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dexar [7]
2 years ago
13

Doug mcdougal is whipping up 72 different snacks for his 12th annual blooper bowl bash he invited 138 guess but only 16 are comi

ng that doesn’t bother dog he still makes 100 bowls of nacho dip he makes twice as many bowls of chili how many guests are coming to the party
Mathematics
1 answer:
Anna71 [15]2 years ago
6 0

The number of guests coming to Doug McDougal party can be calculated to be 23 people.

<h3>What number of people are coming?</h3>

Doug invited 138 guests and out of this, only 1/6 will be coming.

The number of people coming is therefore:

= Number invited x Proportion coming

Solving gives:

= 138 x 1/6

= 23 people

Find out more on problems involving proportions at brainly.com/question/14581282.

#SPJ1

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You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the
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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}}

In which

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

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

98% confidence level

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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Answer: <

6.12 x 10^4 < 612,000

Step-by-step explanation:

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Read 2 more answers
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