Ask question

Ask question

All categories

3 years ago

14

Nine new employees, two of whom are married to each other, are to be assigned nine desks that are lined up in a row. If the assi

gnment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks? (Hint: The event that the couple have nonadjacent desks is the complement of the event that they have adjacent desks.) (Round your answer to the nearest tenth of a percent.)

1 answer:

meriva3 years ago

4 0

Answer:

The probability is 0.8

Step-by-step explanation:

The key to answering this question is considering the fact that the two married employees be treated as a single unit.

Now what this means is that we would be having 8 desks to assign.

Mathematically, the number of ways to assign 8 desks to 8 employees is equal to 8!

Now, the number of ways the couple can interchange their desks is just 2 ways

Thus, the number of ways to assign desks such that the couple has adjacent desks is 2(8!)

The number of ways to assign desks among all six employees randomly is 9!

Thus, the probability that the couple will have adjacent desks would be ;

2(8!)/9! = 2/9

This means that the probability that the couple have non adjacent desks is 1-2/9 = 7/9 = 0.77778

Which is 0.8 to the nearest tenth of a percent

You might be interested in

A trough has ends shaped like isosceles triangles, with width 5 m and height 7 m, and the trough is 12 m long. Water is being pu

Svet_ta [14]

Answer:

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

$V = \dfrac{1}{2}(bh)\times L$

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

$\dfrac{b}{h} = \dfrac{5}{7}$

$b=\dfrac{5}{7}h$ this can be substituted back in the volume equation

$V = \dfrac{5}{14}h^2L$

the rate of the water flowing in is:

$\dfrac{dV}{dt} = 6$

The question is asking for the rate of change of height (m/min) hence that can be denoted as: $\frac{dh}{dt}$

Using the chainrule:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

$V = \dfrac{5}{14}h^2L$

$\dfrac{dV}{dh} = \dfrac{5}{7}hL$

reciprocating

$\dfrac{dh}{dV} = \dfrac{7}{5hL}$

plugging everything in the chain rule equation:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

$\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6$

$\dfrac{dh}{dt}=\dfrac{42}{5hL}$

L = 12, and h = 1 (when the water is 1m deep)

$\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}$

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

6 0

4 years ago

Read 2 more answers

Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.

zvonat [6]

Answer:

The maximum volume of a box inscribed in a sphere of radius r is a cube with volume $\frac{8r^3}{3\sqrt{3}}$ .

Step-by-step explanation:

This is an optimization problem; that means that given the constraints on the problem, the answer must be found without assuming any shape of the box. That feat is made through the power of derivatives, in which all possible shapes are analyzed in its equation and the biggest -or smallest, given the case- answer is obtained. Now, 'common sense' tells us that the shape that can contain more volume is a symmetrical one, that is, a cube. In this case common sense is correct, and the assumption can save lots of calculations, however, mathematics has also shown us that sometimes 'common sense' fails us and the answer can be quite unintuitive. Therefore, it is best not to assume any shape, and that's how it will be solved here.

The first step of solving a mathematics problem (after understanding the problem, of course) is to write down the known information and variables, and make a picture if possible.

The equation of a sphere of radius $r$ is $x^2 + y^2 + z^2=r^2$ . Where $x$ , $y$ and $z$ are the distances from the center of the sphere to any of its points in the border. Notice that this is the three-dimensional version of Pythagoras' theorem, and it means that a sphere is the collection of coordinates in which the equation holds for a given radius, and that you can treat this spherical problem in cartesian coordinates.

A box that touches its corners with the sphere with arbitrary side lenghts is drawn, and the distances from the center of the sphere -which is also the center of the box- to each cartesian axis are named $x$ , $y$ and $z$ ; then, the complete sides of the box are measured $2x$ , $2y$ and $2z$ . The volume $V$ of any rectangular box is given by the product of its sides, that is, $V=2x\cdot 2y\cdot 2z=8xyz$ .

Those are the two equations that bound the problem. The idea is to optimize $V$ in terms of $r$ , therefore the radius of the sphere must be introduced into the equation of the volumen of the box so that both variables are correlated. From the equation of the sphere one of the variables is isolated: $z^2=r^2-x^2 - y^2\quad \Rightarrow z= \sqrt{r^2-x^2 - y^2}$ , so it can be replaced into the other: $V=8xy\sqrt{r^2-x^2 - y^2}$ .

But there are still two coordinate variables that are not fixed and cannot be replaced or assumed. This is the point in which optimization kicks in through derivatives. In this case, we have a cube in which every cartesian coordinate is independent from each other, so a partial derivative is applied to each coordinate independently, and then the answer from both coordiantes is merged into a single equation and it will hopefully solve the problem.

The $x$ coordinate is treated first: $\frac{\partial V}{\partial x} =\frac{\partial 8xy\sqrt{r^2-x^2 - y^2}}{\partial x}$ , in a partial derivative the other variable(s) is(are) treated as constant(s), therefore the product rule is applied: $\frac{\partial V}{\partial x} = 8y\sqrt{r^2-x^2 - y^2} + 8xy \frac{(r^2-x^2 - y^2)^{-1/2}}{2} (-2x)$ (careful with the chain rule) and now the expression is reorganized so that a common denominator is found $\frac{\partial V)}{\partial x} = \frac{8y(r^2-x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} - \frac{8x^2y }{\sqrt{r^2-x^2 - y^2}} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}$ .

Since it cannot be simplified any further it is left like that and it is proceed to optimize the other variable, the coordinate $y$ . The process is symmetrical due to the equivalence of both terms in the volume equation. Thus, $\frac{\partial V}{\partial y} = \frac{8x(r^2-x^2 - 2y^2)}{\sqrt{r^2-x^2 - y^2}}$ .

The final step is to set both partial derivatives equal to zero, and that represents the value for $x$ and $y$ which sets the volume $V$ to its maximum possible value.

$\frac{\partial V}{\partial x} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} =0 \quad\Rightarrow r^2-2x^2 - y^2=0$ so that the non-trivial answer is selected, then $r^2=2x^2+ y^2$ . Similarly, from the other variable it is obtained that $r^2=x^2+2 y^2$ . The last equation is multiplied by two and then it is substracted from the first, $r^2=3 y^2\therefore y=\frac{r}{\sqrt{3}}$ . Similarly, $x=\frac{r}{\sqrt{3}}$ .

Steps must be retraced to the volume equation $V=8xy\sqrt{r^2-x^2 - y^2}=8\frac{r}{\sqrt{3}}\frac{r}{\sqrt{3}}\sqrt{r^2-\left(\frac{r}{\sqrt{3}}\right)^2 - \left(\frac{r}{\sqrt{3}}\right)^2}=8\frac{r^2}{3}\sqrt{r^2-\frac{r^2}{3} - \frac{r^2}{3}} =8\frac{r^2}{3}\sqrt{\frac{r^2}{3}}=8\frac{r^3}{3\sqrt{3}}$ .

6 0

3 years ago

Help me prove this trig identity

alexdok [17]

Hello here is a solution:

3 0

3 years ago

4x^5 in verbal algebraic expression

disa [49]

Answer:

The product of 4 times the quantity x to the fifth power

6 0

3 years ago

Find three consecutive integers such that the product of the first and third is 5 greater than 5 times the second.

ivolga24 [154]

Answer:

Step-by-step explanation:

3 consecutive integers....

1st integer = x

2nd integer = x + 1

3rd integer = x + 2

the product of 1st and 3rd is 5 greater then 5 times the 2nd...

x(x+2) = 5(x + 1) + 5

x^2 + 2x = 5x + 5 + 5

x^2 + 2x = 5x + 10

x^2 + 2x - 5x - 10 = 0

x^2 - 3x - 10 = 0

(x - 5)(x + 2) = 0

x - 5 = 0 x + 2 = 0

x = 5 x = -2

so it will be : 5,6,and 7 or -2,-1, and 0

6 0

3 years ago