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At the beginning of each year Jane puts $2,000 in
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30 Jul 2017, 08:24
30 Jul 2017, 08:24
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42% (02:07) correct
57% (01:28) wrong based on 110 sessions
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At the beginning of each year Jane puts $2,000 in an account that earns 6 % interest, compounded annually. To the nearest dollar, how much money will Jane have in the account at the end of 4 years if she makes no withdrawals?
Round your answer to the nearest whole dollar: $
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$ 9,274
Re: At the beginning of each year Jane puts $2,000 in
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28 Sep 2017, 06:31
28 Sep 2017, 06:31
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Given the formula for the compound interest rate \(A=P(1+\frac{r}{n})^{nt}\), where P is the sum invested, A is the amount after interests, r is the interest rate and n is the number of times the interests are compounded in a year, the solution is pretty straightforward.
The interests are compounded once a year so that the formula reduces to \(A=P(1+0.06)^t\). Then, we just have to compute the amount every year, remembering of adding 2000$ at the beginning of every new year.
Thus, the first year we will have 2000(1+0.06)^1=2120
The second year we will have (2120+2000)*(1+0.06)^1=4367.2
And so on for other two years until we reach our answer, $ 9,274
The interests are compounded once a year so that the formula reduces to \(A=P(1+0.06)^t\). Then, we just have to compute the amount every year, remembering of adding 2000$ at the beginning of every new year.
Thus, the first year we will have 2000(1+0.06)^1=2120
The second year we will have (2120+2000)*(1+0.06)^1=4367.2
And so on for other two years until we reach our answer, $ 9,274
Re: At the beginning of each year Jane puts $2,000 in
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26 May 2019, 09:26
26 May 2019, 09:26
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so at the end of 1st year =2000*1.06=2120
so at the end of 2nd year = 4120*1.06= 4367.2
so at the end of 3rd year = 6367.2*1.06= 6749.232
so at the end of 4th year = 8749.232*1.06= 9274.18592
so our answer is 9274 $
so at the end of 2nd year = 4120*1.06= 4367.2
so at the end of 3rd year = 6367.2*1.06= 6749.232
so at the end of 4th year = 8749.232*1.06= 9274.18592
so our answer is 9274 $
