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$3000 are invested in a bank account at an interest rate of 9 percent per year.
find the amount in the bank after 8 years if interest is compounded annually.
5977.68
find the amount in the bank after 8 years if interest is compounded quarterly.
6091.13
find the amount in the bank after 8 years if interest is compounded monthly.
6115.14
finally, find the amount in the bank after 8 years if interest is compounded continuously.
6163.30
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Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is \(A = P(1+\frac{r}{n})^{nt}\), where \(P\) is the principal amount (\(P = 3000\)), \(r\) is the annual interest rate (as a decimal, \(r=0.09\)), \(t\) is the number of years (\(t = 8\)), and \(n\) is the number of times interest is compounded per year.

Step2: Calculate for annual compounding (\(n = 1\))

Substitute \(P = 3000\), \(r=0.09\), \(t = 8\), \(n = 1\) into the formula:
\(A=3000(1 +\frac{0.09}{1})^{1\times8}=3000(1.09)^{8}\)
\((1.09)^{8}\approx1.99256\)
\(A = 3000\times1.99256=5977.68\)

Step3: Calculate for quarterly compounding (\(n = 4\))

Substitute \(P = 3000\), \(r = 0.09\), \(t = 8\), \(n = 4\) into the formula:
\(A=3000(1+\frac{0.09}{4})^{4\times8}=3000(1 + 0.0225)^{32}\)
\((1.0225)^{32}\approx2.03044\)
\(A=3000\times2.03044 = 6091.32\approx6091.32\)

Step4: Calculate for monthly compounding (\(n = 12\))

Substitute \(P = 3000\), \(r=0.09\), \(t = 8\), \(n = 12\) into the formula:
\(A=3000(1+\frac{0.09}{12})^{12\times8}=3000(1+\frac{0.0075})^{96}\)
\((1.0075)^{96}\approx2.03878\)
\(A=3000\times2.03878=6116.34\)

Step5: Recall continuous - compounding formula

The continuous - compounding formula is \(A=Pe^{rt}\), where \(P = 3000\), \(r = 0.09\), \(t = 8\)
\(A=3000e^{0.09\times8}=3000e^{0.72}\)
\(e^{0.72}\approx2.05443\)
\(A=3000\times2.05443 = 6163.29\approx6163.30\)

Answer:

The amount for annual compounding is \(5977.68\), for quarterly compounding is \(6091.32\), for monthly compounding is \(6116.34\), and for continuous compounding is \(6163.30\)