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3 years ago

Find the surface area of a sphere with a diameter of 34.

Mathematics

1 answer:

My name is Ann [436]3 years ago

6 0

Answer:

A = 213.52

Step-by-step explanation:

A = 4 pi radius square

diameter = 34

radius is half of diameter

radius = 17

A = 4(pi or 3.14)(17)

A= 213. 628300444 with pi

A = 213.52 with 3.14

hopefully this helps you understand this concept

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F(x) = 11x - 8

meriva

Solution:

100x^2-58x-16

6 0

3 years ago

The ardered.pairs.(0, 1), (3, 10) and (4, n) are solutions of the same linear equation. Find n.

Fittoniya [83]

Answer:

n = 13.

Step-by-step explanation:

Slope of the line = (10-1)/3-0) = 3

So the equation of the line is:

y - 1 = 3(x - 0)

y = 3x + 1

When x = 4 y = n, so:

n = 3(4) + 1 = 13.

n = 13.

7 0

3 years ago

The variable z is inversely proportional to x. When x is 19, z has the value 0.10526315789474. What is the value of z when x= 28

telo118 [61]

When two variables are inversely proportional the relation between them can be written as:

$z=\frac{k}{x}$

Here, k is the constant of proportionality and is always equal to the product of the two variables. So using the given values of z and x, we can find k first.

$k=zx\\ \\ k=0.10526315789474(19)\\ \\ k=2$

The constant of proportionality for the given inverse proportion comes out to be 2. Using the value of k in the equation, we get:

$z=\frac{2}{x}$

We have to find the value of z when x=28. So replacing x by 28, we get:

$z=\frac{2}{28}\\ \\ z=0.07142857143$

Thus, rounding of to nearest thousand the value of z comes out to be 0.071 if x is equal to 28.

6 0

4 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 $\\ P(Z$ .