In the next five years there are expected to be over _____ unfilled jobs in the US in computer science.

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

2 years ago

Suppose that you need to maintain a collection of data whose contents are fixed- i.e., you need to search for and retrieve exist

Keith_Richards [23]

Answer:

Option a: a sorted array

Explanation:

Since the expectation on the data structure never need to add or delete items, a sorted array is the most desirable option. An array is known for its difficulty to modify the size (either by adding item or removing item). However, this disadvantage would no longer be a concern for this task. This is also the reason, linked list, binary search tree and queue is not a better option here although they offer much greater efficiency to add and remove item from collection.

On another hand, any existing items from the sorted array can be easily retrieved using address indexing and therefore the data query process can be very fast and efficient.

4 0

2 years ago

add is a method that accepts two int arguments and returns their sum. Two int variables, euroSales and asiaSales, have already b

Oduvanchick [21]

Answer:

// here is code in Java.

// package

import java.util.*;

// class definition

class Main

{

// method that return sum of two sale value

public static int Add(int euroSales,int asiaSales)

{

// return the sum

return euroSales+asiaSales;

}

//main method of the class

public static void main (String[] args) throws java.lang.Exception

{

try{

// variables

int euroSales=100;

int asiaSales=150;

int eurasiaSales;

// call the function

eurasiaSales=Add(euroSales,asiaSales);

// print the sum

System.out.println("total sale is:"+eurasiaSales);

}catch(Exception ex){

return;}

}

Explanation:

Declare and initialize two variables "euroSales=100" and "asiaSales=150". Declare another variable eurasiaSales. Call the method Add() with euroSales and asiaSales as parameter. This method will add both the value and return the sum.This sum will be assigned to variable eurasiaSales.Then print the sum.

Output:

total sale is:250

8 0

2 years ago

In public-key encryption, the two keys–one for coding and one for decoding–are known as ________.

Feliz [49]

Public key and private key - In public key encryption, a pair of keys is used (public key and private key). The public key can be made available publicly, while the private key is only known by the owner. The public key is used to encrypt the message, while the private key is used to decrypt the message.

5 0

2 years ago

Create a class named BaseballGame that contains data fields for two team names and scores for each team in each of nine innings.

m_a_m_a [10]

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

protected String[] names = new String[2];

protected int[][] scores;

protected int innings;

public BaseballGame() {

innings = 9;

scores = new int[2][9];

for(int i = 0; i < 9; i++)

scores[1][i] = scores[0][i] = -1;

}

public String getName(int team) {

return names[team];

}

public void setNames(int team, String name) {

names[team] = name;

}

public int getScore(int team, int inning) throws Exception {

if(team < 0 || team >= 2)

throw new Exception("Team is ut of bounds.");

if(inning < 0 || inning >= innings)

throw new Exception("Inning is ut of bounds.");

return scores[team][inning];

}

public void setScores(int team, int inning, int score) throws Exception {

if(team < 0 || team >= 2)

throw new Exception("Team is ut of bounds.");

if(inning < 0 || inning >= innings)

throw new Exception("Inning is ut of bounds.");

if(score < 0)

throw new Exception("Score is ut of bounds.");

for(int i = 0; i < inning; i++)

if(scores[team][i] == -1)

throw new Exception("Previous scores are not set.");

scores[team][inning] = score;

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

public HighSchoolBaseballGame() {

innings = 7;

scores = new int[2][7];

for(int i = 0; i < 7; i++)

scores[1][i] = scores[0][i] = -1;

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

public LittleLeagueBaseballGame() {

innings = 6;

scores = new int[2][6];

for(int i = 0; i < 6; i++)

scores[1][i] = scores[0][i] = -1;

}

6 0

2 years ago