1. **State the problem:** We analyze the profit $P$ from selling $n$ power bars, given a graph and data points. 2. **Are the data discrete or continuous?** The data are **discrete** because the number of power bars sold, $n$, can only be whole numbers (you cannot sell a fraction of a power bar). 3. **Identify the independent variable:** The independent variable is $n$, the number of power bars sold, because profit $P$ depends on how many bars are sold. 4. **Is the relation a function?** Yes, it is a function because for each number of bars sold $n$, there is exactly one profit value $P$. 5. **Break-even point:** The profit is $0$ when $n=0$ bars sold (the line passes through the origin). This is the $P$-intercept. 6. **Domain and range:** - Domain: $0 \leq n \leq 300$ (since only up to 300 bars are available to sell). - Range: $0 \leq P \leq 200$ (profit ranges from 0 to about 200 as per the graph). 7. **Value of $P(150)$:** From the graph and given point, $P(150) = 0.80 \times 150 = 120$. 8. **Profit of 160:** To find $n$ when $P=160$, use the linear relation $P = 0.80 n$. $$ 160 = 0.80 n $$ Divide both sides by 0.80: $$ \cancel{0.80} n = \frac{160}{\cancel{0.80}} = 200 $$ So, $n=200$ bars must be sold. 9. **Rate of change:** The rate of change is the slope of the line, which is $\frac{\text{change in } P}{\text{change in } n} = 0.80$ profit per bar sold. **Final answers:** - a) Discrete data - b) Independent variable: number of bars sold $n$ - c) Relation is a function - d) Break-even at $n=0$, $P$-intercept - e) Domain: $0 \leq n \leq 300$, Range: $0 \leq P \leq 200$ - f) $P(150) = 120$ - g) $n=200$ bars for $P=160$ - h) Rate of change = $0.80$