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Dennis_Churaev [7]
3 years ago
5

The population of current statistics students has ages with mean muμ and standard deviation sigmaσ. samples of statistics studen

ts are randomly selected so that there are exactly 5454 students in each sample. for each​ sample, the mean age is computed. what does the central limit theorem tell us about the distribution of those mean​ ages?
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
6 0

What does the central limit theorem tell us about the distribution of those mean​ ages?

<span>A.    </span>Because n>30, the sampling dist of the mean ages can be approximated by a normal dist with a mean u and a SD o/sqrt 54,

Whenever n<span>>30 the central limit theory applies.</span>

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Answer:

D) 2

Step-by-step explanation:

Use the following slope formula:

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(x₂ , y₂) = (1 , -5)

Plug in the corresponding numbers to the corresponding variables:

<em>m</em> = (-5 - 1)/(1 - 4)

Simplify. Remember to follow PEMDAS. First subtract, then divide:

<em>m</em> = (-5 - 1)/(1 - 4)

<em>m</em> = (-6)/(-3)

<em>m</em> = 2

D) 2 is your answer.

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Step-by-step explanation:

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Answer:

a) zeros of the function are x = 1 and, x = -1

b) zeros of the function are x = 2 and, x = 4

c) zeros of the function are x = \frac{3}{2}

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Step-by-step explanation:

Zeros of the function are the values of the variable that will lead to the result of the equation being zero.

Thus,

a) f(x)= (x +1)(x − 1)(x² +1)

now,

for the (x +1)(x − 1)(x² +1) = 0

the condition that must be followed is

(x +1) = 0 ..........(1)

or

(x − 1) = 0 ..........(2)

or

(x² +1) = 0 ...........(3)

considering the equation 1, we have

(x +1) = 0

or

x = -1

for

(x − 1) = 0

x = 1

and,

for (x² +1) = 0

or

x² = -1

or

x = √(-1)         (neglected as it is a imaginary root)

Thus,

zeros of the function are x = 1 and, x = -1

b) g(x) = (x − 4)³(x − 2)⁸

now,

for the (x − 4)³(x − 2)⁸ = 0

the condition that must be followed is

(x − 4)³ = 0 ..........(1)

or

(x − 2)⁸ = 0 ..........(2)

considering the equation 1, we have

(x − 4)³ = 0

or

x -4 = 0

or

x = 4

and,

for (x − 2)⁸ = 0

or

x - 2 = 0

or

x = 2        

Thus,

zeros of the function are x = 2 and, x = 4

c) h(x) = (2x − 3)⁵

now,

for the (2x − 3)⁵ = 0

the condition that must be followed is

(2x − 3)⁵ = 0

or

2x - 3 = 0

or

2x = 3

or

x = \frac{3}{2}

Thus,

zeros of the function are x = \frac{3}{2}

d)   k(x) =(3x +4)¹⁰⁰(x − 17)⁴

now,

for the (3x +4)¹⁰⁰(x − 17)⁴ = 0

the condition that must be followed is

(3x +4)¹⁰⁰ = 0 ..........(1)

or

(x − 17)⁴ = 0 ..........(2)

considering the equation 1, we have

(3x +4)¹⁰⁰ = 0

or

(3x +4) = 0

or

3x = -4

or

x = \frac{-4}{3}

and,

for (x − 17)⁴ = 0

or

x - 17 = 0

or

x = 17        

Thus,

zeros of the function are x = \frac{-4}{3}  and, x = 17

7 0
3 years ago
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