Hi there!
Finding the equivalen ratio means to find a equivalent numbers that’s equivalent to 6 and 7 which also means a number that 6 and 7 all goes into...
6,12,18,24,30,36,42
7,14,21,28,35,42
Final Result: 42
3 - 1 = 2
1/4 - 3/16 = 4/16 - 3/16 = 1/16
answer is : 2 1/16 (in mixed number) or 33/16 (in improper fraction)
Answer:
It is impossible to simulate this with a random number generator without knowing what the correct answer choices were
Step-by-step explanation:
Given


Required
How can he select the right answer
Using randint will only generate a random number which could or could not be the answer to the question.
This is so because each of the 5 options for the question has the same probability of 1/5. So, using randint will only generate a random number. This generated random number has 1/5 chance of being the answer and 4/5 of not being the answer to the question.
In a nutshell, he can not make use of a simulator to select the answer to the questions in this scenario, unless he knows the solution.
<em>Hence (a) answers the question.</em>
Answer:
11 pt
Step-by-step explanation:
There are 2 pints in one quart. This means that there are 10 pints in 5 quarts, plus 1 would be 11.
Answer:
Verified
Step-by-step explanation:
Let the 2x2 matrix A be in the form of:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
Where det(A) = ad - bc # 0 so A is nonsingular:
Then the transposed version of A is
![A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26c%5C%5Cb%26d%5Cend%7Barray%7D%5Cright%5D)
Then the inverted version of transposed A is
![(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5ET%29%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20cb%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
The inverted version of A is:
![A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-b%5C%5C-c%26d%5Cend%7Barray%7D%5Cright%5D)
The transposed version of inverted A is:
![(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5E%7B-1%7D%29%5ET%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
We can see that

So this theorem is true for 2 x 2 matrices