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tamaranim1 [39]
3 years ago
10

A business school event invites all of its graduate and undergraduate students to attend. Of the students who attend, male gradu

ate students outnumber male undergraduates by a ratio of 7 to 2, and females constitute 70% of the group. If undergraduate students make up 1/6 of the group, which of the following CANNOT represent the number of female graduate students at the event?
a. 18
b. 27
c. 36
d. 72
e. 180
Mathematics
1 answer:
Mademuasel [1]3 years ago
7 0
B)27
To solve this problem using a double set matrix, first jot down one set of variables as the row headings and The other as column headings, as well as a row and column for “totals”. Now all you need to do is add in the information line by line as you read through the question.
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Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

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}}

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This means that n = 2809, \pi = \frac{1634}{2809} = 0.5817

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So \alpha = 0.02, z is the value of Z that has a pvalue of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

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