When examining the central tendency of a probability distribution, u.pdf

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When examining the central tendency of a probability distribution, under what circumstance would you be indifferent to using the median vs the mean? Solution If distribution is symmetric, then mean and median would coincide, and we would be indifferent. If distribution is skewed, or has outliers, we prefer median because it is not affected by extreme values..

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How does each of the differences between prokaryotes and eukaryotes influence bacterial infection and treatment Solution The differences between prokaryotes and eukaryotes such as cell wall, nucleus, flagella, membrane-bound organelles, genetic makeup, metabolic differences and antigenicity are responsible for the infection and treatment of bacterial diseases. For example, the bacteria contains cell wall, which is absence in animal cells. So, cell wall synthesis inhibition is a strategy in the treatment of bacterial infections..

How does each of the differences between prokaryotes and eukaryotes .pdf

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For MIMO system, (a) Please talk about the advantage and disadvantage of MIMO technique. (b) Please explain why we need Cooperative Diversity to replace MIMO in cellular network. Solution A)MIMO advantages • Capacity scales linearly with number of antennas Channel knowledge/estimation at Rx needed • MIMO offers potential for larger data rate larger spectral efficiency larger number of users improved range/coverage better interference suppression better quality of service (QoS), lower bit-error rate (BER) lower Tx power MIMO disavantages • Hardware complexity: Each antenna needs a radio-frequency (RF) unit Powerful digital signal processing (DSP) unit required • Software complexity: Most signal processing algorithms are computationally intensive Power consumption: Battery lifetime of mobile devices Thermal problems Antennas: Antenna spacing (electromagnetic mutual coupling-e.g. mobile handsets) RF interference and antenna correlation B) Most of the present researches on cooperative network in which the user nodes are equipped with a single antenna or multiple, there have been some new results which exploit the benefits of multiple antenna deployment. Cooperative MIMO technology allows a wireless network system to attain better performance gains than provided by either usual MIMO or cooperative systems. It promised significant improvement in spectral efficiency and network coverage phenomena for different next generation wireless communication systems. In wireless communication, the path towards the various techniques that gives high service quality and data rate has been through the use of the cooperative network provided by the rich scattering wireless channels. Due to their great aspects, MIMO and cooperative systems have found their way into several standards for future wireless communication systems, especially in cellular networks and wireless local area networks (LAN) in this review article we are presenting the comparative analysis..

For MIMO system, (a) Please talk about the advantage and disadvantag.pdf

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Explain why and give examples. \"In OO programming Polymorphism is only possible when using Interface or Inheritance\". Solution Polymorphism is the ability of an object to take on many forms. The most common use of polymorphism in OOP occurs when a parent class reference is used to refer to a child class object. Any Java object that can pass more than one IS-A test is considered to be polymorphic. In Java, all Java objects are polymorphic since any object will pass the IS-A test for their own type and for the class Object. It is important to know that the only possible way to access an object is through a reference variable. A reference variable can be of only one type. Once declared, the type of a reference variable cannot be changed. The reference variable can be reassigned to other objects provided that it is not declared final. The type of the reference variable would determine the methods that it can invoke on the object. A reference variable can refer to any object of its declared type or any subtype of its declared type. A reference variable can be declared as a class or interface type. Polymorphism using inheritance: Polymorphism using interface: class Main { }.

Explain why and give examples. In OO programming Polymorphism is on.pdf

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Eukaryotic cells modify RNA after transcription What critical RNA processing events usually happen to pre-mRNA\'s before they sent to the cytoplasm for translation? What is a 5\' cap? What is a poly A tail? What do these end modifications do for the mRNA transcript? What is RNA splicing? What are introns? What are exons? How are intros spliced out? What is a spliceosome? What is a \"snurp\" (snRNA)? Solution After its synthesis, the eukaryotic mRNA will undergo extensive modification like capping, polyadenylation and splicing to enter into the process of translation. (B) Capping: Here the 5\' end of the mRNA is modified by the addition of 7-methylguanosine (m7G) and the main function of this cap is to protect the 5\' end of the primary RNA transcript from attack by ribonucleases and this 5’ cap will be recognized by eukaryotic initiation factors, so that it can assemble the mature mRNA with the ribosome to start the process of translation. At the 3\' end of the RNA, we will have polyadenylation signal and during transcription itself this sequence will be chopped by an enzyme and another enzyme will add about 100100100 - 200200200 adenine (A) nucleotides to the 3’ end and this will form the poly-A tail. The main function of this poly A tail is proving stability to the transcript and also helping it to get exported from the nucleus to the cytosol.RNA splicing (C )Splicing is the third big RNA processing event and the pre-mRNA will have two sequences, exons and introns. Introns are the non-coding sequences and exons are the coding sequences, here in this step the through splicing, the introns will be removed and exons will be attached together. In RNA splicing, specific parts of the pre-mRNA (introns) will be recognized and removed by a protein-and-RNA complex called the spliceosome. Mature mRNA will have only exons, but no introns. (D)The splicing signal exon/GU-intron-AG/exon will be present in nuclear mRNA precursors and 5\' and 3\' splice sites always have consensus sequences extending beyond GU and AG motifs. During splicing, the exon-intron boundaries will be recognized by snRNA and the consensus sequences within introns will get hybridized and now the proteins other snRNAs will assemble the spliceosome on the transcript, the unpaired A present at 3\' side of the introns will attacks the 5\' exon -intron boundary with the help of 2\' OH and this will give rise to lariat structure. The free 3\' OH of the upstream exon will displace the downstream junctional nucleotide, like this introns will be removed and exons will be attached together. (E)A spliceosome is a large complex formed due to the assembly of snRNAs and protein complexes, and plays an important role in splicing of pre mRNA. snRNPs (snurps) is the small nuclear ribonucleic proteins and this is a RNA-protein complexes and they will form a larger complex with the unmodified pre-mRNA and various other proteins to form the structure called spliceosome.(A)Critical RNA processing events.

Eukaryotic cells modify RNA after transcription What critical RNA pr.pdf

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Driver: import java.util.Scanner; //A class that keeps a fleet of different types of vehicles public class FleetOfVehicles { //An array of vehicles private Vehicle[] fleet; public static final int MAX_VEHICLES = 100; public FleetOfVehicles() { fleet = new Vehicle[MAX_VEHICLES]; } public Vehicle[] getFleet() { return this.fleet; } //Adds a new vehicle to the first empty spot in the fleet array public void addVehicle(Vehicle aV) { for(int i=0;i Solution import java.util.Scanner; //A class that keeps a fleet of different types of vehicles public class FleetOfVehicles { //An array of vehicles private Vehicle[] fleet; public static final int MAX_VEHICLES = 100; public FleetOfVehicles() { fleet = new Vehicle[MAX_VEHICLES]; } public Vehicle[] getFleet() { return this.fleet; } //Adds a new vehicle to the first empty spot in the fleet array public void addVehicle(Vehicle aV) { for(int i=0;i 0) this.cylinders = cylinders; else this.cylinders = 0; } /** * @return the ownersName */ public String getOwnersName() { return ownersName; } /** * @param ownersName * the ownersName to set */ public void setOwnersName(String ownersName) { this.ownersName = ownersName; } @Override public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Vehicle) { Vehicle vehicle = (Vehicle) obj; if ((this.getManuName().equals(vehicle.getManuName())) && (this.getOwnersName().equals(vehicle.getOwnersName())) && (this.getCylinders() == vehicle.getCylinders())) return true; else return false; } else return false; } @Override public String toString() { // TODO Auto-generated method stub return getManuName() + \" \" + getCylinders() + \" \" + getOwnersName(); } } public class Truck extends Vehicle { double loadCap, towCap; public Truck() { // TODO Auto-generated constructor stub super(); setLoadCap(0); setTowCap(0); } /** * @param manuName * @param cylinders * @param ownersName * @param loadCap * @param towCap */ public Truck(String manuName, int cylinders, String ownersName, double loadCap, double towCap) { super(manuName, cylinders, ownersName); setLoadCap(loadCap); setTowCap(towCap); } /** * @return the loadCap */ public double getLoadCap() { return loadCap; } /** * @param loadCap * the loadCap to set */ public void setLoadCap(double loadCap) { if (loadCap > 0) this.loadCap = loadCap; else this.loadCap = 0; } /** * @return the towCap */ public double getTowCap() { return towCap; } /** * @param towCap * the towCap to set */ public void setTowCap(double towCap) { if (towCap > 0) this.towCap = towCap; else this.towCap = 0; } @Override public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Truck) { if (super.equals(obj)) { Truck vehicle = (Truck) obj; if (this.getLoadCap() == vehicle.getLoadCap() && this.getTowCap() == vehicle.getTowCap()) return true; else return false; } else return false; } else return false; } @Override public String toString() { // TODO Auto-generated method stub return super.toString() + \" \" + g.

Driverimport java.util.Scanner;A class that keeps a f.pdf

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Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Solution The idea here is to use the concept of two binary heaps along with the concept of list(doubly linked list). The doubly linked list contains all input items and indexes of corresponding min and max heap nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root node of min heap stores the address of minimum item in doubly linked list. Similarly, root of max heap stores address of maximum item in doubly linked list. Following are the details of operations. 1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a O(1) operation. 1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a O(1) operation. 3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this address to find the node in doubly linked list. From the doubly linked list, we get node of Max Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time. delete() operations for max and min heaps take O(Logn) time. 4) removeMax(): is similar to deleteMin() 5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn) #include #include #include // A node of doubly linked list struct LNode { int data; int minHeapIndex; int maxHeapIndex; struct LNode *next, *prev; }; // Structure for a doubly linked list struct List { struct LNode *head; }; // Structure for min heap struct MinHeap { int size; int capacity; struct LNode* *array; }; // Structure for max heap struct MaxHeap { int size; int capacity; struct LNode* *array; }; // The required data structure struct CDS { struct MinHeap* minHeap; struct MaxHeap* maxHeap; struct List* list; }; // A utility function to create a new List node struct LNode* newLNode(int data) { struct LNode* node = (struct LNode*) malloc(sizeof(struct LNode)); node->minHeapIndex = node->maxHeapIndex = -1; node->data = data; node->prev = node->next = NULL; retu.

Describe a data structure that supports both removeMin() and rem.pdf

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Columbus Incorporated just paid $4.3 per share dividend yesterday (i.e.,D0). The dividend is expected to grow at a constant rate of 6% a year. The required rate of return on the stock, r, is 11%. What is the value per share of the company\'s stock? Round your answer to two decimal places. Solution Recent Dividend, D0 = $4.30 Growth Rate, g = 6% Required Return, rs = 11% D1 = D0 * (1 + g) D1 = $4.30 * 1.06 D1 = $4.558 Value per share = D1 / (rs - g) Value per share = $4.558 / (0.11 - 0.06) Value per share = $4.558 / 0.05 Value per share = $91.16.

Columbus Incorporated just paid $4.3 per share dividend yesterday (i.pdf

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Anatomy Question: Please Answer them all. A. Using a punnett square, show the possible genotypes for John and Andrea\'s children? (John has O blood type and Andrea has A blood type) B. Based on the blood typing results, could Daryl be John and Andrea\'s Son? How Do you know this? ( Daryl has O blood type) Solution A. Andrea :Mother has A blood type , her genotype will be either AO or AA John :Father has O blood type ,his genotype will be OO. O allele is masked by the presence of A allele or B allele. OO blood group is expressed only in presence of OO alleles. Father Father B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood groupOO O blood group.

Anatomy Question Please Answer them all.A. Using a punnett square.pdf

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A person is lost in the desert. (S)he has no water and after a couple of days loses so much water through perspiration that his/her body fluids become hypertonic. What happens to his cells? What happens to his proteins? What happens to his carbohydrates? Solution Early sign of dehydration there is no symptom but later on there is dryness of mouth and tongue, dark yellow urine,dryness of skin cramp in arm and leg. When fluid become hypertonic i.e sodium is lost from cell this result in cell shrinkage. After two days of starvation body will first utilize carbohydrate and then glycogen from liver and muscle for energy. Protein from cell / muscle will be next to be used as source of energy for body.

A person is lost in the desert. (S)he has no water and after a coupl.pdf

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2. Describe an advantage and a limitation to tracing ancestry with both types of DNA. Solution The tracing of ancestral origins using mitochondrial DNA, Y-DNA & autosomal DNA is advantageous to know about the historical evolutionary aspects of humans from thousands of years based on short tandem repeats, radioactive labeling & based on single nucleotide polymorphism. This Genetic genealogy is advantageous to unravel familial relationships but has ethical and moral limitations when examining prenatal ancestral tracing& the information about the ancestral DNA is limited to reveal about the inheritance pattern.

2. Describe an advantage and a limitation to tracing ancestry with b.pdf

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Write your thought on the following: 1. petroski notes that most engineering failures are much more involved than simple technical mis-calculations and involve the failure of the design process or management culture.However, not all engineering failures involve ethical issue.The infamous collapse of the first Tacoma Narrow Brodge, and the losses of the Mars polar lander and Mars Climate Orbiter were technical and design process failures. Solution In the face of new technology, how do we balance public welfare and progress? If Moisseiff had designed a bridge similar to the ones which had already proven their stability, Tacoma Narrows Bridge would never have collapsed costing thousands of dollars and endangering many lives. It would also have been significantly more expensive. On the other hand, if engineers had never tried innovative techniques, suspension bridges may never have been built at all. At the time of their introduction, no one believed that a suspension bridge could safely accommodate trains. Roebling, however, took a gamble, pushed the limits of the current technology, and built a suspension bridge that he believed could safely support rail traffic. Luckily he was correct, and suspension bridges soon became widely accepted (Petroski. Moisseiff also took a gamble, trying to create a longer, sleeker, less expensive bridge, by pushing the limits of technology. He, however, was not as lucky, and what could have been a breakthrough in technology turned into a catastrophic failure. Every time engineers push the limits of technology they risk a similar loss like losses of mas polar lander and mass climate orbiter in which there were technical error like Mass polar lander 1. Inadequate specifications which includes what the software was supposed to do, but no mention of what it must not to do. 2. Flawed review process: Violation of basic engineering safety pracitices in the digital part of the system. Mass Climate Orbiter 1. Units communication and attention to detail failure. The primary cause of this discrepancy was that one piece of ground software supplied by Lockheed Martin produced results in a United States customary unit, contrary to its Software Interface Specification (SIS), while a second system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS. Specifically, software that calculated the total impulse produced by thruster firings calculated results in pound-seconds. The trajectory calculation software then used these results - expected to be in newton-seconds - to update the predicted position of the spacecraft. sometimes even It can cause a loss of life. How much is too much? When is a possible advance worth a risk to public safety? What can the engineering profession do to make the implementation of new technology safer? Do our current peer review and building code committee processes adequately protect public safety?.

Write your thought on the following1. petroski notes that most en.pdf

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You have two cell lines, a macrophage cell line and a CD4 T-cell line. The macrophage cell line is resistant to infection by HIV, whereas the T-cell line is fully susceptible and fully permissive. Why are macrophage cell line resistant to infection by HIV? Why are T-cell fully susceptible and fully permissive? Please explain. Also, A heterokaryon cell made from the fusion of these two cells is able to be infected by HIV. Explain why? Please explain in detail. Solution Macrophages are antigen- presenting cells so they process the viral peptides and present it to uninfected CD4 T-Cell which is the important transmission process. Also HIV-infected macrophages release soluble cytotoxic factors which induce apoptosis in CD4 T-Cells. Macrophages produce certain host restriction factors which inhibits the release of viral progeny from infected cells. So macrophages are resistant to lysis because of its genetic make up and its phenotypes (restriction factors). So when two cell lines are fused to form heterokaryon cell, the genes of macrophages are suppressed or it is recessive or the T cell genes are dominant to make the heterokaryon susceptible to infection..

You have two cell lines, a macrophage cell line and a CD4 T-cell lin.pdf

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With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and females. In X-linked dominant traits, males hemizygous for trait will express the trait, but females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected female will show the trait. For autosomal dominant, two affected parents may produce an unaffected child. Solution Answer: A. True Most X-linked conditions are recessive. This means that in a person with two X chromosomes (most females), both copies of a gene (i.e., one on each X chromosome) must have a change or mutation whereas in a person with one X chromosome (most males), only one copy of a gene must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X chromosome is typically affected with the condition. The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome to the son and other to the daughter. B. False For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a female with two X chromosomes or males with on X chromosome must have a change or mutation for an individual to be affected with the condition. For this reason, X-linked disorders are often seen with similar frequency in males and females. However, since females also have one normal X chromosome as well as an X chromosome with a mutation, the condition is often more “mild.” C.False For X-linked recessive disorders, an affected father who has a mutation in a gene on the X chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his children. If the mother is not affected or a carrier, none of his sons will be affected since they can only inherit a normal X chromosome from their mother and they inherit a Y chromosome from their father. But if the female is affected, it will receive one of the affected X chromosomes, which is enough to express the trait in the son. D. True A person affected by an autosomal dominant disorder has a 50 percent chance of passing the mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50 percent..

With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf

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What will organizations have to do to adapt to the influx of needed Generation Y individuals? Solution simple thing is choose the right person for a work and let them free, they will create miracles. here generation Y people are those who can perform variaous tasks at a time and with more accuracy. these are multi talented, always carries courage, never hides their talent and ready to accept challenges. first to attract those kind of people organisations should change their existing rules, policies and regulations. all they were writen decades ago and still we are follwoing them. they need to redesign the work and the structure. if we are succede at recruiting the correct people for our work, then the remaining changes can automatically comes into the picture. the organisations should create the work environment to the Y generation people There are no precise dates when the generation starts and ends; most researchers and commentators use birth years ranging from the early 1980s to early 2000s. Generation Y people may be known as echo boomers because they are the children of baby boomers, or someone who was born during the period of increased birth rates that occurred between 1946 and 1964. The Generation Y cohort is also referred to as Millennials, the Internet generation, iGen and/or the Net generation.

What will organizations have to do to adapt to the influx of needed .pdf

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What are the main functions of the components of prokaryotic vs. eukaryotic cells? Could either cell type survive without any of them? (Answer must be at least a page long, very detailed and in paragraph form please) Solution Prokaryotes are unicellular living beings that need organelles or other interior film bound structures . In this manner, they don\'t have a core, at the same time, rather, by and large have a solitary chromosome: a bit of roundabout, twofold stranded DNA situated in a territory of the phone called the nucleoid. Most prokaryotes have a cell divider outside the plasma film. Like a prokaryotic cell, an eukaryotic cell has a plasma film, cytoplasm, and ribosomes. Nonetheless, not at all like prokaryotic cells, eukaryotic cells have: a film bound core various film bound organelles (counting the endoplasmic reticulum, Golgi contraption, chloroplasts, and mitochondria) a few bar formed chromosomes Since an eukaryotic cell\'s core is encompassed by a film, it is regularly said to have a \"genuine core. \" Organelles (signifying \"little organ\") have specific cell parts, similarly as the organs of your body have particular parts. They permit diverse capacities to be compartmentalized in various territories of the cell. The piece of the cell divider varies altogether between the areas Bacteria and Archaea, the two spaces of life into which prokaryotes are separated . The structure of their cell dividers additionally varies from the eukaryotic cell dividers found in plants (cellulose) or organisms and bugs (chitin). The cell divider works as a defensive layer and is in charge of the life form\'s shape. A few microorganisms have a container outside the cell divider. Different structures are available in some prokaryotic species, yet not in others. For instance, the case found in a few animal types empowers the living being to connect to surfaces, shields it from lack of hydration and assault by phagocytic cells, and expands its imperviousness to our invulnerable reactions. A few animal categories likewise have flagella utilized for velocity and pili utilized for connection to surfaces. Plasmids, which comprise of additional chromosomal DNA, are likewise present in numerous types of microorganisms and archaea. Commonly, the core is the most conspicuous organelle in a cell. Eukaryotic cells have a genuine core, which implies the cell\'s DNA is encompassed by a layer. Thusly, the core houses the cell\'s DNA and coordinates the blend of proteins and ribosomes, the cell organelles in charge of protein union. The atomic envelope is a twofold layer structure that constitutes the furthest bit of the core. Both the internal and external layers of the atomic envelope are phospholipid bilayers. The atomic envelope is punctuated with pores that control the section of particles, particles, and RNA between the nucleoplasm and cytoplasm. The nucleoplasm is the semi-strong liquid inside the core where we discover the chromatin and the nucleolus. M.

What are the main functions of the components of prokaryotic vs. euk.pdf

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There is a requirement to design a system to sense the presence of glass bottles leaving an automatic washer on a production line conveyor belt. The production line is for food products and the environment is clean and maintained at a temperature of 15 degree C. The system must meet the following criteria: The switching distance will lie between 5 and 20 mm. The output from the sensor is to be of binary form, i.e. logical 1 when the part is present and logical 0 when no part is present. It is necessary to ensure that a change of sensor state occurs closer to the sensor when a bottle approaches than when it moves away. Bottles will pass the sensor at the rate of 10 every second. There is no requirement for protection from contact with objects but it must be proof against water drips. Recommend a suitable sensor and justify your recommendation. Sketch a diagram that shows the functional blocks within the sensor and explain each function. Solution lock Diagram of the system Basic Elements Used MICRO CONTROLLER RESISTORS CAPACITOR INDUCTOR POWER DIODE WATER PUMP IR SENSORS DC MOTOR CONVEYOR BELT Resistore A part from resister and inductor, a capacitor is the other basic component used in electronic circuits. It is a device which 1. Has the ability to store charge which neither a resistor nor an inductor can do; 2. opposes any change of voltage in the circuit in which it is connected; 3. Blocks the passage of direct current through it. CAPACITOR The electronic component (mostly a coil) which opposes the change of current in circuit is called an inductor. The ability of coil or inductor due to which it opposes the change of current flowing through it is known as its inductance. The inductance of an inductor may also be defined as the ability of an inductor to produce induced voltage when the current varies through it. The unit of inductance is Henry which is named after Joseph Henry. INDUCTOR Diodes allow electricity to flow in only one direction. The arrow of the circuit symbol shows the direction in which the current can flow. Diodes are the electrical version of a valve and early diodes were actually called valves. The construction and the fundamental working of the power diode and signal diode are same but the power handling capacity of the power diode is higher than the signal diodes POWER DIODE Sensors IR Sensors work by using a specific light sensor to detect a select light wavelength in the Infra- Red (IR) spectrum. By using an LED which produces light at the same wavelength as what the sensor is looking for, you can look at the intensity of the received light. When an object is close to the sensor, the light from the LED bounces off the object and into the light sensor. This results in a large jump in the intensity, which can be detected using a threshold DRIVER CIRCUIT FOR DC MOTOR We have used the dc motors to drive conveyer belt and to. The driver circuit for dc motors is shown below. COMPARATOR CIRCUIT Comparator circuit is used in thi.

There is a requirement to design a system to sense the presence of gl.pdf

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1. What are the characteristics of financial assets that have highly developed secondary markets? 2. What is the difference between the secondary market in mortgages and the secondary market in mortgage-backed securities? Solution 1. Financial assets in the developed world are more complex and there are many options available for investors. The financial assets include equity and derivative instruments and a fully developed bond market. In developing nations, since most of the assets held by individuals are in form of property and other hard assets, the financial assets are not that well developed. A developing nation may have a stock market in the weak form and will not have a fully developed bond market. 2. A secondary market in mortgages is where the home loans and their servicing rights and bought and sold between lenders and investors. In mortgage backed security, several primary mortgages which are sold in the secondary marked are pooled into large mortgage back investor units and are sold to investors once again..

1. What are the characteristics of financial assets that have highly.pdf

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The tools that cyber criminals often use, including computer viruses, worms, trojan horses and spyware, are called . Solution most of the cyber crimes are committed by the individuals or small groups and their attack techniques ie there are few type of attacks that cyber criminal use to commit crimes ie botnet -a network of software robots or bots that automatically spread malware , fast flux- moving data quickly among the computers in botnet to make it difficult to trace the source of malware or phishing websites , zoombie computer - a computer that has been hacked into and is used to launch malacious attacks or to become part of botnet , social engineering - using lies and manipulation to trick people into revealing their personal information and phishing is form of social engineering and this attack include ie sending a victims an email that claims theirs problem with their account and has a link to fake the website and entering their account information into the site sends it atraight to the cyber criminal and trying to convince victims to open an email attachments that contain malware by claiming it is something that might enjoy or need and pretending to be network or account administrator and asking for the victims password to perfom maintainance and claiming that victim has won prize must give the credit card informaition in order to receive it and asking for the victims password for an internet service and then using the same password to access other accounts and servicessince many people reuse the same password and promising the victim that they wil send the millions of dollarsif they help out sender by giving the money or the bank account information there are some gey hat hackers who wants to come clean and turn away from crime an option is to work for the people they used to torment by becoming security consultant in past they were black hat hackers who use their computer expertise to break into the computer systems and steal information illegally . therefore the tools that cyber criminals often use including computer viruses , worms torjan horses and spyware are called spam denial of service hacking malware. typically a spam is sent to multiple receipients who didnot ask to recieve it and it includes email spam and instant messaging spam , websearch engine spam spam inn blogs and mobile phone messaging spam and it includes misleading advertisements and phishing messages and designed to trick reciepients in giving personal information and financial information .these are the tools that cyber criminals use..

The tools that cyber criminals often use, including computer viruses.pdf

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The Poisson distribution analysis can be used to calculate the titer of a given suspension of virus. I ran an experiment dealing with egg infectious dose using a newly isolated paramyxovirus in suspension. I would like to know the titer of the virus suspension in egg infectious dose units. 100 eggs were inoculated with ten microliter of a 1: million dilution of the virus suspension. After five days of incubation I discovered that ten eggs were not infected. Calculate the titer of my virus in the undiluted suspension. What are the units ? Solution titre(pfu/ml) = number of plaques/(dilution*volume of diluted virus added to the egg) titre= 90/(0.000001*0.01) titre=9*109 pfu/ml.

The Poisson distribution analysis can be used to calculate the titer.pdf

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Save I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont know how to implement the FindshortestPath Method. I was given this as a stragedy to solve the findshortestpathMethod(): Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at distance x from the center. The other will be the list of map objects for cells at distance x + 1. Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the maze. Since this map object refers to the center cell, it does not need to refer to any other MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field to true to indicate that this cell has already been added to the map. Set up a loop to run until the maze entrance is mapped. This loop does the following: Set up a loop to process each map object from the list at distance x. This loop does the following: i.Get the cell from the map object ii.If this cell is the entrance cell, set the entrance and break out of the loop! iii.Check each direction from this cell and for an accessible neighbor cell Get a reference to that neighbor cell If the neighbor cell has not already been visited Set the cell’s visited field to true so its only processed once Create a MazeMap object that refers to this neighbor cell Make this MazeMap object refer to the current map object being processed – ie the cell one step closer to center Add this MazeMap object to the list at distance x + 1 List at distance x has been processed. The list at distance x + 1 is now complete, so make the list at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x + 1. At this point, the entrance points at the map object which refers to cell 1,1. That map object also refers to a map object which is one step closer to the center. Write a loop that traverses this trail of references until it reaches the end (a null reference). For each map object: Get the cell referred to. Call the drawCircle method with a color, the cell’s row and column, and SMALL. Call the delay method with a SHORT delay. This slows down the drawing process so you can watch the path being displayed. Advance to the next map object in the list. here is the code so far********* here is my code: ******mazeApp******* package mazepackage; import javafx.application.Application; import javafx.application.Platform; import javafx.scene.Group; import javafx.scene.Scene; import javafx.scene.canvas.Canvas; import javafx.scene.canvas.GraphicsContext; import javafx.scene.control.TextInputDialog; import javafx.scene.paint.Color; import javafx.stage.Stage; public class MazeApp extends Application { // default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE private static final int DEFAULT_SIZE = 768; private int width = DEFAULT_SIZE; private int height = DEFAULT_SIZE; // The graphics context is needed to enable drawing on the canvas private Graphics.

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