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]
\[
\begin{array}{|c|c|c|}
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{N/A} & 0.15 \, \text{s} \\
\hline
\end{array}
\]
[/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 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.

1. Identify the Means:
- The mean for Team A is 59.32 seconds.
- The mean for Team B is 59.1 seconds.

2. Calculate the Difference in the Means:
- To find the difference in the means, subtract the mean of Team B from the mean of Team A:
[tex]\[
\text{Difference in means} = 59.32 - 59.1 = 0.22
\][/tex]

3. Given Mean Absolute Deviation:
- The mean absolute deviation for Team B is provided as 59.1 seconds.

4. Calculate the Ratio:
- Now, we need to calculate the ratio of the difference in the means to the mean absolute deviation of Team B:
[tex]\[
\text{Ratio} = \frac{\text{Difference in means}}{\text{MAD of Team B}} = \frac{0.22}{59.1} \approx 0.0037
\][/tex]

As a result, the ratio of the difference in the means to the mean absolute deviation of Team B is approximately 0.0037. Based on the options given in the problem, none matches exactly with this value, so it seems there might be an error in the calculation or the provided options.