Who is your favorite smite god in Hi-Rez’s “Smite”

Answer:

1. Logical

2.=

3.IF

Explanation:

just did the assignment

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.

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.

Answer:

Title page should be the first page of a report.

hope it helps!

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);

Answer:

Check the explanation

Explanation:

SDLC

SDLC stands for Software Development Life Cycle

It provides steps for developing a software product

It also checks for the quality and correctness of a software

The main aim of SDLC is to develop a software, which meets a customer requirements

SDLC involves various phases to develop a high-quality product for its end user

Phases of SDLC

           There are seven phases in a software development life cycle as follows,

Requirement Analysis   Feasibility Study   Design    Coding    Testing Deployment      Maintenance

kindly check the attached image below

Requirement analysis

This is the first stage in a SDLC.

In this phase, a detailed and precise requirements are collected from various teams by senior team members

It gives a clear idea of the entire project scope, opportunities and upcoming issues in a project

This phase also involves, planning assurance requirements and analyze risk involved in the project

Here, the timeline to finish the project is finalized

Feasibility Study

In this stage, the SRS document, that is “Software Requirement Specification” document is defined, which includes everything to be designed and developed during the life cycle of a project

Design

In this phase as the name indicates, software and system design documents are prepared, based on the requirement specification document

This enable us to define the whole system architecture

Two types of design documents are prepared in this phase as follows,

High-Level Design

Low-Level Design

Coding

In this phase, the coding of project is done in a selected programming language

Tasks are divided into many units and given to various developers

It is the longest phase in the SDLC

Testing

In this stage, the developed coding is tested by the testing team

The testing team, checks the coding in the user point of view

Installation/Deployment

In this stage, a product is released by the company to the client

Maintenance

Once the product is delivered to the client, the maintenance phase work will start by fixing errors, modifying and enhancing the product.

Employee Payroll System

Requirement analysis

Gather information from various team for requirement

Analyze the number of employees to handle the project

Timeline to finish the project, eg.1 month

Feasibility Study

The system requirements such as how many systems are required

Decide which software to be installed to handle the project.

For example, if we are going to develop the software using C language, then software for that has to be installed

The SRS document has to be prepared

Design

In this the HLD and LLD is prepared, the overall software architecture is prepared

The modules involved in coding are decided

Here the modules can be employee, payroll manager, Employee Log time, account details

The input and output are designed, such as employee name is the input and output is the salary for him, based on working hours

Coding

Here the coding is divided into various sections and given to 2 or more employees

One may code employee detail, one will code working hours of employees and one may code the banking details of employee

Testing

The coding is tested for syntax, declaration, data types and various kinds of error

The testing team will test the coding with various possible values

They may even check with wrong values and analyze the output for that

Installation

Now the software is installed in the client system and the client is calculating the payroll for an employee based on working hours in a month

Maintenance

If any error occurs, the team will clear the issue.

When a new employee joins, then the employee data will added to the database and the database is updated

Also if the client asks for any new features, it will done in this phase.

Answer:

See explaination

Explanation:

//ReadFile.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class ReadFile

{

public static void main(String[] args)

{

int elementsCount=0;

//Create a File class object

File file=null;

Scanner fileScanner=null;

String fileName="sample.txt";

try

{

//Create an instance of File class

file=new File(fileName);

//create a scanner class object

fileScanner=new Scanner(file);

//read file elements until end of file

while (fileScanner.hasNext())

{

double value=fileScanner.nextInt();

elementsCount++;

}

//print smallest value

System.out.println(elementsCount+" data values are read");

}

//Catch the exception if file not found

catch (FileNotFoundException e)

{

System.out.println("File Not Found");

}

}

}

Sample output:

sample.txt

10

20

30

40

50

output:

5 data values are read

Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.

Explanation:

Answer:

Bits

Explanation:

Answer:

Following are the code to method calling

backwardsAlphabet(startingLetter); //calling method backwardsAlphabet

Output:

please find the attachment.

Explanation:

Working of program:

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

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

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))

Answer:

I don't remember much on this stuff but I think it was B

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:

Answer:

The java program for the given scenario is as follows.

import java.util.*;

//interface with method area

interface ObjectWithTwoParameters

{

double area (double d1, double d2);

}

class RectangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return d1*d2;

}

}

class TriangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return (d1*d2)/2;

}  

}

class CylinderClass implements ObjectWithTwoParameters

{

public double area (double d1, double d2)

{

return ( 2*3.14*d1*d1 + d2*(2*3.14*d1) );

}

}

public class Test

{

public static void main(String[] args)

{

//area displayed for all three shapes

ObjectWithTwoParameters r = new RectangleClass();

double arear = r.area(2, 3);

System.out.println("Area of rectangle: "+arear);

ObjectWithTwoParameters t = new TriangleClass();

double areat = t.area(4,5);

System.out.println("Area of triangle: "+areat);

ObjectWithTwoParameters c = new CylinderClass();

double areac = c.area(6,7);

System.out.println("Area of cylinder: "+areac);

}

}

OUTPUT

Area of rectangle: 6.0

Area of triangle: 10.0

Area of cylinder: 489.84

Explanation:

1. The program fulfils all the mentioned requirements.

2. The program contains one interface, ObjectWithTwoParameters, three classes which implement that interface, RectangleClass, TriangleClass and CylinderClass, and one demo class, Test, containing the main method.

3. The method in the interface has no access specifier.

4. The overridden methods in the three classes have public access specifier.

5. No additional variables have been declared.

6. The test class having the main() method is declared public.

7. The area of the rectangle, triangle and the cylinder have been computed as per the respective formulae.

8. The interface is similar to a class which can have only declarations of both, variables and methods. No method can be defined inside an interface.

9. The other classes use the methods of the interface by implementing the interface using the keyword, implements.

10. The object is created using the name of the interface as shown.

ObjectWithTwoParameters r = new RectangleClass();

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;

}

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.

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

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.

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.

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

Answer:

A. 0.0450

B. 4

C. 0.25

D. 37.68

E. 6Hz

F. -0.523

G. 1.5m/s

H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

I. -1.67m/s.

Explanation:

Given the equation:

y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)

Standard wave equation:

y(x, t)=Asin(kx−ωt+ϕ)

a.) Amplitude = 0.0450

b.) Wave number = 1/ λ

λ=2π/k

From the equation k = 25.12

Wavelength(λ ) = 2π/25.12 = 0.25

Wave number (1/0.25) = 4

c.) Wavelength(λ ) = 2π/25.12 = 0.25

d.) Angular frequency(ω)

ωt = 37.68t

ω = 37.68

E.) Frequency (f)

ω = 2πf

f = ω/2π

f = 37.68/6.28

f = 6Hz

f.) Phase angle(ϕ) = -0.523

g.) Wave propagation speed :

ω/k=37.68/25.12=1.5m/s

h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.

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:

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

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

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

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.