Respuesta :
Answer:
(i)[tex]M \cup P=\{2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24,26,28,29,30\}[/tex]
(ii)M-T={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
(iii)P∪(M∩T) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
(iv)P’U(M∩T’)={2,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30}
Step-by-step explanation:
Given the sets:
ε = {x: 2 ≤ x ≤30, x is an integer}
M = {even numbers}={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
P = {prime numbers}={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
T = {odd numbers} ={3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}
(i)This is the union of sets M and P. (Do not repeat same elemnts)
[tex]M \cup P=\{2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24,26,28,29,30\}[/tex]
(ii)M-T: This is the set M less elements in set T.
Since [tex]M \cap T =\{\}[/tex], the set M-T=M.
M-T={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
(iii) P∪(M∩T)
[tex]M \cap T =\{\}[/tex]
Therefore:
P∪(M∩T) = P
P∪(M∩T) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
(iv) P’U(M∩T’)
Since T is the set of odd numbers, its complement will be the set of even numbers.
T'=M={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
M∩T’=M={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
[tex]P' =\{4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30\}[/tex]
Therefore:
P’U(M∩T’) = {4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30} U {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
={2,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30}
