The daily number of accidental disconnects with a server follows a Po.pdf

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The daily number of accidental disconnects with a server follows a Poisson distribution. On live days 2 5 3 3 7 accidental disconnects are observed. Obtain the maximum likelihood estimate of lambda. Find the maximum likelihood estimate of the probability that 3 or more accidental disconnects occur. Solution Average = 20/4 =5 Hence MLE of lemda = 5 P(X>=3) = 0.8753.

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5 000-0 SolutionEquity multiplier = Total Assets Equityor, .pdf

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1sequences and sampling. Suppose we went to sample the x-axis from Xmin to Xmax using a step size of step A)Draw a picture of what is going on. B) Write a expression for n the total number of samples involved (in terms of Xmin, Xmax and step) C) Write out the sequence of x-samples D) Write a direct and general expression for xi that captures the sequence E) Write a recursive expression for the sequence F) Write a program to compute and store the x-samples over the range -5x5 using a step size of 0.1 do everything in main () 2 . We talked about the following string functions that are available in C (as long as you include string.h): int strlen(char str[]) void strcpy(char str1[], char str2[]) void strcat(char str1[], str2[]) Write your own versions of these functions; for example: int paul_strlen(int char str[]). Hint: for your version of the strlen function, start at the first character in the array and keep counting until you find the ‘\\0’ character (use a while loop for this). Note: Use your version of the strlen function in the strcpy and strcat functions. 9. We want to insert a number into an array. (a) Formulate the problem mathematically with two sequences: x and y. (b) Write a function of the form: insertNumIntoArray(int n, int array[], int num, int index) The function inserts num into the array at the specified index. The rest of the array then follows. For example, if num = 9 and index = 3 and array = [7 2 8 8 3 1 2] then the function will produce: array = [7 2 8 9 8 3 1 2] Note: assume that array is properly dimensioned to have at least 1 extra space for storage. 10. Repeat #2 by for the delete operation; that is, we want to delete a single element (at a specified index) from an array; for example, suppose index = 3 and array = [50 70 10 90 60 20], then the result will be array: [50 70 10 60 20] 11. Repeat #2 by for an insert operation where we are inserting several values into the array. The function should be of the form: int insertArrayIntoArray(int n, int inArray[], int nInsert, int insertArray[], int outArray[], int index) The dimension of outArray is returned (explicitly). For example: inArrayarray: [7 2 8 6 3 9] insertArray: [50 60 70] index: 2 outArray: [7 2 50 60 70 8 6 3 9] Assume that outArray is large enough to hold all n + nInsert values. Solution #include //Simulates strlen() library function int paul_strlen(char str[]) { int l; for(l = 0; str[l] != \'\\0\'; l++) ; return l; } //Simulates strcpy() library function void paul_strcpy(char str1[], char str2[]) { int c; for(c = 0; str1[c] != \'\\0\'; c++) str2[c] = str1[c]; str2[c] = \'\\0\'; printf(\"\ Original String: %s\", str1); printf(\"\ Copied String: %s\", str2); } //Simulates strcat() library function void paul_strcat(char str1[], char str2[]) { int i, j; for(i = 0; str1[i] != \'\\0\'; i++) ; for (j = 0; str2[j] != \'\\0\'; i++, j++) { str1[i] = str2[j]; } str1[i] = \'\\0\'; printf(\"\ Concatenated String: %s\", str1); } int main() { char data1[20], data2[20]; pri.

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4 (4 points). How would you describe the difference in cell structures between a thermophile and a psychrophile? Include mention of cell membrane construction and enzyme/protein structure at least. Solution Thermophilic bacteria have more long chain fatty acids or lipids which maintain the membrane fluidity. They have more tetraether lipids that increase the membrane stability. Hence they survive in high temperatures. Thermophiles also have a high level of K+ salts which protect the DNA and proteins by forming hydrophilic shell. Thermophiles have have acidic aminoacids. psychrophiles grow at very low temperatures. They have high polyunsaturated and methyl- branched fatty acids, and/or a shorter acyl-chain length which are generally present in cis configuration . This increases membrane fluidity and reduces the interactions between the membrane preventing freezing at low temperature. They have cryoprotectants and antifreeze agents..

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Distinguish between cell fate and cell commitment. How can one assay these aspects of cellular phenotype by experimental embryology? Solution Cell fate--It means the development of the cell into an embryo and the events which govern the development of the egg cell throught the process of embryogenesis.The factors which determine the cell fate include cell signaling pathways like notch pathway etc, regulation and control by pathways like mRNA,siRNA inside the cell,development of different cavity inside the embryo, development of ectoderm,mesoderm and endoderm, and the regulation of cell fate determination by release of various controlling hormones. cell commitment--it involves the orientation and decision by the cell to undergo differentiation and follow a particular developmental pathway.It involves the release of various growth promoting factors by stem cells which lead to the differentiation of stem cells into mature cells by triggering and activating a particular organ development pathway example--blood cells. There are several aspects of determination of the cellular phenotype by experimental embryology----1)sex determination pathways in insects and dosage compensation 2)regulation of different genes in different climates,temperature, and background environment determines the activation and inhibition of development genes in caterpillar larvae of insects3) the division of fertilized egg into morula, blastula, gastrula ,and development of different layers by cell regulatory protein factors 4)role of cell adhesion protein molecules during cell differentiation pathway 5)development of cell symmetry and asymmetry of the insects 6)development of embryo cavity and germline layers of development after fertilization 6)development of cells to tissues to organs with the help of secreting chemicals, protein factors and hormones which activate the differentiation pathway..

Distinguish between cell fate and cell commitment. How can one assay.pdf

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Differentiate the processes of oogenesis and spermatogenesis. Are there similarities between the two processes? Differences? Consider the cells, their structures and functions. Solution Comparison between Spermatogenesis and Oogenesis Characteristics Spermatogenesis Oogenesis Definition The process of formation of sperm cells is called spermatogenesis Formation of ovum is called oogenesis Primary organ Testis (in males) Ovary (in females) The source cell/ structure that undergoes gametogenesis Spermatogonium Primordial follicle Source cell is preserved Yes, Spermatogonium is a stem cell. No Type of cell division Meiosis Meiosis Onset of gametogenesis At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty. During embryonic development. Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive cycles. Continuous or cyclic Continuous. Millions of sperms are produced each day. Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days). Occurrence of gametogenesis in life Begins at puberty Continues throughout life Begins at menarche Ends at menopause No. of functional gametes in one cycle 4 1 Polar cells (that degenerate later) produced No Yes, one polar cell in each of meiosis I and II. In spite of major difference of these two processes, they both share a common path and fate- they meiotic divisions that produce haploid gametes. Characteristics Spermatogenesis Oogenesis Definition The process of formation of sperm cells is called spermatogenesis Formation of ovum is called oogenesis Primary organ Testis (in males) Ovary (in females) The source cell/ structure that undergoes gametogenesis Spermatogonium Primordial follicle Source cell is preserved Yes, Spermatogonium is a stem cell. No Type of cell division Meiosis Meiosis Onset of gametogenesis At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty. During embryonic development. Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive cycles. Continuous or cyclic Continuous. Millions of sperms are produced each day. Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days). Occurrence of gametogenesis in life Begins at puberty Continues throughout life Begins at menarche Ends at menopause No. of functional gametes in one cycle 4 1 Polar cells (that degenerate later) produced No Yes, one polar cell in each of meiosis I and II..

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Define induction and give an example of deductive reasoning? Solution There are two methods of reasoning. One is Deductive approach and another is Inductive approach. Deductive reasoning:-This kind of reasoning works from the general to more specific. It is also referred to as \"Top-down\" approach. Examples:- 1) If X = Y and Y = Z, then by deductive reasoning X = Z. 2) Snakes are reptiles and reptiles are cold-blooded, by deductive reasoning it can be said that snakes are cold-blooded. Inductive reasoning:- Inductive reasoning as opposed to deductive reasoning work from specific observations to broader generalizations and theories. It is also referred to as \"bottom up\" approach in informal sense. Example:- 1) A just moved here from Hongkong. A has red hair, therefore by inductive reasoning, it can be said that People from Hongkong have red hair..

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A report says that the between-subjects factor of participants sal.pdf

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City GUI Write a Java GUI program which reads data about US cities. The data file is different, but is in the same format as in Lab #01. The GUI provides a combo box with the cities in the data file. As cities are selected in the combo box, information about the city is displayed. Requirements 1. The program must have a graphical user interface similar to the one on the bottom of the Requirements section. It must have a text field, a button, a combo box, and labels. 2. One or more layout managers must be used. This example was created using a border layout. The east and west panels each used a grid layout in addition. 3. A message must be displayed in the GUI after the data has been successfully read. A message should also be displayed if the file is not found. The program should not exit in that case, but allow the user to enter a different file name. 4. The combo box must be populated by the names of the cities in the data file in alphabetical order. Note that this means the GUI starts with an empty combo box. Items are added to it as or after the data file is read. 5. The program must provide an ActionListener for the Read button. It must also provide either an ActionListener or an ItemListener for the combo box, which displays the state, zip code, and time zone for that city. 6. The information for each city must be stored in its own object. Another object must be used to manage the collection of cities. This is similar to Lab #01. For this assignment you may assume that there will be no more than 100 cities. 7. The time zone should be displayed as the correct abbreviation for that city’s time zone in July. Here are the abbreviations. If the city observes daylight savings time, the S should be changed to a D. Note that some cities in the data file do not observe daylight savings time. Hours relative to GMT -4 -5 -6 -7 -8 -9 -10 Standard Time Abr. AST EST CST MST PST AKST HST I have attached an executable .jar file to the assignment that you can run to see how this program should work. Copy it to your computer. Copy the data file into the same directory as the .jar file. Suggestions 1. Create class diagrams first. This will help you think through and organize your program. There are many ways to do this. Start with the class diagrams from Lab #01. Both classes will require some additions. 2. Create the GUI and handle the IO in a separate class which extends JFrame. The main method can be included in this class. 3. Don’t worry about making your implementation efficient or using wonderful abstract data types (ADTs). You will be required to do that soon enough. 4. This program will take you several days to complete. Break it down into smaller tasks. This is the order I used, yours may be different. *Modify the class diagrams from Lab #01 to include new requirements that you identify now. You can add additional functionality later as you discover the need. *Implement those changes in the classes. * Create a GUI class. Implement a non-functiona.

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What is the role of sulfur chemoautotrophs in the sulfur cycle? Decomposition reduction oxidation assimilation Reduction of CO2 to generate carbohydrates can occur when coupled to the oxidation of reduced sulfur compounds (for examples, H2S). This latter process is carried out by what type of sulfate oxidizers sulfate reducers Solution 10 Oxidation Sulfur chemotrophs oxidize sufur to derive energy for the synthesis of ATP.

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What is the PDU at Layer 4 called? A. Data B. Segment C. Packet D. Frame E. Bit Solution B is correct. The Layer 4 PDU is called a segment. A, C, D, and E are incorrect. A is incorrect because the data is the PDU for the top three layers. C is incorrect because the packet is the PDU at the Network layer. D is incorrect because the frame is the PDU for the Data Link layer. E is incorrect because the bit is the PDU at the Physical layer..

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Briefly describe the contributions to the quality movement made by each of the following pioneers. Shewhart Dodge and Romig Sarasohn Deming Juran Solution Walter Shewhart - The Grandfather of Total Quality Management. His Ideas The original notions of Total Quality Management and continuous improvement trace back to a former Bell Telephone employee named Walter Shewhart. One of W. Edwards Deming\'s teachers, he preached the importance of adapting management processes to create profitable situations for both businesses and consumers, promoting the utilization of his own creation -- the SPC control chart. Dr. Shewhart believed that lack of information greatly hampered the efforts of control and management processes in a production environment. In order to aid a manager in making scientific, efficient, economical decisions, he developed Statistical Process Control methods. Many of the modern ideas regarding quality owe their inspirtation to Dr. Shewhart. He also developed the Shewhart Cycle Learning and Improvement cycle, combining both creative management thinking with statistical analysis. This cycle contains four continuous steps: Plan, Do, Study and Act. These steps (commonly refered to as the PDSA cycle), Shewhart believed, ultimately lead to total quality improvement. The cycle draws its structure from the notion that constant evaluation of management practices -- as well as the willingness of management to adopt and disregard unsupported ideas --are keys to the evolution of a successful enterprise. Harold French Dodge (January 23, 1893 in Lowell, Massachusetts – December 10, 1976) was one of the principal architects of the science of statistical quality control. He is universally known for his work in originating acceptance sampling plans for putting inspection operations on a scientific basis in terms of controllable risks. From 1917 to 1958 worked at quality assurance department at Bell Laboratories with Walter Shewhart, George Edwards, Harry Romig, R. L. Jones, Paul Olmstead, E.G.D. Paterson, and Mary N. Torrey. At that time the basic concepts of acceptance sampling was developed, such as : · Consumer\'s Risk, · Producer\'s Risk, · double sampling, · lot tolerance percent defective (LTPD), and · average outgoing quality limit (AOQL). Also he originated several types of: · acceptance sampling schemes, · CSP type continuous sampling plans, · chain sampling plans, and · skip-lot sampling plans. During World War II, Dodge served as a consultant to the Secretary of War, and was chairman of the American Standards Association (now the American National Standards Institute) War Committee Z1, which prepared the Z1.1, Z1.2, and Zl.3 quality control standards. After he retired from Bell Labs in 1958, Dodge became a professor of applied mathematical statistics at Rutgers. He also served as a consultant to NASA and the Sandia Corporation. The American Society for Testing and Materials honors Harold Dodge\'s memory with the Harold F. Dodge Award. Ho.

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Background Angiosperms (flowering plants) are the largest Phylum in the plant kingdom Plantae, (Note that the Phylum A may also called Anthophyta or Magnoliophyta in different classification systems.) These plants have true roots, stems, leaves, and flowers. The roots grow into the soil to anchor the plant in place and take up water and nutrients. The leaves are above ground and act as the main organs for photosynthesis. Stems provide above ground support for the plant and operate as conduits to move nutrients and water throughout the plant. Flowers contain the male and female reproductive organs of the plant. Assignment Details Part 1 Plants develop specialized roots, stems, leaves, and flowers to make them better adapted to their environment. People exploit many modifications as vegetables. An example of this is the taproot of carrots. The carrot plant stores sugars in the taproot to supply energy for the formation of flowers in later development. People dig up and eat the taproots while the sugar content is still high. Complete the following: Describe the anatomy (structure) and physiology (function) of a plant. Is the structure of your plant a modification from the standard root, stem, leaf, or flower? Explain how this structure differs from the standard root, stem, leaf, or flower. Compare the plant that you have chosen, its structure, and its function with 1 or more of your classmates\' choices. Part 2 Animals are important for humans in ways other than providing food. Medicines, clothing, beauty, fuel, air quality, building materials, and industrial products are a few of the ways in which humans exploit animals. Complete the following: Choose l product or benefit that people get from animals (besides food), describe it, and explain how people benefit. Which do you think is the most important benefit of animals? Select 1 benefit and explain. List 3 adaptations that animals have that plants do not have. Provide references in APA format. This includes a reference list and in-text citations for references used throughout the assignment Solution 1. the normal function of roots is fixation of the plant in the soil absorption of water and Minerals from the soil and conduction of the materials to the stem but in some plants roots perform special functions like storage of food materials(carrot,raddish) respiration(rhizopora) vegetative reproduction (sweet potato)support(betel vine) photosynthesis(orchids). Betel vine: Weak Climbing Plants produce some advantitious roots which cling to the supports and help the plants in climbing they sometimes they secrete sticky substance or a disc like structures for this purpose. Betel Vine is a plant having climbing roots these adventitious roots are produced from their nodes which help them to climb over a support another example is money plant. These roots are aerial roots which twist and clasp the support to help plant to vlimb. Normal rootsd show positive geotropism but these climbing roots show negativ.

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Write a Pep/8 Assembly program that reads in and stores two integers from the user and prints out their sum in the form: Total = Submit: Pep/8 formatted program in a text file Solution Answer: Assembly Language Program : .zero 1 .LC0: .string \"Enter two integers: \" .LC1: .string \"Total\" .LC2: .string \" = \" main: push rbp mov rbp, rsp sub rsp, 16 mov esi, OFFSET FLAT:.LC0 mov edi, OFFSET FLAT:std::cout call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*) lea rax, [rbp-8] mov rsi, rax mov edi, OFFSET FLAT:std::cin call std::basic_istream >::operator>>(int&) mov rdx, rax lea rax, [rbp-12] mov rsi, rax mov rdi, rdx call std::basic_istream >::operator>>(int&) mov edx, DWORD PTR [rbp-8] mov eax, DWORD PTR [rbp-12] add eax, edx mov DWORD PTR [rbp-4], eax mov esi, OFFSET FLAT:.LC1 mov edi, OFFSET FLAT:std::cout call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*) mov esi, OFFSET FLAT:.LC2 mov rdi, rax call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*) mov rdx, rax mov eax, DWORD PTR [rbp-4] mov esi, eax mov rdi, rdx call std::basic_ostream >::operator<<(int) mov eax, 0 leave ret __static_initialization_and_destruction_0(int, int): push rbp mov rbp, rsp sub rsp, 16 mov DWORD PTR [rbp-4], edi mov DWORD PTR [rbp-8], esi cmp DWORD PTR [rbp-4], 1 jne .L5 cmp DWORD PTR [rbp-8], 65535 jne .L5 mov edi, OFFSET FLAT:std::__ioinit call std::ios_base::Init::Init() mov edx, OFFSET FLAT:__dso_handle mov esi, OFFSET FLAT:std::__ioinit mov edi, OFFSET FLAT:std::ios_base::Init::~Init() call __cxa_atexit .L5: nop leave ret push rbp mov rbp, rsp mov esi, 65535 mov edi, 1 call __static_initialization_and_destruction_0(int, int) pop rbp ret.

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Why is it important that the duration of cell-cycle phases be controllable in adult multicellular organisms? b. What is MFP? Describe how this complex affects the cell cycle. Solution SOME BRIEF DESCRIPTION WHICH HELP YOU TO UNDERSTAND QUESTION AND IT S ANSWER FOR MY KNOWLEDGE IT HELPS YOU NOTE ;------ (The cell grows . The duration of these cell cycle phases varies considerably in different kinds of cells. For a typical rapidly proliferating human cell with a total cycle time of 24 hours, the G1 phase might last about 11 hours, S phase about 8 hours, G2 about 4 hours, and M about 1 hour.) The cell cycle or cell-division cycle is the series of events that take place in a cell leading to its division and duplication of its DNA (DNA replication) to produce two daughter cells. In bacteria, which lack a cell nucleus, the cell cycle is divided into the B, C, and D periods. The B period extends from the end of cell division to the beginning of DNA replication. DNA replication occurs during the C period. The D period refers to the stage between the end of DNA replication and the splitting of the bacterial cell into two daughter cells. In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into three periods: interphase, the mitotic (M) phase, and cytokinesis. During interphase, the cell grows, accumulating nutrients needed for mitosis, preparing it for cell division and duplicating its DNA. During the mitotic phase, the chromosomes separate. During the final stage, cytokinesis, the chromosomes and cytoplasm separate into two new daughter cells. To ensure the proper division of the cell, there are control mechanisms known as cell cycle checkpoints. The cell-division cycle is a vital process by which a single-celled fertilized egg develops into a mature organism, as well as the process by which hair, skin, blood cells, and some internal organs are renewed. After cell division, each of the daughter cells begin the interphase of a new cycle. Although the various stages of interphase are not usually morphologically distinguishable, each phase of the cell cycle has a distinct set of specialized biochemical processes that prepare the cell for initiation of cell divisions. Cell cycle checkpoints are control mechanisms in eukaryotic cells which ensure proper division of the cell. Each checkpoint serves as a potential point along the cell cycle, during which the conditions of the cell are assessed, with progression through the various phases of the cell cycle occurring when favorable conditions are met. Currently, there are three known checkpoints: the G1 checkpoint, also known as the restriction or start checkpoint or (Major Checkpoint); the G2/M checkpoint; and the metaphase checkpoint, also known as the spindle checkpoint. Effects of MULTIFOLIATE-PINNA, AFILA, TENDRIL-LESS and UNIFOLIATA genes on leafblade architecture in Pisum sativum In order to dissect the genetic regulation of leafblade morphogenesis, 16 genotypes of pea, constructed.

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How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression? How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression? Solution The RB(Retinonlastoma) is a tumor-suppressor protein and it has capability to inhibit progressing the cell cycle meticulously at the G1 checkpoint, Hyperphosphorylation of RB trigger release of “E2F” factor. G1 cyclin-cyclin dependent kinases (CDK) complex is mainly involved in the phosphorylation of RB to activate E2F result in initiation of transcription rate to promote DNA synthesis. These events are leading higher DNA synthesis to promote cell division in cell cycle result in tumor cells formation.

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Would you use physical segmentation or VLANs. Why or why not and where would you use each? In the Diagram below where would you put them. HU HUB Roud er Router Switeh Sreh Routen pe Wineless access point DMZ Zone auess point out ey acess poinf rele 70adba Broadbond wrreless access auuess point server access point access Solution Vlans and physical segmentation are both different things which both are used when we want to divided network. Basically physical segmentation means to have many different network on the basis of physical connectivity and VLANs may be defined as the division of network into logical parts and to have different networks with different nodes. Suppose if you have switch then it is better to have vlans rather than physical segmentation since as we know that switch is broadcasting frams and each interface of switch is present into different collision domain. So in order to have different broadcast domian we have to configure vlans into the switch. Which can certainly provide us security on the basis of different network. In above diagram we can have vlans on the every switch where each node connected to the switch would be placed into different network thus saves us from the collisions and broadcasting to the other nodes unnecessarily which may cause unnecessary overheads and less security in the network..

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Which plants produce leaves? (Select all that apply.) a.Gametophytes of bryophytes b.Gametophytes of seedless vascular plants c.Sporophytes of bryophytes d.Sporophytes of seedless vascular plants e.None of the above a.Gametophytes of bryophytes b.Gametophytes of seedless vascular plants c.Sporophytes of bryophytes d.Sporophytes of seedless vascular plants e.None of the above Solution c. Sporophytes of bryophytes d. Sporophytes of seedless vascular plants The sporophyte is associated with the green parts of the plant roots, stems, leaves and flowers. The gametophytes are the haploid structures, which are present in the multicellular stage of plants, which give rise haploid gametes (both male and female gametes)..

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Why are there so many blood vessels adjacent to the serosa? What is their function? Solution serosa is a smooth membrane that consists of a thin connective tissue layer and a thin layer of cells that secrete serous fluid. Serous membranes line and enclose several body cavities, known as serous cavities, where they secrete a lubricating fluid to reduce friction from muscle movements. Serosa is not to be confused with adventitia , a connective tissue layer that binds together structures rather than reduces friction between them. Each serous membrane is composed of a secretory epithelial layer and a connective tissue layer underneath. The epithelial layer, known as mesothelium, consists of a single layer of avascular flat nucleated cells (simple squamous epithelium) that produce the lubricating serous fluid. This fluid has a consistency similar to thin mucus . These cells are bound tightly to the underlying connective tissue. The connective tissue layer provides the blood vessels and nerves for the overlying secretory cells, and also serves as the binding layer that allows the whole serous membrane to adhere to organs and other structures. For the heart, the surrounding serous membranes include: the outer, inner, parietal pericardium, and visceral pericardium (epicardium). Other parts of the body may also have specific names for these structures. For example, the serosa of the uterus is called the perimetrium. The pericardial cavity (surrounding the heart), pleural cavity (surrounding the lungs) and peritoneal cavity (surrounding most organs of the abdomen) are the three serous cavities within the human body. While serous membranes have a lubricative role to play in all three cavities, in the pleural cavity it has a greater role to play in the function of breathing . The serous cavities are formed from the intraembryonic coelom and are basically an empty space within the body surrounded by a serous membrane. Early in embryonic life, visceral organs develop adjacent to a cavity and invaginate into the bag-like coelom. Therefore each organ becomes surrounded by a serous membrane—they do not lie within the serous cavity. The layer in contact with the organ is known as the visceral layer, while the parietal layer is in contact with the body wall. Function- The function of a serous membrane is to secrete a lubricant, appropriately called serous fluid . Why? Well, imagine rubbing your dry hands together over and over again. What would happen? First, they would heat up due to friction. Then, they would start to turn red. The more you keep doing it, the more painful it will become. Not long after (and please don\'t actually do this), you\'ll notice the skin begins to shred and break apart due to all the friction. Had you used a little lubrication prior to rubbing your hands together, this wouldn\'t have happened. Now imagine the internal body organs within, say, the abdomen. The serosa that lines the abdominal cavity and its organs is called the periton.

Why are there so many blood vessels adjacent to the serosa What is t.pdf

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Write a program CountWords in Java that reads text from a file and breaks it up into individual words. Insert the words into a tree set. At the end of the input file, print all words, followed by the size of the resulting set. This program determines how many unique words a text file has. If the file is not found, give the user a chance to select another file. Solution Code: package chegg; import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.util.StringTokenizer; import java.util.TreeSet; import javax.swing.text.html.HTMLDocument.Iterator; public class Test { public static void main(String args[]) throws IOException { BufferedReader br = null; String sCurrentLine; TreeSet ts = new TreeSet(); br = new BufferedReader(new FileReader(\"C:/Users/ansharm5/Desktop/chegg\")); while ((sCurrentLine = br.readLine()) != null) { StringTokenizer st = new StringTokenizer(sCurrentLine); while(st.hasMoreTokens()) { ts.add(st.nextToken()); } } System.out.println(\"Total no. of words: \"+ ts.size()); java.util.Iterator iterator = ts.iterator(); while (iterator.hasNext()){ System.out.println(iterator.next() + \" \"); } } }.

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Where would you NOT find autonomic ganglia? unpaired, anterior to spinal cord paired, beside spinal cord within wall of organ served or close to organ within spinal cord Solution Autonomic ganglia is a group of neuronal cell bodies and their dendrites trnasmitting sensory signals. There are two types of autonomic ganglia- para-sympathetic and sympathetic. They are close to and located on either side of the spinal cord. Thus, the most apt option would be (D) within spinal cord..

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