Liberty Corporation reported the following figures- (Click the icon to.docx

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Liberty Corporation reported the following figures: (Click the icon to view the figures.) Liberty?s inventory turnover during 2016 was (amounts rounded) A. 6 times. B. 7 times. C. 8 times. D. not determinable from the data given. Solution Calculation of Inventory turnover Ratio: Inventory turnover Ratio = Cost of Goods sold / Average inventory Cost of Goods sold = $7055 Average inventory = ( Current year inventory + Last year inventory) / 2 =(1324 + 1060 ) / 2 = $1192 Hence Inventory turnover Ratio = 7055 / 1192 = 5.92 = 6 Times .

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List all of the different intermolecular interactions (IMFs) that are possible in a solution of CaCl 2 in methanol (CH 3 OH). Explain how/why these are possible Solution Answer: CaCl 2 has inonic bond within the molecule. Methanol (CH 3 OH) has hydrogen bonding between another molecule while within the molecule it has covalent linkage. When CaCl 2 in dissolved in methanol the solubility is due to formation of ion-dipole forces. This is due to fact that this force develops because of the interaction of ions and polar molecule. The slightly negative Oxygen of methanol is attarcted towards Ca 2+ while slightly positive Hydrogen gets attracted to Cl - . .

Liquidity ratios are an indication of a company-'s ability to effectiv.docx

sthomas232

Lines present in atomic emission spectra are - translucent Solut.docx

sthomas232

Lines present in atomic emission spectra are - Solution The emission lines are specific wavelengths where the atom or molecule has emitted photons. Since light is quantized, atoms and molecules can have energy levels that correspond to specific photon wavelengths. When a particle transitions from a higher energy to a lower energy a photon is emitted. These are the emission lines. These energy emmited from element are bright where electron emits some specific amount of energy when it moves from Upper energy level to low energy level. Nd the area where lights are absirbed seen as dark line in emmision spectra. . .

Link-Local addresses Loopback addresses Broadcast addresses Addresses.docx

sthomas232

Link-Local addresses Loopback addresses Broadcast addresses Addresses for self-identification Solution Ans- Link-Local addresses Link-local addresses are designated for use on a single local network. Link-local addresses are automatically configured on all interfaces. This automatic configuration is comparable to the 169.254.0.0/16 APIPA automatically assigned IPv4 addressing scheme. The prefix used for a link-local address is fe80::/64. On a single-link IPv6 network with no router, link-local addresses are used to communicate between devices on the link. .

licl dissolved in 100mL of water calculate molar concentration licl.docx

sthomas232

licl dissolved in 100mL of water calculate molar concentration licl dissolved in 100mL of water calculate molar concentration Solution Molar concentration ( Molarity ) = No. Of moles of the substance Ã· vol. Of solution in L No. Of moles = mass of substance Ã· molar mass of substance . Molar mass of LiCl = 42.394 g/mol. By knowing the mass of LiCl no . Of moles can be calculated and then molarity can be calculated as. .. Molarity = No. Of moles Ã· 0.100 L .

let S - v1 -v2-v3- v4 - where v1 - -121-^T v2 ---1-1 1 - ^T v3 - - -1.docx

sthomas232

let S = v1 ,v2,v3, v4 , where v1 = [121]^T v2 =[-1-1 1 ] ^T v3 = [ -1 1 7 ]^T v4 = [-2 -4 -4 ] ^T find every subset of S that is a basis for R^3 the answers are [v1, v2 , v3 ] , [v1,v2,v4] and [v1,v3,v4] I some what get it but im stuck so if u could explain it to me step by step i would be very grateful. thank u :) Solution basis is the collection of the subspaces,that uniquely define the vector space. for example,for v1,v2,v3 to be a basis, there should exist unique a ,b and c such that a*v1+b*v2+c*v3=v4 now the three resulting equations are: a-b-c=-2 2*a-b+c=-4 a+b+7*c=-4 solving for a,b,c we get a=0 b=3 c=-1 so as a,b,c exist uniquely,[v1,v2,v3] consist a basis. similar analysis can be done for other sets. .

Let M be a plane in Rn and lets assume that the plane has a null point.docx

sthomas232

Let M be a plane in R n and lets assume that the plane has a null point. What is the mirror image of the point (1,2,3) in the plane M that has the form x 1 -2x 2 +x 3 = 0 ? Solution FInd the foot of the perpendicular from the point (1,2,3) to the plane x 1 -2x 2 +x 3 = 0 Hope you know how to find foot of the perpendicular, assuming it to be (d,e,f) let the image of the given point be (a,b,c) now we have (a+1)/2 = d (b+2)/2 = e (c+3)/2 = f, Knowing d,e,f values you can find a,b,c which gives you the image point .

Let F -- L be a finite extension of fields- and let L - F-a- for some.docx

sthomas232

This document discusses a finite field extension L over a field F, where L = F[a] for some element a in L. It shows that the number of intermediate fields between F and L, denoted F(L/F), is finite by the following logic: Since L is a finite extension of F generated by a single element a, the degree of the extension F(a)/F is at least 1. Therefore, the number of intermediate fields F(L/F), which is equal to F(F(a)/F), is bounded above by 1 and is thus finite rather than infinite.

Many intuitively plausible beliefs conflict with science. For example, the belief that the Sun moves around a stationary Earth conflicts with current astronomy. Give at least one other intuitively plausible belief that conflict with either Copernican astronomy or Newtonian physics or both and discuss why it is anomolous. Solution It was earlier believed that the universe was static, i.e. it neither contracts nor expands. But it was found by Hubble that the objects which were farther away were moving more quickly away than the nearby ones. Recently, in 1999 it was found that universe was expanding and this expansion is accelerating. .

maom an learning What is the concentration of an HBr solution if 28-5.docx

sthomas232

maom an learning What is the concentration of an HBr solution if 28.5 mL of a 0.200 M NaOH solution are needed to titrate 55.5 mL of the acid? Number Solution Before dilution After dilution M1 = 0.2M M2 = V1 = 28.5ml V2 = 55.5ml M1V1 = M2V2 M2 = M1V1/V2 = 0.2*28.5/55.5 = 0.1027M final concentration = 0.1027M .

Let be unknown constants and consider the linear system- This syst.docx

sthomas232

Let be unknown constants and consider the linear system: This system has a unique solution whenever . h does not equal ____? If is the (correct) value entered above, then the above system will be consistent for how many value(s) of ? A. a unique value B. infinitely many values C. no values Solution the system will have a unique solution when the determinant of the coefficient matrix ? 0 So according to me A is correct option .

Many companies such as McDonalds have embraced supply chain management.docx

sthomas232

made a year end adjusting entry to accrue interest on a note payable t.docx

sthomas232

made a year end adjusting entry to accrue interest on a note payable that has the interest rate stated separately from the principal amount made a year end adjusting entry to accrue interest on a note payable that has the interest rate stated separately from the principal amount Solution Answer: Year end adjusting entry for Accrued interest: Interest Expense Debit Interest payable Credit (Being interest on notes accrued) .

List some advantages of having a well developed international code of.docx

sthomas232

List some advantages of having a well developed international code of ethics that would guide international marketing and operations of a company? Conversely, what are some disadvantages that you can think of? Solution advantages-it forces a large number of people in the firm to think through, in a fresh way, their mission and the important obligations they as a group and as individuals have to the firm, to each other, to their clients and customers, and to society as a whole. Second, once adopted a code can be used to generate continuing discussion and possible modification to the code. Third, it could help to inculcate in new employees at all levels the perspective of responsibility, the need to think in moral terms about their actions, and the importance of developing the virtues appropriate to their position. disadvantages-inequality in basic social goods such as income and wealth distribution. .

List several of the major differences between manufacturing and servic.docx

sthomas232

List several of the major differences between manufacturing and service operations and then articulate the importance of a strong manufacturing base to the overall economy. Solution Manufacturing starts from desingning a part and producing it. So it includes all the things such as Raw material, R and D , logistics. Service operaion does not provide any tangible objects . They are basicaly doing service such as Barber, police etc The importance of strong manufacturing is that the country is self sufficient in manufacturing the products and technology is highly developed. This decreases the imports and increases the export which is good for economy .

Mack and Joan are married but Mack moved out of the family home in Oct.docx

sthomas232

Mack and Joan are married but Mack moved out of the family home in October 2014. They have two children, May, age 9, and Milt, age 3. Mack and Joan together paid more than half of the support for the children and the children lived with Joan all year. There is no legal separation or written agreement that states who is entitled to the children Solution The correct answer is: C.Mack and Joan must file married filing separately since they cannot agree to file married filing jointly. The children are qualifying children for both parents so they need to agree on how to claim the children. They can each take one of the children as a dependent and the child tax credit and /or the additional child tax credit or they can agree for one parent to claim both children .

luonbcator assignment-take uonor ior SPeCIes war acuAS OUTIOS ouy acce.docx

sthomas232

luonbcator assignment-take uonor ior SPeCIes war acuAS OUTIOS ouy acceptor for species that act as acceptors only both for species that act as both donors and acceptors neither for species that act neither as donors nor as acceptors. H3C Si 2req CH3 Hydrogen bonding behavior 2req NH Hydrogen bonding behavior both Hydrogen bonding behavior CH3 CH3 Hydrogen bonding behavior acceptor Solution Hydrogen bond donor is the atom (H) attached to more electronegative atoms like F, O and N. While acceptor is the one which has more electronegativity with lone pair of electrons on hetero atoms like F, N and O etc. 1. Neither, since it don\'t have Hydrogen and lone pair of electrons on hetero atom Si. 2. Both, donor and acceptor 3. Acceptor only since lack of hydrogen on oxygen 4. Acceptor onlt .

Luther recently received $30-000 of compensation for some consulting w.docx

sthomas232

Luther recently received $30,000 of compensation for some consulting work (paid in cash). David recently received $30,000 of interest income from City of Ft. Lauderdale bonds. Both taxpayers report no taxable income from these transactions. Is this considered tax avoidance or tax evasion? What is the difference, if any, between the two? Solution Both of these are Tax evasion. Reason - The income is reportable to the IRS and the taxpayers are not reporting it. This is illegal and tax evasion. Difference between tax evasion and tax avoidance Tax Avoidance - This is legal. It is simply minimizing your taxable income by taking all deductins and using legal avenues to avoid paying taxes using legitimate tools and deduction methods. Tax Evasion - This is illegal. It involves not paying taxes that you owe. Tax Evasion involves tax payers who deliberately are not reporting income or over stating their deductions. .

Let A be an invertible matrix such that - where L and L1 are unit low.docx

sthomas232

Let A be an invertible matrix such that , where L and L1 are unit lower triangular matrices, and U and U1 are upper triangular matrices. Show that L = L1 and U = U1. Solution A = L1U1 = L2U2 are two LU-factorizations of the nonsingular matrix A.The equation L1U1 = L2U2 can be written in the form L12 L1 = U2U11 , where by lemmas 1.2-1.4 L1 2 L1 is unit lower triangular and U1 2 U1 is upper triangular. But then both matrices must be diagonal with ones on the diagonal. We conclude that L12 L1 = I = U1U21 which means that L1 = L2 and U1 = U2. .

LS P Explain how one can be certain whether an unknown contains the I.docx

sthomas232

Let A and B be rings- and f - A -- B a homomophism-Prove that f is inj.docx

sthomas232

Let A and B be rings, and f : A -> B a homomophism.Â Â Prove that f is injective iff its kernal is equal to {0}. Solution (Note that 0 is always in the kernel of f since f is a homomorphism) One direction is really easy. If ker f =/= {0} then there exists another element a in A such that f(a) = 0. So we have a =/= 0, but f(a) = f(0) = 0. Thus f is not injective. Now, assume ker f = {0}, and suppose that f(x) = f(y). We will show that x = y. f(x) = f(y) f(x) - f(y) = 0 Since f is a homomorphism we have, f(x-y) = 0 But ker f = {0}, so this implies that x - y = 0 Or x = y Thus f is injective. .

Lollipop Inc- completed transactions during 2013- journal transactions.docx

sthomas232

Lollipop Inc. completed transactions during 2013, journal transactions: Feb.9Â Â Â Declared a cash dividend on the 10,000 shares of $1.50, no-par preferred stock. Declared a $.20 per share dividend on the 10,000 shares of common stock outstanding. The date of record is Feb. 16 and the payment date is Feb.28Â Â Paid the cash dividends Mar.21Â Â Split common stock 2 for 1. Before the split, Lollipop had 10,000 shares of $10 par common stock outstanding. April18Â Â Distributed a 10% stock dividend on the common stock. The market value of the common stock was $27 per share June18Â Â Â Purchased 2,000 shares of treasury stock at $25 per share Dec.22Â Â Â Â Sold 1,000 shares of treasury stock for $28 per share Solution x. .

List three qualities of software that make it vulnerableSolutionThe th.docx

sthomas232

List three qualities of software that make it vulnerable Solution The three qualities of the software which make it vulnerable are : 1 .Process : processes utilize some data like metrics, test results etc to check whether the security needs are met or exceed the policy compliance 2.Metrics :These metrics are used to provide some bench mark or base line for the measurment of the application security. 3.Testing Activities :Test the application software which is been developed. It includes cross-site scripting tests etc. these activities identify any vulnerabilities injected inside the software. If any of the above qualities are not met with the standards will make the software vulnerable. .

List the main documents that are created during the architecture and u.docx

sthomas232

List the main documents that are created during the architecture and user design processes? Explain the purpose of the documents, how they are created, and why they are important? Solution The main documents that are created during the architecture and user design processes A) User Requirements Specification B) User Interface C) System Requirements Specification D) System Test Cases E) User Acceptance Testing F) Defect Track Log The Purpose of documentation and importance: After all bugs are fixed second cycle of STC and UAT is conducted to check if everything is fine now and fulfill user expectation. And finally the software is ready to use .

List the circumstances under which land would be classified under the.docx

sthomas232

List the circumstances under which land would be classified under the following balance sheet classifications: 1. Current assets 2. Investments (noncurrent) 3. Property, plant, and equipment 4. Other non-current assets Solution Hi Please see below 1. Current assets : Land held for short-term speculation would be a short-term investment. Land could also be held as inventory by, for example, a real estate developer. 2. Investments (non-current) : Land held for long-term speculation would be a long-term investment. 3. Property, plant, and equipment : The most common use of land would be under this category when it is actively used in operations. 4. Other non-current assets : Land in idle use or held as a future building site could be included in this category Thank you .

List the matters an auditor should consider when developing an audit p.docx

sthomas232

List the matters an auditor should consider when developing an audit plan Solution MATTERS TO BE CONSIDERED WHILE AN AUDIT 1.The terms of engagement and any statutory responsibilities. 2.Nature and timing of reports or other communication. 3.Effect of new accounting or audit pronouncements on the audit. 4.Identification of significant audit areas. 5.Nature timing and extent of audit evidence. 6.Applicable of legal & statutory requirements. 7.Involement of other auditors. 8.Involvement of experts. 9.Allocation of work between joint auditors. 10.Reliance to be placed on accounting systems & internal controls. .

List and define the factors that are important to successful implement.docx

sthomas232

List and define the factors that are important to successful implementation effort. Solution Several factors are important to the successful implementation of a system, including management support, the involvement of users, and user expectations. Additionally, commitment to the project, commitment to change, and extent of project definition and planning are important: Commitment to the project refers to managing the project so that the problem being solved is well understood and the target system actually solves the problem. Commitment to change means being willing to change behaviors and procedures. The premise of project definition and planning is that more extensive planning is more effective than less extensive planning. .

List at least seven security best practices that you should implement.docx

sthomas232

List at least seven security best practices that you should implement to secure your multi-function device and wireless network. Solution 1. Enterprise mobility solution risk analysis and mitigation recommendation 2. System Engineering and integration of mobile and wireless solutions 3. Design and implementation of cost-effective and managable multi-platform mobility solutions that provide enterprise and personal data sepration 4. Security requirement and policy development for mobile and remote access solutions 5. Mobile device operating system and application security evaluations 6. Bluetooth, WLAN and remote access solution auditing, vulnerability assesssments,and penetration tests 7. FISMA compilance assistance for mobile solutions .

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This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications. A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function. Healing is the body’s response to injury in an attempt to restore normal structure and functions. Healing can occur in two ways: Regeneration and Repair There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc. Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.

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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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