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GenaCL600 [577]
3 years ago
13

I did the no question and 86 through 88

Mathematics
2 answers:
EleoNora [17]3 years ago
8 0
86. D
87. H
88. Quadrant II
kolbaska11 [484]3 years ago
4 0
86. D. -10
87. H. -6, -3, 0, 2
88. Quadrant 1
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20 business cards are placed in a hat, 10 belong to women, 10 belong yo men If 2 cards are drawn at once. what's the probability
erik [133]

Answer:

First choice:

               \large\boxed{\large\boxed{5/19}}

Explanation:

<em>The probability that the first is a man's card and the second, a woman's card</em> is calculated as the product of both probabilities, taking into account the fact that the second time the number of cards  in the hat has changed.

In spite of it is said that the cards are drawn at once, since it is stated a specific order for the cards (first is a man's card and the second, a woman's card) you can model the procedure as if the cards were drawn consecutively, instead of at once.

<u>1. Probability that the first is a man's card</u>

  • Number of cards in the hat = 20 (the 20 business card)

  • Number of man's card in the hat: 10

  • Probability = favorable oucomes / possible outcomes = 10/20 = 1/2.

<u />

<u>2. Probability that the second is a woman's card</u>

  • Number of cards in the hat = 19 (there is one less card in the hat)

  • Number of wonan's card in the hat: 10

  • Probability = favorable oucomes / possible outcomes = 10/19.

<u>3. Probability that the first is a man's card and the second, a woman's card</u>

<u />

  • (1/2) × (10/19) = 5/19

That is the first choice.

5 0
3 years ago
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There are five red marbles eight blue marbles and 12 Green marbles in a bag. What is the theoretical probability of randomly dra
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Answer:

where is the question

Step-by-step explanation:

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Would the answer be <br> 40, 10 ?
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Answer:

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3 years ago
I need help with geometry:
nikklg [1K]

Answer:

x = 5

Step-by-step explanation:

6/18 = x/(20-x)

1/3 = x/(20-x) Multiply both sides by 20-x

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20-x = 3x Add x in both sides

20 = 4x Divide both sides by 4

5 = x

3 0
3 years ago
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