1.2.3.4.SolutionThe number of observations per cell was 3To.pdf

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1.2.3.4. Solution The number of observations per cell was: 3 Total df=23 Total observation = 24 Interaction Df=3 Df for amplifier and position are (1 and 3) any order Number of factors for amplifier and position = (2 and 4) any order Observations per cell = 24/(2*4) =3 Two-factor ANOVA with replication The two factors are PVC and Temperature. Each cell having 3 observations or replications..

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This week you surveyed many different groups of living things. What Domain of living things is thought to have been the first form of life on Earth? Name one piece of evidence that exists for the claim that these were the earliest living things? Solution Archaea are found in the harshest environments on Earth, and are the oldest known organisms on Earth, appearing in the fossil record over 3.6 BILLION years ago Archaea display unique capacities, such as methanogenesis and survival at temperatures higher than 90 °C, that make them crucial for understanding the nature of the biota of early Earth..

This week you surveyed many different groups of living things. What D.pdf

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Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains 10 yellow marbles and 6 brown marbles. Urn III contains 7 yellow marbles and 11 brown marbles. The probability that Urn I will be selected is 0.42, th e probability that Urn II will be selected is 0.45, and the probability that Urn III will be selected i s 0.13. Find each probability: D. The probability that the marble selected was from Urn II, if it is known that the marble was yellow. Solution.

Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains.pdf

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The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequence. On average, how often would this enzyme cut a random sequence of DNA? How many times would the enzyme cut the E. coli genome? (The E. coli genome contains 4.6 x 106 bp). Solution There are 4 types of bases so the chances of each base to cut in the sequence is (1/4) (1/4) (1/4) (1/4) (1/4) (1/4)= (1/4)6 base pairs=1/4096. Now as the DNA sequence is completely random and we can expect that the enzyme may cut the DNA once every 4096 bases. Now the E.coli genome contains 4.6 x 106 base pairs of sequence. Then the cutting site will be in the whole E.coli genome is (4.6*106 /4096)= 1123 times..

The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequenc.pdf

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The vertical motion of a mass attached to a spring is described by t.pdf

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There are several types of bacterial toxins that are categorized as spreading factors. One of those is leukocidins. Do the 3 species of staphylococci you\'ve described produce leukocidins? What is their function? How do they work? Solution Leukocidins are toxins secreted by Staphylococcus aureus, Streptococcus pyogenes, and S. pneumoniae Among the three important Staphylococci, viz., S. aureus, S.epidermidis, and S.saprophiticus, leukocidins play a major role in the pathogenesis of S. aureus. Leukocidins are toxins that kill leukocytes (lit., Leuko: for leukocytes and caedo: killing) with or without lysis. In Streptococcus aureus, they attacks neutrophils, disrupt their membrane that leads to the release of the lysosomal contents, thereby, killing them..

There are several types of bacterial toxins that are categorized as .pdf

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the life span of the general population. is the distribution uni.pdf

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A volume of 0.0125 m3 at a temperature of 1300C and pressure at 90kP.pdf

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Please help!!! I can change a link layer protocol and the application layer protocol does not need to know or be made aware of the change. Explain why? Solution Answer :- The protocols are like a pile of building blocks stacked one upon another. Because of this appearance, the structure is often called a stack or protocol stack. A layer does not define a single protocol it defines a data communications function that may be performed by any number of protocols. Therefore, each layer may contain multiple protocols, each providing a service suitable to the function of that layer. For example, a file transfer protocol and an electronic mail protocol both provide user services, and both are part of the Application Layer. 1) Application Layer :- The Application Layer is the level of the protocol hierarchy where user-accessed network processes reside. In this text, a TCP/IP application is any network process that occurs above the Transport Layer. This includes all of the processes that users directly interact with as well as other processes at this level that users are not necessarily aware of. 2) Presentation Layer :- For cooperating applications to exchange data, they must agree about how data is represented. In OSI, the Presentation Layer provides standard data presentation routines. This function is frequently handled within the applications in TCP/IP, though TCP/IP protocols such as XDR and MIME also perform this function. 3) Session Layer :- As with the Presentation Layer, the Session Layer is not identifiable as a separate layer in the TCP/IP protocol hierarchy. The OSI Session Layer manages the sessions (connections) between cooperating applications. In TCP/IP, this function largely occurs in the Transport Layer, and the term “session” is not used; instead, the terms “socket” and “port” are used to describe the path over which cooperating applications communicate. 4) Transport Layer: Much of our discussion of TCP/IP is directed to the protocols that occur in the Transport Layer. The Transport Layer in the OSI reference model guarantees that the receiver gets the data exactly as it was sent. In TCP/IP, this function is performed by the Transmission Control Protocol (TCP). However, TCP/IP offers a second Transport Layer service, User Datagram Protocol (UDP), that does not perform the end-to-end reliability checks. 5) Network Layer :- The Network Layer manages connections across the network and isolates the upper layer protocols from the details of the underlying network. The Internet Protocol (IP), which isolates the upper layers from the underlying network and handles the addressing and delivery of data, is usually described as TCP/IP’s Network Layer. 6) Data Link Layer :- The reliable delivery of data across the underlying physical network is handled by the Data Link Layer. TCP/IP rarely creates protocols in the Data Link Layer. Most RFCs that relate to the Data Link Layer discuss how IP can make use of existing data link protocols. 7) Physi.

Please help!!!I can change a link layer protocol and the applicati.pdf

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4. (a) Negate the statement Sarah has a spaceship and has three f.pdf

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Match the bones in column A with the features in column B. Place the letter of your choice in the space provided. (Some answers are used more than once.) a. Ethmoid bone b. Frontal bone c. Occipital bone d. Parietal bone e. Sphenoid bone f. Temporal bone Solution correct answers a) Ethmoid bone - 2) Cribriform plate b) Frontal lobe - 8) middle nosal concha c) Occipital bone - 5) foramen magnum d) Parietal bone - 1) form sagittal, coronal , squamous and lambdiod sutures e) Sphenoid bone - 10) Sella turcica f) Temporal bone - 11) Styloid process.

Match the bones in column A with the features in column B. Place the .pdf

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1. Prepare and analize the common statement for Anandam Manufacturing Company. Show all calculations. 2. Prepare and analize the cash flow statement for Anandam Manufacturing Company. Show all calculations. 3. Calculate the ratios based on case Exhibit 3. Show all calculations. Based on financial analysis of financial statements, would a loan officer grant a loan to Anandam financial company? EXHIBIT 3: INDUSTRY AVERAGE OF KEY RATIos Sector Average Ratio 2:301 Current ratio Acid test ratio (quick ratio) 120 Receivable tunover ratioimes turnover 52 days Receivable days Inventory turmover ratio Inventory days Long-term detttototal debt Debt-to-equity ratio Gross profit rato Net profit ratio Retun on equity Retum on total assets 4.85 times 75 days 24% 35% 40% 18% 22% 10% Total asset turnover ratio Ti Fixed asset turnover ratio Current asset tunover ratio3 interest coverage ratio (tmes 10 nterest earned Warking captal smover ratio Retu on ed sset Solution 1….Common -size Income statement 2012-13 % to Total sales 2013-14 % to Total sales 2014-15 % to Total sales Analysis of % to Total sales proportion Sales Cash 200 10.00% 480 10.00% 800 10.00% Credit 1800 90.00% 4320 90.00% 7200 90.00% Total Sales 2000 100.00% 4800 100.00% 8000 100.00% COGS 1240 62.00% 2832 59.00% 4800 60.00% COGS has reduced in the last 2 yrs. Gross profit 760 38.00% 1968 41.00% 3200 40.00% so, G/P has increased Operating Expenses: Gen.adm.& sell.exp. 80 4.00% 450 9.38% 1000 12.50% % to sales has doubled in 2013-14 & increased by more than 25% in 2014-15 Depreciation 100 5.00% 400 8.33% 660 8.25% Has increased by 3% in the last 2 yrs. Int.exp.(on borrowings) 60 3.00% 158 3.29% 340 4.25% Has slightly increased in 2014-15 Profit before tax(PBT) 520 26.00% 960 20.00% 1200 15.00% PBT% has reduced gradually in both the yrs.due to increase in all operating expenses esp.gen admn.& sell exp. Tax at 30% 156 7.80% 288 6.00% 360 4.50% Decrease % due to decrease in profit % Profit after tax(PAT) 364 18.20% 672 14.00% 840 10.50% PAT% has reduced gradually in both the yrs.due to increase in all operating expenses esp.gen admn.& sell exp. Common-size Balance Sheet Assets % to Total % to Total % to Total Fixed assets(net of dep) 1900 74.22% 2500 44.64% 4700 51.33% Assets have decreased to the total Current assets Cash & Cash equivalents 40 1.56% 100 1.79% 106 1.16% Slight variation in ratio to total Accounts Receivables 300 11.72% 1500 26.79% 2100 22.94% Has increased in 2013-14 & again decreased in 2014-15 Inventories 320 12.50% 1500 26.79% 2250 24.57% Has increased in 2013-14 & again decreased in 2014-15 Total 2560 100.00% 5600 100.00% 9156 100.00% Equity & Liabilities Equity share capital($ 10) 1200 46.88% 1600 28.57% 2000 21.84% Decreased steadily in all the 2 yrs. Reserves & surplus 364 14.22% 1036 18.50% 1876 20.49% Increased steadily in both 2 yrs.due to increase in $ net income in those yrs. Long-term borrowings 736 28.75% 1236 22.07% 2500 27.30% Decreased in 2013-14 & again increased i.

1. Prepare and analize the common statement for An.pdf

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Julia primarily uses her minivan transport her grandchildren, Vera, Chuck, and Dave, to and from school. She, 25/l00 bodily injury$50, 000 property damage coverage. Julia has a $50 deductible. Her driver-rating factor is 1.75. Her is classified as B, 12. What is Julia\'s annual premium? Solution you have deducted 50 twice.That is mistake as she has either collision deductible or comprehensive deductible.so it is to be summed only once..

Julia primarily uses her minivan transport her grandchildren, Vera, C.pdf

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I have an interface Peron that has firstnamestring, lastnamestring.pdf

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How do condition variables and semaphores differ when sending a signal? Solution conditional varaibles: 1) these are the variable which are combined with mutexes and used for implementing primitives which are synchronized 2)these variable allows the thread for wait for completion by using wait after the completion of job then it invoke other by using signal method semaphore: 1)here the shared resource can accesible by only one thread at a time 2)here at first n threads can access a resource concurrently and then it provide signal mechanism by using of wait and signal methods.

How do condition variables and semaphores differ when sending a sign.pdf

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Function of yolk sac in human embryos? Forms blood cells shock absorber forms the placenta Nutrition. Which cells allow embryos to implant in the Uterus? Syncytriophoblast trophoblast E- cadherin\'s goner cell mass Solution Option D. Yolk sac is responsible for the process of viteline circulation in which it provides nutrition to the developing embryo through the primitive aorta. 2. Option B. As trophoblasts are the cells which make the placenta. Placenta connects the foetus to the uterine wall which helps in transfer of nutrients etc to the developing foetus..

Function of yolk sac in human embryos Forms blood cells shock absorb.pdf

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Explain the basic characteristics of a partnership. Solution There must be at least two or more persons to form Partnership. There must be agreement between partners and such agreement can be Oral or Written. The agreement forms the basis of Mutual rights and duties of Partners. Partnership can be formed for the purpose of carrying on some business with the intension of earning some profit. The Agreement must include how the profit or loss to be shared. If the persons joins firm for some Charity then it will not be called partnership and if the persons is deprived of his right to share the profit then he/she will not be called partner. And it is not necessary that all partner share loss also, it may be agreed between partners that some oo one of the partners will not be liable for losses. All the persons have Unlimited liability i.e. if the assets of the company is not sufficient to pay the liabilities then the Creditors can come after the personal property of the Partners. In the eyes of Law, A partner and Partnership firm are same, i.e. Partnership firm doesn’t have its own Legal identity. The business must be Legal..

Explain the basic characteristics of a partnership.SolutionThe.pdf

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Explain the functions of the different types of meristems found in plants. Be able to distinguish between secretory tissues. Solution In plants, meristems are resent broadly of three types namely: 1. Apical meristem 2. intercalary Meristem 3. Lateral meristem 1. APICAL MERISTEM This is present at the root and shoot apex of the plant and is actively dividing zone. Shoot apcal meristem causes plant to grow up and root apical meristem causes thr roots to grow down in the soil. Apical meristems keep dividing and result in the primary growth of plant and the tissues formed as their division as named as Protoderm and Procambium. 2. INTERCALARY MERISTEM Intercalary meristem is found between the mature tisues like nodes and internodes. It functions to regulate the regenration of Tissues when the leaves and grasses are grazed by herbivores. This meristem is found mainly in monocot plants. 3. LATERAL MERISTEM This type of meristems is developed late in the plant life cycle and thus it is also known as secondary meristem. It functions in the secondary growth of plant where plant grows in girth not in height. Cork cambium fascisular cambium and interfascicular cambium are types of lateral meristems..

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Crossing-over occurs during prophase I of meiosis.True or False.pdf

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A single card is drawn from a standard deck of cards. What is the pr.pdf

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Consider the security of a personal computer. List a few of the attackers who might break into the computer, their incentives, and the associated costs and risks to the attacker. Solution A top-down approach to identifying threats and countermeasures, CISOs should consider a threat modeling technique allows the target application to be decomposed to reveal its attack surface and subsequently its relevant threats, associated countermeasures, and finally, its security control gaps and design flaws. applications are also a source of potential risks especially when new or different technologies are integrated within applications. As applications evolve by offering new services to citizens, clients, customers and employees, it is also necessary to plan for mitigation of new vulnerabilities introduced by the adoption and implementation of new technologies such as mobile devices, web 2.0 and new services such as cloud computing. Adopting a risk framework to evaluate the risks introduced by new technologies is essential to determine which countermeasures to adopt to mitigate these new risks. threat agents seek financial gain such as by attacking applications to compromise users\' sensitive data and company’s proprietary information for financial gain, fraud as well as for competitive advantage the risks posed by these threat agents, it is necessary to determine the risk exposure and factor the probability and the impact of these threats as well as to identify the type of application vulnerabilities that can be exploited by these threat agents. The exploit of some of these application vulnerabilities might severely and negatively impact the organization and personal computer. data from legal in relation to law suits and regulatory fines and fraud data that includes amount of money losses incurred because of online fraud. All this type of information is essential to determine the overall impact. In absence of this data, the best the CISO can do is to use data breach incident data from public sources and data breach incident reports. In part I of this guide, we provided some examples of how this data can be used to estimate impact. We documented what are the critical factors to estimate impacts of data breaches: these as the value of the data assets liability for the organization in case these assets are lost. Once the potential business impact of a data breach is estimated, the next step is to determine how much should be spent to mitigate the risk. At high level, this is a risk strategy decision that depends on the organization risk culture and the organization priorities for mitigating risks..

Consider the security of a personal computer. List a few of the atta.pdf

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