## College Algebra (10th Edition)

Set $A$ consists of males 18 years old and older who are married, and set $B$ consists of males 18 years old and older who are widowed, and set $C$ consists of males 18 years old and older who are divorced. The first question asks for the union of sets $B$ and $C$ Using the Counting Formula for sets: $n(B\cup C)=n(B)+n(C)-n(B\cap C)$ According to the survey, there isn't anybody who is both widowed and divorced; therefore: $n(B\cap C)=0$ $n(B\cup C)=3.1+10.9-0=14$ 14 million males 18 years old and older are widowed or divorced. The second question asks for the union of sets $A$, $B$, and $C$. Using the Counting Formula for sets: $n(A\cup B\cup C)=n(A)+n(B+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$ According to the survey there isn't anybody who is in either intersection of these sets; therefore: $n(A\cup B\cup C)=n(A)+n(B+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)=65.3+3.1+10.9-0-0-0+0=79.3$ 79.3 million males 18 years old and older are married, widowed, or divorced.