I think it’s A. If the first event is 3/5 and the second event is 3/5×2/5, the third will be 3/5(2/5)²=0.096, and the fourth will be 3/5(2/5)³=0.0384 so the 5th will be 3/5(2/5)⁴=0.01536.
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Answer:
27°
Step-by-step explanation:
The inscribed angle is half the measure of the arc it intercepts.
∠M = 1/2 arc QR = 1/2(54°)
∠M = 27°
Answer:
Her utility bill decreased
Step-by-step explanation:
$-9.66 is closer to 0 but before her bill was -$25.66 so that's farther away so her bill decreased.
Tan² B = 4
(tan B)² = 4
tan B = ±2
Basic Angle
= tan^-1 2
= 63.4° (1dp)
B = 63.4°, 116.6°, 243.2°, 296.6°
Assuming you're actually supposed to present your answers in 0dp, then all answers apply.
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
