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Drupady [299]
4 years ago
5

How do you simplify x^2+y^2?

Mathematics
2 answers:
Doss [256]4 years ago
7 0
The other responder is correct. There is nothing more to be done.

You may think this is a trivial response, but it is a good one. It teaches you not to destroy plus signs by combining unlike terms. You should treat this as a "can you combine apples and oranges" problem. 

In other words, is there any similarity between apples and oranges other than both are fruits? You should answer no there is no similarity.

Same with xs and ys. There is no similarity. until you are told there is.

Answer<<<<< nothing can be done.
iVinArrow [24]4 years ago
3 0
Nothing further can be done, its x^2+y^2

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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}

X - 128 = 1.28*23

X = 157.44

So

157.44mmHg

6 0
3 years ago
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