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Orlov [11]
3 years ago
9

If square LMNO is dilated by a scale factor of two about the center of the square to create square L'M'N'O', dilated line P'Q' w

ill (5 points) Question 6 options: 1) be parallel to line PQ and pass through point N 2) contain the points P and Q 3) be perpendicular to line PQ and pass through point L 4) shift two units to the right

Mathematics
2 answers:
allochka39001 [22]3 years ago
7 0

Line PQ is vertical thru the middle of the original square. The new square is dilated with a scale factor of 2 at the center, so the dilated line P'Q' would also be vertical running through the middle, so it would contain the points P and Q.


The answer is 2) contain the points P and Q

Genrish500 [490]3 years ago
6 0

Solution:

As given Square L M NO is dilated by  a scale factor of two about the center of the square to create square L'M'N'O'.

Original line of Dilation = Along P Q

New Dilated line = P'Q'

As scale factor > 1

1. Image Size > Pre image size

2. The two images will be similar.

3. Length of Dilated Line P' Q' = 2 × Length of PQ

As you can see from the diagram drawn below, Dilated line P'Q'  will contain the point P and Q.

All four points P,Q,Q',P' are collinear , lie in the same line.

Option (2) dilated line P'Q' will contain the points P and Q is true.



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Answer:

z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

\hat p=0.021 estimated proportion of interest

p_o=0.012 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:  

Null hypothesis:p \leq 0.012  

Alternative hypothesis:p > 0.012  

When we conduct a proportion test we need to use the z statisitic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

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Read 2 more answers
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