MCQOPTIONS
Saved Bookmarks
| 1. |
The Boolean expression \({\rm{F}}\left( {{\rm{x}},{\rm{y}},{\rm{z}}} \right) = {\rm{\;\bar xy\;\bar z}} + {\rm{\;x\;\bar y\bar z}} + {\rm{\;x\;y\;\bar z}} + {\rm{\;x\;y\;z}}\) is converted into the canonical product of sum (POS) form is |
| A. | \(\left( {{\rm{x\;}} + {\rm{\;y\;}} + {\rm{\;z}}} \right)\left( {{\rm{x\;}} + {\rm{\;y\;}} + {\rm{\;\bar z}}} \right)\left( {{\rm{x\;}} + {\rm{\;\bar y}} + {\rm{\;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{y}} + {\rm{\bar z}}} \right)\) |
| B. | \(\left( {{\rm{x\;}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;y\;}} + {\rm{\;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;\bar y}} + {\rm{\;\bar z}}} \right)\) |
| C. | \(\left( {{\rm{x\;}} + {\rm{\;y\;}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;y\;}} + {\rm{\;\bar z}}} \right)\left( {{\rm{x\;}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\bar y}} + {\rm{\bar z}}} \right)\) |
| D. | \(\left( {{\rm{x\;}} + {\rm{\;\bar y}} + {\rm{\;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;y\;}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{x}} + {\rm{y}} + {\rm{z}}} \right)\) |
| Answer» B. \(\left( {{\rm{x\;}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;y\;}} + {\rm{\;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;\bar y}} + {\rm{\;z}}} \right)\left( {{\rm{\bar x}} + {\rm{\;\bar y}} + {\rm{\;\bar z}}} \right)\) | |