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stich3 [128]
2 years ago
10

Positive integers greater than -6 but less than +21​

Mathematics
1 answer:
Nina [5.8K]2 years ago
7 0

Step-by-step explanation:

<u>Inequality to describe problem:</u><u> </u>

- 6 < x < 21

x represents the unknown integer(s).

Integers are positive or negative numbers that <u>don't</u> have a fraction, or a decimal.

<u>Positive integers that work in the inequality:</u>

<u>- 6 < 1< 21</u>

- 6 < 2 < 21

- 6 < 3 < 21

- 6 < 4 < 21

- 6 < 5 < 21

- 6 < 6 < 21

- 6 < 7 < 21

- 6 < 8 < 21

- 6 < 9 < 21

- 6 < 10< 21

- 6 < 11 < 21

- 6 < 12 < 21

- 6 < 13 < 21

- 6 < 14 < 21

- 6 < 15 < 21

- 6 < 16 < 21

- 6 < 17 < 21

- 6 < 18 < 21

- 6 < 19 < 21

- 6 < 20 < 21

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 work out as positive integers.

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<u><em>Answer:</em></u>

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"-3x" + 5 &lt; "-10" Inequality 2: 7x + x - 4 &gt; 28 Which of the following values makes BOTH inequalities TRUE? Select all tha
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

Assuming that \bar X =5.5 and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

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Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that t_{\alpha/2}=2.33

Now we have everything in order to replace into formula (1):

5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139    

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So on this case the 95% confidence interval would be given by (5.139;5.861)    We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

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