...Group Work – Chapter 7 1. Assume McGregor Inc. has the following account balances at December 31, 2014 – before adjusting entries: Refer to the information above for McGregor a. If McGregor uses the Balance Sheet Approach and estimates its bad debt for the year to be 6% of the Accounts Receivable balance, prepare the adjusting journal entry as of December 31, 2014. [54,000 * .06 = 3,240] DR CR Bad Debt Expense 3,740 Allowance For Doubtful Accounts 3,740 b. What is the Net Realizable Value of Accounts Receivable that will be reported on the December 31, 2014 Balance Sheet after the adjusting entry is posted? Accounts Receivable 54,000 Allowance for Doubtful Accounts (3,240) Net Realizable Value of AR 50,760 c. Assume, on January 8, 2015, JD Rhimes Inc. contacts McGregor Inc and says they have declared bankruptcy and will not be able to pay their Accounts Receivable debt of $2,000. Prepare the journal entry to write-off this customer’s account. DR CR Allowance For Doubtful Accounts 2,000 Accounts Receivable 2,000 d. What is the Net Realizable Value of Accounts Receivable on January 8, 2015 after the customer write off? [Assume no other activity has affected those accounts since December 31, 2014.] Accounts Receivable 52,000 Allowance for Doubtful Accounts (1,240) Net Realizable Value of AR 50,760 e. Does the write off affect the Net...
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...Accounting Skills Brief Skills • Experience with budgets, forecasting, payroll, accounts payable and receivable • Detail-oriented, accurate, general accounting data processing skills • Skilled at developing and maintaining professional relationships with clients Sample Tasks • Prepare, examine, or analyze accounting records, financial statements, or other financial reports to assess accuracy, completeness, and conformance to reporting and procedural standards. • Report to management regarding the finances of establishment. • Establish tables of accounts and assign entries to proper accounts. • Develop, implement, modify, and document recordkeeping and accounting systems, making use of current computer technology. • Compute taxes owed and prepare tax returns, ensuring compliance with payment, reporting or other tax requirements. • Maintain or examine the records of government agencies. • Advise clients in areas such as compensation, employee health care benefits, the design of accounting or data processing systems, or long-range tax or estate plans. • Develop, maintain, and analyze budgets, preparing periodic reports that compare budgeted costs to actual costs. • Provide internal and external auditing services for businesses or individuals. • Analyze business operations, trends, costs, revenues, financial commitments, and obligations, to project future revenues and expenses or to provide advice. • Generate budgets and forecasts on a quarterly basis and present...
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...being un-supervised. Other people in the community speed over the limit posted and children play there and may be hit or injured. Even though the children play in the street it cause a safety risk to drivers that don’t speed or drive wreck less do to the children running in and out of the streets without looking for cars. The children are our future and they need to be protected. When a person is looking to buy a future home they want to see a nice neighborhood that looks respectable and not as if it’s a bad neighborhood. The appearance of a home in my belief can say a lot about the people who live there. Someone who takes care of their home for example pressure washing, not trash in yard, paint nicely applied. Also there are some resident that don’t have the money but the residents should do as much as they can. To give an example of this a resident in my neighborhood has fixed the garage door which is tilted half way closed and it’s not very appealing. The appearance of a residents lawn as well as home its self make a community. A yard or lawn care should at the minimum have the grass mowed. The plants should be not over grown or deceased. Living in Florida the grass isn’t always...
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...The poor people usually depends of this places for basic things like food or sleep,without that help some of them don`t survive winter. Most of these institutions exist thanks to donations made it by average people and not government assistance. Every day there is more people living on the streets and the help is simply not enough, even when these kind of institutions exist. The people which suffer of poverty even when we see them like a kind of dog in the street, they are human and they want a decent life like normal people. Is normal to say that those poor people are living on the streets by their own choice, but like I said before, that is ignorance. That people suffer a lot of things, even things that we never going to feel, for example the hungry . The sad of these situations is that most of these people is...
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...Search Quiz ICS171 Name___________ ID____________ No text, no notes, no questions. Do the best that you can on each question. No questions will be answered about the quiz questions. If you think a question is ambiguous, write your interpretation and answer your modified question. Be reasonable. The following abbreviations are used: BF = branching factor, DFS = depth first search, BFS = breadth first search, IDS = iterative deepening search, A* = A* search, LI = local improvement search, HC = hill-climbing search. [pic] 1. For the 8-tile puzzle, what is the average branching factor, assuming the blank is equally likely to occur in any position. Show your work. (4*2 + 4*3 +1*4)/9 = 24/9 = 2 & 2/3. 2. Suppose that you are solving the 8-tile puzzle where it has solution. Which of the methods (DFS, BFS, IDS) is guaranteed to find a solution, assuming no computational limits are reached. List all that are correct. DFS, BFS, IDS 3. For the same puzzle, which of the methods (DFS,BFS,IDS) is guaranteed to find the shortest solution? List all that are correct. BFS, IDS 4. For the same puzzle, which methods are guaranteed to use no more than O(BF * length of solution) amount of memory. DFS, IDS 5. Suppose you apply the A* algorithm to the same problem. You decide to let f = current cost of the path. Would it be appropriate to let h = 0 for all states? Yes or no and why. ...
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...D2 evaluate the adequacy of accounting ratios as a means of monitoring the state of the business in a selected organisation, using examples: The importance of ratios in any business is very important because it gives the business a better understanding of the financial data. By using ratios the business is able to compare data from the current year with the previous years. From this the business will know and be able to identify if they are making more profit or a loss or if they just broke even. There are four types of ratios that a business has to calculate. These are things like investment ratio, financial ratio, profitability ratio and utilization ratio. From the results shown you can see if the business profits are increasing which is a positive thing and shows the company is on a rise and they are making money, this is a steady rise in the company’s money. If the results show a decrease it means that the company is making a loss. another one you can evaluate is the net profit margin as this is helpful as it shows how profitable the company has been throughout the years and whether it is increasing in profit or not. Current ratio is the most effective because it shows you whether you can have enough money to pay back your debts. The ratio shows that they have more liabilities it means they would struggle to pay off their debts. A figure below one means that the business does not have enough current assets to pay off its debts, a figure between 1 and 1.5 means that...
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...Data Structures & Algorithms Coursework Assignment 1 Q1. (a) Algorithm swap(x, y): Node n head While (n.getNext () != x ) do n n.getNext() Node v y.getNext () n.setNext(y) y.setNext(x) x.setNext(v) (b) Algorithm swap Doubly(x, y): DNode n x.getPrev() DNode v y.getPrev() n.setNext(y) y.setPrev(n) y.setNext(x) x.setPrev(y) x.setNext(v) v.setPrev(x) (c) The run time complexity for the singled linked algorithm is O (n) and for the doubly linked algorithm is O (1). Doubly linked list has the best time complexity. Time complexity in singly linked list take more time because we have to move from head to the node before x Q2. (b) RedBlueStack implements Stack{ protected Object A[]; Int capacity; int top = -1; RedBlueStack(int cap) { A = new Object [capacity]; capacity = cap; } int size() { return (top + 1); } void push(Object obj) throws FullStackException { if (size() == capacity) throws new FullStackException("Stack is full."); A[++top] = obj; } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Boolean isEmpty() { return (top < 0); } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Object pop() throws EmptyStackException { Object elem; if (isEmpty()) throws new EmptyStackException("Stack...
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...Task 1 LO 1.1 All high-level programming languages support the concept of data types. A data type defines a set of values that a variable can store along with a set of operations. Data types are used to store various types of data which is managed by program. Data type attaches with variable to determine the number of bytes to be allocate to variable and valid operations which can be performed on that variable. Although C has several built-in data types, it is not a strongly typed language, as are Pascal and Ada. C supports various data types and here some common data type as character, integer and floating-point types. C defines five foundational data types as defined below: ▪ character ▪ integer ▪ floating-point ▪ double floating-point ▪ valueless These are declared as by char, int, float, double, and void, respectively. These types form the basis for some other types also. The extent and choice of these and data types may contrast amongst processor natures and compilers. However, in all belongings an object of type char is 1 byte. C stores character type inside as an integer. Each character has 8 bits so, we can have 256 different characters values (0-255). Character set is used to map between an integer value and a character. The size of an int is ordinarily the same as the word length of the implementation setting of the program. C has 3 classes of integer storage namely short int, int and long int. All of these data types...
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...Write the following queries in SQL: Display a list of all instructors, showing their ID, name, and the number of sections that they have taught. Make sure to show the number of sections as 0 for instructors who have not taught any section. Your query should use an outer join, and should not use scalar subqueries. By using the university schema provided by db-book.com the following queries were done on the university database. The first query uses an outer join which works similar to the join operation but it keeps the rows that don’t match between the two tables that would be lost in a join operation. There are three forms of outer join: a) Left outer join displays the results from the left table even if the condition does not find any matching record in the right table. b) Right outer join will displays the results from the right table regardless if there is matching data in the left table. c) Full outer join will retain all rows from both tables, regardless if the data matches or not. The group by clause when used in a select statement collects data from multiple records and groups the results into one or more columns. The below is the query using left outer join operation: select ID, name, count(sec_id) as Number_of_sections from instructor natural left outer join teaches group by ID, name; The next query was written by using a scalar subquery without using an outer join operation. A scalar subquery is where the output of a subquery returns only...
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...i SASTRA UNIVERSITY Shanmugha Arts, Science, Technology and Research Academy Thirumalaisamudram Thanjavur – 613 402 BCSCCS705 PARALLEL COMPUTING AND ALGORITHMS LAB B.TECH (COMPUTER SCIENCE AND ENGINEERING) 7TH SEMESTER ii List of Exercises: 1. Basic arithmetic operations in parallel 2. Find out factorial of a number 3. Generation of Fibonacci series, finding prime numbers in an interval. 4. Evaluate the integral of a function 5. Merging of two sorted lists 6. Parallel tree traversals 7. Matrix multiplication 8. Enumeration sort 9. Odd-Even transposition sort 10. Bitonic merge 11. Quick sort 12. Single source shortest path iii Exercise 1 Aim: Basic Arithmetic Operations in parallel To perform a set of arithmetic operations in parallel using a cluster of computers Procedure: Step 1: Identify a set of numbers over which arithmetic operations are to be done Step 2: Identify the rank in the cluster Step 3: Split the set of numbers into domains and assign each domain to their corresponding processor Step 4: Get the result from each computer in the cluster and assimilate them at the master Step 5: Verify the time taken for the completion of each task. Algorithm:function sum(+,identity,a) = if #a == 1 then [identity] else let e = even_elts(a); o = odd_elts(a); s = scan_op(op,identity,{op(e,o): e in e; o in o}) in interleave(s,{op(s,e): s in s; e in e}); iv Input:- An array of integers Output: - The sum 3,2,7,6 P1 18 0,5,4,8 P2 17 62 2,0,1,5 P3 8...
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...*****Library for OleDb command and Connection Imports System.Data.OleDb ** *****Declaration for command, connection and Reader Public dcom As OleDbCommand Public dcon As OleDbConnection Public dbrd As OleDbDataReader Public sql As String ** ******Connection String *******2000 – 2003 Version (.mdb format) **A. dcon = New OleDbConnection("provider=microsoft.jet.oledb.4.0;data source = " & Application.StartupPath & "\maindata.mdb") dcon.Open() **B. dcon = New OleDbConnection("provider=microsoft.jet.oledb.4.0;data source = C:\Folder\DSource.mdb") dcon.Open() *******2007-2013 Version (.accdb format) dcon = New OleDbConnection ("Provider=Microsoft.ACE.Oledb.12.0;Data Source= " & Application.StartupPath & "\maindata.accdb") dcon.Open() ******2007-2003 Version (.xlsx format or excel) dcon = New OleDbConnection ("Provider=Microsoft.ACE.Oledb.12.0;Data Source= C:\Folder\DSource.xlsx"; Extended Properties=""Excel 12.0;HDR=Yes;IMEX=1"";") dcon.Open() SQL Query ******Add sql = "insert into TableName (Field1, Field2, Field3) values ('" & ValueIsString & "', #" & ValueIsDate & "#," & ValueIsNumber & ")" dcom = New OleDbCommand(sql, dcon) dcom.ExecuteNonQuery() *****Update sql = "update TableName set Field1= '" & ValueIsString & "', Field2= #" & ValueIsDate & "#, Field3= "#," & ValueIsNumber & " where ID = '" & ID & "'" dcom = New OleDbCommand(sql, dcon) dcom.ExecuteNonQuery() *****Delete sql = "DELETE FROM TableName WHERE...
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...Public Class RBTree Public root As RBTNode Dim null As RBTNode Public Sub New(ByVal data As Integer) null = New RBTNode("-1") null.color = "black" root = New RBTNode(data) root.color = "black" root.Left = null root.Left.parent = root root.right = null root.right.parent = root End Sub Public Sub New() End Sub Public Sub Insert(ByVal data As Integer) null = New RBTNode("-1") null.color = "black" 'step one, insert a red node in the tree If root Is Nothing Then root = New RBTNode(data) root.color = "black" root.Left = null root.Left.parent = root root.right = null root.right.parent = root Else Dim newNode As New RBTNode(data) newNode.Left = null newNode.Left.parent = newNode newNode.right = null newNode.right.parent = newNode Dim current As RBTNode current = root Dim parent As RBTNode While (True) parent = current If (data < current.data) Then current = current.Left If (current.data < 0) Then parent.Left = newNode newNode.parent = parent Exit While End...
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...Table Name : Employee EMPLOYEE _ID FIRST_NA ME LAST_NA ME SALA RY JOINING_D ATE DEPARTME NT 1 John Abraham 1000000 01-JAN-13 12.00.00 AM Banking 2 Michael Clarke 800000 01-JAN-13 12.00.00 AM Insurance 3 Roy Thomas 700000 01-FEB-13 12.00.00 AM Banking 4 Tom Jose 600000 01-FEB-13 12.00.00 AM Insurance 5 Jerry Pinto 650000 01-FEB-13 12.00.00 AM Insurance 6 Philip Mathew 750000 01-JAN-13 12.00.00 AM Services 7 TestName1 123 650000 01-JAN-13 12.00.00 AM Services 8 TestName2 Lname% 600000 01-FEB-13 12.00.00 AM Insurance Table Name : Incentives EMPLOYEE_REF_ID INCENTIVE_DATE INCENTIVE_AMOUNT 1 01-FEB-13 5000 2 01-FEB-13 3000 3 01-FEB-13 4000 1 01-JAN-13 4500 2 01-JAN-13 3500 SQL Queries Interview Questions and Answers on "SQL Select" 1. Get all employee details from the employee table Select * from employee 2. Get First_Name,Last_Name from employee table Select first_name, Last_Name from employee 3. Get First_Name from employee table using alias name “Employee Name” Select first_name Employee Name from employee 4. Get First_Name from employee table in upper case Select upper(FIRST_NAME) from EMPLOYEE 5. Get First_Name from employee table in lower case Select lower(FIRST_NAME) from EMPLOYEE 6. Get unique DEPARTMENT from employee table select...
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...the rows that are in the first query but not the second by removing the rows that are only found in the second query. There are three primary SQL commands involved when implementing a Union, Intersection and difference relational operators. As you may know, SQL data manipulation commands are set-oriented which are involved in operating over entire sets of rows and columns in tables at once. The UNION, INTERSECT, and MINUS statements make sure these operations occur. Union, Intersect and Minus only work properly if relations are Union-Compatible, which is based on the names of the relation attributes that must be the same and their data types must be alike. Being compatible does not mean the data types have to be exactly the same. For example, both data types can be used to store numeric values such as NUMBER and SMALLINT as well as character (string) values such as VARCHAR and CHAR. An SQL IN statement can be used with where clause to list a set of matching records of a table. We can use SQL IN query with a NOT combination to filter out some records. We can link or collect data from two different tables using...
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...Cursors A cursor is a variable that runs through the tuples of some relation. This relation can be a stored table, or it can be the answer to some query. By fetching into the cursor each tuple of the relation, we can write a program to read and process the value of each such tuple. If the relation is stored, we can also update or delete the tuple at the current cursor position. The example below illustrates a cursor loop. It uses our example relation T1(e,f) whose tuples are pairs of integers. The program will delete every tuple whose first component is less than the second, and insert the reverse tuple into T1. 1) DECLARE /* Output variables to hold the result of the query: */ 2) a T1.e%TYPE; 3) b T1.f%TYPE; /* Cursor declaration: */ 4) CURSOR T1Cursor IS 5) SELECT e, f 6) FROM T1 7) WHERE e < f 8) FOR UPDATE; 9) BEGIN 10) OPEN T1Cursor; 11) LOOP /* Retrieve each row of the result of the above query into PL/SQL variables: */ 12) FETCH T1Cursor INTO a, b; /* If there are no more rows to fetch, exit the loop: */ 13) EXIT WHEN T1Cursor%NOTFOUND; /* Delete the current tuple: */ 14) DELETE FROM T1 WHERE CURRENT OF T1Cursor; /* Insert the reverse tuple: */ 15) INSERT INTO T1 VALUES(b, a); 16) END LOOP; /* Free cursor used by the...
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