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GalinKa [24]
3 years ago
6

Pls help math problem

Mathematics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

angle plm+angle lpm+angle pml=180°[by angle sum property]

21°+angle lpm+38°=180°

59°+ lpm=180°

lpm=180°-59°

lpm=121°

lpm+npl=180°[by limear pair]

121°+npl=180°

npl=180°-121°

npl=59°

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It seems to you that fewer than half of people who are registered voters in the City of Madison do in fact vote when there is an
Ad libitum [116K]

Answer:

a) Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

b) The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

c) We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

d) The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

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 interval 1-\alpha, we have the following confidence interval of proportions.

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

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

(a) How might a simple random sample have been gathered?

Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

(b) Construct an 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

You take an SRS of 200 registered voters in the City of Madison, and discover that 122 of them voted in the last non-presidential election. This means that n = 200, \pi = \frac{122}{200} = 0.61.

We want to build an 80% CI, so \alpha = 0.20, z is the value of Z that has a pvalue of 1 - \frac{0.20}{2} = 0.90[tex], so [tex]z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 - 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.5533

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 + 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.6667

The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

(c) Interpret the interval you created in part (b).

We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

(d) Based on your CI, does it seem that fewer than half of registered voters in the City of Madison vote in non-presidential elections? Explain.

The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

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Find the area of a playing field whichs area is 15 meters
uranmaximum [27]
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artcher [175]
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Three muffin pans with six in each pan what is the fraction for this
irakobra [83]

Answer:

1/18 ?

Step-by-step explanation:

I'm not sure if you're written the whole question?

3 pans with 6 muffins in each pan, so total number of muffins = 3 x 6 = 18

So each muffin is 1/18

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