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2 years ago

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A small company plans to invest in a new advertising campaign. There is a 20% chance that the company will lose $5,000, a 50% ch

ance of a break even, and a 30% chance of a $10,000 profit. Based on this information, what should the company do?

1 answer:

GREYUIT [131]2 years ago

5 0

Are you asking what they should do because of the profit ?

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What can be multiplied to get 5 and added to get 4

bonufazy [111]

It's not possible nothing but 1 times 5 is the only thing that multiplies to 5 and 1+5 doesn't equal 4

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3 years ago

The function w=30-3/4n represents the weight w in pounds of the papers in tyrones newspaper delivery bag after he delivers n new

adoni [48]

Answer:

domain (−∞,∞)

range (−∞,∞)

n = 40 and it refers to the fact that after delivering 40 newspapers, there will be no paper left in the bag

Step-by-step explanation:

To find the zero, we equate the linear equation to zero

w = 30 - 3n/4

3n/4 = 30

3n = 120

n = 120/3

n = 40

What this mean in this context is that the the bag becomes empty after he has delivered 40 newspapers or we can say that the maximum number of papers the bag can take is 40

The range refers to the number on the y-axis

Here the range is (−∞,∞)

The domain refers to the x axis values

We have (−∞,∞)

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2 years ago

31,059 is rounded by ten thousands in math

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What is the weight of a 32 kg child

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32 kgs ?

Step-by-step explanation:

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Read 2 more answers

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

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2 years ago