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3 years ago

What is the relative frequency of 90 out of 200

Mathematics

1 answer:

Nataliya [291]3 years ago

5 0

Answer:

9/20

Plz mark as brainliest have a blessed day and I hope this helps you

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A bicycle store costs $3300 per month to operate. The store pays an average of $75 per bike. The average selling price of each

Klio2033 [76]

S=Selling price 135
V=Variable cost 75
F=Fixed cost 3300
Let quantity be Q
The formula to break even is
135Q-75Q-3300=0
Solve for Q
60Q-3300=0
60Q=3300
Q=3300/60
Q=50
So the store must sell 50 bicycles to break even

Hope it helps!

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4 years ago

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If f(x) = 3x - 1 and g(x) = x + 2, find (f+ g)(x).

aivan3 [116]

Answer:

the answer is a

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3 years ago

-7-(-7)=? Plz help plz plz

Reil [10]

Answer:

the answer is 0

Step-by-step explanation:

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3 years ago

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(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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3 years ago

Arianna runs up 4 flights of stairs and then runs down 4 flights of stairs. Does this situation represent additive inverses?

SVETLANKA909090 [29]

No because she runs up yes but she also ran down witch means she went exactly 0

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3 years ago

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