FINITE MATH Help.?

Question

Let U be a universal set with subsets A and B such that n(U) = 122, n(A′) = 44, n(B′) = 64, and n(A∩B) = 25. Find n(A∪B).

n(A∪B)

I am stuck on this problem. i know they share 25 elements and the total number of elements in the set is 122. How do I set up the venn diagram or find the numbers for this formula to the problem: n(A∪B)= n(A)+n(B)-n(A∩B). any type of help would be appreciated!!

Answer 1

You should know this formula:
n(A∪B) = n(A) + n(B) - n(A∩B)
You should also know that
n(A') = n(U) - n(A)
The same goes for B (or any other set as well)
n(A') = 44 and n(U) = 122, so
44 = 122 - n(A) which means n(A) = 78

n(B') = n(U) - n(B)
64 = 122 - n(B), so n(B) = 58

Going back to the first formula
n(A∪B) = 78 + 58 - 25
n(A∪B) = 111

Answer 2

Do you mean
n(AUB) = n(A) + n(B) - n(A@B)?
I'm not familiar with the boxlike symbol you use on the left side of the equation, but this formula looks like the usual formula for the number in the union of two sets. The only wrinkle here is finding the number in A and B when you know only the number in their compliments. But this is easy enough when you realise that the number in a set is equal to the number in the universe minus the number in the compliment of the set. So...
n(A) = n(U) - n(A') = 78
and n(B) = n(U) - n(B') = 58
so n(AUB) = 78 + 58 - 25 = 111

Answer 3

44 elements are not in A, so there must be (122 - 44 = ) 78 elements in A. So n(A)=78
Similarly n(B)= 122 - 64= 58

Now: n(AuB) = n(A)+n(B)-n(A∩B)
So: n(AuB) = 78 + 58 - 25
So n(AuB) = 11

Answer 4

those are 2 diferent math classes. I took both one in all them. Finite arithmetic is extra of a survey route, while files is extra particular. particular math classes are continually harder. once you get previous multi-variable calculus, I right here that math is all on a similar aspect of difficulty.

Answer 5

n(A) = n(U) - n(A').

This is all you need. Good luck.