Answer:
Option E
10 kHz
Explanation:
[tex]F_o=\frac {F_i}{2^{N}}[/tex]
Where [tex]F_o[/tex] is output frequency, [tex]F_i[/tex] is input frequency and N is the number of flip-floop
Substituting 80 kHz for [tex]F_i[/tex] and 3 for N we obtain
[tex]F_o=\frac {80 kHz}{2^{3}}=10 kHz[/tex]
Therefore, the output frequency is equivalent to 10 kHz