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Zina [86]
2 years ago
13

What is the value of the variable named myNum after the statements that follow are executed? var myNum = 14; var yourNum = 4; my

Num++; yourNum-- myNum = myNum * yourNum;
Computers and Technology
1 answer:
Lostsunrise [7]2 years ago
8 0

Answer:

45

Explanation:

Initially, myNum is equal to 14 and yourNum is equal to 4

Then, myNum is incremented by 1 and becomes 15

Also, yourNum is decremented by 1 and becomes 3

Finally, myNum is set to myNum x yourNum, 15 x 3 = 45

You might be interested in
Write a program that has the following String variables: firstName, middleName, and lastName. Initialize these with your first,
Licemer1 [7]

Answer:

The programming language is not stated; However, I'll answer your question using C++ programming language.

Comments are used for explanatory purpose

Program starts here

#include<iostream>

#include <string>

using namespace std;

int main()

{

//Declare Variables

string firstName, middleName, lastName;

char firstInitial, middleInitial, lastInitial;

/*Initialize firstName, middleName and lastName (Replace values with your details)*/

firstName = "First Name";

middleName = "Middle Name";

lastName = "Last Name";

 

// Get Initials

firstInitial = firstName.at(0);

middleInitial = middleName.at(0);

lastInitial = lastName.at(0);

 

//Print Results

cout<<"Lastname: "<<lastName<<endl;

cout<<"Firstname: "<<firstName<<endl;

cout<<"Middlename: "<<middleName<<endl;

cout<<"Last Initial: "<<lastInitial<<endl;

cout<<"First Initial: "<<firstInitial<<endl;

cout<<"Middle Initial: "<<middleInitial<<endl;

return 0;

}

3 0
2 years ago
8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101
Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

           = 64 + 32 + 16 + 8 + 4 + 2 + 1

           = 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

               = 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

               = 128 + 64 + 32 + 16

               = 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

               = 64 + 32 + 4 + 2 + 1

               = 103

01111101   = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

               = 64 + 32 + 16 + 8 + 4 + 1

               = 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

             = 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                 = 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

              = 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 10101111 11000000 11111000 00011101  to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 223

10110000 =  1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                = 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

              = 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

            = 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 247

11000111 =  1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:                

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0
1 year ago
An administrator has been working within an organization for over 10 years. He has moved between different IT divisions within t
rodikova [14]

Answer:

User roles and access authorization

(Explanation of this question is given below in the explanation section.

Explanation:

As noted in question that a terminated employee comeback in office and installed some malicious script on a computer that was scheduled to run a logic bomb on the first day of the following month. And, that script will change administrator passwords, delete files, and shut down over 100 servers in the data center.

The following problem can be with that account

  • Unauthorized user access and privileges

The user may be given unauthorized access to system resources. If a user has unauthorized access to system resource then he can modify or destroy the system function easily.

  • The firewall may be disabled to detect the malicious script

The firewall should be enabled to detect such types of attack from unauthorized access to the system

  • Role

May be user role is not defined properly. If role would be defined according to user role then there are very fewer chances in doing unauthorized changes in the system that will result in an unexpected outage.

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