Answer:
See explaination
Explanation:
PhoneLookup.java
import java.io.FileReader;
import java.io.IOException;
import java.util.Scanner;
public class PhoneLookup
{
public static void main(String[] args) throws IOException
{
Scanner in = new Scanner(System.in);
System.out.println("Enter the name of the phonebook file: ");
String fileName = in.nextLine();
LookupTable table = new LookupTable();
FileReader reader = new FileReader(fileName);
table.read(new Scanner(reader));
boolean more = true;
while (more)
{
System.out.println("Lookup N)ame, P)hone number, Q)uit?");
String cmd = in.nextLine();
if (cmd.equalsIgnoreCase("Q"))
more = false;
else if (cmd.equalsIgnoreCase("N"))
{
System.out.println("Enter name:");
String n = in.nextLine();
System.out.println("Phone number: " + table.lookup(n));
}
else if (cmd.equalsIgnoreCase("P"))
{
System.out.println("Enter phone number:");
String n = in.nextLine();
System.out.println("Name: " + table.reverseLookup(n));
}
}
}
}
LookupTable.java
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
/**
A table for lookups and reverse lookups
*/
public class LookupTable
{
private ArrayList<Item> people;
/**
Constructs a LookupTable object.
*/
public LookupTable()
{
people = new ArrayList<Item>();
}
/**
Reads key/value pairs.
atparam in the scanner for reading the input
*/
public void read(Scanner in)
{
while(in.hasNext()){
String name = in.nextLine();
String number = in.nextLine();
people.add(new Item(name, number));
}
}
/**
Looks up an item in the table.
atparam k the key to find
atreturn the value with the given key, or null if no
such item was found.
*/
public String lookup(String k)
{
String output = null;
for(Item item: people){
if(k.equals(item.getName())){
output = item.getNumber();
}
}
return output;
}
/**
Looks up an item in the table.
atparam v the value to find
atreturn the key with the given value, or null if no
such item was found.
*/
public String reverseLookup(String v)
{
String output = null;
for(Item item: people){
if(v.equals(item.getNumber())){
output = item.getName();
}
}
return output;
}
}
Item.java
public class Item {
private String name, number;
public Item(String aName, String aNumber){
name = aName;
number = aNumber;
}
public String getName(){
return name;
}
public String getNumber(){
return number;
}
}
input.txt
Abbott, Amy
408-924-1669
Abeyta, Ric
408-924-2185
Abrams, Arthur
408-924-6120
Abriam-Yago, Kathy
408-924-3159
Accardo, Dan
408-924-2236
Acevedo, Elvira
408-924-5200
Acevedo, Gloria
408-924-6556
Achtenhagen, Stephen
408-924-3522
Note: Replace all the "at" with at symbol
Carrie is looking at the Form Properties Sheet. Which tab will control how the data is displayed and the data source it is bound to?
Format wrong
Data
Event
Other
Answer:
B
Explanation:
Identify the following as True or False: (1) Naive Bayes is a linear classifier. (2) SVMs are only usable when the classes are linearly separable in the feature space.(3) Adding training data always results in a monotonic increase in the accuracy of a Naive Bayes classifier.(4) When sufficient data is available, SVMs generally perform as well or better than other common classifiers such as KNN. (5) With enough training data, the error of a nearest neighbor classifier always goes down to zero.
Answer:
1. True: Naive Bayes is a linear classifier.
2. False: SVMs are only usable when the classes are linearly separable in the feature space.
3. False: Adding training data always results in a monotonic increase in the accuracy of a Naive Bayes classifier.
4. True: When sufficient data is available, SVMs generally perform as well or better than other common classifiers such as KNN.
5. False: With enough training data, the error of a nearest neighbor classifier always goes down to zero.
Explanation:
Naive Bayes is a linear classifier that leads to a linear decision boundary. It can be applied to a linearly separable problems and when the elements are independent i.e the occurrence of an element doesn't affect the occurrence of another. It can be used for making multi class predictions in artificial intelligence.
The Support Vector Machine (SVM) on the other hand, can either be a non-linear classifier (with RBF kernel) or a linear classifier (with linear kernel). It maximizes the margin of a decision boundary in its mode of operation.
Hence, the SVMs can be used for regression or classification problems.
For example, determining whether an e-mail is a spam or not.
What is the most common propulsion system in modern aircraft?
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.
Consider the markets for monitors, USB drives, central processing units, and Microsoft’s Windows. Assume monitor manufacturers use advertising to differentiate their products, the market for USB drives is controlled by many firms selling similar products, only a few firms control a large portion of the market for CPUs, and Microsoft owns the copyright on Windows. Classify the market for each of the following computer goods and services as either monopoly, oligopoly, monopolistic competition, or perfect competition. 1. Monitors
2. USB drives
3. Central processing units (CPUs)
4. Microsoft's Windows
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.
You are building an L1 data cache for a 32-bit ARMv8 processor. It has a total capacity of 512 MB bytes. It is 4-way set associative with a block size of 16 bytes. Give your answers to the following questions in terms of these parameters. (a) Which bits of the address are used to select a word within a block? (b) Which bits of the address are used to select the set within the cache? (c) How many bits are in each tag? (d) How many tag bits are in the entire cache? (e) What is the total size of the cache in bytes?
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.
providing incentives for customers to learn more about your service is known as?
A) branding
B) advertising
C) permission marketing
D) search engine optimization
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.
TCP has a 32-bit sequence number field and 16-bit advertised window field. Assume that RTT is 512 (2 9 ) ms, transmission speed is 1 Gbps (2 30 bps) and each segment transmitted is 1B byte. Note that Since not two identical sequence numbers can be unacknowledged in the pipe, half of the sequence numbers can used (2 31). (a) How long does it take for the sequence numbers to warp around? 5 pts (b) Now, instead of sending 1 B segment, let’s send a 16 B segment. How long does it take for the sequence numbers to warp around? 5 pts (c) What is the drawback in using large segments? 5 pts (d) What is the maximum achievable throughput?
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
Research the recommended core elements of a single-entity MPI and a multi-enterprise MPI through the Internet and professional journals like the Journal of AHIMA. Write up the recommended core elements for the new eMPI, and how they differ from single-entity MPI core elements, list any changes that might need to be made in the elements at existing facilities for the new eMPI, and cite all your references.
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.
Explanation: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.
The function below takes a string argument sentence. Complete the function to return the second word of the sentence. You can assume that the words of the sentence are separated by whitespace. If the sentence contains less than two words, return None. If the last character of the word you are returning is a comma, semi-colon, period, exclamation point, or question mark, remove that last character.
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.
Develop Java methods to perform the following operations on an unordered list (using linked lists). What is the complexity of each of these algorithms. 1. Insert an element at the start (front) of the list 2. Insert an element at the end (rear) of the list 3. Insert an element at the middle of the list following an element already in the list. 4. Remove an element from the middle of the list
Answer:
Check the explanation
Explanation:
Java Program to Implement Singly Linked List
import java. util. Scanner;
/* Class Node */
class Node
{
protected int data;
protected Node link;
/* Constructor */
public Node()
{
link = null;
data = 0;
}
/* Constructor */
public Node(int d,Node n)
{
data = d;
link = n;
}
/* Function to set link to next Node */
public void setLink(Node n)
{
link = n;
}
/* Function to set data to current Node */
public void setData(int d)
{
data = d;
}
/* Function to get link to next node */
public Node getLink()
{
return link;
}
/* Function to get data from current Node */
public int getData()
{
return data;
}
}
/* Class linkedList */
class linkedList
{
protected Node start;
protected Node end ;
public int size ;
/* Constructor */
public linkedList()
{
start = null;
end = null;
size = 0;
}
/* Function to check if list is empty */
public boolean isEmpty()
{
return start == null;
}
/* Function to get size of list */
public int getSize()
{
return size;
}
/* Function to insert an element at begining */
public void insertAtStart(int val)
{
Node nptr = new Node(val, null);
size++ ;
if(start == null)
{
start = nptr;
end = start;
}
else
{
nptr. setLink(start);
start = nptr;
}
}
/* Function to insert an element at end */
public void insertAtEnd(int val)
{
Node nptr = new Node(val,null);
size++ ;
if(start == null)
{
start = nptr;
end = start;
}
else
{
end. setLink(nptr);
end = nptr;
}
}
/* Function to insert an element at position */
public void insertAtPos(int val , int pos)
{
Node nptr = new Node(val, null);
Node ptr = start;
pos = pos - 1 ;
for (int i = 1; i < size; i++)
{
if (i == pos)
{
Node tmp = ptr. getLink() ;
ptr. setLink(nptr);
nptr. setLink(tmp);
break;
}
ptr = ptr. getLink();
}
size++ ;
}
/* Function to delete an element at position */
public void deleteAtPos(int pos)
{
if (pos == 1)
{
start = start. getLink();
size--;
return ;
}
if (pos == size)
{
Node s = start;
Node t = start;
while (s != end)
{
t = s;
s = s. getLink();
}
end = t;
end. setLink(null);
size --;
return;
}
Node ptr = start;
pos = pos - 1 ;
for (int i = 1; i < size - 1; i++)
{
if (i == pos)
{
Node tmp = ptr. getLink();
tmp = tmp. getLink();
ptr. setLink(tmp);
break;
}
ptr = ptr. getLink();
}
size-- ;
}
/* Function to display elements */
public void display()
{
System.out. print("\nSingly Linked List = ");
if (size == 0)
{
System. out. print("empty\n");
return;
}
if (start. getLink() == null)
{
System. out. println(start. getData() );
return;
}
Node ptr = start;
System. out. print(start. getData()+ "->");
ptr = start. getLink();
while (ptr. getLink() != null)
{
System. out. print(ptr. getData()+ "->");
ptr = ptr. getLink();
}
System. out. print(ptr. getData()+ "\n");
}
}
/* Class SinglyLinkedList */
public class SinglyLinkedList
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
/* Creating object of class linkedList */
linkedList list = new linkedList();
System. out. println("Singly Linked List Test\n");
char ch;
/* Perform list operations */
do
{
System. out. println("\nSingly Linked List Operations\n");
System. out. println("1. insert at the start");
System. out. println("2. insert at the rear");
System. out. println("3. insert at the middle of the list");
System. out. println("4. Remove from the middle of list");
int choice = scan. nextInt();
switch (choice)
{
case 1 :
System. out. println("Enter integer element to insert");
list. insertAtStart( scan. nextInt() );
break;
case 2 :
System. out. println("Enter integer element to insert");
list. insertAtEnd( scan. nextInt() );
break;
case 3 :
System. out. println("Enter integer element to insert");
int num = scan. nextInt() ;
System. out. println("Enter position");
int pos = scan. nextInt() ;
if (pos <= 1 || pos > list. getSize() )
System. out. println("Invalid position\n");
else
list. insertAtPos(num, pos);
break;
case 4 :
System. out. println("Enter position");
int p = scan. nextInt() ;
if (p < 1 || p > list. getSize() )
System. out. println("Invalid position\n");
else
list. deleteAtPos(p);
break;
default :
System. out. println("Wrong Entry \n ");
break;
}
/* Display List */
list. display();
System. out. println("\nDo you want to continue (Type y or n) \n");
ch = scan. next(). charAt(0);
} while (ch == 'Y'|| ch == 'y');
}
}
The Beaufort Wind Scale is used to characterize the strength of winds. The scale uses integer values and goes from a force of 0, which is no wind, up to 12, which is a hurricane. Write a script that generates a random force value between 0 and 12, with use of randi command, Then, it prints a message regarding what type of wind that force represents. The answers should follow this pattern:
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
g Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the middle element of the array Sort the arrays using pivot as the median of the first, last, and middle elements of the array Sort the arrays using pivot as the middle element of the array. However,, when the size of any sub-list reduces to less than 20, sort the sub-list using insertion sort. Sort the array using pivot as the median of the first, last and middle elements of the array. When the size of any sub-list reduces to less than 20, sort the sub-list using insertion sort. Calculate and display the CPU time for each of the preceding four steps.
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.
What is the available vector for each resource and the need matrix for each thread. Note that an available vector depicts the quantities of resources of each type that are available in the system after current allocation. The need matrix depicts of the quantities of resources of each type still needed by each thread, with each column corresponding to a thread and each row corresponding to a resource and an entry in location [i, j] of the matrix showing the quantity of a resource rj still needed by a thread ti.
Answer:
If request granted then T₁ and T₂ are in deadlock.
Explanation:
See attached image
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. Sound F/X sentence(one, two, 3); sentence(two, three, 14); sentence(three, three, number + 1); sentence(three, two, 1); sentence("eight", three, number / 2);
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
Consider a system with a total of 150 units of memory, allocated to three processes as follows: Process Maximum Hold 1 70 45 2 60 40 3 60 15 Apply the banker’s algorithm to determine whether it would be safe to grant each of the following requests. Explain your answer. (a) A fourth process arrives, with a maximum memory need of 60 and an initial need of 25 units. (b) A fourth process arrives, with a maximum memory need of 60 and an initial need of 35 units. (c) A fourth process arrives, with a maximum memory need of 50 and an initial need of 30 units.
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:
Sure, it is safe. After assigning 25 units for P4, there'll be enough RAM to ensure termination of either P1 or P2.Following that, students can finish the other three main tasks in any sequence they want.Nope, it is not safe. When allocating 15 units for P4, there will also be insufficient RAM to ensure the termination of any process.Learn more:
brainly.com/question/24097231
What techniques are required to accept
constructive criticism? Select all that apply
3. not interrupting or speaking until the other
person is finished making a point
2. filing a formal complaint with human
resources to make sure your point of view is
documented
1. not frowning, sneering, or defensively folding
arms in front of you
Answer:
1 and 3
Explanation:
Implement a C program that in a loop listens on a port for incoming TCP requests from clients. For each accepted incoming request it forks a child to read and process the request. The parent process continues to listen and accept incoming TCP requests in an endless loop. The program accepts 2 command line parameters: the port number to listen on, the pathname to a directory that serves as root to all requested files or directories. For example: % ./z123456 9001 www The requests received by the program are of the form: GET pathname where the pathname refers to a file or directory to be se
To implement a C program that listens on a port for incoming TCP requests, accepts two command line parameters (the port number and the root directory pathname), and forks a child process to handle each request, you will need to utilize socket programming and process control in C. The parent process will perpetually listen for incoming connections, while each child process will read and process individual client requests. Use network APIs like socket, bind, listen, and accept for setting up the server, and fork to create child processes.
Your server program should begin by initializing a socket using the socket() function, binding it to the specified port using bind(), and starting to listen for connections with listen(). Upon accepting a new connection with accept(), the program should fork a new process. The child will process the request such as reading the pathname, and if valid, sending back the requested file contents. The parent, on the other hand, continues to accept new connections. The exact API functions you'll need to look at include recv or read for receiving the request, and write or send to send a response back. It's also important to handle errors and edge cases, like ensuring the requested file is within the specified root directory and properly handling the termination of child processes.
9-9. Your computer sends a DNS request message to your local DNS server. After an unusually long time, your computer receives a DNS response message that the host name in your request message does not exist. This is a host you use every day. a) List problems that may have happened and comment on their likelihood in broad terms like low. Give an explanation for your rating. (Draw the picture.)
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:
Internet issues may occur due to censure, cyber warfare, catastrophic events, efforts on the part of police and security utilities. It may also be known as the underwater communications, in which cable breakdowns cause disruptions to large parts. A power failure is also the most frequent cause of server failure. If you don't even have a backup generator, storms, natural catastrophes, and statewide power failures may shut your server off. It may be overloaded, that can trigger device crashes. Essentially, it's when too many users access the server simultaneously.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.
Show the output waveform of an AND gate with the inputs A, B,
and C indicated in the figure below.
Answer:
Please see the attached image
Explanation:
.Assume the following schedule for a set of three jobs, A, B, and C, with a single CPU. Assume that there is no i/o. --A runs first (for 10 time units) but is not yet done. --B runs next (for 10 time units) but is not yet done --C runs next (for 10 time units) and runs to completion --A runs to completion (for 10 time units) --B runs to completion (for 5 time units) Which scheduling disciplines could allow this schedule to occur? Explain. (a) FIFO (b) Round Robin. (c) STCF (Shortest Time to Completion First)
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:
FIFO (First In, First Out): This scheduling discipline executes jobs in the order they arrive. Based on the sequence provided, FIFO could potentially allow this schedule to occur since A starts first, followed by B and then C, each running to completion in their second turn.Round Robin: This scheduling discipline allocates a fixed time unit or quantum for each job to run. The schedule provided fits a Round Robin approach if the time quantum is set to 10 time units. Each job gets a chance to run for 10 units before the CPU switches to the next job in sequence, which matches the described scenario.STCF (Shortest Time to Completion First): This discipline schedules jobs based on the shortest completion time. The provided schedule does not explicitly align with STCF since it requires knowledge of each job's total execution time beforehand, and the sequence doesn't prioritize finishing B before A, even though B takes less total time to complete.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.
When a Python program is reading a file, the data is input as plain ASCII characters in a string. What is the following code doing assuming a file is opened with file object inputFile? r_in = inputFile.readline() t = 0 while r_in != '': r = r_in.rstrip('\n') r_list = r.split('|') n = float(r_list[2]) t += n r_in = inputFile.readline()
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.
In this code, in the file, each line is the format in field 1 and field 2. The code puts its fields into the list first and then converts into the last dimension to the numerical type before applying to a sum.Your co-worker is at a conference in another state. She requests that you
send her the current draft of the report you are working on. You create an
email, attach the file, and click Send. A few seconds later, you receive an error
report: The attachment was too large to go through. What should you do
next?
A. Print a hard copy of the report, and then mail it to your colleague.
O
B. Delete sections of the file that are less important, and then try
emailing it again.
O
C. Upload the file to an FTP site and then email the colleague with a
link to the site.
D. Transfer the file to a USB flash drive, and then mail the drive to
your colleague.
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.
What is meant by an Error report?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.
To learn more about Email attachments, refer to the link:
https://brainly.com/question/28348657
#SPJ2
QUESTIONS / TASKS • Which class is the superclass? Which is the subclass? What does it mean that the Cat class extends the Animal class? • The Cat class cannot directly access the fields it inherits from Animal. Why not? • 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); • 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 ); • 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.
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.
Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the middle element of the array Sort the arrays using pivot as the median of the first, last, and middle elements of the array Sort the arrays using pivot as the middle element of the array. However,, when the size of any sub-list reduces to less than 20, sort the sub-list using insertion sort. Sort the array using pivot as the median of the first, last and middle elements of the array. When the size of any sub-list reduces to less than 20, sort the sub-list using insertion sort. Calculate and display the CPU time for each of the preceding four steps. Example of the median of the first, last and middle elements: 1 2 3 4 5 6 7 8 0 (median of 1, 5, 0 is 1) 8 0 1 2 3 4 5 6 7 (median of 8, 3, 7 is 7) To calculate the CPU time, use the header , and clock_t type. Depends on the CPU of your computer, your number would not be the same as in the sample output below.
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;
}
Consider designing a program where you need to store information about every student ASU. You need to be able to quickly determine which students are graduating this semester. Would you use an array or linked list? Analyze the problem, design a choice, and justify the choice
Answer:
Check the explanation
Explanation:
Both Arrays and Linked List can be used to store linear data of similar types, but they both have some advantages and disadvantages over each other. To store information about every student we can use an array
Reason:
In the array the elements belong to indexes, i.e., if you want to get into the fourth element you have to write the variable name with its index or location within the square bracket.
In a linked list though, you have to start from the head and work your way through until you get to the fourth element.
Accessing an element in an array is fast, while Linked list takes linear time, so it is quite a bit slower.
Example: store the data of students and have faster access and this program in example also do sorting
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// struct person with 3 fields
struct Student {
char* name;
int id;
char age;
};
// setting up rules for comparison
// to sort the students based on names
int comparator(const void* p, const void* q)
{
return strcmp(((struct Student*)p)->name,
((struct Student*)q)->name);
}
// Driver program
int main()
{
int i = 0, n = 5;
struct Student arr[n];
// Get the students data
arr[0].id = 1;
arr[0].name = "bd";
arr[0].age = 12;
arr[1].id = 2;
arr[1].name = "ba";
arr[1].age = 10;
arr[2].id = 3;
arr[2].name = "bc";
arr[2].age = 8;
arr[3].id = 4;
arr[3].name = "aaz";
arr[3].age = 9;
arr[4].id = 5;
arr[4].name = "az";
arr[4].age = 10;
// Print the Unsorted Structure
printf("Unsorted Student Records:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
// Sort the structure
// based on the specified comparator
qsort(arr, n, sizeof(struct Student), comparator);
// Print the Sorted Structure
printf("\n\nStudent Records sorted by Name:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
return 0;
}
Output:
Unsorted Student Records: Id = 1, Name = bd, Age = 12 Id = 2, Name = ba, Age = 10 Id = 3, Name = bc, Age = 8 Id = 4, Name = aaz, Age = 9 Id = 5, Name = az, Age = 10
Final answer:
For a program tracking students' graduation statuses at ASU, a linked list is the most suitable data structure due to its efficient handling of frequent insertions and deletions, as well as its ability to maintain a sorted list for quick identification of graduating students.
Explanation:
When considering how to store information about every student at ASU and quickly determine which students are graduating this semester, it's important to analyze the requirements and choose the appropriate data structure. An array is best suited for scenarios where memory usage is predictable and access to data by an index is frequent. However, a linked list is preferred when the dataset is dynamic, and you need to frequently add or remove records, as is likely the case with a constantly updating list of graduating students.
In this scenario, using a linked list would be beneficial since students will be added or removed as they meet or no longer meet the graduating criteria throughout the semester. This structure allows for efficient insertions and deletions without the need to reallocate or reorganize the entire data structure. Additionally, you can maintain a sorted list based on graduation status, which expedites the process of identifying graduating students.
Therefore, a linked list is the recommended choice for a program tracking graduating students due to its flexibility in handling frequent changes and requiring less overhead for managing dynamic datasets. To further expedite the process, you could keep two linked lists, one for graduate students and one for other students, and move students between the two as their graduation status changes. This design optimizes for quick access to the list of graduating students, aligning with the programs' requirements.
Write two scanf statements to get input values into birthMonth and birthYear. Then write a statement to output the month, a slash, and the year. End with newline. The program will be tested with inputs 1 2000 and then with inputs 5 1950. Ex: If the input is 1 2000, the output is: 1/2000 Note: The input values come from user input, so be sure to use scanf statements, as in scanf("%d", &birthMonth), to get those input values (and don't assign values directly, as in birthMonth = 1).
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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True or false :User intent refers to what the user was trying to accomplish by issuing the query
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.
Write a method called rearrange that accepts a queue of integers as a parameter and rearranges the order of the values so that all of the even values appear before the odd values and that otherwise preserves the original order of the queue. For example, if the queue stores [3, 5, 4, 17, 6, 83, 1, 84, 16, 37], your method should rearrange it to store [4, 6, 84, 16, 3, 5, 17, 83, 1, 37]. Notice that all of the evens appear at the front followed by the odds and that the relative order of the evens and odds is the same as in the original. Use one stack as auxiliary storage.
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());
}
}
Given the following schedule, show the locks that will occur and the subsequent schedule. Assume that strict 2PL is in effect, with no deadlock prevention and no starvation prevention. Assume that upgrading locks are allowed, downgrades are not. Assume that requests will be satisfied in the order of arrival if possible.
T1:R(X), T2:R(Q), T3:R(Q), T(2):R(X), T3:R(X), T2:W(X), T2:Commit, T3:Abort, T1:W(X), T1:Commit
Answer:
see explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
The Domain Name System (DNS) provides an easy way to remember addresses. Without DNS, how many octets for an Internet Protocol (IP) address would have to be memorized?
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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