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LUCKY_DIMON [66]

3 years ago

5

A set of data included the numbers 19, 81, 75, 42, 33 57, and 60. What number could be added to the data to cause the range to b

e 90?

1 answer:

dsp733 years ago

6 0

Answer:

109

Step-by-step explanation:

81-19 is 62. if you add 109 then you could do 109-19 is 90

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Someone help me with this please

anzhelika [568]

for question 1

Area of Base B =45×25 = 1125 cm²

perimeter of base P = 2(45+25) = 140 cm

Height = 15 cm

Using the formula

Surface area = 2B + Ph

A = 2×1125 + 140×15= 4350 cm²

Voume = Bh = 1125×15= 16875 cm³

For question 2,

Area of base = 7× 5/2= 17.5 cm²

Perimeter of base = (7+7+7) = 21 cm

Surface area = 2B + Ph

Surface area of prism = 2×17.5+ 21×10 = 245cm²

Volume = Bh = 17.5×10 = 175 cm³

6 0

3 years ago

Read 2 more answers

Multiplying a number by 4/5 then dividing by 2/5 is the same as multiplying what number?

prisoha [69]

Multiplying a number by 4/5 and then dividing by 2/5 is the same as multiplying by 2

5 0

3 years ago

Kevin is 3 times as old as Daniel. 4 years ago, Kevin was 5 times as old as Daniel.

rusak2 [61]

Answer:

24

Step-by-step explanation:

4 years ago, Daniel would've been 4 and Kevin would be 20, so Kevin would've been 5 times as old as Daniel. And 8 x 3 = 24. So, Kevin is 24.

4 0

3 years ago

Read 2 more answers

Use the table to work out the values of a , b , c , and d .

slavikrds [6]

Answer:

a = 10, b = 6, c = 2, d = 0

Step-by-step explanation:

Substitute the appropriate values of x into the equation and evaluate

x = - 3 : y = 4 - 2(- 3) = 4 + 6 = 10 → a

x = - 1 : y = 4 - 2(- 1) = 4 + 2 = 6 → b

x = 1 : y = 4 - 2(1) = 4 - 2 = 2 → c

x = 2 : y = 4 - 2(2) = 4 - 4 = 0 → d

5 0

3 years ago

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

4 0

3 years ago