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Ksenya-84 [330]
3 years ago
13

Use the digits 5,4 and 8 once each to write an odd number greater than seven hundred

Mathematics
1 answer:
natali 33 [55]3 years ago
4 0
845 because 800 is greater than 700 and 845 is an odd number because it ends in 5 (which is an odd number).
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pav-90 [236]
20-24 ...................
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3 years ago
Idk why I can’t get this ?! number two plz
Arturiano [62]
The only possible outcome to this would be any number greater or less than 361 in the 300’s
7 0
3 years ago
What does p equal when p=210e^(0.0069*20)?
Korolek [52]
<span>p=210e^(0.0069*20)
'e' is a mathematical constant equal to 2.718281828459
</span>
<span>0.0069*20 = .138

e^.138 =
</span>
<span>2.718281828459^.138 =
</span>
<span> <span> <span> 1.1479755503 </span> </span> </span>
210 * <span>1.1479755503 =
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<span> <span> <span> 241.074865563 </span> </span> </span>
5 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
Aeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
m_a_m_a [10]

Answer:o

Step-by-step explanation:

5 0
3 years ago
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