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3 years ago

The file type ____ identifies a word 2013 document.

1 answer:

5 0

I believe the answer would be ".docx". It is the file extension used in recent version of microsoft word. These type of files are XML based and are able to have texts, objects and other additions which are stored in separate files and are compressed as one single file.

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A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slo

krok68 [10]

Answer:

The answer is below

Explanation:

Given that:

Frame transmission time (X) = 40 ms

Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request

a) Probability that there is success on the first attempt = $e^{-G}G^k$ but k = 0, therefore Probability that there is success on the first attempt = $e^{-G}=e^{-2}=0.135$

b) probability of exactly k collisions and then a success = P(collisions in k attempts) × P(success in k+1 attempt)

P(collisions in k attempts) = [1-Probability that there is success on the first attempt]^k = $[1-e^{-G}]^k=[1-0.135]^k=0.865^k$

P(success in k+1 attempt) = $e^{-G}=e^{-2}=0.135$

Probability of exactly k collisions and then a success = $0.865^k0.135$

c) Expected number of transmission attempts needed = probability of success in k transmission = $e^{G}=e^{2}=7.389$

6 0

3 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

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
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).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
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

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

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

3 years ago

Purpose: This application provides experience with user input and interaction in the Console, writing files to disk, working wit

pishuonlain [190]

Answer:

the answer is c#/.net

Explanation:

5 0

3 years ago

When you make a cell phone call, which type of electromagnetic wave is most likely transmitting the signal?

Olegator [25]

They use Radio waves

6 0

3 years ago

Implement function hex2dec that takes a hex number hex_num as a string argument and prints out the corresponding decimal number.

DerKrebs [107]

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void hex2dec(string hex_num){

int n = 0;

//Loop through all characters in string

for(int i=0;i<hex_num.size();i++){

//take ith character

char c = hex_num[i];

//Check if c is digit

if(c>='0' && c<='9'){

n = 16*n + (c-48);

}

//Convert c to decimal

else{

n = 16*n + (c-55);

}

cout<<hex_num<<" : "<<n<<endl;

}

int main()

{

hex2dec("EF10");

hex2dec("AA");

return 0;

}

The Output can be seen below :

4 0

3 years ago