Answer: add 11
Step-by-step explanation:
132+11= 143
143+11= 154
and so on
Answer:
Let’s denote X to be the number of white chips in the sample and E be the event that exactly half of the chips are white. Then,
a) Find α
α = P (reject H0 | H0 is true) = P (X ≥ 2|E)
= P (X = 2|E) + P (X = 3|E),
We took two case, as we can draw only only three chips with two or more white to reject H0, it means we can only take 2 white chips or 3, not more, we get solution
= (5C2 * 5C1)/10C3 + (5C3 * 5C0)/10C3
= 0.5
So, α = 0.5
b) Find β
i) Let E1 be the event that the urn contains 6 white and 4 red chips. (As given)
β = P (accept H0 | E1) = P (X ≤ 1|E1)
= (6C0 * 4C3)/10C3 + (6C1 * 4C2)/10C3
= 1/3
= 0.333
So, β = 0.333
i) Let E2 be the event that the urn contains 7 white and 3 red chips. (As given)
β = P (accept H0 | E2) = P (X ≤ 1|E2)
= (7C0 * 3C3)/10C3 + (7C1 * 3C2)/10C3
= 11/60
= 0.183
So, β = 0.183
Answer:
Yessir all the opps tryna put me on a t shirt
Step-by-step explanation:
If there are multiple operations at the same level on the order of operations off from left to right and you work like this first noticed that there are no parentheses or exponents so we moved to multiplication and division with any sense of parentheses
I believe the answer would be this:
-y^2 - 21
