A $10,000 deposit earns 4.8% interest for 3 years. What is the final account balance if the interest is compounded semiannually? Quarterly? Monthly?
A $10,000 deposit earns 4.8% interest for 3 years. What is the final account balance if the interest is compounded semiannually? Quarterly? Monthly?
The balance at semi-annual compound =$11,529.22
The balance at quarterly compound =$11,538.95
The balance of monthly compound =$11,545.52
+1904 To get the three answers, you need to know what the interest rate formula is. For this math problem, the interest rate formula is:
\(A=P{(1+\frac{r}{n})}^{nt}\)
A = amount of money earned after investment
P = amount of money invested
r = interest rate per period
n = number of times the interest is compounded each year
t = number of years the amount of money is deposited for
\(A=$10,000{(1+\frac{0.048}{2})}^{2\times3}\)
\(A=$10,000{(1+\frac{0.048}{2})}^{6}\)
\(A=$10,000{(1+0.024)}^{6}\)
\(A=$10,000{(1.024)}^{6}\)
\(A= $11,529.21504606846976\)
\(A≈ $11,529.22\)
Depositing $10,000 into an account that earns 4.8% for 3 years semiannually wil give you approximately $11,529.22.
\(A=$10,000{(1+\frac{0.048}{4})}^{4\times3}\)
\(A=$10,000{(1+\frac{0.048}{4})}^{12}\)
\(A=$10,000{(1+0.012)}^{12}\)
\(A=$10,000{(1.012)}^{12}\)
\(A=$10,000(1.153894624182586)\)
\(A=$11,538.94624182586\)
\(A≈$11,538.95\)
Depositing $10,000 into an account that earns 4.8% for 3 years quarterly wil give you approximately $11,538.95.
\(A=$10,000{(1+\frac{0.048}{12})}^{12\times3}\)
\(A=$10,000{(1+\frac{0.048}{12})}^{36}\)
\(A=$10,000{(1+0.004)}^{36}\)
\(A=$10,000{(1.004)}^{36}\)
\(A=$10,000(1.1545524338367405)\)
\(A= $11,545.524338367405\)
\(A≈$11,545.52\)
Depositing $10,000 into an account that earns 4.8% for 3 years monthlty wil give you approximately $11,545.52.
