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You are competing in a race. The table shows the times from last year's race you want your time to be last year's median time with an absolute

deviation of at most 5 minutes. Complete the inequality to represent the time in minutes) you hope to achieve​

You are competing in a race. The table shows the times from last year's race you want-example-1

2 Answers

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To find the inequality range for race times, we must first arrange the times and find the median, which is 32 minutes. The desired race time with an absolute deviation of at most 5 minutes from the median results in the inequality 27 ≤ x ≤ 37.

Step-by-step explanation:

To solve this problem, we need to arrange the given race times in order and find the median. Once we find the median, we can set up an inequality that represents a race time within 5 minutes more or less than this median time.

First, we arrange the race times in ascending order: 20, 31, 27, 32, 25, 42, 48, 28, 40, 38, 32. After putting them in the proper order (20, 25, 27, 28, 31, 32, 32, 38, 40, 42, 48), we find that the median, which is the middle value, is 32 minutes since there are 11 times, and the 6th time is the median.

Next, we want to have a race time with an absolute deviation of at most 5 minutes from the median (32 minutes). This gives us the inequality 27 ≤ x ≤ 37. So, the time you hope to achieve in the race is between 27 and 37 minutes.

User Sibaz
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8.3k points
3 votes

Answer:


27\leq x \leq37

Step-by-step explanation:

Let's arrange the numbers from least to greatest:

20, 25, 27, 28, 31, 32, 32, 38, 40, 40, 42, 48

The median is the "middle" number when arranged from least to greatest.

There are 12 numbers, so the middle number would be "average" of 6th and 7th terms.

6th term = 32

7th term = 32

Average = (32+32)/2 = 32

Hence, median = 32

Absolute deviation of 5 minutes means 5 minutes less or greater than the median, 32.

So, 32 - 5 = 27 & 32 + 5 = 37

Hence
27\leq x \leq37

User Drudge Rajen
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8.3k points