Auniversal set u consists of 16 elements. if sets a, b, and c are proper subsets of u and n(u) = 16, n(a ∩ b) = n(a ∩ c) = n(b ∩ c) = 6, n(a ∩ b ∩ c) = 2, and n(a ∪ b ∪ c) = 14, determine each of the following. a) n(a ∪ b) b) n (a' ∪ c )c) n (a ∩ b )'

a) 14

b) 12

c) 10

Step-by-step explanation:

The easiest way to do this is with a Venn diagram. So we put a 2 in the intersection between A,B AND C (n(A ∩ B ∩ C) = 2). And we are going to called that region as V.

n(A ∩ B) = 6, so the sum of the region (that we can see in the picture) V and ii should be 6. V+ii= 6; ii= 4

The same happen with n(A ∩ C) = n(B ∩ C). Region iv and vi =4

n(A ∪ B ∪ C) = 14 and the sum of all the other region is already 14. So the region i, iii and vii are going to be 0. But n(U)=16, so there will be another region, called viii=2 (16-14)

n(A U B)=n(A)+n(B)-n(A ∩ B) --> region (A)= i, ii, v, iv + region (B) = iii, ii, v, vi- region (A ∩ B)= v, ii= 4+2+4+4+2+4-6=14

n(A' U C)--> regions of n(A')= region vi, iii, vii, viii + regions of n(C)= region vii, vi, v, iv= region iii, iv, v, vi, vii, viii=0+4+2+4+0+2=12

n(A ∩ B)'=n(U)-n(A ∩ B)--> n(U)=region i, ii, iii, iv, v, vi, vii, viii-n(A ∩ B)=region v, ii=16-6=10

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

what's the question?

step-by-step explanation:

there is no question.