The matrix is not properly formatted.
However, I'm able to rearrange the question as:
![\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C-2%263%265%7C3%5C%5C3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
Operations:
Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.
Answer:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The first operation:
This means that the new second row (R2) is derived by:
Multiplying the first row (R1) by 2; add this to the second row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by 2
![2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]](https://tex.z-dn.net/?f=2%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&5&7|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%265%267%7C1%5Cend%7Barray%7D%5Cright%5D)
The second operation:
This means that the new third row (R3) is derived by:
Multiplying the first row (R1) by -3; add this to the third row
The row 1 elements are:
Multiply by -3
![-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]](https://tex.z-dn.net/?f=-3%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Hence, the new matrix is:
Answer:
Step-by-step explanation:
The lady gets paid $8 per hour that the shop is opened, so multiply the pay rate ($8) by the amount of hours the shop is opened
$8 * 8 = $64
Now add the $64 onto the expense of running the shop
$64 + $240 = $304
Now take that amount away from Friday's sales
$530 - $304 = $ 226
Profit($226) = Income($530) - Expenses($304)
$226 = $226
=)
Answer:
x=1
Step-by-step explanation:
4x+1=5
4x=5-1
4x=4
x=4/4=1
This happens to be a whole number.
Answer:
15.87%
Step-by-step explanation:
Notice that the mean of 0.35 inches with a standard deviation of 0.01 gives you when you add (to the right of the distribution), exactly 0.36. Since you want to find the probability (or percentage) of the bolts that have diameter LARGER than 0.36 in, that means you want to estimate the area under the Normal distribution curve from 0.36 to the right). See attached image.
We can use the tables of Z distribution for that, or the standard normal tables:
P(x>0.36) = P(z>(0.36-0.35)/0.01) = P(Z>1) = 0.1587 = 15.87%
Answer:
Hey! So I think the answer is A! I'm so sorry if its wrong, I'm trying to get points so could you mark me as brainliest? Bye and stay safe!! <3
Step-by-step explanation:
