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2 years ago

15

One marble is picked at random and NOT replaced from a box that has 3 red marbles, 4 green marbles 2 blue marbles and one yellow

marble. Find the probability of selecting a red and a red marble.

1 answer:

yKpoI14uk [10]2 years ago

6 0

Answer:

3/9

Step-by-step explanation:

because there are total 9 marbles. 3 red, 4 green and 2 blue.so the probability is 3/9.

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ASAP please help!!!

Serggg [28]

Y+4<u><</u>9
subtract 4
y<u><</u>5
y is smaller than and equal to 5
so you shade from 5 to the negative end to infinity (to the left) and shade the 5 to show that it is included (attachment says A)

6 less than (-6) 2 times a number (2 time x) is greater than (>) 8 (8)

-6+2x>8
add 6
2x>14
divide 2
x>7
so
x is bigger than 7
shade from 7 to the positive end to infinity (to the right) and don't shade 7 but put a circle around it to show that it is not included (attachment says B )

I have included pictures of the number lines

4 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

What does (2+3) +4^2 x (5-3) equal

givi [52]

Answer:

37

Step-by-step explanation

Do what is in the parenthesis first

(5)+4^2x(2)

Exponents next

4^2=16

5+16x2

Multiplication next

16x2=32

You're left with the equation 5+32=37

6 0

3 years ago

How can you tell whether the parabola had a vertical or horizontal shift?

Schach [20]

If the x is squared, the parabola is vertical (opens up or down). If the y is squared, it is horizontal (opens left or right). If a is positive, the parabola opens up or to the right. If it is negative, it opens down or to the left.

3 0

2 years ago

Positive fractions less than 1,with interval of 1/8 between each pair of fractions

Sati [7]

Yes it is keep up the good work

4 0

3 years ago