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olasank [31]
3 years ago
15

Please answer this!!

Mathematics
2 answers:
777dan777 [17]3 years ago
4 0

Answer:

C, 5/12

Step-by-step explanation:

The tangent of an angle is defined as the side opposite to that angle divided by the side adjacent to that angle. The tangent of angle A would be equal to the value of side BC divided by side AB. The value of side BC is 5, and the value of side AB is 12. The answer is 5/12.

earnstyle [38]3 years ago
4 0

Answer: ∠A=\frac{5}{12}

Step-by-step explanation:

Tangent is opposite over adjacent.

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a twelve sided die with side labeled 1-12 will be rolled once. What is the probability of rolling a number less than 9?
puteri [66]
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1 year ago
Find the value of x. Show all of your work!
Georgia [21]

Answer:

145⁰

Step-by-step explanation:

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7 0
3 years ago
A segment has endpoints (-9,-20) and (14,12). What is the midpoint of this segment
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3 years ago
a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske
Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

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

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1600, \pi = 0.4

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}} = 0.4 - 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.3685

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.4315

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

7 0
3 years ago
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