8) a) The new fastest time is 56.7 seconds, b) The <em>overall percentage</em> improvement is 19 %.
9) The height after the first increase is 1.375 meters and after the second increase is 1.54 meters.
<h3>How to apply percentages in real life applications</h3>
In this problem we have two examples from real life, in which percentages are used. The first is the distance traveled by a runner in a given time and the second is the growth of a person in time. Change in variables based on percentages are described by the following formula:
C' = C · (1 + r / 100) (1)
Where:
- C - Initial value
- C' - Final value
- r - Percentage of change.
Now we proceed to resolve on each case. 8) a) Please notice that the runner must cover the trail in less time and percentage time must be negative. If we know that C = 70 and r = - 10, then his new fastest time is:
C' = 70 · (1 - 10 / 100)
C' = 70 · 0.90
C' = 63
And again: (C' = 63, r = - 10)
C'' = 63 · (1 - 10 / 100)
C'' = 63 · 0.90
C'' = 56.7
The new fastest time is 56.7 seconds.
b) And the overall percentage improvement is:
r = (|C'' - C| / C) × 100 %
r = (|56.7 - 70| / 70) × 100 %
r = 19 %
The <em>overall percentage</em> improvement is 19 %.
9) If we know that C = 1.25 and r = 10, then the height next year is:
C' = 1.25 · (1 + 10 / 100)
C' = 1.375
And again: (C' = 1.375, r = 12)
C'' = 1.375 · (1 + 12 / 100)
C'' = 1.54
The height after the first increase is 1.375 meters and after the second increase is 1.54 meters.
To learn more on percentages: brainly.com/question/13450942
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