The table below shows the average annual cost of health insurance for a single individual, from 1999 to 2019, according to the Kaiser Family Foundation.YearCost1999$2,1962000$2,4712001$2,6892002$3,0832003$3,3832004$3,6952005$4,0242006$4,2422007$4,4792008$4,7042009$4,8242010$5,0492011 $5,4292012$5,6152013$5,8842014$6,0252015$6,2512016$6,1962017$6,4352017$6,8962019$7,186(a) Using only the data from the first and last years, build a linear model to describe the cost of individual health insurance from 1999 onward. Use t to represent years after 1999 (treating 1999 as year 0).Pt = (b) Using this linear model, predict the cost of insurance in 2030.$ (c) = According to this model, when do you expect the cost of individual insurance to reach $12,000? Give your answer as a calendar year (ex: 2020)..

[tex]\begin{gathered} 1999\rightarrow0\rightarrow2196 \\ 2019\rightarrow20\rightarrow7186 \\ \text{Havng} \\ x_1=0,y_1=2196 \\ x_2=20,y_2=7186 \\ \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}_{} \\ \text{The model will be:} \\ P_t=249.5t+2196 \end{gathered}[/tex]

b) The cost of insurance in 2030

[tex]\begin{gathered} P_t=249.5t+2196 \\ t=2030-1999=31 \\ \text{The cost of insurance in 2030 therefore will be:} \\ =249.5(31)+2196 \\ =7734.5+2196 \\ =\text{ \$9930.5} \end{gathered}[/tex]

c) When do we expect the cost to reach $12,000

[tex]\begin{gathered} P_t=249.5t+2196 \\ 12,000=249.5t+2196 \\ 12000-2196=249.5t \\ 9804=249.5t \\ \frac{9804}{249.5}=\frac{249.5t}{249.5} \\ 39.2946=t \\ Since\text{ t = year -1999} \\ 39.2946+1999=\text{year} \\ 2038.2946=\text{year} \\ Since\text{ we are to give our answer as an exact year} \\ \text{The year will be }2039. \end{gathered}[/tex]