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Lena [83]
3 years ago
5

Please help timed test

Mathematics
2 answers:
beks73 [17]3 years ago
7 0

Answer:

e

Step-by-step explanation:

bixtya [17]3 years ago
6 0
The answer is E i hope this helped
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Construct a 96% confidence interval if a sampling distribution has a mean of 20, a standard deviation of 5, and a size of 100.
MAXImum [283]

Using the t-distribution, the 96% confidence interval is given as follows:

(18.96, 21.04).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 96% confidence interval, with 100 - 1 = 99 df, is t = 2.0812.

The parameters are given as follows:

\overline{x} = 20, s = 5, n = 100

Hence the bounds of the interval are:

\overline{x} - t\frac{s}{\sqrt{n}} = 20 - 2.0812\frac{5}{\sqrt{100}} = 18.96

\overline{x} + t\frac{s}{\sqrt{n}} = 20 + 2.0812\frac{5}{\sqrt{100}} = 21.04

More can be learned about the t-distribution at brainly.com/question/16162795

#SPJ1

6 0
2 years ago
The equation of (0, 3 (4, 2)
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Answer:

4x+y=12

Step-by-step explanation:

its the only one that is positive

and this is guess so don’t hate me if its wrong

hope this is right

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3 years ago
Finding the missing value to the nearest hundredth. tan__=73
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A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys. One part of
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Answer:

Step-by-step explanation:

In statistics, about 68 percent of values come in one standard deviation of the mean by using a standard normal model.   Approximately 95% of the data were all within two standard deviations from the mean. Almost all of the data are in the range of three standard deviations of the mean (roughly 99.7%).

The 68-95-99.7 law, also known as the Empirical Rule, is based on this evidence. 68 percent of the data values of a naturally distributed data collection of small children with a mean of 8.2 and a standard deviation of 10.8 would be between -2.2 and 19.0.

Within a mean of 14.1 as well as a standard deviation of 8.2, 68 percent of the data values in a usually distributed data collection of older children would be between 5.9 and 22.3.

However, we cannot conclude that the data is naturally distributed since the real actual data vary from the usual normal curve computed above.

Hence, various measures like either goodness of fit or theory testing, would be used for this.

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