<img src="https://tex.z-dn.net/?f=%28x%20-%203%29%20%7B2%7D" id="TexFormula1" title="(x - 3) {2}" alt="(x

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Nat2105 [25]

2 years ago

ddle" class="latex-formula">

Mathematics

2 answers:

ch4aika [34]2 years ago

4 0

The answer is = x^2-6x-9

a_sh-v [17]2 years ago

3 0

Answer:

Step-by-step explanation:

( x - 3)²

= x² - 2 * x * 3 + 3²

= x² - 6x + 9

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3 years ago

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