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Solnce55 [7]
3 years ago
8

Vladimir says that the equation of the line that passes through points above Robyn says that the line passes through the points

above Who is correct?​

Mathematics
1 answer:
aleksley [76]3 years ago
8 0

Answer:

Both are correct

Step-by-step explanation:

The Line has the following equation: y= 1+4/5 x

So we need to know which points: (-5,-3) and (10,9), (-10,-7). (-15, -11) are part of the line

For instances, lest start by evaluating (-15,-11) in y=1+4/5 x

this leaves us to the following operation: -11=1+4/5 * (-15)

-11=1-4*3 resulting in a equality -11=-11, this point does belong to the line

Repeat the above for all of the remaining points and you will get the same conclusion, that is we Vladimir and Robyn are correct.

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A right prism is made up of two identical, parallel bases on the ends, and the faces are perpendicular to the bases. The formula for finding surface area is 2B + hP (where B is the area of one of the bases, h is the prism's height and P is the perimeter of the base)

Step-by-step explanation:

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3 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
kobusy [5.1K]

Answer:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

Step-by-step explanation:

Previous concepts

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".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

4 0
3 years ago
Select the correct answer.
Oduvanchick [21]
I think answer should be thousands please give me brainlest I hope this helps let me know if it’s correct or not okay thanks
6 0
2 years ago
Read 2 more answers
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