Answer:
19
Step-by-step explanation:
Let T be the set of positive integers less than 100 with a 9 in the ten's place.
T = {90, 91, 92, 93, 94, 95, 96, 97, 98, 99}
n(T) = 10
Let O be the set of positive integers less than 100 with a 9 in the one's place.
O = {9, 19, 29, 39, 49, 59, 69, 79, 89, 99}
n(O) = 10
The common element is 99.
[tex]T\cap O={99}[/tex]
[tex]n(T\cap O)=1[/tex]
The union of both sets is
[tex]n(T\cup O)=n(T)+n(O)+n(T\cap O)[/tex]
[tex]n(T\cup O)=10+10-1[/tex]
[tex]n(T\cup O)=19[/tex]
Therefore, 19 positive integers are less than 100 those have at least one 9.
