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irga5000 [103]
3 years ago
10

After a rotation of 90° about the origin, the coordinates of the vertices of the image of a triangle are A'(6, 3), B'(–2, 1), an

d C'(1, 7). What are the coordinates of the vertices of the pre-image?
Mathematics
2 answers:
Licemer1 [7]3 years ago
4 0
Assuming the original was rotated counterclockwise, and this only works for a rotation about the origin, a rotation of 90 degrees switches the x and y coordinate and depending on which part of the graph (quadrant) it is in the signs may change as well.  First, and the trickiest part is finding how the signs change. 

There are four quadrants, first, second third and fourth starting with the upper right than upper left than lower left then lower right.Since none of the veracities have a 0 in them we don't have to worry about the x or y axis, though it is worth realizing each axis is 90 degress from the others.  So something  at 2 on the x axis (2,0) moves 90 degrees counterclockwise would then be at 2 on the y axis (0,2) and moves one more time would be at (-2,0).

A' is in the first quadrant (upper right) so the previous quadrant 90 degrees clockwise is the fourth quadrant (lower right).  The signs of the coordinates then will go from (+,+) to (+,-).

Now that we have the signs we just switch the coordinates so (6,3) goes to (3,6) and the sign change makes it (3,-6).  Just follow this pattern with the other two points (-2,1), and (1,7) find the signs by figuring out the previous quadrant than switch the coordinates and apply the sign changes
iragen [17]3 years ago
3 0

A=(3,-6) B=(1,2) C=(7,-1)

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Answer:

a

The null hypothesis is  H_o : \mu  =  13.4000

The alternative hypothesis is  H_a :  \mu \ne  13.4000

The null hypothesis is rejected

b

The  99% confidence level is   13.3930  < \mu  < 13.3994

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  10

   The  population mean is  \mu =  13.4 000 \  angstroms

   The level of significance is  \alpha =  0.05

   The  sample data is  

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

        \= x =  \frac{13.3987+ 13.3957\cdots +13.4002 }{10}

=>     \= x = 13.3962

Generally the sample standard deviation  is mathematically represented as

    \sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }

=> \sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 +  (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2  }{10} }

=>  \sigma =0.0039

The null hypothesis is  H_o : \mu  =  13.4000

The alternative hypothesis is  H_a :  \mu \ne  13.4000

Generally the test statistics is mathematically represented as

      z  =  \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }

=>    z  =  \frac{13.3962 -  13.4000}{\frac{0.0039}{\sqrt{10} } }

=>   z =  3.08

Generally the p-value is mathematically represented as

       p-value  = 2P( z   >  3.08)

From the z-table  P(z >  3.08)= 0.001035

So

       p-value  = 2* 0.001035

      p-value  = 0.00207

So from the obtained value we see that

     p-value  < \alpha

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is  99%  then the level of significance is

    \alpha =  (100 -99)\%

=> \alpha =  0.01

Generally from the normal distribution table critical value  of  \frac{\alpha }{2} is  

    Z_{\frac{\alpha }{2} } =  2.58

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>   E = 2.58  *  \frac{0.0039}{\sqrt{10} }

=>    E = 0.00318

Generally the 99% confidence interval is mathematically represented as

     13.3962   - 0.00318  < \mu  < 13.3962   +  0.00318

=>   13.3930  < \mu  < 13.3994

 

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