$3\mathrm{fl}$. oz. of a $90\mathrm{\%}$ alcohol solution was mixed with $6\mathrm{fl}$. oz. of a $36\mathrm{\%}$ 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 $total\mathrm{\_}alcohol=v1\times c1+v2\times c2=4.86$ fl. oz.

Step 4 ：The total volume of the mixture is the sum of the volumes of the two solutions, which is $total\mathrm{\_}volume=v1+v2=9$ 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 $concentration=\frac{total\mathrm{\_}alcohol}{total\mathrm{\_}volume}=0.54$.

Step 6 ：So, the concentration of the new mixture is $\boxed{{\displaystyle 54\mathrm{\%}}}$.