maria takes out a loan of $5000 to purchase some new equipment for her business. the loan accrues interest…

Explanation:

Step1: Recall compound - interest formula

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), $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested or borrowed for in years. In Maria's case, $P = 5000$, $r=0.09$ (since $9%=0.09$), and $n = 12$ (compounded monthly). So the equation that models the total amount Maria will owe on the loan over time $t$ is $A(t)=5000(1 +\frac{0.09}{12})^{12t}$. Here, $5000$ is the initial loan amount, $\frac{0.09}{12}$ is the monthly interest rate, $12$ is the number of compounding periods per year, and $t$ is the number of years.

Step2: Calculate the value of $(1+\frac{r}{n})$ for part (b)

First, calculate $1+\frac{r}{n}=1+\frac{0.09}{12}=1 + 0.0075=1.0075$. This is the growth factor per compounding period.

Step3: Calculate the amount owed after 1 year

Substitute $t = 1$, $P = 5000$, $r=0.09$, and $n = 12$ into the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$. $A=5000(1+\frac{0.09}{12})^{12\times1}=5000\times(1.0075)^{12}$. $(1.0075)^{12}\approx1.093807$. $A = 5000\times1.093807=5469.035\approx5469.04$.

Answer:

The equation is $A(t)=5000(1+\frac{0.09}{12})^{12t}$. The $b -$value (the growth factor per compounding period) is $1.0075$. The amount Maria will owe after one year is $$5469.04$.