+109 An investment is advertised as returning 0.6% every month, compounded monthly. If $10,000 is invested, the growth can be modeled by the equation A(t) = 10,000(1.006)^12t. What is the equivalent annual growth rate for this investment (rounded to the nearest tenth of a percent) and what is it worth (rounded to the nearest thousand dollar) after 20 years?
Hint: Find the value of 1.006^12 on your calculator.
A.) 5.0% and $42,000
B.) 7.2%, and $11,000
C.) 7.4% and $42,000
D.) 7.2% and $40,000
please help im bad at this :(
+129852 We need to solve this, stephdp......
(1.006)^(12 *20) = (1 + r)^20
(1.006)^240 = (1 + r)^20 take the log of both sides
log (1.006)^240 = log (1 + r)^20 and we can write
240*log(1.006) = 20 log(1 + r) divide both sides by 20
12*log(1.006) = log(1 + r)
And this says that
10^[12*log(1.006)] = 1 + r subtract 1 from both sides
10^[12*log(1.006)] - 1 = r = about 7.4%
And in 20 years, we will have
10000(1 + .074)^20 = about $41, 695 [round to $ 42000]
So....it looks like (C) is the correct choice
