What are the possible types of interfund transactions What are.pdf

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What are the possible types of interfund transactions? What are the possible types of interfund transactions? Solution Two types of interfund transactions are Quasi-External Transactions and Reimbursements QUASI-EXTERNAL TRANSACTIONS These are transactions that would be treated as revenues, expenditures or expenses if they involved organizations external to the State. They are accounted for as ordinary revenues, expenditures or expenses of the fund involved. Examples are payments by a State agency to the Office of General Services and Centralized Services Fund billings. REIMBURSEMENTS These are transactions that constitute reimbursements of a fund for expenditures or expenses initially made from it which are properly applicable to another fund. They are recorded as expenditures or expenses of the reimbursing fund and as reductions of the expenditure or expense originally charged in the fund that is reimbursed. However, interfund reimbursements do not include situations whereby interfund receivables or payables have been set up..

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8-1. Explain the distinction(s) between agency funds and trust funds. What financial statements are prepared for each? With Agency funds and trust funds there are some clear distinctions, like fiduciary which are agency funds. Generally, trust funds are more complicated than agency funds, requiring greater representation and development of the beneficiary’s interest. Agency funds are fiduciary funds, also called conduit or clearinghouse funds. They are established to account for assets received for and paid to other funds, individuals, or organizations. These assets are offset by liabilities equal in the amount and no fund net position exists; therefore, Agency funds are not included in the statement of changes in fiduciary net position 8-2.Identify the different types of trust funds and explain the purpose of each type. For reporting purposes GASB classifies trust funds in three types: investment trusts, private-purpose trusts and pension trusts. An investment trust fund is used to account for and report the fund equity held by fund participants who are external to the government operating the fund and its own money. Deductions in this fund are withdrawals and investment expenses incurred. Private-purpose trust funds record and report principal and/or interest managed by a government for the benefit of an individual, private organization or another government at the present time or future date. The principal or corpus of the trust is the fair value of assets placed into the trust. The principle can be expendable or nonexpendable. The distinguishing characteristic is that the party benefiting from the trust must be external to the government operating the trust. 8-3. Describe the basic activities conducted by a tax agency fund. What are some of the issues that make tax agency fund accounting complex? Basic activity conducted of tax agency fund is collection of all the taxes (property taxes etc.) of the county or other area (jurisdiction) Then all proceeds are distributed into each county’s government. It is used to combine tax levies from multiple governments in the same geographic area. This will avoid duplicating assessment and collections processes. Then each government continues to maintain their own tax records individually. ...

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Which of the following are true when comparing TCP/IP to the OSI Reference Model? (Choose two.) A. The TCP/IP model has seven layers, while the OSI model has only four layers. B. The TCP/IP model has four layers, while the OSI model has seven layers. C. The TCP/IP Application layer maps to the Application, Session, and Presentation layers of the OSI Reference Model. D. The TCP/IP Application layer is virtually identical to the OSI Application layer. Solution B and C are correct. The TCP/IP model has four layers. The Application layer maps to the top three layers of the OSI Reference Model. A and D are incorrect. A and D are incorrect because they are contrary to B and C..

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What is the reason for having most of the blood volume distributed within the venous vessels at any given time? Solution The thin, non-elastic wall of veins (as compared to the thick, elastic, muscular walls of arteries) allows greater volumes of blood to be accommodated within the lumen of the veins. This accounts for most of the blood volume distributed within the venous vessels at any given time..

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What is eutectic reaction? What is eutectoid reaction? What is perifectic reaction? Are these transformations reversible? in the following phase diagrams (figure QB) these points from points A, B, C, D, E, F, G, H, K, and write down the reactions at these points. For example point A is perifectic reaction at this point can be written as Lipid and delta-iron transform to gamma-iron during cooling. Solution Eutectic reaction: A eutectic reaction is a reaction when one a liquid cools down into a mixture of two solids. Elements have different miscibility at different temperatures. So, when a liquid containing two elements soluble in one another cools down to a level where the solutbility is limited, then the excess element is sandwiched between the zones containing the mixture. This reaction is Eutectic reaction. In the Iron-Carbide diagram above, it is represented by the point D. Eutectoid reaction: A eutectoid reaction is a reaction where one solid divides into a mixture of two solids. It has the same concept as eutectic reaction but here, instead of a liquid splitting up into two solids, a solid breaks down into two solids. For example, B is an eutectoid reaction Peritectic reaction: A peritectic reaction is where the mixture of a solid and liquid combines into a single solid. Here, the miscibility is low when there is a phase-equilibrium between the liquid and the solid but as the temperature decreases, the solubility increases. Hence, it is completely miscible in the solid. For example, A shows the peritectic reaction happening. Yes, since all the solutions are in a state of pahse equilibrium, they can be interchanged and the reaction can go either ways. It is completely reversible..

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What are Sociocultural issues when it come to body image? Solution Answer: Sociocultural issues such as dissatisfaction in women and men each they are obese due to eating disorders & media aspects and cultural aspects often have meticulous influence on individual body image finally may lead to “development of depressive disorders”. Therefore, unrealistic images from media & family pressure and society pressure do generate consciousness on physical beauty via body image. Therefore, “perfect bodies” often maintained by various people via performing gymnastics, swimming etc. Individual mental health in association with sociocultural factors such as westernization hypothesis & abnormal eating attitudes can exert influence on inappropriate body image.

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Water can absorb and store a large amount of heat while increasing only a few degrees in temperature. Why? The heat must first be used to break the hydrogen bonds rather than raise the temperature. The heat must first be used to break the ionic bonds rather than raise the temperature. The heat must first be used to break the covalent bonds rather than raise the temperature. An increase in temperature causes an increase in adhesion of the water. An increase in temperature causes an increase in cohesion of the water. A chemical compound that releases H^+ into a solution is called: a proton. A base. an acid. a hydroxide ion. a hydrogen ion. Which of the following foods is not a significant source of complex carbohydrates? fresh fruit rice pasta oatmeal All of the above are significant sources of complex carbohydrates Solution Thanks for the question 5.A The hydrogen bonds are reponsible for the high boiling points of water. 6.C an acid is a substance which donates H. 7.E All of the above mentioned foods are good source of complex carbohydrates.

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True or false 1 Anabolism results in increased production of biomass. 2 Oxygenic phototrophs can use hydrogen sulfide as an electron donor. 3 Oxidation is the process of accepting electrons. 4 Water appeared on the earth as a result of bombardment by asteroids. 5 The highest abundance of microbial life in soils appears in the O horizon. 6 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during the day and an increase in pH values at night. 7 Freshwater represents about 10% of the total water on Earth. 8 Oligotrophic organisms thrive in low nutrient conditions. 9 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level phosphorylation. 10 Siderphores are used to increase availability of manganese in aerobic environments. Solution 1. Anabolism: The process of formation of biomolecules during metabolism such as proteins, polysaccharides, lipids, and nucleic acids etc is known as anabolism, this results in an increase in the production of biomass. Therefore, option is: True 2.Oxygenic phototrophs use oxygen as electron donar, while as anoxygenic phototrophs uses hydrogen sulfide as electron donor. Therefore, option is: False 3. Oxidation is a process in which a substance looses one or more electrons, while as reduction is a process of accepting electrons. Therefore, option is: False 4. The earth got most of the water about 4.6 billion years ago from the asteroid impacts. Therefore, option is: True.

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The diploid number of the hypothetical animal Geneticus introductus is 2n= 44. Each diploid nucleus contains 5 ng of DNA in G1. Part A : What amount of DNA is contained in each nucleus at the end of S phase? What amount of DNA is contained in each nucleus at the end of phase? a)There are 22 chromosomes containing 2.5 ng of DNA. b)There are 44 chromosomes containing 5 ng of DNA. c)There are 44 chromosomes containing 10 ng of DNA. d)None of the above. Part B: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of Telophase I of meiosis (after cytokinesis). Express your answer as two numbers separated by a comma (example: 12, 4). Part C: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of Anaphase of mitosis. Part D: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of Telophase II of meiosis (after cytokinesis). Solution The cell cycle of actively dividing eukaryotic cell consists of gap phases (G 1 and G2), S phase and M phase. In G 1 phase the cell prepares for cell division .In S phase DNA replication occurs and the DNA content doubles. In G 2 phase, the cell gets ready for mitosis. In mitosis, two daughter cells with same genetic content as parent are produced. Mitosis occurs through five stages: prophase, prometaphase, metaphase, anaphase and telophase. Prophase: This stage extends over half of mitosis. The nuclear membrane disintegrates. The centrosome of chromosome duplicates to form two daughter centrosomes and migrates to opposite poles of the cell. The centrosomes organise the production of microtubules, from which the spindle fibres arise. The chromosomes condense and consist of two identical chromatids (or sister chromatids) held together at the centromere. Prometaphase: The chromosomes, move to the middle of the cell. The spindle fibres bind to centromere of each chromosome at kinetochore. Metaphase: The chromosomes align along the metaphase plate of the spindle apparatus. Anaphase: This is the shortest stage of mitosis. The centromeres divide, and the sister chromatids of each chromosome are separated and move to the opposite poles of the cell. The separated sister chromatids ordaughter chromosomesreceives a copy of each parental chromosome. Telophase: This is the final stage of mitosis. The nuclear membrane reforms, the chromosomes uncoil, and the spindle fibres disappear. Cytokinesis: the membrane appears to divide the two daughter cells with same chromosome number and the DNA content as the parent cell. The cell then enters interphase - the interval between mitotic divisions. Meiosis is the cell division occuring in sex cells or gametes (which contain a single copy of each chromosome). The process involves one DNA replication followed by two successive nuclear and cellular divisions (Meiosis I and Meiosis II). Similar to mitosis, meiosis is preceded by a process of DNA replication that converts each chromosome into two sister chrom.

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The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For temperatures near the normal boiling point: a. Derive an expression that gives the behavior of Hvap as a function of temperature which is thermodynamically consistent with the Antoine equation. Solution The Antoine equation is a mathematical expression of the relation between the vapour pressure and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french researcher, in 1888. The basic form of the equation is: lnP = A - B / ( T + C ) and it can be transformed into this temperature-explicit form: T = B ( A - lnP ) - C where: P is the absolute vapor pressure of a substance T is the temperature of the substance A, B and C are substance-specific coefficients (i.e., constants or parameters log is typically either log10 or loge A simpler form of the equation with only two coefficients is sometimes used: lnP = A - ( B \\ T ) which can be transformed to: T = B ( A - lnP ) The Antoine equation is a semi-empirical equation which expresses vapour pressure as a function of temperature. A new, rapid and highly accurate method for obtaining its three constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic substances where p is the vapour pressure, T is temperature and A, B and C are component- specific constants. Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure curve from the triple point to the critical point, because it is not flexible enough. Therefore, multiple parameter sets for a single component are commonly used. A low-pressure parameter set is used to describe the vapour pressure curve up to the normal boiling point and the second set of parameters is used for the range from the normal boiling point to the critical point. The Antoine Equation is a vapor pressure equation and describes the relation between vapor pressure and temperature for pure water between 0°C and 373.946°C. For T in Kelvins and P in kPa: 273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724 373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665 Antoine(T) = exp(A B / (T + C)) For temperatures above the critical temperature (373.946°C / 647.096K), where water is a superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e. dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can write P(T Tc) = Antoine(T) P(T > Tc) = 235.88 (T Tc) + Antoine(Tc) Hence, the behaviour of delta hvap as a function of temperature which is termodynamically consistent with the the antoine equation is derived ..

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Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa Cnidaria Pha An tea cnidarians-- jellyfish, hyvha. Radar 4 nerve Platyhelminthes ema toda (flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with gastrulation. Note that deuterostomes product of developmental in which leads threetissue layers Solution 3. Bilateral symmetry 5.Value of body cavity: a) Pseudocoelomic fluid is rich in protein and other dissolved substances which is responsible for turgescent state of nematode body. b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic skeleton. It transports food, waste and other materials..

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Question 12 (5 points) The reason that mitosis can result in the production of haploid gametes in alternation of generations is because Question 12 options: The sporophyte and gametophyte are both haploid Eggs and sperm are produced directly by the diploid sporophyte via meiosis Eggs and sperm are produced by the diploid gametophyte via mitosis Eggs and sperm are produced by the haploid gametophyte via mitosis Save Question 13 (5 points) In the fern life cycle, how does fertilization occur? Question 13 options: The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf and find each other on the ground to fuse The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming through water The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The sperm reach the eggs by swimming through water There is no sexual reproduction in ferns Save Question 14 (5 points) The earliest land plants were most likely to have lived in which of the following habitats? Question 14 options: arid desert environments Salt water marshes Tropical rainforests fresh water wetlands Save The sporophyte and gametophyte are both haploid Eggs and sperm are produced directly by the diploid sporophyte via meiosis Eggs and sperm are produced by the diploid gametophyte via mitosis Eggs and sperm are produced by the haploid gametophyte via mitosis Solution Q12 Eggs and sperms are produced by haploid gametophyte by mitosis. The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells undergo meiosis to create haploid cells. However, these cells do not immediately unite with others to recreate the diploid state. Instead, they are shed as spores and germinate into small haploid organisms. Because the diploid organism creates spores, it is called the sporophyte generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called the gametophyte generation of the life cycle. The male gametes (sperm) are then released and swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing the life cycle. Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The sperm reach the eggs by swimming through water The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through water; they do not travel great distances and are only released when free water is available. Only a continuous film of water is needed. They antherozoids are chemically attracted to the developing flask-like female organs or archegonia where fertilization of the single egg cell embedded in each archegonium takes place. The sporophyte has a double set of chromosomes (i.e. diploid) and i.

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peripheral circuit functions built-in to the HC12 microcontroller, that are not part of the CPU itself or the memories. Access to these is via external pins on the Microcontroller IC package. What are some of these built-in circuit functions called, and what do they do? Solution The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS. The bus is a set of shared communication lines that serves as the central nervous system of the computer. Data, address, and control signals are shared by all system components via the bus. Each component connected to the bus communicates information to and from the bus via its own bus controller. The data lines,address lines, and control lines allow a specific component to access data addressed to that component. The data lines are used to communicate words to and from data. registers in the various system components such as memory, CPU, and input/output (I/O) peripherals. The address lines are used to select devices on the bus or specific data locations within memory. Devices usually have a combinational logic address decoder circuit that identifies the address code and activates the device..

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Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding. In Mendez’s first four years of operation, its board of directors paid the following cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000. Determine the dividends per share and total cash dividends paid to the preferred and common stockholders during each of the four years. Solution The dividends per year on preferred stock = 10,000 x $ 100 x 7 % = $ 10,000 x $ 100 x 0.07 = $ 70,000 Dividend distribution: Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000 Explanation: First dividend is paid to Preferred stock and then excess amount goes to common share holder. For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent years and is added to that year\'s dividend. Preferred Dividends unpaid in 2011 = $70,000 Preferred Dividends on 2012 = $70,000 + $ 70,000 = $ 140,000 Preferred Dividends paid in 2012 = $ 120,000 Preferred Dividends shortage on 2012 = $ 20,000 As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any dividend during 2011 & 2012. Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000 Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000 Preferred Dividends paid in 2014 = $ 70,000 Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000 Computation of Dividends per share: 2011 2012 2013 2014 Preferred Stock (Dividends per share) $0 $12.00 $9.00 $7.00 Common Stock (Dividends per share) $0 $0 $1.00 $1.40 Explanation: Dividends per share = Dividend paid/No. of shares Year 2012: Dividends per share for Preferred Stock = $120,000/10,000 = $ 12 Dividend per share for Common Stock = $ 0 Year 2013: Dividends per share for Preferred Stock = $90,000/10,000 = $ 9 Dividends per share for Common Stock = $ 50,000/50,000 = $ 1 Year 2014: Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7 Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40 Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000.

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If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes would the F_2 progeny show? If the two genes are on different chromosomes and therefore independently assorting, how many phenotypes would you see in the F_2 worms? List the phenotype(s). Strawberry plants have varieties that flower when the days are still short, and other varieties that flower only when the days get longer. The long-day allele is dominant to the short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion. When you cross between heterozygous plants and count the offspring in the F_1 generation, this is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short days What would you expect the ratio of phenotypes be? What is your Chi squared value? Show your work What does this mean? Solution 2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs independent assortment occurs. No new characters appear. Where as gegenes are on different chromosomes, they are non linked. Crossing over or independent assortment takes place. New characters appear along with parental characters. So dpy-11 & unc-31 present on same chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the said genes present on different chromosomes, they give 4 different phenotypes 3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between these genes Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short. Chi square value can be calculated by using the formula. (O-E)^2/E O = observed phenotypes; E = Expected phenotypes.

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How are health services paid for? Provide a definition for the term \"third-party payer\", discuss the different payers currently operating in the market, and assess the importance of each. Please use a table format to list the information. Solution There are two main payers types: single payer and third-party payer. Single payer is when the person that obtains the health service is the same which pay for it, here relationship is direct taker and provider of health services. However, it is more common the presence of a third party payer, which is an external agent that covers the other health expenses or part of them, they are insurance companies or public social protection systems. Single Payer Third-Party Payer Type of relation with provider Direct Indirect Capability to compare Prices/Features Relatively High Almost Null Payment proportion Full Low percentage Single Payer Third-Party Payer Type of relation with provider Direct Indirect Capability to compare Prices/Features Relatively High Almost Null Payment proportion Full Low percentage.

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Given the information regarding mitosis and meiosis, clearly and convincingly explain in your own words how an error in MITOSIS could possibly lead to a gamete containing new genetic information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points) Solution The process of mitosis is divided into two phases: Interphase and M-phase. Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S- phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start properly. Now, during S phase, the ‘2n’ nucleus converts into 4n. So, the cell entering M-phase is already ‘4n’. M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases, prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In prophase, the entire DNA converts into chromosomes; the process is known as condensation of DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is not correct, then anaphase cannot be started due to metaphase check point. In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell; then two gametes, ‘3n’ and n will be formed. This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to their ‘2n’ counterparts. Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division). Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end; then two gametes will be formed; one having all normal chromosomes + one double chromosome with 2c DNA, and the other having no DNA. Please note: normally gametes are 1n (haploid) and 1c. Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”. This condition if occurring with chromosome 21 is known as Down\' syndrome. It is important to note that errors in Mitosis result in non-disjunction; which further results in many Mendelian disorders; as well as ploidy defects. ** Hope it helps..

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First, look to determine what is the earliest era that fossils have been found here in Georgia. Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of the fossil, when it was found, and where the fossil was found. Give a brief description of the organism that was fossilized. Solution Trilobites are the oldest fossils found in georgia.They have been found in Georgia’s Appalachian Plateau which is present the Coosa River in Floyd County. These fossils are considered to be arthropods,Moreover they had to shed their shells to grow.As a result the fossils found in these places usually shed shells as bits of individuals or as complete individuals or as imprints of individuals..

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explain why right and why wrong Indentify configuration (below) which I s not a correct representation of H and I. Solution option C is not correct one. Explanation: If we stretch all right hand fingers in the direction of current direction given then the thumb gives us into the board (into symbol) or out of the board (dot symbol). In the option C the field should be out of the board, so it is not correct one..

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