a) A model of the provided data in the form y=ab^{x}. b) The function

a) Given:
The monthly costs for a small company to do business has been increasing over time due to inflation. The below table gives the monthly cost y(in $) for the month of January for selected years. The variable t represents the number of years since 2008.
$\begin{array}{|cc|}\hline Year(t=0is2008)& Monthlycosts(\$)y\\ 0& 12,000\\ 1& 12,400\\ 2& 12,800\\ 3& 13,300\\ \hline\end{array}$
Explanation:
Consider the provided table below:
$\begin{array}{|cc|}\hline Year(t=0is2008)& Monthlycosts(\$)y\\ 0& 12,000\\ 1& 12,400\\ 2& 12,800\\ 3& 13,300\\ \hline\end{array}$
Use graphic utility calculator Ti-83 for modeling an equation of provided data:
Step1: Press STAT key then press 1 in Ti-83 calculator. A table will arise. Fill above values of years and monthly cost in that table.
Step2: After filling the above data in table press CALC key and then press 0 to select ExpReg to calculate the model function of the provided data.
The regression equation is ${\displaystyle y=11988{\left(1.034\right)}^t}$
b) Given: The function obtained in part (a) is:
${\displaystyle y=11988{\left(1.034\right)}^t}$
Formula used:
Calculation:
Consider the function obtained in part (a):
${\displaystyle y=11988{\left(1.034\right)}^t}$
Apply the property that ${\displaystyle e^{\left(\ln b\right)t}=b^t}$
${\displaystyle y={11988}^{\left(\ln 1.034\right)t}}$
${\displaystyle y=11988e^{0.034t}}$
The equivalent function of ${\displaystyle y=11988{\left(1.034\right)}^t}$ using base e is ${\displaystyle y=11988e^{0.034t}}$
c) Given: The function obtained in part (a) is, ${\displaystyle y=11988{\left(1.034\right)}^t}$ and the year 2015.
Calculation:
Consider the provided function:
${\displaystyle y=11988{\left(1.034\right)}^t}$
Substitute the value ${\displaystyle t=7}$.
${\displaystyle y=11988{\left(1.034\right)}^7}$
${\displaystyle \approx 15,150}$
The monthly cost for January in the year 2015 would be approximately $15,150.