Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
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
Answer:
A. 0
Explanation:
The technician should configure the RAID 0 for Joe.
RAID 0 also referred to as the striped volume or stripe set is configured to allow the fastest speed and the most storage capacity by splitting data evenly across multiple (at least two) disks, without redundancy and parity information.
Also, RAID 0 isn't fault tolerant, as failure of one drive will cause the entire array to fail thereby causing total data loss.
Answer:
antivirus.......hope this answers your question
Answer:
# include <iostream.h>
# include <stdio.h>
# include <string.h>
using namespace std;
class citizen
{
int i;
public string name[30];
public long int phonenumber[30];
public void addindividual(string name1)
{
If (i<=30)
{ int flag=0;
for(int j=0; j<=i;j++)
{
if (strcmp(name[i], name1)
{
flag=1;
}
else
{
flag=0;
}
}
If (flag)
{
if (i<30)
{
for(j=i+1;j<=30; j++)
{
cout<<"Enter the name:"; getchar(name[j]);
cout<<"Enter the phone number:"; cin>>phonenumber[j];
i++;
}
else
{
cout<<"The person already exists";
exit();
}
}
else
{
cout<<"array is full:";
exit();
}
}
}
Void main()
{
string str;
cout<<" Enter name:";
getline(cin, str); ;
citizen c1=new citizen();
c1.addindividual(name1);
}
Explanation:
With a little more effort you can make the program allow the user to enter any number of details, but less than 30 overall. We have used here flag, and as a programmer we know why we use the Flag. It is used to check whether certain Boolean condition is fulfilled or not. Here, we are checking whether a given name is present in the array of names, and if it is not present, we add that to the list. And if the name is present, we print, it already exist.
