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Artyom0805 [142]
3 years ago
12

2. C = 2πr C ≈ 2 (__ 22 7 ) ( ) C ≈ cm

Mathematics
1 answer:
Kamila [148]3 years ago
5 0
P is just a definition: it's the ratio between a circle's diameter and its circumference. In other words, any circle circumference is approximately 3.14 (p) times longer (around) than its diameter length. Some old Greek dude discovered it a few thousand years ago... Let's do the first 3 problems together - by then I think you'll have confidence you can do the remaining 3! 1. If C = 2pr, then we substitute C = 2*3.14*6, = 37.68 2. If C = 2pr, then let's rearrange that so we can find r: divide both sides (to keep it equal) by 2p to get r by itself: C/2p = 2pr/2p, so r = C/2p. Now substitute: r = 24p/2p = 12 3. If C = 2pr, then r = C/2p (see #2 above), then r = d/2. If A = pr2, then A = p (C/2p)2. Substitute: A = p (32p/2p)2, or A = p (16)2 A = 3.14 * 256 = 803.84 Now can you complete #3 - #6? I bet you can... :-)
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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]

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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

8 0
3 years ago
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