El precio de un automovil $696.000 y se puede pagar de dos maneras distintas.

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago

Different hotels in a certain area are randomly selected, and their ratings and prices were obtained online. Using technology, w

Savatey [412]

Answer:

a. $y= 130x + 350 + \epsilon$

b. The symbol y-bar represents the average price of hotels in the area

Step-by-step explanation:

a. The regression line equation is given by:

$y=\alpha x + \beta + \epsilon$

Where Alpha represents the slope, Beta represents the intercept with the Y axis, and epsilon represents the random component of the model. Thus,

$\alpha = 130$ y $\beta = 350$ Obtaining:

$y= 130x + 350 + \epsilon$ .

b. In a simple linear regression model the value of y-bar represents the average of all values of y observed. Thus, since y represents the prices of the hotels, then y-bar is the average price of hotels in the area.

5 0

3 years ago

A cave contains 2 brown bats and 3 gray bats. The bats will fly out of the cave one at a time. What is the probability that the

Marianna [84]

The answer would be 1/5

4 0

3 years ago

What is the product??

tangare [24]

Answer:

The answer to your question is the third choice

Step-by-step explanation:

To solve this problem, just multiply 6 by each of the numbers of the matrix and simplify.

6 | 4 -2 1 | = | (6 x 4) (6 x -2) (6 x 1) |

| 7 3 0| | (6 x 7) (6 x 3) (6 x 0) |

= | 24 -12 6 |

| 42 18 0 |

The first choice is wrong because only multiply by six the first row.

The second and fourth rows are incomplete.

7 0

3 years ago

PLEASEEEEE HELP BIT FINALS TEST!!!! PLSSS :(

boyakko [2]

Answer:

The 1st, 3rd, and the 4th

8 0

3 years ago