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olchik [2.2K]
4 years ago
8

The probability that a given urine sample analyzed in a pathology lab contains protein is 0.73 (event A). The probability that t

he urine sample contains the protein albumin is 0.58 (event B). The probability that a sample contains protein, given that albumin is detected, is 1.
Which statement is true?

Events A and B are dependent because P(A|B) = P(A).

Events A and B are dependent because P(A| B) P(B).

Events A and B are dependent because P(A|B) = P(A) + P(B).

Events A and B are dependent because P(A| B) P(A).

Events A and B are dependent because P(A|B) = P(A).
Mathematics
1 answer:
KATRIN_1 [288]4 years ago
6 0
The correct answer among the choices provided is the second option. Events A and B are dependent because P(A|B) P(B). And from conditional probability, we have P(A|B) = P(AintersectionB) / P(B).
Therefore, P (AintersectionB) is equal to P(A|B) × P(B).
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If 3t - 7 = 5, then 6t=
Alenkinab [10]
3t - 7 = 5

3t = 12

t = 4

6t = 6.4 = 24

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4 0
3 years ago
Read 2 more answers
Mean absolute deviation of 75, 89, 145, 85, 80, 92, 104, 90, 100?
Mama L [17]
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Next compute the absolute deviations for each datapoint:
|75-95.555|=20.555
|89-95.555|=6.555
|145-95.555|=49.445
|85-95.555|=10.555
|80-95.555|=15.555
|92-95.555|=3.555
|104-95.555|=8.445
|90-95.555|=5.555
|100-95.555|=4.445

Next we'll take the mean of these deviations: \frac{20.555+6.555+49.445+10.555+15.555+3.555+8.445+5.555+4.445}9=\frac{124.665}9=13.8517

The mean absolute deviation of your data is 13.8517
7 0
3 years ago
assume the probability of having a boy or a girl is the same. what is the probability of have a boy then a girl then another gir
NNADVOKAT [17]

Answer:

\dfrac{1}{8}

Step-by-step explanation:

If the probability of having a boy or a girl is the same, then

the probability of having a boy = a probability of having a girl = 1/2.

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The sample space of having three children is:

BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.

The probabilities of each of those events is

Pr(BBB)=Pr(BBG)=Pr(BGB)=Pr(GBB)=\\ \\=Pr(BGG)=Pr(GBG)=Pr(GGB)=Pr(GGG)=\\ \\=\dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}=\dfrac{1}{8}.

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4 years ago
What are like terms?
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4 years ago
Solve by completing the square: x^2-2x-1=0
NemiM [27]
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