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]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times}} \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 1.5 \, \text{s} & \text{(Missing Data)} \\
\hline
\end{array}
\]
[/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

Sure! Let's work through the problem step-by-step:

1. Understand the Problem: We're looking to calculate the ratio of the difference in the means of two teams to the mean absolute deviation of Team B.

2. Identify the Given Data:
- 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 1.5 seconds.

3. Calculate the Difference in Means:
- Subtract the mean of Team B from the mean of Team A:
[tex]\[
\text{Difference in Means} = \text{Mean of Team A} - \text{Mean of Team B} = 59.32 \, \text{s} - 59.1 \, \text{s} = 0.22 \, \text{s}
\][/tex]

4. Calculate the Ratio:
- Divide the difference in means by the mean absolute deviation of Team B:
[tex]\[
\text{Ratio} = \frac{\text{Difference in Means}}{\text{MAD of Team B}} = \frac{0.22}{1.5} \approx 0.15
\][/tex]

5. Conclusion:
- The ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is approximately 0.15.

Therefore, the correct answer to the question is 0.15.