Question

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If P dollars is deposited in a savings account that pays interest at a rate r% per year compounded continuously, find the balance after t years. (Round your answer to the nearest cent)

P = 60, r = 4 1/2, t = 10​

Answer 1

$90.10 (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 = $60
• r = 4.5% = 0.045
• t = 10 years

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

`\implies \sf A=60 \cdot e^(0.045 \cdot 10)`

`\implies \sf A=60 \cdot e^(0.45)`

`\implies \sf A=60 \cdot 1.56831218...`

`\implies \sf A=94.0987311...`

Therefore, the balance of the account after 10 years is $90.10 (nearest cent).

Answer 2

• $94.09

Explanation:

Use continuous compound equation:

• 
  `A = P*e^(rt)`, where A- future amount, P- invested amount, t- time, r- rate

Given

• P = $60,
• r = 4 1/2% = 4.5% = 0.045,
• t = 10 years.

Plug in and calculate

• 
  `A = 60*e^(0.045*10)=60*e^(0.45)=94.09`