Answer:
90 + 3 - 180 = 87
Step-by-step explanation:
I remembered doing a problem like this in Algebra, but i think this is right
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
The answer is b i hope this right to u
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.
∴N(c₁) =

N(c₂) =

∴N(c₁c₂) =

∴ Number of positive integers that are not divisible by 4 or 6,
N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
The decimal form of 4/5 is .8, which can also be written as 0.8 or 0.80.Fractions can be converted into decimals using a calculator or by doing the math manually