Raj encoded a secret phrase using matrix multiplication. Using A = 1, B = 2, C = 3, and so on, he multiplied the clear text code
for each letter by the matrix to get a matrix that represents the encoded text. The matrix representing the encoded text is . What is the secret phrase? Determine the location of spaces after you decode the text.
2 answers:
Answer:
THE CORN IS YUMMY
Step-by-step explanation:
Understanding the question:
Raj encoded a secret phrase using matrix multiplication. For that he first converted each letter in that phrase to integers using the following codes:
A=1,B=2,C=3,D=4,E=5,F=6,G=7,H=8,I=9,J=10,K=11,L=12,M=13,N=14,O=15,P=16,Q=17,R=18,S=19,T=20,U=21,V=22,W=23,X=24,Y=25,Z=26
We suppose that Raj had a matrix X formed by letters of that phrase encoded in integers/numbers. And he multiplied that matrix X with matrix C to get an encoded matrix, let us say 'encoded X'.
Where; C =
and encoded X =
What we need now is X.
<u>Raj got from the phrase to encoded matrix by:</u>
1. Encoding each letter into numbers and forming a matrix X.
2. Multiplying that matrix with another Matrix C to obtain the given encoded matrix.
<u>To find the secret phrase we will reverse the steps as:</u>
1. Multiplying the encoded matrix by 'Inverse of Matrix C' to get matrix X.
2. Converting the elements of X into alphabets.
Inverse of C = (Adjoint of C)/Determinant(C)
Adjoint of =
Determinant of C = |C| = ad - bc = 2*2 - 5*1 = -1
Inverse of C = / (-1)
=
Multiplying Inverse(C) and encoded X =
=
Converting these elements back to alphabets, we get:
=
Hence, it is obvious that the secret phrase is "THECORNISYUMMY"
The location of spaces are between every word ( THE CORN IS YUMMY).
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Answer:
a) 0.7287
b) 0.9663
c) 0.237
d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.5 million tons of cargo per week
Standard Deviation, σ = 0 .82 million tons
We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.
Formula:
a) P( port handles less than 5 million tons of cargo per week)
P(x < 5)
Calculation the value from standard normal z table, we have,
b) P( port handles 3 or more million tons of cargo per week)
Calculating the value from the standard normal table we have,
c)P( port handles between 3 million and 4 million tons of cargo per week)
d) P(X=x) = 0.85
We have to find the value of x such that the probability is 0.85.
P(X > x)
Calculation the value from standard normal z table, we have,
Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.