1. **Stating the problem:** We are given a graph showing the plumber's fees as a function of hours worked. The graph is a straight line passing through points (0,30), (3,105), and (5,150). 2. **Formula and rules:** The fee structure appears linear, so we use the linear equation formula: $$y = mx + c$$ where $y$ is the fee, $x$ is the hours worked, $m$ is the rate per hour, and $c$ is the fixed call-out fee. 3. **(i) Find the call-out fee:** The call-out fee is the fee when $x=0$ hours: From the graph, at $x=0$, $y=30$, so $$c = 30$$ 4. **(ii) Fee for 3.5 hours:** First, find the rate $m$ using points (0,30) and (3,105): $$m = \frac{105 - 30}{3 - 0} = \frac{75}{3} = 25$$ So the fee for 3.5 hours is: $$y = 25 \times 3.5 + 30 = 87.5 + 30 = 117.5$$ 5. **(iii) Hours worked if fee is 135:** Set $y=135$ and solve for $x$: $$135 = 25x + 30$$ $$25x = 105$$ $$x = \frac{105}{25} = 4.2$$ So, the plumber worked 4.2 hours. 6. **(iv) Are fees linear?** Yes, the fees form a straight line on the graph, indicating a linear relationship between hours worked and fees charged. 7. **(v) Rate per hour excluding call-out fee:** The rate per hour is the slope $m=25$ euros per hour. 8. **(vi) Is the slope equal to the hourly rate?** Yes, the slope of the line represents the hourly rate charged by the plumber. 9. **(vii) Fee for 10 hours:** Calculate using the formula: $$y = 25 \times 10 + 30 = 250 + 30 = 280$$ **Final answers:** (i) Call-out fee = 30 (ii) Fee for 3.5 hours = 117.5 (iii) Hours worked for 135 fee = 4.2 (iv) Fees are linear (v) Hourly rate = 25 (vi) Slope equals hourly rate (vii) Fee for 10 hours = 280