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4vir4ik [10]
3 years ago
10

The first class had a kids in it, the second had b kids in it, and the third class had c kids in it. Kids from all three classes

were equally divided between two buses. How many kids were on each bus?
Mathematics
2 answers:
astra-53 [7]3 years ago
5 0
There are no numbers and only variables so I can't tell you the amount of kids on each bus but here is the equation:
(a*b*c)/2
kramer3 years ago
3 0
(a+b+c)/2. this is your answer. if it had the numbers i would solve it fully, but it only has variables.
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The following data were collected by counting the number of operating rooms in use atTampa general Hospital over a 20-day period
arlik [135]

Answer:

a)

Operating room x    1         2      3       4

P(x)                          0.15  0.25  0.4   0.2

b)

The graph of probability distribution is in attached image.

c)

All the probabilities lies in the range (0 to 1) and the probabilities add up to 1 so, the computed probability distribution is the valid probability distribution.

Step-by-step explanation:

Number of operating rooms   Days

1                                                  3

2                                                 5

3                                                 8

4                                                 4

a)

Sum of frequencies=1+2+3+4=10

Operating rooms x   frequency f  Relative frequency

1                                    3                    3/20=0.15

2                                   5                   5/20=0.25

3                                   8                   8/20=0.4

4                                   4                   4/20=0.2

So, using the relative frequency approach, a discrete probability distribution for number of operating rooms in use on any given day is

Operating room x    1         2      3       4

P(x)                          0.15  0.25  0.4   0.2

b)

The graph of probability distribution is in attached image.

c)

There are two conditions for a probability distribution to be valid

1. All probability must ranges from 0 to 1.

2. All probabilities must add up to 1.

We can see that all the probabilities lies in the range (0 to 1), so, condition 1 is satisfied.

For condition 2,

sum[p(x)]= 0.15+0.25+0.4+0.2=0.4+0.6=1.

As the probabilities add up to 1, so the condition 2 is also satisfied.

Thus, the computed probability distribution is the valid probability distribution.

7 0
3 years ago
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Leya [2.2K]
H=-5(t^2 - 16t)

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10–{22–[(−9)+(−11)]} help i'M HAVING A bad day :( fix it
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Answer:

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Step-by-step explanation:

10–{22–[(−9)+(−11)]}

Work inside out

10–{22–[(-20)]}

Subtracting a negative is adding

10–{22+20}

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Answer:

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Step-by-step explanation:

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0.02 this isa the answer


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