I'll give brainliest if you show your answer. plz answer before thirty minutes. the means and mean absolute deviations of the individual times of members of two relay swim teams are shown in the table below. means and mean absolute deviations of individual times of members of 4*200- meter relay swim teamsteam a team bmean 127.9 s 127.4 smean absolute deviation 0.26 s 0.23 sthe difference of the means is found and then compared to each of the mean absolute deviations. which is true? the difference between the mean times is about equal to the mean absolute deviation of the data sets. the difference between the mean times is about 2 times the mean absolute deviation of the data sets. the difference between the mean times is about 5 times the mean absolute deviation of the data sets. the difference between the mean times is about 16 times the mean absolute deviation of the data sets.

The difference between the mean times is about 16 times the mean absolute deviation of the data sets.

We are given the mean and mean absolute deviation of two teams in a table as:

                                              Team A           Team B

             Mean                        127.9 s             127.4 s

Mean Absolute Deviation     0.26 s             0.23 s

The difference of mean between two team is: 0.5 s

( Since 127.9-127.4=0.5 s)

Also, the difference of Mean absolute deviation is: 0.03 s

( Since, 0.26-0.23=0.03 s)

Now let the difference of the mean times be equal t n times the difference  of the mean absolute deviation of two teams

i.e.

      0.5=n×0.03

⇒ n=0.5/0.03

⇒ n=16.6667

which is approximately equal to 16.

Hence, The difference between the mean times is about 16 times the mean absolute deviation of the data sets.

c. (0.6,-1.4)

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