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3 years ago

What is an expression for the distance between the origin and a point P(x,y)

Mathematics

1 answer:

kirza4 [7]3 years ago

5 0

Answer:

$d = \sqrt{ {x}^{2} + {y}^{2} }$

Step-by-step explanation:

Distance between origin (0, 0) and point (x, y) is given as:

$d = \sqrt{ {(x - 0)}^{2} + {(y - 0)}^{2} } \\ \\ \red{ \bold{ d = \sqrt{ {x}^{2} + {y}^{2} } }}$

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Ine AB contains points A (−3, 5) and B (−3, 3). Line AB has a slope that is

lesantik [10]

Slope = Y2 - Y1 / X2 - X1

Slope =(3-5) / (-3-[-3])
Slope = -2 / 0

The answer is infinity because anything divided by 0 = infinity.

6 0

3 years ago

Read 2 more answers

A distribution of values is normal with a mean of 60 and a standard deviation of 16. From this distribution, you are drawing sam

professor190 [17]

Answer:

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

A distribution of values is normal with a mean of 60 and a standard deviation of 16.

This means that $\mu = 60, \sigma = 16$

Samples of size 25:

This means that $n = 25, s = \frac{16}{\sqrt{25}} = 3.2$

Find the interval containing the middle-most 76% of sample means.

Between the 50 - (76/2) = 12th percentile and the 50 + (76/2) = 88th percentile.

12th percentile:

X when Z has a p-value of 0.12, so X when Z = -1.175.

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

By the Central Limit Theorem

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

$-1.175 = \frac{X - 60}{3.2}$

$X - 60 = -1.175*3.2$

$X = 56.24$

88th percentile:

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

$1.175 = \frac{X - 60}{3.2}$

$X - 60 = 1.175*3.2$

$X = 63.76$

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%7B%20%5Ccos%5E%7B4%7D%5Calpha%7D%20%5C%3A%20%2B%20%20%20%5Csin%5E%7B2%7D%20%5Calpha%20%5C%

timurjin [86]

Answer:

Proved

Step-by-step explanation:

$cos^4\alpha +sin^4\alpha =\frac{1}{4}(3+cos4\alpha )\\\\$

Take the Left Hand Side:

$cos^4\alpha +sin^4\alpha\\\\(cos^2\alpha )^2+(sin^2\alpha )^2\\\\(\frac{1+cos2\alpha }{2})^2+(\frac{1-cos2\alpha }{2})^2 \\\\\frac{1+2cos2\alpha +cos^22\alpha }{4}+\frac{1-2cos2\alpha +cos^22\alpha }{4} \\\\$

$\frac{1+2cos2\alpha +cos^22\alpha +1-2cos2\alpha +cos^22\alpha }{4} \\\\\frac{2+2cos^22\alpha }{4} \\\\\frac{1+cos^22\alpha }{2} \\\\\frac{1}{2}(1+cos^22\alpha )\\\\$