Answer:
[tex]\sf 22. \quad \left(Y \cap X'\right) \cup Z = \{b,c,d,e,f\}[/tex]
[tex]\sf 28. \quad Y\;\!'-X = \{d, f\}[/tex]
Step-by-step explanation:
Set notation
[tex]\begin{array}{|c|c|l|} \cline{1-3} \sf Symbol & \sf N\:\!ame & \sf Meaning \\\cline{1-3} \{ \: \} & \sf Set & \sf A\:collection\:of\:elements\\\cline{1-3} \cup & \sf Union & \sf A \cup B=elements\:in\:A\:or\:B\:(or\:both)}\\\cline{1-3} \cap & \sf Intersection & \sf A \cap B=elements\: in \:both\: A \:and \:B} \\\cline{1-3} \sf ' \:or\: ^c & \sf Complement & \sf A'=elements\: not\: in\: A \\\cline{1-3} \sf - & \sf Difference & \sf A-B=elements \:in \:A \:but\: not\: in \:B}\\\cline{1-3} \end{array}[/tex]
Given sets:
- X = {a, c, e, g}
- Y = {a, b, c}
- Z = {b, c, d, e, f}
Therefore,
- Universal set = {a, b, c, d, e, f, g}
- X' = not in X = {b, d, f}
- Y' = not in Y = {d, e, f, g}
Question 22
[tex]\begin{aligned}\sf \left(Y \cap X'\right) \cup Z & = \sf \left(\{a,b,c\} \cap \{b,d,f\}\right) \cup \{ b,c,d,e,f \}\\& = \sf \{b\} \cup \{b,c,d,e,f\}\\& = \sf \{b,c,d,e,f\}\end{aligned}[/tex]
Question 28
[tex]\begin{aligned}\sf Y'-X & = \sf \{d,e,f,g\} - \{a,c,e,g\}\\& = \sf \{d,f\}\end{aligned}[/tex]
