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givi [52]
3 years ago
15

If triangle ABC is defined by the coordinates A(3, 6), B(7, 6), C(7, 3) is dilated by a scale factor of 3, with resulting vertex

A' at (9, 18). What is the center of the dilation?
A) (0,0)

B) (3,6)

C) (9,18)

D) (6,12)
Mathematics
2 answers:
san4es73 [151]3 years ago
8 0
Are all these the choices??
None of these can actually be the answer
mars1129 [50]3 years ago
5 0

Answer:

the answer would be A trust me

Step-by-step explanation:

0,0 is the center of dilation

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Suppose data made available through a health system tracker showed health expenditures were $10,348 per person in the United Sta
nignag [31]

Answer:

a) 30.08% probability the sample mean will be within $100 of the population mean.

b) 0% probability the sample mean will be greater than $12,600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 10348, \sigma = 2500, n = 100, s = \frac{2500}{\sqrt{100}} = 250

a. What is the probability the sample mean will be within ±$100 of the population mean?

This is the pvalue of Z when X = 100 divided by s subtracted by the pvalue of Z when X = -100 divided by s. So

Z = \frac{100}{250} = 0.4

Z = -\frac{100}{250} = -0.4

Z = 0.4 has a pvalue of 0.6554, Z = -0.4 has a pvalue of 0.3556

0.6554 - 0.3546 = 0.3008

30.08% probability the sample mean will be within $100 of the population mean.

b. What is the probability the sample mean will be greater than $12,600?

This is 1 subtracted by the pvalue of Z when X = 12600. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{12600 - 10348}{250}

Z = 9

Z = 9 has a pvalue of 1.

1 - 1 = 0

0% probability the sample mean will be greater than $12,600

5 0
3 years ago
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