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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?

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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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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What is the value of 7P7? <br><br> A. 1<br> B. 14<br> C. 49<br> D. 5040

sp2606 [1]

The answer is
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an employee adds 160 fluid ounces of chemical to a feature that holds 120,000 gallons of water. did the employee add yhe correct

Hitman42 [59]

We are given a volume of 160 fluid ounces of chemical which is added to a container that holds 120,000 gallons of water. Assuming that the chemical has the same density as water, we just need to convert 120,000 gallons to ounces.

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The coordinates of the midpoint of line GH are M(−132,−6) and the coordinates of one endpoint are G(−4, 1). The coordinates of t

Darina [25.2K]

Answer:

Since, the coordinates of the midpoint of line GH are M(\frac{-13}{2}, -6)(

2

−13

,−6) .

The coordinates of endpoint G are (-4,1)

We have to determine the coordinates of endpoint H.

The midpoint of the line segment joining the points (x_1, y_1)(x

1

,y

1

) and (x_2, y_2)(x

2

,y

2

) is given by the formula (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

) .

Here, The endpoint G is (-4,1) So, x_1 = -4 , y_1=1x

1

=−4,y

1

=1

Let the endpoint H be (x_2,y_2)(x

2

,y

2

)

The midpoint coordinate M is (\frac{-13}{2}, -6)(

2

−13

,−6) .

So, \frac{-13}{2} = \frac{-4+x_2}{2}

2

−13

=

2

−4+x

2

{-13} = {-4+x_2}−13=−4+x

2

{-13}+4 = {x_2}−13+4=x

2

{x_2}=9x

2

=9

Now, -6 = \frac{1+y_2}{2}−6=

2

1+y

2

-12 = {1+y_2}−12=1+y

2

y_2= -13y

2

=−13

So, the other endpoint H is (-9,-13).

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