Step 1
Plot a graph of t representing the years where the first year 1997 is taken as 0 and other years follow consecutively against the spending in billions of dollars.
From the graph, the quadratic model that best fits the data with all coefficients rounded to 2 decimal places is
![P_t=-0.05t^2_1+0.91t_1+6.22](https://img.qammunity.org/2023/formulas/mathematics/college/s0mcgocp3nrmidvjvw9ez4mawtqa0h0oxi.png)
Step 2
Based on the model, find how much you would expect to be spent on movie theater tickets in 2008
![\begin{gathered} t_1=11 \\ P_t=-0.05(11)^2+0.91(11)+6.22 \\ P_t=-(121)/(20)+(1001)/(100)+6.22 \\ P_t=(509)/(50) \\ P_t=10.18\text{ billions of dollars} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/swu8u2giqxztst5osmdk5nvdzvecyv2qvv.png)
Step 3
Find when you would expect movie theater tickets to fall to $5 billion dollars.
![\begin{gathered} P_t=5 \\ 5=-0.05t^2+0.91t+6.22 \\ 5(\: 100)=-0.05t^2(\: 100)+0.91t(\: 100)+6.22(100) \\ 500=-5t^2+91t+622 \\ -5t^2+91t+622=500 \\ -5t^2+91t+622-500=500-500 \\ -5t^2^{}+91t+122=0 \\ Solve\text{ with the quadratic formula} \\ t_(1,\: 2)=\frac{-91\pm\sqrt[]{91^2-4\mleft(-5\mright)(\: 122)}}{2\mleft(-5\mright)} \\ \sqrt[]{91^2-4\mleft(-5\mright)(\: 122})=√(10721) \\ t_1=(-91+√(10721))/(2\left(-5\right)),\: t_2=(-91-√(10721))/(2\left(-5\right)) \\ t_1=-(-91+√(10721))/(10),\: t_2=(91+√(10721))/(10) \\ t_1=-1.25422619,t_2=19.45422619 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hl0t8sw0xb6fuwtxu00oq1pcr1i0d51f18.png)
Since the value of t cannot be negative t is taken as 19.45422619
This means movie theater tickets will fall to $5 billion dollars in 19.45422619 years as required 19.45422619 years as a calendar year based on the question is during the year 2016
Answer to part C; During the year 2016