Answer:
The image of ![\left[\begin{array}{c}4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D)
through T is ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We know that 
→ is a linear transformation that maps 
into 
⇒
And also maps 
into 
⇒
We need to find the image of the vector
We know that exists a matrix A from 
(because of how T was defined) such that :
for all x ∈
We can find the matrix A by applying T to a base of the domain ().
Notice that we have that data :
{
}
Being 
the cannonic base of
The following step is to put the images from the vectors of the base into the columns of the new matrix A :
![T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%265%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
![T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%261%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-2%267%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
Writing the matrix A :
![A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now with the matrix A we can find the image of ![\left[\begin{array}{c}4&-4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
such as :
⇒
![T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
We found out that the image of through T is the vector
-10w - 400 = o
w= weeks
o = new outstanding balance
For a probability distribution the expected value is the summation of product of probabilities with their respective data values. Let x be the probability that Jackson goes gym for 2 days and y be the probability that he goes gym for 3 days.
For the given case we have following values and their probabilities:
0 : 0.1
2 : x
3 : y
So the expected value will be = 0(0.1) + 2(x) + 3(y)
Expected value is given to be 2.05. So we can write the equation as:
2x + 3y = 2.05 (Equation 1)
Also for a probability distribution, the sum of probabilities must always equal to 1. So we can set up the second equation as:
0.1 + x + y = 1
x + y = 0.9 (Equation 2)
From Equation 2 we can write the value of x to be x = 0.9 - y. Using this value in equation 1, we get:
2(0.9 - y) + 3y = 2.05
1.8 - 2y + 3y = 2.05
1.8 + y = 2.05
y = 0.25
Using the value of y in equation 2 we get value of x to be 0.65
Therefore we can conclude that:
The probability that Jackson goes to gym for 2 days is 0.65 and the probability that he goes to gym for 3 days is 0.25
