How has Linux affected the market for proprietary software?​

Answer:

The description of this capability is:

d. Horizontal scaling

Explanation:

The capability that Brandon has achieved is horizontal scaling, which involves the increasing of the resources of cloud applications to meet his increasing demand for cloud services.  Vertical scaling involves the addition or subtraction of power to a cloud server, which practically upgrades the memory, storage, or processing power.  But to scale horizontally, more resources are added or subtracted.  The purpose of horizontal scaling is to spread out or in the workload of existing resources and either increase or decrease overall performance and capacity.

Answer:

Most of the tools are used to perform simple repetitive tasks that take a lot of time if done manually.

1. Sending Connection requests  

2. Running Direct messaging campaigns  

3. Endorsing people’s skills.  

4. Collecting leads data  

5. Extracting data from LinkedIn  

With automation, it may seem like you’ve been dealt the ultimate hand.    

In addition, some of the latest LinkedIn automation tools run personalized campaigns that result in increasing:

1. Increasing networks  

2. Connections  

3. Leads  

4. Sales

A desktop computer is a type of mobile device: B. False.

A desktop computer simply refers to an electronic device that is designed and developed to receive data in its raw form as an input and processes these data into an output that's usable by an end user.

Generally, desktop computers are fitted with a power supply unit (PSU) and designed to be used with an external display screen (monitor) unlike mobile device.

In conclusion, a desktop computer is not a type of mobile device.

Learn more about desktop computer here: brainly.com/question/959479

#SPJ9

Answer:

The program is as follows:

float x

float y

float z

x = Get next input

y = Get next input

z = Get next input

Put RaiseToPower(x,y) to output with 5 decimal places

Put "\n" to output

Put RaiseToPower (x,RaiseToPower (y,z)) to output with 5 decimal places

Put "\n" to output

Put AbsoluteValue(x) to output with 5 decimal places

Put "\n" to output

Put SquareRoot(RaiseToPower (x * y,z)) to output with 5 decimal places

Explanation:

This declares all variables

float x

float y

float z

This gets input for all variables

x = Get next input

y = Get next input

z = Get next input

This prints x^y

Put RaiseToPower(x,y) to output with 5 decimal places

This prints a new line

Put "\n" to output

This prints x^(y^z)

Put RaiseToPower (x,RaiseToPower (y,z)) to output with 5 decimal places

This prints a new line

Put "\n" to output

This prints |x|

Put AbsoluteValue(x) to output with 5 decimal places

This prints a new line

Put "\n" to output

This prints sqrt((x * y)^z)

Put SquareRoot(RaiseToPower (x * y,z)) to output with 5 decimal places

Answer:

All enterprise platforms are cloud based.

Answer:

a high speed central network that connects other networks in a distance spanning up to several miles.

Explanation:

A backbone network functions just like the human backbone providing support for network systems by offering a network infrastructure that allows small, high speed internet connectivity. It is a principal data route between large interconnected networks which offers connection services spanning several miles. Most local area networks are able to connect to the backbone network as it is the largest data connection on the internet. This backbone networks are mainly utilized by large organizations requiring high bandwidth connection.

Answer:  

Assumption: Only 1 job can be taken at a time  

This becomes a weighted job scheduling problem.  

Suppose there are n jobs

   Sort the jobs according to fj(finish time)

   Define an array named arr to store max profit till that job

       arr[0] = v1(value of 1st job)

       For i>0. arr[i] = maximum of arr[i-1] (profit till the previous job) or wi(weight of ith job) + profit till the previous non-conflicting job

   Final ans = arr[n-1]

The previous non-conflicting job here means the last job with end timeless than equal to the current job.  

To find the previous non-conflicting job if we traverse the array linearly Complexity(search = O(n)) = O(n.n) = O(n^2)  

else if we use a binary search to find the job Complexity((search = O(Logn)) = O(n.Log(n))

Answer:

def multiply_by_20(num):

   print(num * 20)

Explanation:

def multiply_by_20(num):

   print(num * 20)

# Testing the function here. ignore/remove the code below if not required

multiply_by_20(2)

multiply_by_20(7)

multiply_by_20(5)

Answer:

Code:  

import os   # os module to create directory  

event_file = open("event_details.txt", "r") # Getting event file  

event_details = ""  # event details to be stored  

for row in event_file:  # traversing through the event_file  

   event_details += row    # appending event details  

os.mkdir("./invitations")   # make directory in the same parent directory  

names = open("guest_list.txt", "r") # getting names of the people  

for name in names:  # traversing through names in guest_list file

   name = name.replace('\n', '')   # removing the ending '\n' from the name  

   invitation_msg = "Hi! " + name + ", You are heartly invited in the Ceremony.\nAt " + event_details # Generating the invitation message  

   file_name = '_'.join(name.split(' '))   # Spliting name in space and joining with the '_'  

   file_path = "./invitations/" + file_name + ".txt" # Generating each file path  

   invite_file = open(file_path, "w")  # Creating the file for each name  

   invite_file.write(invitation_msg)   # Write invitation to file

Output:

Answer:

Is an electronic device used for storing and processing data.

Answer:

Explanation:

260 cost units, Big O(n) complexity for a push

Answer:

def main():

   singleMarkerPrice = 0.80

   packageOfMarkersPrice = 3.50

   tax = 0.065

   shipping = 0.050

   userInput = int(input("How many markers do you want? "))

   amountOfPackages = 0

   singleMarkers = 0

   if userInput % 5 == 0:

       amountOfPackages = userInput / 5

   else:

       amountOfPackages = userInput // 5

       singleMarkers = userInput % 5

   # Just for syntax so if the single marker amount is one, it prints package instead of packages

   if amountOfPackages == 1:

       print("You have " + str(int(amountOfPackages)) + " complete package")

   else:

       print("You have " + str(int(amountOfPackages)) + " complete packages")

   # Just for syntax so if the single marker amount is one, it prints marker instead of markers

   if singleMarkers == 1:

       print("You have " + str(int(singleMarkers)) + " single marker")

   else:

       print("You have " + str(int(singleMarkers)) + " single markers")

   totalAmountBeforeTax = (amountOfPackages * packageOfMarkersPrice) + (singleMarkers * singleMarkerPrice)

   print("The total amount before tax comes out to " + str(float(totalAmountBeforeTax)))

   costOfShipping = float(round((totalAmountBeforeTax * shipping), 2))

   costOfTax = float(round((totalAmountBeforeTax * tax), 2))

   print("The cost of shipping is " + str(costOfShipping))

   print("The cost of tax is " + str(costOfTax))

   totalAmount = totalAmountBeforeTax + costOfShipping + costOfTax

   print("The total amount comes out to " + str(round(totalAmount, 2)))

main()

Explanation:

This should be correct. If it isn't let me know so I can fix the code.

Answer:

A security stance that allows all communications except those prohibited by specific deny exceptions

A plan to restore the mission-critical functions of the organization once they have been interrupted by an adverse event

A security guideline, procedure, or recommendation manua

lAn administrative rule whereby no single individual possesses sufficient rights to perform certain actions

Answer:

Explanation:

The following code is written in Java. It creates the SimpleCar class with the variables for position, milesForward, and milesReverse. It contains the constructor, Honk, reportStatus, and setter methods needed and as requested. The scanner object is created in the main method and asks the user for the number of miles forward as well as the number of miles in reverse. A test case was created and the output can be seen in the attached image below.

import java.util.ArrayList;

import java.util.Scanner;

class Brainly {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("How many miles forward did the car drive?");

       int forward = in.nextInt();

       System.out.println("How many miles in reverse did the car drive?");

       int reverse = in.nextInt();

       SimpleCar beep = new SimpleCar(forward, reverse);

       beep.Honk();

       beep.reportStatus();

   }

}

class SimpleCar {

   int position;

   int milesForward;

   int milesReverse;

   public SimpleCar(int milesForward, int milesReverse) {

       this.milesForward = milesForward;

       this.milesReverse = milesReverse;

       this.position = 0;

   }

   public void Honk() {

       System.out.println("HONK");

   }

   public void reportStatus() {

       position = milesForward - milesReverse;

       System.out.println("Car is " + position + " miles from starting point.");

   }

   public void setMilesForward(int milesForward) {

       this.milesForward = milesForward;

       this.position += milesForward;

   }

   public void setMilesReverse(int milesReverse) {

       this.milesReverse = milesReverse;

       this.position -= milesReverse;

   }

}

Answer:

def main():

   # input

   num1 = int(input("Type in a number: "))

   num2 = int(input("Type in another number: "))

   num3 = int(input("Type in another number: "))

   list1 = [num1, num2, num3]

   # sorts the array

   list1.sort()

   # list1[-1] prints the first element in the array

   print("The largest number is: " + str(list1[-1]))

   # -len(list1) takes the length of the list, in this case 3, and makes it a negative number.  

   # That negative number is then used as an index for the list1 array in order to print the lowest number.

   print("The smallest number is: " + str(list1[-len(list1)]))

main()

Explanation:

Hope this helped :) I left some comments so you know what's going on. You can also use max(list1) and min(list1) but I chose indexing because indexing is the better way of doing stuff like this.

Have a good day!

Answer:

The program is as follows:

import random

print("Rock\nPaper\nScissors")

points = int(input("Points to win the game: "))

player_point = 0; computer_point = 0

while player_point != points and computer_point != points:

   computer = random.choice(['Rock', 'Paper', 'Scissors'])

   player = input('Choose: ')

   if player == computer:

       print('A tie - Both players chose '+player)

   elif (player.lower() == "Rock".lower() and computer.lower() == "Scissors".lower()) or (player.lower() == "Paper".lower() and computer.lower() == "Rock".lower()) or (player == "Scissors" and computer.lower() == "Paper".lower()):

       print('Player won! '+player +' beats '+computer)

       player_point+=1

   else:

       print('Computer won! '+computer+' beats '+player)

       computer_point+=1

print("Player:",player_point)

print("Computer:",computer_point)

Explanation:

This imports the random module

import random

This prints the three possible selections

print("Rock\nPaper\nScissors")

This gets input for the number of points to win

points = int(input("Points to win the game: "))

This initializes the player and the computer point to 0

player_point = 0; computer_point = 0

The following loop is repeated until the player or the computer gets to the winning point

while player_point != points and computer_point != points:

The computer makes selection

   computer = random.choice(['Rock', 'Paper', 'Scissors'])

The player enters his selection

   player = input('Choose: ')

If both selections are the same, then there is a tie

   if player == computer:

       print('A tie - Both players chose '+player)

If otherwise, further comparison is made

   elif (player.lower() == "Rock".lower() and computer.lower() == "Scissors".lower()) or (player.lower() == "Paper".lower() and computer.lower() == "Rock".lower()) or (player == "Scissors" and computer.lower() == "Paper".lower()):

If the player wins, then the player's point is incremented by 1

       print('Player won! '+player +' beats '+computer)

       player_point+=1

If the computer wins, then the computer's point is incremented by 1

   else:

       print('Computer won! '+computer+' beats '+player)

       computer_point+=1

At the end of the game, the player's and the computer's points are printed

print("Player:",player_point)

print("Computer:",computer_point)

Answer:

TDMA: Time-division multiple access (TDMA) will operate effectively.

CSMA: Carrier-sense multiple access (CSMA)  will NOT operate properly.

Slotted Aloha: Slotted Aloha will NOT perform effectively.

Token passing: Token passing will operate effectively.

Explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Suppose you were charged with putting together a LAN to support IP telephony (only) and that multiple users may want to carry on a phone call at the same time. Recall that IP telephony digitizes and packetizes voice at a constant bit rate when a user is making an IP phone call. How well suited are these four protocols for this scenario?

TDMA:

CSMA:

Slotted Aloha:

Token passing:

Provide a brief explanation of each answer.

The explanation of the answers is now provided as follows:

TDMA: Time-division multiple access (TDMA) will operate effectively in this situation because it provides a consistent bit rate service of one slot every frame.

CSMA: Because of collisions and a changing amount of time to access the channel, Carrier-sense multiple access (CSMA)  will NOT operate properly in this situation. Also, the length of time it takes to access a channel is not limited.

Slotted Aloha: Just like CSMA, Slotted Aloha will NOT perform effectively in this situation because of collisions and a different amount of time to access the channel. Also, the length of time it takes to access a channel is limitless.

Token passing: Token passing will operate effectively in this case because each station has a turn to transmit once per token round, resulting in a service with an effectively constant bit rate.

Answer:

However, if the dataset is relatively small, every data point counts. In these situations, a missing data point means loss of valuable information. In any case, generally missing data creates imbalanced observations, cause biased estimates, and in extreme cases, can even lead to invalid conclusions.

It makes data analysis to be more ambiguous and more difficult.

Missing data is simply the same as saying that there are values and information that are unavailable. This could be due to missing files or unavailable information.

A dataset set with missing data means more work for the analyst. There needs to be a transformation in those fields before the dataset can be used.

Generally speaking missing data could lead to bias in the estimation of data.

A data scientist is a data expert who is in charge of data. He performs the job of data extraction, data analysis, data transformation.

read more at https://brainly.com/question/17578521?referrer=searchResults

Answer:

Explanation:

The Rectangle class was not provided in this question but after a quick online search I was able to find the class code. Due to this I was able to create a test code in the main Method of my program to create a Rectangle object and call it's display() method. The Test code and the output can be seen in the attached image below. While the code below is simply the Main Test Program as requested.

class Brainly {  

   public static void main(String[] args) {

       Rectangle rectangle = new Rectangle(20, 8);

       rectangle.display();

   }

}

Answer:

D. changing the font of a label

Answer:

The first computer brought in Nepal was IBM 1401 which was brought by the Nepal government in lease (1 lakh 25 thousands per month) for the population census of 1972 AD (2028 BS). It took 1 year 7 months and 15 days to complete census of 1crore 12.5 lakhs population.

Answer:

perdonnosee

Explanation:

Complete Question:

Departmental information for the four departments at Samoa Industries is provided below. Total Cost Cost Driver Square Feet Number of employees Janitorial $150,000 square footage serviced 200 Cafeteria 50,000 Number of employees 20,000 Cutting 1,125,000 4,000 120 Assembly 1,100,000 16,000 The Janitorial and Cafeteria departments are support departments Samoa uses the sequential method to allocate support department costs, first allocating the costs Department to the Cafeteria, Cutting and Assembly departments Determine the proportional (percentage) usage of the Janitorial Department by the a. Cafeteria Department b. Cutting Department c. Assembly Department

Answer:

Samoa Industries

Proportional (percentage) usage of the Janitorial Department:

Cafeteria = 50%

Cutting = 10%

Assembly = 40%

Explanation:

a) Data and Calculations:

                 Total Cost         Cost Driver             Square Feet       Number of

                                                                                                    employees

Janitorial   $150,000 Square footage serviced          200                40

Cafeteria      50,000 Number of employees        20,000                 12

Cutting     1,125,000                                                 4,000               120

Assembly 1,100,000                                               16,000                 40

Proportional (percentage) usage of the Janitorial Department:

Cafeteria = 50% (20,000/40,000 * 100)

Cutting = 10% (4,000/40,000 * 100)

Assembly = 40% (16,000/40,000 * 100)

Cost Allocation of the Janitorial Department:

Cafeteria = 50% * $150,000 = $75,000

Cutting = 10% * $150,000 =        15,000

Assembly = 40,% * $150,000 = 60,000

Total =                                    $150,000

Answer:

explain all background information first.

Explanation:

Now if you are to deliver a message and you have suspicions that the person who is to read it might be uninterested, unwilling, hostile or displeased, you should put the main idea later in the message. That is, it should come after you have provided details given explanations or evidence.

It is not right to start with bad news. As a matter of fact bad news should not be shared through mails, IM or texts.

Answer:

Explanation:

Given

Answer:

The points according to the given question are provided below.

Explanation:

Answer:

aaaa

Explanation: