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Lera25 [3.4K]
3 years ago
8

Simplify 1. 4(y - 3) + 2y 2. -2(b - 6) 3. 6x + 2 ­- 3x ­- 5

Mathematics
2 answers:
garri49 [273]3 years ago
5 0
Answers in order

6y-12
-2b+12
3x-3
Dafna11 [192]3 years ago
5 0
1.) 4(y-3)+27
4y-12+27
final answer:4y+15

2.)-2(b-6)
final answer:-2b+12

3.)6x+2-3x-5
6x-3x+2-5
final answer:3x-3

^_^
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It would be 6.0x4 because the square is 6.0 on each side
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3 years ago
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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.

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3 years ago
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Simplify.<br><br><br><br> Enter your answer, in simplest radical form, in the box.
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A survey found that​ women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. The survey also
Effectus [21]

Answer: 99.51%

Step-by-step explanation:

Given : A survey found that​ women's heights are normally distributed.

Population mean : \mu =63.2  \text{ inches}

Standard deviation: \sigma= 2.4\text{ inches}

Minimum height = 4ft. 9 in.=4\times12+9\text{ in.}=57\text{ in.}

Maximum height = 6ft. 2 in.=6\times12+2\text{ in.}=74\text{ in.}

Let x be the random variable that represent the women's height.

z-score : z=\dfrac{x-\mu}{\sigma}

For x=57, we have

z=\dfrac{57-63.2}{2.4}\approx-2.58

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z=\dfrac{74-63.2}{2.4}\approx4.5

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

P(-2.58

Hence, the percentage of women meeting the height requirement = 99.51%

8 0
3 years ago
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