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2 years ago

What are the coordinates of R' for the dilation D(0.5. p)(PQRS)?

Mathematics

1 answer:

RideAnS [48]2 years ago

6 0

Answer:

Step-by-step explanation:

Three step process.

Step 1, convert Q, R, P, and S to coordinates measured from point P rather than from the origin by subtracting P from each point:

Q' = Q-P = (-3-(-3),2-(-3)) = (0,5)

R' = R-P = (1-(-3),2-(-3)) =(4,5)

P' = P-P = (-3-(-3),-3-(-3)) =(0,0)

S' = S-P = (1-(-3),-3-(-3)) = (4,0)

Step 2, Apply the dilation factor (0.5):

Q'' = (0.5)Q' = (0.5*0, 0.5*5) = (0,2.5)

R'' = (0.5)R' = _____

P''= (0.5)P' = _____

S''= (0.5)S' = _____

Step 3, Convert coordinates back to the origin (vice point P) by adding P to each point. These are the dilated points.

Q''' = Q'' + P = (0+(-3),2.5+(-3)) = (-3, -0.5)

R''' = R'' + P = _____

P''' = P'' + P = _____

S''' = S'' + P = _____

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hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a

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Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

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2 years ago

Aileen can read 1.5 pages for every page her friend can read. Aileen's mom was very excited and she said to Aileen: "So, if your

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No, Aileen's mom is incorrect. Aileen can read 30 pages in the same time period as her friend reads 20 pages.

To get this answer, you multiply 20 by 1.5

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The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

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Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

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