We have the following:
let s is speed, d is distance and t is time, therefore:
[tex]\begin{gathered} s=\frac{d}{t} \\ d=s\cdot t \\ d=2\frac{1}{9}\cdot6 \\ d=\frac{18+1}{9}\cdot6=\frac{19}{9}\cdot6 \\ d=\frac{114}{9}=\frac{38}{3} \\ d=12\frac{2}{3} \end{gathered}[/tex]Therefore, the answer is the third option 12 2/3 miles
