The median is often a better representative of the central value of a data set when the data set: source is bimodal. has a high
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Answer:
Step-by-step explanation:
First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is
and we can use that in a d = rt table:
d = r * t
Nina .145
Jo
Now we can fill in the distance which is 6 for both, since that is the distance where they met:
d = r * t
Nina 6 = .145
Jo 6 =
Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:
d = r * t
Nina 6 = .145 * t + 3
Jo 6 = r * t
As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then
6 = .145(t + 3) and
6 = .145t + .435 and
5.55 = .145t so
t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then
6 = r(38.379) and
r = .16 km/min
Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:
6 should equal .145454545( 38.37931034 + 3)
6 ?=? .145454545(41.37931034)
6 ?=? 6.0 so
Jo's rate is .16 km/min rounded