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strojnjashka [21]
3 years ago
6

Help me with this??​

Mathematics
1 answer:
Mandarinka [93]3 years ago
7 0
The answer would be 2 to the 2nd power
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How many sixths are in 4<br><br><br>how many two-thirds are in 2​
sveticcg [70]

Answer:

24 sixths in 4 and 3 two-thirds

Step-by-step explanation:

6=24 sixths in 4

6 0
3 years ago
Read 2 more answers
Ansel has two family members who travel for work. Ansel’s aunt travels every 12 days. Ansel’s cousin travels every 8 days. Today
Marat540 [252]
24 days,
12 & 8 have a least common multiple of 24 so that is when they would both travel
8 0
3 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
X+9=13352643-2x answer get brainliest
Ymorist [56]

Answer:

4450878

Step-by-step explanation:

7 0
3 years ago
ava pours 3 gallons of paint equally into 5 cans. how many gallon of paint will there be in each can?​
IrinaK [193]

Answer:

15 gallons of paint

Step-by-step explanation:

5 x 3 = 15

equal groups !

6 0
2 years ago
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