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AGummiBear55
How do I use the formula 6000=2884.25(1+1.9%/12)^(12)(t)... how do I put 38 years and 7 months into the equation?
It depends on what t is supposed to be in the formula. If t is in months: put 38*12+7= 463 in. If t is in years: 38+7/12 = 38.58333 in.
Here is the question: Chloe deposited $2,884.25 into a savings account with an interest rate of 1.9% compounded monthly. About how long will it take for the account to be worth $6,000? A) 9 years, 1 month B) 21 years, 8 months C) 38 years, 7 months D) 38 years, 11 months I think the answer is 38 years, 7 months... but I'm not sure and I don't know how to plug in 38 years, 7 months into the equation to make sure I have the right answer.
The formula to calculate the the amount is\[A(t)=2884.25\left(1+\frac{ 0.019 }{ 12 }\right)^{12t}\]Because of the "12t" I think you have to put in years:\[A(38.58333)=(calculator)=5994.36\]It is easier to replace the "12t"in the formula by the number of months = 463:\[2884.25\left(1+\frac{ 0.019 }{ 12} \right)^{463}=6000.054\approx 6000\]
So that would mean the answer is 38 years, 7 months... correct? (: