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3 years ago

A subject or object's ability to use, manipulate, modify, or affect another subject or object is known as ___________.

Computers and Technology

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer:

The correct answer to the following question will be "Access".

Explanation:

Registered users have legal access to a network, while hackers may access a system illegally. The capacity is controlled by access controls. Access is defined as the right of a subject or object to use, manipulate, modify, or influence another subject or object.
Access control was the selective limitation of access to a location or other property while the mechanism is represented by access management. Access act may mean use, entry, or use. Permission is called authorization to access a resource.

Therefore, Access is the right answer.

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PLEASE HELP ASAP i will mark brilliant

Ray Of Light [21]

1. Cyberbullying
2.Confide in an adult you trust, such as a teacher or a parent
3.It is easy to miscommunicate and escalate the situation in an online environment

Cyberbully hurts people, so NEVER post something that could hurt someone's feelings

Hope this helps!

3 0

3 years ago

Read 2 more answers

(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

%(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

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

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

5 0

2 years ago

Suppose we have two String objects and treat the characters in each string from beginning to end in the following way: With one

Simora [160]

Answer:

If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.

Explanation:

Lets take an example of two strings abc and cba which are reverse of each other.

string1 = abc

string2 = cba

Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.

Push abc each character on a stack in the following order.

Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.

Insert cba each character on a stack in the following order.

a b c

First c is added to queue then b and then a.

Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.

First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match

c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.

6 0

3 years ago

Assume the following variable definition appears in a program:

AnnZ [28]

Answer:

cout << setprecision(2)<< fixed << number;

Explanation:

The above statement returns 12.35 as output

Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.

The statement starts by setting precision to 2 using setprecision(2)

This is immediately followed by the fixed manipulator;

The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;

Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.

The fixed manipulator is then followed by the variable to be printed.

See code snippet below

#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

// Initializing the double value

double number = 12.3456;

//Print result

cout << setprecision(2)<< fixed << number;

return 0;

}

4 0

3 years ago

Is Bob Lazar telling a lie/withholding information?

Anon25 [30]

Answer:

I think he may be a nut, but I'm not sure, there's little evidence of what he says to be true.

6 0

3 years ago