Answered

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A universal set U consists of elements. If sets​ A, B, and C are proper subsets of U and ​n(U)​, ​n(A ​B)​n(A ​C)​n(B ​C)​, ​n(A B ​C)​, and​ n(A B ​C)​, determine each of the following.

a. n(AUB)
b. n(A'UC)
c. n(AnB)'

Sagot :

MrRoyal

Answer:

[tex]n(A\ u\ B) = 10[/tex]

[tex]n(A'\ u\ C) = 10[/tex]

[tex]n(A\ n\ B)' = 6[/tex]

Step-by-step explanation:

Given

[tex]n(U) = 12[/tex]

[tex]n(A\ n\ B) =n(A\ n\ C) = n(B\ n\ C) =6[/tex]

[tex]n(A\ n\ B\ n\ C)=4[/tex]

[tex]n(A\ u\ B\ u\ C)=10[/tex]

Required

Solve a, b and c

There are several ways to solve this; the best is by using Venn diagram (see attachment for diagram)

Solving (a):

[tex]n(A\ u\ B)[/tex]

This is calculated as:

[tex]n(A\ u\ B) = n(A) + n(B) - n(A\ n\ B)[/tex]

From the attachment

[tex]n(A) = 0+2+4+2 = 8[/tex]

[tex]n(B) = 0+2+4+2 = 8[/tex]

[tex]n(A\ n\ B) = 4 +2 = 6[/tex]

So:

[tex]n(A\ u\ B) = 8 + 8 - 6[/tex]

[tex]n(A\ u\ B) = 10[/tex]

Solving (b):

[tex]n(A'\ u\ C)[/tex]

This is calculated as:

[tex]n(A'\ u\ C) = n(A') + n(C) -n(A'\ n\ C)[/tex]

From the attachment

[tex]n(A) = n(U) - n(A) = 12 - 8 = 4[/tex]

[tex]n(C) = 0+2+4+2 = 8[/tex]

[tex]n(A'\ n\ C) = 2[/tex]

So:

[tex]n(A'\ u\ C) = 4 + 8 - 2[/tex]

[tex]n(A'\ u\ C) = 10[/tex]

Solving (c):

[tex]n(A\ n\ B)'[/tex]

This is calculated as:

[tex]n(A\ n\ B)' = n(U) - n(A\ n\ B)[/tex]

[tex]n(A\ n\ B)' = 12- 6[/tex]

[tex]n(A\ n\ B)' = 6[/tex]

View image MrRoyal
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