Question

NO LINKS!! How much money, invested at an interest rate of r% per year compounded continuously, will amount to A dollars after t years? (Round your answer to the nearest cent.) Part 3v​

Answer 1

$10,433.87 (nearest cent)

Explanation:

Continuous Compounding Formula

\large \text{$ \sf A=Pe^{rt} $}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).

Given values:

A = $14,000

r = 4.9% = 0.049

t = 6 years

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

\implies \sf 14000=P \cdot e^{0.049 \cdot 6}⟹14000=P⋅e

0.049⋅6

\implies \sf 14000=P \cdot e^{0.288}⟹14000=P⋅e

0.288

\implies \sf P=\dfrac{14000}{e^{0.288}}⟹P=

e

0.288

14000

\implies \sf P=10433.8708...⟹P=10433.8708...

Therefore, the principal amount invested was $10,433.87 (nearest cent).

Answer 2

$10,433.87 (nearest cent)

Explanation:

Continuous Compounding Formula

`\large \text{$ \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).

Given values:

• A = $14,000
• r = 4.9% = 0.049
• t = 6 years

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

`\implies \sf 14000=P \cdot e^(0.049 \cdot 6)`

`\implies \sf 14000=P \cdot e^(0.288)`

`\implies \sf P=(14000)/(e^(0.288))`

`\implies \sf P=10433.8708...`

Therefore, the principal amount invested was $10,433.87 (nearest cent).