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3 years ago

If y = 2x + 7 were changed to y = 12x + 7, how would the graph of the new function compare with the original?

2 answers:

6 0

The <u>correct answer</u> is:

C) It would be steeper.

Explanation:

These equations are in slope-intercept form.

Slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept.

The change from y=2x+7 to y=12x+7 is that 2 is changed to 12. Both 2 and 12 are in the place of m, meaning they are the slope.

This means the y-intercept stays the same; they start in the same position.

The slope, or rate of change, changes from 2 to 12. Since both are positive, the line will continue to point up (increase). However, an increase in slope means the line will increase more quickly, which results in a steeper line.

dangina [55]3 years ago

4 0

The x value governs the steepness of the line, so 12x is steeper than 2x

So the answer is C

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a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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

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

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

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

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We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

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

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$