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3 years ago

© Suppose you did not know that x had a normal distribution. Would you be

Mathematics

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

No.

Step-by-step explanation:

The central Limit Theorem states that the sample size must be greater than 30 for an approximate normal distribution.

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Please help me and answer both? tysm and enjoy your day! hope this helps!

IceJOKER [234]

The answer is C and b

8 0

3 years ago

What is the solution to the system of equations?<br>- (8. –1)<br>0 (s. 2)<br>19 (3:5)

Oksana_A [137]

Answer:

Step-by-step explanation: You have to do the distributive property to get the answer.

3 0

4 years ago

I need help, please.

Usimov [2.4K]

2 rooms would be 62+42 which is 104

3 rooms would be 104+42 which is 146

4 rooms would be 146+42 which is 188

5 rooms would be 188+42 which is 230

6 rooms would be 230+42 which is 272

but these are the prices including the 20$

without the 20$ the prices would be

2 rooms is 84

3 rooms is 126

4 rooms is 168

5 rooms is 210

6 rooms is 252

4 0

3 years ago

Read 2 more answers

How can we use the distributive property to find an expression equivalent to

koban [17]

Answer: multiply 10 to both terms and subrtact products

Step-by-step explanation:

10(5 - 2x)

multipy 10 to both terms

50 20x

subtract products

50 - 20x

7 0

2 years ago

Read 2 more answers

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

8 0

3 years ago