Let A, B, and W denote the sets of students apply to Addis Ababa Uni (A), Bahir Dar Uni (B), or Wachemo Uni (W). Let U denote the universal set of all students in the class.
We're given the cardinalities of several sets:
• total number of students:
• A applicants:
• B applicants:
• W applicants:
• A and B applicants:
• A and W applicants:
• B and W applicants:
• non-applicants:
The last cardinality tells us students applied anywhere at all.
We want to find , the number of students that applied to each of the three universities.
By the inclusion/exclusion principle,
That is, we count up all the students in the sets A, B, and W, then subtract the number of students in each pairwise intersection to not double-count, then add back the number of students in the intersection of all three sets since it was removed in the previous step.
Now solve.
Simplify the question so it uses x's and y's or any two letters instead of egg/juice (not necessary)
egg = x
juice = y
x + 2y = 2.30
2x + y = 3.40
Make x the subject
x = 2.30 - 2y
Add pervious step to 2x + y = 3.40 then solve for y
2(2.30 - 2y) + y = 3.40
4.60 - 4y + y = 3.40
4.60 - 3y = 3.40
-3y = -1.20
3y = 1.20
y = 0.40
Now solve for x
x + 2y = 2.30
x + 2(0.40) = 2.30
x + 0.80 = 2.30
x = 1.50
One egg costs $1.50
One juice costs $0.40
Therefore, one egg plus one juice costs $1.90
Answer:
61, 63, 65
Step-by-step explanation:
n + n+2 + n+4 = 189
3n = 183
n = 61
I got 224. 7(4)^2 = 112 •2 =224
Need a little more info for the fist part but 34-17 is 17