For the first question, all you need to know is that the (amount you want) divided by (the amount of cards total) is the probability of getting the amount you want at random. for example, if there are 3 red marbles in a bag of ten marbles, you divide 3 by 10 (0.30) the probability of picking up a red marble is 30%. try adding up how many cards you want to pick up by the number of cards total.
If the coefficient matrix has a pivot in each column, it means that it is shaped like this:
![A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D)
So, the correspondant system
will look like this:
![\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%5C%5Cx_2%5C%5Cx_3%5C%5Cx_4%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Db_1%5C%5Cb_2%5C%5Cb_3%5C%5Cb_4%5Cend%7Barray%7D%5Cright%5D)
This turn into the following system of equations:
The last equation is solvable for 
: we easily have
Once the value for is known, we can solve the third equation for 
:
(recall that is now known)
The pattern should be clear: you can use the last equation to solve for . Once it is known, the third equation involves the only variable . Once
Answer:
Step-by-step explanation:
No. Each side must be less than the sum of the remaining two sides.
10 ≮ 5+1
Answer:
3
Step-by-step explanation:
You would start by dividing
3 by 4. Then multiply 3/4 and 4. The 4s would cancel out leaving you with 3. 3/ 4×4=3
hope this helps
Answer:
A & B
Step-by-step explanation:
A) V = pi×r²×h
= 3.142 × 8² × 5 = 1005.44 cm³
B) V = ⅓ × 18×16 × 12 = 1152 cm³
C) V = ½×5×3 × 7 = 52.5 cm³
