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antiseptic1488 [7]
3 years ago
12

What is a good comparison of the estimated sum and the actual sum of 7 9/12+2 11/12

Mathematics
1 answer:
earnstyle [38]3 years ago
4 0
10 8/12 because you get an improper fraction when you add 9/12 and 11/12 (20/12) so you see how many times 12 goes into 20 which is only once. you see how many are left over from when you take 12 from 20. then you have you fraction which is 8/12. now you have to add the numbers which is 7+2=9 but 12 goes into 20 once so you add 1 to your number to make it 10. That's the answer, 10 8/12.
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Answer:

The 99% confidence interval for p in this case is (0.3317, 0.5883).

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 level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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This means that:

n = 100, \pi = \frac{46}{100} = 0.46

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.3317

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 + 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.5883

The 99% confidence interval for p in this case is (0.3317, 0.5883).

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