Answer:
Answered below
Explanation:
public ArrayList<Integer> removeRange(ArrayList<Integer> X, int min, int max){
int i; int j;
//Variable to hold new list without elements in //given range.
ArrayList<Integer> newList = new ArrayList<Integer>();
//Nested loops to compare list elements to //range elements.
if(max >= min){
for( i = 0; i < x.length; I++){
for(j = min; j <= max; j++){
if( x [i] != j){
newList.add( x[i] );
}
}
}
return newList;
}
}
5. All sites are required to have the following reference materials available for use at VITA/TCE sites in paper or electronic format: Publication 17, Publication 4012, Volunteer Tax Alerts (VTA), and Quality Site Requirement Alerts (QSRA). AARP Foundation Tax Aide uses CyberTax Alerts in lieu of VTAs and QSRAs. What other publication must be available at each site and contains information about the new security requirements at sites
Answer:
Publication 5140.
Explanation:
The acronym VITA stands for Volunteer Income Tax Assistance and TCE stands for Tax Counseling for the Elderly. VITA/TCE is a certification gotten from the Internal Revenue Service in which the holders of such certification are trained on how to help people that have disabilities or that their incomes earners that are low. These volunteers help these set of people to prepare for their tax return.
In order to make the volunteers to be able to perform their work with high accuracy, the Department of Treasury, Internal Revenue Service gave out some aids for Quality Site. One of the aids which is the one contains information about the new security requirements at sites is given in the Publication 5140.
Last one, this is fashion marketing and I need to know the answer to get the 80%
Answer:
D. It requires that each item of stock is counted and recorded to determine stock levels
Explanation:
A perpetual system of inventory can be defined as a method of financial accounting, which involves the updating informations about an inventory on a continuous basis (in real-time) as the sales or purchases are being made by the customers, through the use of enterprise management software applications and a digitized point-of-sale (POS) equipment.
Under a perpetual system of inventory, updates of the journal entry for cost of goods sold or received would include debiting accounts receivable and crediting sales immediately as it is being made or happening.
Hence, an advantage of the perpetual system of inventory over the periodic system of inventory is that, it ensures the inventory account balance is always accurate provided there are no spoilage, theft etc.
The following are the characteristic of a perpetual inventory system;
I. It is the most popular choice for modern business.
II. Inventory is updated in near real-time.
III. It provides up-to-date balance information and requires fewer physical counts.
You've created a new programming language, and now you've decided to add hashmap support to it. Actually you are quite disappointed that in common programming languages it's impossible to add a number to all hashmap keys, or all its values. So you've decided to take matters into your own hands and implement your own hashmap in your new language that has the following operations:
insert x y - insert an object with key x and value y.
get x - return the value of an object with key x.
addToKey x - add x to all keys in map.
addToValue y - add y to all values in map.
To test out your new hashmap, you have a list of queries in the form of two arrays: queryTypes contains the names of the methods to be called (eg: insert, get, etc), and queries contains the arguments for those methods (the x and y values).
Your task is to implement this hashmap, apply the given queries, and to find the sum of all the results for get operations.
Example
For queryType = ["insert", "insert", "addToValue", "addToKey", "get"] and query = [[1, 2], [2, 3], [2], [1], [3]], the output should be hashMap(queryType, query) = 5.
The hashmap looks like this after each query:
1 query: {1: 2}
2 query: {1: 2, 2: 3}
3 query: {1: 4, 2: 5}
4 query: {2: 4, 3: 5}
5 query: answer is 5
The result of the last get query for 3 is 5 in the resulting hashmap.
For queryType = ["insert", "addToValue", "get", "insert", "addToKey", "addToValue", "get"] and query = [[1, 2], [2], [1], [2, 3], [1], [-1], [3]], the output should be hashMap(queryType, query) = 6.
The hashmap looks like this after each query:
1 query: {1: 2}
2 query: {1: 4}
3 query: answer is 4
4 query: {1: 4, 2: 3}
5 query: {2: 4, 3: 3}
6 query: {2: 3, 3: 2}
7 query: answer is 2
The sum of the results for all the get queries is equal to 4 + 2 = 6.
Input/Output
[execution time limit] 4 seconds (py3)
[input] array.string queryType
Array of query types. It is guaranteed that each queryType[i] is either "addToKey", "addToValue", "get", or "insert".
Guaranteed constraints:
1 ≤ queryType.length ≤ 105.
[input] array.array.integer query
Array of queries, where each query is represented either by two numbers for insert query or by one number for other queries. It is guaranteed that during all queries all keys and values are in the range [-109, 109].
Guaranteed constraints:
query.length = queryType.length,
1 ≤ query[i].length ≤ 2.
[output] integer64
The sum of the results for all get queries.
[Python3] Syntax Tips
# Prints help message to the console
# Returns a string
def helloWorld(name):
print("This prints to the console when you Run Tests")
return "Hello, " + name
Answer:
Attached please find my solution in JAVA
Explanation:
long hashMap(String[] queryType, int[][] query) {
long sum = 0;
Integer currKey = 0;
Integer currValue = 0;
Map<Integer, Integer> values = new HashMap<>();
for (int i = 0; i < queryType.length; i++) {
String currQuery = queryType[i];
switch (currQuery) {
case "insert":
HashMap<Integer, Integer> copiedValues = new HashMap<>();
if (currKey != 0 || currValue != 0) {
Set<Integer> keys = values.keySet();
for (Integer key : keys) {
copiedValues.put(key + currKey, values.get(key) + currValue);
}
values.clear();
values.putAll(copiedValues);
currValue = 0;
currKey = 0;
}
values.put(query[i][0], query[i][1]);
break;
case "addToValue":
currValue += values.isEmpty() ? 0 : query[i][0];
break;
case "addToKey":
currKey += values.isEmpty() ? 0 : query[i][0];
break;
case "get":
copiedValues = new HashMap<>();
if (currKey != 0 || currValue != 0) {
Set<Integer> keys = values.keySet();
for (Integer key : keys) {
copiedValues.put(key + currKey, values.get(key) + currValue);
}
values.clear();
values.putAll(copiedValues);
currValue = 0;
currKey = 0;
}
sum += values.get(query[i][0]);
}
}
return sum;
}
Approximately how many numeric IP addresses are possible with IPv4?
4 billon
Answer:
4,294,967,296 (~4.3B)
Explanation:
IPv4 uses 32-bits for representing addresses, thus you can have 2^32 total combinations.
Implement the method countInitial which accepts as parameters an ArrayList of Strings and a letter, stored in a String. (Precondition: the String letter has only one character. You do not need to check for this.) The method should return the number of Strings in the input ArrayList that start with the given letter. Your implementation should ignore the case of the Strings in the ArrayList. Hint - the algorithm to implement this method is just a modified version of the linear search algorithm. Use the runner class to test your method but do not add a main method to your U7_L4_Activity_One.java file or your code will not be scored correctly.
Answer:
Answered below
Explanation:
//Program is implemented in Java
public int countInitials(ArrayList<string> words, char letter){
int count = 0;
int i;
for(i = 0; i < words.length; i++){
if ( words[i][0]) == letter){
count++
}
}
return count;
}
//Test class
Class Test{
public static void main (String args[]){
ArrayList<string> words = new ArrayList<string>()
words.add("boy");
words.add("bell");
words.add("cat");
char letter = 'y';
int numWords = countInitials ( words, letter);
System.out.print(numWords);
}
}
developed the first compiler and conducted work that led to the development of COBOL ?
Answer:
Grace Hopper.
Explanation:
Grace Hopper was a US Naval Rear Admiral and an American computer scientist who was born on the 9th of December, 1906 in New York city, United States of America. She worked on the first commercial computer known as universal automatic computer (UNIVAC), after the second World War II.
In 1953 at the Remington Rand, Grace Hopper invented the first high-level programming language for UNIVAC 1 by using words and expressions.
Additionally, the high-level programming language known as FLOW-MATIC that she invented in 1953 paved the way for the development of common business-oriented language (COBOL).
Hence, Grace Hopper developed the first compiler and conducted work that led to the development of COBOL.
A structure that organizes data in a list that is commonly 1-dimensional or 2-
dimensional
Linear Section
Constant
No answertet provided
It intro technology
Answer:
An array.
Explanation:
An array can be defined as a structure that organizes data in a list that is commonly 1-dimensional or 2-dimensional.
Simply stated, an array refers to a set of memory locations (data structure) that comprises of a group of elements with each memory location sharing the same name. Therefore, the elements contained in array are all of the same data type e.g strings or integers.
Basically, in computer programming, arrays are typically used by software developers to organize data, in order to search or sort them.
Binary search is an efficient algorithm used to find an item from a sorted list of items by using the run-time complexity of Ο(log n), where n is total number of elements. Binary search applies the principles of divide and conquer.
In order to do a binary search on an array, the array must first be sorted in an ascending order.
Hence, array elements are mainly stored in contiguous memory locations on computer.
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest values that m could have
Answer:
Explanation:
From the given information:
[tex]a_{n-1} , a_{n-2}...a_o[/tex] in binary is:
[tex]a_{n-1}\times 2^{n-1} + a_{n-2}}\times 2^{n-2}+ ...+a_o[/tex]
So, the largest number posses all [tex]a_{n-1} , a_{n-2}...a_o[/tex] nonzero, however, the smallest number has [tex]a_{n-2} , a_{n-3}...a_o[/tex] all zero.
∴
The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times
i.e.
[tex](11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}[/tex]
[tex]= \dfrac{1(2^n-1)}{2-1}[/tex]
[tex]\mathbf{=2^n -1}[/tex]
[tex](1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}[/tex]
[tex]\mathbf {= 2 ^{n-1}}[/tex]
Hence, the smallest value is [tex]\mathbf{2^{n-1}}[/tex] and the largest value is [tex]\mathbf{2^{n}-1}[/tex]
