Answer:
The software can't locate the device
Explanation:
The device wasn't properly setup in order for the software to locate it.
(a) What is the difference between a compare validator and a range validator?
(b) When would you choose to use one versus the other?
Answer:
Check the explanation
Explanation:
a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.
(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.
Explanation:
a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.
(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.
#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
"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.
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))
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:
Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\ Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Answer:
Check the explanation
Explanation:
1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Note : Log (m *n) = Log m + Log n : this is property of logarithm
3) f(n) = [tex]O( n^2 )[/tex], since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n
for n times, this makes the complexity – n * n = n2
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
How can u refer to additional information while giving a presentation
Answer:
There are various ways: Handing out papers/fliers to people, or presenting slides.
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".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;
}
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.
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);
User documentation _____. Group of answer choices allows users to prepare overall documentation, such as process descriptions and report layouts, early in the software development life cycle (SDLC) describes a system’s functions and how they are implemented contains all the information needed for users to process and distribute online and printed output consists of instructions and information to users who will interact with the system
Answer:
And user documentation consists of instructions and information to users who will interact with the system
Explanation:
user documentation will not allow user to prepare overall documentation because it is already prepared so not A.
user documentation does not need implementation, So not B.
user documentation may not be a printed one . So not C
And user documentation consists of instructions and information to users who will interact with the system, so D is the right option
Any set of logic-gate types that can realize any logic function is called a complete set of logic gates. For example, 2-input AND gates, 2- input OR gates, and inverters are a complete set, because any logic function can be expressed as a sum of products of variables and their complements, and AND and OR gates with any number of inputs can be made from 2-input gates. Do 2-input NAND gates form a complete set of logic gates? Prove your answer.
Answer:
Explanation:
We can use the following method to solve the given problem.
The two input NAND gate logic expression is Y=(A*B)'
we can be able to make us of this function by using complete set AND, OR and NOT
we take one AND gate and one NOT gate
we are going to put the inputs to the AND gate it will give us the output = A*B
we collect the output from the AND and we put this as the input to the NOT gate
then we will get the output as = (A*B)' which is needed
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:
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;
}
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!
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.
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.
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.Use a vector to solve the following problem. Read in 20 numbers, each of which is between 10 and 100, inclusive. As each number is read, validate it and store it in the vector only if it isn't a duplicate of a number already read. After reading all the values, display only the unique values that the user entered. Begin with an empty vector and use its push_back function to add each unique value to the vector.
Answer:
The following are the program in the C++ Programming Language.
//header files
#include <iostream>
#include <vector>
//using namespace
using namespace std;
//define a function
int find(vector<int> &vec, int n)
{
//set the for loop
for(int i = 0; i < vec.size(); ++i)
{
//check the elements of v is in the num
if(vec[i] == n)
{
//then, return the values of i
return i;
}
}
return -1;
}
//define the main method
int main()
{
//declare a integer vector type variable
vector<int> vec;
//print message
cout << "Enter numbers: ";
//declare integer type variable
int n;
//set the for loop
for(int i = 0; i < 20; ++i)
{
//get input in the 'num' from the user
cin >> n;
//check the following conditions
if(n >= 10 && n <= 100 && find(vec, n) == -1)
{
//then, assign the values of 'num' in the vector
vec.push_back(n);
}
}
//print the message
cout << "User entered: ";
//set the for loop
for(int i = 0; i < vec.size(); ++i)
{
//print the output
cout << vec[i] << " ";
}
cout << "\n";
return 0;
}
Output:
Enter numbers: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 65 48 46 28 75 56
User entered: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 48 46 28 56
Explanation:
The following are the description of the program:
Firstly, set the required header files and required namespace.Define an integer data type function 'find()' and pass two arguments that is integer vector type argument 'vec' and integer data type argument 'n'.Then, set the for loop that iterates according to the size of the vector variable 'vec' inside the function, after that set the if conditional statement to check the following condition and return the variable 'i'.Define the main method that gets the input from the user then, check that the following inputs are greater than 10 and smaller than 100 then assign these variables in the vector type variable 'vec' and print all the elements.You work at a cheeseburger restaurant. Write a program that determines appropriate changes to a food order based on the user’s dietary restrictions. Prompt the user for their dietary restrictions: vegetarian, lactose intolerant, or none. Then using if statements and else statements, print the cook a message describing how they should modify the order. The following messages should be used: - If the user enters "lactose intolerant", say "No cheese." - If the user enters "vegetarian", say "Veggie burger." - If the user enters "none", say "No alterations."
Answer:
See explaination
Explanation:
dietary_restrictions = input("Any dietary restrictions?: ")
if dietary_restrictions=="lactose intolerant":
print("No cheese")
elif dietary_restrictions == "vegetarian":
print("Veggie burger")
else:
print("No alteration")
write the steps to insert picture water mark
Answer:
Is there a picture of the question?
Explanation:
Your job as a researcher for a college is going well. You have gained confidence and skill at pulling data and you are not making as many rookie mistakes. The college executives are begging to take notice of your skills. The college has had serious problems with reporting in the past for several reasons. One problem is there was no centralized department for numbers. Human Resources did some reporting, financial aid another, and the rest was covered by the registrar’s office. It was difficult to determine simple things like number of students enrolled in the fall semester because different departments would generate different. The higher ups want one consistent number they can rely on and they think your department should be in charge of generating that number.
As the first report as the official office numbers they want you to generate a report that tracks student demographics over time (a longitudinal study). Your college has a large percentage of its student body who are affiliated with the military (active duty military, retired military, military spouse, military dependent). For this study the college executives want to see how they stack up to other colleges that have a large percentage of military students.
After doing some research you find a field in the database that named mil_status. The documentation you have on the field says this is the field you are looking for. Since you need to determine when the student was in the military to generate this report you look for a start and end date associated with mil_status. Sure enough the table also has a mil_status_start and a mil_status_end field. These fields when combined with enrollment data allow you to determine if a student was in the military when they were a student. You query the data to check for bad dates and discover a serious issue. Most of the start dates and end dates are NULL. You once again make some quick phone calls to find out what’s going on. It seems this field is not populated unless the college receives official paperwork from the military listing the soldier’s enlistment date. In addition to this you find out that students are not required to update their information ever semester. This means once their mil status is set it is unlikely to ever change. Based on this information prepare a post that addresses the following:
1) What recommendation(s) would you make to the college’s executives to address this issue?
2) Collecting this data will have tangible and intangible costs associated with it. For example requiring official paper work adds an extra burden on registration and on the student. Students may decide to go elsewhere instead of supplying the paperwork or may stop identifying themselves as military. The executives want the data but they don’t want its collection to impact enrollment or place an undue burden on registration (the line is long enough already). How would you respond to these concerns?
3) You noticed something odd in the mil_Status field. The possible values are "Active Duty Military", "Military spouse", "Retired military", "Military Reserves", and "Military dependent". In what way is this field problematic? How could you fix it?
Answer:
Check the explanation
Explanation:
Following are the things to be implemented immediately
Database should be maintained centralized manner so that any one can access data. Number should be given at the time of admission it should contain year of admission,course of admission …etc in the form of code eg. 2015FW1001 means year 2015 Fashion studies Winter admission no 1001. At the time of admission only they have to give the details of military status. Department which taking care of this issue should only have the permission to modify and delete the data. All the departments should be connected to the server by internet/intranet/network. Students should be given user id and password
Same database should be connected and given access through internet also with limited privileges so that user can update their status of military by submitting proof. By these information uploaded centralized administrative department update the status if they satisfied with the documents uploaded.
We can overcome the problem of mil status by giving the freedom to the student update their military status by uploading the documents.
the smallest unit of time in music called?
Answer:
Ready to help ☺️
Explanation:
A tatum is a feature of music that has been defined as the smallest time interval between notes in a rhythmic phrase.
Answer:
A tatum bc is a feature of music that has been variously defined as the smallest time interval between successive notes in a rhythmic phrase "the shortest durational value
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
A cloud provider is deploying a new SaaS product comprised of a cloud service. As part of the deployment, the cloud provider wants to publish a service level agreement (SLA) that provides an availability rating based on its estimated availability over the next 12 months. First, the cloud provider estimates that, based on historical data of the cloud environment, there is a 25% chance that the physical server hosting the cloud service will crash and that such a crash would take 2 days before the cloud service could be restored. It is further estimated that, over the course of a 12 month period, there will be various attacks on the cloud service, resulting in a total of 24 hours of downtime. Based on these estimates, what is the availability rating of the cloud service that should be published in the SLA?
Answer:
99.6
Explanation:
The cloud provider provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.
The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.
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] << " ";
}
}
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.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
