a. Dipole b. Set of a finite number of point charges c. A uniformly c.pdf

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a. Dipole b. Set of a finite number of point charges c. A uniformly charged rod d. A uniformly charged ring e. A uniformly charged disk f. A solid sphere uniformly charged throughout its volume g. A spherical shell with a uniformly charged surface Solution its better to choose A solid Sphere uniformly charged throughout its volume. because, it just acts as a point charge. and there will be no change in either Electric field, Potential or Force etc except at the distance from centre to R..

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Verified the following trig identities and show all work. csc4 t-cot4 t = csc2t+cot2 t cosB/1-sinB = secB+tanB 1/csc y-cot y = csc y + cot y Cot 4u-1/cot 4u+1 = 1-tan 4u/1+tan 4u (sec t + tan t)2 = 1+sin t / 1-sin t Cos3x –sin3x /cos x-sin x = 1+sin x cos x Csc x /1+csc x – csc x/ 1-csc x= 2sec2 x (1-tan2 theta)2 = sec4 theta – 4 tan2 theta Solution 1)csc4t-cot4t a2-b2=(a-b)(a+b) =(csc2t-cot2t)(csc2t+cot2t) =1*(csc2t+cot2t) =csc2t+cot2t 2)cosB/1-sinB multiply and divide by (1+sinB) =cosB(1+sinB)/((1-sinB)(1+sinB)) =cosB(1+sinB)/(1-sin2B) =cosB(1+sinB)/(cos2B) =(1+sinB)/cosB =(1/cosB)+(sinB/cosB) =secB+tanB.

Water is essential to all living organisms. Discuss THREE properties .pdf

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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. The article “Reproduction Study of Toxaphene in the Rat” (J. of Environ. Sci. Health, 1988: 101-126) reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was 32 g and for 20 female low-dose rats was 54 g. Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05 For this problem, make a two-sided t interval & perform an F test for the ratio of variance, & create a confidence interval for the ratio of the variance. Solution.

tree to answer question 2. n Arabidopsis arenosa Arabidopsis halleri .pdf

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The number of passengers (including the driver) in cars that cross a certain bridge is a random variable Y having mean E(Y) = 2.7 and variance V(Y) = 1.2. The toll on the bridge is $3.00 per car plus $0.50 per person in the car. The mean and variance of the total amount of money that is collected from a car that crosses the bridge are: (a) mean = $8.60, variance = $.30 (b) mean = $4.35, variance = $3.30 (c) mean = $8.60, variance = $.60 (d) mean = $4.35, variance = $.30 (e) mean = $1.35, variance = $.30 Solution The amount they pay in terms of Y is A = 3.00 + 0.50Y Thus, E(A) = E(3.00 + 0.50Y) = 3.00 + 0.50E(Y) = 3.00 + 0.50*2.7 E(A) = 4.35 Also, Var(A) = Var(3.00 + 0.50Y) = 0.50^2 Var(Y) = 0.50^2 * 1.2 Var(A) = 0.30 Thus, it is OPTION D. [ANSWER, D].

The plant pathogen Xylella fastidiosa is a gram negative that attach.pdf

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the projected number of households( in millions) with digital video records thats allow viewers to record shows onto a server and skip commercials are in the following 5. The projected number of households (in millions) with digital video recorders that allow viewers to record shows onto a server and skip commercials are given in the following table: Year Households 31.2 a) Let t = 0 correspond to the beginning of 2006, and use QuadReg to find a second- 2010 130.2 2006 2008 71.6 2009 97.0 2007 49.0 degree polynomial regression model based on the given data. (2 pts) b) Using your regression equation, find the number of households in 2008. (2 pts) c) Using your regression equation, find the number of households in 2013. (2 pts) Solution a) Using technology to get the best quadratic fit, y^ = 2.4t^2 + 15t + 31.4 [answer] where y^ = predicted number of housholds. b) Plugging in t = 2 to y^ above, y^ = 2.4(2)^2 + 15(2) + 31.4 = 71 [answer] c) Plugging in y = 7 above, y^ = 2.4(7)^2 + 15(7) + 31.4 = 254 [answer].

The region labeled with the letter B is transcribed into mRNA True.pdf

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The plant pathogen Xylella fastidiosa is a gram negative that attaches to surfaces and forms a biofilm in host plants. Which of the following typically triggers an immune response? a) Pili b) Teichoic acid c) LPS d) EPS Solution Ans--(C) : LPS Lipopolysaccharide (LPS) is the predominant macromolecule displayed on the cell surface of Gram-negative bacteria and thus mediates bacterial interactions with the environment and potential hosts Utilizing a mutant that produces truncated O antigen we present results that link this LPS structural alteration to a significant decrease in the attachment of X,Xylella fastidiosa is a plant pathogenic bacterium that lives inside the host xylem vessels where it forms biofilm believed to be responsible for disrupting the passage of water and nutrients..

The positron decay of 15 o goes directly to the ground state of,15 N;.pdf

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a) What is the Chromosome Theory of Inheritance and how was it determined? (SHOW WORK & EXPLAIN) b) Describe the contributions of Sutton, Boveri, McClung, Stevens, Wilson and Morgan to determining that genetic information is carried on chromosomes. (SHOW WORK & EXPLAIN) c) Describe the XX-XO, XX-XY and ZZ-ZW mechanisms of sex determination and give examples of organisms that use each mechanism. (SHOW WORK & EXPLAIN) d) How does the environment affect sexual phenotype in turtles and alligators? (SHOW WORK & EXPLAIN) Solution a) Chromosomal Theory of Inheritance is a fundamental theory of genetics.Thomas Hunt Morgan establishes the Chromosomal theory of heredity.In 1907 Morgan was based on experiments by using Drosphila melanogaster or fruit flies. Drosphila have red eyes.Red eye colour is dominant.Morgan discovered a recessive mutation that caused white eyes. B) Boveri-Sutton chromosome theory identifies chromosomes as the carriers of genetic material which states that chromosomes carry the unit of heredity. McClung : A zoologist studied the behaviour of chromosomes in the sex cells and discovered the sex determining chromosome. Stevens: Stevens was interested in the process of sex determination. Morgan proposed the Chromosomal Theory of Heredity by using a common fruit fly. C) Sex determination is a biological system that determines the development of sexual characteristics. XX-XO types : In this type females have two homomorphic sex chromosomes XX ,male have one sex chromosome XO. This type of sex determination seen in cockroaches. XX-XY : This sex determination occurs in human beings.Females have two identical homologous sex chromosome XX and males have XY. ZZ-ZW : Females carry two types of sex chromosome ZW heterogametic and males carry ZZ homogametic. This type of sex determination occurs in fishes reptiles. D) Sex determination in reptiles are classified into two categories that is genotypic sex determination and temperature dependent. Environmental sex determination in which temperature is a crucial factor in embryonic development. In many turtle species eggs from cooler nests hatch males and from warmer nests hatch the females..

a) What is the Chromosome Theory of Inheritance and how was it deter.pdf

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QuestionProblem 1 In a source file carddeck.cpp, provide an impl.pdf

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Prove that the chromatic number of a cycle graph_n of odd length equals 3. Solution Chromatic Number: The chromatic number of a graph G is the least k for which a k-coloring of G exists. First note that the chromatic number must be at least 2 for any graph which has an edge in it, including all cycles. We now prove the theorem by induction on n. We will consider two base cases, C3 and C4. C3 is isomorphic to K3 which has chromatic number 3. C4 can be colored with two colors by giving opposing corners of the square the same color. For n > 4, we can take a coloring of Cn-2 and insert 2 adjacent vertices and edges and then color the new vertices appropriately to get a coloring of Cn. Thus the chromatic number of Cn is not greater than that of Cn-2. In the case where n is odd, note that if Cn had chromatic number 2, we could remove two adjacent vertices and edges to get a 2-coloring of Cn-2 which contradicts the inductive hypothesis since n-2 must be odd if n is odd. Therefore for odd numbers the chromatic number will be 2+1 = 3..

provide a brief definition of self assemblySolutionMolecular s.pdf

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Many non-scientists have focused their criticism of evolution on the fact that even scientists have criticized Darwin’s theory. Is this true? In what respect did Darwin get it wrong, or maybe in what way has Darwin’s theory been modified? Solution The \"Origin of Species\" of Darwin described about the heritable change in an organism over a period of time due to natural selection. Scientists also criticized of Danwin\'s opinion on use and disuse theory. At the time of Origin of species, Darwin do not know about the existed of microorganisms and his theory of tree of life became wrong. Pangenesis theory of Darwin also became wrong as he do not know anything about genetics..

Match the Statements below with bacteriological tests. Eosin-methyle.pdf

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M2-13 Identifying Transactions and Preparing Journal Entries [LO 2-3] J.K. Builders was incorporated on July 1 a. Received $83,000 cash invested by owners and issued common stock. b. Bought an unused field from a local farmer by paying $73,000 cash. As a construction site for smaller projects, it is estimated to be worth $78,000 to J.K. Builders c. Alumber supplier delivered lumber supplies to J.K. Builders for future use. The lumber supplies would have normally sold for $23,000, but the supplier gave J.K. Builders a 10 percent discount. J.K. Builders has not yet received the $20,700 bill from the supplier d. Borrowed $38,000 from the bank with a plan to use the funds to build a small workshop in August. The loan must be repaid in two years. e. One of the owners sold $23,000 worth of his common stock to another shareholder for $24,000 Prepare journal entries for the above transactions from the first month of business. (If no entry is required fora transactionlevent, selec \"No Journal Entry Required\" in the first account field.) Solution Ref Account Debit Credit a Cash 83000 Common stock 83000 b Land 73000 Cash 73000 c Lumber supplies 20700 Accounts payable 20700 d cash 38000 Notes payable 38000 e No entry.

Many non-scientists have focused their criticism of evolution on the.pdf

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iQ2 In a study about children?s pocket money, two samples of schoolchildren were asked how much money they received each week. In one sample, of size 80, the mean is ?6.65. In the other sample, of size 41, the mean is pound 8.21. Calculate the mean based on the combined sample. Give your answer correct to two decimal places. (Do NOT include the pound sign in your answer.) Solution n1=80 x1=6.65 n2=41 x2=8.21 combined mean = (n1*x1+n2*x2)/( n1+n2) =( 80*6.65+41*8.21) /( 80+41) =868.61/121 =7.1785 Combined mean =7.18 ( two decimals).

It has been observed that F_st increases with the distance between po.pdf

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Let f(x), g(x), h(x), epsilon F[x] such that f(x) does not equal 0. Prove that: If f(x) | g(x) and f(x) | [g(x)^2 +h(x)], then f(x). Hint: g(x)^2 = g(x) * g(x) Solution given f(x) | g(x) =>g(x)=k.f(x) =>g(x)2=k2.f(x)2 =>g(x)2=l.f(x) eq1 where l=k2.f(x) and f(x) | [g(x)^2 +h(x)] =>[g(x)^2 +h(x)]=m.f(x) eq2 eq2-eq1 =>h(x)=(l-m)f(x)=p.f(x) where p=l-m there fore f(x)|h(x) then f(x). Hint: g(x)^2 = g(x) * g(x).

Let R be an integral domain and F a field such that R is a subring of.pdf

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Judith, Martha, and Sophia are organizing their backpacks. While org.pdf

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A population of 10,000 students spends an average of $8.50 a day on dinner. The standard deviation of the expenditure is $4. A simple random sample of 100 students is taken. What is the standard deviation of the sampling distribution of the sample mean? Round your answer to 3 decimal places and no dollar sign. Solution As s(X) = s/sqrt(n) Then, as s = 4, n = 100, s(X) = 4/sqrt(100) = 0.4 [ANSWER].

A population of 10,000 students spends an average of $8.50 a day on .pdf

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a local fan club surveyed 67 of its members regarding their viewing habits last weekend, and the 44 following information was obtained: 30 members watched an episode of Naruto, 44 watched an episode of Death Note, 23 watched an episode of Inuyasha, 20 watched both Naruto and Inuyasha, 5 watched Naruto and Inuyasha but not Death Note, 15 watched both Death Note and Inuyasha, and 23 watched onlyDeath Note. Draw a Venn Diagram. Solution.

a local fan club surveyed 67 of its members regarding their viewing .pdf

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A gas is a fluid. It has less molecular space and freedom of motion than liquids as well negligible cohesive intermolecular forces. Shear thinning and thickening are types of non- Newtonian fluids. Gage pressure = Absolute Pressure + 14.7 4. The specific weight of an unknown object is 75 lbf/ft^3. The object is placed in water. The object will float within the water but not float to the water surface. The center of mass is the point through which the resultant force acts on a submerged inclined plane. Solution 1.TRUE...ie correct explaination 2.TRUE SHEAR THICKENING IS DILATENT...SHEAR THINING IS PSEUDOPLASTIC 3.FALSE....ABSOLUTE PRESSURE=GAUGE+ ABSOLUTE PRESSURE 4.FALSE..IT IS ALWAYS FLOAT ON THE WATER 5. TRUE...

A gas is a fluid. It has less molecular space and freedom of motion t.pdf

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A man accidentally let his parakeet out the door. The bird flew up to a power line and sat there. Unable to coax the bird down, the man stands on an aluminium ladder and reaches for the bird with an aluminium pole. Is there any potential danger in this situation? If so, what, and if not, why not? (Figure adapted from: Irwin, J. D. & Kerns, D. v. (1995). Introduction to Electrical Engineering. Upper Saddle River, NJ: Prentice-Hall.) Solution Yes there a potentially fatal scenario because here the man may provide a path to ground from the power line through his body. Since aluminium pole is a conductor so as soon as the man touches power line there there will be a electrical path produced through him and power passes from pole to man and then to ground..

A man accidentally let his parakeet out the door. The bird flew up to.pdf

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A culture of yeast was grown in a media containing sucrose labeled by 14C at 2nd position of fructose. Determine the localization of the label in palmitic acid synthesized by the yeast. Solution GLYCOLYSIS result in synthesis of acetyl CoA acetyl CoA formed from fructose will be labeled by 14C as 2nd position of fructose is labeled by 14C this will result in sub cellular localisation of pamitic acid synthesis......by acetyl CoA which is labeled by 14C.

A culture of yeast was grown in a media containing sucrose labeled b.pdf

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5. When a phosphate group is found in water: a) The partially-negative oxygen atoms of the water molecules surround the oxygen atoms of the phosphate group. b) The partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group. c) The partially-positive oxygen atoms of the water molecules surround the oxygen atoms of the phosphate group. d) Hydrophobic forces push the phosphate groups together to minimize their disruption on hydrogen bonding between water molecules. e) There is more than one correct answer above. Solution ANS: b) The partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group. Explanation: In the water molecule two polar covalent bonds are present. Among this oxygen molecule had a partial negative charge and both hydrogen atoms partial positive charge. The positive and negative ends of the bond cause charge separation or a dipole. So water molecule had 2 polar covalent bonds and 2 bond dipoles. In phosphate group two oxygen molecules had partial negative charges, so the partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group..

5. When a phosphate group is found in water a) The partially-negati.pdf

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Implement a minimum spanning tree using Boruvka\'s algorithm. Implement a minimum spanning tree using Boruvka\'s algorithm. Solution The psuedo code is as follows: 1 Graph input = /* read input */; 2 UnionFind uf (input.getNodes ()); 3 Workset ws = input.getNodes(); 4 foreach (Node n : ws) { 5 Edge e = minWeight (input.getOutEdges (n)); 6 Node rep1 = uf.find (e.src); 7 Node rep2 = uf.find (e.dst); 8 if (rep1 != rep2) { 9 e.markInMST(); 10 Node rep = uf.union (rep1, rep2); 11 ws.add (rep) 12 } 13 }.

Implement the additional 5 methods as indicated in the LinkedList fi.pdf

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