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zheka24 [161]
3 years ago
8

The probability that a random smoker will develop a sever lung condition during his or her lifetime is 0.3. we will choose a ran

dom sample of 120 smokers and let x be the number of these smokers that will develop a severe lung condition.
a.in our sample, find the mean and standard deviation of the number of smokers that will develop a sever lung condition. give your answers to two decimal plac
Mathematics
1 answer:
postnew [5]3 years ago
8 0
The mean is 36 and the standard deviation is 5.02.

The mean is given by
μ = np = 120*0.3 = 36.

The standard deviation is given by
σ = √(n*p*(1-p)) = √(120*0.3*0.7) = √25.2 = 5.02.
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All the input values are being multiplied by themselves by the output values.

This creates a congruent and linear correlation and congruence. 

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Anuta_ua [19.1K]

Answer:

1)  -120, -240, -360

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Answer:

2π - 4√5

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If you want an approximate value it is

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