Simulation was used to estimate the empirical probability that eleph.pdf

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Simulation was used to estimate the empirical probability that elephants born today will have a genetic defect that causes severe arthritis by their 14th year of life. What statement is true about the situation? 1. simulations cannot be used to estimate empirical probability 2. simulations require that one know the exact theoretical probability 3. simulations can only be used to estimate the empirical probability of very rare events 4. simulations are appropriate when the expense or time to observe an event is large Solution 2. simulations require that one know the exact theoretical probability.

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Customers arrive at a bank teller\'s booth at a rate of 2 per minute. What is the probability (to four decimal places) that 4 customers will arrive at the bank teller\'s booth within the next minute (assume a Poisson distribution) Solution Possion Distribution PMF of P.D is = f ( k ) = e- x / x! Where = parameter of the distribution. x = is the number of independent trials P( X = 4 ) = e ^-2 * 2^4 / 4! = 0.0902.

Customers arrive at a bank tellers booth at a rate of 2 per minute.pdf

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Create a JAVA program that performs file IO and database interaction via SQL. The program needs to read data from the provided file: \"Project.csv\" and insert the data into a database. Then the program needs to create a report from the database sorted by price descending. The report should be in the format demonstrated below. id (primary key - generated by the database) cpuname performance price Project.csv contents:CPU NamePerformancePrice (USD)Intel Core i7-3770K @ 3.50GHz9,556$560.50Intel Core i7-3770 @ 3.40GHz9,327$335.55Intel Core i7-3820 @ 3.60GHz8,990$404.38AMD FX-8350 Eight-Core8,940$149.99Intel Core i7-2600K @ 3.40GHz8,501$379.97Intel Core i7-2600 @ 3.40GHz8,242$214.99Intel Core i7-4720HQ @ 2.60GHz8,046NAAMD FX-8320 Eight-Core8,008$145.99Intel Core i7-6700HQ @ 2.60GHz7,997$1509Intel Core i7-4710HQ @ 2.50GHz7,826NAIntel Core i5-6600K @ 3.50GHz7,762$239.99Intel Core i7-4700HQ @ 2.40GHz7,754$383.00Intel Core i7-4700MQ @ 2.40GHz7,736$467.40Intel Core i5-4690K @ 3.50GHz7,690$239.99AMD FX-8150 Eight- Core7,619$165.99Intel Core i7-3630QM @ 2.40GHz7,604$304.49Intel Core i5-4670K @ 3.40GHz7,598$249.99Intel Core i5-4690 @ 3.50GHz7,542$224.99Intel Core i7-3610QM @ 2.30GHz7,460$399.99Intel Core i5-4670 @ 3.40GHz7,342$226.99Intel Core i5-4590 @ 3.30GHz7,174$199.99Intel Core i7-4702MQ @ 2.20GHz7,146NAIntel Core i5-3570K @ 3.40GHz7,130$477.23 Solution import java.io.IOException; import java.nio.file.Files; import java.nio.file.Paths; import java.util.Arrays; public class Main { /** * This the main function that runs at the start * param args - input arguments from the command line */ static public void main(String[] args) { CPUList cpuList = new CPUList(); //The CPUList used to retrieve data from the fiile and store in the db CPUList cpuListRetrieved = new CPUList(); //The CPUList used to retrieve data from the database CpuDb cpuDb = new CpuDb(); //The database object used to move data to and from the CPU Lists try { //Read in the file and store each line into the CPU objects in a list Files.lines(Paths.get(\"Project04Data.csv\")) .map(line -> line.split(\"\ \ \")) // Stream .flatMap(Arrays::stream) // Stream .forEach(line -> cpuList.AddCpu(line)); //Clear the list table for the new listing cpuDb.Clear(); //Insert the Cpu List into the database cpuDb.SetCpuList(cpuList); //Retrieve the Cpu List into a different CPU List object from the database cpuDb.GetCpuList(cpuListRetrieved); //Show the report from the new list that was retrieved from the database cpuListRetrieved.ShowReport(); } catch (IOException e) { e.printStackTrace(); } } } CPUList.java import java.util.ArrayList; import java.util.Collections; import java.util.Iterator; public class CPUList { ArrayList theList = new ArrayList<>(); /** * Default constructor for the CPU */ public void CPUList() { } /** * param strInputLine Input line to be used in creating the CPU object */ public void AddCpu(String strInputLine) { theList.add(new CPU(strInputLine)); } /** * param tempCPU - A CPU object to.

Create a JAVA program that performs file IO and database interaction.pdf

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Wolves were hunted and poisoned to extinction in Sweden and Norway by the mid-20th century. Three wolves migrated from Russia and founded a new Scandinavian wolf population in the early 1980s. That population now consists of approximately 100 animals. How large do you think the Scandinavian wolf population should be to maintain genetic variation that may be important for the viability (continued persistence) and continued evolution of this population? There is some possibility of gene flow into this population from Russian wolves. Do you think that it is important that gene flow between the Scandinavian and the Russian populations be maintained in the future? Why? Solution There is no cut-off formula for calculating the length of population to maintain genetic diversity. It is important to note that even after bottleneck effect populations are able to survive and are able to maintain genetic diversity. Here, there are 100 animals; and these 100 animals are heterozygous, not pure lines; because they have already mixed up with the Scandinavian population. So, the population is large enough to maintain genetic variation. Yes, there is further possibility of mixing of this population with Russian wolves further also; but that possibility is very rare. This is so because wolves have already been extinct over there; as per the question. No, it is no more compulsion for gene flow between the two populations. This is so because the populations have already diverged. And a subsequent amount of genetic variability has been attained..

Wolves were hunted and poisoned to extinction in Sweden and Norway b.pdf

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Describe the differences between OpenPGP, PGP and GPG. Solution All the three mentioned are encryption method and stands for PGP - Pretty Good Privacy, GPG(GNU Privacy Guard) The difference between three could be below: Availability in market - PGP is a licensed product whereas OpenPGP and GPG are freewares. Functionality - PGP is a encyrption method that uses hashing,compression, cyrptography concepts, OpenPgP is an encryption method which uses processes that are interoperatable with PGP and GPG is a encryption method which provides a graphical inerface to encrypt files..

Describe the differences between OpenPGP, PGP and GPG.Solution.pdf

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Declaring a Pointer To define a pointer, use an asterisk, (*), in the declaration to specify the variable will be a pointer to the specified data type. Recall that the name of an array holds the memory address of the first element of the array. The second statement then would store the memory address of the first value in the variable valuePtr. The name of a pointer is an identifier and must follow the rules for defining an identifier. Some programmers place the letters ptr, a common abbreviation for pointer, at the end of the variable name, others prefix a pointer with p_ or ptr_, and there are those who do nothing to distinguish a pointer by its name. The naming of a pointer variable is a matter of programming style. Once a pointer has been declared and initialized, it can be used to access the data to which it points. In order to access this value, the dereference operator, *, must be used to prefix the name of the pointer. From the code above, valuePtr holds the address of the first element of the values array; therefore, *valuePtr will access the value 325. Add the following to main and execute the program. The output generated should look something like the following. Where 0xbfad43e8 is a memory address displayed in hexadecimal. Obtaining the Memory Address of a Variable The address operator, &, is used to determine the memory address of a variable that is not an array. Add code necessary to use fixed and setprecision, then add the following to main and run the program to confirm the use of the address operator. Using a Pointer to Alter Data Just as the dereference operator is used to retrieve data, so too is it used to store data. Add the following to function main and run the program. The pointer, ratePtr, is used to change the data stored in payRate. The output of payRate confirms the change. Using a Pointer in an Expression As previously shown, the value pointed to by a pointer variable can be retrieved by dereferencing the pointer. In the above code, the retrieved value was simply displayed to the screen; however, it can also be used in an expression. Add the following to main and execute the program. Pointer Arithmetic It is possible to add to or subtract from a memory address stored in a pointer. These actions can be accomplished using the addition operator,+, subtraction operator, - and the increment and decrement operators, ++ and --. This is helpful when accessing arrays via pointers. The increment operator has been used to add the value 1 to a variable. When used with a pointer, the increment operator adds the size of the data referenced to the memory address stored in the pointer, effectively moving the pointer to the next data value. Building on the code from above, add the following to main and execute the program. We initialized valuePtr to the memory address of the first element of the array values. Incrementing valuePtr instructs the computer to add four bytes, the size of an integer, to the memory addressed stored in .

Declaring a PointerTo define a pointer, use an asterisk, (), in t.pdf

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Who identified three key economic “advantages” that firms should hav.pdf

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Which of these is the largesta. The surface zoneb. The mixed la.pdf

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Which of the following concepts is exemplified by the story of Parus major and the island of Vlieland? Gene flow reduces the genetic differences between populations. Gene flow can affect how well populations are adapted to local environmental conditions. Genetic drift is significant in small populations. Genetic drift can lead to a loss of genetic variation within populations. Relative fitness is not the same as fitness when calculated on a 0-1 scale. Solution ANS: E. Relative fitness is not the same as fitness when calculated on a 0-1 scale. Explanation: Ex: the great tit (Parus major) on the Dutch island of Vlieland. In Parus major high immigrated central region birds have a lower fitness, when compare with the low immigration birds which are present in east. This phenomenon clearly explains low immigrated birds had higher fitness. Definitions of Other Terminology: Darwin\'s fitness: It is the ability of an individual to survive and reproduce in its environment. Fitness of an organism is measured by how many offspring it produces compared to other individuals of a species. Adaptation refers to the trait that increases organism fitness relative to individuals without the trait. Genetic drift: It is random and happens in every population due to chance. Because of this smaller population are more susceptible to the effect of genetic drift. Gene pool: Gene pool is a reservoir of all genes, which was present in a population. Gene pool corresponds with a number of unique alleles present in a population..

Which of the following concepts is exemplified by the story of Parus .pdf

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You will be implementing the following functions. You may modify the main to test the functions, but this programwill work as a starting point. //********** Function Prototypes ****************** //function isPrime //input parameter - positive integer greater than 1 //returns true if the number is prime, otherwise false // bool isPrime (int number); //function findPrimes //input parameter - positive integer //Uses the isPrime method to print a list of prime numbers between 1 and n. void findPrimes (int n); //function findFibo //input parameter - positive integer //returns the nth fibonacci number where //Fib(0) -> 0 //Fib(1) -> 1 //Fib(N) -> Fib(N-2) + Fib(N-1) //Note: You must use a loop in your solution. Also, if passed a 0, return 0. int findFibo (int n); //function findFactors //input parameter - positive integer //prints the prime factors of the given number to standard output (cout) //example output: 52=2*2*13 (or 52=1*2*2*13) or 13=prime //Hint: You will need nested loops for your solution. void findFactors (int number); Solution #include #include using namespace std; //********** Function Prototypes ****************** //function isPrime //input parameter - positive integer greater than 1 //returns true if the number is prime, otherwise false // bool isPrime (int number); //function findPrimes //input parameter - positive integer //Uses the isPrime method to print a list of prime numbers between 1 and n. void findPrimes (int n); //function findFibo //input parameter - positive integer //returns the nth fibonacci number where //Fib(0) -> 0 //Fib(1) -> 1 //Fib(N) -> Fib(N-2) + Fib(N-1) //Note: You must use a loop in your solution. Also, if passed a 0, return 0. int findFibo (int n); //function findFactors //input parameter - positive integer //prints the prime factors of the given number to standard output (cout) //example output: 52=2*2*13 (or 52=1*2*2*13) or 13=prime //Hint: You will need nested loops for your solution. void findFactors (int number); //****************** MAIN ***************************** int main () { int testNum; //test for the isPrime function cout << \"Enter a test number - \"; cin >> testNum; cout << endl; if (isPrime(testNum)) cout << testNum << \" is prime.\" << endl; else cout << testNum << \" is not prime.\" << endl; //test for how many primes cout << \"Enter n to print the prime numbers between 1 and n: \"; cin >> testNum; cout << endl; findPrimes(testNum); cout << endl; //test for Fibonacci number finder cout << \"Which Fibonacci number? \"; cin >> testNum; cout << endl; cout << \"The \" << testNum << \" Fibonacci number is : \" << findFibo(testNum) << endl; cout << endl; //test for prime factors cout << \"Factor what number: \"; cin >> testNum; cout << endl; findFactors(testNum); return 0; } //function isPrime //input parameter - positive integer greater than 1 //returns true if the number is prime, otherwise false // bool isPrime (int number) { int c, f = 0; if(number > 1) { for(c = 2; c < number; c++) { i.

You will be implementing the following functions. You may modify the.pdf

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When Elvis swims with the current, he swims 18km in 2 hours. Against.pdf

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What tarsal is lateral to the medial cuneiform? What tarsal is lateral to the medial cuneiform? Solution Answer: The tarsal bone lateral to the medial cuneiform is themiddle cuneiform (or the intermediate cuneiform) . There are seven tarsal bones and they are the calcaneus, talus, cuboid, navicular, and the medial, middle, and lateral cuneiforms. Among the three cuneiforms, the medial cuneiform (also known as first cuneiform) is the largest cuneiform bone. Lateral to the medial cuneiform is the middle cuneiform and its lateral surface is marked by an inverted L-shaped facet along the posterior and superior margins for the middle cuneiform bone..

What tarsal is lateral to the medial cuneiform What tarsal is .pdf

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What is the definition of memory access time? A. The difference between read and write times B, the time required to store data in the memory array C. The time required to get valid output after a valid address is applied D none of thee above Also is read and times the same on typical computers the same or is one faster? Solution opptin B Memory access time is how long it takes for a character in memory to be transferred to or from the CPU.

What is the definition of memory access timeA. The difference bet.pdf

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What are the main challenges to be dealt within a distributed operating system? 150 words expected. Solution Challenges in the distributed operating systems do not come for free. Some challenges need to be beat in order to get the perfect systems. The challenges in distributed systems are: Heterogeneity: is the variety of the distributed systems in terms of hardware, software and platforms Transparency: Distributed systems designers hide the complexity of the systems as much as they preserve. Addition of abstraction layer is particularly helpful in distributed systems Concurrency: Concurrency is a tricky challenge and then we must keep away from the systems state from becoming unstable when the users compete to view or update the data Security: Every system must consider strong security height. Distributed Systems somehow deals with responsive information.

What are the main challenges to be dealt within a distributed operat.pdf

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What networking challenges does for the Open Ethernet address and how does it improve them? What would be some of the future business demands on data networks that Open Ethernet would satisfy? Solution 1)Answer: challenges: If a networked Layer 2 device such as a bridge or LAN switch could contain a list of all known MAC addresses, then the network node could function in much the same way as a router, forwarding frames instead of packets hop-by-hop through the network from source LAN to destination LAN. However, the MAC address is much larger than the IPv4 address currently used on the Internet backbone (48 bits compared to the 32 bits of IPv4). This poses problems. Also, because the MAC address has no “network organization” like the IPv4 or IPv6 address, an Layer 2 network node must potentially store every conceivable MAC address in memory for next-hop table lookups. Instead of tables of about 125,000 entries, every Layer 2 network node would have to store millions of entries. Improvement: Wireless access points often use MAC addresses for access control. They only allow access for known devices (MAC address is unique and identifies devices) with the correct passphrase. DHCP servers use the MAC address to identify devices and give some devices fixed IP addresses. uses: Hotels, restaurants and bars are increasingly being pushed to offer ubiquitous and free WiFi – in hotel rooms and also in all common areas. Travelers and patrons, now used to having access anywhere, at any time for their multiple smart devices, demand connectivity and choose where to patronize accordingly. Ethernet-based Dedicated Internet Access can provide this feature both costeffectively and reliably. Moreover, multi-site hotel and restaurant chains operating in today’s fastpaced online business environment can upgrade and consolidate services onto a single network that supports voice, data and video traffic alike..

What networking challenges does for the Open Ethernet address and ho.pdf

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What are some experiences you have seen with browsers? What problems do you see with the current state of mobile browsers? Where do you see them going? Solution browser is tool used to access webpages over the worldwide, it provides a better interface for the user, and better way to interact with the webpages..provides good navigation among webpages browsers are user friendly, they store history of our log(visited urls) for further reference, they save authentication passwords for webpages, by saving cookies makes webpage load faster..they provide downloading facility The problems with the current mobile browsers is poor navigation between the tabs, low speed of access...it can\'t show some websites with high end graphics...sometimes make phone crash or hang.

What are some experiences you have seen with browsers What problems.pdf

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True or False 9.An indenture is a bond that is backed by good faith of the company. 10.Default risk is the risk that a company can\'t make its\' interest payments. 11.Subordinated debentures are higher order claimants (i.e. paid out earlier) compared to mortgage bond holders when there is a bankruptcy. 12. A put provision gives the investor the right but not the obligation to sell the bond back to the company. 14. The debenture statement is the legal document that establishes and characterizes a bond\'s features. Solution 9) true Between bond issuers and bondholders, anindenture is a legal and binding contract specifying all the important features of abond, such as its maturity date, timing of interest payments, method of interest calculation and callable or convertible features, if applicable 10) true Default risk is the chance that companies or individuals will be unable to make the required payments on their debt obligations. Lenders and investors are exposed to default risk in virtually all forms of credit extensions. 11) false subordinated debt (also known assubordinated loan, subordinated bond,subordinated debenture or junior debt) isdebt which ranks after other debts if a company falls into liquidation or bankruptcy 12) true A put provision is a provision in some bonds which allows the bondholder to resell a bond back to the bond\'s issuer at par or the face value of the bond before the bond matures 13) true.

True or False9.An indenture is a bond that is backed by good faith.pdf

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This is a digital forensics questions. How can a hard drive be used for potential evidence as documentary evidence. If it cannot, why? Solution Please follow the data and description : Hard Drives as Evidence : In the stream of forensics department there are many evidences that could be considered and taken in to account for the culprit to be caught. In the same way a hard drive can also be presented as an documentary evidence because it contains various records, data, information and the respective pictures showing the perpetrators behaviour patterns leading to the offence. The hard disk can act as a demonstrative evidence to retrieve the important information that offers for the department and the forensics to prove of what has actually took place over the incident or the case. It can also be used to explain the behaviour patterns of the perpetrator using charts, presentations, slides and so on. Hope this is helpful..

This is a digital forensics questions. How can a hard drive be used .pdf

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Trace the path the red blood cell would take as it travels from the stomach to the left lung. Include all chambers, valves, and vessels. You can use arrows between structures. Solution Answer: Systemic circulation: the right auricle through the superior and inferior vena cava (the great vein of the body) receives the deoxygenated blood. From the right ventricle, blood flows into the pulmonary artery and is carried to the lungs The red blood cell would take as it travels from the stomach --> deoxygenated blood from RBC of stomach ---> smaller capillaries ---> veins ---> vena cava (the superior and inferior vena cava) ---> right atrium of the heart ---> Right ventricle of heart pulmonary valve --> pulmonary trunk pulmonary arteries pulmonary arterioles --> left lung.

Trace the path the red blood cell would take as it travels from the .pdf

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/** The LinkedList1 class implements a Linked list. */ class LinkedList1 { /** The Node class stores a list element and a reference to the next node. */ private class Node { String value; Node next; /** Constructor. @param val The element to store in the node. @param n The reference to the successor node. */ Node(String val, Node n) { value = val; next = n; } /** Constructor. @param val The element to store in the node. */ Node(String val) { // Call the other (sister) constructor. this(val, null); } } private Node first; // list head private Node last; // last element in list /** Constructor. */ public LinkedList1() { first = null; last = null; } /** The isEmpty method checks to see if the list is empty. @return true if list is empty, false otherwise. */ public boolean isEmpty() { return first == null; } /** The size method returns the length of the list. @return The number of elements in the list. */ public int size() { int count = 0; Node p = first; while (p != null) { // There is an element at p count ++; p = p.next; } return count; } /** The add method adds an element to the end of the list. @param e The value to add to the end of the list. */ public void add(String e) { if (isEmpty()) { first = new Node(e); last = first; } else { // Add to end of existing list last.next = new Node(e); last = last.next; } } /** The add method adds an element at a position. @param e The element to add to the list. @param index The position at which to add the element. @exception IndexOutOfBoundsException When index is out of bounds. */ public void add(int index, String e) { if (index < 0 || index > size()) { String message = String.valueOf(index); throw new IndexOutOfBoundsException(message); } // Index is at least 0 if (index == 0) { // New element goes at beginning first = new Node(e, first); if (last == null) last = first; return; } // Set a reference pred to point to the node that // will be the predecessor of the new node Node pred = first; for (int k = 1; k <= index - 1; k++) { pred = pred.next; } // Splice in a node containing the new element pred.next = new Node(e, pred.next); // Is there a new last element ? if (pred.next.next == null) last = pred.next; } /** The toString method computes the string representation of the list. @return The string form of the list. */ public String toString() { StringBuilder strBuilder = new StringBuilder(); // Use p to walk down the linked list Node p = first; while (p != null) { strBuilder.append(p.value + \"\ \"); p = p.next; } return strBuilder.toString(); } /** The remove method removes the element at an index. @param index The index of the element to remove. @return The element removed. @exception IndexOutOfBoundsException When index is out of bounds. */ public String remove(int index) { if (index < 0 || index >= size()) { String message = String.valueOf(index); throw new IndexOutOfBoundsException(message); } String element; // The element to return if (index == 0) { // Removal of first item in the list element.

The LinkedList1 class implements a Linked list. class.pdf

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O Clock 8:55 PM 67% Signals and Systems: Analysis Using Transform Methods & MATLAB Chapters Chapter 5 Step 3 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 n20 l, n27 Problem 20 0 otherwise 0 0 otherwise 1.n24 1. n211 Problem 21 0 otherwise0otherwise Problem 22 g[n] ramp[n +1 ramp[n 3]- ramp n 6]+rampn 10] eln) Problem 23 Problem 24 Problem 25 Problem 26 15 Problem 27 I was this solution help&P; Chegg Problem 28 Share Solution %x[n] = u[n]-u[n-7]; h[n] = u[n]-u[n-4]; (Comment line) x = ones(1,6); h = ones(1,3); g = conv(x,h);.

O Clock 855 PM 67 Signals and Systems Analysis Using Transform Met.pdf

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