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Leya [2.2K]
3 years ago
5

If you were writing a story set in our time, what would you include to give readers a sense of our world? Check any that apply.

cell phones DVD and Blu-ray flat-screen televisions the Internet solar panels
Computers and Technology
1 answer:
charle [14.2K]3 years ago
6 0

I would include cell phones , DVD and Blu-ray flat- screen televisions into my story to give the readers a sense of our world.

Explanation:

At this time, the world and its people are connected by cell phones.

Everything that is shown to the world by the media, can be seen clearly and  with a good audio output, by using DVD and Blu ray flat screen televisions, as these are equipped with  the latest technology to display the images clearly.

People use Social platforms ,to spread information to every corner of the world. This is possible, when each and every individual has a cell phone with an internet connection.

So, If I were writing a story , I would like to include these gadgets as part of my storyline, as this would depict the time and the sense of our world ,at that time to the readers.

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Modify the list according to the following: append the smallest value at the end of the list when all numbers in the list are ne
ddd [48]

Answer:

The solution code is written in Python 3:

  1. def modifyList(listNumber):
  2.    posCount = 0
  3.    negCount = 0
  4.    for x in listNumber:
  5.        if x > 0:
  6.            posCount += 1
  7.        else:
  8.            negCount += 1
  9.    
  10.    if(posCount == len(listNumber)):
  11.        listNumber.append(max(listNumber))
  12.    
  13.    if(negCount == len(listNumber)):
  14.        listNumber.append(min(listNumber))
  15.    
  16.    print(listNumber)
  17. modifyList([-1,-99,-81])
  18. modifyList([1,99,8])
  19. modifyList([-1,99,-81])

Explanation:

The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).

To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).

If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).

If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.

At last we can print the listNumber (Line 17).

If we test our function using the three sets of input list, we shall get the following results:

[-1, -99, -81, -99]

[1, 99, 8, 99]

[-1, 99, -81]

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Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

 //Declaration of variables in the program

 double start_organisms;

 double daily_increase;

 int days;

 double updated_organisms;

 //The user enters the number of organisms as desired

 cout << "Enter the starting number of organisms: ";

 cin >> start_organisms;

 //Validating input data

 while (start_organisms < 2)

 {

     cout << "The starting number of organisms must be at least 2.\n";

     cout << "Enter the starting number of organisms: ";

     cin >> start_organisms;

 }

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 cout << "Enter the daily population increase: ";

 cin>> daily_increase;

 //Validating the increase

 while (daily_increase < 0)

 {

     cout << "The average daily population increase must be a positive value.\n ";

     cout << "Enter the daily population increase: ";

     cin >> daily_increase;

 }

 //The user enters number of days

 cout << "Enter the number of days: ";

 cin >> days;

 //Validating the number of days

 while (days<1)

 {

     cout << "The number of days must be at least 1.\n";

     cout << "Enter the number of days: ";

     cin >> days;

 }

 

 //Final calculation and display of results based on formulas

 for (int i = 0; i < days; i++)

 {

     updated_organisms = start_organisms + (daily_increase*start_organisms);

     cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

     

     start_organisms = updated_organisms;

 }

 system("pause");

  return 0;

//end

}

Explanation:

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Answer:

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Explanation:

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Answer: this is a moral question, but here is my view

Explanation:

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