One spring day, London noted the time of day and the temperature, in degrees Fahrenheit. Her findings are as follows: At 6 a. m., the temperature was 58° F. For the next 2 hours, the temperature rose 3° per hour. For the next 6 hours, it rose 1° per hour. The temperature then stayed steady until 6 p. m. For the next 2 hours, the temperature dropped 1° per hour. The temperature then dropped steadily until the temperature was 66° at midnight. On the set of axes below, graph London's data.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

so mark-up is 0.4 multiplied by 180 = 72