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kipiarov [429]
3 years ago
7

PLEASE HELP A bag contains red and blue marbles, such that the probability of drawing a blue marble is 3/8 An experiment consist

s of drawing a marble, replacing it, and drawing another marble. The two draws are independent. A random variable assigns the number of blue marbles to each outcome. What is the probability of drawing one blue marble?
Mathematics
2 answers:
jeka57 [31]3 years ago
4 0

Answer:

15/64

Step-by-step explanation:

3 blue

5 red marbles.

since the marble is put back in, the denominator will not change,

so to draw ONE marble, the chances would be

chances of picking a blue marble 3/8

chances of picking a red marble 5/8

3/8 * 5/8 - 15/64

please mark me brainly.

Kay [80]3 years ago
4 0

Answer:

The probability of drawing one blue marble is 3/8 because you have three blue marbles out of a total of eight marbles. The probability of drawing a second blue marble is also 3/8 but since you have to draw both marbles consecutively you have to multiply the probabilities of drawing each marble. So,

3/8•3/8 = 9/64.

Now you need to change it to a percent.

We do this by

9/64 x 100 = 14% approx

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3 years ago
Find the range of each function for the domain {-4, -2, 0, 1.5, 4}. f(x) = 5x^2 + 4
andrew11 [14]
Substitute x with the members of the domain.
f(x) = 5x² + 4

Substitute with the domain of -4
f(x) = 5x² + 4
f(-4) = 5(-4)² + 4
f(-4) = 5(16) + 4
f(-4) = 80 + 4
f(-4) = 84

Substitute with the domain of -2
f(x) = 5x² + 4
f(-2) = 5(-2)² + 4
f(-2) = 5(4) + 4
f(-2) = 20 + 4
f(-2) = 24

Substitute with the domain of 0
f(x) = 5x² + 4
f(0) = 5(0)² + 4
f(0) = 5(0) + 4
f(0) = 0 + 4
f(0) = 4

Substitute with the domain of 1.5
f(x) = 5x² + 4
f(1.5) = 5(1.5)² + 4
f(1.5) = 5(2.25) + 4
f(1.5) = 11.25 + 4
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Substitute with the domain of 4
f(x) = 5x² + 4
f(4) = 5(4)² + 4
f(4) = 5(16) + 4
f(4) = 80 + 4
f(4) = 84

The range of the function for those domain is {4, 24, 15.25, 84}
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3 years ago
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Answer:

Step-by-step explanation:

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2x - 3*(6x + 11) = 7
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Answer:

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