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Sagot :
To solve the given problem, we need to use principles from set theory and properties such as union, intersection, and complement of sets. Let's go step by step.
Given:
- Universal set [tex]\( U \)[/tex] has 23 elements: [tex]\( n(U) = 23 \)[/tex]
- [tex]\( n(A \cap B) = 9 \)[/tex]
- [tex]\( n(A \cap C) = 9 \)[/tex]
- [tex]\( n(B \cap C) = 9 \)[/tex]
- [tex]\( n(A \cap B \cap C) = 4 \)[/tex]
- [tex]\( n(A \cup B \cup C) = 19 \)[/tex]
### Part (a) [tex]\( n(A \cup B) \)[/tex]
Since [tex]\( n(A \cup B \cup C) = 19 \)[/tex], this is already provided and pertains to the union of sets [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex].
To find [tex]\( n(A \cup B) \)[/tex], we use the formula for the union of two sets:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]
However, since the original question specifies [tex]\( A \cup B = 19 \)[/tex] directly:
[tex]\[ n(A \cup B) = 19 \][/tex]
### Part (b) [tex]\( n(A' \cup C) \)[/tex]
The complement [tex]\( A' \)[/tex] (also written as [tex]\( A^c \)[/tex]) means the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]. Here's the formula we will use for the union involving a complement:
[tex]\[ n(A' \cup C) = n(U) - n(A \cap C') = n(U) - n(A) + n(A \cap C) - n(C) \][/tex]
Since [tex]\( n(A \cap B \cap C) = 4 \)[/tex], applying the formula: [tex]\( n(A \cap C) \)[/tex] includes elements in their common intersection [tex]\(n(A \cap B \cap C)\)[/tex], so:
[tex]\[ n(A) + n(C) - n(A \cap C) + n(A' \cap C) = n(U) \][/tex]
We know from De Morgan's laws:
[tex]\[ A' \cup C = (A \cap C') \cup C = U - A \][/tex]
Thus:
[tex]\[ n(A' \cup C) = n(U) - n(A) + n(C) - n(A \cap B \cap C)\][/tex]
Since :
[tex]\[ n(U) = 23 - 19=4\][/tex]
### Part (c) [tex]\( n(A \cap B)' \)[/tex]
The complement of [tex]\( n(A \cap B) \)[/tex] is the total number of elements not in [tex]\( A \cap B \)[/tex], given by:
[tex]\[ n(A \cap B)' = n(U) - n(A \cap B) \][/tex]
Given:
[tex]\[ n(U) = 23 \][/tex]
[tex]\[ n(A \cap B) = 9 \][/tex]
So:
[tex]\[ n(A \cap B)' = 23 - 9 = 14 \][/tex]
To summarize:
[tex]\[ \begin{align*} a) \quad &n(A \cup B) = 19 \\ b) \quad &n(A' \cup C) = 4 \\ c) \quad &n(A \cap B)' = 14 \end{align*} \][/tex]
These are the simplified answers to the given questions.
Given:
- Universal set [tex]\( U \)[/tex] has 23 elements: [tex]\( n(U) = 23 \)[/tex]
- [tex]\( n(A \cap B) = 9 \)[/tex]
- [tex]\( n(A \cap C) = 9 \)[/tex]
- [tex]\( n(B \cap C) = 9 \)[/tex]
- [tex]\( n(A \cap B \cap C) = 4 \)[/tex]
- [tex]\( n(A \cup B \cup C) = 19 \)[/tex]
### Part (a) [tex]\( n(A \cup B) \)[/tex]
Since [tex]\( n(A \cup B \cup C) = 19 \)[/tex], this is already provided and pertains to the union of sets [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex].
To find [tex]\( n(A \cup B) \)[/tex], we use the formula for the union of two sets:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]
However, since the original question specifies [tex]\( A \cup B = 19 \)[/tex] directly:
[tex]\[ n(A \cup B) = 19 \][/tex]
### Part (b) [tex]\( n(A' \cup C) \)[/tex]
The complement [tex]\( A' \)[/tex] (also written as [tex]\( A^c \)[/tex]) means the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]. Here's the formula we will use for the union involving a complement:
[tex]\[ n(A' \cup C) = n(U) - n(A \cap C') = n(U) - n(A) + n(A \cap C) - n(C) \][/tex]
Since [tex]\( n(A \cap B \cap C) = 4 \)[/tex], applying the formula: [tex]\( n(A \cap C) \)[/tex] includes elements in their common intersection [tex]\(n(A \cap B \cap C)\)[/tex], so:
[tex]\[ n(A) + n(C) - n(A \cap C) + n(A' \cap C) = n(U) \][/tex]
We know from De Morgan's laws:
[tex]\[ A' \cup C = (A \cap C') \cup C = U - A \][/tex]
Thus:
[tex]\[ n(A' \cup C) = n(U) - n(A) + n(C) - n(A \cap B \cap C)\][/tex]
Since :
[tex]\[ n(U) = 23 - 19=4\][/tex]
### Part (c) [tex]\( n(A \cap B)' \)[/tex]
The complement of [tex]\( n(A \cap B) \)[/tex] is the total number of elements not in [tex]\( A \cap B \)[/tex], given by:
[tex]\[ n(A \cap B)' = n(U) - n(A \cap B) \][/tex]
Given:
[tex]\[ n(U) = 23 \][/tex]
[tex]\[ n(A \cap B) = 9 \][/tex]
So:
[tex]\[ n(A \cap B)' = 23 - 9 = 14 \][/tex]
To summarize:
[tex]\[ \begin{align*} a) \quad &n(A \cup B) = 19 \\ b) \quad &n(A' \cup C) = 4 \\ c) \quad &n(A \cap B)' = 14 \end{align*} \][/tex]
These are the simplified answers to the given questions.
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