What are vitaminsSolutionVitaminsVitamins are organic nitr.pdf

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What are vitamins? Solution Vitamins: Vitamins are organic nitrogenous compounds that are required in very small quantities for carrying out various biochemical reactions. They are not synthesized in the human body and hence should be supplemented in diet by including different varieties of fruits, vegetables and food in diet. Their deficiency in diet leads to characteristic deficiency disorders. Classification of Vitamins: There are about 15 vitamins are essential for humans. They are classified as.. A. Fat soluble vitamins: Vitamin A, D, E and K. B. Water soluble vitamins: It again divides into 2 groups. 1. Non B-Complex: vitamin C 2. B-Complex: Vitamin B1, B2, B3, B5, B6, B7, B9, B12..

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Which of the following is true about asymmetric reflexive associations? They have to be unidirectional They have to be labeled using different role names They have to be labeled using association names They connect a class to its super class They represent a mutually exclusive relationship between classes Solution A. They have to be unidirectional B. They have to be labeled with different role names.

Which of the following is true about asymmetric reflexive association.pdf

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What is your stance on vaccination? Do you believe that vaccination should be optional given its potential link to autism and other diseases; why or why not? Provide a brief (half-page) response. Also, provide at least one future recommendation for how you would tackle this issue from a more global health perspective. Solution Vaccines are biological preparations contains pathogrn8c material either active or killed, which when inoculated in persons body activates immune system by stimulating production of antibodies against that pathogen. Vaccines have been one of the remarkable breakthroughs in the biological world. Diseases which were once dreadful and life threatening have been wiped out. Small pox was the first diseases erad8cated from the world by vaccination. Nowadays efforts are going on to produce vaccine against AIDS and cancer. In the ancient times when vaccines were not designed, many of the diseases like small pox, polio, measles , chicken pox etc were very life threatening and in most of the cases childrens were at the receivibg end. We still can see how diseases like polio has hindered life of children in many under developed and developing nations due to non availability of vaccines. These days, where new pathogens and diseases causing agents are evilvung at a very rapid rate, it is highly important and mandatory for us to keep ourselves and our kids safe as much as we can and if vaccines have proved helpful in this safety measure we should opt it..

What is your stance on vaccination Do you believe that vaccination .pdf

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Help Save & Exit Exercise 1-14 Analysis of return on assets LO A2 Swiss Grou 2017, assets had grown to $198,000. p reports net income of $40,000 for 2017. At the beginning of 2017, Swiss Group had $148.000 in assets. By the end of What is Swiss Group\'s 2017 return on assets? Solution Choose Numerator: / Choose Denominator: = Return on Assets Net Income / Average Total Assets = Return on Assets $40000 / 148000+198000 2 = 173000 = 23.12% Choose Numerator: / Choose Denominator: = Return on Assets Net Income / Average Total Assets = Return on Assets $40000 / 148000+198000 2 = 173000 = 23.12%.

Help Save & Exit Exercise 1-14 Analysis of return on assets LO A2 Swi.pdf

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What kind of micro organisms should be present in a dark plant infusion, a peppercorn infusion, and a Bacillus culture? What kind of micro organisms should be present in a dark plant infusion, a peppercorn infusion, and a Bacillus culture? Solution Dark plant infusion contains a lotamount of humus so it can easily attract mosquitos. These mosquito are carrier of some species of virus. Bacterias are also easily grow in these. Peppercorn infusion contains microorganisms like Trichomonas sp. And Enterobius. Bacillus culture contains Bacillus substilis. Because this species can soon and easily groun in any kind of environment..

What kind of micro organisms should be present in a dark plant in.pdf

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What is a cell What is a cellSolutionCell is a biological.pdf

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What are constructs in relationship to health behavior theory? 1. Concepts developed or adopted for use in a particular theory. 2. Operational forms 3. They draw on a number of theories to help understand a particular problem in a certain setting or context. 4. Building blocks—the primary elements—of a theory. Solution Correct option (A). A construct is a variable or hypothetical object that is used in a particular theory, but are not directly observable..

What are constructs in relationship to health behavior theory 1. Co.pdf

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Using Method of Substitution, solve the following recurrence relation T(N) = 2T(N-1) + 1 With T(l) = 0 [Some of the popular finite series are: 1 + 2 + 3 + ..+ N = N(N+l)/2 1 + r + r^2 + r^3 + ... + r^N = (r^N + 1 - 1)/(r-1) (r is any positive number) 1 + 3 + 5 + ...+(N-2) + N = Solution The given series is in arthematic progression, with first term a = 1 Common difference d = 2 Therefore sum of N terms = a+(N-1)d = 1+(N-1)2 =. 1+2N-2 = 2N-1 Therefore,. 1+3+5+.....(N-2)+N =. 2N-1.

Using Method of Substitution, solve the following recurrence relation.pdf

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Use Theorem 14.1 to show that for all odd n>= 3 we have D2n is isomorphic to Dn x Z2. Theorem 14.1: Suppose A and B are subgroups of G s.t. i) A is normal to G, B is normal to G ii) AB = G iii)A n B = {e} Then G is isomorphic to A x B Solution When k 3, the dihedral group D2 k has one normal subgroup of each index except for three normal subgroups of index 2. The “exceptional” normal subgroups hr 2 , si and hr 2 , rsi in Dn for even n 4 can be realized as kernels of explicit homomorphisms Dn Z/(2). In Dn/hr 2 , si we have r 2 = 1 and s = 1, so r a s b = r a with a only mattering mod 2. In Dn/hr 2 , rsi we have r 2 = 1 and s = r 1 = r, so r a s b = r a+b , with the exponent only mattering mod 2. Therefore two 8 KEITH CONRAD homomorphisms Dn Z/(2) are r a s b 7 a mod 2 and r a s b 7 a+b mod 2. These functions are well-defined since n is even and their respective kernels are hr 2 , si and hr 2 , rsi. We can also see that these functions are homomorphisms using the general multiplication rule in Dn: r a s b · r c s d = r a+(1)b c s b+d . We have a + (1)b c a + c mod 2 and a + (1)b c + b + d (a + b) + (c + d) mod 2.

Use Theorem 14.1 to show that for all odd n= 3 we have D2n is isomo.pdf

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The Zygomycota produce... (mark all that apply) septate hyphae with a central pore multi- nucleate zygosporangia asexual spores that are the most desiccation-resistant stage of the life cycle coenocytic hyphae zygosporangia that are produced asexually septate hyphae with solid septae two different types of spores one through asexual processes and one through sexual processes Solution The zygomycota produce.

The Zygomycota produce... (mark all that apply) septate hyphae with .pdf

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The probability that individual VI-1 would have this disease is: a. 1/4 b. 1/9 c. 1/18 d. 1/24 e. 1/36 f. 1/64 ODFOODOHO L11110 Solution Answer is option f. 1/64 .There are 6 generations and circle represents female whereas square represents male. The horizontal line between them represents mating whereas vertical line represents children/offspring. The double line represents consanguineous (related by blood) marriages. So the coloured squares/circles represents diseased/affected individuals. The diseased persons are found in the fourth generation and are offsprings of a consanguineous marriage. Since their parents are unaffected they are unaffected carriers, and the recurrence risk of the disease for each future child is 25%, while the risk of a child becoming a carrier like the parents is 50%. The fourth generation unaffected individuals will have probability of getting disease of 1/4. The fifth generation children mate with other individuals and will carry the risk of inheritance of disease probability as 1/4 x 1/4 = 1/16. Again in the fifth generation there is a consanguineous marriage and the VI 1 individual is found unaffected. So his probability of getting disease is 1/16 x 1/4 = 1/64.

The probability that individual VI-1 would have this disease isa..pdf

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Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension) / larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? Solution We have given that ,Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension) / larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Consider alpha = level of significance = 0.05 Here hypothesis for the test is, H0 : µ1 = µ2 Vs H1 : µ1 µ2 The test is two sided. Here first we have to check whether the two variances are equal or not. Hypothesis for the test is, H0 : The two variances are equal. H1 : The variances are not equal. The test statistic is, F = Larger variance / Smaller variance We have given that, n1 = number of early works = 37 n2 = number of later works = 65 X1bar = 1.74 X2bar = 1.746 s1 = 0.11 s2 = 0.1064 F = 0.11^2 / 0.1064^2 = 1.0688 critical value for F distribution we can calculate by using EXCEL. syntax : =FINV(probability , d.f.1, d.f.2) probability = alpha / 2 = 0.05 / 2 = 0.025 d.f.1 = n1 - 1 = 37 - 1 = 36 d.f.2 = n2 - 1 = 65 - 1 = 64 Critical value = 1.7523 F < critical value Fail to reject H0 at 5% level of significance. Conclusion : Variances are equal. Now we test two means using pooled variance. Pooled variance (S) = sqrt [ (n1-1) * s1^2 + (n2 - 1) * s2^2 / (n1 + n2 -2) ] S = sqrt [ (37-1) * 0.11^2 + (65-1) * 0.1064^2 / 100 ] = sqrt(1.1601 / 100) S = 0.10771 The test statistic is, t = (X1bar - X2bar) / [ S * sqrt(1/n1 + 1/n2) ] = ( 1.74 - 1.746) / [ 0.10771 * sqrt ( 1 / 37 + 1 / 65 ) ] t = - 0.006 / 0.0222 = - 0.27049 t = - 0.27049 We can find P-value for taking decision. EXCEL syntax : =tdist ( x,d.f.,tails) x is the test statistic value d.f. = n1 + n2 - 2 = 37 + 65 - 2 = 100 tails = 2 P-value = 0.787341 P-value > alpha Fail to reject H0 at 5% level of significance. Conclusion : The two means are equal..

Elizabeth Mjelde, an art history professor, was interested in whethe.pdf

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Discrete Math: How many ways are there for a man to distribute 12 indistinguishable gold coins to his 5 children if each child must recieve at least one coin? Solution Number of indistinguishable gold coins (n) = 12 No. of children (k)= 5 No. of ways of distributing n indistinguishable objects among k children such that each child receive at least one coin is (n-1)! / (k-1)!(n-k)! ie., 11!/4! 7! = 330 ways.

Discrete Math How many ways are there for a man to distribute 12 in.pdf

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Refer to the drawing in Figure 13.2 of a single pair of homologous chromosomes as they might appear during various stages of either mitosis or meiosis, and answer the following question (28). Which diagram represents chromosomes in prophase I of meiosis? I II IV V VI Which diagram represents anaphase II of meiosis? I III IV V Solution Chromosomes in prophase 1 makes \"bouquet stage\" , that is they situated beside each other. And also crossing over takes place in the Pachytene stage in prophase 1 . So , here picture no -IV clearly depicts a crossings over and situated beside each other. So , pic -IV is a picture of prophase 1. In anaphase 2 both chromosome get segregated from each other. And they have an Anaphasic movement. picture no. -II clearly depicts that. So picture no.-II describes anaphase -II. [ note- in pic no 1 the chromosomes are not segregated and in pic no -V one set of chromosome is missing ].

Refer to the drawing in Figure 13.2 of a single pair of homologous ch.pdf

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Organisms that survive in varied temperatures adapt to changes by using supercoiling. hyperosmotic molecules. countercurrent circulation. isozymes. Coral bleaching occurs when coral exoskeletons begin to break down. is due to a lack of dissolved nutrients from the surrounding water. is a temporary process that is usually reversed within days. occurs when algae are expelled from coral. The primary cause of coral bleaching is decreased water temperature. increased water temperature. decreased salt concentrations. increased water pH. decreased water pH Solution 17. Option C. Counter current circulation. Heat exchange or conservation can be achieved by counter current circulation. 18. Option D. Occurs When algae are expelled from coral 19. Option B. Increased water temperature When the water temperature increases, algae are expelled from the corals making them to appear white or bleached.

Organisms that survive in varied temperatures adapt to changes by usi.pdf

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Describe the basic properties of DNA that are responsible for fiber formation when alcohol is added to a solution that contains native (double-stranded) DNA. Describe the type of precipitate that is formed when alcohol is added to denatured (single stranded) DNA. Offer an explanation as to why this precipitate is different from that which is observed with native DNA. Describe the action of DNAse I on DNA and relate this effect to the results of your experiment in Section III. Solution 1). The highly polar phosphate backbone of the DNA and the sugar molecules makes the DNA to dissolve in water because water is highly polar, “like dissolves like.” In contrast to water, ethanol is less polar and the polarity decreases with increasing carbon number. So, when ethanol and positive ions are added to the solution, the charges are disrupted and the phosphate ions become free. The negatively charged phosphate ions combine with the positive ions and form ionic bonds, which is visible as DNA precipitation..

Describe the basic properties of DNA that are responsible for fiber f.pdf

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John, Rick, and Molli paint a room together. A. Pick a reasonable amount of time the three can paint the room together. Also pick a reasonable amount of time that John can paint it alone and a reasonable amount of time that Rick can paint it alone. B. What is the hourly rate for John, Rick, and Molli (when working together)? C. What is the hourly rate for John? What is the hourly rate for Rick? Refer back to the amount of time that you determined John and Rick can paint the room alone. Use rooms/hours as the units for your rates. D. Write an equation comparing the group rate to the sum of the individual rates. How should the group rate and the sum of the individual parts compare? Use parts (b) and (c) to assist in writing the equation. E. What is the least common denominator for the equation you found in (d)? F. Solve the equation and determine how long it will take Molli to paint the room alone. Solution Let John, Rick and Molli be able to paint the room alone in x, y and days respectively. Then John, Rick and Molli can paint 1/x, 1/y and 1/z of the room alone, respectively in 1 day. A. The three together can paint (1/x ) + (1/y) + (1/z) = (yz + xz + xy)/ xyz portion of the room in 1 day. THerefore, the three together can paint the room in xyz/ (yz + xz + xy) days. Jojn can paint it alone in x days and Rick can paint it alone in y days. B. We have assumed that John, Rick and Molli can paint 1/x, 1/y and 1/z of the room alone, respectively in 1 day i.e. in 1 hour, they can paint 1/24x, 1/24y and 1/24z of the room respectively in 1 hour. Then, the hourly rate for John, Rick, and Molli (when working together) is k/24x, k/24y and k/24z respectively where k is the rate for painting the room C. The hourly rate for John alone is k/24x. The hourly rate for Rick alone is k/24y. D. The group rate is k( 1/24x + 1/24y + 1/24z) = k/24 (yz + xz + xy)/ xyz = k (yz + xz + xy)/24xyz . The individual rates are k/24x , k/24y , and k/24z. Thus, the required equation is k (yz + xz + xy)/24xyz = k/24x+k/24y+ k/24z. The group rate and the sum of individual rates are equal. E.The least common denominator is 24xyz. F. Molli will takez days to paint the room alone. Note: In the absence of any other data, this is the only way to proceed..

John, Rick, and Molli paint a room together.A. Pick a reasonable a.pdf

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INCENTIVE TO OVERPRODUCE INVENTORY The absorption of fixed overhead costs as part of the cost of inventory on the balance sheet presents ethical challenges because it provides the opportunity to manipulate reported income. This classic case is based on an actual company’s experience.* Brandolino Company uses an actual-cost system to apply all production costs to units produced. The plant has a maximum production capacity of 40 million units but during year 1 it produced and sold only 10 million units. There were no beginning or ending inventories. The company’s absorption-costing income statement for year 1 follows: BRANDOLINO COMPANY Income Statement For Year 1 Sales (10,000,000 units at $6) $ 60,000,000 Cost of goods sold: Direct costs (material and labor) (10,000,000 at $2) $20,000,000 Manufacturing overhead 48,000,000 68,000,000 Gross margin $ (8,000,000) Less: Selling and administrative expenses 10,000,000 Operating income (loss) $(18,000,000) The board of directors is upset about the $18 million loss. A consultant approached the board with the following Page 345offer: “I agree to become president for no fixed salary. But I insist on a year-end bonus of 10 percent of operating income (before considering the bonus).” The board of directors agreed to these terms and hired the consultant as Brandolino’s new president. The new president promptly stepped up production to an annual rate of 30 million units. Sales for year 2 remained at 10 million units. Here is the resulting absorption-costing income statement for year 2: BRANDOLINO COMPANY Income Statement For Year 2 Sales (10,000,000 units at $6) $60,000,000 Cost of goods sold: Costs of goods manufactured: Direct costs (material and labor) (30,000,000 at $2) $60,000,000 Manufacturing overhead 48,000,000 Total cost of goods manufactured $108,000,000 Less: Ending inventory: Direct costs (material and labor) (20,000,000 at $2) $40,000,000 Manufacturing overhead (20/30 × $48,000,000) 32,000,000 Total ending inventory costs $72,000,000 Cost of goods sold 36,000,000 Gross margin $24,000,000 Less: Selling and administrative expenses 10,000,000 Operating income before bonus $14,000,000 Bonus 1,400,000 Operating income after bonus $12,600,000 The day after the year 2 statement was verified, the president took his check for $1,400,000 and resigned to take a job with another corporation. He remarked, “I enjoy challenges. Now that Brandolino Company is in the black, I’d prefer tackling another challenging situation.” (His contract with his new employer is similar to the one he had with Brandolino Company.) What do you think is going on here? How would you evaluate the company’s year 2 performance? Using variable costing, what would operating income be for year 1? For year 2? (Assume that all selling and administrative costs are committed and unchanged.) Compare those results with the absorption-costing statements. INCENTIVE TO OVERPRODUCE INVENTORY The absorption of fixed overhead costs as part of the c.

INCENTIVE TO OVERPRODUCE INVENTORYThe absorption of fixe.pdf

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1. Please explain the reaction mechanism and kinetics for the treatment of municipal sewage using constructed wetland. (20 points) 2. Please describe the commonly used treatment processes in wastewater reclamation and reuse. (20 points) Please describe the commonly applied control measures for the management of nonpoint source pollution from construction sites. (18 points) 3. 4. What is a TMDL? What is the MOS for TMDL calculation? How the TMDL is allocated among different pollution sources? (17 points) Solution 1. Natural wetlands perform many functions that are beneficial to both humans and wildlife. One of their most important functions is water filtration. As water flows through a wetland, it slows down and many of the suspended solids become trapped by vegetation and settle out. Other pollutants are transformed to less soluble forms taken up by plants or become inactive. Wetland plants also increse the necessary conditions for microorganisms to live there. Through a series of continual processes, these microrganisms also transform and remove pollutants from the water. These are some of the design conditions for construstioc of welands: Construct treatment wetlands should be done on uplands and outside floodplains in order to avoid damage to natural wetlands and other aquatic resources, • Consider the importance of treatment wetlands within the watershed like potential water quality impacts, surrounding land uses and relation to local wildlife in that area • Closely examine factors, such as soil suitability, hydrology, vegetation, and presence of endangered species or critical habitat, when determining an appropriate location • Use water control measures that will allow good results to changes in water quantity, quality, depth and flow. • The flow in that area to be maintained continually with regular inspection. 2. There are three major stages in wastewater treatment Primary:Raw sewage is first subjected to preliminary treatment where it is treated to remove large objects. Primary treatment is the process wherein sludge settles by sedimentation, and all floating material like oil is taken off from the surface. This produces a somewhat homogenous liquid, Secondary: In this stage, aerobic or anaerobic bacteria feed on organic matter, thus removing most of it. After this, if necessary, the water can be disinfected through chlorination which will kill off microorganisms. Secondary treated water can be used for non-potable purposes like toilet- flushing and gardening. Tertiary: In this treatment involves more advanced processes like reverse osmosis (RO) and demineralisation. Dissolved solids and nutrients get further removed during these processes The most used treatment processes are: Activated Sludge Procees ASP: Atmospheric air or pure oxygen is bubbled through primary treated sewage combined with organisms to develop a biological floc which reduces the organic content of the sewage. Extended Areation: This is a modification for ASP as it .

1. Please explain the reaction mechanism and kinetics for the treatme.pdf

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IN C++ language Write a simple word processor that will accept text and commands as input. The commands it should respond to are: Substitute/oldstring/newstring/ Copy # Locate/string/ Delete # Move # Type # Pastes Insert # Replace # Quit The program should be able to open an existing file or create a new one. It should keep track of where the user is in the file at any time. Each command is followed by a number which indicates the number of line (the current line & those following) upon which the command should act. Each command should operate on the specified line or lines of the file. Solution #include #include #include #include using namespace std; const bool DEBUG = false; const int MAX = 100; struct file // This contains the file information and contents { string contents[MAX]; string fileName; int currentLineIndex; int total; bool markupHighlighting; }; int NumberInputReader(string); // Reads the numbers from user input void Help(); // Prints out a help menu void Markup(file&, int); // Does HTML highlighting void MarkupApply(file&, string, int); // Applies the HTML highlighting to the file //void Substitute(file&, string); // substitutes portions of lines of text //void Type(file&, int); // prints text to the screen //void Copy(file, string[], int&); // copies text //void Paste(file&, string[], int, int); // pastes text //void Locate(file &, string); // finds text //void Insert(file &, int); // inputs text to the array //void Delete(file &, int); // deletes text //void Replace(file &, int); // replaces while lines of text text //void Move(file &, int); // moves the current line //void Quit(file); // quits the program, asks the users if they want to save. void Open(file &); // allows the user to open a file void New(file &); // allows the user to create a file void Save(file); // saves the array to a file. void IndexPrint(file); // for debugging, this prints the contents of the array. int main() // Main program { file txtFile; txtFile.currentLineIndex=-1; txtFile.total=0; int inputValue=0, copyNumber; string inputString, openYN, copyQue[MAX]; char inputChar, openYNtemp; // ANIMATION LOOP GOES HERE cout<<\"~~~\\033[1;32m Welcome \\033[0m~~~\ \"; // first use of colors in the program cout<<\"Would you like to open an existing file? \" <\' Markup(txtFile, inputValue); break; case \'>\': // Unhighlight txtFile.markupHighlighting=false; // if false, it\'ll go through the motions Markup(txtFile, inputValue); // again, but revert to the original array break; // afterward. case \'*\': // * saves file to disk Save(txtFile); break; case \'?\': Help(); // \"I seem to be having difficulty with my lifestyle.\" -Arthur Dent break; case\'.\': // default for re-looping break; default: cout<<\"\\033[1;31m\'\"< total: \"< currentLineIndex: \"<=80) // If it takes up the entire default width of { // the window. cout<<\"\\033[1;31m-That\'s a very long command, \" <<\"I\'m just but a humble Line Editor.-\\033[0m \"<=8) { inputColor=0; // Default color cout<<\.

IN C++ languageWrite a simple word processor that will accept text.pdf

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How do animals, plants, and fungi vary in terms of where in the life cycle mitotic divisions (to create a multicellular body) take place? Solution In animals and plants mitosis occurs during cellular growth phase. The cells increase in number or proliferate through mitosis. In fungi, mitosis is the asexual reproduction through which daughter fungal cells are formed. Therefore, mitosis in animals and plants takes place during growth stage and in fungi, mitosis takes place during reproduction stage. The mitosis is carried out in four stages known as prophase, metaphase, anaphase and telophase. Prophase: The nucleolus is disappeared and the nuclear envelop is fragmented. The spindle fibers attached to the chromosomes at the centromere. The chromosomes are pulled to the poles. Metaphase: During this phase the chromosomes are arranged at spindle equator that is the midpoint between the spindle poles. The spindle fibers extend up to opposite poles and the unattached spindle fibers overlap behind the equator. Anaphase: During this phase, the sister chromatids separate apart to become the daughter chromosomes. The attached spindle fibers disassembled and at each pole a daughter chromosome is formed. The spindle fibers move apart so that the separation of chromosomes occurs. Telophase: During this phase, the spindle fibers disappear and the nuclear envelop is formed around each daughter chromosome. Each nucleus consists of same number and type of chromosomes similar to the parent cell. Cytokinesis: After the division of nucleus the division of cytoplasm takes place and the daughter cells are completely apart from the parent cell. In some cases the division of cytoplasm does not take place and only mitosis is continued resulting in a multinucleated cell..

How do animals, plants, and fungi vary in terms of where in the life.pdf

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