Question

Nathaniel invested $2,900 in an account paying an interest rate of 5.4% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest cent, would be in the account after 11 years?

Answer 1

$5,252.53 (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:

• P = $2,900
• r = 5.4% = 0.054
• t = 11 years

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

`\implies \sf A=2900 \cdot e^((0.054 \cdot 11))`

`\implies \sf A=2900 \cdot e^(0.594)`

`\implies \sf A=2900 \cdot 1.81121882`

`\implies \sf A=5252.53457...`

Therefore, assuming no deposits or withdrawals are made, the amount of money in the account after 11 years would be $5,252.53 (nearest cent).

Answer 2

• $5252.53

============================

Given

• Invested amount P = $2900,
• Annual interest rate r = 5.4% = 0.054,
• Time t = 11 years,
• Compound number = continuous.

Find the balance after 11 years

Use equation for continuous compound:

• 
  `P(t) = P_0e^(tr)`,
• where P(t) - final amount, P₀ - initial amount, t - time, r - interest rate

Plug in the values and calculate:

• 
  `P(11) = 2900e^(11*0.054)=5252.53 \ rounded`