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If $10,000 is invested at 8 percent annual interest, compounded semian
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03 Jul 2017, 17:01
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If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars? A. \(10,000(1+ \frac{0.08}{2})^{10}\) B. \(10,000(1+0.08)^5\) C. \(10,000(1+0.08)\) D. \(10,000(10.08)^2\) E. \(10,000*1.08\)
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian
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03 Jul 2017, 18:56
MathRevolution wrote: If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?
A. \(10,000(1+ \frac{0.08}{2})^{10}\) B. \(10,000(1+0.08)^5\) C. \(10,000(1+0.08)\) D. \(10,000(10.08)^2\) E. \(10,000*1.08\) Whenever an amount is compounded semiannually, the interest rate becomes half and TIME becomes two times.. So 1) Rate becomes 0.08/2 2) Time becomes 5*2=10 A is correct
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian
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03 Jul 2017, 23:51
Compound interest = \(Amount * (1 + \frac{ROI}{100})^n\)where ROI is the rate of interest n is the amount of years the amount is invested When an amount is invested and interest is calculated semiannually, the rate of interest is halved and the number of years doubles. The formula for compound interest is : Compound interest = \(Amount * (1 + \frac{ROI}{200})^{2n}\)Coming to the problem in hand, Amount : 10000$ Interest(annual)ROI : 8% = \(\frac{8}{100} = 0.08\) Number of yearsn : 5 Substituting these values, Interest = \(10,000(1+ \frac{0.08}{2})^{10}\) (Option A)
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If $10,000 is invested at 8 percent annual interest, compounded semian
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04 Jul 2017, 03:15
MathRevolution wrote: If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?
A. \(10,000(1+ \frac{0.08}{2})^{10}\) B. \(10,000(1+0.08)^5\) C. \(10,000(1+0.08)\) D. \(10,000(10.08)^2\) E. \(10,000*1.08\) The formula for compound interest can be written in a few ways. IMO, the version that matches the answer choices most closely is A = P \((1+ \frac{r}{n})^{nt}\) A= amount of money accumulated over t years, including interest (what the question asks for) P = initial amount invested r = annual rate of interest in decimal form n = the number of times the interest is compounded annually t = number of years the amount is deposited (or borrowed) for nt = # of times per year interest is compounded times # of years P = $10,000 r = .08 n = 2 (semiannually = twice per year) t = 5 (years) \(\frac{r}{n}\) = \(\frac{.08}{2}\) nt = 2*5 = 10 Using figures above, the equation becomes A = \(10,000(1+ \frac{0.08}{2})^{10}\) Answer A
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian
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05 Jul 2017, 00:03
==> if it is invested at 8 percent annual interest and is compounded semiannually, From 10,000(1+8%\((\frac{1}{2}))^{2*5}= 10,000(1+)^{10}\) the answer is A Answer: A
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian
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05 Jul 2017, 00:30
MathRevolution wrote: ==> if it is invested at 8 percent annual interest and is compounded semiannually, From 10,000(1+8%\((\frac{1}{2}))^{2*5}= 10,000(1+)^{10}\) the answer is A
Answer: A MathRevolution, your final step is incomplete. Please correct the solution!
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian
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10 Sep 2018, 06:37
generis wrote: MathRevolution wrote: If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?
A. \(10,000(1+ \frac{0.08}{2})^{10}\) B. \(10,000(1+0.08)^5\) C. \(10,000(1+0.08)\) D. \(10,000(10.08)^2\) E. \(10,000*1.08\) The formula for compound interest can be written in a few ways. IMO, the version that matches the answer choices most closely is A = P \((1+ \frac{r}{n})^{nt}\) A= amount of money accumulated over t years, including interest (what the question asks for) P = initial amount invested r = annual rate of interest in decimal form n = the number of times the interest is compounded annually t = number of years the amount is deposited (or borrowed) for nt = # of times per year interest is compounded times # of years P = $10,000 r = .08 n = 2 (semiannually = twice per year) t = 5 (years) \(\frac{r}{n}\) = \(\frac{.08}{2}\) nt = 2*5 = 10 Using figures above, the equation becomes A = \(10,000(1+ \frac{0.08}{2})^{10}\) Answer A A slightly modified formula. Just for ease of use. A = P \((1+ \frac{r}{100n})^{nt}\)




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