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alisha [4.7K]
2 years ago
6

In a survey of 1000 adults, 34% found they prefer charcoal to gas grills. The 1000 would be considered a:

Mathematics
1 answer:
Morgarella [4.7K]2 years ago
7 0

Answer:

1000 represents the sample

Step-by-step explanation:

Given

Survey:

Adults = 1000

Proportion = 34\%

Required

What does 1000 represent?

First, we should know that 1000 adults were picked or selected from a larger number of adults.

This implies that:

The larger number of adults is the population and the 1000 selected adults represents the sample.

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soldi70 [24.7K]

Answer:

22

Step-by-step explanation:

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The answer is 2.625$ rounded up  to 2.63$
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: Use The TI Calculator To Answer The Question. Find The Probability That A Z-score Will Be Between 0.7 And 1.4. A) 0.242 O B) 0
ZanzabumX [31]

Answer:

P(0.7

And we can find this probability with the following difference:

P(0.7

We can use the following commands on the ti 84

2nd>VARS>DISTR

And then we look for normalcdf and we input this:

normalcdf(0.7,1.4,0,1)

The other possible code would be:

normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)

And we got:

P(0.7

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

For this case we want to find this probability:

P(0.7

And we can find this probability with the following difference:

P(0.7

We can use the following commands on the ti 84

2nd>VARS>DISTR

And then we look for normalcdf and we input this:

normalcdf(0.7,1.4,0,1)

The other possible code would be:

normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)

And we got:

P(0.7

3 0
3 years ago
Read 2 more answers
Student Debt – Vermont: The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to
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Answer:

The confidence interval is (24116.3878,24883.6122).

Step-by-step explanation:

We are given the following information in the question:

Population mean, \mu = $23,500

Sample mean,\bar{x} = $24,500

Sample standard deviation,s = $2,800

Sample size, n =  146

Confidence interval:

\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}  

Putting the values, we get,

t_{critical}\text{ at}~\alpha_{0.05} = \pm 1.655436

24500 \pm 1.65543(\frac{2800}{\sqrt{146}} ) = 24500 \pm 383.6122 = (24116.3878,24883.6122)

The confidence interval is (24116.3878,24883.6122).

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AnsAnswerwer:

sorry i can't help without the picture of the problem

Step-by-step explanation:

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