$3 \mathrm{fl}$. oz. of a $90 \%$ alcohol solution was mixed with $6 \mathrm{fl}$. oz. of a $36 \%$ alcohol solution. Find the concentration of the new mixture.

Step 1 ：Let's denote the volume of the 90% alcohol solution as \(v1 = 3\) fl. oz., and the volume of the 36% alcohol solution as \(v2 = 6\) fl. oz.

Step 2 ：The concentration of the 90% alcohol solution is \(c1 = 0.9\), and the concentration of the 36% alcohol solution is \(c2 = 0.36\).

Step 3 ：The total amount of alcohol in the mixture can be found by adding the amount of alcohol in the 90% solution to the amount of alcohol in the 36% solution. This gives us \(total\_alcohol = v1 \times c1 + v2 \times c2 = 4.86\) fl. oz.

Step 4 ：The total volume of the mixture is the sum of the volumes of the two solutions, which is \(total\_volume = v1 + v2 = 9\) fl. oz.

Step 5 ：The concentration of the new mixture can be found by dividing the total amount of alcohol by the total volume of the mixture. This gives us \(concentration = \frac{total\_alcohol}{total\_volume} = 0.54\).

Step 6 ：So, the concentration of the new mixture is \(\boxed{54\%}\).