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maksim [4K]
3 years ago
7

Please help hurry I will mark brainliest

Mathematics
2 answers:
sladkih [1.3K]3 years ago
5 0

Answer:

15

Step-by-step explanation:

pretty sure dont get mad if i am wrong tho

sorry if i did make u get it wrong

nataly862011 [7]3 years ago
4 0

Answer:

Step-by-step explanation:

x values: 0,1,2,3,4,5

y=3/4(0)+5

=5

y=3/4(1)+5

=5.75

y=3/4(2)+5

=6.5

y=3/4(3)+5

=7.25

y=3/4(4)+5

=8

y=3/4(5)+5

=8.75

plot: (0,5)(1,5.75)(2,6.5)(3,7.25)(4,8)(5,8.75)

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Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of
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Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

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This means that n = 100, \pi = \frac{42}{100} = 0.42

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.293

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 + 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.547

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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