Let
\[
\begin{array}{l}
U=\{3,15,8,14,16,17,20,18\} \\
A=\{3,20,18\} \\
C=\{15,8,14,16,17\}
\end{array}
\]
Find $A \cup C^{\prime}$.
Choose the correct answer below, and, if necessary, fill in the answer box to complete your choice.
A. $A \cup C^{\prime}=$
(Use a comma to separate answers as needed.)
B. $A \cup C^{\prime}$ is the empty set.
Answer
\(A \cup C'=\boxed{\{3,18,20\}}\) is the final answer
Step 1 ：Define the universal set U as \(U=\{3,15,8,14,16,17,20,18\}\)
Step 2 ：Define set A as \(A=\{3,20,18\}\)
Step 3 ：Define set C as \(C=\{15,8,14,16,17\}\)
Step 4 ：Find the complement of set C, denoted as C', which is the set of all elements in U that are not in C. So, \(C'=\{3,18,20\}\)
Step 5 ：Find the union of set A and the complement of set C, denoted as \(A \cup C'\). This is the set of all elements that are in A or in C'. So, \(A \cup C'=\{3,18,20\}\)
Step 6 ：\(A \cup C'=\boxed{\{3,18,20\}}\) is the final answer
