an investment of $1,000 is earning interest at the rate of 3.8% compounded daily over 3 years. approximately…

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula when compounded $n$ 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), $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P=$1000$, $r = 0.038$ (since $3.8%=0.038$), $n = 365$ (compounded daily), and $t = 3$.

Step2: Calculate the final amount $A$

$A=1000(1 +\frac{0.038}{365})^{365\times3}$. First, calculate the value inside the parentheses: $\frac{0.038}{365}\approx0.00010411$. Then $1+\frac{0.038}{365}=1 + 0.00010411=1.00010411$. Next, $365\times3 = 1095$. So $A = 1000\times(1.00010411)^{1095}$. Using a calculator, $(1.00010411)^{1095}\approx1.12075$. Then $A=1000\times1.12075=$1120.75$.

Step3: Calculate the interest earned

The interest earned $I=A - P$. Substitute $A = 1120.75$ and $P = 1000$ into the formula. $I=1120.75−1000=$120.75$.

Answer:

C. $$120.75$