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LUCKY_DIMON [66]
3 years ago
6

I need help...I don't understand

Mathematics
1 answer:
Andrej [43]3 years ago
3 0
Keep change change [keep the first integer change the opperation and change the integer on your number or fraction ... Rember the answer takes the sign of the larger number .
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4.
Debora [2.8K]
Use desmos hope that helps
4 0
3 years ago
Triangles J K L and M N R are shown.
Dmitry [639]

Answer:

JK ≅ MN

Step-by-step explanation:

SAS states that any two sides of the angle and the angle itself, if , of two triangles are equal the two triangles are equal.

It is given that  ∠J ≅ ∠M and JL ≅ MR   i.e an angle and a side are equal. We need one more side to prove that the two triangles are equal.

If we look at the diagram closely we see that the angles J and M are formed by the sides JK & JL  and MN& MR.

It is given that JL ≅ MR  so we are left with JK ≅ MN

6 0
3 years ago
Read 2 more answers
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)

3 0
3 years ago
Read 2 more answers
(Please ans. fast)
REY [17]
The answer is D. -8/45
8 0
3 years ago
What is the value of y in the equation 4y – 2(1 – y) = –44?
LekaFEV [45]
Y=-7 use distributive property add like terms 


8 0
3 years ago
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