Answer:
1) Is more representative
Step-by-step explanation:
The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability
This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.
If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.
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
Answer: 
Step-by-step explanation:
From the origin we can see to get to the point plotted we have to go down 1 and right three giving us the slope 
These are in the form of y=mx+b
b is the y-intercept and m is the slope
since the y-intercept is 0 b is 0 and isn't need leaving us with y=mx
We can put our slope into the equation giving us the answer in C
Answer:
60 arrangements
Step-by-step explanation:
This is a permutation, since I am assuming that order matters. To solve this you start at the number of items you have and multiply that by 1 less than the max until number of items are used up.
5P3 = 5 * 4 * 3 = 60
Answer
11a = 11
a = 1
Step-by-step explanation:
