Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
I'd love to help but I have no idea how this stuff works, but I must make my exit note-worthy.
Algebraaaaaaaaaaaa isn't coooooooooooool.
And neither are fractioooooooooooooooooooooooooooooonnnnnnnnssssss.
does anyone know how to do it :,(
Can you paste the words into the chat
it wont let me
Here's a useful tip for Venn-diagram-counting problems like this one: Start with the most inclusive sets.
What I mean by that is that you can tackle this question by first considering the set that contains elements that belong to all (or most) of the three given sets, which in this case would be \(A\cap B\cap C\). You know it contains \(5\) elements.
Another set that's very inclusive is \(A'\cap B'\cap C'\), which has elements that don't belong to any of \(A,B,C\). You're told it has \(22\) elements.
So from this info, you can gather the following:
Does that make sense?
Now consider the next most inclusive set - there are a few of these, so let's just pick \(A\cap C\) as an example. We're told that that it contains \(14\) elements. But notice that some of the elements of \(A\cap C\) were already counted in \(A\cap B\cap C\), since \(A\cap C\subseteq A\cap B\cap C\) (subset, if you're not familiar with the symbol). So to avoid double counting, we'll need to subtract.
\[|A\cap C|=n(A\cap C)-n(A\cap B\cap C)=14-5=11\]
where \(|S|\) is used to denote the size of the set \(S\) and \(S\) only. Not to be confused with \(n(S)\), which means the number of elements that belong to \(S\), but might also belong to other sets containing \(S\). (If that's confusing, I'll try to elaborate.)