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Lyrx [107]
3 years ago
12

What kind of function is 3y-2=x+7

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0
I agree with the answer above
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Cassie paid $276.45 for 5 months of cell phone service. At that rate, how much will Cassie pay for a full year of cell phone ser
irinina [24]
It’s 663.48 because if you divide 276.45 by 5 than you get 55.19 and multiply that by 12 and it’s 663.48
3 0
2 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

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3 years ago
(Writing to Explain) There are 12 plots in a community garden. What Information do you need to know if you wanted to know how mu
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3 years ago
You and three friends are going to Splash Water Park. The park has three water rides. THe Coaster, Slide, and Drop cost $5.50, $
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3 years ago
Read 2 more answers
f(x) = x^2 − 5x/( x^2 − x − 20 ), x = 5.5, 5.1, 5.05, 5.01, 5.005, 5.001, 4.9, 4.95, 4.99, 4.995, 4.999
vivado [14]
<span>the answer is 0.1   (have  to type in stuff because of the 20 char limit)</span>
4 0
3 years ago
Read 2 more answers
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