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managerial reporting systems are: Standardized Rigid Flexible managerial reporting systems are: Standardized Rigid Flexible managerial reporting systems are: Standardized Rigid Flexible Solution Managerial accounting systems are Flexible. The budgets, standards are revised according to the situations and to adapt to changes.

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Describe the differences seen in the hydra, planarian, clam, grasshopper, and pig for symmetry, tissue organization, body cavity, digestive system, circulatory system, respiratory system, excretory system, support system and nervous system. Solution Hydra Planarian: Clam: Symmetry radial bilateral bilateral Tissue Organization diploblastic triploblastic triploblastic Type of Body Cavity: gastrovascular cavity acoelomate coelomate: pericardial cavity Digestive Openings one mouth (pharynx extended during feeding) mouth, a nu s Circulatory System none none heart, open c.s. Respiratory Organs: none none incurrent and excurrent siphon, gills Excretory System none protonephridia, flame bulb nephridia Support System : none none nephridia Nervous System Organization: nerve net Organization: brain, ventral nerve cords Hydra Planarian: Clam: Symmetry radial bilateral bilateral Tissue Organization diploblastic triploblastic triploblastic Type of Body Cavity: gastrovascular cavity acoelomate coelomate: pericardial cavity Digestive Openings one mouth (pharynx extended during feeding) mouth, a nu s Circulatory System none none heart, open c.s. Respiratory Organs: none none incurrent and excurrent siphon, gills Excretory System none protonephridia, flame bulb nephridia Support System : none none nephridia Nervous System Organization: nerve net Organization: brain, ventral nerve cordsNervous System Organization: three ganglia.

Describe the differences seen in the hydra, planarian, clam, grasshop.pdf

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Companyhas prot s Price per unit Variable cost per unit Fixed costs p.pdf

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Alice Agent is an employee of Patti Principal. She negligently runs a stop sign in the course of her employment and injures Bart Bikerider. Can Bart sue Patti for his damages from Alice’s Negligence? Why or why not? Solution Yes Bart can sue Patti for his damages because the employer is responsible for actions of employees during the course of their employment. The doctrine is called respondeat superior..

Alice Agent is an employee of Patti Principal. She negligently runs .pdf

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at log kHz Solution You can have several problems: 1) if you are working with transistors sure you\'re not saturating. 2) if diodes are sure they are doing the right contact or try another. 3) Make sure your input source this good. 4) for me the most certain is that you are using an active elemnto and this is in court and why that image, check input sources, the currents that these manipulated are optimal for circuit elements you use or application you\'re doing..

at log kHz SolutionYou can have several problems 1) if you a.pdf

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Bark, which is thicken, toughen periderm, arises from this lateral m.pdf

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Why do you think it is usually preferable to start with the side containing the more complicated expression when establishing a trigonometric identity? Solution 1. Human brains are accustomed to simplify complicated things rather than complicating a simpler one. 2. A simpler/simplest conclusion of a complicated problem in trigonometric identities is only a handful;mere 2 or 3. But if I want to complicate a simpler side, outcomes are numerous. For eg: prove cosx*tanx =sinx Now I know that if I want to simplify LHS, I will try to convert it is terms of Sin or Cos (which we consider simple in the world of trigonometry) So cosx*(sinx/cosx)= cosx which is RHS. But if I want to complicate sinx ,I can write it as 1/cscx or 1-cos^2x or (1-cos2x)/2 . . . . And the list goes on! So, by and large, it is customary simplify complicated things rather than the other way. Hope it clarifies your doubt!.

Why do you think it is usually preferable to start with the side con.pdf

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Why is the mucus thicker than normal in CF patients? A good answer will include a description of how osmosis is affected by CF. Solution Mucus is a substance that coats our Epithelial surface.It keeps surfaces lubricated and helps prevent disease by trapping potential invaders.Invading microbes can become trapped in the mucus.The epithelial cells are lined with cilia that are constantly beating and moving mucus with trapped bacteria up and out of the respiratory tract.Mucus is amixture of predominantly water and glycproteins.The mucins are produced by goblet cell that are interspersed among the respiratory epithelial cell.The composition,water content and volume of mucus are controlled by the secretion and absorption of ions and water across the respiratory epithelium.The salt imbalance resulting from a disfuctional chloride channel could through osmosis lead to the thick mucus characterstic of cystic fibrosis mutation could lead to disease.cystic fibrosis is a thick that lines both the airways and the intestinal tract of affected individual.In the respiratory tract this thick mucus is not removed by the cillia that usually beat and push mucus up into the throat.This thick mucus builds up in the airways and bacteria trapped in the mucus are not removed.Thus bacterial respiratory infections are extremely common and dangerous in CF individuals..

Why is the mucus thicker than normal in CF patients A good answer w.pdf

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Which of the following primordial reproductive structures is NOT ext.pdf

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what kind of accessory protein would you classify IRS1 and 2 as? Docking protein Adaptor protein? Transcription factor? SH2 domain protein Solution Docking protein - The robosomes are of two types based on their occurence in the cell. Free ribosomes and membrane bound (on ER surface). Ever wondered why and how they are of two types? Yes!! Some specific proteins which are found on the surface of certain specific ribosomal particles making them \"dock\" (stick) on to the ER surface are called docking proteins. Adapter protein - An adapter protein is one which assists in signal transduction across membranes. They aid receptor mechanisms in the cells. Transcription factor - The process of copying cell DNA codons in to an mRNA strand in called transcription. This is assisted by certain types of protein called Transcription factors. SH2 domain - These are specific proteins that mediate and fecilitate protein-protein inteactions in the cell. SH2 domain (a structurally special part within a protein molecule) is found within Src oncoprotein (cancer protein) which makes the Src oncoprotein attach to other protein containing phosphorylated tyrosin residues. **IRS1 and IRS2 proteins are classified as adapter proteins.**.

what kind of accessory protein would you classify IRS1 and 2 as.pdf

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What is the main difference between prokaryotic and eukaryotic ribosomes? Prokaryotic ribosomes synthesize RNA whereas eukaryotic ribosomes synthesize proteins Eukaryotic ribosomes are composed of RNA and protons whereas prokaryote ribosomes are Solution None of the above The main difference is that Prokaryotic ribosomes(70S) having two subunits- 30S and 50S and Eukaryotic ribosomes(80S) also having two subunits- 40S and 60S Both are composed of RNA & proteins. 70S having 60% RNA & 40% proteins while 80S having 40%RNA & 60% proteins..

What is the main difference between prokaryotic and eukaryotic riboso.pdf

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what is the cardinality of a sample space? Solution A sample space is a collection of all possible outcomes of a random experiment. A random variable is a function defined on a sample space. We shall consider several examples shortly. Later on we shall introduce probability functions on the sample spaces. A sample space may be finite or infinite. Infinite sample spaces may be discrete or continuous.

what is the cardinality of a sample spaceSolutionA sample spa.pdf

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What is computer information systems in three paragraphs. Solution A computer information system be a system collected of people and computer that process or interpret information. The word is also from time to time used in more limited senses to refer to only the software used to sprint a automated database or to refer to only a computer system. A computer(based) information system, or CBIS, uses computers to gather, procedure, store, examine and allocate information for a specific reason, such as gathering a trade objective. The main mechanism of a CBIS comprise hardware, software, data, events and public. A computer(based) information system be basically an IS using computer knowledge to take out some or all of its intended tasks. The basic mechanism of computer based information system are:- · Hardware these are the plans like the check, processor, printer and keyboard, all of which work jointly to accept, procedure, show information and in sequence. · Software are the program that let the hardware to procedure the data. · Databases are the meeting of connected files or tables contain related data. · Networks are a between system that allow diverse computer to hand out resources. · Procedures are the instructions for combine the mechanism above to development in sequence and produce the ideal output..

What is computer information systems in three paragraphs.Solutio.pdf

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Van der Waals bonding is due to the sharing of electrons between atoms. the transfer of electrons between atoms. unequal charge distribution around neutral molecules. atoms bonding to hydrogen molecules. Solution Covalent bond = the sharing of electrons between atoms. Ionic bond = the transfer of electrons between atoms. Hydrogen bond = atoms bonding to hydrogen molecules. Vanderwaals bonding = unequal charge distribution around neutral molecules..

Van der Waals bonding is due tothe sharing of electrons between at.pdf

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USE TORA TO: Determine the maximum flow and the optimum flow in each arc for the network in the following figure. Clearly show all steps and data input. Exterior Graph Data, Nodes listed outside paranthesis Node 1 (14) to Node 3 (0) Node 3 (10) to Node 5 (0) Node 5(0) to Node 4 (5) Node 4(6) to Node 2 (7) Node 2(0) to Node 1 (8) Interior Graph (Triangle ) Data Node3 (10) to Node2(5) Node3 (9) to Node 4(7) Second Interior Graph Data Node5 (0) to Node2(6) Solution The most successful and simple model is Erlang\'s loss system where the blocking probability is given by Erlang\'s B-formula. The traffic is described by the offered traffic A, the system (only one link) by the number of channels, and the strategy is full accessibility with lost calls cleared. The above-mentioned three elements of the model are each described by only one parameter. Only single-channel calls are considered. The network performance is described by the blocking probability E1,n(A), i.e. the probability that a call attempts is blocked because all n channels are busy. For Erlang\'s loss system time, call, and traffic congestions are equal. The state probabilities are given by the truncated Poisson distribution, and when the number of channels is very large this becomes a Poisson distribution. This model has been very successful for traffic engineering. The background for this success is that the traffic is very well modeled by one parameter only. The underlying mathematical assumption is a Poisson arrival process. This is fulfilled when the traffic is generated by many independent users, which is the case for telephony. If the arrival process is a Poisson process, then the model is insensitive to the holding time distribution, which means that only the mean holding time is of importance. So the model is very robust to the traffic and models the real world extremely well. Improvement function: This denotes the increase in carried traffic when the number of channels is increased by one from n to n + 1: 64 Attention: This is not an ITU publication made available to the public, but an internal ITU Document intended only for use by the Member States of the ITU and by its Sector Members and their respective staff and collaborators in their ITU related work. It shall not be made available to, and used by, any other persons or entities without the prior written consent of the ITU. ITU Telecom Network Planning Reference Manual - Draft version 4.1 January 2007 4.8.4.3 Engset\'s loss system The Poisson arrival process is the most random process, and the calls are generated by a very large number of independent sources, each having an infinitesimal calling rate. In many real systems the number of users is limited, and the arrival process is more regular or smooth than random traffic. This is modeled by Engset\'s loss system where we have a finite number S of users (traffic sources) which alternates between the states off (= idle) and on (= busy). When a source is idle it generates cal.

USE TORA TODetermine the maximum flow and the optimum flow in eac.pdf

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The three ways of promoting ACTIVE transport are: a) Channels, porins and coupled transporters. b) Porins, channels and ATP driven transport. c) Coupled transporters, ATP and LIGHT driven transport. d) Antiporter, symporter and coupled transporters. Solution c. Coupled transporters, ATP and LIGHT driven transport ATP use the energy derived from hydrolysis of ATP to transport solutes from the electrochemical gradient. Coupled transporters use the energy stored in the electrochemical gradient of one solute to transport another. Light-driven transport use energy derived from gentle to drive translocation..

The three ways of promoting ACTIVE transport area) Channels, pori.pdf

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The picture below describes three aspects of using a variable. Identify each. Name the three types of loops and briefly describe the main use of each. Solution 1) A)This one describes the datatype of the varibles, datatType are int,float,char,short,long,double etc B)This one describes the identifier/variable, which is sequence of symbols, which is representation of a memory block. C)value, this is value of the variable, this value is stored in the memory block that is associated to the variable 2) A)for loop: syntax is for(i=0;i.

The picture below describes three aspects of using a variable. Identi.pdf

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The idea of the principle of superposition in stratigraphy is that the layer on the bottom in a stratigraphic sequence is the youngest layer. the layer on the bottom in a stratigraphic sequence is the oldest layer. the decay at a known rate of the radioactive isotope in the stratigraphy makes it possible to determine the date. all layers in a stratigraphic sequence date to the same time. An artifact is a. an object made by humans, such as a stone tool. a person who excavates and studies artifacts. A location with evidence of past human activity, including artifacts and other remains. an object used by humans, but not modified by them, such as plant remains. An event that occurred at the end of the Mesolithic was the first appearance of life on earth. first appearance of reptiles. first appearance of the therapsids. extinction of the dinosaurs. Solution 31.b.The principle of superposition states that rock forms horizontally with the oldest layers of rock lying at the bottom of a formation and newer rock resting on top. 32.a.Any object made by human beings,especially with a view to subsequent use.It is a handmade object. 33.4.extinction of dinosaurs.

The idea of the principle of superposition in stratigraphy is that t.pdf

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The Force is strong among the inhabitants of Tatooine, the birthplace of Luke Skywalker. Not all of its inhabitants are capable of becoming a Jedi, but the probability is higher than among the rest of the members of the Alliance. Information was collected regarding this phenomena and it was determined that 54% of all of the members of the Alliance were born on Tatooine. It was also found that of all the members of the Alliance, 44% had the potential to become a Jedi. Finally, it was determined that 35% of all members of the Alliance were both from Tatooine and capable of becoming a Jedi. Based on this information calculate the following probabilities: 1. What is the probability that a member of the Alliance is randomly selected and found to be from Tatooine and not a Jedi? 2. What is the probability that a member of the Alliance is randomly selected and found to be from Tatooine or a Jedi? 3. What is the probability that a member of the Alliance is randomly selected and found to be neither from Tatooine nor a Jedi? 4. Given that a member of the Alliance is a Jedi, what is the probability that he/she is from Tatooine? Solution a Jedi 1. P(from Tatooine and not a jedi) = P(from Tatooine) - P(from Tatooine and a Jedi) = 0.54-0.35 = 0.19 2. P(from Tatooine or a jedi) = 0.35+0.19+0.09 = 0.63 3. P(neither from Tatooine nor a jedi) = 1 - 0.63 = 0.37 4. P(from Tatooine | a jedi) = 0.35/0.44 = 0.80 a Jedinot a jedifrom Tatooine0.350.54-0.35 = 0.19not from Tatooine0.44-0.35 = 0.091- (0.35+0.19+0.09)= 0.37.

The Force is strong among the inhabitants of Tatooine, the birthplac.pdf

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Suppose that you are given a short fragment ofDNA to sequence. You amplify the fragment with PCR and set up a series of four dideoxy reactions. You then separate the products of the reactions by gel electrophoresis and obtain the following banding pattern: Write out the base sequence of the original fragment that you were given. Original sequence: 5\'-__3^r Solution The sequence of the DNA fragment from 5\'-3\' will be: TACTGATGC these method of DNA sequencing is known as Sanger method or dideoxynucleotide chain termination method of DNA sequencing. In this method first of all the DNA fragment to be sequenced is first of all is denatured by denaturing agents and then is annealed with primers to initiate the copying of DNA, after which it is extended by DNA polymerase enzyme using 1) normal deoxynucleotide triphosphate(dNTPs) and 2)dideoxynucleotide triphosphates (ddNTPs) which are chain terminating nucleotide. Normally in copying of DNA the new dNTP is added to the 3’OH group of the previous dNTP thus the extension of DNA continues. The ddNTPS lacks these 3’OH group thus the next dNTP cannot be added thus the DNA polymerase fails in its work and cannot copy the DNA and thus these leds in the termination of the chain. Thus the copied strand lenth depends on when the ddNTP is incorporated,as these incorporation leads to stopping of chain elongation. as in the above question 4 different testubes have been prepared all of them containing denatured DNA, annealed primer , DNA polymerase and dNTP but they vary in ddNTP containing different types of ddNTP when incorporated stops the chain at that particular ddNTP containing aliquot. thus when such testtubes are run in electrophoresis give varying bands of different length which can be read from 5\'-3\' direction. here, starting from down i.e. T will be the first nucleotidethen the next band to the level just above it here it is A and in the same way other can be read..

Suppose that you are given a short fragment ofDNA to sequence. You am.pdf

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Suppose that species I has 2n=10 and species II has 2n=12. How many chromosomes would be present in the allotetraploid? Solution Allotetraploid is the condition where hybrid must have chromosomes from both species and a total of 4n(chromosome set four times of haploid organism) Species I has 2n=10 (n = 5) and species II has 2n=12 (n = 6) For allotetraploid there are 3 possible combination - 2n from both species I and species II - 10 + 12 = 22 chromosomes 3n from species I + 1n from species II : 3(5) + 1(6) = 21 chromosomes 1n from species I + 3n from species II : 1(5) + 3(6) = 23 chromosomes..

Suppose that species I has 2n=10 and species II has 2n=12. How many .pdf

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