searched (x in the above code) is not present in the array. When x is not present‚ the search () functions compares it with all the elements of arr [] one by one. Therefore‚ the worst case time complexity of linear search would be. AVERAGE CASE ANALYSIS (SOMETIMES DONE): In average case analysis‚ we take all possible inputs and calculate computing time for all of the inputs. Sum all the calculated values and divide the sum by total number of inputs. We must know (or predict) distribution of cases
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Normal Distribution It is important because of Central Limit Theorem (CTL)‚ the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal. Normal P.D.F Now we want to find c This integral has been proved that it cannot have close form solution. However‚ someone gives an idea that looks stupid but actually very brilliant by multiply two of them. reminds the function of circle which we can replace them to polar coordinate Thus Mean
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Homework #4 - SQL Use the following (denormalized) database schema (and the attached tables) to write the queries. Publishers (custid‚ name‚ city‚ phone‚ creditcode) Bookjobs (jobid‚ custid‚ jobdate‚ descr‚ jobtype) POS (jobid‚ poid‚ podate‚ vendorid) Items (itemid‚ descr‚ onhand‚ price) Po_Items (jobid‚ poid‚ itemid‚ quantity) For each question‚ turn in the Oracle SQL query and the output. You should feel free to do these by hand (paper and pencil)‚ or you may actually run them
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conditional probabilities. For example: Suppose there is a certain disease randomly found in one-half of one percent (.005) of the general population. A certain clinical blood test is 99 percent (.99) effective in detecting the presence of this disease; that is‚ it will yield an accurate positive result in 99 percent of the cases where the disease is actually present. But it also yields false-positive results in 5 percent (.05) of the cases where the disease is not present. The following table shows
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LEARNING OBJECTIVE: Understand and perform an integrated order-to-cash cycle .The goal of SAP Sales and Distribution (also referred to as SAP Supply Chain Management-Order Fulfillment) is to provide with a complete knowledge of the Sales Cycle implementation using SAP. Project Management and some background of the SAP Transportation System is also part of the case study. The basic learning is to perform the functional duties of a SAP SD Consultant and develop a strong conceptual and practical knowledge
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3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5‚ find a) P(X < 15); b) the value of k such that P(X < k) = 0.2236; c) the value of k such that P(X > k) = 0.1814; d) P( 17 < X < 21). Ans : X ~ N ( 18‚ 2.52) a) P ( X < 15) P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places) b) P ( X < k) = 0.2236 P ( Z < ( k – 18) / 2.5 ) = 0.2236
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Overview | | | |The Process of Quantitative Modeling | | | |Review of Probability Concepts
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classical and empirical probabilities. a. Classical probabilities are based on assumptions; Empirical probabilities are based on observations. b. Classical probabilities do not require an action to take place; Empirical probabilities have to have been “performed”. 2) Gather 16 to 30 coins. Shake and empty bag of coins 10 times and tally up how many head and tails are showing. Number of coins: 20 * Consider the first toss‚ what is the observed probability of tossing a head? Of
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Study Guide for Probability Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. Which inequality represents the probability‚ x‚ of any event happening? a.||c.|| b.||d.|| ____ 2. Which event has a probability of zero? a.|choosing a letter from the alphabet that has line symmetry|c.|choosing a pair of parallel lines that have unequal slopes| b.|choosing a number that is greater than 6 and is even|d.|choosing a triangle that is both
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of 25 and a variance of 9. Find the probability that a measurement selected at random will be between 19 and 31. Solution: The values 19 and 31 must be transformed into the corresponding z values and then the area between the two z values found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3)‚ we have z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2 From the area between z =±2 is 2(0.4772) = 0.9554 Therefore the probability that a measurement selected at random
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