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}{|r|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of $4 \times 400$-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} & 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 need to find the ratio of the difference in the means of two track teams to the mean absolute deviation (MAD) of Team B.

Here's how you can solve it step-by-step:

1. Find the means of the two teams:
- Mean for Team A = 59.32 seconds
- Mean for Team B = 59.1 seconds

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

3. Find the mean absolute deviation for Team B:
- MAD for Team B = 2.4 seconds

4. Calculate the ratio:
- Ratio = Difference in Means / MAD of Team B
- Ratio = 0.22 / 2.4

5. Calculate the numerical value of the ratio:
- Ratio = 0.091666... (approximately)

Given this calculation, the ratio falls closest to 0.09.

Therefore, the correct answer is 0.09.