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Kipish [7]
3 years ago
10

Que medida tendra un angulo si es complementario a un angulo de 49 grados 15 minutos y 30 segundos

Mathematics
1 answer:
never [62]3 years ago
4 0

Answer: a medida del ángulo sería de 51 grados

los ángulos complementarios suman 90 grados

49+x=90

49+x-49=90-49

x=51

Usé traducir para esto, espero que puedas entender

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Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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3 years ago
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Answer:

-3.7 > -3.1

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Because -3.1 more close to 0 so is supposed to be bigger than -3.7

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2 years ago
I need help please to find the answer
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B! If you draw the line of symmetry diagonally, you have two triangles
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Umnica [9.8K]

Answer:

3.

Step-by-step explanation:

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3 years ago
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Plis help! Will give brainliest!
4vir4ik [10]

Answer: 7) Step 1 8) Step

Step-by-step explanation:

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The answer to 8 is also 11. 18-14/2 is 11.

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