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

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11) Establish each identity. Start from left side of the equation (LHS), and show that it\'s equal to right hand side (RH a) (sin x + cos x)^2 = 1 + sin 2x b) sin^2x + cos^2x=cos^2 x I c) sin Theta(cot Theta +tan Theta)=sec Theta d) csc Theta- cot Theta=sin Theta/a+cos Theta Solution 1. LHS = (sin x + cos x)^2 = sin^2 x + cos^2 x + 2 sin x cos x = 1 + sin 2x = RHS 2. LHS = sin^2 x + cos 2x = sin^2 x + cos^2 x - sin^2 x = cos^2 x =RHS 3. LHS = sin x (cot x + tan x) = sinx (1/tanx + tanx) =sinx / tan x * [1 + tan^2 x] = sinx * cosx / sin x * sec^2 x = cos x / cos^2 x = 1/cos x = sec x 4. LHS = =cosecx - cotx = 1/sin x - cosx/sinx = (1-cos x ) / sin x =[(1-cos x ) * (1+cos x)] / [ sinx (1+cosx)] = [1-cos^2 x] / [sin x *(1+cos x)] = sin^2 x / [sin x *(1+cos x)] =sinx/(1+cos x) =RHS if you have further query, then ask in comment..

11) Establish each identity. Start from left side of the equation (LH.pdf

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Software EvolutionWhy might it sometimes be necessary to bypass th.pdf

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With reference to Penman (2007) and Laux, C., & Leuz, C. (2009), discuss the main issues related to use of the fair value measurements. Solution 1. Fair-value accounting contributes to excessive leverage in growings periods and leads to excessive write-downs in down periods. 2. It is easier to measure liquid assets, but having an measure for illiquid asset can be conflicting. 3. It can create e contagion in financial markets, i.e. if one has to sell his asset at lower than ifs FMV due to some reasons than other person has to take that basis and devalue his asset which can create not true measurements. 4. Another issue to address is the balance between relevane and reliability. As stated by Laux and Leuz, Relevant accounting information is capable of confirming decisions that stakeholders have already made and/or of making, and reliability requires that accounting information to be independently verifiable, unbiased, and an accurately portray the underlying reality of the transactions and events..

With reference to Penman (2007) and Laux, C., & Leuz, C. (2009), dis.pdf

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Which of the following pairs of events are disjoint? (Select all that apply) E1 = [Patient is dead],E2 = [Patient is alive] E1 = [A litter contains 10 mice], E2 = [A litter contains 15 or more mice] E1 = [A patient weighs between 100 and 200 lbs], E2 = [A patient weighs between 80 and 130 lbs] E1 = [A subject has green eyes], E2 = [A subject is male] Solution Disjoint : E1 = [Patient is dead],E2 = [Patient is alive] Disjoint : E1 = [A litter contains 10 mice], E2 = [A litter contains 15 or more mice] Not disjoint : E1 = [A patient weighs between 100 and 200 lbs], E2 = [A patient weighs between 80 and 130 lbs] Not Disjoint : E1 = [A subject has green eyes], E2 = [A subject is male].

Which of the following pairs of events are disjoint (Select all tha.pdf

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Which of the following activities can be simulated using a computer.pdf

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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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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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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..

What are vitaminsSolutionVitaminsVitamins are organic nitr.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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