Answer:
Check the explanation
Explanation:
--Query 1)
SELECT ename, sal, age
FROM Emp;
--Query 2)
SELECT did
FROM Dept
WHERE floot = 10 AND budget<15000;
Zoom Vacuum, a family-owned manufacturer of high-end vacuums, has grown exponentially over the last few years. However, the company is having difficulty preparing for future growth. The only information system used at Zoom is an antiquated accounting system. The company has one manufacturing plant located in Iowa; and three warehouses, in Iowa, New Jersey, and Nevada. The Zoom sales force is national, and Zoom purchases about 25 percent of its vacuum parts and materials from a single overseas supplier. You have been hired to recommend the information systems Zoom should implement in order to maintain their competitive edge. However, there is not enough money for a full-blown, cross-functional enterprise application, and you will need to limit the first step to a single functional area or constituency. What will you choose, and why?
Answer:A TPS focusing on production and manufacturing to keep production costs low while maintaining quality, and for communicating with other possible vendors. The TPS would later be used to feed MIS and other higher level systems.
Explanation:
The relationship between the temperature of a fluid (t, in seconds), temperature (T, in degrees Celsius), is dependent upon the initial temperature of the liquid (T0, in degrees Celsius), the ambient temperature of the surroundings (TA, in degrees Celsius) and the cooling constant (k, in hertz); the relationship is given by: ???? ???? ???????? ???? ???????????? ???? ???????????? ???????????????? Ask the user the following questions: From a menu, choose fluid ABC, FGH, or MNO. Enter the initial fluid temperature, in units of degrees Celsius. Enter the time, in units of minutes. Enter the ambient air temperature, in units of degrees Celsius. Enter the following data into the program. The vector contains the cooling constant (k, units of hertz) corresponding to the menu entries. K Values = [0.01, 0.03, 0.02] Create a formatted output statement for the user in the Command Window similar to the following. The decimal places must match. ABC has temp 83.2 degrees Celsius after 3 minutes. In
Answer:
See explaination
Explanation:
clc;
clear all;
close all;
x=input(' choose abc or fgh or mno:','s');
to=input('enter intial fluid temperature in celcius:');
t1=input('enter time in minutes:');
ta=input('enter ambient temperature in celcius:');
abc=1;
fgh=2;
mno=3;
if x==1
k=0.01;
elseif x==2
k=0.03;
else
k=0.02;
end
t=ta+((to-ta)*exp((-k)*t1));
X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);
disp(X);
Donnell backed up the information on his computer every week on a flash drive. Before copying the files to the flash drive, he always ran a virus scan against the files to ensure that no viruses were being copied to the flash drive. He bought a new computer and inserted the flash drive so that he could transfer his files onto the new computer. He got a message on the new computer that the flash drive was corrupted and unreadable; the information on the flash drive cannot be retrieved. Assuming that the flash drive is not carrying a virus, which of the following does this situation reflect?
a. Compromise of the security of the information on the flash drive
b. Risk of a potential breach in the integrity of the data on the flash drive
c. Both of the above
d. Neither of the above.
Answer:
b. Risk of a potential breach in the integrity of the data on the flash drive
Explanation:
The corrupted or unreadable file error is an error message generated if you are unable to access the external hard drive connected to the system through the USB port. This error indicates that the files on the external hard drive are no longer accessible and cannot be opened.
There are several reasons that this error message can appear:
Viruses and Malware affecting the external hard drive .Physical damage to external hard drive or USB memory .Improper ejection of removable drives.*Sometimes it is difficult to convince top management to commit funds to develop and implement a SIS why*
Step-by-step Explanation:
SIS stands for: The Student Information System (SIS).
This system (a secure, web-based accessible by students, parents and staff) supports all aspects of a student’s educational experience, and other information. Examples are academic programs, test grades, health information, scheduling, etc.
It is difficult to convince top management to commit funds to develop and implement SIS, this can be due to a thousand reasons.
The obvious is that the management don't see the need for it. They would rather have students go through the educational process the same way they did. Perhaps, they just don't trust the whole process, they feel more in-charge while using a manual process.
The cord of a bow string drill was used for
a. holding the cutting tool.
b. providing power for rotation.
c. transportation of the drill.
d. finding the center of the hole.
Answer:
I don't remember much on this stuff but I think it was B
Evaluati urmatoarele expresii
5+2*(x+4)/3, unde x are valoare 18
7/ 2*2+4*(5+7*3)>18
2<=x AND x<=7 , unde x are valoare 23
50 %10*5=
31250/ 5/5*2=
Answer:
A) 22 ⅓
B) 111>18
C) There is an error in the expression
D) 25
E) 62500
Question:
Evaluate the following expressions
A) 5 + 2 * (x + 4) / 3, where x has a value of 18
B) 7/2 * 2 + 4 * (5 + 7 * 3) & gt; 18
C) 2 <= x AND x<= 7, where x has value 23
D) 50% 10 * 5 =
F) 31250/5/5 * 2 =
Explanation:
A) 5 + 2 * (x + 4) / 3
x = 18
First we would insert the value of x
5 + 2 * (x + 4) / 3
5 + 2(18 + 8) / 3
Then we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
= 5 + 2(26) / 3
= 5 + 52/3
= 5 + 17 ⅓
= 22 ⅓
B) 7/2 * 2 + 4 * (5 + 7 * 3) > 18
we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
7/2 * 2 + 4 * (5 + 7 * 3) >18
= 7/2 × 2 + 4× (5 + 7 × 3)>18
= (7×2)/2 + 4× (5+21) >18
= 14/2 + 4(26) >18
= 7 + 104 >18
= 111>18
C) 2 <= x AND x<= 7, where x has value 23
There is an error in the expression
D) 50% of 10 * 5
we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
The 'of' expression means multiplication
= 50% × 10×5
= 50% × 50
50% = 50/100
=50/100 × 50
= 1/2 × 50
= 25
F) 31250/5/5 * 2
The expression has no demarcation. Depending on how it is broken up, we would arrive at different answers. Let's consider:
31250/(5/5 × 2)
Apply BODMAS
= 31250/[5/(5 × 2)]
= 31250/(5/10)
= 31250/(1/2)
Multiply by the inverse of 1/2 = 2/1
= 31250 × (2/1)
= 62500
Create a program that generates a report that displays a list of students, classes they are enrolled in and the professor who teaches that class. There are 3 files that provide the input data: 1. FinalRoster.txt : List of students and professors ( S: student, P: professor) Student row: S,Student Name, StudentID Professor row: P, Professor Name, Professor ID, Highest Education 2. FinalClassList.txt: List of classes and professor who teach them (ClassID, ClassName, ID of Professor who teach that class) 3. FinalStudentClassList.txt: List of classes the students are enrolled in. (StudentID, ClassID) The output shall be displayed on screen and stored in a file in the format described in FinalOutput.txt You will need to apply all course concepts in your solution. 1. Student and Professor should be derived from a super class "Person" 2. Every class should implement toString method to return data in the format required for output 3. Exception handling (e.g. FileNotFound etc.) must be implemented 4. Source code must be documented in JavaDoc format 5. Do not hard code number of students, professors or classes. Submit source Java files, the output file and screenshot of the output in a zip format
Answer:
All the classes are mentioned with code and screenshots. . That class is shown at last.
Explanation:
Solution
Class.Java:
Code
**
* atauthor your_name
* This class denotes Class at the college.
*
*/
public class Class {
private String className,classID;
private Professor professor;
/**
* atparam className
* atparam classID
* atparam professor
*/
public Class(String className, String classID, Professor professor) {
this.classID=classID;
this.className=className;
this.professor=professor;
}
/**
* at return classID of the class
*/
public String getClassID() {
return classID;
}
/**
* Override toString() from Object Class
*/
public String toString() {
return classID+" "+className+" "+professor.getName()+" "+professor.getEducation();
}
}
Person.Java:
Code:
/**
* atauthor your_name
* This class represents Person
*
*/
public class Person {
protected String name;
/**method to fetch name
* at return
*/
public String getName() {
return name;
}
/**method to set name
* at param name
*/
public void setName(String name) {
this.name = name;
}
Professor.java:
Code:
import java.util.ArrayList;
import java.util.List;
/**
* at author your_name
*
*This class represents professors
*
*/
public class Professor extends Person{
private String professorID, education;
private List<Class> classes=new ArrayList<Class>();
/**
* at param name
* at param professorID
* at param education
*/
public Professor(String name,String professorID,String education) {
this.name=name;
this.professorID=professorID;
this.education=education;
}
/**
* at return
*/
public String getEducation() {
return this.education;
}
/**
* at return
*/
public String getprofessorID() {
return this.professorID;
}
/** to add classes
* at param Class
*/
public void addClass(Class c) {
classes.add(c);
}
/**
* Override toString() from Object Class
*/
public String toString() {
String result=this.getName()+" - "+professorID+" - "+education;
for(Class c:classes) {
result+=c.toString()+"\n";
}
return result;
}
}
}
Student.java:
Code:
import java.util.ArrayList;
import java.util.List;
/**
* This class represents students
* at author your_Name
*
*/
public class Student extends Person{
private String studentID;
private List<Class> classes=new ArrayList<Class>();
/**
* atparam name
* atparam studentID
*/
public Student(String name,String studentID) {
this.name=name;
this.studentID=studentID;
}
/**
* atreturn
*/
public String getStudentID() {
return studentID;
}
/**
* atparam c
*/
public void addClass(Class c) {
classes.add(c);
}
/**
* atreturn
*/
public int getClassCount() {
return classes.size();
}
/**
* Override toString() from Object Class
*/
public String toString() {
String result=this.getName()+" - "+studentID+"\n";
for(Class c:classes) {
result+=c.toString()+"\n";
}
return result;
}
}
NOTE: Kindly find an attached copy of screenshot of the output, which is a part of the solution to this question
Select the correct navigational path to create the function syntax to use the IF function.
Click the Formula tab on the ribbon and look in the ???
'gallery
Select the range of cells.
Then, begin the formula with the ????? click ?????. and click OK.
Add the arguments into the boxes for Logical Test, Value_if_True, and Value_if_False.
Answer:
1. Logical
2.=
3.IF
Explanation:
just did the assignment
1. Write a pair of classes, Square1 and Rectangle1. Define Square1 as a subclass of Rectangle1. In addition to setters and getters, provide such methods as computeArea and computePerimeter. Specify the preconditions, postconditions and class invariants, if any, as comments (
Answer:
The java program including classes is shown below.
import java.util.*;
import java.lang.*;
//base class
class Rectangle1
{
//variables to hold dimensions and area
static double l;
static double b;
static double a;
//setter
static void setWidth(double w)
{
b=w;
}
//getter
static double getWidth()
{
return b;
}
//setter
static void setLength(double h)
{
l=h;
}
//getter
static double getLength()
{
return b;
}
//area calculation for rectangle
static void computeArea()
{
a = getLength()*getWidth();
System.out.println("Area of rectangle = " +a);
}
}
//derived class
class Square1 extends Rectangle1
{
//variables to hold dimensions and perimeter
static double s;
static double peri;
//settter
static void setSide(double d)
{
s=d;
}
//getter
static double getSide()
{
return s;
}
//perimeter calculation for square
static void computePerimeter()
{
peri = 2*(getSide()+getSide());
System.out.println("Perimeter of square = "+peri);
}
}
public class MyClass {
public static void main(String args[]) {
//object of derived class created
Square1 ob = new Square1();
ob.setLength(1);
ob.setWidth(4);
ob.setSide(1);
ob.computeArea();
ob.computePerimeter();
}
}
Explanation:
1. The class for rectangle is defined having variables to hold the dimensions of a rectangle.
2. The getters and setters are included for both the dimensions.
3. The method to compute the area of the rectangle is defined.
4. The class for square is defined having variables to hold the side of a square.
5. The getter and setter are included for the dimension of the square.
6. The method to compute the perimeter of the square is defined.
7. The class Square1 is the derived class which inherits the class Rectangle1 through the keyword extends.
8. All the variables, getters, setters and methods are declared as static.
9. Another class is defined having only the main() method.
10. Inside main(), the object of the derived class, Square1, is created.
11. The setters and the methods are called through this object.
12. The value of the dimensions are passed as arguments to the setters.
13. The output is attached in an image.
Write a program whose input is a character and a string, and whose output indicates the number of times the character appears in the string.
Ex: If the input is:
n Monday
the output is:
1
Ex: If the input is:
z Today is Monday
the output is:
0
Ex: If the input is:
n It's a sunny day
the output is:
2
Case matters.
Ex: If the input is:
n Nobody
the output is:
0
n is different than N.
This is what i have so far.
#include
#include
using namespace std;
int main() {
char userInput;
string userStr;
int numCount;
cin >> userInput;
cin >> userStr;
while (numCount == 0) {
cout << numCount << endl;
numCount = userStr.find(userInput);
}
return 0;
}
TRUE OR FALSE! HELP!!
Answer:
True
Explanation:
There's no one law that governs internet privacy.
Help its simple but I don't know the answer!!!
If you have a -Apple store and itunes gift card-
can you use it for in app/game purchases?
Answer:
yes you can do that it's utter logic
Answer:
Yes.
Explanation:
Itunes gift cards do buy you games/movies/In app purchases/ect.
Retail price data for n = 60 hard disk drives were recently reported in a computer magazine. Three variables were recorded for each hard disk drive: y = Retail PRICE (measured in dollars) x1 = Microprocessor SPEED (measured in megahertz) (Values in sample range from 10 to 40) x2 = CHIP size (measured in computer processing units) (Values in sample range from 286 to 486) A first-order regression model was fit to the data. Part of the printout follows: __________.
Answer:
Explanation:
Base on the scenario been described in the question, We are 95% confident that the price of a single hard drive with 33 megahertz speed and 386 CPU falls between $3,943 and $4,987
Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions from a file and create expression trees for each expression, one at a timethe expression in the file must be in "math" notation, for example x+y*a/b.display the preorder traversal of a binary tree as a sequence of strings each separated by a tabdisplay the postorder traversal of a binary tree in the same form as aboveWrite a function to display the inorder traversal of a binary tree and place a (before each subtree and a )after each subtree. Don’t display anything for an empty subtree. For example, the expression tree should would be represented as ( (x) + ( ( (y)*(a) )/(b) ) )
Answer:
Explanation:
Program:
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
//check for operator
bool isOperator(char c)
{
switch(c)
{
case '+': case '-': case '/': case '*': case '^':
return true;
}
return false;
}
//Converter class
class Converter
{
private:
string str;
public:
//constructor
Converter(string s):str(s){}
//convert from infix to postfix expression
string toPostFix(string str)
{
stack <char> as;
int i, pre1, pre2;
string result="";
as.push('(');
str = str + ")";
for (i = 0; i < str.size(); i++)
{
char ch = str[i];
if(ch==' ') continue;
if (ch == '(')
as.push(ch);
else if (ch == ')')
{
while (as.size() != 0 && as.top() != '('){
result = result + as.top() + " ";
as.pop();
}
as.pop();
}
else if(isOperator(ch))
{
while (as.size() != 0 && as.top() != '(')
{
pre1 = precedence(ch);
pre2 = precedence(as.top());
if (pre2 >= pre1){
result = result + as.top() + " ";
as.pop();
}
else break;
}
as.push(ch);
}
else
{
result = result + ch;
}
}
while(as.size() != 0 && as.top() != '(') {
result += as.top() + " ";
as.pop();
}
return result;
}
//return the precedence of an operator
int precedence(char ch)
{
int choice = 0;
switch (ch) {
case '+':
choice = 0;
break;
case '-':
choice = 0;
break;
case '*':
choice = 1;
break;
case '/':
choice = 1;
break;
case '^':
choice = 2;
default:
choice = -999;
}
return choice;
}
};
//Node class
class Node
{
public:
string element;
Node *leftChild;
Node *rightChild;
//constructors
Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}
Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}
};
//ExpressionTree class
class ExpressionTree
{
public:
//expression tree construction
Node* covert(string postfix)
{
stack <Node*> stk;
Node *t = nullptr;
for(int i=0; i<postfix.size(); i++)
{
if(postfix[i]==' ') continue;
string s(1, postfix[i]);
t = new Node(s);
if(!isOperator(postfix[i]))
{
stk.push(t);
}
else
{
Node *r = nullptr, *l = nullptr;
if(!stk.empty()){
r = stk.top();
stk.pop();
}
if(!stk.empty()){
l = stk.top();
stk.pop();
}
t->leftChild = l;
t->rightChild = r;
stk.push(t);
}
}
return stk.top();
}
//inorder traversal
void infix(Node *root)
{
if(root!=nullptr)
{
cout<< "(";
infix(root->leftChild);
cout<<root->element;
infix(root->rightChild);
cout<<")";
}
}
//postorder traversal
void postfix(Node *root)
{
if(root!=nullptr)
{
postfix(root->leftChild);
postfix(root->rightChild);
cout << root->element << " ";
}
}
//preorder traversal
void prefix(Node *root)
{
if(root!=nullptr)
{
cout<< root->element << " ";
prefix(root->leftChild);
prefix(root->rightChild);
}
}
};
//main method
int main()
{
string infix;
cout<<"Enter the expression: ";
cin >> infix;
Converter conv(infix);
string postfix = conv.toPostFix(infix);
cout<<"Postfix Expression: " << postfix<<endl;
if(postfix == "")
{
cout<<"Invalid expression";
return 1;
}
ExpressionTree etree;
Node *root = etree.covert(postfix);
cout<<"Infix: ";
etree.infix(root);
cout<<endl;
cout<<"Prefix: ";
etree.prefix(root);
cout<<endl;
cout<< "Postfix: ";
etree.postfix(root);
cout<<endl;
return 0;
}
A customer seeks to buy a new computer for private use at home. The customer primarily needs the computer to use the Microsoft PowerPoint application for the purpose of practicing presentation skills. As a salesperson what size hard disc would you recommend and why?
Answer:
500gbs of ssd storage. Microsoft also allows for multi platform cloud cross-saving.
Explanation:
Dave owns a construction business and is in the process of buying a laptop. He is looking for a laptop with a hard drive that will likely continue to function if the computer is dropped. Which type of hard drive does he need?
Answer:
Solid state drive
Explanation:
The term solid-state drive is used for the electronic circuitry made entirely from semiconductors. This highlights the fact that the main storage form, in terms of a solid-state drive, is via semiconductors instead of a magnetic media for example a hard disk. In lieu of a more conventional hard drive, SSD is built to live inside the device. SSDs are usually more resistant to physical shock in comparison to the electro-mechanical drives and it functions quietly and has faster response time. Therefore, SSD will be best suitable for the Dave.
Who is your favorite smite god in Hi-Rez’s “Smite”
Answer:
Variety
Explanation:
Double any element's value that is less than minValue. Ex: If minValue = 10, then dataPoints = {2, 12, 9, 20} becomes {4, 12, 18, 20}.
import java.util.Scanner;
public class StudentScores {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_Points = 4;
int[] dataPoints = new int[NUM_POINTS];
int minValue;
int i;
minValue = scnr.nextInt();
for (i = 0; i < dataPoints.length; ++i) {
dataPoints[i] = scnr.nextInt();
}
/* Your solution goes here */
for (i = 0; i < dataPoints.length; ++i) {
System.out.print(dataPoints[i] + " ");
}
System.out.println();
}
}
Answer:
Following are the code to this question:
for(i=0;i<dataPoints.length;++i) //define loop to count array element
{
if(dataPoints[i]<minValue) // define condition that checks array element is less then minValue
{
dataPoints[i] = dataPoints[i]*2; //double the value
}
}
Explanation:
Description of the code as follows:
In the given code, a for loop is declared, that uses a variable "i", which counts all array element, that is input by the user. Inside the loop and if block statement is used, that check array element value is less then "minValue", if this condition is true. So, inside the loop, we multiply the value by 2.#define DIRECTN 100
#define INDIRECT1 20
#define INDIRECT2 5
#define PTRBLOCKS 200
typedef struct {
filename[MAXFILELEN];
attributesType attributes; // file attributes
uint32 reference_count; // Number of hard links
uint64 size; // size of file
uint64 direct[DIRECTN]; // direct data blocks
uint64 indirect[INDIRECT1]; // single indirect blocks
uint64 indirect2[INDIRECT2]; // double indirect
} InodeType;
Single and double indirect inodes have the following structure:
typedef struct
{
uint64 block_ptr[PTRBLOCKS];
}
IndirectNodeType;
Required:
Assuming a block size of 0x1000 bytes, write pseudocode to return the block number associated with an offset of N bytes into the file.
Answer:
WOW! that does not look easy!
Explanation:
I wish i could help but i have no idea how to do that lol
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the customer's loan balance each month until the loan is paid off. b. Modify the Bob's E-Z Loans application so that after the payment is made each month, a finance charge of 1 percent is added to the balance.
Answer:
part (a).
The program in cpp is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>0)
{
if(balance<payment)
{ payment = balance;}
else
{
std::cout << balance <<"\t\t\t"<< payment << std::endl;
balance = balance - payment;
}
}
return 0;
}
part (b).
The modified program from part (a), is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
double charge=0.01;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
balance = balance +( balance*charge );
std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>payment)
{
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
balance = balance +( balance*charge );
balance = balance - payment;
}
if(balance<payment)
{ payment = balance;}
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
return 0;
}
Explanation:
1. The variables to hold the loan balance and the monthly payment are declared as double.
2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.
3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.
4. The computed values are displayed for each month till the loan balance becomes zero.
5. The output for both part (a) and part (b) are attached as images.
"Write pseudocode that outputs the contents of parallel arrays. You do NOT have to write the entire program. The first array will hold phone numbers The second array will hold company names You do NOT need to load the arrays. Your code can assume they are already loaded with data. The arrays are named phone[ ] and company[ ] Output the phone number and associated company name for every entry in the array."
Answer:
Check the explanation
Explanation:
PSEUDO CODE:
for(int i=0,j=0;i<phone.length,j<company.length;i++,j++)
{
print(phone[i]," - ",company[j]);
}
this for loop will print each phone number associated with the company names.
write the steps to insert picture water mark
Answer:
Is there a picture of the question?
Explanation:
In this question, we give two implementations for the function: def intersection_list(lst1, lst2) This function is given two lists of integers lst1 and lst2. When called, it will create and return a list containing all the elements that appear in both lists. For example, the call: intersection_list([3, 9, 2, 7, 1], [4, 1, 8, 2])could create and return the list [2, 1]. Note: You may assume that each list does not contain duplicate items. a) Give an implementation for intersection_list with the best worst-case runtime. b) Give an implementation for intersection_list with the best average-case runtime.
Answer:
see explaination
Explanation:
a)Worst Case-time complexity=O(n)
def intersection_list(lst1, lst2):
lst3 = [value for value in lst1 if value in lst2]
return lst3
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele) # adding the element
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele) # adding the element
print(intersection_list(lst1, lst2))
b)Average case-time complexity=O(min(len(lst1), len(lst2))
def intersection_list(lst1, lst2):
return list(set(lst1) & set(lst2))
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele)
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele)
print(intersection_list(lst1, lst2))
Which of the following should be the first page of a report?
O Title page
Introduction
O Table of contents
Terms of reference
Answer:
Title page should be the first page of a report.
hope it helps!
Caches are important to providing a high-performance memory hierarchy to processors. Below is a list of 32-bits memory address references given as word addresses. 0x03, 0xb4, 0x2b, 0x02, 0xbf, 0x58, 0xbe, 0x0e, 0xb5, 0x2c, 0xba, 0xfd For each of these references identify the binary word address, the tag, and the index given a direct mapped cache with 16 one-word blocks. Also list whether each reference is a hit or a miss, assuming the cache is initially empty.
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem.
Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument. Use conditional operators to allow the macro to adapt to the size of the desired result. Write a program that tests the macro, passing it operands of various sizes.
Answer:
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
Explanation:
For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
6. Why did he choose to install the window not totally plumb?
Answer:
Because then it would break
Explanation:
You achieve this by obtaining correct measurements. When measuring a window, plumb refers to the vertical planes, and level refers to the horizontal planes. So he did not install the window totally plumb
You can use this area to create your resume.
Answer:
YOUR NAME
YOUR CITY, STATE, AND ZIP CODE
YOUR PHONE NUMBER
YOUR EMAIL
Professional Summary
Reliable, top-notch sales associate with outstanding customer service skills and relationship-building strengths. Dedicated to welcoming customers and providing comprehensive service. In-depth understanding of sales strategy and merchandising techniques. Dependable retail sales professional with experience in dynamic, high-performance environments. Skilled in processing transactions, handling cash, using registers and arranging merchandise. Maintains high-level customer satisfaction by smoothly resolving customer requests, needs and problems. Seasoned Sales Associate with consistent record of exceeding quotas in sales environments. Delivers exceptional customer service and product expertise to drive customer satisfaction ratings. Proficient in use and troubleshooting of POS systems.
Skills
Returns and Exchanges
Adaptable and Flexible
Excellent Written and Verbal Communication
Meeting Sales Goals
Strong Communication and Interpersonal Skills
Time Management
Cash Handling
Reliable and Responsible
Work History
March 2020 to September 2021
Goodwill OF YOUR STATE
Retail Sales Associate
Helped customers complete purchases, locate items and join reward programs.
Checked pricing, scanned items, applied discounts and printed receipts to ring up customers.
Folded and arranged merchandise in attractive displays to drive sales.
Greeted customers and helped with product questions, selections and purchases.
Organized store merchandise racks and displays to promote and maintain visually appealing environments.
Monitored sales floor and merchandise displays for presentable condition, taking corrective action such as restocking or reorganizing products.
Balanced and organized cash register by handling cash, counting change and storing coupons.
Trained new associates on cash register operations, conducting customer transactions and balancing drawer.
Answered questions about store policies and addressed customer concerns.
Issued receipts and processed refunds, credits or exchanges.
Maintained clean sales floor and straightened and faced merchandise.
Education
YOUR HIGH SCHOOL THAT YOU ATTEND
Languages
Spanish
Explanation:
THIS IS MY RESUME IF YOU HAVE MORE WORK EXPERIENCE THEN ADD IT AFTER THE GOODWILL. I GOT A 100% ON EDGE.
Write a statement that calls the recursive method backwardsAlphabet() with parameter startingLetter.
import java.util.Scanner; public class RecursiveCalls { public static void backwardsAlphabet(char currLetter) { if (currLetter == 'a') { System.out.println(currLetter); } else { System.out.print(currLetter + " "); backwardsAlphabet((char)(currLetter - 1)); } } public static void main (String [] args) { Scanner scnr = new Scanner(System.in); char startingLetter; startingLetter = scnr.next().charAt(0); /* Your solution goes here */ } }
Answer:
Following are the code to method calling
backwardsAlphabet(startingLetter); //calling method backwardsAlphabet
Output:
please find the attachment.
Explanation:
Working of program:
In the given java code, a class "RecursiveCalls" is declared, inside the class, a method that is "backwardsAlphabet" is defined, this method accepts a char parameter that is "currLetter". In this method a conditional statement is used, if the block it will check input parameter value is 'a', then it will print value, otherwise, it will go to else section in this block it will use the recursive function that prints it's before value. In the main method, first, we create the scanner class object then defined a char variable "startingLetter", in this we input from the user and pass its value into the method that is "backwardsAlphabet".Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a clock rate of 750 MHz. We have a particular program we wish to run. When compiled for computer A, this program has exactly 100,000 instructions. How many instructions would the program need to have when compiled for Computer B, in order for the two computers to have exactly the same execution time for this program
Answer:
Check the explanation
Explanation:
CPI means Clock cycle per Instruction
given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M
Execution time is given by following Formula.
Execution Time(CPU time) = CPI*Instruction Count * clock time = [tex]\frac{CPI*Instruction Count}{ClockRate}[/tex]
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is [tex]=\frac{2.5*100,000}{750*10^6}[/tex]
= 333.33 micro sec
c) Since the system B is slower than system A, So the system A executes the given program in less time
Hence take CPU execution time of system B as CPU time of System A.
therefore
216.67 micro = =[tex]\frac{2.5*Instruction}{750*10^6}[/tex]
Instructions = 216.67*750/2.5
= 65001
hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A
The number of instructions that the program would need to have when compiled for Computer B is; 65000 instructions
What is the execution time?
Formula for Execution time is;
Execution time = (CPI × Instruction Count)/Clock Time
We are given;
CPI for computer A = 1.3
Instruction Count = 100000
Clock time = 600 MHz = 600 × 10⁶ Hz
Thus;
Execution time = (1.3 * 100000)/(600 × 10⁶)
Execution time(CPU Time) = 216.67 * 10⁻⁶ second
For CPU B;
CPU Time = (2.5 * 100000)/(750 × 10⁶)
CPU Time = 333.33 * 10⁻⁶ seconds
Thus, instructions for computer B for the two computers to have same execution time is;
216.67 * 10⁻⁶ = (2.5 * I)/(750 × 10⁶)
I = (216.67 * 10⁻⁶ * 750 × 10⁶)/2.5
I = 65000 instructions
Read more about programming instructions at; https://brainly.com/question/15683939
