\\&=\$20913.02\; \; \sf (2\; d.p.)\end{aligned}[/tex]
[tex]\begin{aligned}t=20 \implies A&=18000e^{0.015(20)}\\&=18000e^{0.3}\\&=\$24297.46\; \; \sf (2\; d.p.)\end{aligned}[/tex]
[tex]\begin{aligned}t=30 \implies A&=18000e^{0.015(30)}\\&=18000e^{0.45}\\&=\$28229.62\; \; \sf (2\; d.p.)\end{aligned}[/tex]
[tex]\begin{aligned}t=40 \implies A&=18000e^{0.015(40)}\\&=18000e^{0.6}\\&=\$32798.14\; \; \sf (2\; d.p.)\end{aligned}[/tex]
[tex]\begin{aligned}t=50 \implies A&=18000e^{0.015(50)}\\&=18000e^{0.75}\\&=\$38106.00\; \; \sf (2\; d.p.)\end{aligned}[/tex]
Completed table:
[tex]\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} t& A\\\cline{1-2} \vphantom{\dfrac12} 10& \$20913.02 \\\cline{1-2} \vphantom{\dfrac12} 20& \$24297.46 \\\cline{1-2} \vphantom{\dfrac12} 30& \$28229.62 \\\cline{1-2} \vphantom{\dfrac12} 40& \$32798.14 \\\cline{1-2} \vphantom{\dfrac12} 50& \$38106.00 \\\cline{1-2} \end{array}[/tex]
