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Firdavs [7]
3 years ago
11

Which statement is NOT true regarding regular expression quantifiers? Group of answer choices They act on the preceding pattern

element They can operate on groups and character classes In order to use a plus sign "+" quantifier as a literal character, you must first 'escape' it using a backslash '\' character The question mark quantifier "?" will match the preceding element exactly one time
Computers and Technology
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

"The question mark quantifier "?" will match the preceding element exactly one time" is the correct answer to the given question .

Explanation:

The Quantifiers are defined as it determine how the several occurrences of a set of symbols, categories, or the characters should be found throughout the input to searching the matches.

  • The * quantifier in the regular expression corresponds the zero or more then zero to the previous item. it is represented by  {0,} quantifier
  • The + quantifier in the regular expression appears to fit in one or more times with the previous item. it equates with {1} quantifier.
  • The? Quantifier in regular expression compares zero or once of a previous item. It equates with {0,1}.
  • All the option are true regarding regular expression quantifiers so these option are incorrect according to the question
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(a) Store last 7 digits of your student ID in a vector (7 element row or column vector). Write a MATLAB code which creates a 7x7
astra-53 [7]

Answer:

MATLAB code explained below with appropriate comments for better understanding

Explanation:

clc

clear all

ID = [1 2 3 4 5 6 7]; % Replace this with your student ID

%(a)

A = zeros(7);

for i=1:7

 

A(i,i)= ID(i);

 

end

fprintf('A =\n');

disp(A);

B = diag(ID);

fprintf('B =\n');

disp(B);

fprintf('Both A and B are same\n');

%(b)

if(mod(A(6,6),2)==0)

fprintf('A(6,6) is even\n');

else

fprintf('A(6,6) is odd\n');

end

%(c)

if(A(3,3)>0)

fprintf('A(3,3) is positive\n');

else if(A(3,3)<0)

fprintf('A(3,3) is negative\n');

else

fprintf('A(3,3)=0\n');

end

end

%(d)

fprintf('\nRequired series : ');

n = 35;

while n>=0

fprintf('%i',n);

if(n>0)

fprintf(', ');

end

n = n - 5;

end

fprintf('\n');

%(e)

n = input('\nInput an integer : ');

fprintf('%i! = ',n);

F = 1;

while n>1

F = F * n;

n = n - 1;

end

fprintf('%i\n',F);

%(f)

clear all

x = [3 7 2 1];

y = [4 3 9 1];

A = 0;

C = x(1);

for i=1:length(x)

A = A + x(i)*y(i);

B(i) = x(i)/y(i);

if(min(x(i),y(i))<C)

C = min(x(i),y(i));

end

end

C = 1/C;

fprintf('\nA = %i\n',A);

fprintf('B = ');

disp(B)

fprintf('C = %0.5g\n',C);

%(g)

clear all

A = randi([5 25],[1,10]);

maxA = A(1);

for i = 2:10

if(maxA<A(i))

maxA = A(i);

end

end

minA = A(1);

i = 2;

while i<11

if(minA>A(i))

minA = A(i);

end

i = i+1;

end

fprintf('\nA = ');

disp(A);

fprintf('maxA = %i\n',maxA);

fprintf('minA = %i\n',minA);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%output

A =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

B =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

Both A and B are same

A(6,6) is even

A(3,3) is positive

Required series : 35, 30, 25, 20, 15, 10, 5, 0

Input an integer : 6

6! = 720

A = 52

B = 0.7500 2.3333 0.2222 1.0000

C = 1

A = 25 14 7 10 13 17 10 17 19 9

maxA = 25

minA = 7

>>

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3 years ago
Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have alre
Fudgin [204]

Answer:

The c++ program to implement the while loop is given below.

#include <iostream>

using namespace std;

int main() {

  // declaration of integer variables

   int k, n, total;

   // initialization of integer variables

   k=1, n=4, total=0;

//  loop executed till value of k becomes equal to value of n

   while( k <= n ){

       // cube of every integer is added to the variable total

       total = total + ( k * k * k );

       // value of k is incremented to go to the next number

k = k + 1 ;

   }  

   return 0;

}  

Explanation:

The program begins with the declaration of integer variables.  

int k, n, total;

This is followed by initialization of these variables.

k=1, n=4, total=0;

The while loop runs over the variable k which is initialized to 1. The loop runs till value of k reaches the value of integer n.

First, cube of k is computed and added to the variable total.

After first execution of the loop, total is initialized to the cube of 1.

Next, value of variable k is incremented by 1 so that k is initialized to next integer.

After first execution of the loop, k is incremented from 1 to 2.

while( k <= n )

{

total = total + ( k * k * k );

k = k + 1 ;

   }

When the value of k reaches the value of integer n, the cube of n is calculated and added to the variable, total.

When k is incremented, it becomes more than n and hence, loop gets terminated.

As the return type of main is int, the program terminates with the statement shown below.

return 0;

No output is displayed as it is not mentioned in the question.

No user input is taken as it is mentioned that integer variables are already initialized.

4 0
3 years ago
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