“Web 2.0 is simply a new label for a range of web technologies and c.pdf

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“Web 2.0 is simply a new label for a range of web technologies and consumer behaviours that have existed since the 1990s. They don’t represent a ‘paradigm shift.’” Is this statement true or false? Be sure to support your answer. Solution Web 2.0 is simply a new label for a range of web technologies and consumer behaviours that have existed since the 1990s. They don’t represent a ‘paradigm shift - FALSE. Explanation :- Web 2.0 is an evolution of technologies and communications approaches which have grown in Importance since 2004-2005..

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Help with my biostats Homework. Please show all work!! Mendel developed his theory of genetics by observing large numbers of offspring from particular parents during plant breeding experiments. However, appropriate statistical methods for a rigorous analysis of such genetic data were only invented about 50 years later, so Mendel basically said \"The data seem to agree with Illy theory.\" Today, we can analyze Mendel\'s data statistically to see if he was right. According to Mendel\'s theory, if pea plants heterozygous for the two traits of seed color and seed texture are crossed, offspring will be produced with phenotypes in the following proportions: Here are the actual data Mendel collected from an experiment crossing two such parents: Use an appropriate statistical analysis to determine whether these data agree with Mendel\'s theory. State the conclusion of your analysis clearly. Solution We use Goodness of Fit Test to do it. The test hypothesis: Ho: these data agree with Mendel\'s theory (i.e. null hypothesis) Ha: these data do not agree with Mendel\'s theory (i.e. alternative hypothesis)observedexpectedO - E(O - E).

Help with my biostats Homework. Please show all work!! Mendel develo.pdf

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genetics q If the offspring of a dihydric testcross are roughly 50% parental type and 50% recombinant, we can conclude that... these genes are likely on different chromosomes. these genes are linked. these genes are likely on the same chromosome. Solution In a dihybrid test cross offspring will be 50% recombinant and 50% parental type , which indicates that these genes are likely on different chromosomes ..

genetics q If the offspring of a dihydric testcross are roughly 50 .pdf

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for fiscal year 2006, the national debt of a country was approximately (8.502 × 1018) dollars. At that time, the number of citizens between the ages of 18 and 65 was known to be (1.5 × 1012). If the national debt of this country were divided evenly among these citizens, about how much debt would be assigned to each individual? (Enter your answer in standard form, rounded to two decimal places.) $ per person Solution national debt of country=8.502*1018 number of citizens between age 18 and 65=1.5*1012 debt assigned to each individual = national debt of country/number of citizens between age 18 and 65 =(8.502*1018)/(1.5*1012) =5.702 dollars/person.

for fiscal year 2006, the national debt of a country was approximate.pdf

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Explain why Linux makes system performance monitoring available to the user. Describe how users can use this information to their advantage. Solution It’s really very tough job for every System or Network administrator to monitor and debug Linux System Performance problems every day. After being a Linux Administrator for 5 years in IT industry, I came to know that how hard is to monitor and keep systems up and running. For this reason, we’ve compiled the list of Top 20 frequently used command line monitoring tools that might be useful for every Linux/Unix System Administrator. These commands are available under all flavors of Linux and can be useful to monitor and find the actual causes of performance problem. This list of commands shown here are very enough for you to pick the one that is suitable for your monitoring scenario. There are several advantages to continuously monitoring system performance. Continuous system performance monitoring can do the following: Successful monitoring involves the following:.

Explain why Linux makes system performance monitoring available to t.pdf

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Discuss the relationships between competitive avantage, istinctive compentencies, resources,an capabilities. Solution Resources The activities and processes of the organization utilize certain assets. These assets are called resources. These resources can be created within the organization. They form the internal resources. Such generated resources are organization-specific. Otherwise they could be obtained externally from the suppliers available in the resource markets. They form the external resources. The externally obtained resources are organization-addressable. In addition resources can be categorised as specific or non-specific. Those resources which can only be used for extremely specialized intentions and are significant to the organization in adding value to goods and services are called specific resources. Resources Resources of the firm can include all assets, capabilities, organizational processes, firm attributes, information and knowledge. In short resources can be considered as inputs that facilitate the organization to perform its activities. All resources that an organization has may not have strategic relevance. Only certain resources are capable of being an input to a value creating strategy which put the organization in a position of competitive advantage. An organization’s resource should have four attributes to provide the potential for competitive advantage. These form the VRIN characteristics. The VRIN characteristics The important features for a resource to be strategically important are as below The VRIN characteristics mentioned above are individually necessary for the resources to be valuable. Non-specific resources are less specific and are less significant in adding value. Also resources can be broadly classified as tangible and intangible. The physical assets that an organization possesses are called tangible resources. The physical resources, human resources and final resources come under this category. The intellectual resources, technological resources and the organizational reputation together form the intangible resources. The patents and copyrights of the organization are typical examples of intellectual resources. The innovation capacity and innovation speed are examples of technological resources. Reputation is basically good-will that the organization has acquired among the customers. It is a critical resource of an organization. Competencies An organization should posses some characteristics in order to have the ability to compete with other organizations in the market place. These characteristics form the competencies of the organization. For any organization to survive in an industry competencies are must. At the same time competencies cannot be useful to an organization when they stand alone. It is when they combine together in the right combination that they help the organization to attain competitive advantage. For instance consider an information technology organization. For this to compete in t.

Discuss the relationships between competitive avantage, istinctive c.pdf

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Determine the intervals of the domain over which each function is continuous. See Solution we would see for any breaks in the graph 11. contious over the entire domain ( -inf , inf) 12 contious over the entire domain ( -inf , inf) 13 continous over the domain [3, inf) 14 continous over the domain (-inf , -1] 15 discontinous at x = 3 So, continous over the interval ( -inf , 3) U( 3, inf) 16 discontinous at x = 1 So, continous over the interval ( -inf , 1) U( 1, inf).

Determine the intervals of the domain over which each function is.pdf

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A storage reservoir contains 200 kg of a liquid that has a specific gravity of 1.5. What is the volume of the storage reservoir? Solution >> As, Specific density =1.5 = Density of Substance/Density of Water = Density/1000 => Density of Substance = 1500 Kg/m3 >> As, m = mass of liquid = 200 kg As, Mass = Volume*Density => Volume = 200/1500 = 0.1333 m3 => Volume of Storage Reservoir= 0.1333 m3. ...... ANSWER......

A storage reservoir contains 200 kg of a liquid that has a specific .pdf

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4. Define: modal split model transportation demand . central vision Solution a) It is used to determine the number of trips on different modes between different pairs of nodes. b) It is the demand for one good or service in one stream occurring as a result of demand from another. c) When our eye focussing the straight path, and the vision is clear and sharp where we can drive comfortably. Such a vision is called central vision..

4. Define modal split model transportation demand . central vision .pdf

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Comparison of dysplasia and hyperplasia? Solution Dysplasia: Dysplasia is a pre cancer, it is a term used in pathology to refer to an abnormality in maturation of cells within a tissue, refers to an abnormal and potentially reversible process where there is disordered growth and maturation of cells and the tissues and organs which they make up.. This generally consists of an expansion of immature cells, with a corresponding decrease in the number and location of mature cells. Dysplasia is often indicative of an early neoplastic process. The term dysplasia is typically used when the cellular abnormality is restricted to the originating tissue, as. For example, Myelodysplastic syndrome, or dysplasia of blood-forming cells, show increased numbers of immature cells in the bone marrow, and a decrease in mature, functional cells in the blood. Dysplasia can still be reversed if the factors that are responsible for causing it are eliminated. Hyperplasia: refer to the proliferation of cells within an organ or tissue beyond that which is ordinarily seen. Hyperplasia may result in the gross enlargement of an organ, the formation of a benign tumor, or may be visible only under a microscope, it is a general term for an increase in the number of the cells of an organ or tissue causing it to increase in size. It may be due to any number of causes such as increased demand, chronic inflammatory response, hormonal dysfunctions, or neoplasia. Hyperplasia is considered to be a physiological response to a specific stimulus, and the cells of a hyperplastic growth remain subject to normal regulatory control mechanisms..

Comparison of dysplasia and hyperplasiaSolutionDysplasia Dys.pdf

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You are required, but not limited, to turn in the following source files: Assignment8.java(More code need to be added) Athlete.java(given by the instructor, it needs to be modified for this assignment) AthleteNameComparator.java MedalCountComparator.java Sorts.java AthleteManagement.java Class Diagram: Athlete The Athlete class implements the \"Serializable\" interface so that its object can be stored. (The Serializable interface is defined in the \"java.io\" package.) AthleteNameComparator The AthleteNameComparator class implements the \"Comparator\" interface (The Comparator interface is in \"java.util\" package.). It needs to define the following method that was an inherited abstract method from Comparator interface: public int compare(Object first, Object second) (Note that you can also define: public int compare(Athlete first, Athlete second) instead by making the class implements Comparator. If the first argument object has a last name lexicographically less than that of the second argument, an int less than zero is returned. If the first argument object has a last name lexicographically larger than that of the second argument, an int greater than zero is returned. If their last names are same, then their first names should be compared. If they have same first and last names, then 0 should be returned. MedalCountComparator The MedalCountComparator class implements the \"Comparator\" interface (The Comparator interface is in \"java.util\" package.). It needs to define the following method that was an inherited abstract method from Comparator interface: public int compare(Object first, Object second) (Note that you can also define: public int compare(Athlete first, Athlete second) instead by making the class implements Comparator. If the first argument object has a larger gold medal count than that of the second argument, an int less than zero is returned. If the first argument object has a smaller gold medal count than that of the second argument, an int greater than zero is returned. If both gold medal counts are same, then their silver medal counts should be compared. If both gold medal counts are same and also silver medal counts are same, then their bronze counts should be compared. If all of gold, silver, and bronze medal counts are same, then 0 should be returned. Sorts The Sorts class is a utility class that will be used to sort a list of Athlete objects. Sorting algorithms are described in the algorithm note posted under Notes section of the course web site. These algorithms sort numbers stored in an array. It is your job to modify it to sort objects stored in an array list (or a vector). The Sorts class object will never be instantiated. It must have the following methods: public static void sort(ArrayList objects, Comparator) Your sort method utilizes the compare method of the parameter Comparator object to sort. You can use one of Selection sort or Insertion Sort. AthleteManagement The AthleteManagement class has a list of Ath.

You are required, but not limited, to turn in the following source f.pdf

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Write a function to merge two doubly linked lists. The input lists have their elements in sorted order, from lowest to highest. The output list should also be sorted from lowest to highest. Your algorithm should run in linear time on the length of the output list. Provide an algorithm for your function Implement and show some samples of your running function Solution The algorithm for LinkedList MergeSort is as follows. The trick in this algorithm is to use a local list to store the value which is pointing to the actual list to be returned as a result of the function Step1: START Step2: GET inputs of list1 and list2 Step3: INITIALIZE mergelist STEP4: SET mylist = mergelist Step5: WHILE list1 OR list2 is EMPTY DO IF (list1 AND list2) are NOT EMPTY IF list1.value > list2.value mylist.next = list2 ELSE mylist.next = list1 ELSE IF list1 is EMPTY mylist.next = list2 ELSE mylist.next = list1 Step6: RETURN mergelist Step7: STOP Sample program for the same implemented in C++ //==================================================================== ======== // Name : DoublyLinkedList.cpp // Author : Kaju // Version : 0.1 // Copyright : This is just an example code // Description : MergeSort in C++, Ansi-style //==================================================================== ======== /* * C++ Program to Implement Doubly Linked List */ #include #include #include /* * Node Declaration */ using namespace std; // struct declaration for node which will hold the integer data with links to previous data and next data struct node { int data; struct node *next; struct node *prev; }; /* Class Declaration */ class DoublyLinkedList { private: struct node *start; // collection of all the nodes public: void create(int value); int peek(); int delete_head(); void display(); int count(); void mergeSort(DoublyLinkedList list1, DoublyLinkedList list2); DoublyLinkedList() { start = NULL; } }; /* * Add data into the list */ void DoublyLinkedList::create(int value) { struct node *s, *temp; temp = new(struct node); temp->data = value; temp->next = NULL; if (start == NULL) { temp->prev = NULL; start = temp; } else { s = start; while (s->next != NULL) s = s->next; s->next = temp; temp->prev = s; } } /* * Deletion of the first element from the list * store the first element in a temp node and then remove it from start. */ int DoublyLinkedList::delete_head() { struct node *tmp, *q; int value; tmp = start; if(start->next!=NULL) { start = start->next; start->prev = NULL; } else start = NULL; value = tmp->data; free(tmp); return value; } /* * return the first element of Doubly Link List */ int DoublyLinkedList::peek() { struct node *q; if (start == NULL) { cout<<\"List is empty\"<data; } } /* * Display elements of Doubly Link List */ void DoublyLinkedList::display() { struct node *q; if (start == NULL) { cout<<\"List is empty\"<data; q = q->next; if(q != NULL) { cout<<\", \"; } } cout<<\"]\"<next; cnt++; } return cnt; } /* * MergeSort - takes two input lists and then based on the.

Write a function to merge two doubly linked lists. The input lists ha.pdf

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Why just one sperm can enter the secondary oocyte? Why just one sperm can enter the secondary oocyte? Solution Sperm fuses with the egg and fertilizes it. Normally only one sperm fuses with an egg and if more than one sperm fuses this is called polyspermy.Development of the zygote usually stops if polyspermy occurs and in most cases, the egg cell blocks polyspermy. In polyspermy,multipolar or extra miotic spindles are formed which stop the fertilized egg development by causing faulty segregation of chromosomes during cell division. The egg plasma prevent polyspermy by depolarising the egg plasma membranes after the first sperm cell fuses with the egg cell. Thus preventing further sperm from fusing, this is the primary block to polyspermy. The secondary block to polyspermy is by the cortical reaction within the egg cell . the fusion of the sperm with an egg plasma membrane causes an increase in cytosolic calcium within the egg. The calcium activates the egg to begin development, which initiates a cortical reaction that releases various enzymes that change the structure of the outside layer of the egg. This hardens the zona pellucida which prevents another sperm from binding to it..

Why just one sperm can enter the secondary oocyteWhy just one s.pdf

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Which plants produce multiple sporangia? (Select all that apply.) Gametophytes of bryophytes Gametophytes of seedless vascular plants Sporophytes of bryophytes Sporophytes of seedless vascular plants None of the above Solution Answer: Sporophytes of seedless vascular plnats Explanation: Each bryophyte sprophyte is unbranched and produces a single sporangium whereas branched sporophytes of vascular plants (ferns & gymnosperms) produce multiple sporangia..

Which plants produce multiple sporangia (Select all that apply.) Ga.pdf

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what is economics Solution \"Economics is the study of people in the ordinary business of life.\" -- Alfred Marshall Economicsis the social science that describes the factors that determine the production, distriution and consumption of goods and services.Economics is the study of how people choose to use resources.Resources include the time and talent people have available, the land, buildings, equipment, and other tools on hand, and the knowledge of how to combine them to create useful products and services. Important choices involve how much time to devote to work, to school, and to leisure, how many dollars to spend and how many to save, how to combine resources to produce goods and services, and how to vote and shape the level of taxes and the role of government. Often, people appear to use their resources to improve their well-being. Well-being includes the satisfaction people gain from the products and services they choose to consume, from their time spent in leisure and with family and community as well as in jobs, and the security and services provided by effective governments. Sometimes, however, people appear to use their resources in ways that don\'t improve their well-being. In short, economics includes the study of labor, land, and investments, of money, income, and production, and of taxes and government expenditures. Economists seek to measure well-being, to learn how well-being may increase over time, and to evaluate the well-being of the rich and the poor. The most famous book in economics is the Inquiry into the Nature and Causes of The Wealth of Nations written by Adam Smith, and published in 1776 in Scotland. Although the behavior of individuals is important, economics also addresses the collective behavior of businesses and industries, governments and countries, and the globe as a whole. Microeconomics starts by thinking about how individuals make decisions. Macroeconomics considers aggregate outcomes. The two points of view are essential in understanding most economic phenomena..

what is economicsSolutionEconomics is the study of people in.pdf

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What are the functions of database management software (DBMS)? Name some common database management software. Solution Functions of Database Management Software are:- 1. Data Dictionary Management - where the DBMS stores definitions of the data elements and their relationships. This function is used by DBMS to look up the required data component structures and relationships. 2. Data Storage Management - This function is used for the storage of the data and related entry forms, report definitions, procedural code, data validation rules, and structures that can handle video and picture formats. 3. Security Management - sets rules that determine specific users that are allowed to access the database through username and password provided to each user. 4. Multiuser Access Control - Data integrity and data consistency are the basis of this function. Multiuser access control is a very useful tool in a DBMS, it enables multiple users to access the database simultaneously without affecting the integrity of the database. Some common DBMS systems are Oracle database, Microsoft SQL, SAP, MySQL, Teradata, informix..

What are the functions of database management software (DBMS) Name .pdf

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Two proteins, HU and H-NS (Histone-like Nucleoid Structuring), are found to be associated with the bacteral genome. You would predict that HU and H-NS would have many ___ amino acid residues. A) aromatic B) aliphatic C) neutral D) acidic E) basic Solution Two proteins HU and H –NS (histone –like nucleoid structure) found to be associated with the bacterial geneome HU and H –NS would have many Neutral aminoacids residues Answer : neutral.

Two proteins, HU and H-NS (Histone-like Nucleoid Structuring), are fo.pdf

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The phylogenetic distribution of multicellular organisms indicates that multicellularity: Question 2 options: distinguishes all eukaryotes from prokaryotes was present in the common ancestor of all lineages arose independently multiple times defines a clear monophyletic group distinguishes all eukaryotes from prokaryotes was present in the common ancestor of all lineages arose independently multiple times defines a clear monophyletic group Solution arose independently multiple times Multicellularity has evolved independently at least 46 times including in some prokaryotes, like cyanobacteria, myxobacteria, actinomycetes, Magnetoglobus multicellularis or Methanosarcina..

The phylogenetic distribution of multicellular organisms indicates t.pdf

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Task #1 Correcting Logic Errors in Formulas Copy and compile the source file NumericTypes.java, run the program, and observe the output. Some of the output is incorrect. You need to correct logic errors in the average formula and the temperature conversion formula. The logic errors could be due to conversion between data types, order of operations, or formula problems. The necessary formulas are average = (score1+score2) / numberOfScores C = 5/9 (F-32) Make sure that the output makes sense before you continue. The average of 95 and 100 should be 97.5 and the temperature that water boils is 100 degrees Celsius Task #2 Using the Scanner Class for User Input 1. Add an import statement above the class declaration to make the Scanner class available to your program. 2. In the main method, create a Scanner object and connect it to the System.in object. 3. Prompt the user to enter his or her first name. 4. Read the name from the keyboard using the nextLine method, and store it into a variable called firstName (you will need to declare any variables you use). 5. Prompt the user to enter his or her last name. 6. Read the name from the keyboard and store it in a variable called lastName. 7. Concatenate the firstName and lastName with a space between them and store the result in a variable called fullName. 8. Print out the fullName. 9. Compile, debug, and run, using your name as test data. 10. Since we are adding on to the same program, each time we run the program we will get the output from the previous tasks before the output of the current task. Task #3 Working with Strings 1. Use the charAt method to get the first character in firstName and store it in a variable called firstInitial (you will need to declare any variables that you use). 2. Print out the user’s first initial. 3. Use the toUpperCase method to change the fullName to uppercase and store it back into the fullName variable. 4. Add a line that prints out the value of fullName and how many characters (including the space) are in the string stored in fullName (use the length method to obtain that information). 5. Compile, debug, and run. The new output added on after the output from the previous tasks should have your initials and your full name in uppercase characters. Task #4 Using Predefined Math Functions Task #4 UsingPredefined Math Functions Add a line that prompts the user to enter the diameter of a sphere. Read in and store the number into a variable called diameter (you will need to declare any variables that you use). The diameter is twice as long as the radius, so calculate and store the radius in an appropriately named variable. The formula for the volume of a sphere is r3 Convert the formula to Java and add a line which calculates and stores the value of volume in an appropriately named variable. Use Math.PI for and Math.pow to cube the radius. Print your results to the screen with an appropriate message. Compile, debug, and run using the following test data and record the resu.

Task #1 Correcting Logic Errors in FormulasCopy and compile the so.pdf

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1 group SIV- chimpanzee V- 1 SIV. chimpanzee SIV- chimpanzee L SIV- chimpanzee SIV chimpanzee HIV. 1 group O HIV- group P SIV. gorilla SIV- chimpanzee SIV-red-capped manga SIV- drill SIV. vervet monkey SIV- tantalus SIV. sooty mangabey HIV- 2 group A HIV-2 group B SIV. sooty mangabey SIV. syke\'s monkey SIV- greater spot-nasal monkey SIV. De Brazzas monkey Solution Choice 4 is correct. Reason: It can be seen from the image that both SIV and HIV emerged at a similar rate and pace in ancient times and their local distribution shares similar levels of complexities. It can also be noted that N type HIV1 is not closely related to O type HIV but represent a divisional subtype of the main HIV-1 type. Also, it can be carefully noted in the initial part of the image that SIV had already emerged in chimpanzees much before evolution of HIV types. However, futher evolution of HIV from SIV can be easily traced back from chimpanzees to humans. Further, evolutionay timelines and the image shows that HIV-1 group O and P are closely related and group A and B of HIV-2 were transmitted to humans. Thus, the information states that choice 4 is most correct..

1 group SIV- chimpanzee V- 1 SIV. chimpanzee SIV- chimpanzee L SIV- c.pdf

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Richards has filed quarterly reports to the SEC based on information supplied by management. A potential concern of such reliance is which of the following Privity only Liability to 3rd parties. Liability to foreseen 3rd parties only. Liability to reasonably foreseen 3rd parties only. Solution Answer: Liability to foreseen 3rd parties only. This is according to the \"middle ground\" approach followed by most states..

Richards has filed quarterly reports to the SEC based on information.pdf

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