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3 years ago

Draw the angle below in standard position. -3pi/4

Mathematics

1 answer:

andriy [413]3 years ago

8 0

First of all, the negative sign means that we will turn clockwise.

Since $\frac{\pi}{4}$ is half a quarter, we have:

$-\frac{\pi}{4}$ brings you at half of quadrant IV
$-\frac{2\pi}{4}$ brings you on the negative side of the y axis
$-\frac{3\pi}{4}$ brings you at half of quadrant III

So, option A is correct.

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9 + |-78| - (-22)<br><br> Answer truthfully please.

GuDViN [60]

Answer:

109

Step-by-step explanation:

9 + |-78| - (-22)

Apply the absolute value which states |-x| = x.

9 + 78 - (-22)

Apply the rule that states x - (-y) = x + y.

9 + 78 + 22

Add 9 and 78.

87 + 22

Add 87 and 22.

109.

4 0

2 years ago

It seems to you that fewer than half of people who are registered voters in the City of Madison do in fact vote when there is an

Ad libitum [116K]

Answer:

a) Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

b) The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

c) We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

d) The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

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 $1 - \frac{\alpha}{2}$ .

(a) How might a simple random sample have been gathered?

Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

(b) Construct an 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

You take an SRS of 200 registered voters in the City of Madison, and discover that 122 of them voted in the last non-presidential election. This means that $n = 200, \pi = \frac{122}{200} = 0.61$ .

We want to build an 80% CI, so $\alpha = 0.20$ , z is the value of Z that has a pvalue of $1 - \frac{0.20}{2} = 0.90[tex], so [tex]z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 - 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.5533$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 + 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.6667$

The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

(c) Interpret the interval you created in part (b).

We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

(d) Based on your CI, does it seem that fewer than half of registered voters in the City of Madison vote in non-presidential elections? Explain.

The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

4 0

3 years ago

Please help me with these 2 questions. I am not good with angles

DIA [1.3K]

They are 1st one is vertical angles, second one is complementary

8 0

3 years ago

Read 2 more answers

Josh is preparing for an exam by completing practice tests that have different numbers of questions. For each practice test, he

asambeis [7]

The residuals are the numbers shown next to each black line. Just total them:

15 + 10 + 11-10 +9 +10 +3-17-12 = 19.