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Semenov [28]
3 years ago
14

Given m n, find the value of X.

Mathematics
1 answer:
Brut [27]3 years ago
7 0

Answer:

x = 62

Step-by-step explanation:

62 and x are alternate exterior angles and alternate exterior angles are equal when the lines are parallel

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In simplest radical form, what are the solutions to the quadratic equation 6 = x2 - 10x?
WARRIOR [948]

Answer:

x = 5 - \sqrt{31}, 5 + \sqrt{31}

Step-by-step explanation:

6 = x^2 - 10x

0 = x^2 - 10x - 6

Let's use the quadratic formula. ( x = \frac{-b +- \sqrt{b^2 - 4ac}}{2a})

x = \frac{10 +- \sqrt{-10^2 - 4(-6)}}{2}

x = \frac{10 +- 2\sqrt{31} }{2}

  = 5 +- \sqrt{31}

x = 5 - \sqrt{31}, 5 + \sqrt{31}

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3 years ago
Is this statement true?
lara [203]

Answer:

yes and no,depends.

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What is the main difference between multiplying fractions and dividing fractions?
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The numbers get smaller in division and larger in multiplying
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3 years ago
Read 2 more answers
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
The 19 members of a football team are trying to raise at least $1266.00 to cover the traveling cost for a holiday tournament. If
SIZIF [17.4K]
To solve this, you need to do 1,266-451. that equals 815. then, divide 815 by 19. you should get 43. hope that helps.
8 0
3 years ago
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