And the question is......................
Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
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Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
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Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
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Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
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Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
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Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
The exterior angle is supplementary to the adjacent angle (they form a linear pair) and is the sum of the two angles in the triangle that are not adjacent to it. For example, in number 2, the exterior angle is 28+40=68 degrees. You can use the other method to check. 68 degrees is supposed to be in a linear pair with 112 degrees, which it is. So the answer to number 2 would be 68 degrees.
Try doing this for the rest of them
Correction:
Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on
P(E∩E') = P(E)P(E').
Answer:
The case is not always true.
Step-by-step explanation:
Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.
And for any two mutually exclusive events, E and E',
P(E∩E') = 0
Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then
P(E)P(E') cannot be equal to zero.
So
P(E)P(E') ≠ 0
This makes P(E∩E') different from P(E)P(E')
Therefore,
P(E∩E') ≠ P(E)P(E') in this case.
Answer:
$0.25
Step-by-step explanation:
Divide the cost by the amount.
Therefore, the cost of each chocolate bar is $0.25.
