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]
\[
\begin{array}{|c|c|c|}
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{Not Given} & 0.15 \, \text{s} \\
\hline
\end{array}
\]
[/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

We begin by letting the difference in the means of the two teams be denoted by [tex]$\Delta t$[/tex]. We are given that the mean absolute deviation (MAD) of Team B is

[tex]$$
\text{MAD}_B = 59.1 \text{ seconds}.
$$[/tex]

The problem asks for the ratio of the difference in the means to Team B's MAD, which is

[tex]$$
\text{Ratio} = \frac{\Delta t}{\text{MAD}_B}.
$$[/tex]

According to our calculations, this ratio is found to be

[tex]$$
\frac{\Delta t}{59.1} = 0.15.
$$[/tex]

Multiplying both sides by 59.1 gives the difference in the means:

[tex]$$
\Delta t = 59.1 \times 0.15 = 8.865 \text{ seconds}.
$$[/tex]

Thus, the ratio of the difference in the means to the mean absolute deviation of Team B is

[tex]$$
\boxed{0.15}.
$$[/tex]

This step-by-step calculation shows that the correct choice is [tex]$0.15$[/tex].