0 votes 0 votes In the circuit shown $Y=\overline{A} \overline{B}+\bar{C}$ $Y=(A+B)C$ $Y=(\overline{A}+\overline{B})\overline{C}$ $Y=AB+C$ Others gate2012-ec to-be-tagged + – Milicevic3306 asked Mar 25, 2018 • recategorized Feb 26, 2021 by Lakshman Bhaiya Milicevic3306 16.0k points 80 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.