Answer:
![x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
According to the given situation, The computation of all x in a set of a real number is shown below:
First we have to determine the 
so that 
![\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%5C0%261%26-3%265%5C%5C2%26-4%264%26-4%5Cend%7Barray%7D%5Cright%5D)
Now the augmented matrix is
![\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C2%26-4%264%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
After this, we decrease this to reduce the formation of the row echelon
![R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_1%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%262%26-6%266%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%204R_2%20%2B5R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%264%26-12%260%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%20%5Cfrac%7BR_2%7D%7B4%7D%2C%20%20R_3%20%3D%20%5Cfrac%7BR_3%7D%7B-4%7D%20%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%261%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B3%20R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B5%20R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%260%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4x_3%263x_3%26x_3%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%20x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
By applying the above matrix, we can easily reach an answer
Total number of students surveyed = 200
Number of male students = 80
Number of female students = 200 - 80 = 120
Number of brown eyed male students = 60
Probability of a brown eyed male student = 60 / 80 = 0.75.
Since, <span>eye color and gender are independent, this means that eye color is not affected by the gender. Thus, we expect a similar probability of brown eye for female as we had for male.
Let the number expected of brown eyed females be x, then x / 120 = 0.75.
Thus, x = 120(0.75) = 90.
Therefore, the number female students surveyed expected to be brown eyed is 90.</span>
Answer:
36 meters²
Step-by-step explanation:
Hassan built a fence around a square yard. It took 48 meters^2 of lumber to build the fence. The fence is 1.5 meters tall.
Let say square yard = x * x meters²
Perimeter of Square yard = 4x meters
Height of fence =1.5 meters
Area of fence = 4x * 1.5 =6x meters²
6x = 48
Divide both sides by 6
x = 8
Area of square yard = 6 × 6= 36 meters²
36 meters² is the area of the yard inside the fence
Answer:
54
Step-by-step explanation:
Total pictures in the roll = 36
half of the roll = 1/2 of the total pictures in the roll = 1/2 * 36 = 18
Given Paul made 2 prints of each of the half pictures
no. of prints made of half of the pictures = 18*2 = 36
No. of prints left after the half of the roll is printed for two prints= 36-18 = 18
No. of prints made of rest of the pictures 1
Therefore , no of prints made of rest half of pictures = 18*1 = 18
Total prints made = 36 + 18 = 54.
Thus, Paul made 54 prints in all.
