Answer:
![A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
message SHOW_ME_THE_MONEY_
Step-by-step explanation:
The matrix
![A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%264%5C%5C-1%26-3%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20%7CA%7C%3D%281%20%5Ctimes%20-3%29-%28-1%5Ctimes%204%29%3D1%5C%5C%5Crightarrow%20A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.
Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table
Encoded message Decoded message message in letters by association
11 52 19 8 S H
-8 -9 15 23 O W
-13 -39 0 13 _ M
5 20 5 0 E _
12 56 20 8 T H
5 20 5 0 E _
-2 7 13 15 M O
9 41 14 5 N E
25 100 25 0 Y _
Then the message decoded is SHOW_ME_THE_MONEY_
Let the require value be x, then
P(z > x) = 0.025
1 - P(z < x) = 0.025
P(z < x) = 1 - 0.025 = 0.975
P(z < x) = P(z < 1.96)
x = 1.96
Therefore, <span>the positive critical value that corresponds to a right tail area of 0.025</span> is 1.96
Answer:
y=-2x-1
Step-by-step explanation:
Considering the given probability distributions, distribution D is valid.
<h3>When a probability distribution is valid?</h3>
A probability distribution is valid if:
- There are no negative probabilities.
- The sum of all probabilities is of 1.
In this problem, only distribution D has a sum of 1, hence it is the only valid distribution.
More can be learned about probability distributions at brainly.com/question/23670007
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