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Marrrta [24]
3 years ago
14

8.

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0
The answer you are looking for is 12! have a great day!:)
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Y=5/2x-20
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A square has a a side lenght of 5 feet what is its area?
goldenfox [79]

Answer:

25 sq feet

Step-by-step explanation:

Length = 5 feet

Area = (Length)²

        = 5²

        = 25 sq feet

7 0
3 years ago
Part A Mr. Bade spends an average of $42.50 during a five-day workweek on lunch.if he eats every day during his 5-day workweek,
trasher [3.6K]

Answer:

Part A: He spends an average of $8.50 per day

Part B:  He would save $17 per week

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
II.) Simplify the following by applying the laws of indices.
Alik [6]
S I took a test and got it right
8 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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