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2 years ago

What is a quotient of 2.5.

Mathematics

1 answer:

Arlecino [84]2 years ago

4 0

Answer:

The quotient is the solution to a division sum.

Step-by-step explanation:

A single number can not have a quotient.

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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

2 years ago

|4x|+6=22?????????????????????

AnnZ [28]

Answer: x=4, x=-4

Step-by-step explanation: because it is absolute value, negative or positive does not matter.

6 0

3 years ago

Madeleine earned $36,000 last

shusha [124]

Owed 5920 in taxes

0.12*16000+0.2*20000=5920

4 0