To determine the domain interval during which Kaitlyn reaches 1.25 miles, let's break down her workout duration and speed.
- Slow Jog (5 minutes): Let's assume she jogs at a constant rate.
- Increased Pace (7 minutes): She speeds up but we don't know the pace yet.
- Break (3 minutes): No running here.
- Run at Constant Pace (7.5 minutes): Paces again unknown for this segment.
- Rest: Any remaining time is for resting.
To find out when she reaches 1.25 miles, we need to consider how far she travels in each segment.
- If we denote her distances as \(d_1\), \(d_2\), and \(d_3\) for each of running phase, we collect the distances as:
-
Jogging Distance Calculation: Let's denote her speed during jogging as \(v_1\). The distance during this phase:
- \(d_1 = v_1 \times 5\) minutes = how far she travels during this segment.
-
Increased Pace Distance Calculation: Say her speed during the faster pace is \(v_2\):
- \(d_2 = v_2 \times 7\) minutes = how far she travels during this segment.
-
Running Distance Calculation:
- The distance of this segment would add to the previous \(d_3 = v_3 \times 7.5\) minutes = additional distance from running this segment.
If we simulate or analyze that her cumulative distance becomes 1.25 miles, it would likely fall between the transition of her moving from slower pace to faster pace and then her final run.
Given that she takes a break before running again, and distances increase overall, you can locate 1.25 after having progressed through some but not all of the continuous jogging and increased speed.
From the options given, the correct interval that fits the scenario where she would reach 1.25 miles during her running session, after potentially crossing the chunk of her workout by combining all distances ran, suggest:
The interval (15 ≤ s ≤ 22.5 minutes) would be the correct resolution where she likely Legally shores up on reaching 1.25 miles considering her increased work pace to cross that distance after engaging both her initial slow jog and then her acceleration.
Thus: The response is: 15 ≤ s ≤ 22.5.