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Triss [41]
3 years ago
13

Can someone help me do this question?

Mathematics
2 answers:
cupoosta [38]3 years ago
8 0

Answer:

Last option.

Step-by-step explanation:

x = -1

y = 6(-1) + 6

y = -6 + 6

y = 0

( -1, 0 )

x = 0

y = -(0)² + 5(0) + 6

y = 0 + 0 + 6

y = 6

( 0, 6 )

Naddik [55]3 years ago
4 0

Answer:

(-1, 0) and (6, 0)

Step-by-step explanation:

y=6x+6\\y=-x^2+5x+6\\\\6x+6=-x^2+5x+6\\6x=-x^2+5x\\x=-x^2

The solutions for these equations are (-1, 0) and (6, 0).

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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.

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In which

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

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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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