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max2010maxim [7]
2 years ago
7

Do you think a sequence of translations across the x- or y- axis and/or reflections on a figure could result in the same image a

s a 90- degree clockwise rotation? Explain why or why not
Mathematics
2 answers:
vlabodo [156]2 years ago
5 0

I think just two reflections would do it.


First we reflect around y = -x, the 45 degree line through the origin and the second and fourth quadrant.


Then we reflect through the y axis, x=0.


The composition of the two reflections is equivalent to a 90 degree clockwise rotation.



faust18 [17]2 years ago
4 0

Answer:

No, it is not possible to get the same image as a 90-degree clockwise rotation using only translations and/or reflections. In the rotation, the x- and y-coordinates are switched. There is no way to reverse the order of the coordinates using only reflections or translations.

Step-by-step explanation:

*It is not possible.

*The coordinates need to be reversed.

*You cannot reverse coordinates using only reflections and translations.

Guest
1 year ago
thx
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Answer:

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A marketing manager samples 150 people and finds that 87 of them have made a purchase on the internet within the past month. a.
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Answer:

a) \hat p = \frac{87}{150}= 0.58

b) We assume for this case a confidence level of 95%

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.58(1-0.58)}{(\frac{0.03}{1.96})^2}=1039.793  

And rounded up we have that n=1040

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".  

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

Solution to the problem

Part a

For this case the estimated proportion of people who have made a purchase on the internet within the past month is given by:

\hat p = \frac{87}{150}= 0.58

Part b

We assume for this case a confidence level of 95%

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 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.58(1-0.58)}{(\frac{0.03}{1.96})^2}=1039.793  

And rounded up we have that n=1040

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