Answer:
im pretty sure its 30
Step-by-step explanation:
The answers are B,C, and F because you have to set up proportions.
The probability that you picked the fair coin given that the outcome of the toss was heads is 1/3.
There is a one in two chance of drawing the fair coin. The chances of flipping heads again are 1 in 2. As a result, there is a 1 in 4 chance that the coin will land on heads.
There is a one in two chance of drawing the trick coin. The probability of flipping heads is then 2 to 1. As a result, there is a 2 in 4 chance that the coin will land on heads.
When each is multiplied by four, the resulting integers are
fair: 1 and trick: 2
and overall results: 3. (fair and tails is not counted)
The likelihood of a fair coin is one in three.
The likelihood that the chosen coin will show heads and be the fair coin is 33.333%.
To learn more about probability
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Answer:
The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.
Step-by-step explanation:
We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.
The probability of one point of being over UCL=104 is:

The probability of one point of being under LCL=96 is:

Then, the probability of exhibit lack of control is:

The probability of having at least one point out of control in the first three points is:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.
Answer:
The quotient of the expression will be (x² - 2x + 3).
Step-by-step explanation:
The given expression is (x³ - 3x² + 5x - 3) ÷ (x - 1)
Now we will use the synthetic division method to solve this question.
We will note down the coefficient of the expression as below
x³ x² x constant
1 (-3) 5 -3
1 1 (1×1-3)=(-2) [1×(-2)+5]=3 (1×3-3)=0
Therefore (x -1) is a perfect zero factor.
Now we can easily say that coefficient of the expression is [1.x² + (-2)x + (3)]
and remainder is 0.
Or in more simplified way coefficient of the expression will be
(x² -2x + 3).