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PilotLPTM [1.2K]
3 years ago
7

A population has a mean mu μ equals = 87 and a standard deviation σ = 24. Find the mean and standard deviation of a sampling dis

tribution of sample means with sample size n equals = 36 mu μx equals = nothing (Simplify your answer.) sigma Subscript x overbar σx equals = nothing (Simplify your answer.
Mathematics
1 answer:
timama [110]3 years ago
7 0

Answer:

The mean  of a sampling distribution of sample means is 87

The standard deviation of a sampling distribution of sample = 4

Step-by-step explanation:

* Lets revise some definition to solve the problem  

- The mean of the distribution of sample means is called μx

- It is equal to the population mean μ

- The standard deviation of the distribution of sample means is

 called  σx

- The rule of σx = σ/√n , where σ is the standard  deviation and n

  is the size of the sample

* lets solve the problem  

- A population has a mean (μ) is 87

∴ μ = 87

- A standard deviation of 24

∴ σ = 24

- A sampling distribution of sample means with sample size n = 36

∴ n = 36

∵ The mean of the distribution of sample means μx = μ

∵ μ = 87

∴ μx = 87

* The mean  of a sampling distribution of sample means is 87

∵ The standard deviation of a sampling distribution of sample

   means σx  = σ/√n

∵ σ = 24 and n = 36

∴ σx = 24/√36 = 24/6 = 4

* The standard deviation of a sampling distribution of sample = 4

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777dan777 [17]

The equations of the three altitudes of triangle ABC include the following:

  1. 3y - 2y - 4 = 0.
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  3. 4y + x - 6 = 0.

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Mathematically, the slope of a straight line can be calculated by using this formula;

Slope, m = \frac{Change\;in\;y\;axis}{Change\;in\;x\;axis}\\\\Slope, m = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}

Also, the point-slope form of a straight line is given by this equation:

y - y₁ = m(x - x₁)

Assuming the following parameters for triangle ABC:

  • Let AM be the altitudes on BC.
  • Let BN be the altitudes on CA.
  • Let CL be the altitudes on AB.

For the equation of altitude AM, we have:

Slope of BC = (2 - 8)/(4 - 0)

Slope of BC = -6/4

Slope of BC = -3/2

Slope of AM = -1/slope of BC

Slope of AM = -1/(-3/2)

Slope of AM = 2/3.

The equation of altitude AM is given by:

y - y₁ = m(x - x₁)

y - 0 = 2/3(x - (-2))

3y = 2(x + 2)

3y = 2x + 4

3y - 2y - 4 = 0.

For the equation of altitude BN, we have:

Slope of CA = (2 - 0)/(4 - (-2))

Slope of CA = 2/6

Slope of CA = 1/3

Slope of BN = -1/slope of CA

Slope of BN = -1/(1/3)

Slope of BN = -3.

The equation of altitude BN is given by:

y - y₁ = m(x - x₁)

y - 8 = -3(x - 0)

y - 8 = -3x

y + 3x - 8 = 0.

For the equation of altitude CL, we have:

Slope of AB = (8 - 0)/(0 - (-2))

Slope of AB = 8/2

Slope of AB = 4

Slope of CL = -1/slope of AB

Slope of CL = -1/4

The equation of altitude CL is given by:

y - y₁ = m(x - x₁)

y - 2 = -1/4(x - 4)

4y - 2= -(x - 4)

4y - 2= -x + 4

4y + x - 2 - 4 = 0.

4y + x - 6 = 0.

In conclusion, we can infer and logically deduce that the equations of the three altitudes of triangle ABC include the following:

  1. 3y - 2y - 4 = 0.
  2. y + 3x - 8 = 0.
  3. 4y + x - 6 = 0.

Read more on point-slope form here: brainly.com/question/24907633

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Answer:

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The average of all grades in his fencing class, including the upcoming one, (all grades summed up divided by the amount of grades) must be equal to 86 (percent), hence C is the right answer for 27.

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The expression is given as:

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Remove brackets in the above expression

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Next, evaluate the like terms

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