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Ray Of Light [21]
2 years ago
10

A group of five people are all working on the same mathematics problem. On the night before it is due, they call each other to d

iscuss their work. Each person talks to all the other people at least once. What is the fewest number of telephone calls that could be made?
Mathematics
1 answer:
UkoKoshka [18]2 years ago
5 0

Answer:

Minimum number of calls = 10

Step-by-step explanation:

Lets name the five people as A,B,C,D and E.

<u>On the night before ,each person talks to every other person atleast once that means A would talk to B,C,D and E atleast once.</u>

Lets start with A. He would talk to other 4 people which means there would be 4 phone calls made.

Now lets take B. He can talk to A,C,D and E. But  A has already talked to C therefore to get minimum number of phone calls , B need not call A again. So he calls only C,D and E.

In case of C using similar logic he need to talk to only D and E.

For D , he talks to E alone.

E does not have to talk to anyone as he has already talked to everyone atleast once.

Total calls = 4 + 3 + 2 + 1

                    = 10

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The population of Champaign, Illinois is given for three years in the table below:
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Answer:

See below

Step-by-step explanation:

a) <u>Using the first two lines to get the equation:</u>

Since t = 0 represents a start point, the y-intercept is 163488

<u>Slope is:</u>

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<u>And the equation:</u>

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b) Prediction of the population in 2012 using the function:

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Build 7 identical towers with a box of 546 blocks. How many blocks are there in each tower?
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1 year ago
Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

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The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

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2 years ago
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