The means and mean absolute deviations of the individual times of members on two [tex]4 \times 400[/tex]-meter relay track teams are shown in the table below.

[tex]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of } 4 \times 400 \text{-meter Relay Track Teams}} \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 1.5 \, \text{s} & 2.4 \, \text{s} \\
\hline
\end{array}
\]
[/tex]

What is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B?

A. 0.09

B. 0.15

C. 0.25

D. 0.65

To solve this problem, we need to find the ratio of the difference in the means of the two teams to the mean absolute deviation (MAD) of Team B. Let's go through the steps carefully:

1. Identify the given values:
- Mean time for Team A is 59.32 seconds.
- Mean time for Team B is 59.1 seconds.
- The mean absolute deviation for Team B is 2.4 seconds.

2. Calculate the difference in means between Team A and Team B:
[tex]\[
\text{Difference in means} = \text{Mean of Team A} - \text{Mean of Team B} = 59.32 - 59.1 = 0.22 \text{ seconds}
\][/tex]

3. Calculate the ratio of the difference in means to the mean absolute deviation of Team B:
[tex]\[
\text{Ratio} = \frac{\text{Difference in means}}{\text{MAD of Team B}} = \frac{0.22}{2.4} \approx 0.0917
\][/tex]

4. Choose the closest answer from the options given:
- 0.09
- 0.15
- 0.25
- 0.65

The calculated ratio is approximately 0.0917, which is closest to 0.09. Therefore, the correct choice is:

0.09