What is the decimal equivalent of (11000 + 10000)/101?

"You are on a service call to fix a customer’s printer when she asks you to install a software package. The software is on a per

pickupchik [31]

Answer:

Ask her to get a genuine software

Explanation:

if i will install, it can cause the following problems:

1. Prated software makes your system Vulnerable to the security attacks because the activities of the software are not monitored by any organization and no one is responsible for anything bad happened to your system.

2. It may stop working anytime because there would not be maintenance patches available for it so that it can work properly.

3. It cannot be updated and may cause problems in core functionalities of it.

4.Serious legal actions can be taken against anyone using them because economy has drastic decrease due ti use of it.

7 0

2 years ago

9.11: Array Expander

vodka [1.7K]

The Array Expander is an illustration of arrays and functions.

Arrays are variables that stores multiple values
Functions are named statements that are executed when called

<h3>The Array Expander program </h3>

The Array Expander program written in C++, where comments are used to explain each action is as follows:

#include <iostream>

using namespace std;

//This declares the Array Expander function

int* ArrayExpander(int *oldArr, int size){

//This declares the new array

int *newArr = new int[size * 2];

//This iteration sets values into the new array

for (int i = 0; i < size * 2; i++) {

if(i < size){

*(newArr+i) = *(oldArr+i);

}

else{

*(newArr+i) = 0;

}

//This returns a pointer to the new array

return newArr;

}

//The main method begins here

int main(){

//This declares the length of the array, N

int N;

//This gets input for N

cin>>N;

int initArr[N];

//If N is between 1 and 50

if(N > 0 && N <=50){

//This iteration gets values for the array

for(int i = 0; i<N; i++){

cin>>initArr[i];

}

//This calls the Array Expander function

int *ptr = ArrayExpander(initArr, N);

//This iteration prints the elements of the new array

for (int i = 0; i < N*2; i++) {

cout << ptr[i] << " ";

}

return 0;

}

Read more abou C++ programs at:

brainly.com/question/27246607

3 0

1 year ago

What two statements about IPv6 addresses are true? This task contains the radio buttons and checkboxes for options. The shortcut

Oliga [24]

Answer:

A and C

Explanation:

Option A:

In IPv6 there is a rule to reduce an IPv6 address when there are two or more consecutive segments of zeros just one time. This rule says that you can change the consecutive zeros for “::”

Here is an example

How to reduce the following IPv6 address?

ff02:0000:0000:0000:0000:0000:0000:d500

Ans: ff02::d500

Example 2:

2001:ed02:0000:0000:cf14:0000:0000:de95

Incorrect Answer -> 2001:ed02::cf14::de95

Since the rule says that you can apply “::” just one time, you need to do it for a per of zero segments, so the correct answer is:

Correct Answer -> 2001:ed02::cf14:0:0:de95

Or

2001:ed02:0:0:cf14::de95

Option C:

Since in IPv6 there are $2^{128}$ available addresses which means 340.282.366.920.938.463.463.374.607.431.768.211.456 (too many addresses), there is no need of NAT solution, so each device can have its own IP address by the same interface to have access through the internet if needed. If not, you can block the access through internet by the firewall.

4 0

3 years ago

You are attempting to print a document from your workstation to the network printer in your office. The print job failed when yo

Helen [10]

Answer:

An Incident

Explanation:

an incident is an unplanned interruption to a service

6 0

2 years ago

(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