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GalinKa [24]
3 years ago
7

34​% of all doctors are family practitioners. If 3 doctors are selected at ​random, what is the probability that all 3 are famil

y​ practitioners? Express your answer to the nearest thousandth.
Mathematics
1 answer:
stich3 [128]3 years ago
7 0

Answer:

0.039

Step-by-step explanation:

(0.34)(0.34)(0.34) = 0.34³

= 0.039304

= 0.039

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Answer:

The area is 21.

Step-by-step explanation:

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7 0
3 years ago
Please help me in this question
charle [14.2K]
Number defective=4
number of non-defective=total-defective
=20-4
=16
the questions asked for 1 defective
so we choose only 1 defective from 4 defective items
4C1

and for 1 non-defective
we choose only 1 non-defective from 16 non-defective items
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we choose 2 items from 20 items
so 20C2

answer:
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6 0
2 years ago
 (6 pts) The average age of CEOs is 56 years. Assume the variable is normally distributed. If the SD is four years, find the pr
Alenkinab [10]

Answer:

The probability that the age of a randomly selected CEO will be between 50 and 55 years old is 0.334.

Step-by-step explanation:

We have a normal distribution with mean=56 years and s.d.=4 years.

We have to calculate the probability that a randomly selected CEO have an age between 50 and 55.

We have to calculate the z-value for 50 and 55.

For x=50:

z=\frac{x-\mu}{\sigma}=\frac{50-56}{4}=\frac{-6}{4}=   -1.5

For x=55:

z=\frac{x-\mu}{\sigma}=\frac{55-56}{4}=\frac{-1}{4}=-0.25

The probability of being between 50 and 55 years is equal to the difference between the probability of being under 55 years and the probability of being under 50 years:

P(50

5 0
3 years ago
Value of x in log5x = 4logx5 is
olchik [2.2K]
Log5 x=4 logx 5
Ln x/ln5=4(ln5 / ln x)                                 logx z=ln x / ln z
least common multiple=ln5 * ln x

ln²x=4ln²5
√(ln² x)=√(4 ln²5)
ln x=2ln 5
x=e^2ln5=25

Answer: x=25.

4 0
3 years ago
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