Answer:
The answer is "[tex]15.38\%[/tex]"
Explanation:
Background[tex]= 13 \ mv\\\\[/tex]
corrected signal[tex]= 91 \ mv-13\ mv= 78\ mv\\\\[/tex]
with attenuator[tex]=25\ mv-13\ mv= 12\ mv\\\\[/tex]
[tex]\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%[/tex]