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]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of } 4 \times 400 \text{-meter Relay Track Teams}} \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{---} & 0.15 \, \text{s} \\
\hline
\end{array}
\]
[/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 the problem of finding the ratio of the difference in the means of the two teams to the mean absolute deviation (MAD) of Team B, follow these steps:

1. Identify the means of each team:
- Mean for Team A is 59.32 seconds.
- Mean for Team B is 59.1 seconds.

2. Calculate the difference in the means of the two teams:
- Difference in means = Mean for Team A - Mean for Team B
- Difference = 59.32 s - 59.1 s = 0.22 seconds

3. Identify the mean absolute deviation of Team B:
- MAD for Team B is 59.1 seconds.

4. Calculate the ratio of the difference in means to the MAD of Team B:
- Ratio = Difference in means / MAD for Team B
- Ratio = 0.22 / 59.1

5. Evaluate the ratio:
- The calculated ratio is approximately 0.0037.

Therefore, the calculated ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is approximately 0.0037. This doesn't match any of the provided options exactly, but if you had to choose the closest option, you might consider reviewing to make sure all calculations and interpretations align with given choices. However, based on the derived numbers, the result as calculated is indeed approximately 0.0037.