Question

you have $4,500 to invest. which plan would generate the most interest after three years? (1 point) 5.2% compounded semi - annually 4.8% compounded quarterly 4.8% compounded monthly 5.1% compounded yearly

Explanation:

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where:

• $A$ is the amount of money accumulated after $n$ years, including interest.
• $P$ is the principal amount (the initial amount of money). Here, $P=\$4500$.
• $r$ is the annual interest rate (in decimal form).
• $n$ is the number of times that interest is compounded per year.
• $t$ is the time the money is invested for in years. Here, $t = 3$ years.

The interest earned $I=A - P$. We will calculate the amount $A$ for each option and then find the interest.

Step 1: Option 1: 5.2% compounded semi - annually

First, convert the percentage to a decimal: $r = 0.052$. Since it is compounded semi - annually, $n = 2$.
Using the compound - interest formula:
\[

$$\begin{align*} A&=4500(1+\frac{0.052}{2})^{2\times3}\\ &=4500(1 + 0.026)^{6}\\ &=4500\times(1.026)^{6} \end{align*}$$

\]
Calculate $(1.026)^{6}\approx1.026\times1.026\times1.026\times1.026\times1.026\times1.026\approx1.1665$
$A\approx4500\times1.1665=\$5249.25$
Interest $I_1=5249.25 - 4500=\$749.25$

Step 2: Option 2: 4.8% compounded quarterly

Convert the percentage to a decimal: $r = 0.048$. Since it is compounded quarterly, $n = 4$.
\[

$$\begin{align*} A&=4500(1+\frac{0.048}{4})^{4\times3}\\ &=4500(1 + 0.012)^{12}\\ \end{align*}$$

\]
Calculate $(1.012)^{12}\approx1.012^{12}\approx1.1539$
$A\approx4500\times1.1539=\$5192.55$
Interest $I_2=5192.55 - 4500=\$692.55$

Step 3: Option 3: 4.8% compounded monthly

Convert the percentage to a decimal: $r = 0.048$. Since it is compounded monthly, $n = 12$.
\[

$$\begin{align*} A&=4500(1+\frac{0.048}{12})^{12\times3}\\ &=4500(1+0.004)^{36}\\ \end{align*}$$

\]
Calculate $(1.004)^{36}\approx1.1549$
$A\approx4500\times1.1549=\$5197.05$
Interest $I_3=5197.05 - 4500=\$697.05$

Step 4: Option 4: 5.1% compounded yearly

Convert the percentage to a decimal: $r = 0.051$. Since it is compounded yearly, $n = 1$.
\[

$$\begin{align*} A&=4500(1+\frac{0.051}{1})^{1\times3}\\ &=4500(1 + 0.051)^{3}\\ &=4500\times(1.051)^{3} \end{align*}$$

\]
Calculate $(1.051)^{3}=1.051\times1.051\times1.051\approx1.1603$
$A\approx4500\times1.1603=\$5221.35$
Interest $I_4=5221.35 - 4500=\$721.35$

Now, compare the interests:

• $I_1=\$749.25$
• $I_2=\$692.55$
• $I_3=\$697.05$
• $I_4=\$721.35$

Since $749.25>721.35 > 697.05>692.55$, the plan with 5.2% compounded semi - annually generates the most interest.

Answer:

5.2% compounded semi - annually