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Basile [38]
3 years ago
12

What is the number 100 before 2,589?

Mathematics
2 answers:
scoray [572]3 years ago
5 0
Uhh yes it is wayyy before 2,589 I believe
marshall27 [118]3 years ago
5 0

Answer:

2489

Step-by-step explanation:

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-7x + 6x + 8 = x - 10<br> Solve for x pls
Art [367]
-x + 8 = x - 10
8 - x = x - 10
8 = x - 10 + x
8 = 2x - 10
8 + 10 = 2x
18 = 2x
18/2 = x
9 = x
x = 9
6 0
3 years ago
Add or Subtract to simplify
exis [7]

Answer:

-5x^5 + 2x^3 - 6x should be your answer

5 0
2 years ago
Laura is saving money to buy a game. The game cost $36, and so far she has saved three- fourths of this cost. How much money has
Andreyy89

Answer:

27

Step-by-step explanation:

3/4*36=27

therefore the answer is

27

5 0
2 years ago
Does the graph represent a function?
Lina20 [59]
The graph does not represent a function
4 0
3 years ago
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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