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Andre45 [30]
2 years ago
13

A local coffee store is curious about the size beverages their customers are ordering. The beverages come in three sizes: small,

medium, and large. An employee runs a simulation of 15 random outcomes using the values 1, 2, and 3 to represent small, medium, and large, respectively. What is the simulated probability that a customer will order the small beverage? 2 2 3 1 1 2 2 3 1 1 3 2 1 1 3
Mathematics
2 answers:
Digiron [165]2 years ago
6 0

Answer:

2 out of 5 or 6 out of 15

Step-by-step explanation:

In this pattern, there is two people ordering it every five times. You can write it out as 6/15 or simplify it as 2/5 (six-fifteenths and two-fifths)

lara31 [8.8K]2 years ago
4 0
<h2>Answer:</h2>

The simulated probability that a customer will order the small beverage is:

                 \dfrac{2}{5}=0.4

<h2>Step-by-step explanation:</h2>

The  1, 2, and 3 to represent small, medium, and large, respectively.

The simulation is as follows:

2   2   3   1   1   2   2   3   1   1    3    2   1    1    3

There are a total outcomes= 15

Also, the number of outcomes that the customer will order small beverage is: 6

( Since,  6 times such that 1 has occurred.

As 1 represent the simulation that the beverage comes in small size )

Hence, the simulated probability is:

       \dfrac{6}{15}=\dfrac{2}{5}=0.4

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Answer:

1) 0.5 or 50%.

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   Order matters: 360,360 permutations.

Step-by-step explanation:

1) The probability of drawing either a purple or a blue chip is the sum of both individual probabilities:

P(B\ or\ P) = P(B) +P(P) = 0.46+0.04\\P(B\ or\ P) = 0.50

the probability that a blue or a purple chip will be drawn is 0.5 or 50%.

2)

a) Choose 5 items from a  total of 15 items when order does not matter:

C(15,5) = \frac{15!}{(15-5)!5!}=\frac{15*14*13*12*11}{5*4*3*2*1}\\C(15,5) = 3,003

b) Choose 5 items from a  total of 15 items when order matters:

P(15,5) = \frac{15!}{(15-5)!}=15*14*13*12*11}\\P(15,5) = 360,360

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Answer:

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Step-by-step explanation:

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y = x^2+2

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