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LekaFEV [45]
4 years ago
10

What is greater 4yd or 13ft

Mathematics
2 answers:
choli [55]4 years ago
8 0
4yds  hope this helped
Lorico [155]4 years ago
5 0
4yd=12 feet

12ft<13ft.

13ft is greater than 4yds.
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Less than $12

Step-by-step explanation:

I changed everything into decimals to make it easier to solve:

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4 years ago
estimate the number of first-year students at an assembly meeting. He surveys 60 students and finds that 24 of them are first-ye
bogdanovich [222]

Answer:

The values needed to calculate a confidence interval at the 95% confidence level are z = 1.96, n = 60, \pi = \frac{24}{60} = 0.4

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).

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 = 60, \pi = \frac{24}{60} = 0.4

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 values needed to calculate a confidence interval at the 95% confidence level are z = 1.96, n = 60, \pi = \frac{24}{60} = 0.4

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.4*0.6}{60}} = 0.276

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.96\sqrt{\frac{0.4*0.6}{60}} = 0.524

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).

6 0
3 years ago
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