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dybincka [34]
2 years ago
10

2. Round off the following numbers to two decimal places:

Mathematics
1 answer:
Viktor [21]2 years ago
6 0

Answer:

I think you mean significant figures

Step-by-step explanation:

  1. 46000
  2. 32000
  3. 560000
  4. 14000
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Step-by-step explanation:

Comment

You are given values for the adjacent side of <x and the hypotenuse of a right triangle. This defines the Cosine of an angle

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Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority
Andre45 [30]

Answer:

The 95% confidence interval of  the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).

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}.

For this problem, we have that:

A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. This means that n = 4000 and \pi = \frac{2250}{4000} = 0.5625

Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 - 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5471

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 + 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5779

The 95% confidence interval of  the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).

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3 years ago
Geometry:
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