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Fed [463]
2 years ago
14

Which measure represents the standard deviation of the sample means and is used in place of the population standard deviation wh

en
the population parameters are unknown?
Mathematics
1 answer:
Ksivusya [100]2 years ago
4 0

The measure represents the standard deviation of the sample means and is used in place of the population standard deviation when the population parameters are unknown is; t-test.

<h3>Which measure is used when the population parameters are unknown?</h3>

A hypothesis test for a population mean when In the case that the population standard deviation, σ, is unknown, carrying out a hypothesis test for the population mean is done in similarly like the population standard deviation is known. A major distinctive property is that unlike the standard normal distribution, the t-test is invoked.

Read more on t-test;

brainly.com/question/6501190

#SPJ1

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A test driver has to drive a car on a 1-mile track for two rounds in a way, that his average speed is 60 mph. On his first round
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to average 60 mph for 2 laps his total speed needs to equal 60 x 2 = 120

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Find the surface area of the cube shown below. units 2 2 squared
ruslelena [56]

The surface area of the cube with a side length of 2 units is 24 units squared

<h3>How to determine the surface area?</h3>

The side length of the cube is given as:

l = 2

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This gives

Surface area = 6 *2^2

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7 0
2 years ago
Help pleaseeeeeeeeeeeeeeeeeeeeeee
bixtya [17]

Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

<u>Step-by-step explanation:</u>

(1)           (12, 18, 27, ...)

The common ratio is:

r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

3 0
3 years ago
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