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
What is the question
there is nothing clear stated here
It's kind of a mix between A and C there are some defragmenting tools that show you how much of your PC is wasted by program files. And in some defragmenting tools, it shows you all the files that are fragmented and gives you the option to defrag them or not. So the best answer would be A.
