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]
\[
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{Means and Mean Absolute Deviations of Individual Times of Members of $4 \times 400$-meter Relay Track Teams} \\
\hline
& Team A & Team B \\
\hline
Mean & 59.32 s & 59.1 s \\
\hline
Mean Absolute Deviation & 1.5 s & 2.4 s \\
\hline
\end{tabular}
\]
[/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

We are given the following information for the two teams:

- Team A has a mean of [tex]$59.32$[/tex] seconds.
- Team B has a mean of [tex]$59.10$[/tex] seconds and a mean absolute deviation (MAD) of [tex]$2.4$[/tex] seconds.

Follow these steps to solve the problem:

1. Compute the difference between the means of Team A and Team B:
[tex]$$
\text{Difference} = 59.32 - 59.10 = 0.22 \text{ seconds}.
$$[/tex]

2. Find the ratio of the difference in means to the mean absolute deviation of Team B:
[tex]$$
\text{Ratio} = \frac{0.22}{2.4}.
$$[/tex]

3. Divide to obtain the ratio:
[tex]$$
\frac{0.22}{2.4} \approx 0.09167.
$$[/tex]

4. Rounding to the nearest hundredth gives approximately [tex]$0.09$[/tex].

Thus, the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is [tex]$0.09$[/tex].