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kherson [118]
3 years ago
8

when a movie is rented it has a due date. If the movie is not returned on time, a late fee is assessed. Katy is charged $5 each

day for a movie that is 4 days late.
Mathematics
1 answer:
Tems11 [23]3 years ago
4 0

Okay, if you pay $5 dollars a day, and you pay for 4 days:

5*4 = 20

Katy will have to pay $20

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In a test of the effectiveness of garlic for lowering​ cholesterol, 4545 subjects were treated with garlic in a processed tablet
n200080 [17]

Answer:

5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637

5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837

And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change in the levels

Step-by-step explanation:

For this case we have the following info given:

n=45 represent the sample size

\bar X= 5.1 represent the sample mean

s= 19.1 represent the sample deviation

We can calculate the confidence interval for the mean with the following formula:

\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

The confidence level is 0.90 then the significance level would be \alpha=0.1 and the degrees of freedom are given by:

df= n-1 = 45-1=44

And the critical value for this case would be:

t_{\alpha/2}=2.015

And replacing we got:

5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637

5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837

And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change or efectiveness in the levels

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About % of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9 and z = 0.9 z = 0.9 (or w
prisoha [69]

Using the normal distribution, it is found that 63.18% of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The area within 0.9 standard deviations of the mean is the <u>p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841)</u>, hence:

0.8159 - 0.1841 = 0.6318 = 63.18%.

More can be learned about the normal distribution at brainly.com/question/4079902

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Which transformation maps the pre-image to the image?
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