...being un-supervised. Other people in the community speed over the limit posted and children play there and may be hit or injured. Even though the children play in the street it cause a safety risk to drivers that don’t speed or drive wreck less do to the children running in and out of the streets without looking for cars. The children are our future and they need to be protected. When a person is looking to buy a future home they want to see a nice neighborhood that looks respectable and not as if it’s a bad neighborhood. The appearance of a home in my belief can say a lot about the people who live there. Someone who takes care of their home for example pressure washing, not trash in yard, paint nicely applied. Also there are some resident that don’t have the money but the residents should do as much as they can. To give an example of this a resident in my neighborhood has fixed the garage door which is tilted half way closed and it’s not very appealing. The appearance of a residents lawn as well as home its self make a community. A yard or lawn care should at the minimum have the grass mowed. The plants should be not over grown or deceased. Living in Florida the grass isn’t always...
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...The poor people usually depends of this places for basic things like food or sleep,without that help some of them don`t survive winter. Most of these institutions exist thanks to donations made it by average people and not government assistance. Every day there is more people living on the streets and the help is simply not enough, even when these kind of institutions exist. The people which suffer of poverty even when we see them like a kind of dog in the street, they are human and they want a decent life like normal people. Is normal to say that those poor people are living on the streets by their own choice, but like I said before, that is ignorance. That people suffer a lot of things, even things that we never going to feel, for example the hungry . The sad of these situations is that most of these people is...
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...Solution Nguyen Van Linh,11A2 Math, Highshool for gifted student, Ha Noi University of Science, Vietnam 4/1/2010 Problem: Given a cyclic pentagon ABCDE.AD ∩ EB = {X}, AD ∩ EC = {L}, AC ∩ EB = {Y }, AC ∩ BD = {Z}, EC ∩ BD = {T }. (EAX) ∩ (ABY ) = {A1 }, (ABY ) ∩ (BCZ) = {B1 }, (BZC) ∩ (DT C) = {C1 }, (DT C) ∩ (ELD) = {D1 }, (ELD) ∩ (EAX) = {E1 }. Prove that A1 B1 C1 D1 E1 is a cyclic pentagon. Solution: First we will express a lemma: Given a pentagon ABCDE.A1 , B1 , C1 , D1 , E1 , A2 , B2 , C2 , D2 , E2 are defined as the figure below. Then AA2 , BB2 , CC2 , DD2 , EE2 are concurrent. A2 A y1 E x3 z3 C1 x2 E2 E3 y 2 O B1 D1 x1 A3 D3 z2 C3 B B2 B3 E1 y3 A1 D2 D C2 z1 C Proof: Denote A3 = AA2 ∩ BE, similar we define B3 , C3 , D3 , E3 . Denote O1 = AA3 ∩ EE3 , O2 = AA3 ∩ BB3 . We will show that O1 ≡ O2 ⇔ Applying Menelaus’s theorem : AO1 EA3 E3 C1 . . =1 A3 O1 EC1 E3 A AO1 AO2 = .(∗) O 1 A3 O2 A3 AO2 A3 B B3 D1 . . = 1. A3 O2 BD1 B3 A EA3 E3 C1 A3 B B3 D1 So (∗) is true iff . = . (∗∗) EC1 E3 A BD1 B3 A Put BD1 = x1 , D1 C1 = x2 , C1 E = x3 , AC1 = y1 , C1 B1 = y2 , B1 D = y3 , CE1 = z1 , E1 D1 = z2 , D1 A = z3 . BA3 A3 C1 x2 (x2 + x1 ) We have A3 D1 .A3 B = A3 C1 .A3 E then = . So we can calculate EA3 = ,similar EA3 A3 D1 x1 + 2x2 + x3 with E3 C1 , E3 A, A3 B, B3 D1 , B3 A. and 1 y2 (y2 + y3 ) z2 (z2 + z1 ) x2 (x2 + x1 ) x2 (x2 + x3 ) + x3 ...
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...Algorithm :- Goal Stack Method. For every operator applicable, we check if prerequisites are satisfied or not. We do so by checking if preconditions are a subset of current state. Then we see if it is a member of plan already, because it would lead to a recursive dead end. If above those are satisfied then we first delete the delete list from current state and then append the add list, that would lead to the next state. We will continue this way until the goal state is a subset of current state. That would give a successful plan. Operators :- go(R, X, Y) :- precond - [landMark(X, R), landMark(Y, R), switch(R, on), posRobo(R, X)] add - [posRobo(R, Y)] delete - [posRobo(R, X)] push(B, X, Y):- precond - [landMark(X, R), landMark(Y, R), box(B, R, X), posRobo(R, X), switch(R, on)] add - [posRobo(R, Y), box(B, R, Y)] delete - [posRobo(R, X), box(B, R, X)] turnOn(R):- precond - [posRobo(R, _), switch(R, off)] add - [switch(R, on)] delete - [switch(R, off)] turnOff(R):- precond - [\+ posRobo(R, _), switch(R, on)] add - [switch(R, off)] delete - [switch(R, on)] roboCtoR(R, D):- precond - [posRobo(corridor, D), landMark(D, R), landMark(D, corridor), switch(R, off)] add - [posRobo(R, D), switch(R, on)] delete - [posRobo(corridor, _), switch(R, off)] roboRtoC(R, D):- precond - [room(R), landMark(D, R), landMark(D, corridor), posRobo(R, D), switch(R, on)] add - [posRobo(corridor, D), switch(R, off)] delete - [switch(R...
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...owned by the family. The list has 11 items, ranging from ‘electricity connection’ and ‘agricultural land’- to cars and air conditioners We have 12 grades in the new SEC system, ranging from A1 to E3 4 THE NEW SEC SYSTEM DISTRIBUTION OF HOUSEHOLDS India ( urban + rural ) New SEC System 18.4 14.7 11.4 9.7 7.5 3.2 1.8 0.4 A1 A2 A3 B1 B2 C1 C2 D1 D2 E1 E2 E3 7.9 5.3 15.4 4.3 *Figures in percentage This round of analysis has been conducted using data from IRS 2008 (round 20)*. We have drawn a sub-sample of 39,441 5 THE NEW SEC SYSTEM DISTRIBUTION OF HOUSEHOLDS Urban New SEC System 12.9 11.6 10.2 7.6 5.1 8.6 13.2 12.6 9.8 4.7 2.6 1.1 A1 A2 A3 B1 B2 C1 C2 D1 D2 E1 E2 E3 Current Urban SEC system The current urban SEC system has 8 grades, and is based on occupation and education of chief earner 24 20.5 18.2 11.9 6.6 7.9 8 2.8 A1 6 A2 B1 B2 C D E1 E2 THE NEW SEC SYSTEM DISTRIBUTION OF HOUSEHOLDS Rural New SEC System 22.2 20.1 15.6 12.8 10.7 5.1 0.04 A1 0.4 A2 1.3 A3 2.3 3.2 6.2 B1 B2 C1 C2 D1 D2...
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...INDUCTIVE APPLICATION SKETCHES : INDUCTIVE PROXIMITY SWITCH IPS IPS IPS RAILROAD YARD POSITATION SENSING NUT PLACEMENT ON TRANSFORMER SORT FERROUS & NON FERROUES CAN TOPS DETECTING INCORRECT SHAPE OF TARGET MATERIAL CUTTER M/MIN DOG IPS M8 IPS XXX SLOT IPS MIN BLK IPS MIN BLK POSITION SENSOR SPEED SENSING OF MACHINES Trigger SENSOR TABLET COUNTING POSITIONING OF BOTTLES Metal Parts Sensors SENSORS Work to be dipped Sensor VALVE POSITION CONTROL POSITION SENSOR FOR GALVANISING PLANT DRILL BREAK SENSOR PART COUNTER INDUCTIVE PROXIMITY SWITCH - SERIES IPS Flush Mounting: Fixing of switches in the metal up to their sensing face. Non-Flush Mounting: These switches are to be mounted keeping metal free zone around its sensing face. SENSING DISTANCE Sensing Distance of inductive Proximity Switch depends on, Size of the Proximity Switch/ Sensing Face and Material to be sensed. OBJECT The sensing distance specified here is for object made of mild steel, having thickness of 1 mm. Size of the object should be at least equal to the diameter of switch / sensing face. For object other than M.S. apply suitable correction factor CORRECTION FACTOR : Apply following correction factor (%) when object is other than Mild Steel. MS 100 SALIENT FEATURES : 1) 2) 3) 4) 5) 6) Repeat Accuracy Hysteresis Grade of Protection Enclosure LED Indication Ambient temp : : : : : : 0.02 mm 3-10% IP 67. For block type: IP 55 Metallic (Nickel-Chrome...
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...Basal Ganglia Pathology in Schizophrenia: Dopamine Connections and Anomalies WORD COUNT: 1100 As unique and complex organisms, we encounter challenges which may have a profound impact on our cells and essentially the quality of life. Schizophrenia affects 1.1% of the world population and has a detrimental impact on society, a mental disease where a cure and predisposition is yet to be established. Primary and secondary literature are distinct informative sources which enable us to better understand scientific concerns in the world and provide effectual reasoning. The review article identifies pathways or systems involving dopamine within the basal ganglia and how fluctuations or absence of this neurotransmitter can cause implications in brain functioning- evident in schizophrenic patients. The Journal of Neurochemistry comprises of peer-reviewed articles that can appeal to a spectrum of individuals, ranging from emerging scientific researchers and health professionals to the general society. The extensive citation and depth may indicate the credibility of the journal and numerous publications demonstrate the success in communicating the nature and extent of current issues in the field of neuroscience. The review article investigates the relation between dopamine levels in regions of the brain and how abnormalities contribute to poor cognitive behavior which can be seen in schizophrenic patients. The most relevant discipline to complement the article would therefore be...
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...many association studies that compared the frequencies of alleles of the dopamine D2 receptor (DRD2) gene between alcoholics and control groups have produced results, but some have been equivocal. Dopamine D3 receptor genes (DRD3) are in the same class as DRD2 but with different pharmacological properties. So we compared the distribution of genotypes and frequencies of Bal I polymorphism of the DRD3 gene in alcoholics and controls to assess the role of the DRD3 gene in Korean alcoholism. For this study, 67 male probands from alcoholics and 67 age-matched normal male controls were engaged. No evidence for an allelic association was found between the A1 allele of DRD3 and alcoholism in a Korean population. These results suggest that any role played by this receptor may account for only part of the variation in susceptibility to alcoholism. Keywords: * sediment compression; * geotechnical; * elevation correction; * index point correction; * decompaction; * Forth valley; * Scotland Abstract Fine-grained sediments usually suffer post-depositional compression (compaction), which reduces bed thickness and lowers the elevations of sample positions and marker horizons. This reduction has been calculated for two general cases using geotechnical theory, and a correction (decompaction factor) is presented that can be applied to many field situations. Its use is illustrated by two examples from the clay sediments of the Claret Formation...
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...chemical named dopamine, which is released in the brain as part of the normal response to accomplishment. When it reaches another part of the brain called the dopamine receptors, it produces a sensation of pleasure, relaxation, and happiness. This is the body's natural mechanism for encouraging itself to further efforts, by making the completion of tasks pleasant. Unfortunately, in a few people, this reaction does not proceed as it should. The dopamine receptors in the brain are classified into five groups, D1 to D5, with a variety of functions. In these people, the dopamine D2 receptor is compromised or entirely fails to function. As a result, the person concerned will feel less of a reward sensation when accomplishing tasks. In some cases, when the D2 receptors are very badly damaged, the person will feel no satisfaction at all. The consequences to such a person vary with the extent of the damage to the D2 receptors. When these are merely dulled, not...
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...Search Quiz ICS171 Name___________ ID____________ No text, no notes, no questions. Do the best that you can on each question. No questions will be answered about the quiz questions. If you think a question is ambiguous, write your interpretation and answer your modified question. Be reasonable. The following abbreviations are used: BF = branching factor, DFS = depth first search, BFS = breadth first search, IDS = iterative deepening search, A* = A* search, LI = local improvement search, HC = hill-climbing search. [pic] 1. For the 8-tile puzzle, what is the average branching factor, assuming the blank is equally likely to occur in any position. Show your work. (4*2 + 4*3 +1*4)/9 = 24/9 = 2 & 2/3. 2. Suppose that you are solving the 8-tile puzzle where it has solution. Which of the methods (DFS, BFS, IDS) is guaranteed to find a solution, assuming no computational limits are reached. List all that are correct. DFS, BFS, IDS 3. For the same puzzle, which of the methods (DFS,BFS,IDS) is guaranteed to find the shortest solution? List all that are correct. BFS, IDS 4. For the same puzzle, which methods are guaranteed to use no more than O(BF * length of solution) amount of memory. DFS, IDS 5. Suppose you apply the A* algorithm to the same problem. You decide to let f = current cost of the path. Would it be appropriate to let h = 0 for all states? Yes or no and why. ...
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...Data Structures & Algorithms Coursework Assignment 1 Q1. (a) Algorithm swap(x, y): Node n head While (n.getNext () != x ) do n n.getNext() Node v y.getNext () n.setNext(y) y.setNext(x) x.setNext(v) (b) Algorithm swap Doubly(x, y): DNode n x.getPrev() DNode v y.getPrev() n.setNext(y) y.setPrev(n) y.setNext(x) x.setPrev(y) x.setNext(v) v.setPrev(x) (c) The run time complexity for the singled linked algorithm is O (n) and for the doubly linked algorithm is O (1). Doubly linked list has the best time complexity. Time complexity in singly linked list take more time because we have to move from head to the node before x Q2. (b) RedBlueStack implements Stack{ protected Object A[]; Int capacity; int top = -1; RedBlueStack(int cap) { A = new Object [capacity]; capacity = cap; } int size() { return (top + 1); } void push(Object obj) throws FullStackException { if (size() == capacity) throws new FullStackException("Stack is full."); A[++top] = obj; } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Boolean isEmpty() { return (top < 0); } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Object pop() throws EmptyStackException { Object elem; if (isEmpty()) throws new EmptyStackException("Stack...
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...D2 evaluate the adequacy of accounting ratios as a means of monitoring the state of the business in a selected organisation, using examples: The importance of ratios in any business is very important because it gives the business a better understanding of the financial data. By using ratios the business is able to compare data from the current year with the previous years. From this the business will know and be able to identify if they are making more profit or a loss or if they just broke even. There are four types of ratios that a business has to calculate. These are things like investment ratio, financial ratio, profitability ratio and utilization ratio. From the results shown you can see if the business profits are increasing which is a positive thing and shows the company is on a rise and they are making money, this is a steady rise in the company’s money. If the results show a decrease it means that the company is making a loss. another one you can evaluate is the net profit margin as this is helpful as it shows how profitable the company has been throughout the years and whether it is increasing in profit or not. Current ratio is the most effective because it shows you whether you can have enough money to pay back your debts. The ratio shows that they have more liabilities it means they would struggle to pay off their debts. A figure below one means that the business does not have enough current assets to pay off its debts, a figure between 1 and 1.5 means that...
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...Task 1 LO 1.1 All high-level programming languages support the concept of data types. A data type defines a set of values that a variable can store along with a set of operations. Data types are used to store various types of data which is managed by program. Data type attaches with variable to determine the number of bytes to be allocate to variable and valid operations which can be performed on that variable. Although C has several built-in data types, it is not a strongly typed language, as are Pascal and Ada. C supports various data types and here some common data type as character, integer and floating-point types. C defines five foundational data types as defined below: ▪ character ▪ integer ▪ floating-point ▪ double floating-point ▪ valueless These are declared as by char, int, float, double, and void, respectively. These types form the basis for some other types also. The extent and choice of these and data types may contrast amongst processor natures and compilers. However, in all belongings an object of type char is 1 byte. C stores character type inside as an integer. Each character has 8 bits so, we can have 256 different characters values (0-255). Character set is used to map between an integer value and a character. The size of an int is ordinarily the same as the word length of the implementation setting of the program. C has 3 classes of integer storage namely short int, int and long int. All of these data types...
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...Write the following queries in SQL: Display a list of all instructors, showing their ID, name, and the number of sections that they have taught. Make sure to show the number of sections as 0 for instructors who have not taught any section. Your query should use an outer join, and should not use scalar subqueries. By using the university schema provided by db-book.com the following queries were done on the university database. The first query uses an outer join which works similar to the join operation but it keeps the rows that don’t match between the two tables that would be lost in a join operation. There are three forms of outer join: a) Left outer join displays the results from the left table even if the condition does not find any matching record in the right table. b) Right outer join will displays the results from the right table regardless if there is matching data in the left table. c) Full outer join will retain all rows from both tables, regardless if the data matches or not. The group by clause when used in a select statement collects data from multiple records and groups the results into one or more columns. The below is the query using left outer join operation: select ID, name, count(sec_id) as Number_of_sections from instructor natural left outer join teaches group by ID, name; The next query was written by using a scalar subquery without using an outer join operation. A scalar subquery is where the output of a subquery returns only...
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...i SASTRA UNIVERSITY Shanmugha Arts, Science, Technology and Research Academy Thirumalaisamudram Thanjavur – 613 402 BCSCCS705 PARALLEL COMPUTING AND ALGORITHMS LAB B.TECH (COMPUTER SCIENCE AND ENGINEERING) 7TH SEMESTER ii List of Exercises: 1. Basic arithmetic operations in parallel 2. Find out factorial of a number 3. Generation of Fibonacci series, finding prime numbers in an interval. 4. Evaluate the integral of a function 5. Merging of two sorted lists 6. Parallel tree traversals 7. Matrix multiplication 8. Enumeration sort 9. Odd-Even transposition sort 10. Bitonic merge 11. Quick sort 12. Single source shortest path iii Exercise 1 Aim: Basic Arithmetic Operations in parallel To perform a set of arithmetic operations in parallel using a cluster of computers Procedure: Step 1: Identify a set of numbers over which arithmetic operations are to be done Step 2: Identify the rank in the cluster Step 3: Split the set of numbers into domains and assign each domain to their corresponding processor Step 4: Get the result from each computer in the cluster and assimilate them at the master Step 5: Verify the time taken for the completion of each task. Algorithm:function sum(+,identity,a) = if #a == 1 then [identity] else let e = even_elts(a); o = odd_elts(a); s = scan_op(op,identity,{op(e,o): e in e; o in o}) in interleave(s,{op(s,e): s in s; e in e}); iv Input:- An array of integers Output: - The sum 3,2,7,6 P1 18 0,5,4,8 P2 17 62 2,0,1,5 P3 8...
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