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Svetllana [295]
3 years ago
6

I've been stuck on this question for a while now, I need help!

Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0
In order to solve it you must make each component negative. Multiply them by negative one!

M < 2

Now which numbers would satisfy it? M is less than two... so, m could be one, or any negative number, yeah?

Multiply by negative one. - M and -2 are what you have. If it were one... negative one is less than negative two. Turns out M is less than two.

M < 6

M is less than six. It could be 1, 2, 3, 4, or 5.

-M and -6

Is negative 3 less than -6? All of the numbers are! So, m is actually less than 6


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There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am
Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

n = 479, \pi = \frac{130}{479} = 0.2714

Hence the bounds of the interval are found as follows:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0
2 years ago
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