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alukav5142 [94]
2 years ago
12

A bag contains 8 red marbles, 9 yellow marbles, and 7 green marbles. how many additional red marbles must be added to the 24 mar

bles already in the bag so that the probability of randomly drawing a red marble is 3/5
Mathematics
2 answers:
Mice21 [21]2 years ago
8 0

Answer:

16 red marbles

Step-by-step explanation:

8 red

9 yellow

7 green

total = 8 + 9 + 7

p(red) = 8/24 = 1/3

Add r number of red marbles to the total.

8 + r red

9 yellow

7 green

total = 8 + r + 9 + 7 = r + 24

p(red) = (8 + r)/(r + 24) = 3/5

Cross multiply.

40 + 5r = 72 + 3r

2r = 32

r = 16

Answer: 16 red marbles

EleoNora [17]2 years ago
6 0

Answer:

16

Step-by-step explanation:

bag red: 8 + x

yellow:9

green 7/24

3/5 = 8+ x/24+ x

5(8 + x) = 3 (24 + x)

40 + 5x

-40 -3x = 72 + 3x

-40 -3x    -40 - 30

2x/2 = 32/2  so,x = 16

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What are three other expressions that are equivalent to 8x-12
Kisachek [45]

Answer:

2(4x-6)

4(2x-3)

8(x-3/2)

Step-by-step explanation:

find all factors of 8 other than 1, you should have 3

divide the whole equation by the factor then multiply by the factor

2(8x-12)/2 = 2(4x-6)

4(2x-3)

8(x-3/2)

3 0
3 years ago
After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only five women among the last
Georgia [21]

Answer:

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317; we can use this result to support her charge of gender discrimination.

Step-by-step explanation:

If we are assuming that the women and the men are equally qualified, then the probability for each employee that is hired the probability for it to be a women should be 1/2. Note that the fact that more men that women are hired in a sample might not be disctrimination: for example, if 2 men are hired out of 2 employees, that can happen with probability 1/4, so it is quite common. In order to support her charge for gender discrimination, we need at least a probability less that 0.05 that 5 (or less) women are hired out of 19 employees.

Since each configuration is equally probable, we will count the total amount of possible cases that 5 or less women are hired, and dividide it by the total amount of cases, 2¹⁹.

  • 0 women hired: one possible case: every employee is male
  • 1 women hired: 19 possible cases
  • 2 women hired: {19 \choose 2} = 171 possible cases
  • 3 women hired: {19 \choose 3}  = 969 possible cases
  • 4 women hired: {19 \choose 4} = 3876 possible cases
  • 5 women hired: {19 \choose 5} = 11628 possible cases

Thus, there are a total of 11628+3876+969+171 = 16644 possible cases out of 2¹⁹ ones. All of them with seemingly equal probability. As a consequence, the probability of 5 or less women to be hired out of 19 employees, assuming that the probability to hire 1 is 1/2, is

16644/2¹⁹ = 0.0317 < 0.05

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317. Since the probability is so low, we can conclude that for the employer, a woman equally qualified as a man is less likely to be hired, therefore, we can support her charge of gender discrimination.

6 0
3 years ago
El perímetro de un rectángulo es 58 y su base excede en 10 a su ancho, ¿Cuánto mide la base? AYUDAAAA
kobusy [5.1K]

Answer:

The base is 19.5.

Step-by-step explanation:

The given question is, "The perimeter of a rectangle is 58 and its base exceeds its width by 10, how long is the base?"

Perimeter = 58

Base, l = 10+b

The perimeter of a rectangle is :

P = 2(l+b)

58 = 2(10+b+b)

29 = (10+2b)

29-10 = 2b

19 = 2b

b = 9.5

Base, l = 10 + 9.5

= 19.5

Hence, the base is 19.5.

5 0
3 years ago
Priya's cat weighs 5 1/2 pounds and her dog weighs 8 1/4 pounds.
Lilit [14]

Answer: The cat is 2/3 times as heavy as the dog.


Step-by-step explanation:

The weight of Priya's cat =5\frac{1}{2}\ pounds

Now, 5\frac{1}{2}=\frac{11}{2}\ pounds

The weight of her dog =8\frac{1}{4}\ pounds

8\frac{1}{4}=\frac{33}{4}\ pounds

Let cat is n times heavy as dog.

Then n=\text{cat's weight}\div\text{dog's weight}

n=\frac{{11}{2}\div\frac{33}{4}\\\Rightarrow\ n=\frac{11}{2}\times\frac{4}{33}\\\Rightarrow\ n=\frac{2}{3}

Hence, The cat is 2/3 times as heavy as the dog.

5 0
3 years ago
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steposvetlana [31]

Answer:

a) (-2,2), (-1,-1), (1,-1), (3,7)

b) x=1.4

8 0
3 years ago
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