An investment is advertised as returning 3.1% every 3 months (quarterly), compounded quarterly. If $30,000 is invested, the growth can be modeled by the equation A (t) = 30,000 (1.031) 4t. What is the equivalent annual growth rate for this investment (rounded to the nearest hundredth of a percent) and what is it worth (rounded to the nearest thousand dollar) after 15 years?

Hint: Find the value of 1.0314 on your calculator.

12.99% and $187,000

12.47% and $173,000

7.82% and $187,000

9.37% and $43,000

Question

Mathematics

Autumn Benitez

16 August, 17:36

An investment is advertised as returning 3.1% every 3 months (quarterly), compounded quarterly. If $30,000 is invested, the growth can be modeled by the equation A (t) = 30,000 (1.031) 4t. What is the equivalent annual growth rate for this investment (rounded to the nearest hundredth of a percent) and what is it worth (rounded to the nearest thousand dollar) after 15 years?

Hint: Find the value of 1.0314 on your calculator.

12.99% and $187,000

12.47% and $173,000

7.82% and $187,000

9.37% and $43,000

+5

Answers (1)

Know the Answer?

Keshawn Mathis · Answer 1

The answer would be 12.99% and $187,000.

Solution:

(1.031^4 (15) = (1+r) ^15

60 log 1.031 = 15 log (1 + r)

4log (1.031) = log (1 + r)

10^[4log (1.031) ] = r + 1

10^[4log (1.031) ] - 1 =

r = 0.1299

= about 12.99%