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2 years ago

Written as a simplified polynomial in standard form, what is the result when

Mathematics

2 answers:

o-na [289]2 years ago

5 0

Answer:

9x^2 - 10x -4

Step-by-step explanation:

It is set up like this

9x^2 - 8x

- (x-2)2

First solve (x-2)2

the answer is (2x-4)

so now you can do this instead

9x^2 - 8x

- 2x - 4

----------------------------------

9x^2 - 10x +4

9x^2 minus zero is 9x^2

-8x-2x is -10x

0- -4 is 4 because that is like saying 0 + 4 because double negatives equal a positive

A simpler way is to change all the numbers to the opposite sign and then change the minus to a plus

like this

9x^2 - 8x

+ -2x + 4

----------------------------------

9x^2 - 10x +4

EleoNora [17]2 years ago

5 0

Answer:

8x^2-4x-4

Step-by-step explanation:

On delta math

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2 years ago

We would like to create a confidence interval.

Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

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 $1-\alpha$ , we have the confidence interval with a margin of error of:

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In which

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

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of $\sqrt{\pi(1-\pi)}$

a.A 99% confidence level and a sample size of 50 subjects.

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99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}$

b.A 90% confidence level and a sample size of 50 subjects.

$n = 50$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}$

c.A 90% confidence level and a sample size of 300 subjects.

$n = 300$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}$

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

$n = 300$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}$

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