Lisa agrees to pay $157.28 per month for her car. How much will she pay in 9 months?

Monica wants to save at least $399 to buy a new laptop computer. She has $284 saved already. Which options below will allow her

german

Answer:

She is given $205 and spends $57.

She earns $580 and spends $422.

She is given $38 and earns $42.

She earns $56 and is given $59.

Step-by-step explanation:

You just have to do the math.

4 0

3 years ago

The probability that a 25-year-old female in the U.S. will die within one year is about 0.000514. An insurance company is prepar

zhannawk [14.2K]

Answer:

Step-by-step explanation:

Expected payment= 0.000514*55000= 28.27

which would round to 29

6 0

3 years ago

I need help Please help.

asambeis [7]

Answer:

Step-by-step explanation:

Number of students 10

Problem 1. $625 for the bus hire per friday, So 625*4=$2500

Problem 2. 2500/25=$100 each for the whole 4 weeks

Problem 3.

10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each

20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each

30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each

41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each

Hope this is correct

8 0

3 years ago

Problem 8

SpyIntel [72]

Answer:

Explanation:

1. x represent the copybook while y represent the pen that is bought.

2. 2x + 3y = 9000 means that in order to buy 2 copybooks and 3 pen you have to pay 9000 LL.

3. Solve for the price of x and y

2x + 3y = 9000 (1)
4x + 2y = 8000 (2)

Multiply (1) by -2:

-2(2x + 3y = 9000)
4x + 2y = 8000
——————————
-4x - 6y = -18000
4x + 2y = 8000
—————————
-4y = -10000
y = -10000/-4
y = 2500

Substitute y value into (2):

4x + 2(2500) = 8000
4x + 5000 = 8000
4x = 3000
x = 3000/4
x = 750

Therefore, 1 copy book = 750LL and 1 pen = 2500 LL

3 0

3 years ago

(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

4 0

3 years ago