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]\[
\begin{array}{|c|c|c|}
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 1.5 \, \text{s} & 2.4 \, \text{s} \\
\hline
\end{array}
\][/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'll go through the steps needed to find the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B.

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

2. Calculate the difference in the means:
[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]\[
\text{Ratio} = \frac{\text{Difference in means}}{\text{MAD of Team B}} = \frac{0.22}{2.4}
\][/tex]

4. Perform the division:
[tex]\[
\text{Ratio} = 0.0917 \text{ (approximately)}
\][/tex]

5. Round to the nearest of the choices given:
- The choices are 0.09, 0.15, 0.25, and 0.65.
- The calculated ratio, which is approximately 0.0917, is closest to 0.09.

Therefore, the correct answer is 0.09.