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Phoenix [80]
3 years ago
5

On Wednesday, Miguel's bank account balance was -$55. On Thursday, his balance was less than that. Use absolute value to describ

e Miguel's balance on Thursday as a debt. In this situation, -$55 represents a debt of _________. On Thursday, Miguel had a debt of _____________than $55.
Mathematics
1 answer:
Juliette [100K]3 years ago
4 0

In this situation, the -$55 represents the debt of $55, since this happens when Miguel does not have enough money in his account to cover payments or transactions. On Thursday, he has a debt of more than $55. The explanation behind this is if an account balance is less than -55 dollars, it exemplifies a debt bigger than 55 dollars. 

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The sum of two consecutive integers is 4,031. What are the integers?
Sati [7]
They are 2015 and 2016.

divide 4031 by 2, and get rid of the decimal:
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2015+2015 = 4030
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8 0
2 years ago
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10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0
1 year ago
An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim
Flura [38]

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population mean, when the population standard deviation is not provided is:

CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}

The sample selected is of size, <em>n</em> = 50.

The critical value of <em>t</em> for 95% confidence level and (<em>n</em> - 1) = 49 degrees of freedom is:

t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000

*Use a <em>t</em>-table.

Compute the sample mean and sample standard deviation as follows:

\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}

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Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

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