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Rudiy27
3 years ago
15

30 POINTS AND BRAINLIEST PLEASE HELP!!! Dogs in the GoodDog Obedience School win a blue ribbon for learning how to sit, a green

ribbon for learning how to roll over, and a white ribbon for learning how to stay. There are 100 dogs in the school. 62 have blue ribbons, 55 have green ribbons, and 63 have white ribbons. 32 have a blue ribbon and a green ribbon; 31 have a green ribbon and a white ribbon; 38 have a blue ribbon and a white ribbon. 16 have all three ribbons. How many dogs have not learned any tricks?
Mathematics
1 answer:
Vlada [557]3 years ago
4 0

Answer:

What are the list of answers for this question? That would kind of help narrow some of my answers down.

Step-by-step explanation:

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The slope is y=0.59x so 0.59

The y intercept is 0

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Identify the relationship between the graphs of
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parallel

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Because the 2 line have same slopes but different y int, they are parallel

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2. List the like terms in each of the following<br> i) 4x2 , -5x , 6 , 7x , -2x2 , -3
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Make r the subject of the formula
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r=\frac{-a+p}{a+p}

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You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

8 0
3 years ago
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