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arlik [135]
2 years ago
7

You have implemented a network where hosts are assigned specific roles, such as for file sharing and printing. Other hosts acces

s those resources but do not host services of their own.
What type of network do you have?
Computers and Technology
1 answer:
yulyashka [42]2 years ago
5 0

<u>Client-server</u> implemented a network where hosts are assigned specific roles, such as for file sharing and printing. Other hosts access those resources but do not host services of their own.

<u>Explanation:</u>

The client-server can be utilized on the web just as on a neighborhood (LAN). Instances of customer server frameworks on the web incorporate internet browsers and web servers, FTP customers and servers, and the DNS. Different hosts get to those assets yet don't have administrations of their own. Since it permits arrange permits numerous PCs/gadgets to interface with each other and offer assets.

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1. Write a set of routines for implementing several stacks and queues within a single array. Hint: Look at the lecture material
JulsSmile [24]

Answer:

Out of the two methods The 2nd method is a better option for implementing several stacks and arrays using the single array system this is because Method 2 makes efficient use of the available space

Explanation:

A) Method 1 ( Divide the given single array in the size of n/k.)

  • How to use method 1 to implement several stacks involves :

i)  To implement several stacks through an array (x) is by dividing the array in n/k parts.

ii) The k represents the slots in which the different stacks will be placed and the n represents the size of the array x.

iii) If you need to implement at least two stacks place the first stack in the slot of a [0] to a [n/k - 1], and another stack in the slot of a[n/k] to a[2n/k-1].

note: The  disadvantage of this method is that the use of the space of the array is not much efficient. therefore method 2 is better

  • How to use method 1 to implement queues involves :

i) queues can also be implemented through an array. applying the same method above

ii) Divide the array in slots and place the queues in that slots.

This method has a problem with  the efficient utilization of the space.therefore method 2 is preferred

B) Method 2 ( uses the space efficiently ) uses two more arrays to implement stacks which are : Top_array and Next_array

  • how to use method 2 to implement several stacks

i) Store the indexes of the next item that will also be stored in all stacks in this initial stack

ii)The initial actual array is x[] and this will store the stacks.

iii)Simultaneously with several stacks, the stack which contain the free slots in the array x[] will also be maintained.

iv)  The entries of the Top_array[] will be initialized to -1. This implies that all the stacks are empty.

v)  Firstly the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

vi) Initialize The top of the free stack that is maintaining the free slots  as 0.

vii)  The complexity of push () (method to insert an element) and pop () (method to delete an element) operations by using this method is O (1).

  • How to use method 2 to implement several queues

The same method applicable to implementing several stacks is used here but with a difference is the presence of three extra arrays which are :

Front_array[] = indicates the number of queues. This array stores the indexes of the front elements of the stacks.

Rear_array[] = determines the sizeof the array k . This array stores the indexes of the last elements of the stacks.

Next_array[] = The array n indicates the size of the single array say x. This array stores the indexes of the next items that is being pushed.

i) The initial actual array is a[] which will store the queues. The free slots will also be maintained.

ii)The entries of the Front_array[] will be initialized to -1. This means  that all the queues are empty initially

ii) Initially the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

iv) Apply The complexity of enqueue ()  and dequeue ()  by using this method is O (1).

4 0
2 years ago
Sam’s password is known to be formed of 3 decimal digits (0-9) in a given order. Karren and Larry are attempting to determine Sa
prohojiy [21]

Answer:

100

Explanation:

7 0
2 years ago
Read 2 more answers
What does the hard disk drive do? It stores all of the information on a computer. It controls a computer’s operating system. It
umka21 [38]

The hard disk drive, OR HDD Stores all the information on the computer.

5 0
2 years ago
Read 2 more answers
Write a class named Employee that has private data members for an employee's name, ID_number, salary, and email_address. It shou
mario62 [17]

Solution :

class Employee:

   #Define the

   #constructor.

   def __$\text{init}$__($\text{self, nam}e$,  ID_number, $\text{salary}$, email):

       #Set the values of

       #the data members of the class.

       $\text{self.nam}e$ = name

       $\text{self.ID}$_number = ID_number

       $\text{self.salary}$ = salary

       self.email_address = email

#Define the function

#make_employee_dict().

def make_employee_dict(list_names, list_ID, list_salary, list_email):

   #Define the dictionary

   #to store the results.

   employee_dict = {}

   #Store the length

   #of the list.

   list_len = len(list_ID)

   #Run the loop to

   #traverse the list.

   for i in range(list_len):

       #Access the lists to

       #get the required details.

       name = list_names[i]

       id_num = list_ID[i]

       salary = list_salary[i]

       email = list_email[i]

       #Define the employee

       #object and store

       #it in the dictionary.

       employee_dict[id_num] = Employee(name, id_num, salary, email)

   #Return the

   #resultant dictionary.

   return employee_dict

6 0
1 year ago
Write a SQL query to find the population of the planet named 'Caprica' -- 10 points Find the first name, last name, and age of p
Artemon [7]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Please find the attached question.

Answer:

a) SQL Query:

SELECT population

FROM bsg_planets

WHERE name='Caprica';

b) SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE lname!='Adama';

c) SQL Query:

SELECT name, population

FROM bsg_planets

WHERE population > 2600000000

d) SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE age IS NULL;

Explanation:

a) Write a SQL query to find the population of the planet named 'Caprica'

Syntax:

SELECT  Column

FROM TableName

WHERE Condition;

For the given case,

Column = population

TableName  = bsg_planets

Condition = name='Caprica'

SQL Query:

SELECT population

FROM bsg_planets

WHERE name='Caprica';

Therefore, the above SQL query finds the population of the planet named 'Caprica' from the table bsg_planets.

b) Find the first name, last name, and age of people from bsg_people whose last name is not 'Adama'

Syntax:

SELECT  Column1, Column2, Column3

FROM TableName

WHERE Condition;

For the given case,

Column1 = fname

Column2 = lname

Column3 = age

TableName  = bsg_people

Condition = lname!='Adama'

SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE lname!='Adama';

Therefore, the above SQL query finds the first name, last name and age of people whose last name is not 'Adama' from the table bsg_people.

c) Find the name and population of the planets with a population larger than 2,600,000,000

Syntax:

SELECT  Column1, Column2

FROM TableName

WHERE Condition;

For the given case,

Column1 = name

Column2 = population

TableName  = bsg_planets

Condition = population > 2600000000

SQL Query:

SELECT name, population

FROM bsg_planets

WHERE population > 2600000000

Therefore, the above SQL query finds the name and population of the planets with a population larger than 2,600,000,000 from the table bsg_planets.

d) Find the first name, last name, and age of people from bsg_people whose age is NULL

Syntax:

SELECT  Column1, Column2, Column3

FROM TableName

WHERE Condition;

For the given case,

Column1 = fname

Column2 = lname

Column3 = age

TableName  = bsg_people

Condition = age IS NULL

SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE age IS NULL;

Therefore, the above SQL query finds the first name, last name and age of people whose age is NULL from the table bsg_people.

6 0
2 years ago
Read 2 more answers
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