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iren [92.7K]
3 years ago
13

In a fictional study, a pretest-posttest design was used to examine the influence of a television program on children's aggressi

veness. The number of aggressive responses was measured during an observation period both before and after the television program. Determine if there is a difference in the number of aggressive behaviors in children after having viewed the television program. Before After 6 9 4 3 12 11 9 12 10 14 2 6 14 12 NOTE: If you need the standard deviations for any calculations, they are sBefore
Mathematics
1 answer:
ioda3 years ago
4 0

Answer: nationwide

Step-by-step explanation: is on ur side haha sorry not smart

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An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an
Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

  • 28 students in the Spanish class,
  • 26 in the French class,
  • 16 in the German class,
  • 12 students that are in both Spanish and French,
  • 4 that are in both Spanish and German,
  • 6 that are in both French and German,
  • 2 students taking all 3 classes.

So,

  • 2 students taking all 3 classes,
  • 6 - 2 = 4 students are in French and German, bu are not in Spanish,
  • 4 - 2 = 2 students are in Spanish and German, but are not in French,
  • 12 - 2 = 10 students are in Spanish and French but are not in German,
  • 16 - 2 - 4 - 2 = 8 students are only in German,
  • 26 - 2 - 4 - 10 = 10 students are only in French,
  • 28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

\dfrac{50}{100} =0.5

B. If a student is chosen randomly,  the probability that he or she is taking exactly one language class is

\dfrac{8+10+14}{100}=0.32

C. If 2 students are chosen randomly,  the probability that both are not taking any language classes is

0.5\cdot 0.5=0.25

So,  the probability that at least 1 is taking a language class is

1-0.25=0.75

3 0
3 years ago
It takes 36 minutes for 7 people to paint 4 walls. How long does it take 9 people to paint 7 walls?
algol [13]
It will take 49 minutes for 9 people to paint 7 walls
7 0
3 years ago
The equation of line f is y = 1/4x -1. line g includes the point (1,-6) and is perpendicular to line f . What is the equation of
Gwar [14]
Y=-4x+b
-6=-4+b
-2=b
y=-4x-2
7 0
3 years ago
The probability that a certain hockey team will win any given game is 0.3723 based on their 13 year win history of 385 wins out
Whitepunk [10]

Answer:

0.1505 = 15.05% probability that the hockey team wins 6 games in November

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that a certain hockey team will win any given game is 0.3723

So p = 0.3723

12 games in November

So n = 12

What is the probability that the hockey team wins 6 games in November?

This is P(X = 6)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{12,6}.(0.3723)^{6}.(1-0.3723)^{6} = 0.1505

0.1505 = 15.05% probability that the hockey team wins 6 games in November

4 0
3 years ago
Draw on a number line and show the direction of the inequalities<br> y ≥ 4
LekaFEV [45]

refer to the image given below:

4 0
2 years ago
Read 2 more answers
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