write 3 3 slide on China and Germany. Individual work (1) Choose a c.pdf

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write 3 3 slide on China and Germany. Individual work (1) Choose a city, a region or a country and a time whe it was especially successtul economically Produce one PowerPoint slide for each of the following topics: Explain your choice and present basic facts Explain what was enabled this place to achicve success Explain its decline . Send the PowerPoint presentation before Jan. 31 11:00pm Solution Chinese Economic growth miracle Snapshot of Chinese economy: Reasons behind success of Chinese economy: First: Rural areas (1978-1984) Non-farm enterprises are allowed to establish by government, which encourages households’ savings (Spencer, 2012). It led to the increase in production and efficiency in the agriculture system as well. Urban Economic reforms (1985-1992) It provided greater operational autonomy (Daniel & Germa, 2012) to the national businesses. It made business more competitive under the presence of foreign businesses. Administrative reform (1995) Reforms in the rural-urban migration, banking system, tax system; international trade and foreign investment took place. Recent decline growth rate: German Post War II Economic growth German economy after WW-II: High Growth Story: From boom to burst:.

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Harrison works in a cubicle at a window next to Karen Ravenwoods cu.pdf

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Hi there I am having difficulty in finalizing my Tetris game , below I posted the codes for the classes and I need the rest six pieces to be done. Can anybody help me here. It is a java code and must use the code I posted bellow. Thanks, Completing Tetris [Description: shapes] Overview You will be completing the game Tetris. The features needed are 1. Adding the other 6 pieces 2. Adding the ability to rotate the pieces Concepts The purpose of this assignment is to gain experience with the following new concepts: inheritance/class hierarchy dynamic dispatch algorithms If you were not able to complete homework 1, please come see me to get a working solution. Implementation Requirements OTHER PIECES: The first step in the assignment is to add the other six pieces. Lets think a little about the planning though; we don\'t want the Game class to have to know about all 7 different piece types (we wouldn\'t want Game to have 7 different instance variables, 1 for each type. How would it keep track of which one was current?) Instead, Game should know about 1 type (a super type) and let dynamic dispatch do the work for us. So to start, we need to redesign (or refactor) the current code. To do this, we want a super class that contains everything common to all pieces; then sub classes for the individual pieces and their individual needs. So what is the same about all the pieces? the state currently handled by LPiece is common to all pieces all the behaviors currently in LPiece are common to all pieces (except the constructor of course) What is different about each piece: How each individual playing piece is constructed The implementation of how each rotates (the second feature for this assignment) 1) Therefore, start by breaking up the LPiece.java class into a super class and sub class. At this point, test your program. It should run as it did before. 2) Now its time to add the other game pieces. You will need to figure out how to initialize the pieces. This will be similar to how the L-shaped piece was done, and, in fact, you may find it helpful to start each new class by copying the code from a previous shape and modifying it. The pieces should be initialized in the following orientations: [Description: shapes] In the Game class, the piece instance variable must have a new type. What do you think it should be? You\'ll need to modify the Game class so that it doesn\'t always instantiate an LPiece piece, but randomly chooses which shape to instantiate. (I strongly recommend creating 1 new shape at a time and testing.) ROTATION: The current tetris piece needs to be able to rotate in quarter turns clockwise each time the \'r\' key is pressed. As an example see the figure below illustrating how the L-shaped piece rotates: [Description: Lrotate] Here is where dynamic dispatch will come in handy since each shape class must have its own rotate implementation. Update the existing classes (super and sub classes) to enable dynamic dispatch to work properly. Modify.

Hi there I am having difficulty in finalizing my Tetris game , below.pdf

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I need help creating a parametized JUnit test case for the following class. import java.util.Arrays; import java.util.Collections; import java.util.LinkedList; import java.util.List; import static java.lang.Integer.valueOf; public class IntegerSet { /** * The only instance variable in the class. No other variables allowed. */ private List list; public int size() { return list.size(); } public boolean isEmpty() { return list.isEmpty(); } /** * Initializes the empty set. */ public IntegerSet() { list = new LinkedList(); } /** * Initializes a set with the numbers in the argument. */ public IntegerSet(Integer[] integers) { this(); if (integers == null) { throw new NullPointerException(\"The argument cannot be null\"); } else { for (Integer i : integers) { if (!exists(i)) { this.insertElement(i); } } } Collections.sort(list); } /** * Inserts an integer to the set if the integer does not exist in the set */ public void insertElement(int i) { if (!list.contains(i)) { list.add(i); } } /** * Inserts to the set all the integers in the argument which do not exist in * the set. */ public void insertAll(Integer[] data) { if (data == null) { throw new NullPointerException(\"The argument cannot be null\"); // throw new // IllegalArgumentException(\"The argument cannot be null\"); } else { Arrays.sort(data); for (Integer num : data) { if (!exists(num)) { insertElement(num); } } } } /** * Deletes an integer from the set if it exists in the set. */ public void deleteElement(int i) { list.remove(i); } /** * Deletes all the elements in the set. */ public void deleteAll() { list.clear(); } /** * Returns true if the argument exists in the set, false otherwise. */ public boolean exists(int i) { return list.contains(i); } public String toString() { char delimiter = \' \'; String str = \"[\"; for (Integer i : list) { str += i; str += delimiter; } return str + \"]\"; } /** * Finds the union of two sets. A null pointer is considered an empty set. * * Return \"new IntegerSet()\" when the result is an empty set. */ public static IntegerSet union(IntegerSet set1, IntegerSet set2) { IntegerSet union = new IntegerSet(); if (set1.isEmpty() && set2.isEmpty()) { return union; } if (!set1.list.isEmpty()) { for (Integer num : set1.list) { if (!union.list.contains(num)) { union.list.add(num); } } } if (!set2.isEmpty()) { for (Integer num : set2.list) { if (!union.list.contains(num)) { union.list.add(num); } } } return union; } /** * Finds the intersection of two sets. A null pointer is considered an empty * set. * * Return \"new IntegerSet()\" when the result is an empty set. */ public static IntegerSet intersection(IntegerSet set1, IntegerSet set2) { IntegerSet intersection = new IntegerSet(); // check for null pointers if (set1.list.isEmpty() || set2.list.isEmpty()) { return intersection; } for (Integer num : set1.list) { if ((!intersection.list.contains(num)) && set2.list.contains(num)) { intersection.list.add(num); } } return intersection; } /** * Builds an array with the ele.

I need help creating a parametized JUnit test case for the following.pdf

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How do the genomes of Archaea and Bacteria compare? Drag and drop the phrases on the left to complete the sentences on the right. Part D Genome size of Archaea versus Bacteria How do the genomes of Archaea and Bacteria compare? Drag and drop the phrases on the left to complete the sentences on the right. Reset Help is/are larger than 1. Just by knowing that the Thaumarchaeota are Archaea, it isare smaller than those of s possible to predict that their genomes islare less complex than may contain a lower most Bacteria. proportion of coenzyme genes than 2. Just by knowing that N. equitans is a parasite, it is may contain more than those of unidentified genes than possible to predict that its genome most other hyperthermophiles. 3. Just by knowing that a bacterium is a autotroph, it is than those of possible to predict that its genome average heterotrophs. Hints My Answers Give Up Review Part Submit Genomic analyses can be extremely useful in learning a great deal about microbes and may eventually lead to more definite conclusions about differences between Archaea and Bacteria. These analyses can also be used at a larger scale. Metagenomics involves studying all of the genetic material in an environment, not just that in individual species. Solution (1) Archaea contain more unidentified genes than bacteria. Therefore, Just by knowing that the Thaumarchaeota are Archaea, it is possible to predict that their genomes (may contain more unidentified genes than) those of most Bacteria. (2) Parasites are smaller in size than hyperthermophiles. Therefore, Just by knowing tha N. equitans is a parasite, it is possible to predict that its genome (is/are smaller than) those of most other hyperthermophiles. (3) Genome of an autotroph is larger than a heterotroph. Therefore, Just by knowing that a bacterium is an autotroph, it is possible to predict that its genome ( is/are larger than) those of average heterotrophs..

How do the genomes of Archaea and Bacteria compare Drag and drop th.pdf

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General question: How would the technology affect who lives and who dies? Solution due to the technology there will be both good and bad also who lives ..? technology became the man\'s sophisticated tool ,with out technology man will lag , how man will get benifited from the technology.? (iam giving some of the example how man is living with the help of technologies) some of the major area\'s are 1)Medical --there is lot improvement in the medical sciencs which help the human to restore his health 2)transport--the travelling time of the man decreases due to the technology improvement in transportation 3)chemical -- due to the improvement in the technology of the chemical industries like creating more sophistictaed material which will help man 4)computer science-- man is getting benifited in most possible ways by the computer technologies who dies..? due to the increasing of the technology there is an effect on the both human and wild life ,sea creatures also 1)as technology increases the pollution also increases due to the pollution many sea ccreature are dying as well as wild life also 2)there is an effect onn human also because as the days going on the life span of the human being is decreases and getting new diseases due to the pollution 3)man is gettig obesity due to the increasing of comforts.

General question How would the technology affect who lives and who .pdf

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Describe how partial diploids can be produced in E. coli. Solution Patial diploids can be produced by the transfer of genetic material of Hfr cell to an F- cell through conjugation.When Hfr mates with F- cell through conjugation the chromosome acts as a rolling circle transferring part of its genetic material across a conjugation bridge.Very rarely whole portion of DNA is transferred,resulting very low chance of passing of F factor to the F- cell.After that F- cell will contain part of Hfr,the exogene and its own genomic material the endogene,at this stage the cell is called partial diploid..

Describe how partial diploids can be produced in E. coli.Solutio.pdf

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4.2 Why is there waiting in an infinite-source queuing system? multiple-phase processing slow service variability in arrival and service rates low utilization poor scheduling of servers Solution The population or source in a queuing theory is divided into two parts including finite and infinite population. In infinite source queuing system, the number of customers in the system has no effect on the arrival pattern of customers. Besides it also experiences huge variability in the arrival rate of customers as well as the service rates. Due to this reason there is waiting in infinite-source queuing system. Hence, the correct option is variability in arrival and service rates..

4.2Why is there waiting in an infinite-source queuing systemmul.pdf

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A single layer of gold atoms lies on a table. The radius of each gold atom is about 1.5×1010 m, and the radius of each gold nucleus is about 7.0×1015 m. A particle much smaller than the nucleus is shot at the layer of gold atoms. Roughly, what is its chance (probability p) of hitting a nucleus and being scattered? (The electrons around the atom have no effect.) Solution So we can estimate the probability neglecting the detailed crystal structure of the gold layer but only from consideration of a single atom. The probability will be proportional to area of the obstacle, so assuming circularity p=area of nucleus/area of the atom=(radius of the nucleus)^2/(that of the atom)^2=2.2*10^(-9).

A single layer of gold atoms lies on a table. The radius of each gol.pdf

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