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DerKrebs [107]
3 years ago
15

According to a survey conducted by Deloitte in 2017, 0.4702 of U.S. smartphone owners have made an effort to limit their phone u

se in the past. In a sample of 54 randomly selected U.S. smartphone owners, what is the probability that between 22 and 29 (inclusively) will have attempted to limit their cell phone use in the past?
1) 0.0241.
2) 0.1703.
3) 0.8779.
4) 0.0483.
5) 0.1221.
Mathematics
1 answer:
musickatia [10]3 years ago
5 0

Answer: 0.628

Step-by-step explanation:

Probability of U.S. smartphone owners have made an effort to limit their phone use in the past : p = 0.4702

Sample size : n= 54

Mean : \mu =(54)(0.4702)=25.39

Standard deviation:

\sigma=\sqrt{(54)(0.4702)}=\sqrt{25.39}\\\\\approx5.039

The probability that between 22 and 29 (inclusively) will have attempted to limit their cell phone use in the past will be :

P(22\leq x\leq29)=P(21

Hence, the required probability = 0.628

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Answer:

Step-by-step explanation:

If we choose chairs having odd number in the row

no of chairs from which selection is made = 10

no of chairs to be selected = 5

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similarly if we choose hairs having even numbers only ,

similar to above , no of ways

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Total no of ways

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6 0
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Answer:

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2 years ago
Given g(x)=x^2 and h(x)= x-3 find (g ∘ h)(x)
DerKrebs [107]

Answer:

<h2>(g ∘ h)(x) = x² - 6x + 9</h2>

Step-by-step explanation:

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To find (g ∘ h)(x) substitute h(x) into g(x) that's for every x in g(x) replace it with h(x)

That's

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We have the final answer as

<h3>(g ∘ h)(x) = x² - 6x + 9</h3>

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