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3 years ago

A ray is a defined term because

Mathematics

1 answer:

Simora [160]3 years ago

4 0

Because it can be described using two undefined terms, point and line.

You might be interested in

Share $600 between Anna and Raman in the ratio 3:7 and 1:4

Ulleksa [173]

Answer:

3 : 7 =>

Anna = $180

Raman = $420

1 : 4 =>

Anna = $120

Raman = $480

Step-by-step explanation:

Shares when ratio is 3 : 7

Anna : Raman = 3 : 7

Sum of ratio 10

Anna's share=

$\frac{3}{10} \times 600 = \$ 180$

Raman 's share=

$\frac{7}{10} \times 600 = \$ 420$

Shares when ratio is 1 : 4

Anna : Raman = 1 : 4

Sum of ratio = 5

Anna 's share =

$\frac{1}{5} \times 600 = \$ 120$

Raman's share =

$\frac{4}{5} \times 600 = \$ 480$

8 0

3 years ago

2x-8y=6 y=-7- x what are the coordinates

goblinko [34]

2x-8(-7-x)=6
2x+56+8x=6
10x=-50
x=-5

y=-7-(-5)
y=-2

(-5,-2)

5 0

3 years ago

What is the coefficient of xy^2 in the expansion of (x+y)^3?

Basile [38]

Step-by-step explanation:

Using Binomial Expansion,

(x + y)³

= 3C0 * x³ + 3C1 * x²y + 3C2 * xy² + 3C3 * y³.

Therefore the coefficient of xy² is 3C2 = 3.

5 0

3 years ago

Read 2 more answers

i have to make an expression that has a value of 10 out of the numbers 5,2,1,1 and the format it has to go in is like ( - )+( x

soldi70 [24.7K]

Answer:

(1-1)+(5×2)

hope this helps you

6 0

3 years ago

A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G

dezoksy [38]

Answer:

a) The mean of a sampling distribution of $\\ \overline{x}$ is $\\ \mu_{\overline{x}} = \mu = 20$ . The standard deviation is $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

b) The standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

c) The standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

d) The probability $\\ P(\overline{x}$ .

e) The probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

f) $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

Step-by-step explanation:

We are dealing here with the concept of a sampling distribution, that is, the distribution of the sample means $\\ \overline{x}$ .

We know that for this kind of distribution we need, at least, that the sample size must be $\\ n \geq 30$ observations, to establish that:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the distribution of the sample means follows, approximately, a normal distribution with mean, $\mu$ , and standard deviation (called standard error), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a standard normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .

$\\ Z \sim N(0, 1)$

The variable Z is

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of $\\ \overline{x}$ is the population mean $\\ \mu = 20$ or $\\ \mu_{\overline{x}} = \mu = 20$ .

The standard deviation is the population standard deviation $\\ \sigma = 16$ divided by the root square of n, that is, the number of observations of the sample. Thus, $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

Part b

We are dealing here with a random sample. The z-score for the sampling distribution of $\\ \overline{x}$ is given by [1]. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{-4}{\frac{16}{8}}$

$\\ Z = \frac{-4}{2}$

$\\ Z = -2$

Then, the standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

Part c

We can follow the same procedure as before. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{3}{\frac{16}{8}}$

$\\ Z = \frac{3}{2}$

$\\ Z = 1.5$

As a result, the standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

Part d

Since we know from [1] that the random variable follows a standard normal distribution, we can consult the cumulative standard normal table for the corresponding $\\ \overline{x}$ already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is two standard units below the mean (because of the negative value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is $\\ P(Z$ .