A sun-like star evolves off the main sequence onto the giant branch .pdf

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A sun-like star evolves off the main sequence onto the giant branch once 10% of its hydrogen has fused into helium. calculate the total amount of mass destroyed in the sun by the time it leaves the main sequence. Solution A: The Sun has a mass of 2x10^30 kg, ten percent of this amounts to 2x10^29 kg of hydrogen. When this hydrogen fuses into helium a fraction of 0.7% of the mass is converted to energy. This amounts to 1.4 x x10^27 kg.

$A sun-like star evolves off the main sequence onto the giant branch once 10% of its hydrogen has fused into helium. calculate the total amount of mass destroyed in the sun by the time it leaves the main sequence. Solution A: The Sun has a mass of 2x10^30 kg, ten percent of this amounts to 2x10^29 kg of hydrogen. When this hydrogen fuses into helium a fraction of 0.7% of the mass is converted to energy. This amounts to 1.4 x x10^27 kg$

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Frequency .java /** * Word frequency counter */ package frequency; import java.util.Iterator; /** * * @author UMD CS */ public class Frequency implements Iterable{ private Node first; private int N; Frequency(){ N = 0; first = null; } @Override public Iterator iterator() { return new ListIterator(); } /** * * List iterator * */ private class ListIterator implements Iterator{ private Node current; private int index ; ListIterator(){ current = first; index = 0; } @Override public boolean hasNext() { return current != null; } public String next() { if(!hasNext()){ return null; } String word = current.key; int count = current.count; String r = \"(\"+word + \",\" + Integer.toString(count)+\")\"; current = current.next; return r; } @Override public void remove() { } } /** * * Node class * */ private class Node { private String key; private int count; private Node next; Node(String item){ key = item; count = 1; next = null; } @Override public String toString(){ return \"(\"+key +\",\"+count+\")\"; } } /* * Inserts a word into the linked list. If the word exists, increment the * count by q. */ public void insert(String word){ if(word.equals(\"\")){ return; } //TODO /* * implement here */ } /** * * @param str input string * This method splits a string into words and pass the words to insert method * */ public void insertWords(String str){ String delims = \"[ .,?!\'\\\"()}{;/<>&=#-:\\\\ _]+\"; String[] words = str.split(delims); for(String s: words){ s = s.toLowerCase(); insert(s); } } /** * prints the word frequency list */ public void print(){ Node c = first; while(c != null){ System.out.print(\"(\"+c.key + \",\" + c.count+\")\"); c = c.next; } System.out.print(\"\ \"); } } -------------------------------------------------------------------------------------------------------------------- ------------------------------------------------------ WordFrequency.java package frequency; import utils.In; import utils.Stopwatch; /** * * @author UMD CS */ public class WordFrequency { /** * @param input source * @param * This method receives input resource and return the word frequency string * DO NOT CHANGE THIS METHOD. */ public static String freq(String inputSource, int maxLines){ In in; // Frequency class can only count the frequency of strings. DO NOT CHANGE THIS. Frequency freq = new Frequency(); int MAX = 100; String inputSourceName = inputSource; try { in = new In(inputSourceName); while (!in.isEmpty()) { String s = in.readLine(); //System.out.println(s); freq.insertWords(s); } } catch (Exception e) { e.printStackTrace(); } StringBuilder strFrequency = new StringBuilder(); int cnt = 0; for(String s: freq){ strFrequency.append(s); strFrequency.append(\",\"); cnt++; if(cnt >= maxLines){break;} } return strFrequency.toString(); } /** * */ public static void main(String[] args) { In in; Frequency freq = new Frequency(); int MAX = 100; // String inputSourceName = \"http://www.cs.umd.edu/class/summer2015/cmsc132/projects/P3_WordFrequency/test1.html\"; String .

Frequency .java Word frequency counter package frequ.pdf

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Suppose the continuous random variable X has probability density function f(x) = x^2/3 for -2 LT xLT 1, 0 else where. Find P{-2 LTX LT -1). Find E(X). Find V(X). Find the cumulative distribution function (cdf) of X, F(x). Be sure to define the cdf for all x (-infinity, infinity). Show F(x) using clear notation. The most common mistake for the cdf is forgetting the lower bound of the support set, which is -2 here. Solution.

Suppose the continuous random variable X has probability density func.pdf

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Some bacteria have one or more flagella. What is true about these structures? 1. Bacteria use flagella to move around in their environment 2. Bacteria use flagella to attach themselves to another bacterial cell 3. Bacterial flagella are molecularly similar to those found in eukaryotes and archaea 4. Bacterial flagella are built from multiple proteins that in the past had different functions 5. Flagella are always found concentrated on one part of the cella) Only 1 Solution Answer -0nly 1 Bacteria use flagella for locomotion Flagella madeup of only one protein.

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our company has an opening for a junior database administrator. As a senior database administrator, your manager has developed the following questions to ask potential new hires and he wants you to provide the answers to the following questions so they can be forwarded to Human Resources: Explain Data Manipulation Language (DML). What is the difference between Data Definition Language (DDL) and Data Control Languages (DCL)? Provide an example of an SQL query that creates and executes four individual tables; Student, Class, InstructorID, and Room Number with not null and default constraints. Solution Data Manupulation Language(DML): -->It permits the users to manipulate data in database. -->manupulation involves in inserting data into database tables,retrieving existing data, deleting data from eisting tables and modifying existing data. -->It is mostly incorporated in SQL databases. -->SELECT,UPDATE,INSERT INTO,DELETE commands comes under DML. ------------------------------------------------------------------------------------------------------ DDL: Data definition language -->It is used to define the database structure or schema. 1)CREATE-It is used to create objects in database 2)ALTER - It is used to alters the structure of the database 3)DROP - It is used to delete objects from the database 4)TRUNCATE -It is used to remove all records from a table, including all spaces allocated for the records are removed 5)RENAME - It is used to rename an object DCL:data control language -->It is used to control the data in database 1)GRANT - It gives users access privileges to database. 2)REVOKE - It withdraw access privileges given with the GRANT command ---------------------------------------------------------------------------------------------------------- 1)CREATE TABLE Student{ Student_name varchar(50) NOT NULL DEFAULT \'Mouni\', Student_Id int NOT NULL DEFAULT 10 } CREATE TABLE Class{ Class_name varchar(15) NOT NULL DEFAULT \'CSE\' InstructorId int NOT NULL DEFAULT \'100\' } CREATE TABLE InstructorID{ InstructorID int NOT NULL DEFAULT \'100\' Instructor_name varchar(15) NOT NULL DEFAULT \'Thomson\' } CREATE TABLE Room{ Room_number int NOT NULL DEFAULT \'12A\' Room_naem varchar(10) DEFAULT \'omega\' }.

our company has an opening for a junior database administrator. As a.pdf

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Part A: Briefly, explain the current hypotheses about how brown algae, red algae, chlorophytes and charophytes (couple of groups of green algae), and land plants are related. What are the characteristics of the land plants that distinguish it from the most closely related algal group, and (if applicable) how do each of these characteristics benefit plants in a terrestrial environment? Part B: What are the differences between plant cells, animal cells, and bacterial cells. And what are the characteristics of different plant cell types. Solution (1)Evolution of plants has facilitated the survival of life on earth and these land plants has their monophyletic lineage embedded with that of green algae and these green algae are believed to be the oldest eukaryotic lineages. Hypothesis states that land plants are related to brown algae, red algae, chlorophytes and charophytes. There are so many evidences to prove that these land plants share similar characteristics with that of brown algae, red algae, chlorophytes and charophytes. Plants and algae are eukaryotic and both posses the chlorophyll, which allows them to make their own food, both plants and algae have similar cell walls, stores energy in the form of starch, follows two stage life cycle. Sporopollenin, chlorophyll a, cellulose, chlorophyll b are common to both charophytes and land plants, but plants has lignin and is absent in case of charophytes. Some green algae exhibit alternation of generations like plants. All land plants exhibit alternation of generations. But no charophytes exhibit alternation of generations. Charophytes are very similar to plants as they both have homologous chloroplasts (chlorophyll b, beta carotene is present, thylakoid are arranged as grana, DNA as genetic material), with respect to biochemical similarity, both has cellulose cell wall, similar enzymes in peroxisomes, when it comes to mitosis and cytokinesis, follow similar mechanisms like disappearance of nuclear envelope and till cytokinesis the spindle will remain. Based on the above features the Charophytes are considered as the closest living algal relatives of land plants Most of the varieties of algal species live in extremely damp terrestrial environment, whereas most of the plants live on land and some plants can live in water also, but algae cannot survive on dry soil and they are very adapted to it. In case of alga, the surrounding water supports it and the whole alga is involved in helping it in the process of photosynthesis, absorption of water and minerals, holdfast will provides the anchoring. But in case of plants, each and every organ is specialized for different functions, like leaves for photosynthesis, cuticle to prevent water loss, stomata for gaseous exchange, stem to support, root to provide anchorage. But alternation of generation is absent in charophytes, but it is found in all plants providing them extra benefit to survive. 2) Animal Cell Plant cell Bacteria These are eukaryotes These ar.

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Please show the whole code... I\'m very confused. Matlab code for earth orbit Using a software package such as MATLAB, create a plot of an Earth orbit with a = 31, 800 km and e = 0.6 for 0 lessthanorequalto theta lessthanorequalto 2 pi. The origin of the plot should be Earth. Perigee should be in the direction of the positive x axis (therta = 0). On this plot, you should label (by hand) the semimajor axis, semiminor axis, radius of perigee, radius of apogee, and the semiparameter (semilatus rectum) of the ellipse. Solution --------------------------------------------------------------------------------------------------------------------- ------------------------------------------------- Elliptic orbit: %periapsis rp = 500e3 + getradius(\'earth\'); %eccentricity ec = .7; %semimajor axis a = getsemimajoraxis(rp,ec); %apoapsis ra = 2*a-rp; %true anomaly theta = [0:.1:360]; orbit = ellipse(theta,a,ec,\'earth\'); figure(1), plot(orbit.x,orbit.y); axis equal Re = getradius(\'earth\'); x = Re*cosd(theta); y = Re*sind(theta); hold on plot(x,y,\'color\',[0 1 1]) hold off figure(2), plot(theta,orbit.V/1000) xlabel(\'\\theta\'), ylabel(\'V (km/s)\') %current time c = clock; %reset hour and second c(4) = 12; %hour c(5) = 0; %minute c(6) = orbit.tau; %seconds %ellapsed time datetime(c) %returns 14-Feb-2015 21:34:54, so it\'s about 9.5 hours later still on %Monday 2 ans = 22-Feb-2015 21:34:54 -----------Elliptic orbit %periapsis rp = 500e3 + getradius(\'earth\'); %eccentricity ec = .7; %semimajor axis a = getsemimajoraxis(rp,ec); %apoapsis ra = 2*a-rp; %true anomaly theta = [0:.1:360]; orbit = ellipse(theta,a,ec,\'earth\'); figure(1), plot(orbit.x,orbit.y); axis equal Re = getradius(\'earth\'); x = Re*cosd(theta); y = Re*sind(theta); hold on plot(x,y,\'color\',[0 1 1]) hold off figure(2), plot(theta,orbit.V/1000) xlabel(\'\\theta\'), ylabel(\'V (km/s)\') %current time c = clock; %reset hour and second c(4) = 12; %hour c(5) = 0; %minute c(6) = orbit.tau; %seconds %ellapsed time datetime(c) %returns 14-Feb-2015 21:34:54, so it\'s about 9.5 hours later still on %Monday 2 ans = 22-Feb-2015 21:34:54 3 The circular orbits at perigee and apogee are 7.6167 and 3.1997 km/s, respectively. -----Parabolic orbit: Sphere of influence is actually a chapter 5 concept, but is the distance at which the gravitational forces between two objects is equal. Thus, While an approximation, if we just estimate this as the point between the sun and the earth where these forces balance, then if the orbital radius of Earth is , then , and thus solution for r_E in gives r_soi = getsphereofinfluence(\'earth\',\'sun\'); display([\'Radius from center of earth to sphere of influence is approximately \' num2str(r_soi)]); %perifocal distance rp = 500e3 + getradius(\'earth\'); %eccentricty ec = 1; %always so for a a parabola theta = -180:1:180; orbit = parabola(theta,rp,\'earth\'); figure(3), plot(orbit.x,orbit.y); axis equal Re = getradius(\'earth\'); x = Re*cosd(theta); y = Re*sind(theta); .

Please show the whole code... Im very confused. Matlab code for ea.pdf

Please show the whole code... Im very confused. Matlab code for ea.pdf

Please show the whole code... Im very confused. Matlab code for ea.pdf

Monomeric single pass trans-membrane proteins generally span the bilayer membrane with an alphahelical domain. Which of the three sequences below is most likely to be a trans-membrane segment. A. I T L I Y F G N M S S V T Q T I L L I S B. I L L I F F G V M A L V I V V I L L I A C. L L K K F F R D M A A V H E T I L E E S EXPLAIN YOUR ANSWER DO NOT JUST ANSWER THE QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Solution A Trans-membrane Protein (TP) is a type of integral membrane protein. A Trans-membrane Protein is called so because the polypeptide chain is exposed on both sides of the biological membrane. It spans the entire biological membrane or plasma membrane and is attached to it permanently. They function as gateways to permit the transport of specific substances across the plasma membrane. For this, these proteins (TP) undergo a conformational change that is a temporary change in their structure. Trans-membrane Segment: It usually denotes a single trans membrane alpha helix of a trans membrane protein, also known as an integral protein. This segment is thermodynamically stable in the membrane. We all know that the secondary structure of proteins refers to the spirally coiled polypeptide chain called alpha helical form or alpha helix. Of the three 20 amino acid trans membrane sequences; Sequence B is most likely to be a trans- membrane segment because it is composed primarily of hydrophobic amino acids, therefore can be stably integrated into lipid bilayer. Sequence A: It contains many polar amino acids therefore it cannot be integrated into the hydrophobic interior of lipid bilayer of plasma membrane. Sequence C: It contains many charged amino acids which again is disfavored into the hydrophobic interior of lipid bilayer..

Monomeric single pass trans-membrane proteins generally span the bil.pdf

Monomeric single pass trans-membrane proteins generally span the bil.pdf

Monomeric single pass trans-membrane proteins generally span the bil.pdf

Match the following to one or more layers of the Internet protocol stack: Provide user services such as e-mail and web browsing Ensure reliable transmission of packets Solution Ans: b - Transport Layer Transport layer: Responsible for reliable transmission of data and service specification between hosts. The major responsibility of this layer is data integrity--that data transmitted between hosts is reliable and timely..

Match the following to one or more layers of the Internet protocol st.pdf

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Kent and Emily have a baby boy Dustin by in vitro fertilization. Dustin does not resemble either one of his parents, so they suspect that the clinic made a mistake during the procedure and implanted the wrong embryo. At Dustin\'s first well-baby checkup they insist that they get his blood type. Baby Dustin is type B-negative. Kent is A-positive and Emily is AB-positive. Given this information answer the following questions. (a) What are Dustin\'s possible genotypes? __________ (b) What are Kent\'s possible genotypes? __________ (c) What are Emily\'s possible genotypes? __________ (d) Is it possible that Dustin is Kent and Emily\'s son? __________ (e) Explain your answer in (d) __________ Carrie is AB-positive and her father is AB-negative. Carrie marries Ryan, who is AB-negative. (f) What is Carrie\'s genotype? __________ (g)What is Ryan\'s genotype? __________ (h) What are the phenotypic ratios of blood types (for both the ABO gene and Rh gene) that would be expected among their children? __________ (Show your work in a Punnett square below!) Solution Answer: a). Dustin’s possible genotypes: IA IB : Rh+ Rh+ or IA IB : Rh+ Rh- b). Kent’s possible genotypes: IA IA : Rh+ Rh+ or IA IA : Rh+ Rh- or IA i : Rh+ Rh+ or IA i : Rh+ Rh- c). Emily’s possible genotypes: IB IB : Rh- Rh- or IB i : Rh- Rh- d). Yes Dustin belongs to Kenta and Emily’s son. e). The gamete, IA: Rh+ of Kent’s and the gamete, IB: Rh- of Emily’s fuse to produce IA IB : Rh+ Rh- (AB-positive). Rh+ is dominant over Rh- IA & IB are codominant. f). Carrie’s genotype (AB-positive): IAIB: Rh+Rh- g). Ryan’s genotype (AB-negative): IAIB: Rh-Rh- IAIB: Rh+Rh- x IAIB: Rh-Rh- IARh- IBRh- IARh+ IAIA Rh+Rh- (A+) IAIBRh+Rh- \\ (AB+) IA Rh- IAIA Rh-Rh- (A-) IAIB Rh-Rh- (AB-) IARh- IBRh- IARh+ IAIA Rh+Rh- (A+) IAIBRh+Rh- \\ (AB+) IA Rh- IAIA Rh-Rh- (A-) IAIB Rh-Rh- (AB-).

Kent and Emily have a baby boy Dustin by in vitro fertilization. Dust.pdf

Kent and Emily have a baby boy Dustin by in vitro fertilization. Dust.pdf

Kent and Emily have a baby boy Dustin by in vitro fertilization. Dust.pdf

Isabel Lopez started Biz Consulting, a new business, and completed the following transactions during its first year of operations a. I. Lopez invests $69,000 cash and office equipment valued at $37,000 in the company in exchange for its common stock b. The company purchased a $309,000 building to use as an office. Biz paid $41,000 in cash and signed a note payable promising to pay the $268,000 balance over the next ten years. c. The company purchased office equipment for $5,400 cash. d. The company purchased $3,500 of office supplies and $1,700 of office equipment on credit. e. The company paid a local newspaper $960 cash for printing an announcement of the office\'s opening. f. The company completed a financial plan for a client and billed that client $4,200 for the service g. The company designed a financial plan for another client and immediately collected an $8.100 cash fe h. The company paid $1,600 cash in dividends to the owner . The company received $3.200 cash as partial payment from the client described in transaction f payment of $850 cash on the equipment purchased in transaction d k. The company paid $2.400 cash for the office secretary\'s wages for this period. Required: 2. Enter the amount of each transaction on individual items of the accounting equation. (Enter reductions to account balances with a minus sign.) Assets Office Office Accounts Solution Answer Assets Liabilities Equity Cash Accounts receivables Office Supplies Office Equip ments Buildings Accounts Payable Notes Payable Common Stock Dividends Revenues Expenses a 69000 37000 106000 b -41000 309000 268000 28000 0 0 37000 309000 0 268000 106000 0 0 0 c -5400 5400 22600 0 0 42400 309000 0 268000 106000 0 0 0 d 3500 1700 5200 22600 0 3500 44100 309000 5200 268000 106000 0 0 0 e -960 960 21640 0 3500 44100 309000 5200 268000 106000 0 0 960 f 4200 4200 21640 4200 3500 44100 309000 5200 268000 106000 0 4200 960 g 8100 8100 29740 4200 3500 44100 309000 5200 268000 106000 0 12300 960 h -1600 1600 28140 4200 3500 44100 309000 5200 268000 106000 1600 12300 960 i 3200 -3200 31340 1000 3500 44100 309000 5200 268000 106000 1600 12300 960 j -850 -850 30490 1000 3500 44100 309000 4350 268000 106000 1600 12300 960 k -2400 2400 Bal 28090 1000 3500 44100 309000 4350 268000 106000 1600 12300 3360 Assets Liabilities Equity Cash Accounts receivables Office Supplies Office Equip ments Buildings Accounts Payable Notes Payable Common Stock Dividends Revenues Expenses a 69000 37000 106000 b -41000 309000 268000 28000 0 0 37000 309000 0 268000 106000 0 0 0 c -5400 5400 22600 0 0 42400 309000 0 268000 106000 0 0 0 d 3500 1700 5200 22600 0 3500 44100 309000 5200 268000 106000 0 0 0 e -960 960 21640 0 3500 44100 309000 5200 268000 106000 0 0 960 f 4200 4200 21640 4200 3500 44100 309000 5200 268000 106000 0 4200 960 g 8100 8100 29740 4200 3500 44100 309000 5200 268000 106000 0 12300 960 h -1600 1600 28140 4200 3500 44100 309000 5200 268000 106000 1600 12300 960 i 3200 -3200 31340 1000 3500 44100 .

Isabel Lopez started Biz Consulting, a new business, and completed th.pdf

Isabel Lopez started Biz Consulting, a new business, and completed th.pdf

Isabel Lopez started Biz Consulting, a new business, and completed th.pdf

Indicate all persons/investigators participating in the Program and explain the role of each participant in the Program. To test the mir-22 function of cancer patients (blood sampling). Note that the type of study is basic research Solution Answer: The participants/investigators participating in a program includes the PI (Principal Investigator), Co-PI (Co-Principal Investigator), faculty participants and human participants. Role of each of them have been discussed below: 1. PI-The primary person to conduct a public service project or sponsored project in accordance with the laws and regulations of the institution concerned. 2. Co-PI- They have responsibilities similar to the PI and has obligations to monitor as to whether the projects are conducted in compliance with the rules and regulations. 3. Faculty participant- They are not directly involved in the research projects, but may act as a key personnel with lesser responsibilities than the PI and Co-PI. 4. Human participant- These are individuals upon whom the investigators conduct research to obtain data either through direct intervention or obtaining individual information about the participant which are not to be made public. Before collecting blood for sampling to study mir-22 function in cancer patients, the human participants should be fully informed (by the investigators) about the facts, benefits, risks and discomforts they might face during the program. These factors are part of the ethical and legal aspects as these might change the participants willingness to participate in the program. The research protocols includes diet, survey, experimental drug intervention and exercise studies. For children (below 18 years), prior consent of the parents are standard requirements. Consents in most cases are to be documented, although it might be waived in special cases..

Indicate all personsinvestigators participating in the Program and .pdf

Indicate all personsinvestigators participating in the Program and .pdf

Indicate all personsinvestigators participating in the Program and .pdf

In the context of materials science & engineering, what is the difference between Curie temperature and Debye temperature? Solution Curie temperature (TC), or Curie point, is the temperature at which certain materials lose theirpermanent magnetic properties, to be replaced by induced magnetism. The Curie temperature is named after Pierre Curie, who showed that magnetism was lost at a critical temperature. Debye temperature is thetemperature of a crystal\'s highest normal mode of vibration, i.e., the highest temperature that can be achieved due to a single normal vibration.

In the context of materials science & engineering, what is the diffe.pdf

In the context of materials science & engineering, what is the diffe.pdf

In the context of materials science & engineering, what is the diffe.pdf

How does the shape of cellulose make possible its function? Note the arrangement of individual cellulose molecules and bonds as part of your answer and mention something about both covalent and noncovalent bonds. If you really want to see the H bonds, right click on the jsmol figure, choose \"style\" \"hydrogen bonds\" and then set their diameter to 30 Angstroms. You can also choose \"color\" and make them some other color to make them easier to see! Solution Cellulose is the most abundant organic compound on earth. It is a major component of tough cell walls that surround plant cells, and is what makes plant stems, leaves, and branches so strong. Its unique, rigid structure serves an important function in plants. Its molecules are joined together using hydrogen bonds and are arranged parallel to one another, forming a long, cable-like structure that merges with other cellulose molecules. This creates the strong support structure that allows plants to stand upright. Note: Image is missing in the question.

How does the shape of cellulose make possible its function Note the .pdf

How does the shape of cellulose make possible its function Note the .pdf

How does the shape of cellulose make possible its function Note the .pdf

22. Mrs. Jones gave birth prematurely to her first child. At birth, the baby weighed 2 lb 8 oz. Within a few hours, the baby had developed severe dyspnea and was becoming cyanotic. Therapy with a positive pressure ventilator was prescribed. Answer the following questions related to the situation just described. Place your responses in the answer blanks. 1. The infant\'s condition is referred to as infant Res 2. It occurs because of a relative lack of 3. The function of the deficient substance is to 4. Explain what the positive pressure apparatus accomplishes. Solution 1. The condition is referred to as infant respiratory distress syndrome (IRDS) also called neonatal respiratory distress syndrome. 2. It occurs due to relative lack of pulmonary surfactant. It is a developmental insufficiency and can also be caused due to structural immaturity in lungs. 3. The function of deficient substance (Surfactant) is to prevent collapse of terminal air spaces during the normal cycle of inhalation and exhalation. 4. Positive pressure keeps the alveoli open at the end of expiration and used to treat IRDS. It decreases the requirement of endotracheal ventilation in infants ..

22. Mrs. Jones gave birth prematurely to her first child. At birth, t.pdf

22. Mrs. Jones gave birth prematurely to her first child. At birth, t.pdf

22. Mrs. Jones gave birth prematurely to her first child. At birth, t.pdf

First year students: Mean 125, SD?, size 350 Third year students: Mean 112, SD?, size 150 Significance level = 0.02 Solution Let group 1 = first year group 2 = third year Formulating the hypotheses Ho: p1^ - p2^ >= 0 Ha: p1^ - p2^ < 0 Here, we see that pdo = 0 , the hypothesized population proportion difference. Getting p1^ and p2^, p1^ = x1/n1 = 0.357142857 p2 = x2/n2 = 0.746666667 Also, the standard error of the difference is sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.043783715 Thus, z = [p1 - p2 - pdo]/sd = -8.896545395 As significance level = 0.02 , then the critical z is zcrit = -2.053748911 Also, the p value is P = 2.8806E-19 Thus comparing z and zcrit (or p and significance level), then we REJECT THE NULL HYPOTHESIS. Thus, there is significant evidence that the proportion of third year students who buy used textbooks is greater than the proportion of first year student who buy used textbooks. [CONCLUSION].

First year students Mean 125, SD, size 350Third year students M.pdf

First year students Mean 125, SD, size 350Third year students M.pdf

First year students Mean 125, SD, size 350Third year students M.pdf

Explain clearly how pseudonyms become credentials in VANETs? Explain clearly how pseudonyms become credentials in VANETs? Solution Each vehicle is equipped with a set of short-term certificates, termed pseudonyms, each with a corresponding short-term private key to sign outgoing messages. Example: Vehicle A digitally signs outgoing messages (time- and geo-stamped) with the private key, kiv , corresponding to the pseudonym Piv({Piv}(PCA) represents the pseudonym signed by the pseudonym issuer) and is attached to messages to facilitate verification on the receiver side. Receiving vehicles B and C verify the pseudonym {Piv}(P CA) and validate the signature (assuming they trust the pseudonym issuer discussed below). This process ensures the authenticity and integrity of the message and enables further validation based on its content. At the same time, transmissions by vehicle A do not reveal its identity (as the short-term certificates are anonymised), and messages signed under different pseudonyms (with different private keys) are, in principle, unlinkable. Vehicles switch from one pseudonym to another (not previously used) to achieve unlinkability of authorities with distinct roles: the Root Certification Author- ity (RCA), the Long Term Certification Authority (LTCA), the Pseudonym Certification Authority (PCA), and the Resolution Authority (RA). Different proposals may refer to these entities with various names, e.g., CAMP refers to the LTCA as the Enrollment Certification Authority (ECA). The RCAs are the highest-level authorities certifying LTCAs, PCAs, and RAs. An LTCA is responsible for registering vehicles and issuing Long Term Certificates (LTCs). A PCA issues sets of pseudonyms for the registered vehicles. An RA can initiate a process to resolve a pseudonym, i.e., identify the long-term identity of the vehicle that used (in a nonrepudiable manner) its short-term keys and credentials..

Explain clearly how pseudonyms become credentials in VANETsExpl.pdf

Explain clearly how pseudonyms become credentials in VANETsExpl.pdf

Explain clearly how pseudonyms become credentials in VANETsExpl.pdf

Explain the use and importance of the scientific name, genus and species. Explain the use and importance of the scientific name, genus and species. Solution Scientific Names(May be called Binomial Nomenclature) is the system of nomenclature in which two terms (mostly Latin forms,sometimes others are also used)are used to denote a species of living organism(The 1 st one is for Genus and the 2nd one is for species). e.g-Escherichia coli, Lactobacillus acidophilus IMPORTANCE- 1.Universal- Everyone in the world use the specific name to identify the specific organism. 2.organisms are easily categorized. 3.The chance of nomenclature is reduced. IMPORTANCE OF GENUS- The genus name forms the first part of the binomial nomenclature for each species. In the hierarchy of biological classification genus comes above species and below family. IMPORTANCE OF SPECIES- Basic unit of biological hierarchy It forms the second part of the biological binomial nomenclature. there are many species under a common genus..

Explain the use and importance of the scientific name, genus and .pdf

Explain the use and importance of the scientific name, genus and .pdf

Explain the use and importance of the scientific name, genus and .pdf

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