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Shkiper50 [21]
2 years ago
10

the average age of my sample is 21.2 years. the average age of the target population is 19.4. the difference is probably due to

Mathematics
1 answer:
tamaranim1 [39]2 years ago
5 0

The difference in the average age of the sample and the population is probably due to sampling error

<h3>What is sampling error?</h3>

The sampling error in a statistical data is simply an error that arises when a sample that represents the population is not properly selected or analyzed

From the question, we have the following parameters:

\bar x = 21.2 -- the sample average

\mu = 19.4 -- the population average

Notice that:

\bar x \ne \mu

i.e. sample and the population averages are not equal.

This difference is often as a result of sampling error

Read more about sampling errors at:

brainly.com/question/19264268

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PLZ HELP I WILL GIVE Brainliest answer
Phoenix [80]
Hi! I believe someone has already asked the same question and received an answer. :) brainly.com/question/2952634


8 0
3 years ago
Solve the equation<br>algebra 2<br><br>-6=3x-6y<br>4x=4+5x
Talja [164]
3x = 6y - 6
x = 2y - 2

4 ( 2y - 2 ) = 4 + 5 ( 2y - 2 )
8y - 8 = 4 + 10y - 10
2y = - 2
y = - 1

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

i am a mathematics teacher. if anything to ask please pm me
6 0
3 years ago
HELPP WITH MATH PLEASSS 11-12
evablogger [386]

Answer:

11. Not enough information

12. You can say DB = DC by SAA congruence rule and CPCTC (Corresponding parts of congruent triangles are congruent)

Step-by-step explanation:

For 12, you can say ΔABD≅ΔACD because they have a side in common (AD) and they have 2 congruent angles. Thus, you can use the SAA congruence criterion. Then, you can use CPCTC (Corresponding parts of congruent triangles are congruent) to say that DB = DC.

Hope this helps :)

4 0
2 years ago
Can you guys please help me? Thanks!
olga2289 [7]
you are correct, it is one to two
4 0
3 years ago
Write the equation of a Line that is perpendicular to y=5/2x+3 and passes
koban [17]

Answer:

Step-by-step explanation:

Slope of perpendicular lines = -1

y = 5/2x + 3

m_{1}=\frac{5}{2}\\\\m_{1}*m_{2}=-1\\\\

        m_{2}= -1 ÷ m_{1}

               = -1*\frac{2}{5}=\frac{-2}{5}

(-3 , -5)

Equation of the required line: y - y₁ = m(x -x₁)

                                                y - [5] = \frac{-2}{5}(x - [-3])\\

                                               y+5=\frac{-2}{5}x + 3*\frac{-2}{5}\\\\y+5=\frac{-2}{5}x-\frac{6}{5}\\\\   y =\frac{-2}{5}x-\frac{6}{5}-5\\\\   y = \frac{-2}{5}x-\frac{6}{5}-\frac{5*5}{1*5}\\\\ y = \frac{-2}{5}x-\frac{6}{5}-\frac{25}{5}\\\\\\ y=\frac{-2}{5}x-\frac{31}{5}

5 0
3 years ago
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