1. **Stating the problem:** We have two options for total cost depending on the number of visits. Option 1 increases linearly from about $30 at 0 visits to $100 at 14 visits. Option 2 is constant at $70 regardless of visits. 2. **Formulas and interpretation:** - Option 1 cost can be modeled as a linear function: $$y_1 = mx + b$$ where $x$ is the number of visits. - From the graph, at $x=0$, $y_1=30$, so $b=30$. - At $x=14$, $y_1=100$, so slope $m = \frac{100-30}{14-0} = \frac{70}{14} = 5$. - Thus, Option 1 cost: $$y_1 = 5x + 30$$. - Option 2 cost is constant: $$y_2 = 70$$. 3. **Check statement A:** For 4 visits, Option 2 is $15 cheaper than Option 1. - Calculate Option 1 cost at $x=4$: $$y_1 = 5(4) + 30 = 20 + 30 = 50$$. - Option 2 cost is $70$. - Difference: $$70 - 50 = 20$$, so Option 2 is actually $20 more expensive, not $15 cheaper. 4. **Check statement B:** For fewer than 8 visits, Option 2 is cheaper than Option 1. - Find where costs are equal: $$5x + 30 = 70$$. - Solve for $x$: $$5x = 70 - 30$$ $$5x = 40$$ $$x = \frac{40}{5} = 8$$. - For $x < 8$, Option 1 cost is less than 70, so Option 2 is more expensive, not cheaper. 5. **Check statement C:** For 11 visits, Option 1 is $10 more expensive than Option 2. - Calculate Option 1 cost at $x=11$: $$y_1 = 5(11) + 30 = 55 + 30 = 85$$. - Option 2 cost is $70$. - Difference: $$85 - 70 = 15$$, so Option 1 is $15 more expensive, not $10. 6. **Check statement D:** For more than 7 visits, Option 1 is more expensive than Option 2. - Since costs are equal at $x=8$, for $x > 8$, Option 1 cost is greater than 70. - For $7 < x < 8$, Option 1 cost is less than 70. - So for more than 7 visits (meaning $x > 7$), Option 1 is more expensive only after $x=8$, not immediately after 7. **Final answers:** - A is false. - B is false. - C is false. - D is false if interpreted strictly; Option 1 becomes more expensive only after 8 visits. Hence, none of the statements are exactly correct as given.