11. raven deposited $5600 into a savings account in which interest is compounded daily at a rate of 3.9%…

Explanation:

Step1: Recall compound - interest formula

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

Problem 11

1. First, identify the values:
   • $P = 5600$, $r=0.039$ (since $3.9%=0.039$), $n = 365$ (compounded daily), and $t = 7$.
   • Calculate $(1+\frac{r}{n})^{nt}$:
     • $\frac{r}{n}=\frac{0.039}{365}\approx0.00010685$.
     • $nt=365\times7 = 2555$.
     • $(1+\frac{0.039}{365})^{365\times7}=(1 + 0.00010685)^{2555}$.
     • Using a calculator, $(1 + 0.00010685)^{2555}\approx1.3017$.
   • Then find $A$:
     • $A = 5600\times(1+\frac{0.039}{365})^{365\times7}\approx5600\times1.3017 = 7289.52$.
   • Calculate the interest $I$:
     • $I=A - P$.
     • $I=7289.52-5600=1689.52$.

Problem 12

1. First, convert the time to years:
   • $t=\frac{21}{12}=1.75$ years, $r = 0.033$ (since $3.3%=0.033$), $n = 4$ (compounded quarterly), and $A = 4369.20$.
2. Use the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$ to solve for $P$:
   • $(1+\frac{r}{n})^{nt}=(1+\frac{0.033}{4})^{4\times1.75}$.
     • $\frac{0.033}{4}=0.00825$.
     • $4\times1.75 = 7$.
     • $(1 + 0.00825)^{7}\approx1.0597$.
   • Since $A = P(1+\frac{r}{n})^{nt}$, then $P=\frac{A}{(1+\frac{r}{n})^{nt}}$.
     • $P=\frac{4369.20}{(1 + 0.00825)^{7}}$.
     • $P=\frac{4369.20}{1.0597}\approx4123$.

Answer:

11. $1689.52$
12. $4123$