Why do the soft technologies open more opportunities for women To w.pdf

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Why do the soft technologies open more opportunities for women? To what extent have these technologies impacted the perceptions of men’s and women’s roles in the economy, within marriage, and in society as a whole? Solution Now a days Woman is playing a very important role in every field of the economy.Either Agriculture or education or any part of the manufacturing area. I am also agree from this statement, beacuse after marriage so many responsibilities increase its libilities through these soft techniqe she can hendle her all task very easily. Technology is a vital element for any community. It plays a significant role in food security, agriculture and small scale production. Women use technical skills and knowledge in their daily activities; they continually innovate and adapt technologies in response to things they face in their everyday lives. However, their role and technical skills are often overlooked and undervalued. There are four main reasons why women are less visible in the application of technology than men:.

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How did the tail of the swordtail fish evolve? Solution The swordtail fish (Xiphophoroushellerii) has distinguishing features in male fishes and female fishes. The male fishes have a sword-like tail and an organ called gonopodium. This tail is an extension of the caudal fin and is ornamental. The female fishes do not have the characteristic tail and lack the gonopodium. The evolution of tail in swordtail fishes is guided by sexual selection and not natural selection. It has been hypothesized that ancestrally the female fishes preferred to mate with swordtail bearing fishes rather that the swordless counterparts. Female preference drove the evolution of the tail in males and the tail-less trait eventually disappeared..

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Helen Bowers was stumped. Sitting in her office at the plant, she pondered the same questions she had been facing for months: how to get her company’s employees to work harder and produce more. No matter what she did, it didn’t seem to help much. Helen had inherited the business three years ago when her father, Jake Bowers, passed away unexpectedly. Bowers Machine Parts was founded four decades ago by Jake and had grown into a moderate-size corporation. Bowers makes replacement parts for large-scale manufacturing machines such as lathes and mills. The firm is headquartered in Kansas City and has three plants scattered throughout Missouri. Although Helen grew up in the family business, she never understood her father’s approach. Jake had treated his employees like part of his family. In Helen’s view, however, he paid them more than he had to, asked their advice far more often than he should have, and spent too much time listening to their ideas and complaints. When Helen took over, she vowed to change how things were done. In particular, she resolved to stop handling employees with kid gloves and to treat them like what they were: the hired help. In addition to changing the way employees were treated, Helen had another goal for Bowers. She wanted to meet the challenge of international competition. Japanese firms had moved aggressively into the market for heavy industrial equipment. She saw this as both a threat and an opportunity. On the one hand, if she could get a toehold as a parts supplier to these firms, Bowers could grow rapidly. On the other, the lucrative parts market was also sure to attract more Japanese competitors. Helen had to make sure that Bowers could compete effectively with highly productive and profitable Japanese firms. From the day Helen took over, she practiced an altogether different philosophy to achieve her goals. For one thing, she increased production quotas by 20 percent. She instructed her first-line supervisors to crack down on employees and eliminate all idle time. She also decided to shut down the company softball field her father had built. She thought the employees really didn’t use it much, and she wanted the space for future expansion. Helen also announced that future contributions to the firm’s profit-sharing plan would be phased out. Employees were paid enough, she believed, and all profits were the rightful property of the owner—her. She also had private plans to cut future pay increases to bring average wages down to where she thought they belonged. Finally, Helen changed a number of operational procedures. In particular, she stopped asking other people for their advice. She reasoned that she was the boss and knew what was best. If she asked for advice and then didn’t take it, it would only stir up resentment. All in all, Helen thought, things should be going much better. Output should be up and costs should be way down. Her strategy should be resulting in much higher levels of productivity and profits. .

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Dr. A. conducts a study on the effect of distraction on learning. Dr. A goes to a local mall and asks shoppers to participate in her study. Those who agree to participate are then randomly assigned to one of three groups. The first group is placed in a quiet room; the second group is placed in a room with a moderate level of background noise; the third group is placed in a room with a high level of background noise. All participants are then given a learning task to complete. The type of sampling strategy used by Dr. A is best characterized as: a. Random b. Stratified c. Convenience d. Snowball Solution a. Random.

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Discuss the function and importance of conceptual modeling. Solution A conceptual model is a symbol of a system, complete of the work of art of concepts which are used to assist people know, appreciate, or simulate a topic the model represent. Some models are physical objects. For example, a toy representation which may be assemble, and may be complete to work like the thing it represents. To stand for data in an comprehensible way. 1. improve an individual’s understanding of the catalog. 2. Facilitate efficient delivery of database details connecting database stakeholders 3. Provide a point of orientation for database designer to extract file specifications 4. Document the catalog for future reference and give a means for association.

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Dave threw a ball into the air while standing on a 10 foot tall table. The height, h(t), of the ball t seconds after it is tossed is can be modeled by the function h(t) = 16t 2 + 32t + 10 How long was the ball in the air? Solution h(t) = 16t^2 + 32t + 10 we need to find time t at which h(t) = 0 h(t) = 16t^2 + 32t + 10 = 0 16t^2 -32t - 10 = 0 8t^2 -16t - 5 = 0 solving for t we get, t= 2.27 s and t = -0.27 s since t can\'t be negative t = 2.27 s Answer: 2.27 s.

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Darwin was not the first to consider evolution as a processes but he did come up with the first effective explanation for how it happens. Describe Darwin\'s theory of evolution by natural selection. Explain how this theory was a major advance over prior ideas as to how organisms changed over time. Give evidence in support of evolution and describe the driving forces for evolutionary change. Darwin was not the first to consider evolution as a processes but he did come up with the first effective explanation for how it happens. Describe Darwin\'s theory of evolution by natural selection. Explain how this theory was a major advance over prior ideas as to how organisms changed over time. Give evidence in support of evolution and describe the driving forces for evolutionary change. Solution Darwin theory was based on the study of size of small birds beaks which were found on the Galapoes island.. He found that with time and areas the size of the bird\'s beak were different in size and shape ie their size increases. . This were due to the fact that the due to to the avalaibilty of their different foods their beaks size also changes and increased in size..!! It was due to the evolution which was further carried out to their future generations..!! The driving force are those forces which results in evolution.. The factors are Mutation Natural selection. Gene flow. Genetic drift..

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Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease. Only individuals that are either affected or normal are shown. In other words, potential heterozygotes are NOT indicated. Characterize each of the following modes of inheritance as: impossible, unlikely, or probable. Justify your answers. Autosomal dominant Autosomal recessive X-linked dominant X-Iinked recessive Y linked Solution Single gene herediary diseases are mainly caused by a change of any one gene in a particular DNA.Its divided into diiferemt categories like dominant ,recessive and X linked.Square indicates a male and circle female.According to the diagram affected male is crossed with an unaffected female.The progeny obtained is affected daughters and unaffected sons. Autosomal dominant: in this case, the father — has a 50 percent chance of having an affected child with one mutated gene (dominant gene) and a 50 percent chance of having an unaffected child with two normal genes (recessive genes).But, in our case none of the sons are affected which means its criss cross inheritance that is from father to daughter.Hence , this case is unlikely Autosomal recessive:To have an autosomal recessive disorder, you inherit two mutated genes, one from each parent. These disorders are usually passed on by two carriers.The cross given in the question is not between the two carriers.The male parent is affected for sure hence this case is impossible X linked dominant:As in autosomal dominant inheritance, only one copy of a disease allele on the X chromosome is required for an individual to be susceptible to an X-linked dominant disease.Both males and females can be affected, although males may be more severely affected because they only carry one copy of genes found on the X chromosome. Some X-linked dominant disorders are lethal in males.When a female is affected, each pregnancy will have a one in two (50%) chance for the offspring to inherit the disease allele. When a male is affected, all his daughters will be affected, but none of his sons will be affected.So, there are probable chance for this type of inheritance X linked recessive:For X-linked recessive disorders, an affected father who has a mutation in a gene on the X chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his children. If the mother is not affected or a carrier, none of his sons will be affected since they can only inherit a normal X chromosome from their mother and they inherit a Y chromosome from their father. Each daughter will have a 50% chance of being an unaffected carrier and a 50% chance of both X chromosomes being normal.In our cross all daughters are affected which means this type of inheritance is impossible Y linked:This type of inheritance is impossible in this case as all female children are getting affected.If it would have been Y linked then only sons should have been affected as Y chromosome is present only in males.Hence this.

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A mutant strain of E. coli is found that produces both beta-galactosidase and permease at high levels whether lactose is present or not. (a) What are two possible genotypes for this mutant? Another mutant is isolated that produces no beta-galactosidase at any time but produces permease if lactose is present in the medium. A merodiploid strain is formed from these two mutants. in the absence of lactose, neither beta-galactosidase nor permease is made. When lactose is added, the merodiploid strain makes both beta-galactosidase and permease. (b) What is the genotype of the merodiploid strain? Solution A mutant of E. coli is found that produces both beta - galactosidase and permease at high levels in the presence or absence of lactose a) The two possible genotypes for this mutant is given as-: F+ Z+ Y+ Now another mutant is isolated that produces no beta-galactosidase at any time but produces permease if lactose is present in the medium. The genotype for this mutant is given as-: F- Z- Y- Given a merodiploid strain is formed from these two mutants. Merodiploid is partially diploid bacterium having its own chromosome and a fragment that is introduced either by conjugation, transformation or transduction. It is given by-: F+ Z+ Y+ x F- Z- Y- =F\' Z+ - Y- b) The genotype for the merodiploid strain is F\' Z+ - Y- For Merodiploid strain in the absence of lactose, neither beta-galactosidase nor permease is formed but in the presence of lactose both beta-galactosidase and permease is produce. As mutant 1 have the ability to make beta-galactosidase in the absence or presence of lactose but the mutant 2 is unable to make beta-galactosidase at any time so in the new mutant the gene for beta galactosidase is expressed only in the presence of lactose. Also the permease producing gene of 2 mutant is dominant that is the reason the merodiploid strain produces permease in the presence of lactose but not in the absence of lactose..

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An unknown individual launches a series of attacks against the Web sites of Prime Sales Corporation. The attacks significantly slow the sites, leading to $100 million in damage in terms of lost work time, lost revenue, site repair costs, and other expenses. The attacker does not intend to profit from the onslaught and in fact does not realize any financial benefit from the effects. How is this attack most likely orchestrated? Who is most likely to engage in this act—that is, whose habits and limitations are clearly suited to such conduct—and why? Solution Most probably the website is slowed by a DDOS (Distributed Denial of Service) attack. In this attack the hacker sends multiple fake requests from different servers with the intention of slowing down the service to legitimate users. The most likely suspect is an anonymous hacking group. These groups are formed by professional hackers who come together for a common cause. The most likely reason might to gain attention or the group considers something the company is doing as against their beliefs..

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You\'ve completed an interrupted mating experiment in E. coli, the leu gene enters before the aln gene. For this experiment, you crossed an Hfr that is leu^+ gln^+ with an F- that is leu^- gln^-. Exconjugant cultures are plated on minimal medium supplemented with leonine but not glutamine, and 250 total colonies were observed. Each of those colonies was then plated on minimal medium containing no leonine or glutamine; only 200 of the cultures grew under these conditions. What are the genotypes of the two different cultures? So, what likely happened at the cellular/molecular level to allow some cells to grow without glutamine supplementation, but not all? Make sure you explain/support your answer. (lessthanorequalto 4 sentences) Solution a) leu and gln gene\'s are the genotypes of the two different cultures. b) because of this minimal medium supplemented and Hfr , F for combination of any two moleucles or ions the opposite signs of its bonding makes stable for a molecule. so, this molecules will allow.this makes cellular/moleculer level to allow some cells to grow without glutamine supplementation..

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Write the following Scheme functions in Java: Solution import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Neshted { public static String neshted_list(List list){ String output = \"\"; float count =0; for(int i : list){ if(count == list.size()/2){ if(list.size() %2 == 0) output += \"()\"; else output += \"(\"; } if(count=list.size()/2) { output += i+\")\"; } count ++; } return output; } public static void main (String args[]){ Integer[] values = {4,5,3,2,8}; List list = Arrays.asList(values); System.out.println(neshted_list(list)); } }.

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Write a C program to enter a first name, middle name and last name in separate variables. After entering each name at the keyboard, call a function named convert that will capitalize the first letter of the name and set the rest of the name to lowercase (for example, jOhN becomesJohn). The convert function will be called three times-- once for the first name, once for the middle name, and once for the last name. Combine the properly capitalized names into one string and output the string (make sure you have a space between each part of the name). Count the number of characters in the total name (including spaces between names) and output that total also. Include descriptive comments in your source code. Solution #include #include #include void Convert(char string[]) { int i; int len = strlen(string); //compute length of string string[0] = toupper(string[0]); // capitalize first character of the string for (i=1;i.

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What are the difference between the testes of a frog and a rat in terms of tunica albuginea, interstitial cells, seminiferous tubules, basement membrane, sertoli cells Solution STRUCTURES FROG RAT TUNICA ALBUGINEA Thick fibrous sheet of connective tissue. Composed of collagen fibers and occasional smooth muscle cells. INTERSTITIAL CELLS Endocrinal in nature and secrete mainly androsterone and testosterone Major source of testosterone and also participate in steroid biosynthesis. SEMINIFEROUS TUBULES Lined by single layer of cuboidal cells. These cuboidal cells form the germinal epithelium Germinal epithelial component is absent SERTOLI CELLS Involved In providing nutrients to mature spermatozoa Capable of steroid biosynthesis and secretion of inhibin and androgen binding protein STRUCTURES FROG RAT TUNICA ALBUGINEA Thick fibrous sheet of connective tissue. Composed of collagen fibers and occasional smooth muscle cells. INTERSTITIAL CELLS Endocrinal in nature and secrete mainly androsterone and testosterone Major source of testosterone and also participate in steroid biosynthesis. SEMINIFEROUS TUBULES Lined by single layer of cuboidal cells. These cuboidal cells form the germinal epithelium Germinal epithelial component is absent SERTOLI CELLS Involved In providing nutrients to mature spermatozoa Capable of steroid biosynthesis and secretion of inhibin and androgen binding protein.

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Which of the following occurs with disruptive selection a. Natural selection favors phenotypes of one extreme, resulting in a population mean shift over several generations b. Natural selection favors phenotypes of both extremes, resulting in an increase in intermediate phenotypes over time c. Natural selection favors phenotype values near the population mean, resulting in no shift in mean trait value over time and a more narrow bell curve d. Natural selection selects against trait values that are close to the population mean, resulting in a bimodal distribution of trait values Solution Disruptive selection happens when conditions favor individuals at both extremes of a phenotypic range over individuals with intermediate phenotypes. one example is a population of black- bellied seedcracker finches in Cameroon whose members show two distinctly different beak sizes. small-billed birds feed generally on soft seeds, whereas large-billed birds specialize in cracking hard seeds. it seems that birds with intermediate-sized bills are comparatively inefficient at cracking both types of seeds and thus have lower relative fitness..

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What is wrong with the following definition of headInsert? struct Node { int item; Node* link; }; typedef Node* NodePtr; void headInsert(NodePtr& head, int data) { NodePtr tmp = new Node; tmp->item = data; head->next = tmp; tmp->next = head->next; } NodePtr head; headInsert(head, 4); A. tmp should be declared to be a Node not a NodePtr. B. head->next is pointing to NULL. C. Nothing is wrong. D. If there were any nodes following head they are now lost. Solution B. head->next is pointing to NULL. Because of which tmp is not getting stored in the linked list Correct Code: struct Node { int item; Node* link; }; typedef Node* NodePtr; void headInsert(NodePtr& head, int data) { NodePtr tmp = new Node; tmp->item = data; // next of tmp should be assigned head tmp->link = head; head = tmp; }.

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What are two specialized epidermal cells and what do they do for the plant? Solution The two specialized epidermal cells in the plant are 1. Guard cells 2. Trichome. Guard cells are kidney shaped cells with rich green chloroplasts.Guard cells occur in form of pairs. A pair of guard cells and pore are referred as stomate. Guard cells perform two different functions. At daytime, guard cells open small pores in the plant epidermal surface allowing carbon dioxide to diffuse from atmosphere into the chlorenchyma tissue required for the photosynthesis process. It also regulates oxygen exchange. At night-time guard cells close the pores in the epidermal cells preventing water loss from leaves. Guard cells can alter the size of stomata by changing its structure in response to water exchange. Trichomes are branched hairy projections of modified epidermal cells. Trichomes are multi cellular with spine shape or hook shaped. Trichomes are branched or unbranched, unicellular or multicellular. Trichomes function to prevent water loss through evapouration and retains moisture. It secretes a sticky substance that repels insects..

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What are the basic differences and commonalties between Structured Analysis (process- oriented). Information Engineering (data-oriented), and Object-Oriented approaches for software development? Solution Structured Analysis is technique used in software development to analyse the designed system to its correctness. The analysis is made to see to what extent the designed system meets the requirement of the user. The structured analysis uses data flow diagrams to analyze the designed structure. It is believed that if the structure is good then the data will be good, but doesnot pay attention on data in detail. Data oriented analysis is used where data plays the utmost role. The diagrams that are used for analysing the data oriented design is ER diagrams. Data oriented design establishes the relations that are required for the development of the system on various data.The data analysis concentrates on incorporating additional features in process to ensure the data integrity. Object oriented analysis unlike of the above two concetrates on the objects, which are the instance of class that consists of both data and process or logic. The data oriented anaylsis cannot work on the structured process. similalry structures analysis cannot be performed on data oriented design.The process oriented and object oriented analysis are widely in use when compared to data oriented analysis..

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