Answer:
hello your question is incomplete below is the missing parts
(a) A\ (A\B) = B\(B\A)
(b) A\ (BA) = B\(A\B)
answer : A\ (A\B) = B\(B\A) = always true
A\ (BA) = B\(A\B) = sometimes true and sometimes false
Step-by-step explanation:
(a) A\ (A\B) = B\(B\A). = ALWAYS TRUE
using de Morgan's law to prove this
A\ (A\B) = A\ ( A ∩ B^c )
= A ∩ ( A^C ∪ B )
= ( A ∩ A^C ) ∪ ( A ∩ B )
= Ф ∪ ( A ∩ B )
= ( A ∩ B )
ALSO : B\(B\A) = attached below is the remaining parts of the solution
B) A\ (BA) = B\(A\B) = Sometimes true and sometimes false
attached below is the prove using De Morgan's law
a. Notice that
1/(1*2) = 1/2 = 1 - 1/2
1/(2*3) = 1/6 = 1/2 - 1/3
1/(3*4) = 1/12 = 1/3 - 1/4
and so on, which suggests the n-th term of the sum can be written as
Then the sum itself is telescoping:
b. The proof is trivial:
so the formula found in (a) is correct.
Answer:
42%
Step-by-step explanation:
Given: P(M) = 0.52, P(A) = 0.33, and P(M and A) = 0.27.
Find: P(not M and not A).
P(not M and not A) = 1 − P(M or A)
P(not M and not A) = 1 − (P(M) + P(A) − P(M and A))
P(not M and not A) = 1 − (0.52 + 0.33 − 0.27)
P(not M and not A) = 1 − 0.58
P(not M and not A) = 0.42