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vlabodo [156]
3 years ago
13

Luis counted 7 birds in a tree he saw 3 bluebirds in the tree the rest were blackbirds which shows the fraction of birds that we

re blackbirds
Mathematics
2 answers:
Doss [256]3 years ago
6 0
? does a picture come with this?


SCORPION-xisa [38]3 years ago
4 0
There are 4 black birds so the fraction would be 4/7
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Pepper Jackie has 225 square inches for the area for her square mosaic.
Marta_Voda [28]

Answer:

ugh man do I hate iready!!

225/4=56.25

square root of 225 equals 15

225 / 2 equals 112.5

5 0
3 years ago
A markdown that is greater than 99% but less than 100%
GenaCL600 [577]
It is 99.5% because 99.5% is in the middle of 99% and 100%
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5 0
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
Joe's lunch at a restaurant costs $13.00 without tax. He leaves the waiter a tip of 17% of the cost of the lunch, without tax. W
choli [55]

Answer: $15.21

Step-by-step explanation:

Given

Joe's lunch cost $13.00 without tax

He tipped the waiter with 17% of bill without tax

So, the amount of tip is

\Rightarrow 13\times 17\%\\\Rightarrow 13\times 0.17\\\Rightarrow \$2.21

Total amount that food costs without tax is

\Rightarrow 13+2.21\\\Rightarrow \$15.21

7 0
3 years ago
after working out for just 8 minutes,ron has burned 56 calories. how many minutes must he work out if he wants to burn 392 calor
dangina [55]
Ron would have to work out 56 minutes to burn 392 calories
7 0
3 years ago
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