a deposit of $40,000 is placed in a scholarship fund that earns an annual interest rate of 3.25% compounded…

Explanation:

Step1: Identify compound - interest formula

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

Step2: Convert values to appropriate form

Given $P = 40000$, $r=0.0325$ (since $3.25%=0.0325$), $n = 365$ (compounded daily), and $t = 3$.

Step3: Substitute values into formula

$A=40000(1 +\frac{0.0325}{365})^{365\times3}$. First, calculate $\frac{0.0325}{365}\approx0.0000890411$. Then $1+\frac{0.0325}{365}=1 + 0.0000890411=1.0000890411$. Next, $365\times3 = 1095$. So $A = 40000\times(1.0000890411)^{1095}$.

Step4: Calculate the final value

$(1.0000890411)^{1095}\approx1.100777$. Then $A=40000\times1.100777 = 44031.08$.

Answer:

$44031.08$