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yulyashka [42]
3 years ago
8

**ANSWER ASAP WILL GIVE BRAINLIEST**

Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

1. given

2. definition of angle bisector

3. reflexive property

4. given

5. definition of angle bisector

6. i have no idea since there are no labels

Step-by-step explanation:

I don't think anyone will answer this since the triangles have no label on what is A, B, C, D and on

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A shose box measures 15in.by 7in.by 4 an half
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The volume is 472.5in^3.

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In quadrilateral LMNO, LO ∥ MN.
wlad13 [49]

The additional information which would be sufficient to conclude that LMNO is a parallelogram is; ML ∥ NO, LO ≅ MN, and ML ≅ LO.

<h3>What information renders LMNO a parallelogram?</h3>

The condition for a quadrilateral to be a parallelogram is that; the opposite pairs must be parallel and consequently opposite pairs are congruent as they have equal length measures.

On this note, it can be concluded that the additional information which would be sufficient are; ML ∥ NO, LO ≅ MN, and ML ≅ LO.

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1 year ago
The Ohio Department of Agriculture tested 203 fuel samples across the state
Rus_ich [418]

Answer:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

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 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The proportion estimated would be:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

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3 years ago
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3 x -6 + 4 x -5
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Read 2 more answers
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