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Anit [1.1K]
2 years ago
15

Is there a relationship between the distance and the sum? Is there a relationship between the distance and the difference? A 5-c

olumn table with 3 rows. Column 1 is labeled a with entries 1, 4, negative 6. Column 2 is labeled b with entries 2, negative 1, negative 3. Column 3 is labeled a + b with entries 3, 3, negative 9. Column 4 is labeled a minus b with entries negative 1, 5, negative 3. Column 5 is labeled Distance with entries 1 unit, 5 units, 3 units. Which describes the relationship between the distance and the difference? The distance is always the opposite of the difference. The distance is exactly the difference. The distance is the absolute value of the difference. The distance is not related to difference.
Mathematics
1 answer:
Genrish500 [490]2 years ago
3 0

Answer:

  • C) The distance is the absolute value of the difference

Step-by-step explanation:

We are interested in the last two columns.

If we compare them we see that each value of the distance column is the absolute value of corresponding value of difference:

  • a - b        = 1, 5, - 3
  • distance  = 1, 5, 3

Corect choice is C

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Answer:

a) 654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

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Step-by-step explanation:

Part a

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

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In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

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Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that t_{\alpha/2}=2.01

Replacing we got:

654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

654.16+2.01\frac{164.55}{\sqrt{52}}=700.03    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

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