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tresset_1 [31]
3 years ago
12

The principal of Oceanside middle school recorded the number of students in each class. What was the upper qualities of the clas

s side?

Mathematics
1 answer:
LekaFEV [45]3 years ago
5 0

Given:

The given box plot represents the class size of Oceanside middle school.

To find:

The upper quartile of the class side.

Solution:

In a box plot the starting the end or box present the first quartile and third quartile respectively. The line inside the box represents the median. The left and right end points of the segments are the minimum and maximum values respectively.

From the given box-plot it is clear that,

Minimum value = 22

First quartile or lower quartile= 24

Second quartile or median = 25

Third quartile or upper quartile= 26

Maximum value = 29

Therefore, the upper quartile of the class size is 26 student.

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Answer: The predicted difference in sleep is 1251437 minutes.

Step-by-step explanation:

Since we have given that

Number of years of education = 12 years

Number of minutes a week = 2000

As we have

In case of individual A:

7 days = 2000 minutes

1 year = 365 days = \dfrac{2000}{7}\times 365=104285

12 years = 104285\times 12=1251420\ minutes

In case of individual B:

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16 years = 156428\times 16=2502857 minutes

So, the predicted difference in sleep between individuals A and B is given by

2502857-1251420\\\\=1251437\ minutes

Hence, the predicted difference in sleep is 1251437 minutes.

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3 years ago
The reference desk of a university library receives requests for assistance. Assume that a Poisson probability distribution with
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Answer:

a) 0.125

b) 7

c) 0.875 hr

d) 1 hr

e) 0.875

Step-by-step explanation:l

Given:

Arrival rate, λ = 7

Service rate, μ = 8

a) probability that no requests for assistance are in the system (system is idle).

Let's first find p.

a) ρ = λ/μ

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Probability that the system is idle =

1 - p

= 1 - 0.875

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probability that no requests for assistance are in the system is 0.125

b) average number of requests that will be waiting for service will be given as:

λ/(μ - λ)

= \frac{7}{8 - 7}

= 7

(c) Average time in minutes before service

= λ/[μ(μ - λ)]

= \frac{7}{8(8 - 7)}

= 0.875 hour

(d) average time at the reference desk in minutes.

Average time in the system js given as: 1/(μ - λ)

= \frac{1}{(8 - 7)}

= 1 hour

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{7}{8}

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