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postnew [5]
3 years ago
9

I need help on this plz

Mathematics
1 answer:
balandron [24]3 years ago
3 0

Answer:

(2) angle1 = 139, angle2 = 41, angle4 = 41, angle5 = 139, angle6 = 41, angle7 = 139, angle8 = 41

(3) angle1 = 150, angle2 = 30, angle4 = 30, angle5 = 150, angle6 = 30, angle7 = 150, angle8 = 30

Step-by-step explanation:

All angle are either equal to each other or supplementary.  I use corresponding angles and vertical angles to prove each of the above.

For number 3, angle 3 and angle 8 are supplementary, so they add up to 180:

8x +70 + (4x - 10) = 180

12x + 60 = 180

12x = 120

x = 10

So if x = 10, then 8x + 70 = 8(10) + 70 = 150

That means all angles are either 150 or 30 for number 3.

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Step-by-step explanation:

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P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

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c)   P(win first game)  = 0.4

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P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

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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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