Math Problem Statement

Please give me the correct answer with accurate calculation.

You must round to the nearest cent as​ needed!

Solution

To find the future value of an ordinary annuity, we use the formula:

FV=PMT×((1+r)n1r)FV = PMT \times \left( \frac{(1 + r)^n - 1}{r} \right)

Where:

  • PMTPMT is the periodic payment
  • rr is the interest rate per period
  • nn is the number of periods

Given:

  • Periodic payment (PMTPMT) = $1,000
  • Annual interest rate = 5.5%
  • Number of periods (years) = 9

First, we find the factor from the table for 9 periods and 5.5% interest rate.

From the table provided in the image, for 9 periods at 5.5% interest rate:

  • The factor is 11.02711.027.

Using this factor, we can find the future value (FV) by multiplying the periodic payment by this factor:

FV=PMT×factorFV = PMT \times \text{factor} FV=1,000×11.027FV = 1,000 \times 11.027 FV=11,027FV = 11,027

Next, we calculate the total interest earned by subtracting the total payments from the future value:

Total payments = PMT×nPMT \times n Total payments=1,000×9=9,000\text{Total payments} = 1,000 \times 9 = 9,000

Total interest = Future value - Total payments Total interest=11,0279,000\text{Total interest} = 11,027 - 9,000 Total interest=2,027\text{Total interest} = 2,027

Answers:

  • The future value of the ordinary annuity is $11,027.
  • The total interest of the annuity is $2,027.

Would you like more details or have any questions? Here are some additional questions you might find interesting:

  1. How is the present value of an ordinary annuity calculated?
  2. What is the difference between an ordinary annuity and an annuity due?
  3. How do varying interest rates affect the future value of an annuity?
  4. What are the implications of compound interest on long-term investments?
  5. How do different payment frequencies impact the future value of an annuity?

Tip: When working with financial tables, ensure you correctly identify the period and interest rate to find the correct factor for accurate calculations.