determine the effective annual yield for each investment. then select the better investment. 3.95%…

Explanation:

Step1: Recall effective - annual - yield formula

The formula for the effective annual yield (EAR) is $EAR=(1 + \frac{r}{n})^{n}-1$, where $r$ is the annual interest rate (in decimal form) and $n$ is the number of compounding periods per year.

Step2: Calculate EAR for the first investment

The first investment has an annual interest rate $r_1 = 0.0395$ and is compounded semiannually, so $n_1=2$. $EAR_1=(1+\frac{0.0395}{2})^{2}-1=(1 + 0.01975)^{2}-1=1.01975^{2}-1=1.04028 - 1=0.04028\approx4.03%$

Step3: Calculate EAR for the second investment

The second investment has an annual interest rate $r_2 = 0.0394$ and is compounded monthly, so $n_2 = 12$. $EAR_2=(1+\frac{0.0394}{12})^{12}-1$. Let $x=\frac{0.0394}{12}\approx0.003283$. Then $(1 + 0.003283)^{12}-1\approx1.0399 - 1=0.0399\approx3.99%$

Step4: Compare the effective - annual yields

Since $4.03%>3.99%$, the first investment is better.

Answer:

A. The first investment, 3.95% compounded semiannually, is the better investment because the effective annual yield for the first investment is $4.03%$ and the effective annual yield for the second investment is $3.99%$.