the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is…

the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is accurate?\nxavier is traveling at 1.5 feet per second.\nmoises is going faster than xavier.\nthe difference in their rates of change is 1 foot per second.\nif both boys remain on the treadmill for 10 minutes, xavier will have traveled a greater distance.

the distance, in feet, two boys travel per second on a treadmill is shown to the left. which comparison is accurate?\nxavier is traveling at 1.5 feet per second.\nmoises is going faster than xavier.\nthe difference in their rates of change is 1 foot per second.\nif both boys remain on the treadmill for 10 minutes, xavier will have traveled a greater distance.

Answer

Explanation:

Step1: Calculate Xavier's rate

Use the slope formula (m=\frac{y_2 - y_1}{x_2 - x_1}). For Xavier, using points ((2,15)) and ((4,30)): (m=\frac{30 - 15}{4 - 2}=\frac{15}{2}=7.5) feet per second. So the first option is wrong.

Step2: Calculate Moises's rate

Using points ((4,20)) and ((8,40)) for Moises: (m=\frac{40 - 20}{8 - 4}=\frac{20}{4}=5) feet per second. Since (7.5>5), Moises is not going faster than Xavier, so the second option is wrong.

Step3: Calculate the difference in rates

The difference is (7.5 - 5 = 2.5) feet per second. So the third option is wrong.

Step4: Analyze the distance after 10 minutes

Since Xavier's rate ((7.5) ft/s) is greater than Moises's rate ((5) ft/s), and distance (d = rt) (where (t) is the same for both, (t = 10\times60=600) seconds), using (d = rt), Xavier will have a greater distance.

Answer:

If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance.