The following asci were obtained in Neurospora a crassa when a wild-t.pdf

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The following asci were obtained in Neurospora a crassa when a wild-type strain (thi^+ 1eu^+) was crossed to a double mutant strain that cannot grow in the absence of thiamine or leucine (thi^- leu^-). Only one member of each spore pair produced by the final mitosis is shown since the two cells in a pair have the same genotype. What is the approximate map unit distance between thi and leu? 25.0 35.0 17.5 8.3 They are unlinked Using the data set from Question 9 above, what is the approximate map unit distance between thi and the centromere? 50.0 8.3 17.5 35.0 16.7 Solution 9) Identify Paretal Ditype (PD) , Non Parental Ditype(NPD) and Tetrad PD have the same genotype as the paretal class NPD have different genotype that the paretal class Tetrad- show all the possible combinations therefore, A is PD, B is T, C is T , D is NPD and E is T Since PD=NPD, they are unlinked . Ans- e 10) identify the first division segregation and the Second division segregation. first division segregation (FDS)- no crossover therefore all the pattern same as the centromeric pattern Second division segregation (SDS)- crossover occured therefore, all the ptterns are different. From the above data , map distance between thi and centromere =1/2* (no of SDS / total no of asci )*100 SDS for thi is asci B and E- 18+2=20 therefore, map distance between thi and centromere=1/2*(20/120)*100=8.3 Ans-b.

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Let us roll a die twice. The number from the first throw is denoted by X1 and the number from the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2. * Make a table that shows all cases of Y with their probabilities. * What is the expectation value of Y, i.e. E[Y]? * Obtain the variance of Y, V[Y]. Solution Let us roll a die twice. The number from the first throw is denoted by X1 and the number from the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2. * Make a table that shows all cases of Y with their probabilities. * What is the expectation value of Y, i.e. E[Y]? * Obtain the variance of Y, V[Y]. We can use the following: Sum of two die Probability 2 1/36 3 2/36 = 1/18 4 3/36 = 1/12 5 4/36 = 1/9 6 5/36 7 6/36 = 1/6 8 5/36 9 4/36 = 1/9 10 3/36 = 1/12 11 2/36 = 1/18 12 1/36 Total 36/36 sum of dice y p y^2 y*p y^2*p 2 9 1/36 81 0.25 2.25 3 4 1/18 16 0.222222 0.888889 4 1 1/12 1 0.083333 0.083333 5 0 1/9 0 0 0 6 1 5/36 1 0.138889 0.138889 7 4 1/6 16 0.666667 2.666667 8 9 5/36 81 1.25 11.25 9 16 1/9 256 1.777778 28.44444 10 25 1/12 625 2.083333 52.08333 11 36 1/18 1296 2 72 12 49 1/36 2401 1.361111 66.69444 sums: 1 9.833333 236.5 To get the y column, subtract 5 from the sum column and then square. The second column and the third column of this table show the possible values of y and their probabilities. E(Y) and V(Y) are calculated using the sums in the last row of this table. E(Y) = 9.8333 Sum of two die Probability 2 1/36 3 2/36 = 1/18 4 3/36 = 1/12 5 4/36 = 1/9 6 5/36 7 6/36 = 1/6 8 5/36 9 4/36 = 1/9 10 3/36 = 1/12 11 2/36 = 1/18 12 1/36 Total 36/36.

Let us roll a die twice. The number from the first throw is denoted .pdf

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Is there relationship between having a high school education and knowledge of the orbit of the earth, which one is independent variable and which one is dependent variable? Solution There is relationship between having a high school education and knowledge of the orbit of the earth. While hving high school education is an independent variable, knowledge of the orbit of earth is dependent upon the former information. If one has high school education then only he can know about the orbit of the earth..

Is there relationship between having a high school education and kno.pdf

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In a survey, 600 mothers and fathers were asked about the importance.pdf

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in html 1: A Header with font size ranging from h4 to h1 with at least your name and class included. Follow the example on the printout I gave in class. 2. Two horizontal lines separating the head and body – spanning the screen and of size 5. 3. A background image chosen judiciously so that fonts will be displayed clearly. If you wish not to use an image, you could use a different background color. 4. An image to the left and one to the right, aligned with each other and enough space in between that you could add some written comments, description, etc. It should be very similar to the example shown. 5. Keep adding more comments, etc. below the images that must now stretch across the page. Somewhere in all of this writing, create a link that will take the user down somewhere near the bottom of the page. After reading at the bottom of the page, there must be another link that will take the user back to the point where he/she left. 6. Make all links blue and visited links red. 7. (a) Now create a bulleted list (unordered with square as the bulleted symbol). (b) Inside of the bulleted list, you must have a sublist that is ordered (using numbers). The font face of this sublist must be different than that of the rest of the document. (c) Inside of this sublist, create an unordered sublist with disc as the bulleted symbols. Look at the example given. 8. Create a table with the following properties: (a) A width of 80% and centered. (b) 10 rows and 6 columns (c) The background of the table must be different from rest of document. (d) The first row must consist of an image not more than 2 inches in height. (e) The label for the first column can be for example NAMES, centered, bold and take up two rows (f) The second row must span 5 columns and can have as its header TEST SCORES or WEEK DAYS or ..., centered and bold. (g) The third row must consist of 5 individual columns each with a header that is centered and bold as shown in the example. (h) Fill in the table with appropriate data as shown in the example making sure that the names are left aligned and the scores are centered but not bold. 9. After the table, follow it up with some written material then include some sort of quote as shown in the example. Make sure you use italics for the quote, different color font and for the author of the quote also use different color font. 10. Use a horizontal line to separate this part of the document from what is to follow. 11. Create a link of some web site that you like. Make sure that the link is live i.e. I can activate the link and be able to go to that web site. 12. Create a link to another page that you have created as demonstrated in the example. (A) Only the written words are active. (B) Both the image and the written words are active. (C) Only the image is active. 13. Add your email address below. Again, make sure that it is active so that I can email you. Also experiment by using an image for the email rather than words, as shown in the example Solut.

in html 1 A Header with font size ranging from h4 to h1 with at l.pdf

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Identify these monomers (general classification). Are the monomers hydrophilic, or amphipathic? Write your answer in the space above each monomer. How many functional groups can you identify? Circle and name them. Show how you would make these monomers into a polymer. Solution A and B. a.Cysteine (Cys, C) - Hydrophobic and aliphatic R group b. Phenylalanine (Phe, F) - Hydrophobic and aromatic R group C. Functional groups Cysteine - COOH - Carboxyl - NH2 - Amino - SH - Sulphydryl Phenylalanine - COOH - Carboxyl - NH2- Amino - Aromatic.

Identify these monomers (general classification). Are the monomers h.pdf

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How is the Internet come about ? How was the World Wide Web ( WWW ) developed ? How is the Internet come about ? How was the World Wide Web ( WWW ) developed ? Solution It was in late 1960s the initial concepts of internet came into exostence, with ARPANET( Advanced Research Projects Agency Networ), which was originally founded by US Department of Defence. They introduced packet switching to communicate between systems over a single network.Later on in 1970s the TCP/IP was got introduced, so the ARPANET adopted TCP/IP in 1983, and from there the network of network model came into existencce, which later known as Internet. WWW (World Wide Web) is an aspect of Internet, like email. Coomonly WWW is being misinterpreted as the so called Internet. WWW is an information space in which one can locate any kind of information through a URL(Uniform Resource Locator). It was in 1989, Tim Berners-Lee invented the Web. He developed a web browser which he was an employee at CERN in Switzerland. Lter on the commercial potential in web makes the growth of web in an exponential paced manner..

How is the Internet come about How was the World Wide Web ( WWW .pdf

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Explain the answers in sufficient details i.e show the work ace? What is the probability of drawing a red ace? Are these events (\"ace\" and \"red\") mutually exclusive? Are they independent? e. List two events that are mutually exclusive for a single draw from a deck of cards. What is the probability of drawing a red f. king? What is the probability of drawing a face card in hearts? Are these two events mutually exclusive? Are they independent? 20. The human genome is composed of the four DNA nucleotides: A, T, G, and C. Some regions of the human genome are extremely G-C rich (i.e., a high proportion of the DNA nucleotides there are guanine and cytosine). Other regions are relatively A-T rich (i.e., a high proportion of the DNA nucleotides there are adenine and thymine). Imagine that you want to compare nucleotide sequences from two regions of the Solution a) Let we have two regions, Region1 and Region2. It is given that 60% nucleotides in 1st region is A&T (30% each) and remaining 40% is of G&C nucleotides(20% each). Probability of the nucleotides in Region1: Similarly for region2, Probability of the nucleotides in Region2: Probability of getting both are same nucleiods,we must multiply the individual probabilities. Probability of getting both has type A nucleoids=P(A1)*P(A2) Probability of getting both has type T nucleoids=P(T1)*P(T2) Probability of getting both has type G nucleoids=P(G1)*P(G2) Probability of getting both has type C nucleoids=P(C1)*P(C2) Required probability=P(A1)*P(A2)+ P(T1)*P(T2)+ P(G1)*P(G2)+ P(C1)*P(C2) =0.25 b) It is given that two regions are independent. To find probability to select three nucleoids sequence (triplets) from each of the two regions are same. All combinations made with the nucleoids A, T, G and C taking three at a time 4C3. The number of all combinations of 4 things taken 3 at a time is 4C3=4 Possible selections are ATG, TGC, GCA and CAT Suppose we get outcome as ATG from Region1, so Probability of getting it isNucleotidesP (N)A10.2T10.2G10.3C10.3.

Explain the answers in sufficient details i.e show the work ace Wha.pdf

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Each figure is an isometry image of the figure at the right. Tell whe.pdf

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E. None of the above. E. Actually, they differ in all of these ways. E. Actually, all of these are accurate. E. None of the above.1.Bacteria are good experimental organisms for genetic studies for all of the following reasons EXCEPT:A. they have large numbers of progeny.B. they reproduce rapidly.C. they are diploid.D. they reproduce asexually.E. Actually, all of these are reasons bacteria are useful in genetic studies. Solution 1. Bacteria are good experimental organisms for genetic studies for all of the following reasons except they have a relatively small, diploid genome that is easier to characterize. So, the correct answer is c. 2. A plasmid is a small, circular, double-stranded DNA molecule that is distinct from a cell\'s chromosomal DNA. So, the correct answer is plasmid DNA is circular like bacterial DNA. Option A. 3. Option B (Transference). Transference is the unconscious transfer to others of feelings, and attitudes which were originally associated with important figures in one\'s early life. 4. Option C (minutes ) is the correct answer. It is a unit of meaure in baterial conjugation experiment..

E. None of the above.E. Actually, they differ in all of these ways.pdf

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Describe the difference between a. The pulmonary and systemic circulation b. An artery and a vein c. An atrium and a ventricle Solution a. Pulmonary and systemic circulation The cardiovascular system comprise of two circulatory pathsways namely pulmonary and sytemic circulation.The pulmonary circulation circuits through the lungs carrying the oxygenated blood whereas the systemic circulation circuit through the rest of the body to provide oxygenated blood to various body parts. In the pulmonary circulation, blood travels through capillaries on the alveoli, air sacs in the lungs allowing the gas exchange. Here, as blood flows, the size of the vessel decreases from artery to arteriole, from vein to venuleand finally to capillaries,which serve as vessels for gas and nutrient exchange.In the systemic circulation,the movement of blood takes place from the heart through the body providing oxygen and nutrients to the tissues of the body while bringing back the deoxygenated to the heart. Systemic and pulmonary circulation transition to the opposite to each other. Systemic circulation is a much larger and higher pressure system than pulmonary circulation. Thus systemic circulation is the larger path that blood takes on, routing oxygenated blood around the tissues of the body and deoxygenated blood from them back to the heart , while pulmonary circulation is the smaller loop that blood takes in the lungs. b. An artery and a Vein Blood is circulated in the body through arteries, veins and capillaries in a closed circulatory system. All blood vessels leading from heart carrying oxygenated blood (except pulmonary artery) to various organs of the body are called as arteries. Whereas, vessels which carry deoxygenated blood (except pulmonary vein) from various organs of the body to the heart are called as veins. Arteries have thick elastic muscular walls and valves are absent, whereas, Veins have thin non- elastic valves and valves are present to prevent bacl flow of blood. c. An atrium and a Ventricle There are 4 chambers in the heart; the atria refer to the upper chambers while the ventricles refer to the lower chambers. Atria (atrium) refers to the upper chambers of the heart that receive the impure blood from the veins to send it to the ventricles. Ventricles are small cavities or chambers that are present within the left chamber of the heart that accepts blood from the arteries (left atrium) and then contracts to force into the aorta. Both right and the left part of our heart has an atrium and one ventricle. The walls of the ventricles are thicker, while that of the atria are thinner. They ventricles contain valves to pump the blood in and out of the heart..

Describe the difference between a. The pulmonary and systemic circula.pdf

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Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is the probable phenotype ratio? Show punnet square and all work. Solution Dihybrid cross Homozygous running & heterozygous black -mouse x heterozygous black & waltzing mouse Parental gametes --> RRBb x rrbb RB Rb RB Rb rb RrBb Rrbb RrBb Rabb rb RrBb Rrbb RrBb Rrbb rb RrBb Rrbb RrBb Rrbb rb RrBb Rrbb RrBb Rrbb F1-phenotypes Running black mouse Running brown mouse.

Cross a homozygous running, heterozygous black mouse with a waltzing.pdf

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Describe some effects that cybertechnology has had so far for our sense of self. Questions to think about as you write: In the age of information how are we beginning to conceive ourselves in relation to nature? How is cybertechnology changing the way he view ourselves and others? Does cybertechnology bring us closer together or create a rift? Are humans becoming more computer-like? How? MUST BE AT LEAST 300 WORDS IN TOTAL. Solution Information technologies are in a stable level of change and break-through.We can see that information technologies have brought great social changes compared to the past years before they appeared.In the past days of disconnect,we plugged into each other\'s nervous system by passing our books,on discussions with others and by sharing our thoughts and dreams.But connectedness made one become others databases snd servers learning on eachother\'s memories.The expanded technology increased the lenghts of people\'s thoughts with this increasing world.It brought people\'s imaginations and thoughts into existence. The effects of technology has to do with whether humans desire greater simplicity or complexity as they progress. Many technology creators acclaim the value of convergence. While that might severely reduce the number of ‘devices’ that we need and use, it also adds complexity to the task of using a device.Alternately, people have written about how we desire purity, and will therefore look for technologies that will do one thing, and do it extremely well. The challenge before technology innovators, clearly, is to enable technologies to communicate with each other, and sort problems out, the way humans do.Technology will enable diverse cultures to collaborate more efficiently, in every sphere. It will bring people and organizations together, closer. Technology will allow people to live multiple lives, assume and live out multiple identities.Every day, as millions of people interact over the internet, they navigate worlds that are simulated, they create virtual personalities, and forge online relationships. Fantasy and role-play, both natural and important functions of child development, are fostered by adult chatrooms and virtual reality. In a distributed, yet isolated world, people desire connections. In doing so, they think not of identity but identity crises. Technological world is a brilliant opportunity for global connections and it gives the exprience of leading different lives and cultures.We must believe that internet brings us together as without internet it is not so easy to communicate with eachother.But on the other hand, it does have some negative effects on people, tearing them apart in way of face to face communication making such relations undervalued. Computers, or operating systems to be exact, are lacking in the the fundamental capabilities possessed by all sentient systems: the ability to reason, to learn from observation and experience, to grow intelectually over time. As of today, computers .

Describe some effects that cybertechnology has had so far for our se.pdf

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Choose a deployment platform that allows for the implementation of SOA. Describe the architectural layers that are included in the platform and analyze the flexibility of the platform for supporting different types of services. Determine whether this platform supports loose coupling or if it is strongly implementation dependent. Choose a deployment platform that allows for the implementation of SOA. Describe the architectural layers that are included in the platform and analyze the flexibility of the platform for supporting different types of services. Determine whether this platform supports loose coupling or if it is strongly implementation dependent. Solution Service-oriented architectures (SOA) are based on the software services, which are high-level software components that include web services like (Rest , Soap). Implementation of an SOA requires tools and run-time infrastructure software. This is collectively referred to as a service- oriented architecture implementation framework or (SOAIF). Application development and deployment : In the traditional software development process, translating the requirements into working distributed systems is time-consuming and difficult, requiring several stages of manual development and then its deployment. This complex, error-friendlytask can be effectively done using a higher-level, component-based SOAIF. The SOAIF incorporates tools that let processes that are developed, using standards such as Business Process Execution Language (BPEL), can be easily translated into distributed, high-level services, which are easy to develop, manipulate, and debug and hence more effecient . These services are easily composed of implementation- level data flows without the user or developer having to track complex middleware concepts, such as topics or queues. Further, the implementation-level services can run on any machine across the network due to the built-in dynamic deployment support SOAIF provides. Architectural Layers Enterprise service bus :core infrastructure of an SOAIF is typically provided by an enterprise service bus (ESB), which points out the challenges in composing, deploying, and managing distributed, service-based enterprise applications. The ESB incorporates systems connectivity using web services, Java EE, .NET Framework, and other standards also ESB also do not provide any guaranty of loose coupling to meet evolving Service-oriented needs . Implementing ESB to meet SOA requirements require the addition of extra functionality to compose for business Services and provide secure Service interactions. Business process management : Business process management (BPM) considers systems and IT assets as activities or tasks that participate in well-coordinated and centrally organised Business processes. Traditionally, the challenge of BPM is that while it is possible to construct any processes that achieve integration goals, enterprises typically use BPM tools only at the time of designing , modeling.

Choose a deployment platform that allows for the implementation o.pdf

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Abstract Base Class (C++ Program) Create an abstract base class called Animal which has the following pure virtual member functions: talk( ), move( ) and the following attribute: animalType (which is a pointer to a dynamically allocated string.) Then create the following derived classes (from the animal class): reptile, bird, and mammal. Here are the concrete classes: From the reptile class derive lizard and snake. From bird derive eagle and chicken. From mammal derive bear, hyena and lion. Each concrete class needs to override the talk() and move() functions. Also each concrete class will have the operator<<( ) defined. The output of cout << smokey; (assuming smokey is a bear) will be the animalType , the sound he makes (talk( ) ) and how he moves ( move( )). Here is an example: int main() { bear smokey; // using the concrete class cout << smokey // prints out: bear, growl, walk animal *aptr = new bear; // I could have used an animal pointer cout << *aptr; // prints the same results: bear, growl, walk return 0; } Create an array of animal pointers called zoo. Make it 10 in size. Here is what the menu should look like: ----Zoo Cages---- 1 [empty] 2 [empty] 3 [empty] 4 [snake] 5 [eagle] 6 [bear] 7 [empty] 8 [empty] 9 [lion] 10 [hyena] ----Menu---- 1) Add 2) Remove 3) Display all animals 4) Display animal count 5) Exit If the user selects 1) then ask for a cage number and the animal he wishes to put in that cage. Be sure to check the cage for an existing animal first. The user cannot put an animal in a cage that is already occupied. If the user selects 2) then ask for the cage number and remove the animal from that cage (use the destructor). If there is no animal in the cage, issue an error to the user. If the user selects 3) then display the contents of each cage. The display function will show each cage number as well as the animalType of the animal, the sound it makes and how it moves, in other words, use the operator<<( ). All this information must be on a single screen. The output screen is up to you. If the user selects 4) then display the count of the number of animals in your zoo. In the above example, you will have 5. You should use a static data member in the animal class to keep track of the number of animals. You should increase that count each time the animal constructor is called and decrease the count when the animal destructor is called. ***Use Virtual Functions and the animal base class MUST be abstract Solution main.cpp #include //input output #include //strings #include //Randoms #include //Randoms #include //For input #include \"lizard.h\" #include \"snake.h\" #include \"chicken.h\" #include \"eagle.h\" #include \"hyena.h\" #include \"bear.h\" #include \"lion.h\" using namespace std; //declare our static variable from the animal class int animal::numAnimals; int main() { //array of animal pointers animal *zoo[10] = { NULL }; //initialize our static variable to zero animal::numAnimals = 0; //To test if cage is empty, it is nu.

Abstract Base Class (C++ Program)Create an abstract base class cal.pdf

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A range of values estimated to have a high probably of containing the population value is called: Point estimate Standard error of the mean Confidence interval Standard deviation Point estimate Standard error of the mean Confidence interval Standard deviation Solution It is OPTION C: CONFIDENCE INTERVAL. [ANSWER] *********** The confidence interval is the range of values in which the true population parameter value is there..

A range of values estimated to have a high probably of containing th.pdf

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A thrown lump of clay hits a block in mid air and sticks to it. what happens to the total kinetic energy of the ball+clay system during this inetraction, ignoring gravity? Solution As the lump of clay sticks to th block, it is an example of inelasti collision. During the inelastic collision, the momentum will be conserved for ball+clay system but Kinetic energy will not. Total Kinetic Energy of ball+clay system will decrease during the interaction..

A thrown lump of clay hits a block in mid air and sticks to it. what.pdf

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Answer die following questions with short answers: Explain the difference between stream ciphers and block cipher. Explain the difference between passive and active attacks. Describe the difference between authentication and non-repudiation Explain the difference between a MAC and message digest. Give one advantage and one disadvantage of public-key cryptography over symmetric-key cryptogra Why is CBC mode considered preferable to ECB mode? Solution a) stream cipher and block cipher differ in whitch the plain text is encoded and decoded. the basic difference is how the logic used for the encryption and decryption. the idea of block cipher is partition the plain text into relatively larger blocks and further each block will be encoded. and same key is used for both encryption and decryption.example DES,IDEA the idea of stream cipher is plain text is encrypted and transfered bit by bit.but key property of stream cipher is different key is generated for each block or bit.example -ISAAC,FISH b) passive attack is indirect attack.the attacked host is completely unaware about the attack. hence it is called passive attack like some body listens the packet transformation or observing the host in network. active attack is direct attack.its nothing but the attacked one gets aware of the attack. like somebody listens the id and password and stop the services(denial of service) or logic bomb. both are harmful. c) authentication and non repudiation are two different things. authentication verifies the who you are(user-id).where the non-repudiation verifies the what you did(messages). authentication is done through a central instance where non repudiation can done by anyone d) message digest is a hash of a message.its output of a cryptographic hash function appilied to input data it refered to as message. mac is a piece of information that proves the integrity of a message. message digest is take only single input and produces a message digest(aka hash) if the message change it produce different hash code. mac (message authentication code) is take input and security code to check the integrity of message. HMAC is one of this kind. e) advantage of public key cryptography over the symmetric key cryptography is increased security and convenience. and no need of key distribution(transmitted to anyone).and it provide digital signatures that cannot be repudiated. the disadvantage of public key cryptography over symmetric cryptography is speed.they are many symmetic key methods runs faster than the public key cryptography. f) the reason why cbc is considered better than ecb has nothing to do with situations involving an attacker with partial cipher text. problem with ECB is that it leaks the information like when two blocks have same plain text then the corresponding cipher blocks also be same where in case of the CBC it randomly independent of plain text to produce the cipher text..

Answer die following questions with short answers Explain the d.pdf

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A single breeding pair of rabbits is introduced to Australia in 1900. Assume (for simplicity) that each pair can produce 4 offspring each year (2 male, 2 female). Also assume, that all births happen in the last hour of the last day each year and the parents die immediately after offspring are born. a) How many rabbits are there after 10 years? b) How many rabbits are there after 25 years? If they have 20 offspring each year, c) How many rabbits are there after 10 years? d) How many rabbits are there after 25 years? Solution Each pair can produce 4 offsprings each year and parents die after giving birth to offsprings. So, offsprings makes the population and parents don\'t add to population. After first year total population is 4. Next year these 4 rabbits make 2 pairs and give birth to 4 off springs each. Second year population becomes 8, multiplies by 2 from last year. This continues till 10th year and then till 25th year. At the end of 10th year, the population of rabbits will be 2048 and at the end of 25th year the population of rabbits will be 33554432. Now when each year 20 offsprings are produced, after the end of first year total population is 20. At the end of second year, 20 rabbits make 10 pairs and give birth to 20 rabbits each taking total population to 10*20=200. At the end of 3rd year 200 rabbits make 100 pairs and give birth to 20 each taking total population to 100*20=2000. In this way at the end of each year population gets multiplied by 10.Therefore, at the end of 10 year total population will be 2*1010 and at the end of 25 years total rabbit population is will be 2*1025..

A single breeding pair of rabbits is introduced to Australia in 1900.pdf

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Write an application that counts the total number of spaces contained in a movie quote entered by the user. Save the file as CountMovieSpaces2.java Solution CountMovieSpaces2.java import java.util.Scanner; public class CountMovieSpaces2 { public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.println(\"Enter the movie quote: \"); String quote = scan.nextLine(); int countSpaces = 0; for(int i=0; i.

Write an application that counts the total number of spaces containe.pdf

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Write a java program using Dialog boxes that prompt the user to input five floating point decimal numbers. the program should then add the five numbers, find the average, convert sum and average to the nearest integer, and print the numbers, sum and average. Solution The java program using Dialog boxes that prompt the user to input five floating point decimal numbers :- import java.lang.*; import javax.swing.JOptionPane; class Example { public static void main (String[ ] args) { String value1,value2,value3,value4,value5 ; float f1,f2,f3,f4,f5 ; float sum ; float Average ; // Read input String from dialog box value1 = JOptionPane.showInputDialog(\"Enter the floating point decimal number\"); f1 = Float.parseFloat(value1); // Convert String to float value2 = JOptionPane.showInputDialog(\"Enter the floating point decimal number\"); f2 = Float.parseFloat(value2); // Convert String to float value3 = JOptionPane.showInputDialog(\"Enter the floating point decimal number\"); f3 = Float.parseFloat(value3); // Convert String to float value4 = JOptionPane.showInputDialog(\"Enter the floating point decimal number\"); f4 = Float.parseFloat(value4); // Convert String to float value5 = JOptionPane.showInputDialog(\"Enter the floating point decimal number\"); f5 = Float.parseFloat(value5); // Convert String to float sum = f1+f2+f3+f4+f5 ; Average = sum/5 ; System.out.println(f1) ; System.out.println(f2) ; System.out.println(f3) ; System.out.println(f4) ; System.out.println(f5) ; System.out.println(\"Math.round(\" + sum + \")=\" + Math.round(sum)); System.out.println(\"Math.round(\" + Average + \")=\" + Math.round(Average)); } }.

Write a java program using Dialog boxes that prompt the user to inpu.pdf

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