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EleoNora [17]
2 years ago
10

Teresa is maintaining a campfire.She has kept the fire steadily burning for 4 hours using 6 logs. The number of logs needed to k

eep the fire burning is proportional to the number of hours it burns. She wants to know how many logs she needs to keep the fire burning for 18 hours.
Select the equationis Teresa can use to determine the number of logs she needs, x, to maintain the fire for 18 hours.

A. x/6 = 4/18
B. x/6 = 18/4
C. x/18 = 6/4
D. x/4 = 18/6
E. 4/6 = 18/x
Mathematics
1 answer:
Alexeev081 [22]2 years ago
4 0

Answer:

E. 4/6 = 18/x

Step-by-step explanation:

4 hour using 6 logs

18 hours using x logs

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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
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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

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3 years ago
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