Write a small program that asks the user how many asterisks it should print out. The user responds with a number, then the program prints that number of asterisks.

Answer:

The key technology/ innovation advantage and disadvantage can be defined as follows:

Explanation:

Following are the 5 innovations and technology, which promote the other development goals, like renewable energy, quality of jobs, and growth of economic with the good health and very well-being:

1) The use of crop monitoring drone technology promotes sustainable farming.  

2) The production of plastic brick including highways, floors, and houses.  

3) The new banking market or digital banking.  

4) E-commerce site.    

5) Renewable energy deployment such as solar panels.  

Advantage:

Disadvantage:

Answer:

Using java

//assuming that hashmap object name is ChangeMap

ChangeMap. replace("short", "long");

System.out.println("New HashMap: "

+ ChangeMap.toString());

Explanation:

From the above we have used the replace method to replace the value of the "short" key in the hashtable with "long" instead of the previous value "tall". We have used the printed the hashtable to the console using println and the ".toString()" method that we added to the function's parameter.

Answer:

(a) values[0].length

Explanation:

In programming languages such as Java, an array is a collection of data of the same type. For example, a and b below are an example of an array.

a = {5, 6, 7, 9}

b = {{2,3,4}, {3,5,4}, {6,8,5}, {1,4,6}}

But while a is a one-dimensional array, b is a regular two-dimensional array. A two-dimensional array is typically an array of one-dimensional arrays.

Now, a few thing to note about a two-dimensional array:

(i) The number of rows in a 2-dimensional array is given by;

arrayname.length

For example, to get the number of rows in array b above, we simply write;

b.length which will give 4

(ii) The number of columns in a 2-dimensional array is given by;

arrayname[0].length

This is with an assumption that all rows have same number of columns.

To get the number of columns in array b above, we simply write;

b[0].length which will give 3

Therefore, for a regular two-dimensional array named values, the number of columns is represented by: values[0].length

Answer:

Explanation:

In such a situation like this one, the first step would be to identify the brand and model of the video card itself. Once you find the brand and model you can simply go over to the brand manufacturer's website and go to the driver downloads section. There you simply search for the model series of the video card in your computer and download/install the latest version. The brand manufacturer's official website always has the most up to date drivers for all of their hardware.

Answer:

The program written in python is as follows:

import math

def max_magnitude(user_val1, user_val2):

     if abs(user_val1)>abs(user_val2):

           return user_val1

     else:

           return user_val2

val1 = int(input("Value 1: "))

val2 = int(input("Value 2: "))

print(max_magnitude(val1,val2))

Explanation:

This line imports the math module in the program

import math

This line declares the function with two parameters

def max_magnitude(user_val1, user_val2):

The if condition gets the absolute value of both integers and compares them; The integer with greater magnitude is returned, afterwards

     if abs(user_val1)>abs(user_val2):

           return user_val1

     else:

           return user_val2

The main method starts here

The next two lines prompt user for input

val1 = int(input("Value 1: "))

val2 = int(input("Value 2: "))

This line gets the integer with higher magnitude

print(max_magnitude(val1,val2))

Answer:

B. Device boot order

Explanation:

The Device boot order makes a list of all the possible devices a system should check to in the operating system's boot files, as well as the proper sequence they should be in. Removable devices, hard drives, and flash drives are some devices that can be listed in the device boot order.

For the user whose computer displays a 'non-bootable drive' error, the device boot order would check all the devices listed to attempt booting from any of them. These devices might be in the order of removable discs, CD-ROM, hard drive. If all the options do not work, the computer would then do a Network boot. The order in which the devices are listed can be altered by the user.

Answer:

Given address = 94EA6[tex]_{16}[/tex]

tag = 0 * 94  ( 10010100 )

line = 0 * 1 D 4 ( 111010100 )

word position = 0*6 ( 110 )

Explanation:

using the direct mapping method

Number of lines = 512

block size = 8 words

word offset = [tex]log ^{8} _{2}[/tex]  = 3 bit

index bit = [tex]log^{512}_{2}[/tex]  = 9 bit

Tag = 20 - ( index bit + word offset ) = 20 - ( 3+9) = 8 bit

Given address = 94EA6[tex]_{16}[/tex]

tag = 0 * 94  ( 10010100 )

line = 0 * 1 D 4 ( 111010100 )

word position = 0*6 ( 110 )

Answer:

a. 255.255.255.0 (class C)

b. 255.255.255.224

Explanation:

Here, we want to give the implied subnet mask of the given classful IPV4 address

We proceed as follows;

Given IPv4 address: 200.200.200.200

Classes

Class A : 0.0.0.0 to 127.255.255.255

Class B: 128.0.0.0 to 191.255.255.255

Class C: 192.0.0.0 to 223.255.255.255

Class D: 224.0.0.0 to 239.255.255.255

so 200.200.200.200 belongs to Class C and it's subnet mask is 255.255.255.0

In CIDR(Classless Inter Domain Routing)

subnet /27 bits means 27 1s and 5 0s. so subnet

i.e 11111111.11111111.11111111.11100000 which in dotted decimal format is 255.255.255.224 .

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

   const int WEIGHT_MIN   =    0,

             WEIGHT_MAX   =   20,

             DISTANCE_MIN =   10,

             DISTANCE_MAX = 3000;

   float package_weight,

         distance,

         total_charges;

   cout << "\nWhat is the weight (kg) of the package? ";

   cin >> package_weight;

   if (package_weight <= WEIGHT_MIN ||

       package_weight > WEIGHT_MAX)

   {

       cout << "\nWe're sorry, package weight must be\n"

            << " more than 0kg and less than 20kg.\n"

            << "Rerun the program and try again.\n"

            << endl;

   }

   else

   {

       cout << "\nDistance? ";

       cin >> distance;

       if (distance < DISTANCE_MIN ||

           distance > DISTANCE_MAX)

       {

           cout << "\nWe're sorry, the distance must be\n"       << "within 10 and 3000 miles.\n"

                << "Rerun the program and try again.\n"

                << endl;

       }

       else

       {

            if (package_weight <= 2)

               total_charges = (distance / 500) * 1.10;

           else if (package_weight > 2 &&

                   package_weight <= 6)

               total_charges = (distance / 500) * 2.20;

           else if (package_weight > 6 &&

                   package_weight <= 10)

               total_charges = (distance / 500) * 3.70;

           else if (package_weight > 10 &&

                   package_weight <= 20)

               total_charges = (distance / 500) * 4.80;

           cout << setprecision(2) << fixed

               << "Total charges are $"

               << total_charges

               << "\nFor a distance of "

               << distance

               << " miles\nand a total weight of "

               << package_weight

               << "kg.\n"

               << endl;

       }

   }

Explanation:

Answer:

[tex]P_{10} =0.1[/tex]

Explanation:

Given

[tex]0.1, 0.1, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.6, 0.7, 0.7, 0.7, 0.8, 0.8[/tex]

Required

Determine [tex]P_{10}[/tex]

[tex]P_{10}[/tex] implies 10th percentile and this is calculated as thus

[tex]P_{10} = \frac{10(n+1)}{100}[/tex]

Where n is the number of data; n = 14

[tex]P_{10} = \frac{10(n+1)}{100}[/tex]

Substitute 14 for n

[tex]P_{10} = \frac{10(14+1)}{100}[/tex]

[tex]P_{10} = \frac{10(15)}{100}[/tex]

Open the bracket

[tex]P_{10} = \frac{10 * 15}{100}[/tex]

[tex]P_{10} = \frac{150}{100}[/tex]

[tex]P_{10} = 1.5th\ item[/tex]

This means that the 1.5th item is [tex]P_{10}[/tex]

And this falls between the 1st and 2nd item and is calculated as thus;

[tex]P_{10} = 1.5th\ item[/tex]

Express 1.5 as 1 + 0.5

[tex]P_{10} = (1 +0.5)\ th\ item[/tex]

[tex]P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item[/tex]

From the given data; [tex]1st\ item = 0.1[/tex] and [tex]2nd\ item = 0.1[/tex]

[tex]P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item[/tex] becomes

[tex]P_{10} =0.1 +0.5(0.1 - 0.1)[/tex]

[tex]P_{10} =0.1 +0.5(0)[/tex]

[tex]P_{10} =0.1 +0[/tex]

[tex]P_{10} =0.1[/tex]

Answer:

Here is the python method:

def is_power(n1, n2): # function that takes two positive integers n1 and n2 as arguments

   if(not n1>0 and not n2>0): #if n1 and n2 are not positive integers

       print("The number is not a positive integer so:") # print this message if n1 and n2 are negative

       return None # returns none when value of n1 and n2 is negative.

   elif n1 == n2: #first base case: if both the numbers are equal

       return True #returns True if n1=n2

   elif n2==1: #second base case: if the value of n2 is equal to 1

       return False #returns False if n2==1

   else: #recursive step

       return is_divisible(n1, n2) and is_power(n1/n2, n2) #call divisible method and is_power method recursively to determine if the number is the power of another

Explanation:

Here is the complete program.

def is_divisible(a, b):

   if a % b == 0:

       return True

   else:

       return False

def is_power(n1, n2):

   if(not n1>0 and not n2>0):

       print("The number is not a positive integer so:")  

       return None  

   elif n1 == n2:

       return True

   elif n2==1:

       return False

   else:

       return is_divisible(n1, n2) and is_power(n1/n2, n2)  

print("is_power(10, 2) returns: ", is_power(10, 2))

print("is_power(27, 3) returns: ", is_power(27, 3))

print("is_power(1, 1) returns: ", is_power(1, 1))

print("is_power(10, 1) returns: ", is_power(10, 1))

print("is_power(3, 3) returns: ", is_power(3, 3))

print("is_power(-10, -1) returns: ", is_power(-10, -1))  

The first method is is_divisible method that takes two numbers a and b as arguments. It checks whether a number a is completely divisible by number b. The % modulo operator is used to find the remainder of the division. If the remainder of the division is 0 it means that the number a is completely divisible by b otherwise it is not completely divisible. The method returns True if the result of a%b is 0 otherwise returns False.

The second method is is_power() that takes two numbers n1 and n2 as arguments. The if(not n1>0 and not n2>0) if statement checks if these numbers i.e. n1 and n2 are positive or not. If these numbers are not positive then the program prints the message: The number is not a positive integer so. After displaying this message the program returns None instead of True of False because of negative values of n1 and n2.

If the values of n1 and n2 are positive integers then the program checks its first base case: n1 == n2. Suppose the value of n1 = 1 and n2 =1 Then n1 is a  power of n2 if both of them are equal. So this returns True if both n1 and n2 are equal.

Now the program checks its second base case n2 == 1. Lets say n1 is 10 and n2 is 1 Then the function returns False because there is no positive integer that is the power of 1 except 1 itself.

Now the recursive case return is_divisible(n1, n2) and is_power(n1/n2, n2)  calls is_divisible() method and is_power method is called recursively in this statement. For example if n1 is 27 and n2 is 3 then this statement:

is_divisible(n1, n2) returns True because 27 is completely divisible by 3 i.e. 27 % 3 = 0

is_power(n1/n2,n2) is called. This method will be called recursively until the base condition is reached. You can see it has two arguments n1/n2 and n2. n1/n2 = 27/3 = 9 So this becomes is_power(9,3)

The base cases are checked. Now this else statement is again executed  return is_divisible(n1, n2) and is_power(n1/n2, n2) as none of the above base cases is evaluated to true. when is_divisible() returns True as 9 is completely divisible by 3 i.e. 9%3 =0 and is_power returns (9/3,3) which is (3,3). So this becomes is_power(3,3)

Now as value of n1 becomes 3 and value of n2 becomes 3. So the first base case elif n1 == n2: condition now evaluates to true as 3=3. So it returns True. Hence the result of this statement print("is_power(10, 2) returns: ", is_power(10, 2))  is:                                                                                                

is_power(27, 3) returns:  True

Following are the program to the given question:

Program Explanation:

Program:

def is_divisible(a, b):#defining a method is_divisible that takes two parameters

   return a % b == 0#using return keyword that modulas parameter value and checks its value equal to 0

def is_power(a, b):#defining a method is_power that takes two parameters  

   if a == 1:#defining if block that checks a value equal to 1  or check odd number condition

       return True#return value True

   if b ==  1:#defining if block that checks b value equal to 1  or check odd number condition

       return False#return value False

   if not is_divisible(a, b):#defining if block that check method is_divisible value

       return False##return value False

   return is_power(a/b, b)#using return keyword calls and return is_power method                          

print("is_power(10, 2) returns: ", is_power(10, 2))#using print method that calls is_power which accepts two parameter

print("is_power(27, 3) returns: ", is_power(27, 3))#using print method that calls is_power which accepts two parameter

print("is_power(1, 1) returns: ", is_power(1, 1))#using print method that calls is_power which accepts two parameter

print("is_power(10, 1) returns: ", is_power(10, 1))#using print method that calls is_power which accepts two parameter

print("is_power(3, 3) returns: ", is_power(3, 3))#using print method that calls is_power which accepts two parameter

Output:

Please find the attached file.

Learn more:

brainly.com/question/24432065

Answer:

while True:

   number1 = input("Enter the number1: ")

   number2 = input("Enter the number2: ")

   if number1.isnumeric() and number2.isnumeric():

       break

number1 = int(number1)

number2 = int(number2)

print("The sum is: " + str(number1 + number2))

Explanation:

*The code is in Python.

Create an infinite while loop. Inside the loop, Get the number1 and number2 from the user. Check if they are both numeric values. If they are, stop the loop (Otherwise, the program keeps asking for the numbers).

When the loop is done, convert the numbers to integer numbers

Calculate and print their sum

Answer:

I am writing a C program.

#include <stdio.h> // for using input output functions

#include <stdbool.h> // for using a bool value as data type

int main() { // start of the main() function body

 int count=0; //count the number of entries

 double gallons, miles, MilesperGallon, combined_avg, sum; //declare variables

 while(true) {// takes input gallons and miles value from user and computes avg miles per gallon

      printf( "Enter the gallons used (-1 to stop): \n" ); //prompts user to enter value of gallons or enter -1 to stop

      scanf( "%lf", &gallons );//reads the value of gallons from user

   if ( gallons == -1 ) {// if user enters -1

     combined_avg = sum / count; //displays the combined average by dividing total of miles per drives to no of entries

     printf( "Combined miles per gallon for all tankfuls:  %lf\n", combined_avg ); //displays overall average value  

     break;} //ends the loop

     printf( "Enter the miles driven: \n" ); //if user does not enter -1 then prompts the user to enter value of miles

     scanf( "%lf", &miles ); //read the value of miles from user

MilesperGallon = miles / gallons; //compute the miles per gallon

printf( "The miles per gallon for tankful:  %lf\n", MilesperGallon ); //display the computed value of miles per gallon

  sum += MilesperGallon; //adds all the computed miles per gallons values

   count += 1;  } } //counts number of tankfuls (input entries)

Explanation:

The program takes as input the miles driven and gallons used for each tankful. These values are stored in miles and gallons variables. The program calculates and displays the miles per gallon MilesperGallon obtained for each tankful by dividing the miles driven with the gallons used. The while loop continues to execute until the user enters -1. After user enters -1, the program calculates and prints the combined miles per gallon obtained for all tankful. At the computation of MilesperGallon for each tankful, the value of MilesperGallon are added and stored in sum variable. The count variable works as a counter which is incremented to 1 after each entry. For example if user enters values for miles and gallons and the program displays MilesperGallon then at the end of this iteration the value of count is incremented to 1. This value of incremented for each tankful and then these values are added. The program's output is attached.

Answer:

def find_max(num_1, num_2):

   max_val = 0.0

   if (num_1 > num_2): # if num1 is greater than num2,

       max_val = num_1 # then num1 is the maxVal.

   else: # Otherwise,

       max_val = num_2 # num2 is the maxVal

   return max_val

max_sum = 0.0

num_a = float(input())

num_b = float(input())

num_y = float(input())

num_z = float(input())

max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)

print('max_sum is:', max_sum)

Explanation:

I added the missing part. Also, you forgot the put parentheses. I highlighted all.

To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)

Answer:

The summary including its given subject is mentioned in the portion here below explanatory.

Explanation:

Answer:

def word_count(words):

   return len(words.split(" "))

print(word_count("This is Python."))

Explanation:

*The code is in Python.

Create a function called word_count that takes one parameter, words

Split the words using split function, use " " for the delimiter, and return the length of this using len function.

Note that split function will give you a list of strings that are written with a space in the words. The len function will just give you the length of this list.

Call the word_count function with a string parameter and print the result.

Answer:

The output is seen bellow

Explanation:

public class BigInteger {

  public class Node{

      private char data;

      private Node next;

      public Node(char d){

          this.setData(d);

          this.setNext(null);

      }

      public char getData() {

          return data;

      }

      public void setData(char data) {

          this.data = data;

      }

      public Node getNext() {

          return next;

      }

      public void setNext(Node next) {

          this.next = next;

      }      

  }

  private Node root;

  public BigInteger(String s){

      this.root=null;

      for(int i=s.length()-1;i>=0;i--){

          Node node=new Node(s.charAt(i));

          node.setNext(root);

          root=node;

      }

  }

  public String toString(){

      String str="";

      Node cur=this.root;

      int i=0;

      while(cur!=null){

          if(i==0 && cur.getData()=='0'){

          }

          else{

              i=1;

              str+=cur.getData();

          }

          cur=cur.getNext();

      }

      return str;

  }

  public String reverse(String s){

      String ret="";      

      for(int i=s.length()-1;i>=0;i--){

          ret+=s.charAt(i);

      }      

      return ret;

  }

  public BigInteger add(BigInteger s){

      String f=this.reverse(this.toString());

      String l=this.reverse(s.toString());

      int max=(f.length()>l.length())?f.length():l.length();

      String ret="";

      int carry=0;

      for(int i=0;i<max;i++){

          int sum=carry;

          if(i<f.length()){

              sum+=(f.charAt(i)-'0');

          }

          if(i<l.length()){

              sum+=(l.charAt(i)-'0');

          }

          ret+=(char)(sum%10+'0');

          carry=(sum/10);

      }

      if(carry!=0){

          ret+='1';

      }

      BigInteger bi=new BigInteger(this.reverse(ret));

      return bi;

  }

  public boolean bigNum(String s1,String s2){

      if(s1.length()>s2.length()){

          return true;

      }

      else{

          if(s1.length()<s2.length()){

              return false;

          }

          else{

              for(int i=0;i<s1.length();i++){

                  if(s1.charAt(i)!=s2.charAt(i)){

                      if(s1.charAt(i)>s2.charAt(i)){

                          return true;

                      }

                      else{

                          return false;

                      }

                  }

              }                  

          }

      }

      return false;

  }

  public BigInteger subtraction(BigInteger s){

      String f=this.toString();

      String l=s.toString();

      boolean b=this.bigNum(f, l);

      if(b==false){

          String tmp=f;

          f=l;

          l=tmp;

      }

      f=this.reverse(f);

      l=this.reverse(l);

      int max=(f.length()>l.length())?f.length():l.length();

      String ret="";

      int borrow=0;

      for(int i=0;i<max;i++){

          int sum=borrow;

          if(i<f.length()){

              sum+=(f.charAt(i)-'0');

          }

          if(i<l.length()){

              sum-=(l.charAt(i)-'0');

          }

          if(sum<0){borrow=-1;sum=10+sum;}

          else{borrow=0;}

          ret+=(char)(sum%10+'0');

      }

      if(b==false){

          ret+="-";

      }

      BigInteger bi=new BigInteger(this.reverse(ret));

      return bi;

  }

  public BigInteger multiplication(BigInteger s){

      String f=this.toString();

      String l=s.toString();

      int len=l.length();

      BigInteger bi=new BigInteger("");

      for(int i=len-1;i>=0;i--){

          //System.out.println(l.charAt(i));

          BigInteger r=new BigInteger(f);

          for(int j=(l.charAt(i)-'0');j>1;j--){

              r=r.add(new BigInteger(f));

              //System.out.print(r+" " );

          }

          //System.out.println();

          bi=bi.add(r);

          f=f+"0";

      }

      return bi;

  }

  public BigInteger division(BigInteger s){

      BigInteger t=this;

      BigInteger bi=new BigInteger("");

      int i=0;

      t=t.subtraction(s);

      String str=t.toString();

      while(str.charAt(0)!='-' && i<40){

          //System.out.println(str+" "+(i+1));

          bi=bi.add(new BigInteger("1"));

          t=t.subtraction(s);

          str=t.toString();  

          i++;

      }

      return bi;    

  }  

}  

-------------------

import java.io.BufferedReader;

import java.io.FileNotFoundException;

import java.io.FileReader;

import java.io.FileWriter;

import java.io.IOException;

import java.io.PrintWriter;

import java.util.Scanner;

public class Driver {

  public static void main(String [] args){

      Scanner sc=new Scanner(System.in);

      String str1="";

      String str2="";

      System.out.print("Enter file name 1 :");

      String file1=sc.next();

      //String file1="datafile1.txt";

      BufferedReader reader1;

      try {

          reader1 = new BufferedReader(new FileReader(file1));

          while((str1=reader1.readLine())!=null){

              break;

          }

          reader1.close();

      } catch (FileNotFoundException e) {

          // TODO Auto-generated catch block

          e.printStackTrace();

      } catch (IOException e) {

          // TODO Auto-generated catch block

          e.printStackTrace();

      }

      System.out.print("Enter file name 2 :");

      String file2=sc.next();

      //String file2="datafile2.txt";

      BufferedReader reader2;

      try {

          reader2 = new BufferedReader(new FileReader(file2));

          while((str2=reader2.readLine())!=null){

              break;

          }

          reader2.close();

      } catch (FileNotFoundException e) {

          // TODO Auto-generated catch block

          e.printStackTrace();

      } catch (IOException e) {

          // TODO Auto-generated catch block

          e.printStackTrace();

      }

Answer:

Chart styles

Explanation:

In PowerPoint, once you have inserted a chart (e.g a bar or pie chart), the format tab would be enabled so that you can format your chart to your taste. In this format tab, there are a bunch of tab groups such as:

(i) Chart styles: This allows you to typically change the component background colors of your chart.

(ii) Data: This contains sub-tabs that allow you to edit and select the data on your chart.

(iii) Type: This contains sub-tabs such as "Change Chart Type" which will allow you to change the type of the chart. For example if you were using a pie chart and you want to change it to a bar chart or any other chart, you can do that here.

(iv) Chart Layouts: This allows to change or modify how the selected chart is laid out.

Answer:

Can be executed on multiple platforms.

Explanation:

A program is portable if it is not platform dependent. In other words, if the program is not tightly coupled to a particular platform, then it is said to be portable. Portable programs can run on multiple platforms without having to do much work. By platform, we essentially mean the operating system and hardware configuration of the machine's CPU.

Examples of portable programs are those written in;

i. Java

ii. C++

iii. C

Answer:

C

Explanation:

A social media specialist does not usually install computer hardware. They will do things like look at engagement data, create new posts to send out to followers, and consult on how to get more people to come into your business, but they wouldn't be doing things like installing new ram.

Answer:

c. a key found on Windows-oriented computer keyboards.

Explanation:

Hope it helps.

Answer:

System.gc()

Explanation:

System.gc() can be defined as the method which can be used to effectively request for garbage collection because they runs the garbage collector, which in turn enables JMV which is fully known as JAVA VIRTUAL MACHINE to claim back the already unused memory space of the objects that was discarded for quick reuse of the memory space , although Java virtual machine often perform garbage collection automatically.

Answer:

The number of execution can't always be determines

Explanation:

The following points should be noted

Variable d relies on variable x (both of double data type) for its value

d is calculated as

[tex]d = \sqrt{x}^2 - x[/tex]

Mere looking at the above expression, the value of d should be 0;

However, it doesn't work that way.

The variable x can assume two categories of values

The range of the above values depend on the system running  the application;

When variable x assumes a small value,

[tex]d = \sqrt{x}^2 - x[/tex] will definitely result in 0 and the loop will terminate immediately because [tex]\sqrt{x}^2 = x[/tex]

When variable x assumes a large value,

[tex]d = \sqrt{x}^2 - x[/tex] will not result in 0  because their will be [tex]\sqrt{x}^2 \neq x[/tex]

The reason for this that, the compiler will approximate the value of [tex]\sqrt{x}^2[/tex] and this approximation will not be equal to [tex]x[/tex]

Hence, the loop will be executed again.

Since, the range of values variable x can assume can not be predetermined, then we can conclude that the number of times the loop will be executed can't be determined.

Answer:

Explanation:

The system will be deadlock free if the below two conditions holds :

Proof below:

Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.

Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.

Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.

Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.

Answer and Explanation:

Computer related errors associated with people that use these computers are errors that are made by human beings in instructing(or programming) the computer to do or complete a certain task. This could involve such things as moving paper based data to the computer(electronic data) where the human could enter incorrect data or inconsistent programs or code.

Computer related errors which directly have to do with the computer consist of error thrown by the system as a result of ineffective code or software related issues or faulty hardware. In either case all computer related errors are inherently caused by humans since a computer would only take instruction given to it.

Answer:

The networks are intelligent and send packets at the highest speed and most efficiently.

Explanation:

Alcatel-Lucent was founded in 1919, it was a French global telecommunications equipment manufacturing company with its headquarter in Paris, France. Alcatel-Lucent provide services such as telecommunications and hybrid networking solutions deployed both in the cloud and on properties.

Alcatel-Lucent’s High Leverage Network (HLN) increases bandwidth and network capabilities while reducing the negative impact on the environment. This high leverage network can handle large amounts of traffic more efficiently because the networks are intelligent and send packets at the highest speed and most efficiently. HLN are intelligent such that it delivers increased bandwidth using fewer devices and energy.

Generally, when the High Leverage Network (HLN) is successfully implemented, it helps telecommunications companies to improve their maintenance costs, operational efficiency, enhance network performance and capacity to meet the bandwidth demands of their end users.

Answer:

All team members

Explanation:

In respect of the question, the or those responsible for tracking the tasks in an agile team comprises of all the team members.

Agile in relation to task or project management, can be refer to an act of of division of project or breaking down of project or tasks into smaller unit. In my opinion, these is carried out so that all team members can be duly involved in the tasks or project.

Answer:

Following are the code to this question:

def average_heart_rate(beats):#defining a method average_heart_rate that accepts list beats  

   total=0 #defining integer variable total,that adds list values  

   count_list=0#defining a count_list integer variable, that counts list numbers  

   for i in beats:#defining for loop to add list values  

       if i>= 100:#defining if block to check value is greater then 100

           total += i#add list values

           count_list += 1# count list number  

   return total//count_list #return average_heart_rate value  

beats=[72,77,79,95,102,105,112,115,120,121,121,125, 125, 123, 119, 115, 105, 101, 96, 92, 90, 85]#defining a list

print("The average heart rate value:",average_heart_rate(beats)) # call the mnethod by passing value

Output:

The average heart rate value: 114

Explanation:

In the given question some data is missing so, program description can be defined as follows:

Answer:

I am writing a C++ program using loops instead of nested if statements.

#include <iostream> // to use input output functions

using namespace std; // to identify  objects like cin cout

void cells(int cells[],int days){ /* function that takes takes one array of integers cells, one integer days representing the number of days to simulate. */

int pos ,num=0; //declares variables pos for position of two adjacent cells and num to iterate for each day

int result[9]; //the updated output array

while (num< days)   { //this loop keeps executing till the value of num is less than the value of number of days

num++;

for(pos=1;pos<9;pos++)   //this loop has a pos variable that works like an index and moves through the cell array

    result[pos]=(cells[pos-1])^ (cells[pos+1]); //updated cell state determined by the previous and next cells (adjacent cells) by bitwise XOR operations

     for(pos=1;pos<9;pos++) //iterates through the array

         cells[pos]=result[pos];    } //the updated cells state is assigned to the cell array simultaneously

   for(pos=1;pos<9;pos++) //iterates through the array and prints the resultant array that contains the updated active and inactive cells values

        cout << result[pos]; }

int main() { //start of the main function body

int j,day;

int output[9];

*/the two cells on the ends (first and last positions of array) have single adjacent cell, so the other adjacent cell can be assumed to be always inactive i.e. 0 */

output[0]=output[9]=0;

for(j=1;j<9;j++) //takes the input array from user

cin >> output[j];

cin >> day;

cells(output,day); } //calls the function cells to print the array with active and inactive cells states.

Explanation:

The program is well explained in the comments mentioned with every statement of the program. I will explain with the help of example:

Suppose the user enters the array = [1,0,0,0,0,1,0,0] and days=1

The while loop checks if the value of num is less than that of days. Here num is 0 and days is 1 So this means that the body of while loop will execute.

In the body of while loop the value of num is incremented by 1. The first loop initializes a variable pos for position of adjacent cells. The statement is a recursive statement result[pos]=(cells[pos-1])^ (cells[pos+1]) that uses previous state for updating the state of other cells. The “^” is the symbol used for bitwise exclusive operator. In bitwise XOR operations the two numbers are taken as operands and XOR is performed on every bit of two numbers. The result of XOR is 1 if the two bits are not same otherwise 0. For example XOR of 1^0 and 0^1 is 1 and the XOR of 0^0 and 1^1 is 0. The second for loop update the cell information of all cells simultaneously. The last loop prints the updated cells states.

The main function takes the input array element from user and the value for the days and calls the cells function to compute the print the active and inactive cells state information.

The screenshot of the program along with its output are attached.

Answer:

Following are the code to this question:

x = int(input())#defining a variable x for user input value  

if(x>=20 and x<=98):#defining an if block that checks value is in between 20 to 98  

   while(not(x%10==x//10)):#defining while loop that seprate numbers and checks it is not equal  

       print(x)#print value

       x-=1# decrease value by subtracting 1

   print(x)# print value

else:#defining else block  

   print("The input value must lie in 20-98")#print message

Output:

36

36

35

34

33

Explanation:

The program that takes in an integer in the range 20-98 as input. The output is a countdown starting from the integer, and stopping when both output digits are identical is as follows:

x = int(input("your number should be between 20 to 98: "))

while 20 <= x <= 98:

   print(x, end='\n')

   if x % 11 == 0:

       break

   x -= 1

else:

   print("your number must be between 20 to 98")

The code is written in python

learn more on python code here: https://brainly.com/question/26104476