Peru is not a part of the Southern South America.SolutionPeru .pdf

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Peru is not a part of the Southern South America. Solution Peru is not a part of the Southern South America..

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The balanced equation is2 H2O2 = 2 H2O + O2SolutionThe bal.pdf

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Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A, and it is a recessive trait. This means that people will have the disease if they have two copies of the defective gene, but they will not have the disease if they have at least one unaffected copy. People with one normal copy and one defective copy are called carriers, because they can pass the disease on to their children. Seizures are sudden altacks of disease, often referring to some type of violent spasms. Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population, about 1 in 250 people carries the gene. People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting in two defective genes that make the body unable to produce Hex-A correctly. People who have only one defective gene are called carriers. Carriers do not have the disease, because they have inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the disease. However, some populations of people include more carriers than others. For example, 1 in 27 people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also have a higher than usual risk of carrying the Tay-Sachs gene. Solution Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A, and it is a recessive trait. This means that people will have the disease if they have two copies of the defective gene, but they will not have the disease if they have at least one unaffected copy. People with one normal copy and one defective copy are called carriers, because they can pass the disease on to their children. Seizures are sudden altacks of disease, often referring to some type of violent spasms. Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population, about 1 in 250 people carries the gene. People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting in two defective genes that make the body unable to produce Hex-A correctly. People who have only one defective gene are called carriers. Carriers do not have the disease, because they have inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the disease. However, some populations of people include more carriers than others. For example, 1 in 27 people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also have a higher than usual risk of carrying the Tay-Sachs gene..

Tay-Sachs disease is caused by a mutation (abnormal change) in the g.pdf

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E.) is less acidic in other A ,B triple bond an.pdf

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Solution : when user sending the email, then user should select either POP3 or IMAP protocol to send the mails. Some of the differences are giving below. POP3 e-mail protocol: IMAP e-mail protocol: SMTP (simple mail transfer protocol): SMTP is not normally used as intermediate mail servers for sending mail. Because SMTP cannot transmit the executable files or any binary objects. Similarly it cannot transmit the text data that contains the national language characters. It will have certain size limit to have an e-mail. So if the mail size is over the limit then it will be rejected..

Solution when user sending the email, then user should select eithe.pdf

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publicclass Date { privatestatic String DATE_SEPARATOR = \"/\"; privatestaticintDAYS_PER_WEEK = 7; //Attributes privateint day; privateint month; privateint year; /** * Default Constructor * Instantiates an object of type Date to 1/1/2000 */ public Date() { this.day = 1; this.month = 1; this.year = 2000; } /** * Constructs a new date object to represent the given date. * @param day * @param month * @param year */ public Date(int day, int month, int year) { if(isValid(day, month, year)) { this.day = day; this.month = month; this.year = year; } else System.out.println(\"Invalid Date.\"); } /** * Returns the day value of this date for example, for the date 2006/07/22, returns 22. * @return */ publicint getDay() { return day; } /** * Returns the month value of this date ,for example, for the date 2006/07/22, returns 7. * @return */ publicint getMonth() { return month; } /** * Returns the year value of this date, for example , the date 2006/07/22, returns 2006. * @return */ publicint getYear() { return year; } /** * @param day the day to set */ publicvoid setDay(int day) { this.day = day; } /** * @param month the month to set */ publicvoid setMonth(int month) { this.month = month; } /** * @param year the year to set */ publicvoid setYear(int year) { this.year = year; } /** * Returns true if the year of this date is a leap year. * A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400. * For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not. * @return */ publicboolean isLeapYear(){ if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0))) returntrue; else returnfalse; } /** * Checks if the date is valid * @param day * @param month * @param year * @return */ publicboolean isValid(int day, int month, int year) { if((month < 1) || (12 < month)) returnfalse; else { if(((month == 1) || (month == 3) || (month == 5) || (month == 7) || (month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day))) returnfalse; elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 < day))) returnfalse; elseif(month == 2) { if(isLeapYear() && ((day < 1) || (29 < day))) returnfalse; elseif((day < 1) || (28 < day)) returnfalse; } } returntrue; } /** * Returns the maximum number of days in a month * @return */ publicint maxMonthDays() { if(this.month == 2) { if(isLeapYear()) return 29; else return 28; } elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) || (this.month == 8) || (this.month == 10) || (this.month == 12)) return 31; else return 30; } /** * Checks if this dat is same as other date * @param other * @return */ publicboolean isEqual(Date other) { if((this.day == other.day) && (this.month == other.month) && (this.year == other.year)) returntrue; else returnfalse; } /** * Moves this Date object forward in time by the given number of days . * @param days */ publicvoid addDays(int days).

$publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf$ $publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf$

publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf

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please give me points as nobody ecen noticed the qusetion as it is u.pdf

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covalent bond .pdf

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Part-I Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach? Give at least three alternative hypotheses that could explain these findings. Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active, chronic gastritis and the presence of the bacteria found in stomachs. hypothesis 1. Dr. Warren noticed a thin blue line on the surface of the tissue. hypothesis 2: Warren found numerous spiral shaped bacteria when stained with a Warthin-Starry silver stain of the histological section. hypothesis 3: Dr. Warren also observed that the number of bacteria observed seemed to correlate with the degree of inflammation of the stomach lining—the more severe the inflammation, the more abundant the bacteria. Question 2.What should Dr. Warren do next? Answer: Dr. Warren started Warthin-Starry silver stain for all the gastric biopsies he examined. Part-II: Question: At this stage of the story, what type(s) of data do Warren and Marshall have on the spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of observational? Answer: Dr. Warren and Marshall had the quantitative data based on the observations made by various researchers, such as Italian pathologist Giulio Bizzozero, Polish clinical researcher, Dr. W. Jaworski. The results that obtained from the previous reports are purely observational, not experimental. Part-III: Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of observational? Answer: Now, at this stage, Warren and Marshall had electron microscopy data of the bacteria that was colonized in the lab culture media, and therefore, the data was now experimentally proven qualitative data. Question: What other information would you like to have about this bacterium now that it can be successfully cultured in the lab? Answer: The bacteria that was isolated from the biopsy of patient\'s stomach for lab culture, was a slow grower, and it took 5-6 days for colonization instead of 2 days which has taken by many identified bacteria. Part IV: Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of observational? Answer: The data was generated by pilot studies from 100 patients and the data was experimental and also statistically analyzed, therefore it was qualitative as well as quantitative data. CANNOT SEE FIGURE 3. UNABLE TO ANSWER FURTHER. Solution Part-I Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach? Give at least three alternative hypotheses that could explain these findings. Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active, chronic gastritis and the presence of the bacteria found in stomachs. hypothesis 1. Dr. W.

Part-IQuestion 1. What is Dr. Warren’s hypothesis regarding the ba.pdf

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Combustion of glucose is respiration. The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from related molecules. A lot of oxygen is required for this process! The sugar AND the oxygen are delivered to your cells via your bloodstream. This process occurs partially in the cytoplasm, and partially in the mitochondria. The mitochondria is another organelle in eukaryotic cells. like the chloroplast, the mitochondria has two lipid bilayers around it, and its own genome (indicating that it may be the result of endosymbiosis long ago). In some ways similar to the chloroplast, the mitochondria also has two main sites for the reactions: The matrix, a liquidy part of the mitochondrion, and the christae, the folded membranes in the mitochondrion. 1: Glycolysis (\"splitting of sugar\"): This step happens in the cytoplasm. One Glucose (C6H12O6) is broken down to 2 molecules of pyruvic acid. Results in the production of 2 ATPs for every glucose. (page 113 of your book goes into painful detail about this process, but all YOU need to know is that glucose is split into to 2 molecules of pyruvate!) 2: Transition Reaction: Pyruvic Acid is shuttled into the mitochondria, where it is converyed to a molecule called Acetyl CoA for further breakdown. 3: The Krebs Cycle, or Citric Acid Cycle: Occurs in the mitochondrial matrix, the liquid-y part of the mitochondria. In the presence of Oxygen gas (O2), all the hydrogens (H2) are stripped off the Acetyl CoA, two by two, to extract the electrons for making ATP, until there are no hydrogens left - and all that is left of the sugar is CO2 - a waste product - and H2O (exhale). The Krebs cycle results in the production of only ~4 ATPs, but produces a lot of NADH, which will go on to the next step... Hans Krebs won the Nobel Prize in 1953 for his discovery of the Citric Acid Cycle. 4: The Electron Transport Chain and Chemiosmosis (\"the big ATP payoff\"). Occurs in the christae of the mirochondria, the folded membranes inside the chloroplast. Electrons from Hydrogen are carried by NADH and passed down an electron transport chain to result in the production of ATP. Results in the production of ~32 ATPs for every glucose. Peter Mitchell won the Nobel Prize in 1978 for his work on energy production in mitochondria, called the Chemiosmotic Theory. Solution Combustion of glucose is respiration. The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from related molecules. A lot of oxygen is required for this process! The sugar A.

Combustion of glucose is respiration. The equatio.pdf

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in truss one...the forces in the member BC and DE have zero.in tru.pdf

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iam giving you entire process of forensc duplication; the response strategy for forensc duplication is Decision of when to perform a forensic duplication based is based, in part, on existing response strategy for the instant situation For example, many organizations have a policy of creating forensic HD duplicates of all PCs used by executives that leave the organization. Forensic Duplicates as Admissible Evidence; Existing legal standards define minimum criteria for an item to be admitted into evidence Collection process usually under scrutiny as well Federal Rules of Evidence ; Federal Rules of Evidence (FRE) 1002 state that the item or information presented in court must be the original. Exceptions: Definitions and Duplicates If data are stored by computer or similar device, any printout or other output readable by sight, shown to reflect the data accurately, is an original. Admissibility of Duplicates A duplicate is admissible to the same extent as an original unless: • A genuine question is raised as to the authenticity of the original, or • In the circumstances it would be unfair to admit the duplicate in lieu of the original first of all what is forensics duplication A file that contains every bit of information from the source in a raw bitstream format. Tools that create forensic duplicates: 1. dd 2. FTK Imager, Access Data 3. Dfcldd, US DOD Computer Forensics Lab version of the dd comand Qualified Forensic Duplicate? A file that contains every bit of information from the source, but may be stored in a altered form. Tools that create qualified forensic duplicate output files: 1. SafeBack 2. EnCase 3. FTK Imager Restored Image; A restored image is what you get when you restore a forensic duplicate or a qualified forensic duplicate to another storage medium. Mismatched drive geometries can cause problems. HD Development; When hard drives grew beyond 512MB, the PC-BIOS needed to be updated (to recognize larger drives). …software emulated a modern BIOS. Software pushed all of the real data on the drive down one sector and stored its program and information in sector 2. The real partition table would be at cylinder 0, head 0, sector2. Safeback, EnCase, FTK Imager, and dd will create a restored image from the qualified forensic duplicate. EnCase and dd images may not need to be restored. Treat images as virtual disks, eliminating the need for restoration. Note, FTK Imager can create images in the EnCase Format Mirror Image ; Created from hardware that does at bit for bit copy from one hard drive to another. Requires two identical hard drives Doesn’t happen very often. Tool Requirements: Forensic Duplication ; Tool must: Create a forensic duplicate or mirror image of the original. Handle read errors in a robust and graceful manner. Not make any changes to source medium. Capable of scientific and peer review. Results must be third party repeatable and verifiable. Legal Issues Tools used for forensic duplication must pass the legal tests for r.

iam giving you entire process of forensc duplication;the response.pdf

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How does traffic analysis work? Internet data packets have two parts: a data payload and a header used for routing. The data payload is whatever is being sent, whether that\'s an email message, a web page, or an audio file. Even if you encrypt the data payload of your communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly, what you\'re saying. That\'s because it focuses on the header, which discloses source, destination, size, timing, and so on. A basic problem for the privacy minded is that the recipient of your communications can see that you sent it by looking at headers. So can authorized intermediaries like Internet service providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic analysis might involve sitting somewhere between sender and recipient on the network, looking at headers. But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts of the Internet and use sophisticated statistical techniques to track the communications patterns of many different organizations and individuals. Encryption does not help against these attackers, since it only hides the content of Internet traffic, not the headers. Solution How does traffic analysis work? Internet data packets have two parts: a data payload and a header used for routing. The data payload is whatever is being sent, whether that\'s an email message, a web page, or an audio file. Even if you encrypt the data payload of your communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly, what you\'re saying. That\'s because it focuses on the header, which discloses source, destination, size, timing, and so on. A basic problem for the privacy minded is that the recipient of your communications can see that you sent it by looking at headers. So can authorized intermediaries like Internet service providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic analysis might involve sitting somewhere between sender and recipient on the network, looking at headers. But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts of the Internet and use sophisticated statistical techniques to track the communications patterns of many different organizations and individuals. Encryption does not help against these attackers, since it only hides the content of Internet traffic, not the headers..

How does traffic analysis work Internet data packets have two parts.pdf

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Br2 is the answer. note for nonpolar molecules,.pdf

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Because leucine is hydrophobic and does not have bulky groups, similar amino acids can be substituted without changing the protein character very much. These amino acids are isoleucine, valine, and alanine. Solution Because leucine is hydrophobic and does not have bulky groups, similar amino acids can be substituted without changing the protein character very much. These amino acids are isoleucine, valine, and alanine..

Because leucine is hydrophobic and does not have .pdf

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Designing affective Human Computer-Interfaces such as Embodied Conversational Agents requires modeling the relations between spontaneous emotions and behaviors in several modalities. There have been a lot of psychological researches on emotion and nonverbal communication. Yet, these studies were based mostly on acted basic emotions. This paper explores how manual annotation and image processing might cooperate towards the representation of spontaneous emotional behavior in low resolution videos from TV. We describe a corpus of TV interviews and the manual annotations that have been defined. We explain the image processing algorithms that have been designed for the automatic estimation of movement quantity. Finally, we explore several ways to compare the manual annotations and the cues extracted by image processing. Solution Designing affective Human Computer-Interfaces such as Embodied Conversational Agents requires modeling the relations between spontaneous emotions and behaviors in several modalities. There have been a lot of psychological researches on emotion and nonverbal communication. Yet, these studies were based mostly on acted basic emotions. This paper explores how manual annotation and image processing might cooperate towards the representation of spontaneous emotional behavior in low resolution videos from TV. We describe a corpus of TV interviews and the manual annotations that have been defined. We explain the image processing algorithms that have been designed for the automatic estimation of movement quantity. Finally, we explore several ways to compare the manual annotations and the cues extracted by image processing..

Designing affective Human Computer-Interfaces such as Embodied Conve.pdf

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Dispersal of population/individuals is described as movement of individuals from one geographical residential location to another where they can occupy the new place, survive, settle and reproduce appropriately. These set of individuals can thus colonize at the new geographical location in a similar comfortable way as they used to colonize in the previous location. This technically does not resemble to fragmentation or generation of a sub-set of parent population because these methods give rise to a smaller subset of individuals which are impaired in some of the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents the high efficacy and survivability of the parent population even under new locations. Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat Solution Dispersal of population/individuals is described as movement of individuals from one geographical residential location to another where they can occupy the new place, survive, settle and reproduce appropriately. These set of individuals can thus colonize at the new geographical location in a similar comfortable way as they used to colonize in the previous location. This technically does not resemble to fragmentation or generation of a sub-set of parent population because these methods give rise to a smaller subset of individuals which are impaired in some of the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents the high efficacy and survivability of the parent population even under new locations. Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat.

Dispersal of populationindividuals is described as movement of indi.pdf

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Decreases..SolutionDecreases...pdf

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ConversionDemo.javaimport java.util.;package for scanner class.pdf

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C. Yes, YN + NY is greater than or equal to 30.SolutionC. Yes,.pdf

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Answer: Total asset turnover = sales/total asset Total asset = current + non current asset =( $100 + $142 ) million = $242 million Total asset turnover = $371/$242 = 1.5331 :1 or 153.31% (answer) Solution Answer: Total asset turnover = sales/total asset Total asset = current + non current asset =( $100 + $142 ) million = $242 million Total asset turnover = $371/$242 = 1.5331 :1 or 153.31% (answer).

AnswerTotal asset turnover = salestotal assetTotal asset = cur.pdf

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