Please help me because i tried but i cant do it righthow do i so.pdf

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Please help me because i tried but i can\'t do it right how do i solve for : sin square theta - 2 sin theta - 3 = 0 Solution (2sin ? + 1)(sin ? - 2) = 0, sin ? = -1/2 or 2. Since sine cannot be greater than 1, sin ? = -1/2 ? = 7?/6 + 2k? and ? = 11?/6 + 2k? where k is any integer. Hope you was helped.

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Please help me and explain 1. If all other agents choose a level of activity e, the payoff of agent i can be expressed as V(ei, e). In this case the optimization problem becomes max ei V(ei, e), Using the optimization problem, show the the conditions for coordination failures. And also discuss coordination failures with help of the game theoretical structures such as the battle of the sexes, prisoners\' dilemma and the stag hunt bu giving the real cases in the economy. Solution Problem of coordination in a shared production process. Let ej be the effort (input) of player i in the production of a public good, c. The production function is c = f(ei,e) with fi > 0 and f2> 0. Agents have identical utility functions defined over consumption and effort U(c,e), with U1 > 0, U2 < 0, and U() quasi-concave. These preferences, together with the production function, generate agent i\'s payoff function: (1) V(ei,e) = U(f(ej,e),ej. Note that V12= U1f2> 0, so the model exhibits positive spillovers. Differentiating the payoff function with respect to ei and e yields (2) V12 = U1f12 + U1lflf2 + U12f2. that preferences between consumption and effort are separable. A necessary condition for strategic complementarity is then that inputs be complementary within the production process (/12 > 0). Increases in e will induce increases in ej if /12 is large enough to offset the reduction in the marginal utility of consumption brought about by f2 > 0. If U() is linear, /12 > 0 is sufficient for strategic complementarities. Note further that - V12 UJ1/2 + Ullfl2 V1l -[U11 (fl)2 + U1f/1 + U22] The necessary condition for multiplicity (p > 1) is (4) U1(/12 + /11) + U1l[(/l)2 + /1/2] _ -U22. Multiple Equilbria are thus more likely when the utility function is not very concave with respect to consumption and effort; when inputs are highly complementary in production; and when the production function is not very concave with respect to own effort..

Please help me and explain 1. If all other agents choose a level of .pdf

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Please help me aswer this...I will rate you life saving!!! Solution Oral cavity Pharynx Esophagus Stomach Duodenum Jejunum Ileum Cecum Ascending Colon Transverse Colon DescendingColon Sigmoid Colon Rectum Anus Feces Human waste Duodenum to the Ileum that isfrom the small intestine Cecum to Rectum is from thelarge intestine Duodenum to the Ileum that isfrom the small intestine Cecum to Rectum is from thelarge intestine.

Please help me aswer this...I will rate you life saving!!! Solu.pdf

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please help and be as detail as you can 1. Variation is a key statistic used in the most analytical studies. what are the implications of the high level of variations vs. low level of variation in a sample data 2. what are the implications Solution 1. In statistics, analysis of variance (ANOVA) is a collection of statistical models, and their associated procedures, in which the observed variance in a particular variable is partitioned into components attributable to different sources of variation. In its simplest form, ANOVA provides a statistical test of whether or not the means of several groups are all equal, and therefore generalizes t-test to more than two groups. Doing multiple two-sample t-tests would result in an increased chance of committing a type I error. For this reason, ANOVAs are useful in comparing two, three, or more means. 2. the act of implicating or the state of being implicated.

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Please help Lot, Xn and Yn have the distributions shown above. Find the expected value and variance of Xn and Yn. What does the Chebyshev inequality\'- tell us about the convergence of Xn and Yn? Is Yn convergent in probability? If so, to what value? Solution 1. E(X) = 0*1/n + 1*1/n = 1/n ; V(X) = E(X^2) - (E(X))^2 = (0*1/n + 1*1/n) - (1/n)^2 = 1/n - 1/n^2 = (n-1)/(n^2) E(Y) = 0*1/n + n*1/n = 1 ; V(Y) = E(Y^2) - (E(Y))^2 = (0*1/n + n^2/n) - (1) = n-1 2.Pr(|X-1/n|>k*sqrt(V(X))) <= 1/(k^2) for k>=1 Pr(|Y-1|>k*sqrt(V(Y))) <= 1/(k^2) for k>=1 3. No its not convergent as Sum of P(X) not converging to 1.

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