Answer:
solution:-We know that for any two finite sets A and B, n(A∪B)=n(A)+n(B)−n(A∩B).
Here, it is given that n(A)=20,n(B)=30 and n(A∪B)=40, therefore,
n(A∪B)=n(A)+n(B)−n(A∩B)
⇒40=20+30−n(A∩B)
⇒40=50−n(A∩B)
⇒n(A∩B)=50−40
⇒n(A∩B)=10
Hence, n(A∩B)=10
Step-by-step explanation:
hope it helps you friend ☺️
A) x * y = 42 (or x = 42/y)
B) x + y = 23
B) 42/y + y = 23
42 + y^2 = 23y
y^2 -23y + 42 = 0
The numbers are 21 and 2
City C (42 and 66 are the two modes they are the most frequent numbers both appearing 3 times)
