Explain the kinds of information that must be maintained in fixed as.pdf

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Explain the kinds of information that must be maintained in fixed asset records during the asset continuance phase. Solution Answer) Here,Fixed assets continuance phase refers to update records accurately throughout their useful life. The information that is maintained is depreciation where different rates and methods may be used for different assets which may be complex job but must be done accurately, any costs that used to updation of asset and qualified for capitalisation, updation of estimated figures like useful life of assets,expected salvage value and keeping track of physical location where the asset is maintained..

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Dr. P has a culture of stem cells and she wants them to specialize into heart muscle cells for treating a patients who suffered a heart attack that weakened his heart muscle. What will she need to do to coax the stem cells into becoming heart muscle cells? A. She will need to add certain chemicals to the culture, stimulating the cells to specialize. B. She will need to expose the cells to human blood. C. She will need to insert the cells into an embryo in the location where the heart would normally develop. D. She will need to zap the culture of cells with electricity, resembling the electrical signals of the heart. E. She will need to insert human heart genes into the nucleus of the stem cells. Solution Question : Dr. P has a culture of stem cells and she wants them to specialize into heart muscle cells for treating a patients who suffered a heart attack that weakened his heart muscle. What will she need to do to coax the stem cells into becoming heart muscle cells? Answer: B. She will need to expose the cells to human blood..

Dr. P has a culture of stem cells and she wants them to specialize i.pdf

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Describe two ecological roles of fungi. Related concepts: decomposers, mycorrhizal symbiont. Solution Fungi as Decomposers Fungi join bacteria as the primary decomposers in different ecosystems. They feed on the organic remains of dead plants. They release special enzymes that break down lignin, a structurally complex substance in wood. Because of their ability to break down wood, fungi are considered the major wood decomposers in forests. To absorb organic matter, fungi release acids to melt organic matter, and then they suck the acid back together with the melted organic matter. Decomposing organic matter is considered by ecologists as the major ecological role of fungi. Plant-Fungi Symbiosis The symbiotic association between a vascular plant root and fungi is called mycorrhiza. In this association, fungi colonize the roots of a plant and gain access to their tissues. The fungi build a connection to the roots in such a way that they can exchange nutrients with each other. Plants provide mycorrhizal fungi shelter and a supply of carbohydrates; in return, mycorrhizal fungi improve the plant’s absorption capabilities for water and nutrients. Mycorrhizal fungi have a larger surface-area ratio compared to roots, so they have higher absorptive capacity for water and minerals. They make it easier for plants to absorb water and minerals in soils that are dry and with low mineral concentration. Without mycorrhiza, it would be difficult for plants to absorb water and minerals from such soils. It is estimated that about 90% of plants depends on mycorrhizal fungi for survival..

Describe two ecological roles of fungi. Related concepts decomposer.pdf

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Differential equations using phase plane analysis. Task 1 (Romeo and Juliet). Let R and J be variables which measure the love/hate that Romeo and Juliet, respectively, have for the other star-erassed lower. We consider their relationship as described by a system of the form The parameters a, b,c, d determine the evalution of their relationship ower time. The signs of the variables a, b determine what kind of lover Romeo is, Of course, the signs of the variables c, d play exactly the same roles for Juliet. For each arrangement of passible signs of a,b explain how Romeo\'s feelings for Juliet change depending on the current state of things. Then make up a descriptive name for the kind of lover Romeo is given those variables. (Remember that these apply to Juliec just as well, so if you would rather, yau may work with J, e, d.) Please remember that the sign of a number is either positive, negative, or zero Explore the different kinds of things that happens when you pair potential Romeos and Juliets. Which kinds of pairs are doomed to terrible fates? Which pairs make for happy coll What interesting things ples? you say? Solution (1) dR/dT = aR+bJ = f(R,J) form here T is time dR/dT = (1)(aR+bJ) dR/(aR+bJ) = dT {let us take J value as constant though it is given as a variable consider nly this equation} integrate on both sides we get 1/a ln(aR+bJ) + k = T similarly in second equation if we take R as constant we get 1/d ln(cR+dJ) + k\' = T we get aR+bJ = e^(a(T-k)) that is an exponential form it comes closer to 0 only when a value is large and negative J value also similarly comes as exponential form e^(d(T-K\')) a(+ve) , b(+ve) love increases ; a(+ve), b(-ve) love increases so much with time a(-ve) , b(+ve) love increases (since in equation -ve on both sides so get cancelled) a(-ve) , b(-ve) love depends on b and J numerical values same thing applies to equation dJ/dT (2) above we can see that in three cases love only increases with time only in 4th case it depends on juliet when he is totally negative for her similarly for juliet only in one case it depends on romeo so we can conclude that love only increases with time and the couple in which love exists at least with one will remain happy.

Differential equations using phase plane analysis. Task 1 (Romeo .pdf

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Develop a stack in c with dynamic memory allocation. Use the following structure to create a stack:t typedef struct Element { char c; struct Element *next; } Element_Type, *Pointer_Type; You need to develop the following functions: 1) push(): demands two arguemnts of which the first nominates a pointer. For example, push(&x, ‘A’) to push ‘A’ onto stack X. You need to allocate memory for a new element. 2) pop(): pops and returns the element on the top of the stack and frees the memory space which was occupied by the element. 3) peep(): simply returns the character value of the top element of the stack. No pointers are disturbed. 4) displayAll(): prints all character values of all elements of the stack from top to bottom. void push(Pointer_Type *q, char ch){……} char pop(Pointer_Type *q){……} char peep(Pointer_Type *q){……} void displayAll(Pointer_Type *q){……} Test your stack and all functions with at least 10 elements. Solution #include #include #include struct Element { char c; struct Element *next; }; struct Element *start=NULL; void push() { char ele; struct Element *new_ele,*temp; int i=0; for(temp=start;temp!=NULL;temp=temp->next) { i++; } if(i==10) { printf(\"Stack Overflow\"); getch(); } else { new_ele=(struct Element *)malloc(sizeof(struct Element)); printf(\"Enter value for Element: \"); scanf(\"%c\",&new_ele->c); new_ele->next=start; start=new_ele; } } void pop() { struct Element *temp,*temp2; int i=0; for(temp=start;temp!=NULL;temp=temp->next) { i++; } if(i==0) { printf(\"Underflow\"); getch(); } else { temp2=start->next; start=temp2; printf(\"\ ***The value has been poped***\ \"); } } void display() { struct Element *temp; printf(\"\ ****Stack Values****\ \"); for(temp=start;temp!=NULL;temp=temp->next) { printf(\"%c\ \",temp->c); } getch(); } void peep() { struct Element *temp; int pos,i=0,j=0; printf(\"Please enter the position number: \"); scanf(\"%d\",&pos); for(temp=start;temp!=NULL;temp=temp->next) { i++; } if(pos>i) { printf(\"\ Over Limit\ \"); getch(); } else { temp=start; for(j=0;jnext; } printf(\"\ The value of position %d is :%c\",pos,temp->c); getch(); } } int main() { int ch; while(ch!=6) { printf(\"\ 1.Push\ \"); printf(\"2.Pop\ \"); printf(\"3.Peep\ \"); printf(\"4.Display\ \"); printf(\"5.exit\ \"); printf(\"Enter your choice:\"); scanf(\"%d\",&ch); switch(ch) { case 1: push(); break; case 2: pop(); break; case 3: peep(); break; case 4: display(); break; case 5: exit(0); } } return 0; }.

Develop a stack in c with dynamic memory allocation. Use the followi.pdf

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Define a struct type that represents a smartphone. The struct should.pdf

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Changes in the economy have determined that for the EZ shipping company, a surcharge will be assessed on all packages that meet certain criteria. The surcharge is a function of the characteristics of the package and the zip code to which it is sent. 1. The regular charges for shipping are calculated as follows: For all weights five pounds and under, the shipping cost is $12.00. For weights over five pounds, the charge is calculated as follows: If length * width * height + weight is: greater than 5 and less than or equal to 15, the shipping cost is 14.00 greater than 15 and less than or equal to 34, the shipping cost is 17.00 greater than 34 and less than or equal to 45, the shipping cost is 21.00 greater than 45 and less than or equal to 60, the shipping cost is 33.00 greater than 60, the shipping cost is 105.00 2. The additional charges are calculated as follows: If the first digit of the zip code is a \"4\" then there is an additional surcharge of 5% on the shipping cost. If the first digit of the zip code is a \"6\" then there is an additional surcharge of 9% on the shipping cost. For all other zip codes there is an additional surcharge of 14% on the shipping cost. 3. Finally, if the zip code is even, then there is an additional surcharge of 2% of the shipping cost. Write a program that allows the user to input (in this order) the zip code, weight, length, width and height of the package then prints out the following: The zip code and dimensions of the package The shipping cost The surcharge The total shipping cost (shipping cost plus surcharge(s)) I have written out the code and have been able to get it to run but when I added in the 2% for even numbers portion of my code every thing went weird with my outputs. Every thing still compiles and works but I am not recieving the outputs I desire. Any help remedying this would be greatly appreciated. import java.util.Scanner; //allows use of scanner for keyboard inputs public class Shipping // Class titling { public static void main(String[] args) { { int zip; double weight; double length; double width; double height; double surcharge; double surcharge2; double shippingCost; //Variables that will need defining for the shipping process Scanner keyboard = new Scanner(System.in); System.out.println(\"Hello, Welcome to EZ Shipping services! \ Please input the important details of your shipment when prompted\"); System.out.println(\"What is the weight of your package? \"); weight = keyboard.nextDouble(); System.out.println(\"What is the length of your package? \"); length = keyboard.nextDouble(); System.out.println(\"What is the width of your package? \"); width = keyboard.nextDouble(); System.out.println(\"What is the height of youre package? \"); height = keyboard.nextDouble(); //Print outs for package detail inputs double packages=length*width*height+weight; //Package variables combined and stored { shippingCost = 12; //Stock sipping cost 12$ } if (weight >5) { if (packages >=5.1 || packages <= 1.

Changes in the economy have determined that for the EZ shipping comp.pdf

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Broadly speaking, the economy consists of households, firms, government, and financial institutions. Name two parts, areas, or types of government that can have influence over the economy Solution Explanation: - Financial policy: Monetary Policy of any government can have major impact on economy of a nation since it has the power to stimulate the economy and change the way economy behave. Financial policies are performing major part in economy and it can make or break economy. How the taxes are levied can majorly impact economy. Financial policies move control the flow of the money and it can inject new money flow in system to revive it. - Fiscal Policy: Another major player in stimulating and impacting economy at it best is fiscal policy. Fiscal policy mainly works to overcome from struggling economy and reduce the inflation in market. Interest rate is popular tool to control the flow money through changing interest rates..

Broadly speaking, the economy consists of households, firms, governm.pdf

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A committee consists of 13 Republicans and 16 Democrats. How many dif.pdf

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2. July 01 Record the dividend received from the foreign subsidiary. 3. December 31 Record the equity in the net income of the foreign subsidiary. 4. December 31 Record the amortization of the differential. On January 1, 20X5, Pirate Company acquired all of the outstanding stock of Ship Inc., a Norwegian company, at a cost of $167,400 Ship\'s net assets on the date of acquisition were 700,000 kroner (NKr). On January 1, 20X5, the book and fair values of the Norwegiarn subsidiary\'s identifiable assets and liabilities approximated their fair values except for property, plant, and equipment and patents acquired. The fair value of Ship\'s property, plant, and equipment exceeded its book value by $18,000. The remaining useful life of Ship\'s equipment at January 1, 20X5, was 10 years. The remainder of the differential was attributable to a patent having an estimated useful life of 5 years. Ship\'s trial balance on December 31, 20X5, in kroner, follows: Debits Credits NKr 160,000 218,000 277,000 609,000 Cash Accounts Receivable (net) Inventory Property, Plant & Equipment Accumulated Depreciatiorn Accounts Payable Notes Payable Common Stock Retained Earnings Sales Cost of Goods Sold Operating Expenses Depreciation Expense Dividends Paid Total NKr 167,000 91,000 209,000 430,000 270,000 745,000 421,000 125,000 61,000 41,000 NKr1,912,000 NKr1,912,000 Solution TotalCash32,000Accounts Receivable43,600Inventory55,400Property, plant, & machinery121,800Cost of Goods Sold84,200Operating expenses25,000Depreciation Expense12,200Dividends Paid8,200Total382,400Total Debits382,400Accumulated Depreciation33,400Accounts Payable18,200Notes Payable41,800Common Stock86,000Retained Earnings54,000Sales149,000 Total382,400.

2. July 01 Record the dividend received from the foreign subsidiary..pdf

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You have isolated a new virus, determined it has a lipid envelope, and wish to characterize the mechanism of entry. Discuss the important research questions you would ask and the types of experiments you might use to address the questions. You have isolated a new virus, determined it has a lipid envelope, and wish to characterize the mechanism of entry. Discuss the important research questions you would ask and the types of experiments you might use to address the questions. Solution Answer: (There may be several questions related to the attachment and entry process, a few have been discussed) The first encounter between a virus and a target cell occurs at the plasma membrane. It is here that viruses first engage binding and entry receptors to initiate the infection process. Specific research questions on the mechanism of entry of this virus may be: 1. Adhesion and attachment: While normally non-specific in nature, different glycosphingolipids on the viral surface have been reported to confer cell specificity: the sialyllactose moiety on human immunodeficiency virus (HIV) membrane gangliosides, for instance, has been suggested to promote uptake of HIV into mature dendritic cells (mDCs). So, it can be asked whether the mechanism of entry is through specific glycosphingolipids and new roles of these glycosphingolipids? Experiments: Using liposomes to mimic the lipid composition and size of the virus, it can be demonstrate whether gangliosides are the key molecules that mediate liposome uptake. Liposome and VLP (virus like particle) capture assays on cultured cells, confocal microscopy and neuraminidase assay with VLP and LUV (large unilamellar vescicles) can be used for the investigation. 2. Lipid-mediated signalling for entry: Low-density lipoprotein receptors, negatively charged phospholipids and gangliosides have all been shown to assist entry of different viruses. Low-density lipoprotein receptors, also known as cholesterol receptors, are used by several members of the Flaviviridae family, including hepatitis C virus (HCV). The interaction is likely to occur through lipoproteins associated with the virion, such as cholesteryl esters and apolipoproteins (ApoB and ApoE) acquired during the assembly and budding of virus. So, it can be determined whether the LDL receptor acts as a receptor for this new virus and whether the endocytosis of the virus correlated with the level of LDL receptors?. Investigation of endocytosis of the virus by cultured cells in vitro can be demonstrated to have LDL receptors with the use of anti-LDL receptor antibody or DiI-LDL uptake. These cell lines can then be inoculated with a high titer virus positive serum. Intracellular viral RNA then can be detected by using ISH. To determine whether endocytosis of the virus correlated with the level of LDL receptor expression on cells, the well known modulatory effect of lipoproteins on the LDL receptor can be used to increase the number of LDL receptors (up-regulate) on cel.

You have isolated a new virus, determined it has a lipid envelope,.pdf

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Write the Java source code necessary to build a solution for the pro.pdf

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Why is agrobacterium tumfaciens useful to Kant biologists? Describe in detail the steps a researcher must take to get the desired DNA to hitch a ride on the back of Agro. Why is agrobacterium tumfaciens useful to Kant biologists? Describe in detail the steps a researcher must take to get the desired DNA to hitch a ride on the back of Agro. Solution Reasons Agrobacterium has a special plasmid called Ti (Tumor inducing) plasmid.It has TDNA (transfer DNA) which has the capability of transferring into the the plant cell and induce tumors. It is naturally attracted towards the wound site by chemicals It can transfer foreign to plants The Vie gene will induce the tumorogenic pathway Method Explants are a small portion of a plant. E.g.: seedling, protocols, leaf disc etc A wound is created in explant and incubated with Agrobacterium and acetosyringone. Acetosyringone is a phenolic product which attracts the Agrobacterium towards the wound. The explant and Agrobacterium are co cultivated and are transferred to a selective medium which contains antibiotic. The presence of antibiotic resistant gene in the Explants will enable the selection and it is transferred to a media which develops the transformed cell to a callus. On adding appropriate plant nutrients, roots and shoots are generated. These are transferred to a pot or field to develop into a adult plant..

Why is agrobacterium tumfaciens useful to Kant biologists Describ.pdf

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Why are the payback and Accounting or Average Rate of Return sometimes refer as unsophisticated methodologies? Solution The payback period is one of the most widely used tools for evaluating capital projects.The payback period represents the number of years it takes for the cash flows from a project to recover the project’s initial investment.Payback period is considered as unsophisticated methodologies because it does not account for time value of money.It does not consider cash flows past the payback period.It is biased against long term projects such as research and development and new product launches. Accounting or Average Rate of Return is considered as unsophisticated methodologies because it also does not account for time value of money.It does not consider the external factors that are also affecting the profitability of project.Accounting or Average Rate of Return cannot be applied in a project where investments are made in parts and in real life, a lot of investmentss are made in parts.So, it is considered as unsophisticated methodologies..

Why are the payback and Accounting or Average Rate of Return sometim.pdf

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Which of the following are potential differences between the Prokaryotic and Eukaryotic ETC? (Choose all that apply) Prokaryotes have Branched ETC Eukaryotes have longer ETC Prokaryotes use different systems to respond to oxygen concentrations All of the proteins in the ETC are shared between Prok. and Euk. The P/O ratio is higher in Prkaryotes The P/O ratio is higher in EukaryotesA. Prokaryotes have Branched ETCB. Eukaryotes have longer ETCC. Prokaryotes use different systems to respond to oxygen concentrationsD. All of the proteins in the ETC are shared between Prok. and Euk.E. The P/O ratio is higher in PrkaryotesF. The P/O ratio is higher in Eukaryotes Solution B)Eukaryotes have longer ETC the electron transport chain is long in eukaryotes ascompared to the prokaryotes..

Which of the following are potential differences between the Prokary.pdf

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When dividing, do we need to report remainder Why When dividi.pdf

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What are socioecology and sociobiology? What can they tell us about primate behavior? Use at least two non-human primate species in your answe Solution Answer: Socioecology focuses on the relationship between social behavior and the natural environment and favors selection for the individual’s daily survival. Sociobiology focuses on the genetic predisposition to behaviors and those behaviors’ enhancement of reproductive success. Socioecology: 1. Primatologists consider the following environmental factors: a. Quality and quantity of different foods b. Distribution of food resources, water, predators and sleeping sites c. Activity patterns (diurnal, nocturnal) d. Relationship with other species e. Impact of human activities 2. Environmental factors such as resource availability and predation have a strong influence on group size and structure. 3. Multimale and multifemale groups have an advantage when predation pressure is high. Adult males may join forces to chase and attack predators. (Savannah baboons have been known to kill domestic dogs and to attack leopards and lions) 4. Solitary foraging may be related to diet and distribution of resources or predator avoidance. Sociobiology: 1. Natural selection acts on behavior in the same way it acts on physical characteristics. 2. Behavior is a phenotypic expression and genes code for specific behaviors. 3. This approach provides an opportunity to study the effects natural selection has had in shaping primate behavior. Nonhuman primates are research subjects because they are so similar to humans. Shared ancestry is a major reason why human and nonhuman primates have many characteristics in common: tool use, long-lasting social relationships, and complex communication systems. By learning about nonhuman primates we may come to learn more about ourselves. Individuals of certain nonhuman primate species, however, are bipedal for some activities. By studying those species of nonhuman primates that are occasionally bipedal, and discovering the circumstances in which they display bipedality, we may gain some understanding of the factors that promoted the evolution of bipedality in humans. Human and nonhuman primates also share physiological characteristics. For example, the way in which the brains of rhesus monkeys and humans are organized is similar. One brain area that has been studied extensively is the visual system. Neuroanatomical studies of the nonhuman primate brain have been extremely useful in helping us to understand how the human brain functions and how we see. In this way, nonhuman primates serve as models of particular processes. Study of nonhuman primates has also contributed to our understanding of basic biological phenomena such as reproduction; to better understanding of diseases such as AIDS. Nonhuman primates are also studied because they are fascinating animals. They live in a wide range of habitats, and show many interesting differences in behavior and life styles. For example, in so.

What are socioecology and sociobiology What can they tell us about .pdf

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Using the code below, answer the questions:For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). Solution 1. Wo is an interface which contains unimplemented methods. It is similar to class and contains collection of abstract method. Here we can see it has one unimplemented methods doThat(). 2. doThat() is implemented in Roo because the Roo class implements Wo , So any class that implements Wo has to compulsory implement doThat() method , otherwise the compiler will throw a compile time error saying implement the unimplemented methods. c) List the methods available to a Roo object. Parent as well as child class methods as well its own implemnted methods..

Using the code below, answer the questionsFor the toolbar, press AL.pdf

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true or false: Strategy formulation and execution is a sequential process Solution Correct answer is True Strategy formulation and executon are the consecutive steps of the process of strategy management. Once the information collection is over and the collected information is alalyzed, the strategy is formulated and the implementation is done for the formulated strategies. It is formulated by the evaluation of the execution process and taking the control measures..

true or false Strategy formulation and execution is a sequential pr.pdf

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The symptoms of depression that occur in a person diagnosed with bipolar disorder are ____. distinct from the symptoms in unipolar disorder the same as the symptoms in major depressive disorder distinct from the symptoms in major depressive disorder the same as the symptoms in obsessive-compulsive disordera. distinct from the symptoms in unipolar disorderb. the same as the symptoms in major depressive disorderc. distinct from the symptoms in major depressive disorderd. the same as the symptoms in obsessive-compulsive disorder Solution The symptoms of depression that occur in a person diagnosed with bipolar disorder are distinct from the symptoms in major depressive disorder ..

The symptoms of depression that occur in a person diagnosed with bip.pdf

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The first signs of niacin deficiency are all of the following EXCEPT.pdf

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