...What is your hypothesis? I believe that the substrate of lactose will work best with pH levels of 7 because that is the pH level best reactive with water (used to bond the two sugars, galactose and glucose) and found within our bodies. Record your data: Table 1: Record your data on the number of product molecules formed per minute obtained from the virtual lab. # Product Molecules/minute at: Amount pH 3 pH 5 pH 7 pH 9 pH 11 of Substrate (Lactose) 0.5 g 19 39 72 45 24 1.0 g 39 81 145 91 49 2.0 g 82 168 300 189 103 4.0 g 96 198 350 223 121 8.0 g 96 198 350 223 121 Graph your data and put the graph here. Now answer the following questions about your results: 1. What substrate amount was required to achieve the maximum reaction rate? 4 grams of lactose was the amount that achieved the maximum rate. 2. At what pH level did the maximum reaction rate occur? pH level 7 was the maximum reaction rate found in all amounts of substrate used. 3. Why was there no increase in the reaction rate with 8.0 g. of substrate as compared to 4.0 g. of substrate? What would you need to add to see an increase in the reaction rate with 8.0 g. of substrate? There was no increase in reaction rate with 8 grams compared to 4 grams because there was not enough lactase solution...
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...User Guide EVGA nForce 780i SLI Motherboard 780i 3-Way SLI Motherboard EVGA ii nForce 780i SLI Motherboard Table of Contents Before You Begin… ..................................................................................................... ix Parts NOT in the Kit .................................................................................................ix Intentions of the Kit ...................................................................................................x EVGA nForce 780i Motherboard..................................................................................1 Motherboard Specifications...................................................................................... 1 Unpacking and Parts Descriptions...............................................................................4 Unpacking ................................................................................................................ 4 Equipment ................................................................................................................ 4 EVGA nForce 780i SLI Motherboard ....................................................................... 5 Hardware Installation ....................................................................................................9 Safety Instructions.................................................................................................... 9 Preparing the Motherboard .........
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...Week 2 Assignment: Understanding Effective Money Management Assessment A, Part 1: Creating a Personal Financial Statement - Assets | 1 point | Car: Bluebook value $1250.00Cash: $378.00Savings Accounts: $826.00 | Assessment A, Part 2: Creating a Personal Financial Statement - Debts | 1 point | Rent: $750.00Electric/ Gas bill: $131.75Cable/ internet/ Phone bill: $80.42Credit Card: $31.00Cell phone bill: $72.37 | Assessment A, Part 3: Identify Money Management Tool | 1 point | Explain to Monica how the money management tools were identified. | Students should explain how they evaluated various cash management products and services. | Assessment A, Part 4: Creating a Personal Financial Statement – Steps | 1 point | Drag the steps listed on the right into their correct sequences on the left. When done click the Send button | Step 1: I got all my financial stuff together – bills, loans, bank statements, etc. | Step 2: I balance my checkbook. | Step 3: I decided what were my assets and what were my debts. | Step 4: I enter my assets in the program. | Step 5: I enter my debts in the program. | Step 6: The program gave me a Net worth figure at the end. | Assessment B: Creating a Monthly Cash Flow Statement ...
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...Diagnostic Algebra Assessment Definitions Categories Equality Symbol Misconception Graphing Misconception Definition Concept of a Variable Misconception Equality Symbol Misconception As algebra teachers, we all know how frustrating it can be to teach a particular concept and to have a percentage of our students not get it. We try different approaches and activities but to no avail. These students just do not seem to grasp the concept. Often, we blame the students for not trying hard enough. Worse yet, others blame us for not teaching students well enough. Students often learn the equality symbol misconception when they begin learning mathematics. Rather than understanding that the equal sign indicates equivalence between the expressions on the left side and the right side of an equation, students interpret the equal sign as meaning “do something” or the sign before the answer. This problem is exacerbated by many adults solving problems in the following way: 5 × 4 + 3 = ? 5 × 4 = 20 + 3 = 23 Students may also have difficulty understanding statements like 7 = 3 + 4 or 5 = 5, since these do not involve a problem on the left and an answer on the right. Falkner presented the following problem to 6th grade classes: 8 + 4 = [] + 5 All 145 students gave the answer of 12 or 17. It can be assumed that students got 12 since 8 + 4 = 12. The 17 may be from those who continued the problem: 12 + 5 = 17. Students with this misconception may also have difficulty with the idea that adding...
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...Team B Calorie Count Tool PRG/211 May 5, 2014 Team B Calorie Count Tool PROBLEM STATEMENT Team B was asked to develop a program which would calculate the user’s daily intake of calories and measure those calories against the overall calories expended. The core purpose of this program will do two primary functions. First, it will record the user intake of calories as acquired through meals throughout the day. Second, the user will record caloric output associated with physical activity. This information will be calculated together to determine the caloric surplus or deficit for the user. In order for the program to execute accurately, and provide customized results, the user will be required to input personal data to include gender, age, weight, and height. This additional information is essential to determine the user’s default caloric burn rate, otherwise known as the basal metabolic rate (BMR). The BMR and the calories burned as a result of physical activity will be calculated against the intake of calories to determine the overall success for the user. As the program is executed it must: * Record user name, age, height, weight to enable more accurate calculations * Record the users specific caloric values entered for each meal * Record the user activity and caloric burn values for that activity * Calculate the basal metabolic rate (BMR) for the individual * Subtotal the total caloric values for the day * Combine the physical activity and...
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...Problem Solving with Computing Homework - WEEK 2 [30 points] This is a review of some of the material from Chapter 2 and lectures from class. No credit for answers that are copies or near verbatim transcripts – please use your own words1 and document sources where appropriate. 1 This will apply to all assignments in this class. Answer the following questions: Chapter 2 1. Short Answers [1 point each, 2 points total] 1. What does a professional programmer usually do first to gain an understanding of a problem? The first thing that a professional programmer usually do first to gain an understanding of a program is to closely relate customer (Interview ) to inquire or gather information about the problem. 2. What two things must you normally specify in a variable declaration? The two things normally specified in a variable declaration are the variable type and identifier. 2. Algorithms / Pseudocode [1 point each, 5 points total] 1. Design an algorithm that prompts the user to enter his or her height and stores the user’s input in a variable named height. Declare height Display “Enter Your Height” Input Height Display “Height” 2. Write assignment statements that perform the following operations with the variables a and b. - Adds 2 to a and stores the result in b. - Subtracts 8 from b and stores the result in a Set b=2+a Set a=b-8 3. Write a pseudocode statement that declares the variable cost so it can hold real numbers. Floating Point-Variable...
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...11108944 Name: ASHWINI KUMAR Roll No. : RE3R02B32 PART- A 1. Ans :- (a) unary and ternary operator Unary operator:- It pecedes an operand . The operand (the value on which the operator operates ) of the unary operator must have arithmetic or pointer type and the result is the value of the argument. Example:- If a=5 then +a means 5 If a=0 then +a means 0. If a=-4 then +a means -4. Ternary operator:- It precedes an operand. The operand of the unary operator must have arithmetic type and the result is the negation of the operand’s value. Example:- If a=5 then –a means -5 If a=0 then –a means 0 If a=-4 then –a means 4. (b) Assignment and equalto operator Assignment operator:- Equal to operator: An assignment operator assigns value In this we put the To a variable. value as it is. Example – Example- a*=5 means a=5*5. Int a; a=5 means a is initialized with 5 if(a==5) { return true; } return false; (c) Expression and statement Expression:- An expression is any valid combination of operators , constants , and variables. Example:- ...
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...Selection statements Selection is used to select which statements are to be performed next based on a condition being true or false. Relational expressions In the solution of many problems, different actions must be taken depending on the value of the data. The if statement in C I used to implement such s decision structure in its simplest form – that of selecting a statement to be executed only if a condition is satisfied. Syntax: if(condtion) statement executed if condition is true When an executing program encounters the if statement, the condition is evaluated to determine its numerical value, which is then interpreted as either true or false. If the condition evaluates to any non-0 value (positive or negative), the condition is considered as a “true” condition and the statement following the if is executed; otherwise this statement is not executed. Relational Operators In C Relational operator | Meaning | Example | < | Less than | age < 30 | > | Greater than | height > 6.2 | <= | Less than or equal to | taxable <= 200000 | >= | Greater than or equal to | temp >= 98.6 | == | Equal to | grade == 100 | != | Not equal to | number !=250 | In creating relational expressions, the relational operators must be typed exactly as given in the above table. Thus, although the following relational expressions are all valid: age > 40 length <= 50 temp >= 98.6 3 < 4 flag == done day != 5 The following are invalid: length =< 50 ...
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...Assignment Title: Paper #1 Forensic Analysis of UEFI by Pui Yee (Pearl) Pang Pearl.pang.py@gmail.com By submitting this assignment I acknowledge that I have read and agree to abide by the Champlain College Academic Honesty Policy. I declare that all work within this assignment is my own or appropriately attributed. I accept that failure to follow the academic honesty policy may result in a failure grade, or expulsion from Champlain College. Date Due: _11/4/2015___ Date Submitted:_11/4/2015__ Abstract The Unified Extensible Firmware Interface (UEFI) Specification is an interface between the operating system (OS) and the platform firmware and is managed through the UEFI forum, a collection of chipset, hardware, system, firmware, and operating system vendors. One of many benefits of using UEFI is that it provides a more secure environment during the boot process by adding several security mechanisms such as secure boot, and update. With the added secure boot feature, the correct bootloader certification is needed and database key authentication is also required before the booting process. As a result, rootkit or other malware program have a hard time hijacking the boot process and concealing itself from the operating system. This paper will focus on the analysis of UEFI's secure boot feature and its implications and challenges for digital investigators conducting computer forensic investigation. Keywords: UEFI secure boot, boot firmware, malware, rootkit. Introduction ...
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...1. Distinguish between and define (see glossary in Maxfield and Brown’s Bebop Bytes Back for the definition of terms not found in Andrews’ A Guide to Managing and Maintaining Your PC): a. Hardware (H/W) b. Software (S/W) c. Firmware (F/W) d. Wetware (W/W) e. Vaporware (V/W) 2. Be able to define or describe: a. I/O device b. I/O controller c. peripheral device d. serial and parallel ports (explain the difference) e. adapter card, expansion card, interface card f. video adapter, video card g. pixel h. keyboard i. mouse j. printer k. BIOS l. device driver m. systemboard, motherboard n. video cable o. drive cable p. ribbon cable q. expansion slot (ISA, EISA, MCA, VL bus, PCI, local bus; what does each of the acronyms stand for?) r. ZIF socket (what does “ZIF” stand for?) s. SIMM (what does “SIMM” stand for?) t. system realtime clock u. jumper v. chipset w. cache memory x. power supply cable y. RAM and ROM z. CPU, microprocessor aa. coprocessor bb. primary storage and secondary storage (give examples of each, and know which is which) cc. volatile vs. nonvolatile memory (know which is which) dd. CMOS configuration chip ee. traces ff. bus gg. power supply 3. Be able to identify all of the items shown in Figures 1-2, 1-3, 1-4, and 1-5 in Andrews’ A Guide to Managing and Maintaining Your PC. 4. What are the principal functions of an Operating System? 5. Distinguish between, and give examples of the use of: a. a command-driven...
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...Exploring the Motherboard and Busses Short Answer Assignment Instructor Haroon Margaret Harm 2/18/2016 1. What are the major components of the motherboard? CPU-Central Processing Unit- also known as the microprocessor or processor. It is the “brain” of the computer. It is responsible for fetching, decoding, and executing program instructions and mathematical and logical calculations. RAM-Random Access Memory- is volatile memory, meaning that it loses its contents once the power is turned off. Basically, it is the work place of the computer where active programs and data are loaded so that any time the processor requests them, it doesn’t have to fetch them from the hard disk which will take longer access time. BIOS-Basic Input Output System- consists of low-level software that controls the system hardware and acts as an interface between the operating system and the hardware. BIOS is all the drivers- it is the link between hardware and software in a system. CMOS-Complementary Metal Oxide Semiconductor- a small separate block of memory made from CMOS RAM chips, supported by a CMOS battery even when the power is off. It is used to store basic information about the PC’s configuration, for example, floppy disk and hard disk drive types, CPU, RAM size, date and time, serial port information, plug and play information, and power saving settings. Cache memory- a small block of high speed memory that enhances PC performance by pre-loading information from the main memory...
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...STUDY HABITS OF SECOND YEAR BS-AVTECH STUDENTS OF PATTS COLLEGE OF AERONAUTICS S.Y 2013-2014 An Undergraduate Research Presented to The Languages Department of PATTS College of Aeronautics In Partial Fulfillment of the Requirements for the course ENGL 211 – Technical Report Writing By Guevarra, Giorgio Martin C Guevarra, Lorenzo Miguel Jang, Jose, Yosalina, Leo Xander March 2014 ACKNOWLEDGEMENT The researcher would like to express our thanks to the lord. Our God for his guidance towards everything we do In life, including this study that we had made, and for being an inspiration for us all to do our best in life. We give our thanks to Ms. Karen M. Millano, our adviser for ENGL 211, for carefully and patiently guiding us so that we may finish the thesis research, and for supporting us and believing in us, that we can accomplish our task finishing the thesis. To the respondents of this study, we express our gratitude because without them, this thesis research would not have been completed, we thank them for allowing us to conduct a survey during their spare time, and their patience and integrity in answering the survey. To our parents, for their support and everlasting patience and understanding for us. And lastly to our classmates, since they have been with us since the beginning of the semester and they had been our companions in everything we do for the subject ENGL 211. ABSTRACT STUDY HABITS OF...
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...SCHOOL OF ENGINEERING YEAR 3 MECHATRONICS ASSIGNMENT LAB REPORT Reading an Analogue Voltage from a Potentiometer to turn a Motor on and off with reading of 40 Assignment 2 Owais Jahanzeb BENG Mechanical Engineering with buissness Lecturer: Dr. Tom Shenton Aim & Objectives The aim of this lab is to develop a functioning program for the PICDEM board to read an analogue signal from a potentiometer and turning a motor on or off if the signal exceeds a certain limit. The program should depict the function that it should turn the motor ON and OFF if the potentiometer reading is less than or equal to 40. The practical uses of such program can be seen in automotive , injection moulding machines, wood processing machines, modern temperature controlled plants, speed control torque operations. Developing Program 1 Figure 1. The schematic circuit & PICDEM board configuration for Program Figure 1. The schematic circuit & PICDEM board configuration for Program The objective of program is to read the correspondent voltage analogous to the potentiometer position and switch the motor on if the reading is over 40 and switch it off if the reading is less than or equal to 40, the value can be adjusted by twisting the screw clockwise and anticlockwise. The program works by implementing the following code. PIC program for Test of potentiometer with value less equal to 40 with comments: include <p16f917.inc> extern DisplayDigit1, DisplayDigit2...
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...Law Written Assignment 3 Case Study 1 Parks, a 7-foot, 265-pound center for the San Diego Slick, objected when his contract was assigned from the ABC Corporation to the XYZ Corporation, the team’s new owner. The change of owners did not cause a change in the composition of the team although a new coach was hired. Parks’s compensation and his responsibilities remained the same. Was this contract assignable? Facts of the Case: 1) Parks contract was assigned from the ABC Corporation to XYZ Corporation. 2) Parks compensation and his responsibilities remained the same. Issues: 1) The reason why we are in court today is to identify if Park’s contract was assignable. Rules of the Law: 1) Personal Service Contract – The parties agree that a personal service contract may be assigned. This allows the trade of an athlete from one team to another team. 2) Notice of Assignment – Assignee is under a duty to notify the obligor that the assignment has been made and performance must be rendered to the assignee. 3) Anti-Assignment Clause – Prohibits the assignment of rights under the contract. 4) Approval Clause – requires that the obligor approves any assignment of contract. Analysis & Conclusion: Since we do not have all the facts we can assume the following: 1) Parks contract did include the Personal service contract. 2) Notice of assignment was made by XYZ Corporation. 3) Parks contract did NOT include Anti-Assignment Clause. ...
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...Management Accounting Individual Assignment 1. Variable manufacturing cost per unit = (323,000,000-160,000,000-24,000,000-100,000,000) 850,000 = $45.88 per unit Fixed manufacturing cost per unit = $100,000,000 850,000units = $117.65 per unit Total manufacturing cost per unit= 45.88+117.65 = $163.53per unit 2. Fixed overhead rate= $100,000,000 800,000units =$125 per unit Production volume variance= (850,000 X 125)-(800,000 X 125) =$106,250,000-$100,000,000 = $6,250,000 Favourable 3. Absorption costing. | $ | $ | SalesLess: Cost of goods soldOpening inventoryProduction(850,000 X (255+125))(-)Ending inventory(30,000 X 405)Gross MarginAdjustments for production variance(850,000-800,000)x125Operating Income | 0323,000,000(11,400,000) | 450,000,000(311,600,000) | | | | | | 138,400,0006,250,000F | | | 144,650,000 | Income Statement for the year ended 31December 2012 4. Variable Costing Income Statement for the year ended 31 December 2012 | $ | $ | SalesOpening inventoryProduction(850,000 X 255)(-)Ending inventoryContribution MarginFixed factory overheadOperating income | 0216,750,000(7650,000) | 450,000,000(209,100,000) | | | 240,900,000(100,000,000) | | | 140,900, 000 | 5. Based on the calculations of absorption costing and variable costing for the year 2012,it would be better to calculate and measure using the absorption costing...
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