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grandymaker [24]
3 years ago
8

At a restaurant mike and his three friends decide to divide the bill evening. If each person paid 13 then what was the total bil

l?
Mathematics
1 answer:
Hatshy [7]3 years ago
8 0
At a restaurant Mike and his three friends that means that there was four people splitting the bill. If each of them paid $13 then the bill was $52

4 * 13 = $52
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A die is rolled 25 times and 12 evens are observed. Calculate and interpret a 95% confidence interval to estimate the true propo
Archy [21]

Answer:

The 95% confidence interval to estimate the true proportion of evens rolled on a die is (0.2842, 0.6758). This means that we are 95% sure that for the entire population of dies, the true proportion of evens rolled on a die is between 0.2842 and 0.6758

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 = 25, \pi = \frac{12}{25} = 0.48

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 - 1.96\sqrt{\frac{0.48*0.52}{25}} = 0.2842

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 + 1.96\sqrt{\frac{0.48*0.52}{25}} = 0.6758

The 95% confidence interval to estimate the true proportion of evens rolled on a die is (0.2842, 0.6758). This means that we are 95% sure that for the entire population of dies, the true proportion of evens rolled on a die is between 0.2842 and 0.6758

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2x+30x=69212

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now divide via by 32

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Answer: Yes they could.

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