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algol13
3 years ago
7

Triangle ABC is ransformed to obtain angle ABC

Mathematics
1 answer:
salantis [7]3 years ago
3 0

Answer:How do you rotate 180 degrees about the origin?

Rotation by 180° about the origin:

The rule for a rotation by 180° about the origin is (x,y)→(−x,−y) .

Step-by-step explanation:

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Which of the following represents the factorization of the binomial below?
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Answer:

D

Step-by-step explanation:

x² - 81 ← is a difference of squares and factors in general as

a² - b² = (a + b)(a - b)

Here a = x and b = 9, thus

x² - 81 = (x + 9)(x - 9) → D

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4 years ago
2/21 x 3 <br> Multiply. Write your answer as a fraction in the simplest form.
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2/7

Step-by-step explanation:

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What is the justification for each step in the solution of the equation 3(2x + 7) - 4 = 53 - 3x?
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4 years ago
Assume that the Richter scale magnitudes of earthquakes are normally distributed with a mean of 1.064 and a standard deviation o
sertanlavr [38]

Answer:

Ok.

Step by Step Explanation: Ok.  

5 0
3 years ago
Suppose x has a distribution with a mean of 50 and a standard deviation of 4. Random samples of size n = 64 are drawn. (a) Descr
olchik [2.2K]

Answer:

(a) \bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)

(b) The <em>z</em>-score for the sample mean \bar x = 51 is 2.

(c) The value of P(\bar X < 51) is 0.9773.

Step-by-step explanation:

The random variable <em>X</em> has mean, <em>μ</em> = 50 and standard deviation, <em>σ</em> = 4.

A random sample of size <em>n </em>= 64 is selected.

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the  distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the  distribution of sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The sample of <em>X</em> selected is, <em>n</em> = 64 > 30.

So, the Central limit theorem can be applied to approximate the distribution of sample mean (\bar x).

\bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)

(b)

The <em>z</em>-score for the sample mean \bar x is given as follows:

z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}

Compute the <em>z</em>-score for \bar x = 51 as follows:

z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}

  =\frac{51-50}{0.5}\\

  =2

Thus, the <em>z</em>-score for the sample mean \bar x = 51 is 2.

(c)

Compute the value of P(\bar X < 51) as follows:

P(\bar X < 51)=P(\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}

                  =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X < 51) is 0.9773.

5 0
3 years ago
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