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Mnenie [13.5K]
4 years ago
7

Suppose you have data that shows that 12% of athletes test positive for steroids. You also know that 11% of athletes test positi

ve for steroids and actually use steroids. What is the probability that an athlete uses steroids, given that he tests positive?
Mathematics
2 answers:
Agata [3.3K]4 years ago
7 0

The probability that an athlete uses steroids, given that he tests positive is 0.92. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

musickatia [10]4 years ago
4 0

Answer:  Probability that an athlete uses steroids given that he tests positive is 91.6%.

Step-by-step explanation:

Let A be the event that athletes test positive for steroids .

Let B be the event that he actually use steroids.

Since we have given that

Probability that athletes test positive for steroids = P(A) = 12%

Probability that athletes test positive for steroids and actually use steroids = P(A∩B) = 11%

So, we need to find the probability that an athletes use steroids given that he tests positive.

For this we will use "Conditional probability":

P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}\\\\P(B\mid A)=\dfrac{11}{12}\\\\P(B\mid A)=0.916\\\\P(B\mid A)=0.916\times 100\%\\\\P(B\mid A)=91.6\%

Hence, Probability that an athlete uses steroids given that he tests positive is 91.6%.

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Pls answer! I don’t know math @ all
Lana71 [14]

The probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is \frac{5}{12}.

Solution:

Total number of outcomes N(S) = 36

Let A be the sum that is a multiple of 6 and

B be the sum that is a multiple of 4.

Sum that is a multiple of 6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)

N(A) = 6

Sum that is a multiple of 4 = (1, 3), (2, 2), (2, 6), (3, 1), (3, 5),

                                              (4, 4), (5, 3), (6, 2), (6, 6)

N(B) = 9

$P(A)=\frac{N(A)}{N(S)}

$P(A)=\frac{6}{36}=\frac{1}{6}

$P(B)=\frac{N(B)}{N(S)}

$P(B)=\frac{9}{36}=\frac{1}{4}

Probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4:

P(A \cup B)=P(A)+P(B)

$P(A \cup B)=\frac{1}{6} +\frac{1}{4}

               $=\frac{2+3}{12} (Make the denominator same)

               $=\frac{5}{12}

Hence the probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is \frac{5}{12}.

8 0
3 years ago
Which will result in a difference of squares?
SIZIF [17.4K]

Answer:

The expression which will result in difference of two squares is:

 (–7x + 4)·(–7x – 4)

Step-by-step explanation:

We know that the formula of the type:

(a-b)(a+b)=a²-b²

i.e. it is a difference of two square quantities. (a^2 and b^2)

since,

a= -7x , b=4

(-7x+4)(-7x-4)= (-7x)² - (4)²

=(7x)² - 4²

So the expression is a difference of two square quantities:

(7x)² and (4)²

Hence the answer is (–7x + 4)·(–7x – 4)....

5 0
3 years ago
Rationalize and simplify completely. <br><br> <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%7B%7D%206%7D%7B%20%5Csqr
lora16 [44]

\frac{ \sqrt{6} }{ \sqrt{5} }   \times  \frac{ \sqrt{5} }{ \sqrt{5} }  =  \frac{ \sqrt{30} }{5}
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3 years ago
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Answer:

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