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LenKa [72]
3 years ago
7

Wood and colleagues (2009) examined the value of self-affirmation. In a typical study, participants either engaged or did not en

gage in self-affirmations. Later, their current self-esteem was assessed. Which alternative CORRECTLY names and identifies the variables in this study?
A) independent variable – self-affirmations; dependent variable – self-esteem scores
B) independent variable – self-esteem scores; dependent variable – self-affirmations
C) experimental variable – self-affirmations; control variable – self-esteem scores
D) experimental variable – self-esteem scores; control variable – self-affirmations
Mathematics
1 answer:
maxonik [38]3 years ago
8 0

Answer:

Option A) independent variable – self-affirmations; dependent variable – self-esteem scores

Step-by-step explanation:

We are given the following in the question:

"Wood and colleagues (2009) examined the value of self-affirmation. In a typical study, participants either engaged or did not engage in self-affirmations. Later, their current self-esteem was assessed."

Independent and Dependent Variable:

  • Dependent variable is the variable whose value depends on the independent variable.
  • Independent variable is the free variable.

For the given scenario, self esteem is assessed based on the fact that participants either engaged or did not engage in self-affirmations.

Thus, the dependent variable is self esteem and the independent variable is engagement in self affirmation.

Thus, the correct answer is

Option A) independent variable – self-affirmations; dependent variable – self-esteem scores

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olga_2 [115]
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8 0
3 years ago
From the start of a trail to the bird lookout point is 3 7/8 mile. From the bird lookout point to the end of the trail is anothe
Juli2301 [7.4K]

Answer:

Distance of the trail from the start to the end is 6\frac{17}{24} miles

Step-by-step explanation:

Distance from the start of a trail to the bird lookout point=3\frac{7}{8} miles

Distance from the bird lookout point to the end of the trail =2\frac{5}{6} miles

We want to find the difference of the trail from the start to the end, that is, how far is the starting point from the end point.

The distance between the start and end points

=3\frac{7}{8}+2\frac{5}{6}\\=\frac{31}{8} +\frac{17}{6}

=\frac{161}{24} =6\frac{17}{24} miles

3 0
3 years ago
This season, Lisa's lacrosse team has won $\frac 23 of their home games (games played at Lisa's school), but just $\frac 25 of t
juin [17]

Answer:

Lisa's team have played 24 home games and 25 away games.

Step-by-step explanation:

Given:

Let x represents games played at home.

also y represent games played away.

Total number of games played = 49

∴ x+y=49 \ \ \ \  equation \ 1

Now according to given data:

Number of home games won = \frac{2}{3}

Number of away game won = \frac{2}{5}

Total games won =26

Hence

\frac{2}{3}x+\frac{2}{5}y=26\\\\\frac{2\times 5}{3\times 5}x+\frac{2\times 3}{5\times 5}y=26\\\\\frac{10}{15}x+\frac{6}{15}y=26\\\\\frac{10x+6y}{15}=26\\\\10x+6y= 26\times 15\\10x+6y = 390\\2(5x+3y)=390\\5x+3y=\frac{390}{5}\\\\5x+3y=195 \ \ \ \ equation \ 2\\

Now multiplying equation 1 by 3 we get,

3(x+y)=3\times49\\3x+3y=147 \ \ \ \ \ equation \ 3\\

Now Subtracting equation 2 by equation 3 we get,

(5x+3y=195)-(3x+3y=147)\\2x=48\\x= \frac{48}{2}=24

Substituting value of x in equation 1 we get.

x+y=49\\24+y=49\\y=49-24=25

Hence Lisa's team have played 24 home games and 25 away games.

3 0
3 years ago
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