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2 years ago

Given the image below, what is the image of C after a dilation with a scale factor of 3

Mathematics

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer: The coordinates of point C after the dilation are (-2, 5)

Step-by-step explanation:

I guess that you want to find where the point C ends after the dilation.

Ok, if we have a point (x, y) and we do a dilation with a scale A around the point (a,b), then the dilated point will be:

(a + A*(x - a), b + A*(y - b))

In this case we have:

(a,b) = (2,1) and A = 3.

And the coordinates of point C, before being dilated, are: (1, 2)

Then the new location of the point C will be:

C' = (1 + 3*(1 - 2), 2 + 3*(2 - 1)) = (1 -3, 2 + 3) = (-2, 5)

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A tailor cuts a piece of thread One-half of an inch long from a piece Nine-sixteenths of an inch long. What is the length of the

lukranit [14]

Answer:

The length of the remaining piece of thread is $\frac{1}{16}.$

Step-by-step explanation:

Given:

A tailor cuts a piece of thread One-half of an inch long from a piece Nine-sixteenths of an inch long.

Now, to find the length of the remaining piece of thread.

Total thread = $\frac{9}{16} \ inch.$

Tailor cuts a piece of thread = $\frac{1}{2}\ inch.$

Now, to get the length of the remaining piece of thread by subtracting tailor cuts a piece of thread from the total thread:

$\frac{9}{16} -\frac{1}{2} \\\\=\frac{9-8}{16} \\\\=\frac{1}{16}$

Therefore, the length of the remaining piece of thread is $\frac{1}{16}.$

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1 year ago

Line jk passes through points j(-4,-5) and k(-6,3 if the equation of the lines is written in slope-intercept form,y=my+b what is

lidiya [134]

Written in 2-point form, the equation of the line is
y = (y2-y1)/(x2-x1)·(x-x1) +y1
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2 years ago

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg

Sati [7]

Answer:

a) Mean blood pressure for people in China.

b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg

f) 157.44mmHg

Step-by-step explanation:

When the distribution is normal, we use 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.

If X is two standard deviations from the mean or more, it is considered unusual.

In this question:

$\mu = 128, \sigma = 23$

a.) State the random variable.

Mean blood pressure for people in China.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the pvalue of Z when X = 135.

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

$Z = \frac{135 - 128}{23}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the pvalue of Z when X = 141.

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

$Z = \frac{141 - 128}{23}$

$Z = 0.565$

$Z = 0.565$ has a pvalue of 0.7140

71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So

X = 125

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

$Z = \frac{125 - 128}{23}$

$Z = -0.13$

$Z = -0.13$ has a pvalue of 0.4483

X = 120

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

$Z = \frac{120 - 128}{23}$

$Z = -0.35$

$Z = -0.35$ has a pvalue of 0.3632

0.4483 - 0.3632 = 0.0851

8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From b), when X = 135, Z = 0.3

Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then

X = 120

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

$1.28 = \frac{X - 128}{23}$