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3 years ago

Which of the following statements is not true about an explanatory variable?

Mathematics

2 answers:

Anna11 [10]3 years ago

4 0

It is a dependant variable

luda_lava [24]3 years ago

3 0

Step-by-step explanation:

Explanatory variable is controlled by the researcher for the given experimental study. The researcher can put some conditions on the variable and the results are calculated. Hence, it is clear that result of the manipulated variable is noted and the manipulated variable is known as explanatory variable. There could be more than one explanatory variable.

For example consider the variables number of hours spent in practicing for a sport and the result in that sport. Here, the number of hours spent practicing influences the performance of the athlete. Hence, the variable number of hours spent is known as explanatory variable.

Also, there is one major difference between an independent variable and explanatory variable. The independent variable is not affected by other independent variables whereas for the explanatory variables, the variables are not independent.

You might be interested in

Simplify the expression:<br> (1 + 6g)(3) =

Irina18 [472]

Answer:

3 + 18g

Step-by-step explanation:

(1+6g)(3)=

Begin by distributing the 3 to each of the terms, I like to re-write it like this before distributing to make sure I remember to multiply all the terms inside the parenthesis.

(3)(1+6g)

3 + 18g

Since we are just simplifying the problem and not solving it, you won't have a number for your answer. It's usually a smaller equation with at least one variable :)

7 0

3 years ago

Please help me i will give brainlist to anyone who answers please help

Lena [83]

Answer:

1. x ≤ 15

2. x +3 < 12

3. x/-2 ≥ 10

4. x - 4 ≥ -6

5. x/13 ≥ 1

6. x - 8 ≤ 5

7. 3 * x > -7

8. x/2 + 5 > 11

9. -6 + x ≤ 15

Hope this helps!

8 0

2 years ago

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A

Zigmanuir [339]

Answer:

The probability that all 4 selected workers will be from the day shift is, = 0.0198

The probability that all 4 selected workers will be from the same shift is = 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

Step-by-step explanation:

Given that:

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews:

The number of selections result in all 4 workers coming from the day shift is :

$(^n _r) = (^{10} _4)$

$=\dfrac{(10!)}{4!(10-4)!}$

= 210

The probability that all 5 selected workers will be from the day shift is,

$\begin{array}{c}\\P\left( {{\rm{all \ 4 \ selected \ workers\ will \ be \ from \ the \ day \ shift}}} \right) = \frac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\\ = \frac{{210}}{{10626}}\\\\ = 0.0198\\\end{array}$

(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:

P( all 4 selected workers will be) $= \dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

where;

$(^{8}_4) } = \dfrac{8!}{4!(8-4)!} = 70$

$(^{6}_4) } = \dfrac{6!}{4!(6-4)!} = 15$

∴ P( all 4 selected workers is ) $=\dfrac{210+70+15}{10626}$

The probability that all 4 selected workers will be from the same shift is = 0.0278

P ( at least two different shifts will be represented among the selected workers) $= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

$=1 - \dfrac{210+70+15}{10626}$

= 1 - 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

(d)What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

$P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

$P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

$P(AUBUC) = \dfrac{1001}{10626}+ \dfrac{1820}{10626}+ \dfrac{3060}{10626}-\dfrac{15}{10626}-\dfrac{70}{10626}-\dfrac{210}{10626} +0$

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

5 0

3 years ago

Can someone pls answer all 6 questions pls? Pretty pls, except 1 and 5 thank you so much!!!

Maksim231197 [3]

2) C
3) B
4) C
6) I'm pretty sure its D but it could be C

7 0

3 years ago

Read 2 more answers

Please answer all questions i will give brainliest

jarptica [38.1K]

Answer:

Answers for 1-126

Step-by-step explanation:

a. x + 8 = 21

-8 -8

x = 13

b. x - 32 = 55

+32. +32

x = 87

c. 3x = 54

÷3. ÷3

x = 18

d. x/5 = 10

×5. ×5

x = 50

Answer for 1-124

1/5 = 2/10 , so 2 of the 10 colors are blue and 3 of the 10 colors are green.

The probability is 5/10 = 1/2

this is all I could do. sorry

4 0

2 years ago

Read 2 more answers