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IRINA_888 [86]
3 years ago
13

What is -3(m+6)-(11+9m)

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
7 0

-3(m+6)-(11+9m)

distribute

-3m -3*6 -11-9m

-3m -18 -11-9m

-12m -29

ELEN [110]3 years ago
3 0

First use the distributive property to multiply the first set of parentheses with the -3 and the second parentheses flip the signs because its negative

(-3m-18)+(-11-9m)

Then remove parentheses because it is addition and combine like terms

-12m - 29

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Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t
bogdanovich [222]

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b) \sigma_{\bar x} = 1.414

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The random sample is of size 50 i.e the population standard deviation  =10

Size of the sample n = 50

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The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

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b) When the population size N= 50000

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Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

c)  When the population size N= 5000

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Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }

\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }

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