well, to do this we first need to find the mean of all values.
we need to find the distance of each value from that mean (subtract the mean from each value, ignore minus signs)
Then find the mean of those distances.
so, after we do all of that with the given numbers, our final result would be 2.333333333..... (It goes on) we round it up to 2.3, and viola! the answer is C.
Answer:
It would be y = 40x
The new equation is y = 45x
Step-by-step explanation:
I did the other question you put ;)
Check what the transformation does for each vector in the standard basis of R²<em> </em>:
<em>T</em> (1, 0) = (5, 2)
<em>T</em> (0, 1) = (-4, 0)
Now compute the weights <em>a</em>, <em>b</em> and <em>c</em>, <em>d</em> such that
<em>a</em> <em>T</em> (1, 0) + <em>b</em> <em>T</em> (0, 1) = (-2, 1)
<em>c</em> <em>T</em> (1, 0) + <em>d</em> <em>T</em> (0, 1) = (-1, 1)
![\left[\begin{array}{cc|c}5&-4&-2\\2&0&1\end{array}\right]\sim\left[\begin{array}{cc|c}1&0&\frac12\\\\0&1&\frac98\end{array}\right]\implies a=\dfrac12,b=\dfrac98](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%26-4%26-2%5C%5C2%260%261%5Cend%7Barray%7D%5Cright%5D%5Csim%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%260%26%5Cfrac12%5C%5C%5C%5C0%261%26%5Cfrac98%5Cend%7Barray%7D%5Cright%5D%5Cimplies%20a%3D%5Cdfrac12%2Cb%3D%5Cdfrac98)
![\left[\begin{array}{cc|c}5&-4&-1\\2&0&1\end{array}\right]\sim\left[\begin{array}{cc|c}1&0&\frac12\\\\0&1&\frac78\end{array}\right]\implies c=\dfrac12,d=\dfrac78](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%26-4%26-1%5C%5C2%260%261%5Cend%7Barray%7D%5Cright%5D%5Csim%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%260%26%5Cfrac12%5C%5C%5C%5C0%261%26%5Cfrac78%5Cend%7Barray%7D%5Cright%5D%5Cimplies%20c%3D%5Cdfrac12%2Cd%3D%5Cdfrac78)
Then the matrix <em>A'</em> is
Rectangle's perimeter formula is p=2(length+width)
2(9x+8x)=68
17x=34
x=2
length is 9(2)=18 feet
width is 8(2)=16feet
