Question

you have $4,500 to invest. which plan would generate the most interest after three years? (1 point) 5.0% compounded semi - annually 4.9% compounded quarterly 4.8% compounded monthly 5.1% compounded yearly

Explanation:

To determine which investment plan generates the most interest, we use the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. The interest earned $I$ is $A - P$. Here, $P=\$4500$ and $t = 3$ years.

Step 1: Analyze Plan 1: 5.0% compounded semi - annually

• First, convert the percentage to a decimal: $r = 0.05$.
• Since it is compounded semi - annually, $n=2$.
• Substitute into the formula: $A_1=4500(1 +\frac{0.05}{2})^{2\times3}=4500(1 + 0.025)^{6}$.
• Calculate $(1 + 0.025)^{6}\approx1.159693$.
• Then $A_1 = 4500\times1.159693\approx5218.62$.
• The interest $I_1=A_1 - 4500=5218.62 - 4500 = 718.62$.

Step 2: Analyze Plan 2: 4.9% compounded quarterly

• Convert the percentage to a decimal: $r = 0.049$.
• Since it is compounded quarterly, $n = 4$.
• Substitute into the formula: $A_2=4500(1+\frac{0.049}{4})^{4\times3}=4500(1 + 0.01225)^{12}$.
• Calculate $(1 + 0.01225)^{12}\approx1.15647$.
• Then $A_2=4500\times1.15647\approx5204.12$.
• The interest $I_2=A_2 - 4500=5204.12 - 4500 = 704.12$.

Step 3: Analyze Plan 3: 4.8% compounded monthly

• Convert the percentage to a decimal: $r=0.048$.
• Since it is compounded monthly, $n = 12$.
• Substitute into the formula: $A_3=4500(1+\frac{0.048}{12})^{12\times3}=4500(1 + 0.004)^{36}$.
• Calculate $(1 + 0.004)^{36}\approx1.15389$.
• Then $A_3=4500\times1.15389\approx5192.51$.
• The interest $I_3=A_3 - 4500=5192.51 - 4500=692.51$.

Step 4: Analyze Plan 4: 5.1% compounded yearly

• Convert the percentage to a decimal: $r = 0.051$.
• Since it is compounded yearly, $n = 1$.
• Substitute into the formula: $A_4=4500(1+\frac{0.051}{1})^{1\times3}=4500(1 + 0.051)^{3}$.
• Calculate $(1 + 0.051)^{3}\approx1.159376$.
• Then $A_4=4500\times1.159376\approx5217.19$.
• The interest $I_4=A_4 - 4500=5217.19 - 4500 = 717.19$.

Answer:

By comparing the interests: $I_1\approx718.62$, $I_2\approx704.12$, $I_3\approx692.51$, $I_4\approx717.19$. The plan with 5.0% compounded semi - annually generates the most interest. So the answer is 5.0% compounded semi - annually.