To solve this problem, let's determine how many boxes of cookies the scout has over time.
- The scout starts with 100 boxes of cookies.
- They sell 25 boxes every 3 hours.
We can set up a model to describe the number of boxes left over time, using \( t \) as the number of hours.
Since the scout sells 25 boxes every 3 hours, the rate of selling boxes is: \[ \text{Rate} = \frac{25 \text{ boxes}}{3 \text{ hours}} = \frac{25}{3} \text{ boxes per hour} \approx 8.33 \text{ boxes per hour} \]
The formula for the number of boxes remaining \( B(t) \) after \( t \) hours can be expressed as: \[ B(t) = 100 - \left(\frac{25}{3} \cdot t\right) \] For \( t = 0 \): \( B(0) = 100 \) For \( t = 3 \): \( B(3) = 100 - 25 = 75 \) For \( t = 6 \): \( B(6) = 100 - 50 = 50 \) For \( t = 9 \): \( B(9) = 100 - 75 = 25 \)
Now let's analyze the graphs based on the coordinates plotted:
- In the first graph, the points (0, 100), (1, 200), (2, 300), and (3, 400) show an incorrect trend because the number of boxes cannot exceed 100.
- The second graph shows points (0, 100), (3, 125), (6, 150), and (9, 175), and also reflects an incorrect upward trend.
- The third graph shows points (0, 100), (3, 75), (6, 50), and (9, 25). This accurately reflects the decrease in boxes sold over time.
- The fourth graph shows an incorrect downward trend for boxes (climbing down from higher values like 10 down to 5), which does not match the selling pattern correctly.
Thus, the correct graph that represents the number of boxes the scout has over time is the third graph, which shows:
- At time t = 0 hours, there are 100 boxes.
- At time t = 3 hours, there are 75 boxes.
- At time t = 6 hours, there are 50 boxes.
- At time t = 9 hours, there are 25 boxes.
So, the answer is the third graph.