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3 years ago

8

The central limit theorem states that sampling distributions are always the same shape as the population distribution from whenc

e the data came. True or False

1 answer:

sergejj [24]3 years ago

3 0

Explanation:

The sample mean is not always equal to the population mean but if we take more and more number of samples from the population then the mean of the sample would become equal to the population mean.

The Central Limit Theorem states that we can have a normal distribution of sample means even if the original population doesn't follow normal distribution, But we have to take a lot of samples.

Suppose a population doesn't follow normal distribution and is very skewed then we can still have sampling distribution that is completely normal if we take a lot of samples.

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Nando is on a rehabilitation program following a car accident. For the first month, he needs to exercise at most five hours per

emmasim [6.3K]

Answer:

See explanation

Step-by-step explanation:

Let x be the number of hours per week Nando is brisk walking and y be the number of hours per week Nando is biking at a moderate pace.

For the first month, he needs to exercise at most five hours per week, then

$x+y\le 5$

Brisk walking burns about 350 calories per hour, then it burns 350x calories per x hours.

Biking at a moderate pace burns about 700 calories per hour, then it burns 700y calories per y hours.

Nando must burn at least 2,000 calories per week, so

$350x+700y\ge 2,000$

You get the system of two inequalities:

$\left\{\begin{array}{l}x+y\le 5\\ 350x+700y\ge 2,000\end{array}\right.$

The attached graph shows the solution set to this system of inequalities. In this diagram, red region represents the solution set to the first inequality, blue region represents the solution set to the second inequality and their intersection is the solution set to the system of two inequalities.

4 0

3 years ago

what ratio was used on a table that has two sides. on one side the numbers are 12,20,28,and 36. on the other side the numbers ar

ad-work [718]

Ratio is used 4:1 for the table that has 2 sides

3 0

3 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this <em>ordinary</em> differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the <em>ordinary</em> differential equation is $-\frac{1}{x}$ .

The <em>particular</em> solution of the <em>ordinary</em> differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ <em>, </em>then the integration constant is $C = -2$ .

The solution of this <em>ordinary</em> differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the <em>ordinary</em> differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The <em>particular</em> solution of the <em>ordinary</em> differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

3 years ago

WILL GIVE BRAINLEIST PLEASE HURRY DUE AT 9 PM SUPER EASY For exercises 1 and 2, simplify each expression.

Ostrovityanka [42]

1. 13x + 21
2. 33 + 6x

6 0

3 years ago

Can someone help me with this please <br> What is the right answer for it?

kondaur [170]

That would be 3 or 4 cm

8 0

3 years ago

Read 2 more answers