Let A$=$a$,$ b$.$ Construct an automaton M which will precisely accept those words over A which end in two b$$s$.[$corresponding Language L$=$L$($r$)$ is regular because is generated by the regular expression r$=($a$\backslash$cup b$)$bb$.$ Really, L$($r$)=($L$($a$\backslash$cup b$))$L$($b$)$L$($b$)]=$xbb$,$ x is any word $($string$)$ in a and b$.$