The concept of a sampling distributionA-Is essential for drawing c.pdf

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The concept of a sampling distribution A-Is essential for drawing conclusions about a population from the results of a sample. B- Is related to the use of probability theory in inferential statistics. C-Helps us use statistics to estimate the corresponding true values in the population. D-All of the above. Solution The concept of a sampling distribution D-All of the above..

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why is there an uneven distribution of photosystem 1 and 2 on the thylakoid membrane? Solution Photosynthesis is carried out by the cyanobacteria, plants involved the absorption of light and the conversion of its energy into chemical energy. The photosynthetic apparatus is made up of 4 multiprotein complexes, which are asymmetrically embedded in the thyalkoid memberane. It is the high structural and functional coordination between these 4 complexes, which enables the efficient coupling of photosynthetic electron transfer and photophosphorylation. The 4 complexes - Photosystem II (PSII), cytochrome b6f complex, PS I and the proton-ATP synthase, are unevenly distributed between the 2 regions of the thylakoids. The PS II complex is situated in the appressed grana-lamella region, the PS I and the ATP synthase complex are located in the non-appressed stroma-lamella region. Light driven electron transfer is intiated with the oxidation of water mole. by the PS II complex and the reduction of quinones associated with the PS II complex. The electrons are then transferred across the memberane to plastocyanin via the Cyt b6f complex. Plastocyanin is a copper protein facing the lumen of the thylakoids, is the secondary electron donor of the PS I complex. In the PS I the electron are translocated across the memberane to ferredoxin which further transfers them to NADP+. The membrane potental and proton gradient resulting from this photochemical electron transport provide the motive force required for the formation of ATP by the ATP synthase. The transport of electron against the electrical potential and the formation of ATP and NADPH are made possible due to the unique ability of the thylakoids to absorb light energy. The 2 pigment complexes (PS II and PS I) are responsible for harvesting the light and performing the charge separation across the thylakoid memberanes. Photosytem II (PS II): 1. PS II is located at the inner surface the grana thyalkoid memberane 2. The photocentre is P680 3. Pigments absorb shorter wavelengths of light (<680nm) 4. Participates only in non cyclic photophosphorylation 5. It is associated with photolysis of water 6. Main functions are ATP synthesis and hydrolysis of water Photosystem I (PS I) 1. PS I is located at the outer surface the grana thylakoid memberane 2. The photocentre is P700 3. Pigments absorb longer wavelength of light (>680nm) 4. Participates in cyclic and non-cyclic photophosphorylation 5. It is not associated with photolysis of water 6. Main function is ATP synthesis This is the purpose the two photosystems (PS) are unevely distributed in thyalkoid memberane..

why is there an uneven distribution of photosystem 1 and 2 on the th.pdf

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Which of the following are examples of cell signaling? Choose all that apply. As a response to increasing densities, bacteria coordinate behaviors and form biofilms. Insulin hormone produced by the pancreas triggers the intake of glucose by liver cells. Carbon dioxide diffuses across the cell membrane into the blood plasma. Chlorophyll absorbs and transfers light energy during photosynthesis. A neuron releases neurotransmitters into a synapse, triggering a target nerve cell. Solution Cell signaling Cell signaling is the communication between the cells through which they coordinate certain activities.

Which of the following are examples of cell signaling Choose all tha.pdf

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What is the purpose of testing a program with different data? Solution The purpose of testing a program functions according to the expectation. The overall goal of testing is to find a bug that might effect the functioning of the program. There are various types of testing and each serve their own purpose. Different data is used for each type of testing. We have to check the whether internal module is functioning properly, how the program responds to invalid data, test how changing one module will effect the whole system etc.. There are many automated tools which can be used for comprehensive testing..

What is the purpose of testing a program with different dataSol.pdf

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what are the factors contributing to the rise in global media? Will global media eventually replace national media? Solution Q1) Media is the mass communication system of letting people informed through broadcasting, publishing, and the internet. It enriches public awareness and knowledge, which is very important in today’s fast moving world. Global media gives the news of whole world, suppose a person sitting in India can get the happenings in the US. Factors are as below: Point 1) The urge of national media: There are so many small national media in various countries. They want to get international news as well but can’t get because of their small infrastructure. Therefore, they are now becoming tied up with big international media agencies in order to get regular updating. This is the reason of rising trend of global media. Point 2) Globalization of economy: Most of the countries have free-trade environment now. They need cheaper rate of factors of production (land, labor, capital, and organization); the global media is a great source of information of it. Point 3) Technological development: The development of internet gives the platform of rising of global media. A person sitting on computer can get to know the whole world. Q2) To some extent it is true. Since national media getting tied up with big global firms, the chance of being acquired may be high in future. The small national media could be acquired by those global media..

what are the factors contributing to the rise in global media Will .pdf

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WACC of Alibaba company WACC of Alibaba companySolutionTh.pdf

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Two particles A and B of equal mass have velocities as shown in the figure to the right. If the particles collide, circle the situation(s) shown 111 the figures below that represent possible velocities for the particles after impact. The velocities in all figures are shown to scale. [0.5 mark for each correct answer] Solution Since thy have equal masses they will exchange velocities both for horizontal and vertical component. horizontal is head on collision so they will have horizontal velocity in opposite direction..

Two particles A and B of equal mass have velocities as shown in the f.pdf

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The forecast equation for mean wind in a turbulent flow is: U_i/t + U_j U_i/x_j = delta_i3 g + f_c epsilon_ij3 U_j - 1/p P/x_i + v ^2 U_i/x^2_j u\'_i u\'_j/x_j Name each term, and give its physical interpretation. Solution solution A is storage of mean momentum B is advection of mean momentum by the mean wind C is gravity to act in the vertical direction only D is the Coriolis effect due to the earth’s rotation E is the mean pressure gradient forces F is the effect of viscous stress on the mean motions G is the effect of Reynolds stress on the mean motions (also called the divergence of turbulent momentum flux).

The forecast equation for mean wind in a turbulent flow is U_it + .pdf

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The image below is of a charged moving in a uniform magnetic field. Is the particle positively or negatively charged? Solution According to Fleming left hand rule, When the thumb,first finger,middle finger are streatched perpendicular to each other and the first finger represents the magnetic field direction ,middle finger represents the direction of motion of positive charged particle then the thumb represents force on charged particle. In figure , magnetic field into the paper and the particle deflect right.So, the charged particle is negative nature.

The image below is of a charged moving in a uniform magnetic field. I.pdf

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The diets common to some cultures emphasize meat. Meat, milk, and bread, but few fruits and vegetables, might be a typical meal for populations in northern latitudes and in desert areas, for example. What ecological factors seem relevant in explaining this human habit? Solution Various studies have shown that populations living in northern latitude and desert areas have dietary patterns higher in animal fat such as meat and dairy products compared to those living in southern latitudes who thrive majorly on fruits and vegetables. This may be because land in northern and desert areas had been less fertile to grow seasonal vegeies, due to extremes of environment and limited rainfalls. People living in these areas over generations have been mostly hunter-gatherers. Such societies had depended for their economic survival primarily on hunting, gathering (wild vegetable and fruits and small land fauna), and fishing. Thus they might have developed the food habit of increased meat and milk consumption (as they are largely dependent on herd of cattle/animals) along with staple diet of breads, over the generations. On the other hand, southern populations residing on more fertile pieces of land relish a vegetarian diet consisting of fresh fruits and vegetables. They are richer in biodiversity due to moderate climates and rainfall distribution throughtout the year. These populations are typically farmers, food growers over several generations and have developed food habits of vegetarian diets..

The diets common to some cultures emphasize meat. Meat, milk, and br.pdf

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aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus and at least one dominant inant gene A pi R P A rose comb is produced when at least and two recessive genes are present at the the first locus at pea comb is produced when two recessive genes are least one dominant gene is at are present at the first and the second locas sNuces 2 for comb shape of the parents in the fol walnut, 1 single, and 1 pea offspring. ngle comb ing cross: Rose 3/. The mutation (bard es is inherited as a maternal effect recessive allele. What is the expected ratio of preho termale and a bad bed male? in offspring cross between bcd normal: mutant 1 3 mutant normal c 3 mutant 1 normal 32 Could the characteristic followed in the pedigree be caused by an autosomal dominant disease? not? Solution 32............is pedigree explaining about autosomal dominant disease/.

aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus.pdf

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t t Deg g olthe year.) 4.9 An engineering construction firm is chartered to provide consulting services, design, and a limited range of construction services. Its designs have developed into registered patents Some designs are licensed by subcontractors, the firm not choosing to use them itself; it does re- ceive income from that business arrangement. These accounts and their balance are summa- rized below or a family-owned firm titled \"Eastwood s are in units of $1000, but ignore r each account title, first identify if it is an asset, liability, net worth, income, expense, Find the income statement and balance sheet f Engineering Services,\" Second Quarter. (Hints: These value that. Fo or tax. Each account will have a spot in either the income or balance-sheet statement.) Solution Income Statement Revenue in $ in $ Fees Recived 73 Sales of design 7 Equipment Lease 6 Royality on Patent 40 Interest On owned Securities 126 Less Salary Expense 28 Utilities Paid 5 Taxes 12 Office Rent 1 Travel 4 Interest On owned Securities 2 Supplies Expense 9 61 Net Income after tax 65 Balance sheet Liability Amount in $ Asset Amount in $ Opening bal of retained earning 27 Equipment Book Value 58 Profit as per income statement 65 Receivable from project 55 Capital Stock 25 Account Payable 30 Patent Asset 11 Bond Owned 12 Cash in Hand 3.

t t Deg g olthe year.) 4.9 An engineering construction firm is .pdf

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Actual costs are determined by plugging the actual level of activity.pdf

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c++ program I need to sort arrays using an insertion sort and a mer.pdf

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Below is my program, I just have some issues when I want to check out a patient and when I want to print a specific patient. Please help. I think there is a minor error, but I can\'t find it. PATIENT ID MUST BE \"LASTNAMEFIRSTNAMEYEAR\" example PattersonMathew2005. Please read instructions help me solve the problem. In this assignment, you will write a struct plus two simple classes and then write a program that uses them all. The first class is a Date class. You are given code for this at the end of this assignment, but you must make some modifications. The other is a Patient class; you are given the class definition and you must write the implementation. The program will keep track of patients at a walk-in clinic. For each patient, the clinic keeps a record of all procedures done. This is a good candidate for a class, but to simplify the assignment, we will make a procedure into a struct. We will assume that the clinic has assigned all care providers (doctors, nurses, lab technicians) with IDs and each procedure (“office visit”, “physical exam”, “blood sample taken”, etc.) also has an ID. Each patient will also have an ID, which is formed by the last name concatenated with the first name and the year of birth. All other IDs (care providers, procedures) will be just integers. When the user runs the program at the start of the day, it first tries to read in from a binary file calledCurrentPatients.dat. It will contain the number of patients(in binary format) followed by binary copies of records for the patients at the clinic . This information should be read and stored in an array of Patient objects. There will be a separate array, initially empty at the start of the program that will store patient records for all patients who are currently checked in at the clinic. It then asks the user to enter the current date and reads it in. It should then presents the user with a simple menu: the letter N to check in a new patient, R for checking in a returning patient, O to check out a patient, I to print out information on a particular patient, P to print the list of patients who have checked in, but not yet checked out, and Q for quitting the program. The following tasks are done when the appropriate letter is chosen. N: The new patient’s first name, last name, and birthdate are asked for and entered. The new patient ID will be the last name concatenated with the first name and the year of birth. (Example: SmithJohn1998) The primary doctor’s ID is also entered. The new patient object is placed in both arrays: one keeping all patients, and one keeping the patients who checked in today. R: The returning patient’s ID is asked for, and the array holding all patients is searched for the patient object. If found, a copy of the patient object is placed in the array for all currently checked in patients. If not found, the user is returned to the main menu after asking them to either try R again, making sure the correct ID was entered, or choose N to enter th.

Below is my program, I just have some issues when I want to check ou.pdf

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Alice uses a biased coin to create a one-time pad to encrypt a message m of 1 bit. The coin comes up 1 (Heads) 60% of the times and 0 (Tails) 40% of the times. Alice sends 0 as a message 75% of the times and 1 as a message 25% of the times. All of this is publicly known. Eve sees cipher c = 1. What was the probability that m = 1? Solution An encryption scheme (E, D) is perfectly secret if there for every set M {0, 1} ` of plaintexts, and for every strategy used by Eve, if we choose at random m M and a random key k {0, 1} n, then the probability that Eve guesses m after seeing Ek(m) is at most 1/|M|. In particular, if we encrypt either “Yes” or “No” with probability 1/2, then Eve won’t be able to guess which one it is with probability better than half. In fact, that turns out to be the heart of the matter: Two to Many Theorem: An encryption scheme (E, D) is perfectly secret if and only if for every two distinct plaintexts {m0, m1} {0, 1} ` and every strategy used by Eve, if we choose at random b {0, 1} and a random key k {0, 1} n, then the probability that Eve guesses mb after seeing Ek(mb) is at most 1/2. Proof: The “only if” direction is obvious— this condition is a special case of the perfect secrecy condition for a set m of size 2 The “if” direction is trickier. We need to show that if there is some set M (of size possibly much larger than 2) and some strategy for Eve to guess (based on the ciphertext) a plaintext chosen from M with probability larger than 1/|M|, then there is also some set M0 of size two and a strategy Eve0 for Eve to guess a plaintext chosen from M0 with probability larger than 1/2. Let’s fix the message m0 for example to be the all zeroes message. Since Eve(Ek(m0)) is a fixed string, if we pick a random m1 from M then it holds that Pr[Eve(Ek(m0)) = m1] 1/|M| while under our assumption, on average it holds that Pr[Eve(Ek(m1)) = m1] > 1/|M| Thus in particular, due to linearity of expectation, there exists some m1 satisfying Pr[Eve(Ek(m1)) = m1] > Pr[Eve(Ek(m0)) = m1] . But this can be turned into an attacker Eve0 such that the probability that Eve0 (Ek(mb)) = mb is larger than 1/2. Indeed, we can define Eve0 (c) to output m1 if Eve(c) = m1 and otherwise output a random message in {m0, m1}. The probability that Eve0 (c) equals m1 is higher when c = Ek(m1) than when c = Ek(m0), and since Eve0 outputs either m0 or m1, this means that the probability that Eve0 (Ek(mb)) = mb is larger than 1/2 (Can you see why?) QED. Exercise: Another equivalent condition for perfect secrecy is the following: (E, D) is perfectly secret if for every plaintexts m, m0 {0, 1} ` , the two random variables {Ek(m)} and {Ek0 (m0 )} (for randomly chosen keys k and k 0 ) have precisely the same distribution. So, perfect secrecy is a natural condition, and does not seem to be too weak for applications, but can it actually be achieved? After all, the condition that two different plaintexts are mapped to the same distribution seems somewhat at odds with th.

Alice uses a biased coin to create a one-time pad to encrypt a messa.pdf

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A paleontologist is studying the history of group of stratigraphic fossils the arthropods. Use the diagram below to answer the questions. The pictures of the organisms represent of two different species that your research team has assigned to the various stratigraphic layers. Which of the fossils has been assigned to the wrong layer? The paleontologist\'s graduate student was unsure which rock layer was the most in important to date so she dated all of them with the results as follows: A: 100 mya, B: 50 mya, C: 10 mya, D: 300 mya When did species 1 go extinct? When did species 2 go extinct? If found a non-fossilized organic sample in the soil above the rock layers, how might my approach to determining its age be different than for the fossilized remains? Solution a) Species in the igneous rock has been assigned to the wrong layer. Because, Igneous rocks do not contain fossils as they form from the molten rocks. These rocks form when the molten rocks become cools and eventually solidify. Instead of this, fossils are generally found in sedimentary rocks which is formed in layers by the depositing and pressing of sediments on top of each other. b) i) species 1 went extinct after 100 million years. ii) Species 2 went extinct after 50 million years. c) If I found a non-fossilized organic sample in the soil above the rock layers then I can use carbon dating or radiometric dating. It is the process of determining the percentage of the radioactive isotope in the sample and using the half life data to ultimately determine the age of the sample. Basic data of radiocarbon dating- Carbon 14 is an isotope of carbon which is naturally formed in the atmosphere. All plants and animals when they are alive, have regular intake carbon. When an animal or plant dies it no longer takes in carbon of any form. That plant or animal most often decays and its constituent parts break down and are eaten by scavengers, or they decay, or are washed away etc. Instead of these events if the animal or plant is encased in some sediment, or frozen, or desiccated, or otherwise preserved quickly so that its structure and constituent parts are preserved. Scientists have ways of counting the carbon 14 atoms in a sample and then they find out the total number of carbon atoms and figure out how old the material is. In this way fairly new organic material’s age is also determined..

A paleontologist is studying the history of group of stratigraphic fo.pdf

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A negative stain is repelled by the negatively charged bacterial cell and allows the observer to see a clear picture of size, shape, and aggregation of the cell. True False Solution The statement is false. Negative stain means during the observation of bacteria or other microbes under microscope, the background of them stained, not the microbes themselves. This allows the microbes visible without staining them. If microbes stain, then it called as Positive stain..

A negative stain is repelled by the negatively charged bacterial cel.pdf

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1. The circulatory system of a whale is considered an organ system because nis composed of different A. cells. B. tissues. C. organs. E. heart. 2. Which of the following lists the terms from simplest to most complex? A. cells, organs, tissues, organ systems, organism B, organs, organ system, organism, cells, tissues C. tissues, organs, organ systems, organism, cells D, cells, tissues, organs, organ systems, organism 3, The smallest unit of life is aln) A tissue. B. organ. C. cell. D. species. transforms solar energy into chemical energy 4. The process of B, homeostasis C. respiration D. photosynthesis 5. Which of the following is matched incorrectly? A. brain organ B. neuron cell C skin organ D. gene organelle living things? 6. Which of the following is not one of the most common elements in A, carbon B. oxygen C, hydrogen D. iron 7 Glucose, CHizo. is best described as A) an element B) an a compound DJan ion isotope Solution Answer: (1) C The circulatory system of a whale is considered an organ system because it is composed of different organs. (2) D Cells, tissues, organs, organ system, organism (Simplest to more complex). (3) C The smallest unit of life is cell. (4) D The process of photosynthesis transforms solar energy into chemical energy. (5) D Gene is not an organelle but is a short portion of DNA which encodes a particular protein. (6) A Carbon is the most abundant element in living organisms. (7) C Glucose is an organic compound composed of carbon, hydrogen and oxygen. (8) A An element that has electrical charge is called ion. (9) B Ionic bonds are formed by transfer of electrons between atoms. (10) B Oxygen is an element. (11) C Deoxyribose is a component of DNA. (12) A Digestion of sucrose results in 2 monomers. (13) B A fat is formed when three fatty acids join a glycerol by a dehydration reaction. (14) B Parts W and Z are hydrophilic since they lie outside the plasma membrane. (15) B Cell wall helps in maintaining cell shape. (16) D It represents cell wall (As evident from the image and its not very clear) (17) B Prokaryotic cell does not have a membrane bound nucleus. (18) B Mitochondria is the site of cellular respiration and produces ATP. (19) A Both chloroplast and mitochondria have their own DNA. (20) C Stack of thylakoids are called grana. (21) A Isotonic solutions have the same concentration. (22) C ATP releases energy when phosphate is removed. (23) C The cell wall protects the cell from bursting in a hypotonic solution. (24) C Solar energy is captured by chloroplasts for photosynthesis. So, solar energy, chloroplast, mitochondria. (25) B The part of the enzyme into which the substrate fits is the active site..

1. The circulatory system of a whale is considered an organ system be.pdf

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Reliability and safety are related but distinct dependability attributes for software. Describe two important distinctions between these attributes. Explain why it is possible for a reliable system to be unsafe and vice versa. Solution Explain why it is possible for a reliable system to be unsafe and vice versa.? ANS:------> Because If the specification does not explicitly exclude dangerous behavior then a system can be reliable but unsafe.RELIABILITYSAFETY1) Reliability involves much more than analyzing test results, estimating remaining faults, and modeling future failure probabilities.1)safety optimizes system safety in the design, development, use, and maintenance of software systems and their integration with safety-critical hardware systems in an operational environment.2)Ensuring system reliability does not necessarily lead to system safety as Reliability is concerned with meeting the system specification (the system \'shall\')2)Safety is concerned with excluding the possibility of dangerous behavior (the system \'shall not\').

Reliability and safety are related but distinct dependability attrib.pdf

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Please answer all questions Define 1. Wild type – 2. Epistasis – Fill in the Blank 3. Two genes may interact effecting one trait and resulting in novel phenotypes with an F2 phenotypic ________________ ratio. 4. Under ________________ conditions, conditionally lethal mutants will die. True or False 5. Penetrance is the degree or intensity with which a particular genotype is expressed in a phenotype. 6. Continuous traits are usually controlled by multiple genes and may be influenced by the environment. Short Answer 7. Describe the inheritance pattern(s) of blood types. 8. When studying a newly discovered flowering plant in Costa Rica, you collect data on its petal color in an effort to determine how this trait may be inherited. You find that 852 flowers are red, 331 are white and 389 are yellow. What are the possible inheritance patterns, and how could you definitely conclude which of these possibilities is the true mode of inheritance? Solution Wild type- wild type phenotype refers to characteristics that occurs naturally in a breeding population. Epistasis - it refers to a phenomena when one gene’s expression is dependent on the presence of another gene(s), called the modifier gene. 9:16 Let\'s take an example of comb varieties in chicken We have two pure breeds : wyandotte whose phenotype is a rose comb and brahmas having the phenotype of a pea comb, when crossed with each other, they create a completely new phenotype of a walnut comb. Let\'s look at the ratio: Parental cross: rose comb x pea comb F1 : all walnut combs F2 : 9 walnut : 3 rose: 3peas : 1 single Hence the ratio is 9:16 Under the restrictive or nonpermissive condition conditions, conditionally lethal mutants will die. Conditionally lethal mutants are those which have an allele that will make the survival of the species difficult only in a specific condition. True. for example if a mutation in a gene gives the species a particular characterisitc which has 97%penetrance, only 97% of those with the mutation will develop that characteristic and 3 % will not. True. multiple genes affect continuous traits. For such complex gene interactions, the traits are influenced by environmental factors too. For example, height and skin color. There are four basic blood types which are A, B, AB and O and together with the Rh factor, these can be +ve or -ve. A and B are dominant, while O is recessive. Blood type A B O A A AB A B AB B B O A B O For Rh factor, Rh (+)ve is dominant. So if the any one or both parents have Rh (+)ve factor, the progeny will be Rh(+)ve. If both parents are Rh(-)ve, the progeny will be Rh (-)ve Father Mother D D d Dd Dd d Dd Dd 100% Rh (+)ve children Father Mother D d d Dd dd d Dd dd 50% Rh(+)ve children Red: 852 White: 331 Yellow:389 According to the given data, red is dominant and white is recessive and yellow is a result of incomplete dominance of both white and red phenotype. It can\'t be codominance, because codominance involves both phenotypes being expressed in di.

Please answer all questionsDefine1. Wild type –2. Epistasis –.pdf

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