| |Issue Date |04/02/2013 |Due Date |26/2/2013 | |Criteria covered in this assignment |P1‚ P2‚ M1& D1 | |Criteria Ref |Assessment Criteria |Achieved Date |Comments/ feedback from assessor
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increasing or decreasing the number of level detector metal strips (L1 through L12) and their associated components. In the circuit‚ diodes D1‚ D2 and D13 form half-wave rectifiers. The rectified output is filtered using capacitors C1 through C3 respectively. Initially‚ when water level is below strip L1‚ the mains supply frequency oscillations are not transferred to diode D1. Thus its output is low and LED1 does not glow. Also‚ since base voltage of transister T1 is low‚ it is in cut-off state and its collector
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| |Assignment 1: Effective Sports Leaders | |Poster (P1‚ M1‚ D1) | |
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prompts that initiate the installation of new or upgraded software P2 Describe the potential risks of installing or upgrading software M1 Explain the advantages and potential disadvantages of installation or upgrade of new software D1 Justify a particular installation or upgrade D2 Evaluate the risks involved in the installation or upgrade of software and explain how the risks could be minimised. Student’s Signature: Date: Assessor’s Signature: Date: Scenario
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Learning Outcome 1:- Understand different types of business information Grading criteria: • P1 explain different types of business information‚ their sources and purposes •M1 analyse different types of business information and their sources •D1 evaluate the appropriateness of business information used to make strategic decisions Scenario You are working as information assistant for a company of financial consultants. The company needs an information sheet for staff to help them understand
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R Q1 Q2 Q3 D1 D2 D3 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 x x 1 0 0 0 0 0 x x 0 0 1 0 0 0 x x 1 0 1 0 1 0 x x 0 1 0 0 0 0 Then‚ we can simplify the logic by using Boolean Algebra. Then we can get: D1=Q1’Q2’Q3’ L D2=Q1 Q2’ Q3 D3=Q1’Q2’Q3’ R We need use 3 D flip flops to build the circuit. From the state table‚ We can see when L is 1 and Q1‚Q2‚Q3 are 0‚ D1 is 1. So we can get D1=Q1’Q2’Q3’ L. D2
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FINS2624 PORTFOLIO MANAGEMENT Week 6 CAPM: The covariance of an assets returns with the market and the required return of the asset. Assumptions: * Investors are price takers * Investors have identical investment horizons * Perfect capital markets * Investors are rational mean-variance optimizers β: Measures how much risk an asset contributes in the market portfolio. * β > 1 asset contributes more risk than the average asset * β < 1 asset contributes less risk
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the firms in both short-run and long-run. First of all‚ customers will lose confidence to the baby milk firms and they will stop buying baby milk which contains melamine. The demand of baby milk drop and demand curve will shift to left from D0 to D1 and D1.Both price and quantity have decreased from P0 to P1 and Q0 to Q1 respectively and therefore new equilibrium E1 achieved. In perfect competition‚ price is equals to marginal revenue (MR) and firm earns profit when marginal revenue is above minimum
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Lecture Notes 1 MSC IN OPERATIONAL RESEARCH LECTURE 8 TRANSSHIPMENT MODEL In generalized transshipment model‚ items are supplied from different sources to different destination. It is sometimes economical if the shipment passes through some transient nodes in between sources and destinations. Unlike in transportation problem‚ where shipments are sent directly to a particular source to a particular destination‚ in transshipment problem‚ the objective is to minimize the total cost of
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stockholders and none comes for debtholders. In this case‚ the company’s balance sheet has net operating assets (NOA) of $8‚000‚ common equity (CE) of $8‚000‚ and zero net financial obligations (NFO). a. Compute D1 for the coming year and the rate of growth in Dt for every year thereafter. • D1 = NI1 – ΔCE1 = 1‚100 – 0.05 * 8‚000 = 700 • D2 = NI2 – ΔCE2 = (1‚100 * 1.05) – 0.05 * (1.05 * 8‚000) = 1.05 * (1‚100 – 0.05 * 8‚000) = 735 = 700 * (1 + 0.05) • D3 = NI3 – ΔCE3 = (1‚100 * 1.052) – 0.05 * (1.052
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