Biconditionals are statements that work both ways.
Some examples:
If it rains, I go out, and if I go out, it must be raining.
This can be stated concisely in mathematical terms as
I go out IF AND ONLY IF it rains.
So looking at the given statements, only the last two work both ways, namely:
If the sun rises in the east, then it is morning, and if it is morning, the sun rises in the east.
Victoria will play outside if and only if the weather is nice.
Answer:
see explanation
Step-by-step explanation:
The bar above the 3 indicates the 3 is being repeated, that is
0.13333....
We require to make 2 equations with the repeating 3 after the decimal point.
let x = 0.13333... , then multiply both sides by 10 then 100
10x = 1.3333 ← equation with repeating 3 after the decimal point
100x = 13.3333 ← second equation with repeated 3 after the decimal point
Subtracting the equations eliminates the repeated 3, that is
100x - 10x = 13.333 - 1.333
90x = 12 ( divide both sides by 90 )
x =
← divide numerator/denominator by 6, hence
0.1333... = 
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Answer:
d is the answer to your question
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