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Vlada [557]
1 year ago
13

An aviation tracking system maintains flight records for equipment and personnel. The system is a critical command and control s

ystem that must maintain a global availability rate of 99%. The entire system is on a cloud platform that guarantees a failover to multiple zones within a region. In addition to the multi-zonal cloud failover, what other solution would provide the best option to restoring data and rebuilding systems if the primary cloud service becomes unavailable?
Computers and Technology
1 answer:
sergeinik [125]1 year ago
5 0

Answer:

offline backup solution

Explanation:

In such a scenario, the best option would be an offline backup solution. This is basically a local and offline server that holds all of the flight record data that the cloud platform has. This offline backup server would be updated frequently so that the data is always up to date. These servers would be owned by the aviation company and would be a secondary solution for the company in case that the cloud platform fails or the company cannot connect to the cloud service for whatever reason. Being offline allows the company to access the database regardless of internet connectivity.

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A computer application such as Microsoft Access that is used to store data and convert it into information is a ________________
lorasvet [3.4K]
All data is stored in table
5 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
QUESTION 9 of 10: Bob charged $200 for a plane ticket last month. When he received his statement, he saw that he could pay the m
IgorLugansk [536]

Answer:

yes cuz 25x8=200

Explanation:

3 0
2 years ago
Read 2 more answers
Develop an EER model for the following situation using the traditional EER notation, the Visio notation, or the subtypes inside
podryga [215]

Answer:

Explanation:

To develop the model for Creating the supertype/subtype relationship three diffrent types of notation is given bellow accrdig to the diagram attached.

1.Traditional EER (Enhance Entity-Relational) notation

2.Microsoft Visio notation

3.Subtypes Inside supertype note

<em><u>Traditional EER(Enhance Entity-Relational)notation for the international school of technology diagram is attached bellow</u></em>

The diagram consist of the following entity

Room Supertype

Media Entity Type

COURSE entity type

Section weak entity type

Schedule associatiative entity type

<u><em>Microsoft Visio Notation for the international school of technology  is ashown in the diagram bellow.</em></u>

The diagram consist of the following entity

Room Supertype

Media Entity Type

Computer entity type

Instructor entity type

Time slot entity type

<u><em>Subtype Inside Supertype note Notation for the international school of technology  is shown in the diagram bellow.</em></u>

The diagram consist of the following entity

Room Supertype

Media Entity Type

Computer entity type

Instructor entity type

Time slot entity type

3 0
2 years ago
Write a program to declare a matrix A[][] of order (MXN) where ‘M’ is the number of rows and ‘N’ is the
Liula [17]

Answer:

import java.io.*;

import java.util.Arrays;

class Main {

   public static void main(String args[])

   throws IOException{

       // Set up keyboard input

       InputStreamReader in = new InputStreamReader(System.in);

       BufferedReader br = new BufferedReader(in);

 

       // Prompt for dimensions MxN of the matrix

       System.out.print("M = ");

       int m = Integer.parseInt(br.readLine());

       System.out.print("N = ");

       int n = Integer.parseInt(br.readLine());

       // Check if input is within bounds, exit if not

       if(m <= 2 || m >= 10 || n <= 2 || n >= 10){

           System.out.println("Matrix size out of range.");

           return;

       }

       // Declare the matrix as two-dimensional int array

       int a[][] = new int[m][n];

 

       // Prompt for values of the matrix elements

       System.out.println("Enter elements of matrix:");

       for(int i = 0; i < m; i++){

           for(int j = 0; j < n; j++){

               a[i][j] = Integer.parseInt(br.readLine());

           }

       }

       // Output the original matrix

       System.out.println("Original Matrix:");

       printMatrix(a);

       // Sort each row

       for(int i = 0; i < m; i++){

         Arrays.sort(a[i]);

       }

       // Print sorted matrix

       System.out.println("Matrix after sorting rows:");

       printMatrix(a);

   }

   // Print the matrix elements separated by tabs

   public static void printMatrix(int[][] a) {

       for(int i = 0; i < a.length; i++){

           for(int j = 0; j < a[i].length; j++)

               System.out.print(a[i][j] + "\t");

           System.out.println();

       }

   }

}

Explanation:

I fixed the mistake in the original code and put comments in to describe each section. The mistake was that the entire matrix was sorted, while only the individual rows needed to be sorted. This even simplifies the program. I also factored out a printMatrix() method because it is used twice.

4 0
1 year ago
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