Premium Essay

Submitted By sharmin1991

Words 359

Pages 2

Words 359

Pages 2

Ordering food in a restaurant via iPad is so 2010. Think beyond the iPad. Imagine a future without waiters, without anyone asking you if you want sparkling or still. Without the condescending glance of a server who thinks they know so much. Imagine sitting down to dinner at a table lit by an overhead projector and interacting with the projections on the table in front of us. Our order? Placed by the swipe of a finger. This is the future of restaurants.

These types of table first introduce London’s “inamo” restaurant called E-Table, the interactive ordering system is designed to give patrons “more control over their dining experience”. A projector, which is installed in the ceiling above each table, projects an interactive menu on the tabletop. Using a touch mouse pad, patrons can navigate the menu, place orders, choose a virtual tablecloth, and project images of each meal on their plates to help them decide which one to get. The system isn’t only for placing orders – patrons can also use it to play video games, call a taxi, or watch the chef prepare their meals via a live webcam! E-Table also offers a lot of perks to the restaurant owners, like reduced staff costs, an increase in customer spending, and a more efficient way to deal with customers.

How it works:

The E-Table system uses overhead projection to deliver a menu and other digital images to a restaurant table-top. Projected images do overlay food dishes in restaurants but the system can be designed as clients want. In fact the restaurants have a clear display no projection over the plate.

There are two touch panels built in to each table, which allow guests to browse the menu, place orders, and interact with the system. This hardware requires custom maid software to operate.

Projection technology means surfaces can be waterproof,…...

Premium Essay

...ISSUES IN ACCOUNTING EDUCATION Vol. 18, No. 3 August 2003 pp. 275-290 Designing Audit Procedures When Evidence Is Electronic: The Case of e-Ticket Travel Revenue A. Faye Borthick and Jack E. Kiger ABSTRACT: The objective of the Fly Airline case is for you to learn to develop audit procedures in a context, e-ticket revenue, in which most of the evidence is available electronically, and in which many tests of controls and substantive tests can be performed using data stored in electronic form. In sequence, you will identify controls and their objectives, match controls to financial statement assertions for revenue, develop tests of controls, develop substantive tests for each assertion, and organize audit procedures into an effective and efficient audit program. Keywords: auditing; e-ticket revenue; collaborative learning; electronic evidence; internal control; monitoring; systems expertise for auditing. OVERVIEW ly Airline's revenue generation begins when a customer selects a flight for a specific date and time. The process is similar whether a ticket agent or the customer interacts with the system to make the reservation. If the customer is paying with a credit card, the system validates the credit card number the customer presents for billing before making the reservation. When it obtains approval of the credit charge (electronically, from the card issuer), the system records the charge and other details of the reservation. The......

Words: 5672 - Pages: 23

Free Essay

...0 f (t)e−st dt, succinctly denoted L(f (t)) in science and engineering literature. The L–notation recognizes that integration always proceeds over t = 0 to t = ∞ and that the integral involves an integrator e−st dt instead of the usual dt. These minor diﬀerences distinguish Laplace integrals from the ordinary integrals found on the inside covers of calculus texts. 7.1 Introduction to the Laplace Method The foundation of Laplace theory is Lerch’s cancellation law ∞ −st dt 0 y(t)e = ∞ −st dt 0 f (t)e (1) L(y(t) = L(f (t)) implies or implies y(t) = f (t), y(t) = f (t). In diﬀerential equation applications, y(t) is the sought-after unknown while f (t) is an explicit expression taken from integral tables. Below, we illustrate Laplace’s method by solving the initial value problem y = −1, y(0) = 0. The method obtains a relation L(y(t)) = L(−t), whence Lerch’s cancellation law implies the solution is y(t) = −t. The Laplace method is advertised as a table lookup method, in which the solution y(t) to a diﬀerential equation is found by looking up the answer in a special integral table. 7.1 Introduction to the Laplace Method 247 Laplace Integral. The integral is called the Laplace N integral of the function g(t). It is deﬁned by limN →∞ 0 g(t)e−st dt and depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t and g(t) = t2 , producing the integral formulas in Table 1. ∞ −st dt 0 (1)e ∞ −st dt 0 g(t)e = −(1/s)e−st = 1/s......

Words: 11519 - Pages: 47

Premium Essay

...EMPLOYEE DATA EMPLOYEE DATA 0 A AND E LIVE AND DRESSED CHICKEN PAYROLL SYSTEM EMPLOYEE ADVANCES EMPLOYEE SALARY ATTENDANCE pHILHEALTH PHILHEALTH CONTRIBUTION TABLE OWNER EMPLOYEE SALARY SUMMARY Social security system (SSS) SSS CONTRIBUTION TABLE SSS CONTRIBUTION PAYMENT National Wage and Productivity Commission WAGE RATE PHILHEALTH CONTRIBUTION PAYMENT HOME DEVELOPMENT MUTUAL FUND (PAG-IBIG FUND) HDMF CONTRIBUTION TABLE HDMF CONTRIBUTION PAYMENT BUREAU OF INTERNAL REVENUE (bir) TAX REPORT 0 A AND E LIVE AND DRESSED CHICKEN PAYROLL SYSTEM EMPLOYEE ADVANCES EMPLOYEE SALARY ATTENDANCE pHILHEALTH PHILHEALTH CONTRIBUTION TABLE OWNER EMPLOYEE SALARY SUMMARY Social security system (SSS) SSS CONTRIBUTION TABLE SSS CONTRIBUTION PAYMENT National Wage and Productivity Commission WAGE RATE PHILHEALTH CONTRIBUTION PAYMENT HOME DEVELOPMENT MUTUAL FUND (PAG-IBIG FUND) HDMF CONTRIBUTION TABLE HDMF CONTRIBUTION PAYMENT BUREAU OF INTERNAL REVENUE (bir) TAX REPORT A AND E LIVE AND DRESSED CHICKEN PROPOSED PAYROLL System Context diagram DFD EMPLOYEE ADVANCES EMPLOYEE SALARY SLIP EMPLOYEE DATA 2 PROCESS EMPLOYEE SALARY ATTENDANCE D1 EMPLOYEE RECORD EMPLOYEE INFORMATION TOTAL HOURS WORKED WAGE RATE EMPLOYEE SALARY SUMMARY D2 ADVANCES RECORD ADVANCES D3 TIME CARD RECORD HOURS WORKED ADVANCES EMPLOYEE INFORMATION SSS CONTRIBUTION TABLE SSS CONTRIBUTION PHILHEALTH CONTRIBUTION TABLE PHILHEALTH CONTRIBUTION HDMF......

Words: 1672 - Pages: 7

Free Essay

...1. a) What is a database? A database is an organized collection of related data. b) How does an RDBMS store data? An RDBMS stores related data in tables. 2. What is one benefit of dividing data into tables? Dividing data into tables eliminates unnecessary data duplication, or data redundancy. 3. a) What is a database schema? A database schema is a description of the data and the organization of the data into tables in a relational database. b) Explain the considerations for dividing data into related groups. Considerations for dividing data into related groups includes grouping information so that there is little or no data redundancy. However, each table must have data related to at least one other table, which will require data duplication. This is not considered data redundancy. A table with no relationship to any another table does not belong in the database. 4. a) What is a field? A column in a table. Used to store data. b) Explain why a field that stores more than one piece of data is considered poor design. A field that stores more than one piece of data limits the sorting and searching capabilities of the database. 5. List four guidelines to follow when choosing field names. 1. Make field names unique. 2. Choose the shortest......

Words: 1840 - Pages: 8

Premium Essay

...WEIGHTED AVERAGE SCORE Table showing weighted average score for elements of work ethics. |Factors |Strongly agree |Agrees |Neutral |Disagree |Strongly disagree| |Performance Appraisal |8 |55 |6 |20 |11 | |Performance Measurement |12 |65 |8 |10 |5 | |Mission statements |10 |33 |15 |40 |2 | |Work culture |35 |40 |5 |18 |2 | |Quality Concerns |5 |28 |1 |32 |34 | Table showing weighted average score for elements of work ethics. |Factors |Strongly agree |Agrees (4) |Neutral (3) |Disagree |Strongly disagree|Total |Rank | | |(5) | | |(2) |(1) | | | |Performance Appraisal |40 |220 |18 |40 |22 |340 = 22.6 |1 | | | | | | | |15 ...

Words: 1188 - Pages: 5

Free Essay

...are conducive to complicated search queries that divide the database further (E) None of the above ANSWER: E 7. Which of the following statements is not correct: (A) In a relational database, all related data is stored as a record (B) In a relational database a primary key is needed to identify the content of a record (C) In relational databases it is often necessary to put facts not about the primary key in a separate table with a different primary key, because the facts in a table are about the primary key (D) In relational databases for a given table its associated tables do not need pointers to indicate their linkage because the validation rules enforce all relationships (E) In relational databases the primary key of any given table is always locally unique ANSWER:: D 8. Which of the following statements is incorrect (A) Relational databases make life difficult for the designer of the database because many possible relationships must be anticipated (B) Relational databases are normally considered hard to modify because adding new concepts involves adding new tables, not simply altering old ones (C) Relational databases make life easy for the user because it is possible to avoid having to enter missing information for variables that are not logically possible (D) Relational databases make life easy for the user data entry and data processing speeds are higher (E) Relational databases make life easy for the user because data can be examined...

Words: 2453 - Pages: 10

Premium Essay

...redundancies lead to anomalies? e) Using two relational database tables, PROJECT and MANAGER, eliminate the redundancies discovered in Problem d. Make sure you use the naming conventions discussed in Section 1.3.3 and connect the two tables through the appropriate link. (Hint: Use Figure 1.11 as an example.) f) Create the relational schema to show how the two database tables in Problem e are linked. 2. Given the file structure shown in Figure P1.7, answer Problems a through g. Figure P1.7 The File Structure for Problems a-g. a) Identify and discuss the serious data redundancy problems exhibited by the file structure shown in Figure P1.7. b) How many different data sources are likely to be used by the file you examined in problem a? c) Given your findings in Problems a and b, how would a relational database environment help eliminate the data redundancy problems? d) Given your answer to Problem c, how many tables would you use to substantially eliminate the data redundancy problems? What table structures would you recommend? e) Given your answers to problem d, show the table contents for each table. f) Identify the types of relationships (1:1, 1:M, or M:N) between the tables you defined in problems d and e. g) Create a relational schema for Problem f. 3. Given the table structure shown in Figure P1.14, what problem(s) might you encounter if you deleted building KOM? Figure P1.14 The Table Structure for......

Words: 1762 - Pages: 8

Premium Essay

...PROBLEM #1: Convert the following CLASS table into 3NF tables. Examples of Course# are: BUS 297D, BUS 240, etc. Examples of Section# are: 1, 2, 3, etc. Examples of Course-Name are: Database Management; VB Programming Examples of Room# are: BBC 003, BBC 301, etc. Examples of Room-Capacity are: 50, 30, 80, etc. CLASS (Course#, Section#, Course-Name, Room#, Room-Capacity) Step 1: List all Functional Dependencies Course# ( Course-Name Room# ( Room-Capacity Course#, Section# ( Room# Step 2: List all Tables COURSE(Course#, Course-Name) ROOMS(Room#, Room-Capacity) CLASS (Course#, Section#, Room#) PROBLEM #2: Convert the following PARTS table into 3NF tables |Part# |Part |Vendor |Vendor |Unit | | |Description|Name |City |Cost | |1234 |Logic Chip|Fast Chips|Cupertino |10.00 | | | |Smart |Phoenix |8.00 | | | |Chips | | | |5678 |Memory |Fast Chips|Cupertino |3.00 | | | |Quality |Austin |2.00 | | | |Chips | | | | | |Smart |Phoenix |5.00 | | | |Chips | | | SOLUTION Step 1: Identify all functional dependencies Part# ( Part-Description Vendor-Name ( Vendor-City Part#, Vendor-Name ( Unit-Cost Step 2: Re-write the functional dependencies as tables PARTS (Part#, Part-Description) VENDORS (Vendor-Name...

Words: 1676 - Pages: 7

Free Essay

...DESIGNING E-LEARNING IN THE BUSINESS SCHOOLS OF KARACHI Research Project 1 Syed Arsalan Ali GR.No: 211018 In partial fulfillment of the requirement of the degree of Bachelors of Business Administration Under the Supervision of Mr. Riaz H. Soomro Presented to Hamdard Institute of Management Sciences HAMDARD UNIVERSITY, KARACHI Spring 2014 ABSTRACT We know that e-learning is very important factor in teaching of the universities, many universities use e-learning methods in their teaching and many not, by using e-learning methods in the teachings the university can satisfy their students and can generate their more and more revenue. For this study the different business schools of Karachi is selected to find out that at what level university is using e-learning methods in their teaching, for that the questionnaire is generated with the help of which the survey is done from different students of different universities of Karachi. In result there are some areas at which the universities of Karachi is not using e-learning methods for that the recommendation is done in order to improve the performance of that universities, that recommendation can be seen in conclusion of this study. CERTIFICATE This is certified that research work contained in this Research Project 1 entitled “DESIGNINGE-LEARNING IN BUSINESS SCHOOLS OF KARACHI” has been carried out and completed by Syed Arsalan Ali under my direct supervision and guidance at the Hamdard......

Words: 6210 - Pages: 25

Free Essay

...Assignment 1 Yang Liu May 5, 2015 Assignment 1, Part 1 (1) Table 0.1: Estimate a logit using solver Product of probability Log likelihood Intercept Eduation coefﬁcient Age coefﬁcient 2.06641E-11 -24.60262143 -11.15550863 0.531907452 0.113507304 (2) M EE d = M E Ag e = βE d e X β (1 + e X β )2 β Ag e e X β (1 + e X β )2 = 1 N βE d (i ) e X βi Σ N 1 (1 + e X βi )2 = 1 N β Ag e(i ) e X βi Σ N 1 (1 + e X βi )2 1 Result: Table 0.2: Add caption Marginal Edu effect Marginal age effect 0.085732302 0.018294992 (3) Table 0.3: linear probability model Coefﬁcients t Stat -1.541134276 0.097414096 0.020646694 Intercept EDUC AGE Std. Err. 0.429125046 0.025225082 0.005153578 -3.591340779 3.861794961 4.00628341 (4) H = −Σi p i (1 − p i )x i x i = X ΩX The diagonal of matrix Ω equal to p i (1 − p i ) The Hessian matrix 8.058949138 105.838015 300.4357025 H = 105.838015 1434.435526 3840.728111 300.4357025 3840.728111 12190.99391 (0.1) The variance-covariance matrix equals to the inverse of Hessian matrix Table 0.4: variance-covariance matrix Constant Constant Edu Age Edu Age 10.2699517 -0.511915922 -0.091816163 -0.511915922 0.029972725 0.003172894 -0.09182 0.003173 0.001345 2 (5) According to Student’s t distribution, under 49 freedom degree, the probabilities that these coefﬁcients are 0s are less than 0.005. We can say......

Words: 856 - Pages: 4

Premium Essay

...growth rate, the deal with laminate quickly becomes accretive. Appendix Table 1: Calculating the Un-levered Beta for Dixon (Assuming Firm’s Debt is Riskfree) |Dixon Corporation Financial Data | | | | | |1975 |1976 |1977 |1978 |1979 |Avg ('75-'79) | |Debt |7314 |6836 |6402 |6138 |6113 |6560.60 | |Total Liab |12029 |13268 |14849 |17382 |20831 |15671.80 | |D/(D+E) |0.61 |0.52 |0.43 |0.35 |0.29 |0.44 | |E/(D+E) |0.39 |0.48 |0.57 |0.65 |0.71 |0.56 | |βD |0.3 |0.3 |0.3 |0.3 |0.3 |0.30 | |βE |1.06 |1.06 |1.06 |1.06 |1.06 |1.06 | |βU = βE * E/(D+E) |0.60 |0.67 |0.73 |0.79 |0.84 |0.73 | Table 2: Calculating the Un-levered Beta for Dixon using only 1978 and 1979......

Words: 2533 - Pages: 11

Premium Essay

...activities as well as Total and Free Slack for each activity. Table 1 presents the project schedule accompanied with the total cost for the project, the duration and both the free and total slack columns. Table 1 Initial project schedule Figure 1 illustrates how the activities from table 1 will be scheduled across time, in the form of a Gantt chart. Figure 1 Gantt chart for initial schedule Figure 1 Gantt chart for initial schedule Figure 2 demonstrates the several “paths” formed by the activities sequence, in a network diagram. Highlighted in red is the critical path, consisted by the critical activities A, D & G. The total duration of the critical path, which at the same time is the total duration of the project, is 12 weeks as seen in table 1 while the total cost of the project is estimated to be at € 61,000.00. Figure 2 Network diagram for initial schedule Table 2 presents the activity paths for the initial schedule. Due to free slack in activities B, E & F, the non-critical paths have a slack of 2 to 4 weeks, which provides a greater flexibility in resource commitment and use for these stated activities. Path 1 | A | D | G | 12 weeks (critical path) | | 3 weeks | 5 weeks | 4 weeks | | Path 2 | B | G | | 10 weeks | | 6 weeks | 4 weeks | | | Path 3 | C | E | G | 8 weeks | | 2 weeks | 2 weeks | 4 weeks | | Path 4 | A | F | | 10 weeks | | 3 weeks | 7 weeks | | | Table 2 Activity paths and respective durations for......

Words: 1897 - Pages: 8

Premium Essay

...probability that the average score of the 36 golfers exceeded 71. 0.0228 2. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. a. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters? 0.2736 using Table E.2 b. What is the probability that the sample mean will be below 0.95 centimeters? 0.0418 using Table E.2 c. Above what value do 2.5% of the sample means fall? 1.057 3. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with µ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected. a. What is the probability that the sample mean will be between 100 and 120 grams? 0.9544 using Table E.2 b. What is the probability that the sample mean will be less than 100 grams? 0.0228 c. What is the probability that the sample mean will be greater than 100 grams? 0.9772 d. So, 95% of all sample means will be greater than how many grams? 101.7757 e. So, the middle 70% of all sample means will fall between what two values? 104.8 and 115.2 4. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. a. What would you expect the standard error of the mean to be? 2.5 minutes b. What is the......

Words: 2499 - Pages: 10

Premium Essay

...probability that the average score of the 36 golfers exceeded 71. 0.0228 2. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. a. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters? 0.2736 using Table E.2 b. What is the probability that the sample mean will be below 0.95 centimeters? 0.0418 using Table E.2 c. Above what value do 2.5% of the sample means fall? 1.057 3. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with µ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected. a. What is the probability that the sample mean will be between 100 and 120 grams? 0.9544 using Table E.2 b. What is the probability that the sample mean will be less than 100 grams? 0.0228 c. What is the probability that the sample mean will be greater than 100 grams? 0.9772 d. So, 95% of all sample means will be greater than how many grams? 101.7757 e. So, the middle 70% of all sample means will fall between what two values? 104.8 and 115.2 4. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. a. What would you expect the standard error of the mean to be? 2.5 minutes b. What is the......

Words: 2499 - Pages: 10

Free Essay

...chi in chi-square is the Greek letter χ, pronounced ki as in kite. Chi-square (χ2) procedures measures the differences between observed (O) and expected (E) frequencies of nominal variables, in which subjects are grouped in categories or cells. There are two basic types of chi-square analysis, the Goodness of Fit Test, used with a single nominal variable, and the Test of Independence, used with two nominal variables. Both types of chi-square use the same formula. The Chi Square Formula The chi-square formula is as follows: Helen C. Ang, “An Analytical Study of the Leadership Style of Selected Academic Administrators in Christian Colleges and Universities as Related to their Educational Philosophy Profile,” (Fort Worth, Texas: Southwestern Baptist Theological Seminary, 1984). 2 3 4 Ibid., 28-29, 46 Ibid., 45 Ibid., 47 1 © 4th ed. 2006 Dr. Rick Yount 2 3-1 Research Design and Statistical Analysis in Christian Ministry IV: Statistical Procedures where the letter O represents the Observed frequency -- the actual count -- in a given cell. The letter E represents the Expected frequency -- a theoretical count -- for that cell. Its value must be computed. The formula reads as follows: “The value of chi-square equals the sum of O-E differences squared and divided by E.” The more O differs from E, the larger χ2 is. When χ2 exceeds the appropriate critical value, it is declared significant. The Goodness of Fit Test The Goodness of Fit Test......

Words: 5119 - Pages: 21