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mariarad [96]
3 years ago
8

What is the "key" to a Caesar Cipher that someone needs to know (or discover) to decrypt the message? a) A secret word only know

n to Caesar. b) The number of characters to shift each letter in the alphabet. c) The letter that occurs most often in the encrypted message. d) The day of the month that the encrypted message was sent.
Computers and Technology
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

The number of characters to shift each letter in the alphabet.

Explanation:

Caeser Cipher is the technique of encryption of data, to make it secure by adding characters between alphabets. These are the special characters that make the message secure while transmitting.

According to the standards, For Decryption, we remove these special characters between alphabets to make message understandable.

<em>So, we can say that,to de-crypt the message, the number of characters to shift each letter in the alphabet.</em>

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ANSWER:

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EXPLANATION:

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4 0
4 years ago
Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,
nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
  3. P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
  4. P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
  5. P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
  3. P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
  4. P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
  5. P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
  6. P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

  1. P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
  2. P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
  3. P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
  4. P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
  5. P5 will not be allocated to any of the available space because none can contain it.
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0
4 years ago
In a C++ program, one and two are double variables and input values are 10.5 and 30.6. After the statement cin &gt;&gt; one &gt;
SSSSS [86.1K]

Answer:

variable one stores 10.5 and two stores 30.6

Explanation:

In c++ language, the cout keyword is used to write to the standard output. The input from the user is taken by using the cin keyword.

For the input from the user, a variable need is declared first.

datatype variable_name;

The input can be taken one at a time as shown.

cin >> variable_name;

The input for more than one variable can be taken in a single line as given by the syntax below.

1. First, two variables are declared.

datatype variable_name1, variable_name2;

2. Both input is taken simultaneously as shown.

cin >> variable_name1, variable_name2;

For the given scenario, two double variables are declared as below.

double one, two;

The question mentions input values for the two double variables as 10.5 and 30.6.

3. The given scenario takes both the values as input at the same time, in a single line as shown below.

cin >> one >> two;

4. After this statement, both the values entered by the user are accepted.

The values should be separated by space.

First value is stored in the variable one.

Second value is stored in the variable two.

5. The user enters the values as follows.

10.5 30.6

6. The value 10.5 is stored in variable one.

7. The value 30.6 is stored in variable two.

The c++ program to implement the above is given below.

#include <iostream>

using namespace std;

int main() {

   double one, two;      

   cout << " Enter two decimal values " << endl;

   cin >> one >> two;

   cout << " The input values are " << endl;

   cout << " one : " << one << " \t " << " two : " << two << endl;

   return 0;

}

OUTPUT

Enter two decimal values  

10.5 30.6

The input values are  

one : 10    two : 530.6

6 0
3 years ago
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