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2 years ago

Anyone know the answer??

Mathematics

1 answer:

meriva2 years ago

6 0

Answer:

X (2, 1)

Y (2, 3)

W (5, 5)

Step-by-step explanation:

The current points are:

X (-2, -1)

Y (-2, -3)

W (-5, -5)

Now, after rotating 180 degrees around the origin, here is what points Y, Y, and W would look like:

X (2, 1)

Y (2, 3)

W (5, 5)

Plot these new points on the graph, and you'll have a new figure which will be rotated to 180 degrees.

I have also provided you with a graph, showing both, the original and the rotated version, below.

You might be interested in

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard

avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Step-by-step explanation:

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.

Central Limit Theorem

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that $\mu = 33208, \sigma = 2503$ .

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when $X = 33208+633 = 33841$ subtracted by the pvalue of Z when $X = 33208 - 633 = 32575$

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57$

X = 33841

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

$Z = \frac{33841 - 33208}{357.57}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

X = 32575

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

$Z = \frac{32575- 33208}{357.57}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0

3 years ago

1. What is 0.078 written in expanded form

ArbitrLikvidat [17]

Answer:

7.8 x $10^{-2}$