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]
\[
\begin{array}{|c|c|c|}
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{(not provided)} & 0.15 \, \text{s} \\
\hline
\end{array}
\]
[/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's times. Here's how you can approach the problem step by step:

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

2. Calculate the Difference in Means:
- The difference in the means is calculated by subtracting the mean of Team B from the mean of Team A:
[tex]\[
\text{Difference in Means} = 59.32 - 59.1 = 0.22 \, \text{seconds}
\][/tex]

3. Use the Mean Absolute Deviation for Team B:
- The mean absolute deviation (MAD) for Team B is given as 59.1 seconds.

4. Calculate the Ratio:
- To find the ratio of the difference in the means to the MAD for Team B, divide the difference by the MAD:
[tex]\[
\text{Ratio} = \frac{0.22}{59.1}
\][/tex]

5. Determine the Result:
- When you perform this division, the resulting value is approximately 0.0037.

Since none of the provided choice options (0.09, 0.15, 0.25, 0.65) match the calculated ratio, it seems there is an error or misunderstanding in the context or initial options. Please verify the context and details or check if there might be another step involved that wasn't initially considered.