Question

Use the compound interest formulas A=P e^rt to solve the problem given. round answers to the nearest cent.

Answer 1

Answer:

a) $20,771.76

b) $20,817.67

c) $20,484.80

d) $20,864.52

Step-by-step explanation:

Compound Interest Formula

$\large \sf A=P\left(1+\frac{r}{n}\right)^{nt}$

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Part (a): semiannually

Given:

• P = $15,000
• r = 5.5% = 0.055
• n = 2
• t = 6 years

Substitute the given values into the formula and solve for A:

$\implies \sf A=15000\left(1+\frac{0.055}{2}\right)^{2 \times 6}$

$\implies \sf A=15000\left(1.0275}{2}\right)^{12}$

$\implies \sf A=20771.76$

Part (b): quarterly

Given:

• P = $15,000
• r = 5.5% = 0.055
• n = 4
• t = 6 years

Substitute the given values into the formula and solve for A:

$\implies \sf A=15000\left(1+\frac{0.055}{4}\right)^{4 \times 6}$

$\implies \sf A=15000\left(1.01375}\right)^{24}$

$\implies \sf A=20817.67$

Part (c): monthly

Given:

• P = $15,000
• r = 5.5% = 0.055
• n = 12
• t = 6 years

Substitute the given values into the formula and solve for A:

$\implies \sf A=15000\left(1+\frac{0.055}{12}\right)^{12 \times 6}$

$\implies \sf A=15000\left(1+\frac{0.055}{12}\right)^{72}$

$\implies \sf A=20484.80$

Continuous Compounding Formula

$\large \sf A=Pe^{rt}$

where:

• A = Final amount
• P = Principal amount
• e = Euler's number (constant)
• r = annual interest rate (in decimal form)
• t = time (in years)

Part (d): continuous

Given:

• P = $15,000
• r = 5.5% = 0.055
• t = 6 years

Substitute the given values into the formula and solve for A:

$\implies \sf A=15000e^{0.055 \times 6}$

$\implies \sf A=20864.52$

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