Asked by apoorva
t(min)x-value:36,38,40,42,44
heartbeats-y value: 2530, 2661, 2806,2948,3080
Graph these points(Each point goes with another so (36,2530), (38,2661)...
The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the heart rate after 42 minutes using the secant line between the points with the given values of t.
they give me t=36 and t=42 for the first problem:
First I found the slope between both points which i got 69.67.
then after i convert it to y=mx+b form. Then do i plug in 42 for x to find the heart rate.
Is that the right steps or no?
heartbeats-y value: 2530, 2661, 2806,2948,3080
Graph these points(Each point goes with another so (36,2530), (38,2661)...
The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the heart rate after 42 minutes using the secant line between the points with the given values of t.
they give me t=36 and t=42 for the first problem:
First I found the slope between both points which i got 69.67.
then after i convert it to y=mx+b form. Then do i plug in 42 for x to find the heart rate.
Is that the right steps or no?
Answers
Answered by
bobpursley
If the slope of the secant line is 69b/min, then x=42, m=69
Yes.
Yes.
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