All human blood can be ABO typed as belonging to one of group A,.pdf

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All human blood can be \"ABO\" typed as belonging to one of group A, B, O, or AB types. The actual distribution varies slightly among different group of people, but for a randomly chosen person from North America, the following are the approximate probabilities:Blood type O, A, B, AB and Probability 0.45, 0.40, 0.11, 0.04. Consider an accident victim with type B blood. She can only receive a transfusion from a person with type B or type O blood. What is the probability that a randomly chosen person will be suitable donor? Solution The probability that the donor is suitable equals the probability that the blood is type B or type O is P(B) + P(O) = .45 + .11 = .56.

Education

After class a student comments to me, “Once a poll has been taken, (the sample proportion with a certain opinion) is a known value, so it cannot have a distribution of possible values. Therefore, I do not understand what you mean by the sampling distribution of.” Write an explanation for this student. Solution A poll is just one sample of the population. If you took other samples (of the same size) you would get different proportions. The distribution of the proportions of all the samples taken is a sampling distribution..

Adjusting entries are required at the end of the period for some acc.pdf

akashenterprises93

Adjusting entries are required at the end of the period for some accounts. (1) Explain why this process is required (15 points) and (2) develop the adjusting entry at the end of the period for salary payable to employees $2400 (10 points). Solution Adjusting entries are required at the end of the period for some accounts, This process is required due to accrual based of accounting i.e Mercantile Accounting , Adjustment entry are passed so as to provide the correct income statement & balance sheet . Adjustment Entry are passed so as to record the Expenses & Revenue accrued or earned in the period in which it lies.Adjustment entry are passed to provide true & fair view of financial statement. (2) Develop the adjusting entry at the end of the period for salary payable to employees $2400 Adjusting EntryAccount Titles and ExplanationDebitCreditSalary & Wages Expenses2400Salary & Wages Payable2400.

advantages and disadvantages of balanced score cardSolution.pdf

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advantages and disadvantages of balanced score card? Solution Advantages: 1) It Helps facilitates better communication, coordination and understanding of goals and strategies accross teams and hierarchy of the organization. 2) Balanced scorecard method enables the firm to focus on all four aspects of a company\'s performance i.e. instead of focusing only on financial health, which helps to get proper view of the company performance. 3) Balanced Scorecards system allows all the stakeholders to determine not only short term, but long term objectives as well. 4) With the help of balanced scorecard, the organization can be sure that any strategic plan implemented is inline with the desired outcome. 5) It helps to integrate innovation and process improvement methods such as six sigma and lean manufacturing to a organizational goal. It also makes sure that the voice of the customer is equally very important. Disadvantages: 1) Balanced Scorecard do not consider direct financial analysis of economic value or risk management. Selection of goal under Balanced Scorecard may not consider opportunity cost. 2) Balanced Scorecard may lead to new type of reporting without improving performance, it may lead to new set of non-value-added reporting. or, may lead to deviation from achieving actual goals. 3) Performance of balanced scorecard system may not be quantifiable. As quality levels, it cannot be quantified except by surveys or management opinion..

Adjusting Entries and Financial StatementsRequired2. Prepare th.pdf

akashenterprises93

Adjusting Entries and Financial Statements Required: 2. Prepare the following financial statements and schedule: a. Income statement Lundberg Company Income Statement For the Year Ended December 31, 20-- Net sales $ Less cost of goods sold: $ $ Cost of goods sold Gross profit $ Operating expenses: $ Total operating expenses Operating income $ Other revenue: $ Other expense: Income before income taxes $ Net income $ b. Schedule of cost of goods manufactured Lundberg Company Schedule of Cost of Goods Manufactured For the Year Ended December 31, 20-- $ Direct materials $ $ Cost of materials used $ Cost of direct materials used $ Total manufacturing costs Total work in process during the period $ Cost of goods manufactured $ c. Statement of retained earnings Lundberg Company Statement of Retained Earnings For the Year Ended December 31, 20-- $ $ $ d. Balance sheet Lundberg Company December 31, 20-- Current assets: $ $ Inventories: $ Total current assets $ Property, plant, and equipment: $ $ $ Total property, plant, and equipment Total assets $ Current liabilities: $ Total current liabilities $ Long-term liabilities: Total liabilities $ $ Total stockholders\' equity Total liabilities and stockholders\' equity $Hide Hint(s) 2. Prepare the following financial statements and schedule: a. Income statement Lundberg Company Income Statement For the Year Ended December 31, 20-- Net sales $ Less cost of goods sold: $ $ Cost of goods sold Gross profit $ Operating expenses: $ Total operating expenses Operating income $ Other revenue: $ Other expense: Income before income taxes $ Net income $ Solution (A). Lundberg Company Income Statement For the year ended December/31/2014 (B). Account ParticularsAmountAmountNet Sales$405Less: Cost of Goods Sold$190Gross Profit$215Operating ExpencessWages Exp$70Ofice Exp$1Bad debt Exp$3Utilities Exp$4$78Operating Profit Before Interest& Tax Expences $137Non-operating ExpencessInterest Exp $7ncome Tax Exp$18$25Non-operating Expencess$112Non-Operating IncomeInerest Revenue$3$3Net Income$115.

Aeroplanes arrive at ChicagoSolution Using a.pdf

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After an effective orientation, the employee should emerge with info.pdf

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All of Ralphs ranch land was divided equally among his six child.pdf

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All of Ralph\'s ranch land was divided equally among his six children. His daughter Lynn\'s portion of the ranch land was divided equally among her four children. How much of Ralph\'s ranch land was inherited by 1 of Lynn\'s children All of Ralph\'s ranch land was divided equally among his six children. His daughter Lynn\'s portion of the ranch land was divided equally among her four children. How much of Ralph\'s ranch land was inherited by 1 of Lynn\'s children Solution ynn received 1/6 of the land. She then divided her 1/6 of the land in 4 parts. So each of her kids received 1/4 of her piece. Therefore, each of Lynn\'s kids received 1/4*1/6 = 1/24.

All numbers are equal Figure out what is wrong with the proof below.pdf

akashenterprises93

All code should be in C++ Using the UnsortedList class (UnsortedList.h file below) write a function sublist which extracts elements that are smaller than a given item from the given list and forms a new list. The precondition of the function is: the list has been initialized and is not empty. The postconditions are: newList contains all the items of the list whose values are less than the given item. Implement the sublist function as a friend function of the UnsortedList class whose declaration is: friend void sublist(const UnsortedList& list, const ItemType& item, UnsortedList& newList); (Hint: The UnsortedList class has private members ItemType list[MAX_LENGTH]; int length; and the member functions getLength, resetList, insert, remove, etc.) //********************************************************** // SPECIFICATION FILE (UnsortedList.h) // This file gives the specification of a basic class // template for unsorted array-based lists. // The list components are not assumed to be in order by // value. //********************************************************** #ifndef UNSORTEDLIST_H #define UNSORTEDLIST_H #include #include // Needed for the exit function using namespace std; const int MAX_LENGTH = 100; // Maximum number of components template // You may also choose to use // typedef statement class UnsortedList { public: // Constructor UnsortedList(); // Post: Empty list has been created. length has been set to zero. // Knowledge responsibilities int getLength() const; // Post: Returns the length of the list bool isEmpty() const; // Post: Returns true if list is empty; false otherwise bool isFull() const; // Post: Returns true if list is full; false otherwise bool isInList(const ItemType& item) const; // Post: Returns true if item is int the list; false otherwise int seqSearch(const ItemType& item) const; // Function to search the list for a given item. // Post: If item is found, returns the index in the array where // item is found; otherwise, return -1. // Action Responsibilities void resetList(); // Post: The list becomes empty. length has been set to zero. void insert(const ItemType& item); // Function to insert item to the end of the list. However, first // the list is searched to see whether the item to be inserted is // already in the list. // Post: list[length] = item and length++. If item is already in // the list or the list is already full, an appropriate message is // displayed. void remove(const ItemType& item); // Function to remove item from the list. // Post: If item is found in the list, it is removed from the list // and length is decremented by one. // Overloaded [] operator declaration. // This function returns a reference to the element in the // array indexed by index. ItemType& operator[](const int& index); // Additional operations void sort(); // Post: list items have been put into ascending order by selection sort void selectionSort(); // Function to sort the items in the list. // Post: list items have been put int.

Algebra Use the given zero to find the remaining zeros of the functi.pdf

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Alice and Ben decide to meet at 4 pm at the library to study. Alice .pdf

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Alice and Ben decide to meet at 4 pm at the library to study. Alice has a class beforehand, so she should arrive between 4:00 and 4:05 pm. Ben comes from off campus and should arrive between 4:00 and 4:10 pm. Let X and Y represent how late Alice and Ben are for their 4 pm meeting. X is uniformly distributed over the [0,5] minutes interval and Y is uniformly distributed over the [0,10] minutes interval. Alice and Ben arrive independently. a. Find the marginal density functions fX(x) and fY(y). b. Find the joint density function fXY(x,y). c. Find E(X), V(X), E(Y), V(Y), Cov(X,Y), Corr(X,Y). d. Find the probability that both Alice and Ben will be late by no more than 2 minutes each. e. Find the probability that Alice will arrive before Ben. Solution (a) fX(x) = 1/5 for x in [0,5] and = 0 for other x fY(y) = 1/10 for y in [0,10] and = 0 for other y (b) Since X and Y are independent fXY(x,y) = fX(x) * fY(y) = 1/50 for x in [0,5] and y in [0,10] = 0 for other x and y (c) for uniform distr. E = (a+b)/2 V = (a-b)^2/12 E(X) = (5+0)/2 = 2.5 E(Y) = 10/2 = 5 V(X) = (5-0)^2/12 = 25/12 V(Y) = (10-0)^2/12 = 25/3 Cov(X,Y) = 0 (since x and y are independent) Cor(X,Y) = 0 (same reason) (d) P(X<2,Y<2) = (double integral) fXY dx dy from x =0 to 2 and y = 0 to 2 = (1/50)*(2-0)*(2-0) = 2/25 (e) P(X.

algebra complex number solve z7 =1SolutionAccording to the f.pdf

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algebra complex number solve z7 =1 Solution According to the fundamental theorem of algebra, the equation z7=1 has 7 roots, called the roots of unity. Only one of them is obvious: z=1. The other six require polar form. First write 1 in polar form: 1 = ePi*in, where n is any integer. Then set z7=ePi*in and take the seventh root: z=ePi*in/7 Now substitute in values for n. You only need the values n=0,1,2,3,4,5, and 6, because the answer will repeat itself for all other n (due to the periodicity of complex exponentials). For n=0, you get the obvious solution, z0=e0=1. For n=1, you get z1=ePi*i/7, which you can write in standard form using Euler\'s identity: z1=cos(Pi/7)+i sin(Pi/7) In this way, you get the seven seventh roots of unity. If you graph them in the complex plane, you will find they all lie equally spaced on the unit circle..

Alan invested his money at 5 ½ compounded quarterly and Paula inve.pdf

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Alan invested his money at 5 ½ % compounded quarterly and Paula invested her money at 6% compounded annually. Which one of the two rates is higher? Solution The formula for compound interest is: A = P(1 + r/n)^(nt) where P is the principal (initial investment), r is the rate written as a decimal, n is the number of times per year you are compounding (so for monthly compounding, say, n=12), t is the number of years the investment has been earning interest, and A is the total amount in the account after t years. For Alan: r = .055, n = 4 for quarterly compounding For Paula: r = .06, n = 1 for annual (once per year) compounding If both Alan and Paula invest P amount of money for t years, we plug these rates into the compound interest formula: AAlan = P(1 + .055/4)^(4t) = P(1.01375)^(4t) = P[(1.01375)^4]^t (multiplying exponents is the same as raising a power to a power) = P(1.05614481)^t APaula = P(1 + .06/1)^(1*t) = P(1.06)^t Since after investing P initial dollars for t years, Alan will have P(1.05614481)^t dollars and Paula will have P(1.06)^t dollars, and since P(1.06)^t > P(1.05614481)^t as both bases are greater than 1 and 1.06 > 1.056...., Paula has the better rate for her investment..

Alberts Company has current earnings of $4.20 per share. The compa.pdf

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Albert\'s Company has current earnings of $4.20 per share. The company intends to use part of its earnings for investments that will generate a return of 20%. The cost of capital of Albert\'s is 15%, and its current stock price is $40. What is the percentage of the earnings Albert\'s intends to reinvest (i.e., the retention ratio)? What is the value of Albert’s growth opportunities? Solution EPS = 4.20 r = 20% Ko = 15% Po = EPS/Ke-g 40 = 4.20/(0.15-g) 0.15 - g = 0.105 g = 4.5% g = b * r 0.045 = b * 0.20 b = 22.5% Retention Ration = 22.5% Dividend Payout Ration = 100%-22.5% = 77.5% Dividend = 4.20 * 77.5% = 3.255 Value of growth opportunity = value of firm with growth - value of firm without growth = D1/(Ke-g) - EPS/Ke = 3.255/(0.15-0.045) - 4.20/0.15 = 31 - 28 = 3.

Airline passengers arrive randomly and independetly at the passenger.pdf

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Ages of the first 6 presidents at inauguration Washington 57 J. .pdf

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After reading MartinSolution A civil court h.pdf

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After reading Martin Solution A civil court hears a transnational dispute between private parties, which, according to the forum’s choice of law rules, is governed by a foreign lex causae. One party submits that it considers the foreign statute applicable to the dispute to be incompatible with the constitutional law of the lex causae and wishes to lead evidence to that effect, or ask the court to inquire into that question, with a view to having the impugned provision disapplied and the dispute decided on the basis of other provisions of the lex causae. How will the court respond to that submission; what does the forum’s private international law require it to do in these circumstances? This question will be addressed in this essay with respect to the laws of England as applied by the German courts on the one hand and those of Germany applied by English courts on the other. The study will deal exclusively with the effects which an incompatibility of primary legislation with constitutional norms of the lex causae creates from a private international law perspective and will therefore not address the questions of ‘unconstitutional’ subordinate legislation and, falling into a very similar category from a European law perspective, national primary legislation in contravention of EC law1. It is conceded that the practical scope of the problem of constitutional review of provisions of the foreign lex causae in the forum courts with respect to the laws of two states that, like England and Germany, by and large, share common values and, as member states of the European Community, show significant convergence or even uniformity in many areas of the law, is indeed a limited one. This finding is reinforced by the distinct ‘homeward trend’ of the English choice of law rules governing important areas of the law such as family law2. The fact, however, that the only case that has - so far - come to the English courts where the court was squarely faced with the question of permissibility of constitutional review of the foreign lex causae provision, related to the corporate law of another EC member state (Italy)3, demonstrates at least a residual practical importance of this matter. Furthermore, a considerable number of aspects that will be discussed in the course of this study will also be applicable to instances where German or English law will have to be applied in fora other than England or Germany. However, what makes the present topic particularly interesting, are the respective peculiarities of English and German constitutional law and their practical effect on transnational civil litigation of the kind in issue. In trying to find a satisfactory answer to the question as to how the English and German forum courts should react when faced with an allegation of ‘unconstitutionality’ of a lex causae statute, jurisprudence and, to a larger extent, influential scholarly writing will be presented and critically examined. In chapter II, the attitude taken by E.

Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...

Diana Rendina

Librarians are leading the way in creating future-ready citizens – now we need to update our spaces to match. In this session, attendees will get inspiration for transforming their library spaces. You’ll learn how to survey students and patrons, create a focus group, and use design thinking to brainstorm ideas for your space. We’ll discuss budget friendly ways to change your space as well as how to find funding. No matter where you’re at, you’ll find ideas for reimagining your space in this session.

Chapter wise All Notes of First year Basic Civil Engineering.pptx

Denish Jangid

Chapter wise All Notes of First year Basic Civil Engineering Syllabus Chapter-1 Introduction to objective, scope and outcome the subject Chapter 2 Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country. Chapter 3 Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements. Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station. Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps. Chapter 4 Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation Chapter 5 Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures. Chapter 6 Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems. Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect Text Books: 1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers. 2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers. 3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House. 4. BCP, Surveying volume 1

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