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3 years ago

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Mathematics

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babymother [125]3 years ago

4 0

Girl what is this ummmmm

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To estimate the average salary among people in a certain district, a marketing team obtains a simple random sample of 100 people

bogdanovich [222]

Answer:

The interpretation is that we are 90% sure that the true average salary of the district is in this interval.

Step-by-step explanation:

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

In this question:

The 90% confidence interval for the average salary of the district is [41500,49750].

This means that we are 90% sure that the true average salary of the district is in this interval.

3 0

3 years ago

If f(1)=-10 write the corresponding order pairs

masya89 [10]

F(1)=-10
f(x), so x=1
f(x)=-10, y=f(x), and y=-10
(x,y)=(1,-10)

3 0

4 years ago

Plz help me with this question

Contact [7]

Answer: KJ becuase it is the first two like on but put backwards.

7 0

3 years ago

Read 2 more answers

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

2 doctors is what percent of 25 doctors?

Luden [163]

2 doctors is 8% of 25 total doctors.

4 0

3 years ago

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