Answer:
Consider the following code.
Explanation:
save the following code in read_and_interp.m
function X = read_and_interp(s)
[m, n] = size(s);
X = zeros(m, 1);
for i = 1:m
if(str2num(s(i, 2:5)) == 9999)
% compute value based on previous and next entries in s array
% s(i, 2:5) retrieves columns 2-5 in ith row
X(i,1) = (str2num(s(i-1 ,2:5)) + str2num(s(i+1,2:5)))/2;
else
X(i,1) = str2num(s(i,2:5));
end
end
end
======================
Now you can use teh function as shown below
s = [ 'A' '0096' ; 'B' '0114' ; 'C' '9999' ; 'D' '0105' ; 'E' '0112' ];
read_and_interp(s)
output
ans =
96.000
114.000
109.500
105.000
112.000
The type of windows I am using is:Windows 8.1
Answer:
James Gleick believes that they immerse themselves in a rich textual world. The answer is C.
From what my teacher taught us it would be c.
A brute force attack. Ciphers should be sufficiently complex for this to be ever successful.
