Of 100 students in a university IT department, 45 are enrolled in Discrete Mathematics, 30 in Linear Algebra, 20 in Calculus, 10 in at least two of the three courses and just 1 in student is enrolled in all the three courses. How many student take at least of one these courses?
Answers
Answer:
Step-by-step explanation:
That’s 53 students in the class that are taking classes: 20+25+8=53. Now this is going to get a little quantum on you. If there are 53 out of 40 students enrolled in the those classes that means there are thirteen students riding on a slip timeline. If we look to see what timeline they are on we will not be able to tell what classes they are taking and if we look at what classes they are taking we won’t be able to calculate which timeline they are on. So back to numbers: if there are 53/40 students taking classes that tells us that we have a slip quotient of 1.325 because 53/40=1.325. So, we take the number of students on slip time and we want to forget the classes for a second and observe what timeline they are on. For any of the 13 students we will see that 50% of the time they exist as a particle participles and the rest of the time they exist in the cloud. AI are intelligent but not humans and therefor not students, though the SJW’s will tell you otherwise. The point being that in the cloud there is only data on servers, no students. Now when we observe what classes they are taking we can’t see if they are present or in the cloud, but since we know the statistical probability of their state we can take the half that are likely present and multiply it by Ine’s number to get the barium separation of the entire classroom: so that’s 6.5 students that are likely present times Ine’s number (3.483309143686502*10^-1) and we get a barium separation of 2.264150943396226. Now take the barium separation of the school and multiply it by the slip quotient you calculated earlier and you get: 2.264150943396226*1.325=3.
So there are 3 students that are taking none of those classes.