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3 years ago

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AN ICE CREAM STORE IS CONDUCTING A SURVEY ASKING CUSTOMERS TO RANK THE TOP 5 OUT OF 10 NEW FLAVORS INTRODUCED THIS YEAR. HOW MAN

Y POSSIBLE ARRANGEMENTS EXIST?

1 answer:

katrin2010 [14]3 years ago

3 0

The problem given above can be solved through the concept of permutation because of the importance of the arrangements. This may be solved through calculators directly by keying in,
10P5
That is, the permutation of 10 taken 5. The answer to this permutation is 30,240.

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Rename 3/8 and 7/10 of fractions with common denominators

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Given that f(x) = 4x - 3 and g(x) 2x-1 divided by 3<br><br> solve for g(f(2))

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2nd option = 3 is the answer.

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The illustration below shows the graph of y as a function of x.

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Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

The moats separating people from the animals are 5 m wide for lions and 4 m wide for the elephants. If the lion’s moat is 4 m de

ZanzabumX [31]

Answer:

<em>The elephants' moat should be 3.2 meters deep.</em>

Step-by-step explanation:

Suppose, the depth of the elephants' moat is $x$ meter.

Width of the lions' moat is 5 meter and width of the elephants' moat is 4 meter.

Given that, the lion’s moat is 4 meter deep.

So, <u>according to the ratio of width and depth</u>, the equation will be.......

$\frac{5}{4}=\frac{4}{x}\\ \\ 5x=4*4\\ \\ 5x=16\\ \\ x=\frac{16}{5}=3.2$

Thus, the elephants' moat should be 3.2 meters deep.

3 0

3 years ago