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Black_prince [1.1K]
3 years ago
7

World Book is a/an A. Social media platform. B. Textbook. C. Almanac. D. Encyclopedia.

Computers and Technology
2 answers:
Nadusha1986 [10]3 years ago
8 0
D. Encyclopedia
Not a b or c
Lady_Fox [76]3 years ago
5 0

Answer:

d

Explanation:

You might be interested in
Rapid development programming languages eliminate the possibility of having bugs in code. True or False
Pachacha [2.7K]

False,


" I am talking about 20-30+ millions lines of code, software at the scale and complexity of Autodesk Maya for example.


If you freeze the development as long as it needs to be, can you actually fix all the bugs until there is simply not a single bug, if such a thing could be verified by computers? What are the arguments for and against the existence of a bug-free system?


Because there is some notion that every fix you make creates more bugs, but I don't think that's true.


By bugs I meant from the simplest typos in the UI, to more serious preventative bugs that has no workaround. For example a particular scripting function calculates normal incorrectly. Also even when there are workarounds, the problem still has to be fixed. So you could say you can do this particular thing manually instead of using the provided function but that function still has to be fixed."

work cited:

https://softwareengineering.stackexchange.com/questions/195571/is-it-possible-to-reach-absolute-zero-bug-state-for-large-scale-software


3 0
3 years ago
Assume you are given a boolean variable named isNegative and a 2-dimensional array of ints that has been created and assigned to
Anon25 [30]

Answer:

#include <iostream>

using namespace std;

int main(){

   int rows, cols;

   bool isNegatives;

   cout<<"Rows: ";

   cin>>rows;

   cout<<"Columns: ";

   cin>>cols;

   int a2d[rows][cols];

   for(int i =0;i<rows;i++){

   for(int j =0;j<cols;j++){

       cin>>a2d[i][j];}    }

   int negatives, others = 0;

   for(int i =0;i<rows;i++){

   for(int j =0;j<cols;j++){

       if(a2d[i][j]<0){

           negatives++;}

       else{

           others++;}}    }

   if(negatives>others){

       isNegatives = true;}

   else{

       isNegatives = false;}

   cout<<isNegatives;  

   return 0;

}

Explanation:

For clarity and better understanding of the question, I answered the question from the scratch.

This line declares number of rows and columns

   int rows, cols;

This line declares the Boolean variable

   bool isNegatives;

This line prompts user for rows

   cout<<"Rows: ";

This line gets the number of rows

   cin>>rows;

This line prompts user for columns

   cout<<"Columns: ";

This line gets the number of columns

   cin>>cols;

This line declares the array

   int a2d[rows][cols];

This line gets user input for the array

<em>    for(int i =0;i<rows;i++){</em>

<em>    for(int j =0;j<cols;j++){</em>

<em>        cin>>a2d[i][j];}    }</em>

This line declares and initializes number of negative and others to 0

   int negatives, others = 0;

The following iteration counts the number of negatives and also count the number of non negative (i.e. others)

<em>    for(int i =0;i<rows;i++){</em>

<em>    for(int j =0;j<cols;j++){</em>

<em>        if(a2d[i][j]<0){</em>

<em>            negatives++;}</em>

<em>        else{</em>

<em>            others++;}}    }</em>

This checks of number of negatives is greater than others

   if(negatives>others){

If yes, it assigns true to isNegatives

       isNegatives = true;}

   else{

If otherwise, it assigns false to isNegatives

       isNegatives = false;}

This prints the value of isNegatives

   cout<<isNegatives;  

<em>See attachment</em>

Download cpp
5 0
3 years ago
. What physical characteristic does a retinal scan biometric device measure?
elena-s [515]

Answer: D)The pattern of blood vessels at the back of the eye.

Explanation: The retinal scan is the techniques which is used in the bio-metric mechanism.The scanning of the retina's blood vessel pattern is capture by the scanner is the unique technique which is used for bio-metric purpose.

Blood vessel pattern is capture because it is present in unique form for every person and will help in distinguishing a person from another person while scanning.

Other options are incorrect because light reflection or reaching near retina and light pattern are not the unique way for identification while scanning a person.Thus the correct option is option(D).

4 0
4 years ago
Write a function called count_occurrences that takes two strings. The second string should only be one character long. The funct
iVinArrow [24]

Answer:

Here is the Python function:

def count_occurrences (string1 , string2):  # method that takes two strings as parameter and returns how many times second string occurs in first

   count = 0  #counts number of occurrence of string2 in string1

   for word in string1:  #iterates through each word of the string1

       for character in word:  #iterates through each character of each word in string1

           if character == string2:  # checks if the character in string1 is equal to the character of string2

               count = count + 1  #adds 1 to the count of string2 in string1

   return count  #returns number of times the string2 occurs in string1

#in order to check the working of the function add the following lines to the code:    

first_str = input("Enter the first string: ")  #prompts user to enter first string

second_str = input("Enter the second string (one character long): ") #prompts user to enter second string which should be one character long

occurrence = count_occurrences(first_str,second_str)  #calls count_occurrences method by passing both strings to it

print(second_str,"occurs",occurrence,"times in",first_str) #prints how many times the second_str occurs in the first_str on output screen

Explanation:

The program is well explained in the comments added to each statement of the above code. I will explain the program with an example:

Suppose

string1 = "hello world"

string2 = 'l'

So the method count_occurrences() should return how many times l occurs in hello world.

count is used to count the number of occurrences of l in hello world. It is initialized to 0.

The first loop iterates through each word of string1. The first word of string1 is "hello"

The second loop iterates through each character/letter of each word. So the first letter of "hello" word is 'h'.

The if condition if character == string2: checks if the character is equal to string2 i.e. l.

character = "h"

string2 = "l"

The are not equal so the program moves to the next character of word hello which is e. The if condition again evaluates to false because e is not equal to l. So the program moves to the next character of word hello which is l.

This time the if condition evaluates to true because

character = "l"

string2 = "l"

character == string2

l == l

So the if part is executed which has the following statement:

count = count + 1

So the count is added to 1. count was 0 previously. Now

count = 1

Next the program moves to the next character of word hello which is l. This time the if condition evaluates to true because the character l matches with string2. So count is again incremented to 1. count = 2.

Next the program moves to the next character of word hello which is o. The if condition evaluates to false as o does not match with string2 i.e. l. So the count remains 2.

Next the first loop moves to the next word of the string1 = "world". The inner (second) for loop iterates through each character of "world" from "w" to "d" and checks if any character is equal to string2. Only one character is equal to string2 in "world". So count is incremented to 1. Hence count = 3

After the loop ends the statement return count returns the number of times string2 occurs in the string1. So the output is:

l occurs 3 times in hello world  

8 0
3 years ago
From your fist impression, write down what first comes to mind as to what is good and bad about the way the device works.
nalin [4]

Answer:

xczczxczx

Explanation:

6 0
3 years ago
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