Answer
a) y | p(y)
25 | 0.8
100 | 0.15
300 | 0.05
E(y) = ∑ y . p(y)
E(y) = 25 × 0.8 + 100 × 0.15 + 300 × 0.05
E(y) = 50
average class size equal to E(y) = 50
b) y | p(y)
25 |
100 |
300 |
E(y) = ∑ y . p(y)
E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3
E(y) = 130
average class size equal to E(y) = 130
c) Average Student in the class in a school = 50
Average student at the school has student = 130
Answer:
(8, -22)
Step-by-step explanation:
The tables each contain four (x,y) points of a straight line. You can see that for every increase of x by 2, y decreases by 8 in the first one (observe 26, 18, 10 2), and decreases by 6 in the second.
If you continue the table with x=4, 6 and 8, you get y=-22 in both cases for x=8. That is the intersection, so the solution is (8,-22).
Added a graph. The equations are y=10-4x and y=2-3x respectively. Hope you understand a bit of this (brief) explanation.
