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Ira Lisetskai [31]
3 years ago
5

Find the difference of 58.2–27.674. a. 30.652 b. 30.526 c. 35.026 d. 35.620

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
7 0

Answer:

B. 30.526

Step-by-step explanation:

Just subtract.

58.20

-27.674

30.526

Pavel [41]3 years ago
4 0

Answer:

B

Step-by-step explanation:

You just subtract

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24 students bought their permission slips to attend the class field trip to the local art museum. If this represented 8 tenths o
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30 in the class as total. 24/3= 8 30/3 =10

Step-by-step explanation:

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3 years ago
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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
3 years ago
Mk, someone knows this answer. I know it! Pwease help!
rosijanka [135]

Answer:

16(base) x 15(height) x 1/2(basically just dividing by 2) = 120

120 x 2 = <em>240</em>

17 x 17 = 289, 289 x 2 = <em>578</em>

17 x 16 = <em>272</em>

Add all given areas:

240 + 272 + 578 = 1,090 in2 is your answer

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3 years ago
Heather works 40 hours a week and she earns $11.50 per hour. When Heather works over 40 hours, her pay is 1.5 times her hourly p
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Answer: it’s(A.) 563.50
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3 years ago
Write 5 1/3% as a fraction in simplest form
goblinko [34]
5+1/3=                                                                                                              15/3+1/3=                                                                                                        16/3
7 0
3 years ago
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