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nydimaria [60]
3 years ago
6

Witold says that 50% of a number will always be greater than 20% of any other number. Complete one inequality to support Witold'

s claim and one to show that he is incorrect.
Mathematics
1 answer:
Lelu [443]3 years ago
4 0
.50 x 2 < .20 x 100
1 < 20

50% of 2 is not greater than 20% of 100.
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B

Step-by-step explanation:

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A therapist wanted to determine if yoga or meditation is better for relieving stress. The therapist recruited 100 of her high-st
schepotkina [342]

Answer:  Choice B

There is not convincing evidence because the interval contains 0.

========================================================

Explanation:

The confidence interval is (-0.29, 0.09)

This is the same as writing -0.29 < p1-p1 < 0.09

The thing we're trying to estimate (p1-p2) is between -0.29 and 0.09

Because 0 is in this interval, it is possible that p1-p1 = 0 which leads to p1 = p2.

Therefore, it is possible that the population proportions are the same.

The question asks " is there convincing evidence of a difference in the true proportions", so the answer to this is "no, there isn't convincing evidence". We would need both endpoints of the confidence interval to either be positive together, or be negative together, for us to have convincing evidence that the population proportions are different.

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105 divided by 3<br> long division estimating please show your work
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35 I had this for a test the answer is

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Evaluate 42,147÷63. Round to the nearest whole number, if necessary.
Harman [31]

Answer:

669 is the answer to this equation

Step-by-step explanation:

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The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A
Aloiza [94]

Answer:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

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

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

\mu=37 and \sigma=28  and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

4 0
3 years ago
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