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4vir4ik [10]
3 years ago
9

Mia enlarged a plan for an outdoor stage. The original plan is shown below. She dilated the outdoor stage by a scale factor of 4

with the center of dilation at the origin. Which ordered pair will be the coordinates of one of the new vertices? A (2,1) B (8,16) C (32,4) D (32,16)

Mathematics
2 answers:
garik1379 [7]3 years ago
5 0
Since you didn't give the original picture, I can only explain the process. Then, you need to select the correct point.

In a dilation, we multiply the coordinates by the scale factor. Take the points that you have and multiply them all by 4. Then, look for a point making your new coordinates.
yawa3891 [41]3 years ago
5 0

Answer:

D (32,16)

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.  Please see the attached photo.

In the photo, the four vertices are:

(-5,4)  by a scale factor of 4 (-20, 16)

(8, 4)  by a scale factor of 4 (32, 16)

(-6, -2)  by a scale factor of 4 (-24, -8)

(8, -2)   by a scale factor of 4 (31, -8)

So only D is correct.

Hope it will find you well.

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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
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