Answer:
A 512 GB Solid State Drive (SSD) will be recommended
Explanation:
Recommended hard disk for the installation of Microsoft PowerPoint application is 3 GB and since the computer is a new one it will be best to buy a hard disc with enough room for expansion, performance speed, durability and reliability
Therefore, a 512 GB Solid State Drive (SSD) is recommended as the price difference is small compared to the spinning hard drive and also there is ample space to store PowerPoint training presentation items locally.
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
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))
Write a program that maintains a database containing data, such as name and birthday, about your friends and relatives. You should be able to enter, remove, modify, or search this data. Initially, you can assume that the names are unique. The program should be able to save the data in a fi le for use later. Design a class to represent the database and another class to represent the people. Use a binary search tree of people as a data member of the database class. You can enhance this problem by adding an operation that lists everyone who satisfi es a given criterion. For example, you could list people born in a given month. You should also be able to list everyone in the database.
Answer:
[tex]5909? \times \frac{?}{?} 10100010 {?}^{?} 00010.222 {?}^{2} [/tex]
For this problem, you will write a function standard_deviation that takes a list whose elements are numbers (they may be floats or ints), and returns their standard deviation, a single number. You may call the variance method defined above (which makes this problem easy), and you may use sqrt from the math library, which we have already imported for you. Passing an empty list to standard_deviation should result in a ZeroDivisionError exception being raised, although you should not have to explicitly raise it yourself.
Answer:
import math def standard_deviation(aList): sum = 0 for x in aList: sum += x mean = sum / float(len(aList)) sumDe = 0 for x in aList: sumDe += (x - mean) * (x - mean) variance = sumDe / float(len(aList)) SD = math.sqrt(variance) return SD print(standard_deviation([3,6, 7, 9, 12, 17]))Explanation:
The solution code is written in Python 3.
Firstly, we need to import math module (Line 1).
Next, create a function standard_deviation that takes one input parameter, which is a list (Line 3). In the function, calculate the mean for the value in the input list (Line 4-8). Next, use the mean to calculate the variance (Line 10-15). Next, use sqrt method from math module to get the square root of variance and this will result in standard deviation (Line 16). At last, return the standard deviation (Line 18).
We can test the function using a sample list (Line 20) and we shall get 4.509249752822894
If we pass an empty list, a ZeroDivisionError exception will be raised.
The Monte Carlo (MC) Method (Monte Carlo Simulation) was first published in 1949 by Nicholas Metropolis and Stanislaw Ulam in the work "The Monte Carlo Method" in the Journal of American Statistics Association. The name Monte Carlo has its origins in the fact that Ulam had an uncle who regularly gambled at the Monte Carlo casino in Monaco. In fact, way before 1949 the method had already been extensively used as a secret project of the U.S. Defense Department during the so-called "Manhattan Project". The basic principle of the Monte Carlo Method is to implement on a computer the Strong Law of Large Numbers (SLLN) (see also Lecture 9). The Monte Carlo Method is also typically used as a probabilistic method to numerically compute an approximation of a quantity that is very hard or even impossible to compute exactly like, e.g., integrals (in particular, integrals in very high dimensions!). The goal of Problem 2 is to write a Python code to estimate the irrational number
Answer:
can you give more detail
Explanation:
Let U = {b1, b2, , bn} with n ≥ 3. Interpret the following algorithm in the context of urn problems. for i is in {1, 2, , n} do for j is in {i + 1, i + 2, , n} do for k is in {j + 1, j + 2, ..., n} do print bi, bj, bk How many lines does it print? It prints all the possible ways to draw three balls in sequence, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints C(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints C(n, 3) lines.
Answer:
Check the explanation
Explanation:
Kindly check the attached image for the first step
Note that the -print" statement executes n(n — I)(n — 2) times and the index values for i, j, and k can never be the same.
Therefore, the algorithm prints out all the possible ways to draw three balls in sequence, without replacement.
Now we need to determine the number of lines this the algorithm print. In this case, we are selecting three different balls randomly from a set of n balls. So, this involves permutation.
Therefore, the algorithm prints the total
P(n, 3)
lines.
Explain possible ways that Darla can communicate with her coworker Terry, or her manager to make sure Joe receives great customer service?
Answer:
They can communicate over the phone or have meetings describing what is and isn't working for Joe. It's also very important that Darla makes eye contact and is actively listening to effectively handle their customer.
Explanation:
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
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;
}
Create a classRationalNumber(fractions, see problem 11.15 p.639) with the following capabilities:a) Create a constructor that prevents 0 on the denominator in a fraction, reduces (or simplifies) fractionsthat are not in reduced form (lowest terms), and avoids negative denominatorsb) Overload the addition, subtraction, multiplication, and division operators for this class.c) Overload the relationaland equality == operators for this class.Test all member functions and overloaded operators.
Answer:
See explaination
Explanation:
#include<iostream>
using namespace std;
class a
{
int i,n,d,sn,sd,cond;
public:
void getdata()
{
cout<<"Enter the numerator"<<endl;
cin>>n;
cout<<"Enter the denominator"<<endl;
cin>>d;
}
void operator+(a a1)
{
if(a1.d!=d)
{
sn=(a1.n*d)+(n*a1.d);
sd=(a1.d)*(d);
}
else
{
sn=a1.n+n;
sd=a1.d;
}
if(sn>sd)
{
cond=sd;
}
else if(sd>sn)
{
cond=sn;
}
}
void sum()
{
for(i=2;i<=cond;i++)
{
if(sn%i==0 && sd%i==0)
{
sn=sn/i;
sd=sd/i;
}
}
cout<<"Sum is "<<sn<<"/"<<sd<<endl;
}
a operator-(a a1)
{
a a3;
if(a1.d!=d)
{
a3.sn=(a1.n*d)-(n*a1.d);
a3.sd=(a1.d)*(d);
}
else
{
a3.sn=a1.n-n;
a3.sd=a1.d;
}
return a3;
}
void sub()
{
if(sn>sd)
{
cond=sd;
}
else if(sd>sn)
{
cond=sn;
}
for(i=2;i<=cond;i++)
{
if(sn%i==0 && sd%i==0)
{
sn=sn/i;
sd=sd/i;
}
}
cout<<"Subtraction is "<<sn<<"/"<<sd<<endl;
}
a operator /(a a1)
{
a a4;
a4.sn=a1.n*d;
a4.sd=a1.d*n;
return a4;
}
void div()
{
if(sn>sd)
{
cond=sd;
}
else if(sd>sn)
{
cond=sn;
}
else if(sn==sd)
{
sn=1;
sd=1;
}
for(i=2;i<=cond;i++)
{
if(sn%i==0 && sd%i==0)
{
sn=sn/i;
sd=sd/i;
}
}
cout<<"Division is "<<sn<<"/"<<sd<<endl;
}
a operator*(a a1)
{
a a5;
a5.sn=a1.n*n;
a5.sd=a1.d*d;
return a5;
}
void mul()
{
if(sn>sd)
{
cond=sd;
}
else if(sd>sn)
{
cond=sn;
}
for(i=2;i<=cond;i++)
{
if(sn%i==0 && sd%i==0)
{
sn=sn/i;
sd=sd/i;
}
}
cout<<"Multiplication - "<<sn<<"/"<<sd<<endl;
}
};
int main()
{
a a1,a2,a3,a4,a5,a6;
a1.getdata();
a2.getdata();
a2+(a1);
a2.sum();
a3=a2-(a1);
a3.sub();
a4=a2/(a1);
a4.div();
a5=a2*(a1);
a5.mul();
return 0;
}
*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.
A university wants to schedule the classrooms for final exams. The attributes are given below:
course : course id, name, and department
section : section id, enrollment, and dependent as a weak entity set on course
room : room number, capacity, and building
exam: course id, section id, room number, and time.
1) Determine the relationships among these entity sets. Clearly specify the arity, the attributes, and the mapping cardinality of these relationships.
2) Construct an E-R diagram to for the scheduling system.
Answer:
See explaination
Explanation:
Clearly the entities are as follows:-
Course
Section
Room
Test
The relationship among entities are as follows:-
Course has Section
Test is conducted for Course
Test is conducted in Section
Test is conducted in Room
Attributes of each entities are as follows:-
Course (CourseID, Name, Department)
Section (SectionID, Enrollment)
Room (RoomNumber, Capacity, Building)
Test (Time)
Section is a week entity as, there may be same sections for different courses, therefore section uses the primary key of course entity as foreign key.
Also entity Test is dependent upon the entities Room,Section and Course, therefore primary keys of these entities will be used as foreign key in the Test entity.
Check attachment for the ER diagram
Write a program that uses the map template class to compute a histogram of positive numbers entered by the user. The map’s key should be the number that is entered, and the value should be a counter of the number of times the key has been entered so far. Use –1 as a sentinel value to signal the end of user input.512355321-1Then the program should output the followings:The number 3 occurs 2 timesThe number 5 occurs 3 timesThe number 12 occurs 1 timesThe
Answer:
See explaination
Explanation:
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <iomanip>
using namespace std;
int main()
{
map<int, int> histogram;
int a=0;
while(a>=0)
{
cout << " enter value of a" ;
cin >>a;
histogram[a]++;
}
map<int,int>::iterator it;
for ( it=histogram.begin() ; it != histogram.end(); it++ )
{
cout << (*it).first << " occurs " << setw(3) << (*it).second << (((*it).second == 1) ? " time." : " times.") <<endl;
}
return 0;
}
Trace the complete execution of the MergeSort algorithm when called on the array of integers, numbers, below. Show the resulting sub-arrays formed after each call to merge by enclosing them in { }. For example, if you originally had an array of 5 elements, a = {5,2,8,3,7}, the first call to merge would result with: {2, 5} 8, 3, 7 ← Note after the first call to merge, two arrays of size 1 have been merged into the sorted subarray {2,5} and the values 2 and 5 are sorted in array a You are to do this trace for the array, numbers, below. Be sure to show the resulting sub-arrays after each call to MergeSort. int[] numbers = {23, 14, 3, 56, 17, 8, 42, 18, 5};
Answer:
public class Main {
public static void merge(int[] arr, int l, int m, int r) {
int n1 = m - l + 1;
int n2 = r - m;
int[] L = new int[n1];
int[] R = new int[n2];
for (int i = 0; i < n1; ++i)
L[i] = arr[l + i];
for (int j = 0; j < n2; ++j)
R[j] = arr[m + 1 + j];
int i = 0, j = 0;
int k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
printArray(arr, l, r);
}
public static void sort(int[] arr, int l, int r) {
if (l < r) {
int m = (l + r) / 2;
sort(arr, l, m);
sort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
static void printArray(int[] arr, int l, int r) {
System.out.print("{");
for (int i = l; i <= r; ++i)
System.out.print(arr[i] + " ");
System.out.println("}");
}
public static void main(String[] args) {
int[] arr = {23, 14, 3, 56, 17, 8, 42, 18, 5};
sort(arr, 0, arr.length - 1);
}
}
Explanation:
See answer
Suppose that a program's data and executable code require 1,024 bytes of memory. A new section of code must be added; it will be used with various values 70 times during the execution of a program. When implemented as a macro, the macro code requires 73 bytes of memory. When implemented as a procedure, the procedure code requires 132 bytes (including parameter-passing, etc.), and each procedure call requires 7 bytes. How many bytes of memory will the entire program require if the new code is added as a procedure? 1,646
Answer:
The answer is 1646
Explanation:
The original code requires 1024 bytes and is called 70 times ,it requires 7 byte and its size is 132 bytes
1024 + (70*7) + 132 = 1024 + 490 +132
= 1646
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
Design a program for the Hollywood Movie Rating Guide, which can be installed in a kiosk in theaters. Each theater patron enters a value from 0 to 4 indicating the number of stars that the patron awards to the Rating Guide's featured movie of the week. If a user enters a star value that does not fall in the correct range, prompt the user continuously until a correct value is entered. The program executes continuously until the theater manager enters a negative number to quit. At the end of the program, display the average star rating for the movie.
Answer:
The program in cpp is given below.
#include <iostream>
using namespace std;
int main()
{
//variables declared to hold sum, count and average of movie ratings
int sum=0;
int count=0;
int rating;
double avg;
//audience is prompted to enter ratings
do{
std::cout << "Enter the rating (0 to 4): ";
cin>>rating;
if(rating>4)
std::cout << "Invalid rating. Please enter correct rating again." << std::endl;
if(rating>=0 && rating<=4)
{
sum=sum+rating;
count++;
}
if(rating<0)
break;
}while(rating>=0);
//average rating is computed
avg=sum/count;
std::cout << "The average rating for the movie is " << avg << std::endl;
return 0;
}
OUTPUT
Enter the rating (0 to 4): 1
Enter the rating (0 to 4): 2
Enter the rating (0 to 4): 0
Enter the rating (0 to 4): 5
Invalid rating. Please enter correct rating again.
Enter the rating (0 to 4): 1
Enter the rating (0 to 4): -1
The average rating for the movie is 1
Explanation:
1. Variables are declared to hold the rating, count of rating, sum of all ratings and average of all the user entered ratings.
2. The variables for sum and count are initialized to 0.
3. Inside do-while loop, user is prompted to enter the rating for the movie.
4. In case the user enters a rating which is out of range, the user is prompted again to enter a rating anywhere from 0 to 4.
5. The loop executes until a negative number is entered.
6. Inside the loop, the running total sum of all the valid ratings is computed and the count variable is incremented by 1 for each valid rating.
7. Outside the loop, the average is computed by dividing the total rating by count of ratings.
8. The average rating is then displayed to the console.
9. The program is written in cpp due to simplicity since all the code is written inside main().
Assume the class Student implements the Speaker interface from the textbook. Recall that this interface includes two abstract methods, speak( ) and announce(String str). A Student contains one instance data, String class Rank. Write the Student class so that it implements Speaker as follows. The speak method will output "I am a newbie here" if the Student is a "Freshman", "I like my school" if the Student is either a "Sophomore" or a "Junior", or "I can not wait to graduate" if the student is a "Senior". The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line. Finally, the class Rank is initialized in the constructor. Only implement the constructor and the methods to implement the Speaker interface.
Answer:
Check the explanation
Explanation:
Based on the above information, this the following resulting code.
The solution has namely one interface named Speaker, one class called Student and one Main class called StudentMain.
The StudentMain class just initialize various instances of the Student class with various classRank values and then subsequently calling in the speak() methods of all.
announce() method is called on one such instance of Student class to demonstrate its functioning.
Comments are added at required place to better help in understanding the logic.
Speaker.java
public interface Speaker {
abstract void speak();
abstract void announce(String str);
}
Student.java
public class Student implements Speaker {
/*
* Write the Student class so that it implements Speaker as follows.
The speak method will output "I am a newbie here" if the Student is a "Freshman",
"I like my school" if the Student is either a "Sophomore" or a "Junior",
or "I can not wait to graduate" if the student is a "Senior".
The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line.
*/
String classRank;
// Based on the requirement taking in classRank String as a constructor parameter
public Student(String classRank) {
super();
this.classRank = classRank;
}
// Using the switch-case to code the above requirment
public void speak() {
switch(classRank) {
case "Freshman":
System.out.println("I am a newbie here");
break;
// This case would output the same result for the 2 case conditions
case "Sophomore":
case "Junior":
System.out.println("I like my school");
break;
case "Senior":
System.out.println("I can not wait to graduate");
break;
default:
System.out.println("Unknown classRank inputted");
}
}
//Based on the requirement, first line is printed first after second which is
// the inputted String parameter
public void announce(String str) {
System.out.println("I am a Student, here is what I have to say");
System.out.println(str);
}
}
StudentMain.java
public class StudentMain {
public static void main(String[] args) {
// Initializing all the Objects with required classRank
Student stu1=new Student("Freshman");
Student stu2=new Student("Sophomore");
Student stu3=new Student("Junior");
Student stu4=new Student("Senior");
Student stu5=new Student("Freshman");
//Calling the speak methods of the all the above create objects
stu1.speak();
stu2.speak();
stu3.speak();
stu4.speak();
stu4.speak();
stu4.announce("Wish u all the best");
}
}
Following is the output generated from the code run.
Perform a Google search, using the Google hacking keywords to locate a page that has the word passwd (note the spelling) on the page but has the phrase "Index of" in the title. First, what was your search term. Second, what is the first URL in your list of results. Third, what exactly would you use this search to find?
Answer:
Following are the step to this question:
i) search "intitle passwd" in google search engine.
ii) Open first link of the page.
iii) use commands to access data.
Explanation:
Google Hacking operates as follows of web engines like google and yahoo to find web pages and sensitive information, that are insecure. It builds on the idea that web pages index a large number of public pages and files and make their discovery easy to create the correct query.
In which, we search the above-given keyword, that will provide many links. In these links we select the first one and open it, it will provide an index, and search in this we use the search command in the database.Describe how DMA is
used to
transfer data
from peripherals.
Answer:
Think of DMA as a co-processor that is used to quickly transfer data between main memory and peripherals without the intervention of the CPU.
Write a program with total change amount as an integer input that outputs the change using the fewest coins, one coin type per line. The coin types are dollars, quarters, dimes, nickels, and pennies. Use singular and plural coin names as appropriate, like 1 penny vs. 2 pennies. Ex: If the input is: 0 or less, the output is: no change Ex: If the input is: 45 the output is: 1 quarter 2 dimes
Answer:.
// Program is written in C++.
// Comments are used for explanatory purposes
// Program starts here..
#include<iostream>
using namespace std;
int main()
{
// Declare Variables
int amount, dollar, quarter, dime, nickel, penny;
// Prompt user for input
cout<<"Amount: ";
cin>>amount;
// Check if input is less than 1
if(amount<=0)
{
cout<<"No Change";
}
else
{
// Convert amount to various coins
dollar = amount/100;
amount = amount%100;
quarter = amount/25;
amount = amount%25;
dime = amount/10;
amount = amount%10;
nickel = amount/5;
penny = amount%5;
// Print results
if(dollar>=1)
{
if(dollar == 1)
{
cout<<dollar<<" dollar\n";
}
else
{
cout<<dollar<<" dollars\n";
}
}
if(quarter>=1)
{
if(quarter== 1)
{
cout<<quarter<<" quarter\n";
}
else
{
cout<<quarter<<" quarters\n";
}
}
if(dime>=1)
{
if(dime == 1)
{
cout<<dime<<" dime\n";
}
else
{
cout<<dime<<" dimes\n";
}
}
if(nickel>=1)
{
if(nickel == 1)
{
cout<<nickel<<" nickel\n";
}
else
{
cout<<nickel<<" nickels\n";
}
}
if(penny>=1)
{
if(penny == 1)
{
cout<<penny<<" penny\n";
}
else
{
cout<<penny<<" pennies\n";
}
}
}
return 0;
}
In this puzzle you have a list of n pitchers, with known capacities, c1, c2, ... cn, of holding water in gallons. You have a faucet that can be used as often and as much as the player needs. The goal is to measure exactly g gallons of water, using nothing other than these pitchers. Suppose the current amounts of water that these pitchers hold is w1, w2, ... wn,. These numbers are assumed to be 0 initially. The puzzle is solved as soon as wi = g for any i. A player can perform the following operations:
Fill pitcher i (from the faucet). The precondition of this operation is: ci > wi ≥ 0. The effect is: wi = ci.
Empty pitcher i. The precondition of this operation is: ci ≥ wi > 0. The effect is: wi = 0.
Pour pitcher i to pitcher j. The precondition of this operation is: (ci ≥ wi > 0) and (cj > wj ≥ 0). In words, pitcher i must have some water to pour, and pitcher j must have some unused capacity to receive it. The (partial) effect is: (wi = 0) or (wj = cj) or both. In words, the pour operation must continue until pitcher i becomes empty (and its content is added to pitcher j's content), or pitcher j becomes full (and pitcher i retains the remainder), whichever occurs first. They may occur simultaneously.
A Sample Run
Select the puzzle to solve:
1. Pitchers
2. Eight puzzle
Your selection: 1
Enter the number of pitchers: 3
Enter the capacities of the 3 pitchers (gallons): 2, 5, 10
Enter the goal (gallons): 1
Current configuration: [0, 0, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 2
3. Fill pitcher 3
Your selection: 2
Current configuration: [0, 5, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 3
3. Empty pitcher 2
4. Pour pitcher 2 to 1
5. Pour pitcher 2 to 3
Your selection: 4
Current configuration: [2, 3, 0]
Please select your next move from the following choices:
1. Fill pitcher 2
2. Fill pitcher 3
3. Empty pitcher 1
4. Empty pitcher 2
5. Pour pitcher 1 to 2
6. Pour pitcher 1 to 3
7. Pour pitcher 2 to 3
Your selection: 3 Current configuration: [0, 3, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 2
3. Fill pitcher 3
4. Empty pitcher 2
5. Pour pitcher 2 to 1
6. Pour pitcher 2 to 3
Your selection: 5 Current configuration: [2, 1, 0]
Great! You have reached the goal in 4 moves. Bye.
Answer:
See explaination
Explanation:
#include <iostream>
#include <vector>
using namespace std;
// function to print configuration for each pitcher
void printConfig(vector<int> w, int n)
{
cout << "Current configuration: [" << w[0];
for(int i=1; i<n; i++)
{
cout << ", " << w[i];
}
cout << "]" << endl;
}
void getOptions(vector<int> c, vector<int> w, int n, vector<pair<int, pair<int, int>>> &options)
{
// options count
int count = 0;
// fill
for(int i=0; i<n; i++)
{
if(c[i] > w[i] && w[i] >= 0)
{
count ++;
cout << count << ". Fill pitcher " << (i+1) << endl;
// add options to list
options.push_back(make_pair(1, make_pair(i, 0)));
}
}
// empty
for(int i=0; i<n; i++)
{
if(c[i] >= w[i] && w[i] > 0)
{
count++;
cout << count << ". Empty pitcher " << (i+1) << endl;
// add options to list
options.push_back(make_pair(2, make_pair(i, 0)));
}
}
// Pour pitcher i to pitcher j
for(int i=0; i<n; i++)
{
if(c[i] >= w[i] && w[i] > 0)
{
for(int j=0; j<n; j++)
{
if(i!=j && c[j] > w[j] && w[j] >= 0)
{
count++;
cout << count << ". Pour pitcher " << (i+1) << " to " << (j+1) << endl;
// add options to list
options.push_back(make_pair(3, make_pair(i, j)));
}
}
}
}
}
void execute(vector<int> c, vector<int> &w, pair<int, pair<int, int>> option)
{
int i = option.second.first;
// for fill
if(option.first == 1)
{
w[i] = c[i];
}
// empty
else if(option.first == 2)
{
w[i] = 0;
}
// pour
else
{
int j = option.second.second;
if(w[i] >= c[j])
{
w[i] = w[i] - c[j];
w[j] = c[j];
}
else
{
w[j] = w[i];
w[i] = 0;
}
}
}
int main()
{
// required variables
int choice, n, temp, g, flag, moves = 0;
// vectors are dynamic sized arrays
vector<int> c, w;
vector<pair<int, pair<int, int>>> options;
// select puzzle
cout << "Select the puzzle to solve:\n";
cout << "1. Pitchers\n";
cout << "2. Eight puzzle\n";
cout << "Your selection: ";
cin >> choice;
if(choice == 1)
{
// input values required
cout << "Enter the number of pitchers: ";
cin >> n;
// array for pitchers
cout << "Enter the capacities of the " << n << " pitchers (gallons): ";
for(int i=0; i<n; i++)
{
cin >> temp;
c.push_back(temp);
w.push_back(0);
}
cout << "Enter the goal (gallons): ";
cin >> g;
// start game
while(true)
{
// print configuration
printConfig(w, n);
// check for goal state
flag = 0;
for(int i=0; i<n; i++)
{
if(w[i] == g)
{
flag = 1;
break;
}
}
if(flag)
{
cout << "Great! You have reached the goal in " << moves << " moves. Bye." << endl;
break;
}
// get and print options
//empty previous options
options.clear();
getOptions(c, w, n, options);
// ask for user's selection
cout << "Your selection: ";
cin >> choice;
// update moves
moves++;
// perform according to the selection
execute(c, w, options[choice-1]);
}
}
return 0;
}
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.
Who is your favorite smite god in Hi-Rez’s “Smite”
Answer:
Variety
Explanation:
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?
Explanation:
The most reliable hard drives are those whose most common size is 3.5 inches, their advantages are their storage capacity and their speed and a disadvantage is that they usually make more noise.
For greater speed, it is ideal to opt for two smaller hard disks, since large disks are slower, but are partitioned so that there are no problems with file loss.
For a person who needs to use content creation programs, it is ideal to opt for a hard drive that has reliability.
Write a functionvector merge(vector a, vector b)that merges two vectors, alternating elements from both vectors. If one vector isshorter than the other, then alternate as long as you can and then append the remaining elements from the longer vector. For example, if a is 1 4 9 16and b is9 7 4 9 11then merge returns the vector1 9 4 7 9 4 16 9 1118. Write a predicate function bool same_elements(vector a, vector b)that checks whether two vectors have the same elements in some order, with the same multiplicities. For example, 1 4 9 16 9 7 4 9 11 and 11 1 4 9 16 9 7 4 9 would be considered identical, but1 4 9 16 9 7 4 9 11 and 11 11 7 9 16 4 1 would not. You will probably need one or more helper functions.19. What is the difference between the size and capacity of a vector
Answer:
see explaination for code
Explanation:
CODE
#include <iostream>
#include <vector>
using namespace std;
vector<int> merge(vector<int> a, vector<int> b) {
vector<int> result;
int k = 0;
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
if (k % 2 == 0) {
result.push_back(a[i ++]);
} else {
result.push_back(b[j ++]);
}
k ++;
}
while (i < a.size()) {
result.push_back(a[i ++]);
}
while(j < b.size()) {
result.push_back(b[j ++]);
}
return result;
}
int main() {
vector<int> a{1, 4, 9, 16};
vector<int> b{9, 7, 4, 9, 11};
vector<int> result = merge(a, b);
for (int i=0; i<result.size(); i++) {
cout << result[i] << " ";
}
}
Components of a product or system must be
1) Reliable
2) Flexible
3) Purposeful
4)Interchangeable
Answer:
The correct answer to the following question will be Option D (Interchangeable).
Explanation:
Interchangeability applies towards any portion, part as well as a unit that could be accompanied either by equivalent portion, component, and unit within a specified commodity or piece of technology or equipment.This would be the degree to which another object can be quickly replaced with such an equivalent object without re-calibration being required.The other three solutions are not situation-ally appropriate, so option D seems to be the right choice.
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".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
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!
