Answer: Internet Protocal (IP)
Answer:
#include <iostream>
using namespace std;
int main() {
int a[4][5];//declaring a matrix of 4 rows and 5 columns.
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
{
if(i==3)//initializing last row as 0.
{
a[i][j]=0;
}
else//initializing last row as 1.
{
a[i][j]=1;
}
}
}
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
cout<<a[i][j]<<" ";//printing the matrix.
cout<<endl;
}
return 0;
}
Output:-
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 0 0 0 0
Explanation:
I have created a matrix of size 4 rows and 5 columns.I have used for loops to fill the array.To fill the last row with 0 i have used if statement.else we are filling it with 1.
Answer:
The correct Answer is 0.0571
Explanation:
53% of U.S. households have a PCs.
So, P(Having personal computer) = p = 0.53
Sample size(n) = 250
np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10
So, we can just estimate binomial distribution to normal distribution
Mean of proportion(p) = 0.53
Standard error of proportion(SE) = 
= 
= 0.0316
For x = 120, sample proportion(p) = 
= 
= 0.48
So, likelihood that fewer than 120 have a PC
= P(x < 120)
= P( p^ < 0.48 )
= P(z < 
) (z=
)
= P(z < -1.58)
= 0.0571 ( From normal table )
Answer:
Part a: The program will print 3.
Part b: The program will print 4
Part c: The program will print 1.
Explanation:
Part a: As the scoping is static, the x integer has a value of 1 and the y integer has a value of 2 so the addition is 2+1=3.
Part b: As the scoping is dynamic with deep binding, the x integer has a value of 2 and the y integer has a value of 2 so the addition is 2+2=4.
Part c: As the scoping is dynamic with shallow binding, the x integer has a value of 1 so it will print 1.
