Asked by Brad
Larry has $2600 to invest and needs $3000 in 11 years. What annual rate of return will he need to get in order to accomplish his goal, if interest is compounded continuously?
Answers
Answered by
Reiny
let
2600(e^(11i) = 3000
e^(11i) = 1.153846
11i = ln(1.153846)
i = .013
a mere 1.3%
2600(e^(11i) = 3000
e^(11i) = 1.153846
11i = ln(1.153846)
i = .013
a mere 1.3%
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