s
average we find the average times for each checkpoint
point 1 Â [tex]t_{avg}[/tex] Â = 2.07 s
point 2 t_{avg} = 3.16 s
point 3 t_{avg} = 4.11 s
point 4 t_{avg} = 4.92 s
given parameters
   * Graph with time for each interval
to calculate
 * average time for each interval (checkpoint)
The average of a magnitude is
     [tex]t_{avg} = \sum \frac{t_i}{n}[/tex]
where [tex]t_{avg}[/tex] in the average time, t_i each time the magnitude is measured and n the number of times the measurement is repeated for the same point.
The table shows the times measured for each interval
measurement    time (s) in interval,
                1   2     3    4
1 Â Â Â Â Â Â Â Â Â Â Â Â 2.02 Â Â 3.17 Â 4.12 Â Â 4.93
2 Â Â Â Â Â Â Â Â Â Â Â Â 2.05 Â Â 3.07 Â 3.98 Â 4.81
3 Â Â Â Â Â Â Â Â Â Â Â Â 2.15 Â Â 3.25 Â 4.23 Â Â 5.01
we look for the average time to get married check point
point 1
     t_{avg} = (2.02 + 2.05 + 2.15) / 3
     t_{avg} = 2.07 s
Point 2
     t_{avg} = (3.17 +3.07 + 3.25) / 3
     t_{avg} = 3.16 s
point 3
     t_{avg} = (4.12 + 3.98 + 4.23) / 3
     t_{avg} = 4.11 s
checkpoint 4
      t_{avg} = (4.93 + 4.81 + 5.01) / 3
      t_{avg} = 4.917 s
We use the criterion of significant figures for the sum, which is stable that the result must be given with the number of decimal places of the figure that I have the least.
      t_{avg} = 4.92 s
the average allows to find the average times for each check point
point 1 tavg = 2.07 s
point 2 tavg = 3.16 s
point 3 tavg = 4.11 s
point 4 tavg = 4.92 s
learn more about average here: Â .com/question/15680250
