Write a program that reads students’ names followed by their test scores. The program should output each student’s name followed by the test scores and the relevant grade. It should also find and print the highest test score and the name of the students having the highest test score. Student data should be stored in a struct variable of type studentType, which has four components: studentFName and studentLName of type string, testScore of type int (testScore is between 0 and 100), and grade of type char. Suppose that the class has 20 students. Use an array of 20 components of type studentType. Your program must contain at least the following functions: A function to read the students’ data into the array. A function to assign the relevant grade to each student. A function to find the highest test score. A function to print the names of the students having the highest test score. Your program must output each student’s name in this form: last name followed by a comma, followed by a space, followed by the first name; the name must be left justified. Moreover, other than declaring the variables and opening the input and output files, the function main should only be a collection of function calls.

Answer:

See explaination

Explanation:

Usable sequence numbers = 2^31

Transmission speed = 2^30 bps

(a)Wrap around time = (2^31 * 8 * 1)/(2^30) = 16 seconds

(b)segment size = 16 Bytes

Wrap around time = (2^31 * 8 * 16)/(2^30) = 256 seconds

(c)if we use large segment size ,which increases size of MTU in data link layer .Maximum transmission unit is the maximum size of a packet or frame that can flow across the network, without being fragmented,which lead to overhead of fragmentation at network layer.

(d)Maximum network throughput equals the TCP window size divided by the round-trip time of communications data packets.

Round trip time = 2^9 ms

window size = 2^16

Throughput = (2^16)/(2^9) = 2^7 KBps = 128 KBps = 1Mbps

Answer:

See explaination

Explanation:

If the computer processor can find the data it needs for its next operation in cache memory, it will save time compared to having to get it from random access memory. L1 is "level-1" cache memory, usually built onto the microprocessor chip itself.

The level 1 cache is a memory cache which is built directly into the microprocessor, which is used for storing the microprocessor's recently accessed information, thus it is also called the primary cache.

See the attached file for detailed and step by step solution suitable for the given problem.

See attachment.

Answer:

1 and 3

Explanation:

The corrected Python code defines a function to return the second word of a sentence, removing specified punctuation, and handles issues for accurate results.

The provided Python code defines a function `return_clean_second_word` that takes a string argument `sentence`. The function aims to return the second word of the sentence, removing certain punctuation marks if present. If the sentence contains less than two words, it returns `None`. However, there are some issues in the code that need correction.

Here is the corrected code:

```python

def return_clean_second_word(sentence):

   l = list(sentence.split())

   if len(l) < 2:

       return None

   else:

       word = l[1]

       char = [',', ';', '!', '?']

       if word != '' and word[-1] in char:

           word = word[:-1]

       return word

print(return_clean_second_word(input("Enter a sentence: ")))

```

Explanation:

1. Removed extra spaces in `split(" ")` to `split()` for accurate word separation.

2. Checked if `word` is not an empty string (`word != ''`) before removing the last character.

3. Used `[:-1]` to remove the last character of the word.

The function now correctly returns the second word of the input sentence, considering the specified conditions.

In summary, the corrected Python code defines a function to return the cleaned second word of a sentence, addressing issues related to word separation and punctuation removal.

Answer: (a). safe    (b). not safe    (c). safe

Explanation:

quite a long way to go about this but it is straightforward nevertheless.

from the question we are given that the values;

(a).

Process             Max             Hold                    Need(Max-Hold)

P1                      70                  45                       25

P2                     60                  40                       20

P3                     60                   15                       45

P4                     60                   25                       35

#

we have the total units of memory to be 150

Our available becomes = 150 - (45+40+15+25) = 25

allocation for P1:

Process P2  has available resources to run = 70 - 60 -20 = 110 units

Process P1  has available resources to run = 70 units

Process P4 has available resources to run = 125 + 60 - 35 = 150 units

Process P3  has available resources to run = 110 + 60 - 45 = 125 units

from this regard we can conclude that it is Safe to grant the request.

(b). we are to add  a fourth process with max memory requirement of 60 and an initial need of 35 units.

Process             Max             Hold                    Need(Max-Hold)

P1                      70                  45                       25

P2                     60                  40                       20

P3                     60                   15                       45

P4                     60                   35                       25

Given our Total units = 150

We have our available units as 150 - (45 + 40 + 15 + 35) =  15

P2 is 20 but available in 15, Thus Minimum need for  a second process

This means that it is Not Safe to grant the request

(c). A fourth process arrives, with a maximum memory need of 50 and an initial need of 30 units

Process             Max             Hold                    Need(Max-Hold)

P1                      70                  45                       25

P2                     60                  40                       20

P3                     60                   15                       45

P4                     50                   30                       20

Given our Total units = 150

We have our available units as 150 - (45 + 40 + 15 + 30) =  20

We can see here that P4 requires 30 units, and thus it is allocated 30 units

From this we can simply state that'

It is Safe to grant request because 20 units will be left after the allocation for P4, so there is sufficient memory to guarantee the elimination of P2

Cheers i hope this helps!!!!

Following are the response to the given points:

Learn more:

brainly.com/question/24097231

Answer:

B

Explanation:

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to the see the step by step explanation to the question above.

Final answer:

Both FIFO and Round Robin scheduling disciplines could allow the schedule provided to occur, with FIFO executing jobs in the order they arrive and Round Robin allocating equal fixed time units for each job. STCF is less likely without additional context.

Explanation:

The scenario provided involves three jobs (A, B, and C) and a single CPU managing these jobs with no IO (input/output) operations. To determine which scheduling discipline could allow this sequence of operations, let's examine each possibility provided:

In summary, both FIFO and Round Robin could allow for the described schedule to occur, considering the context given. STCF is less likely to be applicable without additional context indicating that jobs B and A's total execution times were known and specifically scheduled in the given order for a reason other than their completion times.

Answer:

The three major problem, which arise there are as follows:

Internet issues, server issues, and server down.

Explanation:

All the issues can be described as follows:

Final answer:

DNS issues could range from typos, an outdated or corrupted DNS cache, problems with the local DNS server, or issues with the host itself, each with varying likelihoods.

Explanation:

When your computer sends a DNS request and receives a response that the host name does not exist despite it being a commonly used host, several problems could be at play. One potential issue is a typo in the host name entered which is highly unlikely if the host is frequently visited and the URL was autofilled. Another issue could be that the DNS cache is outdated or corrupted; this is more common and clearing the cache could resolve this problem. Additionally, the local DNS server might be experiencing problems, such as being overwhelmed with requests or having a configuration issue which could be a moderate likelihood. Finally, the host itself may be experiencing issues, such as having moved to a new server or the domain name having expired, though this is less likely if no prior indication of such a change was given.

Answer:

Please see the attached image

Explanation:

Permission marketing is the practice of providing incentives for customers to learn more about a service, which involves the consent of the consumer before sending information, distinguishing it from advertising, SEO, and branding.

Providing incentives for customers to learn more about your service is known as C) permission marketing. This approach involves getting the consent of the consumer before sending them more information about your products or services. While some may confuse this with branding, advertising, or search engine optimization, each of these has distinct functions within the realm of marketing strategies.

Advertising is a tool used to increase the chance that a firm's product or service is considered by consumers and involves creating awareness and desire amongst consumers for new and existing products. Marketing also involves fostering organic word-of-mouth from consumers themselves as opposed to formal advertising alone. Search engine optimization (SEO), on the other hand, focuses on enhancing the visibility of a website in search engine results.

Answer:

Check the explanation

Explanation:

#include<iostream.h>

#include<algorithm.h>

#include<climits.h>

#include<bits/stdc++.h>

#include<cstring.h>

using namespace std;

int partition(int arr[], int l, int r, int k);

int kthSmallest(int arr[], int l, int r, int k);

void quickSort(int arr[], int l, int h)

{

if (l < h)

{

// Find size of current subarray

int n = h-l+1;

// Find median of arr[].

int med = kthSmallest(arr, l, h, n/2);

// Partition the array around median

int p = partition(arr, l, h, med);

// Recur for left and right of partition

quickSort(arr, l, p - 1);

quickSort(arr, p + 1, h);

}

int findMedian(int arr[], int n)

{

sort(arr, arr+n); // Sort the array

return arr[n/2]; // Return middle element

}

int kthSmallest(int arr[], int l, int r, int k)

{

// If k is smaller than number of elements in array

if (k > 0 && k <= r - l + 1)

{

int n = r-l+1; // Number of elements in arr[l..r]

// Divide arr[] in groups of size 5, calculate median

// of every group and store it in median[] array.

int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;

for (i=0; i<n/5; i++)

median[i] = findMedian(arr+l+i*5, 5);

if (i*5 < n) //For last group with less than 5 elements

{

median[i] = findMedian(arr+l+i*5, n%5);

i++;

}

int medOfMed = (i == 1)? median[i-1]:

kthSmallest(median, 0, i-1, i/2);

int pos = partition(arr, l, r, medOfMed);

if (pos-l == k-1)

return arr[pos];

if (pos-l > k-1) // If position is more, recur for left

return kthSmallest(arr, l, pos-1, k);

return kthSmallest(arr, pos+1, r, k-pos+l-1);

}

return INT_MAX;

}

void swap(int *a, int *b)

{

int temp = *a;

*a = *b;

*b = temp;

}

int partition(int arr[], int l, int r, int x)

{

// Search for x in arr[l..r] and move it to end

int i;

for (i=l; i<r; i++)

if (arr[i] == x)

break;

swap(&arr[i], &arr[r]);

// Standard partition algorithm

i = l;

for (int j = l; j <= r - 1; j++)

{

if (arr[j] <= x)

{

swap(&arr[i], &arr[j]);

i++;

}

}

swap(&arr[i], &arr[r]);

return i;

}

/* Function to print an array */

void printArray(int arr[], int size)

{

int i;

for (i=0; i < size; i++)

cout << arr[i] << " ";

cout << endl;

}

// Driver program to test above functions

int main()

{

float a;

clock_t time_req;

int arr[] = {1000, 10, 7, 8, 9, 30, 900, 1, 5, 6, 20};

int n = sizeof(arr)/sizeof(arr[0]);

quickSort(arr, 0, n-1);

cout << "Sorted array is\n";

printArray(arr, n);

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(i*i*i*i);

}

time_req = clock()- time_req;

cout << "Processor time taken for multiplication: "

<< (float)time_req/CLOCKS_PER_SEC << " seconds" << endl;

// Using pow function

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(pow(i, 4));

}

time_req = clock() - time_req;

cout << "Processor time taken in pow function: "

<< (float)time_req/CLOCKS_PER_S

return 0;

}

..................................................................................................................................................................................................................................................................................................................................

OR

.......................

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

}

}

}

// Insertion sort

void insertionSort(long int arr[], long int n)

{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

}

}

// Selection sort

void selectionSort(long int arr[], long int n)

{

long int i, j, midx;

for (i = 0; i < n - 1; i++) {

// Find the minimum element in unsorted array

midx = i;

for (j = i + 1; j < n; j++)

if (arr[j] < arr[min_idx])

midx = j;

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

// increases the size of array by 10000

n += 10000;

}

return 0;

}

Answer:

See explaination

Explanation:

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

}

}

}

// Insertion sort

void insertionSort(long int arr[], long int n)

{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

}

}

// Selection sort

void selectionSort(long int arr[], long int n)

{

long int i, j, midx;

for (i = 0; i < n - 1; i++) {

// Find the minimum element in unsorted array

midx = i;

for (j = i + 1; j < n; j++)

if (arr[j] < arr[min_idx])

midx = j;

// Swap the found minimum element

// with the first element

swap(&arr[midx], &arr[i]);

}

}

// Driver code

int main()

{

long int n = 10000;

int it = 0;

// Arrays to store time duration

// of sorting algorithms

double tim1[10], tim2[10], tim3[10];

printf("A_size, Bubble, Insertion, Selection\n");

// Performs 10 iterations

while (it++ < 10) {

long int a[n], b[n], c[n];

// generating n random numbers

// storing them in arrays a, b, c

for (int i = 0; i < n; i++) {

long int no = rand() % n + 1;

a[i] = no;

b[i] = no;

c[i] = no;

}

// using clock_t to store time

clock_t start, end;

// Bubble sort

start = clock();

bubbleSort(a, n);

end = clock();

tim1[it] = ((double)(end - start));

// Insertion sort

start = clock();

insertionSort(b, n);

end = clock();

tim2[it] = ((double)(end - start));

// Selection sort

start = clock();

selectionSort(c, n);

end = clock();

tim3[it] = ((double)(end - start));

// type conversion to long int

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

// increases the size of array by 10000

n += 10000;

}

return 0;

}

Answer:

The loop in cpp language is as follows.

for(int n=10; n<=90; n++)

{

cout<<n<<" ";

count = count+1;

if(count==10)

{

cout<<""<<endl;

count=0;

}

}

Explanation:

1. The expected output is obtained using for loop.

2. The loop variable, n is initialized to the starting value of 10.

3. The variable n is incremented by 1 at a time, until n reaches the final value of 90.

4. Every value of n beginning from 10 is displayed folllowed by a space.

5. When 10 numbers are printed, a new line is inserted so that every line shows only 10 numbers.

6. This is achieved by using an integer variable, count.

7. The variable count is initialized to 0.

8. After every number is displayed, count is incremented by 1.

9. When the value of count reaches 10, a new line is inserted and the variable count is initialized to 0 again. This is done inside if statement.

10. The program using the above loop is shown below along with the output.

PROGRAM

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   int count=0;

   for(int n=10; n<=90; n++)

   {

       cout<<n<<" ";

       count = count+1;

       if(count==10)

       {

           cout<<""<<endl;

           count=0;

       }

   }

   return 0;

}  

OUTPUT

10 11 12 13 14 15 16 17 18 19  

20 21 22 23 24 25 26 27 28 29  

30 31 32 33 34 35 36 37 38 39  

40 41 42 43 44 45 46 47 48 49  

50 51 52 53 54 55 56 57 58 59  

60 61 62 63 64 65 66 67 68 69  

70 71 72 73 74 75 76 77 78 79  

80 81 82 83 84 85 86 87 88 89  

90

1. In the above program, the variable count is initialized inside main() but outside for loop.

2. The program ends with a return statement since main() has return type of integer.

3. In cpp language, use of class is not mandatory.

Answer:

see explaination

Explanation:

Given that we have;

Defects before testing = Defects planted = 5

Defects after testing = 20

Seeded defects found = 2

Real, non seeded defects = 18

Hence, Total number of defects = (defects planted / seeded defects found) * real, non seeded defects = (5/2)*18 = 45

Therefore, Estimated number of real defects still present = estimated total number of defects - number of real, non seeded defects found = 45-18 = 27

1. `Animal` is the superclass, and `Cat` is the subclass, signifying inheritance.

2. Private fields in `Animal` can't be directly accessed in `Cat`.

3. Created a `Cat` constructor and updated the test program.

4. Added a `toString` method for `Cat` and updated the test program.

5. Implemented an `isSleeping` method for `Cat` and updated the test program.

(Given in the attachment)

The question probable maybe:

PROGRAM

Consider the Animal and Cat classes in the following Java source code files. SAVE ALL 3.

/*

*Animal.java

*

*A simple class representing an animal.

*/

public class Animal {

private String name;

private int numLegs:

public Animal (String name, int numLegs) {

            this.name name;

             this.numlegs =  numLegs;

}

public String getName() {

             return this.name;

}

public int getNumLegs()   {

              return this.numLegs;

}

public boolean isSleeping(int hour, int minute) (

              if (hour> 24 || hour <0 || minute> 60 || minute < 0) {

                 throw new IllegalArgumentException("invalid time

specified");

}

              return (hour > 22 || hour <= 5)

}

public String toString() {

               return this.name;

}

public static void printAnimalName (Animal an) (

           // Output the name of the animal
               System.out.println( an );

      }

}

/*

*Cat.java

*

*A simple class represting a cat.

*/

public class Cat extends Animal {

        private boolean isShortHaired;

        public Cat() {

                    super ("Mycatt", 3);

         }

        public boolean isShortHaired() {

                     return this.isShorthaired;

          }

         public boolean isExtroverted() {

                    return false;

          }

}

/*

*testDriver.java

*

* A simple test program for the Animal class

*

*/

public class testDriver {

           public static void main (String [] argv) {

            }

}

QUESTIONS / TASKS •
1. Which class is the superclass? Which is the subclass? What does it mean that the Cat class extends the Animal class?

2. The Cat class cannot directly access the fields it inherits from Animal. Why not?

3. The subclass constructor typically calls the superclass constructor to initialize the inherited fields. Write a constructor for the Cat class above. It should take as parameters the cat’s name and a boolean indicating whether it is short-haired, and it should call the superclass constructor to initialize the inherited fields.

Update the test program to create an instance of class Cat.

Cat c = new Cat("Kitty", false);

4. To manipulate the inherited fields, the subclass can use the inherited accessor and mutator methods. Write a toString method for the Cat class above. It should return a string consisting of the cat’s name followed by either " (short-haired)" or " (long-haired)".

Update the test program to test your method.

System.out.println( c );

5. The subclass can override an inherited method, replacing it with a version that is more appropriate. Write an isSleeping method for the Cat class. It should reflect the fact that cats seem to sleep all of the time!

Update the test program to test your method.    

Answer:

header.h->function bodies and header files.

#include<iostream>

#include<cstdlib>

#include<ctime>

using namespace std;

/* Partitioning the array on the basis of piv value. */

int Partition(int a[], int bot, int top,string opt)

{

int piv, ind=bot, i, swp;

/*Finding the piv value according to string opt*/

if(opt=="Type1 sort" || opt=="Type3 sort")

{

piv=(top+bot)/2;

}

else if(opt=="Type2 sort" || opt=="Type4 sort")

{

piv=(top+bot)/2;

if((a[top]>=a[piv] && a[top]<=a[bot]) || (a[top]>=a[bot] && a[top]<=a[piv]))

piv=top;

else if((a[bot]>=a[piv] && a[bot]<=a[top]) || (a[bot]>=a[top] && a[bot]<=a[piv]))

piv=bot;

}

swp=a[piv];

a[piv]=a[top];

a[top]=swp;

piv=top;

/*Getting ind of the piv.*/

for(i=bot; i < top; i++)

{

if(a[i] < a[piv])

{

swp=a[i];

a[i]=a[ind];

a[ind]=swp;

ind++;

}

}

swp=a[piv];

a[piv]=a[ind];

a[ind]=swp;

return ind;

}

void QuickSort(int a[], int bot, int top, string opt)

{

int pindex;

if((opt=="Type3 sort" || opt=="Type4 sort") && top-bot<19)

{

/*then insertion sort*/

int swp,ind;

for(int i=bot+1;i<=top;i++){

swp=a[i];

ind=i;

for(int j=i-1;j>=bot;j--){

if(swp<a[j]){

a[j+1]=a[j];

ind=j;

}

else

break;

}

a[ind]=swp;

}

}

else if(bot < top)

{

/* Partitioning the array*/

pindex =Partition(a, bot, top,opt);

/* Recursively implementing QuickSort.*/

QuickSort(a, bot, pindex-1,opt);

QuickSort(a, pindex+1, top,opt);

}

return ;

}

main.cpp->main driver file

#include "header.h"

int main()

{

int n=100000, i;

/*creating randomized array of 100000 numbers between 0 and 100001*/

int arr[n];

int b[n];

for(i = 0; i < n; i++)

arr[i]=(rand() % 100000) + 1;

clock_t t1,t2;

t1=clock();

QuickSort(arr, 0, n-1,"Type1 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type2 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type3 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time and insertion sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type4 sort");

t2=clock();

cout<<"Quick sort time and insertion sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

return 0;

}

Explanation:

Change the value of n in the main file for different array size. Output is in the same format as mentioned, time is shown in milliseconds.

Answer:

The program can be found in the explanation part while the output is attached as files.

Explanation:

The program to display the birth month and year, is given below:

//Include the necessary library files.

#include "stdafx.h"

#include <iostream>

using namespace std;

//Start the main() function.

int main()

{

//Declare the necessary variables.

int birthMonth;

int birthYear;

//Prompt the user to enter the birth month.

cout << "Enter the birth month: ";

//Store the input.

cin >> birthMonth;

//Prompt the user to enter the birth year.

cout << "Enter the birth year: ";

//Store the input.

cin >> birthYear;

//Display the output.

cout << "The birth date is: " << birthMonth << "/" << birthYear << endl;

system("pause");

return 0;

}

The program is a sequential program, and does not require loops or conditional statements

The program in C, where comments are used to explain each line is as follows:

#include <stdio.h>

int main() {  

   //This declares the variables

   int year, month;

   //This gets input for month

   scanf("%d", &month);

   //This gets input for year

   scanf("%d", &year);

   // This prints the inputs, separated by slash

   printf("%d/%d", month, year);

   return 0;

}

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Answer:

See explaination

Explanation:

Script by switch statement

random_force = randi(12) switch(random_force) case 0 disp("There is no wind") case {1,2,3,4,5,6} disp("there is a breeze") case {7,8,9} disp("there is a gale") case {10,11} disp("it is a storm") case 12 disp("Hello Hurricane!") end

First it generates a random force between 1 and 12

Then the switch statement is executed according to the force

in case values of corresponding force are written and their string will be printed according to the case which depends upon the value of random_force

Multiple case values are used because we need multiple values to print the same string

So values {1,2,3,4,5,6} , {7,8,9},{10,11} are combined into one case becuse they all have same value

Script by using nested if-else

random_force = randi(12) //It will genrate a random force value between 1 and 12 which can be used further if random_force == 0 disp("There is no wind") elseif (random_force > 0) && (random_force < 7) disp("there is a breeze") elseif (random_force > 6) && (random_force < 10) disp("there is a gale") elseif (random_force > 9) && (random_force < 12) disp("it is a storm") elseif random_force == 12 disp("Hello Hurricane!") end

First if will check if force is zero and will print its corresponding string

Second,elseif will check if force is between 1 and 6 and will print its corresponding string

third elseif will check if force is between 7 and 9 and will print its corresponding string

Fourth elseif will check if force is 10 or 11 and will print its corresponding string

Last else if will compare value to 12 and return the corresponding string

Explanation:

See attached images

The transition from single-entity and multi-enterprise MPIs to an eMPI focuses on incorporating IPEC's core competencies for interprofessional collaborative practice, necessitating changes in data management, privacy, and interoperability at existing facilities.

The question pertains to researching the recommended core elements of a single-entity Master Patient Index (MPI) and a multi-enterprise MPI, and then discussing the emerging core elements of a new electronic MPI (eMPI). The core elements of an MPI often include the patient's name, date of birth, gender, and a unique identifier. In contrast, an eMPI for interprofessional collaborative practice, as described in the Interprofessional Education Collaborative (IPEC)'s 2017 update, emphasizes core competencies for effective interprofessional collaborative practice. These competencies include values/ethics for interprofessional practice, roles/responsibilities, interprofessional communication, and teams and teamwork. Transitioning to an eMPI implies adjustments in existing facilities to incorporate these competencies, focusing on enhanced interprofessional communication and teamwork for patient-centered care.

Moreover, the transition to an eMPI from traditional MPI systems necessitates changes in data management, privacy protocols, and interoperability standards. Facilities must ensure their systems can securely share and access patient information across different healthcare settings, maintaining data integrity and confidentiality.

Answer:

Four

Explanation:

The domain name system is a naming database that maps the name people use to locate a website to the IP address that is then used by a computer to locate a website. It is connected to the Internet or a private network.

An IP address is a 32-bit binary number, but it's normally written out as 4 octets in decimal form as it is easier for humans to read.

Based on the information given, the number of octets for an Internet Protocol (IP) address is four.

It should be noted that a domain name system simply means the naming database that's useful in mapping the name that people use to locate a website to the IP address

Also, An IP address is a 32-bit binary number, but it's normally written out as 4 octets. Therefore, the number of Internet Protocol (IP) address that would have to be memorized is four.

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Answer:

See explaination

Explanation:

Monopoly is a competition where there is a single seller and a single product in the market and hence no competition.

Monopolistic Competition is a competition where there are a large number of small firms competing with each other.

Oligopoly is a market that has a limited number of buyers and Sellers.

Perfect Competition is a market with large number of buyers and Sellers.

The pairing can be seen below:

1. Monitors is under monopolistic competition as there are a large number of firms competing.

2. USB drives this comes under perfect competition as there a large number of buyersand Sellers.

3. Central processing units (CPUs) this is under oligopoly as there a limited number of buyers and Sellers.

4. Microsoft's Windows this fall under monopoly market structure. They have no competition at all.

The function so that the main() code below can be replaced by the simpler code is given below.

We are given that;

The functions

Now,

The main() code can then be simplified as:

int main() {

  double milesPerHour;

  double minutesTraveled;

  double milesTraveled;

  cin >> milesPerHour;

  cin >> minutesTraveled;

  milesTraveled = MphAndMinutesToMiles(milesPerHour, minutesTraveled);

  cout << "Miles: " << milesTraveled << endl;

  return 0;

}

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Final answer:

The student needs to write a function that converts speed from miles per hour and minutes traveled into miles, which involves programming and applying unit conversions. The MphAndMinutesToMiles() function calculates hours from minutes and then computes distance by multiplying with the provided speed.

Explanation:

The question involves creating a function to facilitate conversion of speed from miles per hour (mph) and travel time in minutes to the distance traveled in miles. This relates primarily to the field of Computers and Technology, particularly in the area of programming and software development, where such a function can be implemented in a variety of programming languages. The student is in college, as the task involves writing a function, which is a concept usually introduced at that level. The task requires understanding of unit conversions and how to apply them within a program.

Write the Function for MphAndMinutesToMiles:

To convert the main code to use the function MphAndMinutesToMiles(), begin by considering the basic formula for distance, which is speed multiplied by time. Since time is given in minutes, it needs to be converted to hours to match the units of miles per hour. This can be done by dividing the minutes by 60. Once the time is in hours, multiplying by the miles per hour speed will give the distance in miles.

The function might look something like:

  double MphAndMinutesToMiles(double mph, double minutes) {
      double hours = minutes / 60.0;
      return mph * hours;
  }

In the main program, you could then get the miles traveled by just calling:

  milesTraveled = MphAndMinutesToMiles(milesPerHour, minutesTraveled);

Answer:

If request granted then T₁ and T₂ are in deadlock.

Explanation:

See attached image

Answer:

Check the explanation

Explanation:

/*Given the following program:

public class MysteryNumbers {

   public static void main(String[] args) {

       String one = "two";

       String two = "three";

       String three = "1";

       int number = 20;

​

       sentence(one, two, 3);

       sentence(two, three, 14);

       sentence(three, three, number + 1);

       sentence(three, two, 1);

       sentence("eight", three, number / 2);

   }

​

   public static void sentence(String three, String one, int number) {

       System.out.println(one + " times " + three + " = " + (number * 2));

   }

}*/

Write the output of each of the following calls.

sentence(one, two, 3);     three times two = 6

sentence(two, three, 14); 1 times three = 28

sentence(three, three, number + 1); 1 times 1 = 42

sentence(three, two, 1);   three times 1 = 2

sentence("eight", three, number / 2); 1 times eight = 20

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

The statement regarding user intent is true; it indicates the user's motive behind a search query. User intent is a crucial element in fields such as SEO and helps provide accurate search results based on the user's objective when issuing a query.

The statement 'User intent refers to what the user was trying to accomplish by issuing the query' is true. The concept of user intent is critical in various fields such as search engine optimization (SEO), user experience design, and information retrieval. It is concerned with understanding the goals or objectives a user has in mind when they input a query into a search engine or another kind of system. User intent can be informational, navigational, or transactional, but fundamentally, it suggests that there is a purpose behind the words and phrases that users choose. Understanding user intent is essential for providing relevant and useful responses or results.

It is also related to the concept of truth-value, which refers to the validity of a sentence as being either true or false. For instance, the assertion itself is a sentence with a truth value. In the context of literature, the intent is often debated in terms of the author's intent versus the reader's interpretation, but in the technological realm, user intent usually refers to the direct objectives sought by the person posing the query.

Answer: C

Explanation:

According to the scenario, the action that will you do next is to upload the file to an FTP site and then email the colleague with a link to the site. Thus, the correct option for this question is C.

An error report may be defined as the methodology of recognizing, monitoring, and announcing errors in software solutions, mobile applications, or web services in order to support companies' elegant both development and deployment.

According to the context of this question, if you compose an email, attach the file, and send it to your colleague but the email is not successfully sent due to the attachment being too large to go through. In this condition, you are required to upload the file to an FTP site and then email the colleague with a link to the site.

Therefore, the correct option for this question is C.

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Answer:

See explaination

Explanation:

import java.util.*;

public class qs {

public static void rearrange(Queue<Integer> q)

{

int n = q.size();

Stack<Integer> st = new Stack<Integer>();

Integer f;

for(int i = 0; i < n ; i++)

{

f = q.poll();

// Even elements are added back to the list

if(f%2==0)

q.add(f);

else

st.push(f);

}

// Odd elements are added to the list in reverse order

while(st.size()>0)

{

q.add(st.pop());

}

// Repeats the above process to correct the order of odd elements

for(int i = 0; i < n ; i++)

{

f = q.poll();

// Even elements are added back to the list

if(f%2==0)

q.add(f);

else

st.push(f);

}

//Order of Odd elements are reversed so as to match the actual order

while(st.size()>0)

{

q.add(st.pop());

}

}

public static void main(String[] args) {

int arr[] = {3, 5, 4, 17, 6, 83, 1, 84, 16, 37};

int n = arr.length;

Queue<Integer> q = new LinkedList<Integer>();

for(int i = 0;i<n;i++)

q.add(arr[i]);

System.out.print("\nOriginal Queue\n");

System.out.println(q.toString());

rearrange(q);

System.out.print("\nReordered Queue\n");

System.out.println(q.toString());

}

}

Final answer:

The most common propulsion system in modern aircraft is the fan-jet or turbofan engine, preferred for its efficiency at high speeds. These engines use thrust to characterize their performance, while propeller-powered aircraft use power. The development from piston engines to turbofans marks significant advancements in aviation technology.

Explanation:

The most common propulsion system in modern aircraft is the fan-jet or turbofan engine. These engines are preferred for their efficiency at the higher speeds required by most commercial aircraft today. While turboprop engines are more efficient than fan-jets and turbojets at lower speeds, the preference for the higher speed travel provided by fan-jets has led to their dominance in the market.

The propulsion performance of jet-powered aircraft is characterized by thrust, while propeller-powered aircraft, which can be driven by internal combustion engines or turbines, are typically discussed in terms of power.

The history of aircraft engines has seen the evolution from the gasoline-powered piston engines of early aviation to the highly efficient turbofan engines used in the majority of today's aircraft. Understanding aircraft propulsion involves basic physical concepts such as conservation of mass and momentum which play a crucial role in evaluating forces and motions in fluid flows.

Analyses become quite complex when considering energy gains and losses in jet engines, which include compressor and turbine blades, combustion of fuel, and flow through internal nozzles. However, for a fundamental understanding, the conservation of momentum provides a relatively straightforward approach to calculating thrust.

Answer:

The given python code mainly add values.

Explanation:

In the given python code, a r_in variable is defined, that reads a file in the next line, a while loop is declared, inside the loop "r_in" variable is used that provides slicing in the code.