The Complexity of Arnold-Chiari Malformation :: Biology Essays Research Papers

The Complexity of Arnold-Chiari Malformation To the medical doctor, Arnold-Chiari Malformation, which may have a genetic link, is characterized by a small or misshapen posterior fossa (the depression in the back of the skull), a reduction in cerebrospinal fluid pathways and a protrusion of the cerebellar tonsils through the bottom of the skull (foramen magnum) into the spinal canal resulting in a multitude of sensory-motor problems and even some autonomous malfunctions (1). These many symptoms can come in a variety of forms which often makes a clinical diagnosis difficult. To the patient this disorder can present not only physical difficulties but also mental distress. Treatment options and their success rates vary widely, and proponents of the cause are demanding more recognition, research, and success. The study of Arnold-Chiari malformations can lead to additional questions and new understandings about the I-function, sensory-motor input/output paths and the general make-up of the brain and nervous system, but a complete und erstanding of the disorder may be a long time coming. Impairment and sometimes loss of motor control of the body and its extremities is one of the many effects of this disorder. Patients may complain of headaches, neck pain, coughing, sneezing, dizziness, vertigo, disequilibrium, muscle weakness, balance problems, and loss of fine motor control (1). The senses (hearing, sight, smell etc.) may also be affected in deleterious ways. On can have blurred vision, decreased sensation of limbs, unable to locate them without looking, decreased sense of taste, ringing of the ears etc. (2). Two ideas about the nervous system that can be better understood from these observations are the concepts of having and locating the I-function. It seems that the I-function here is very often affected in terms of voluntary movement. A person with Arnold-Chiari malformation who has lost the feeling in and control of his arm for example will not be able to move it even upon someone's request and his or her own desire to do so. Some use of the I-function is definitely impaired. However, these observations do not seem to necessarily imply that some part of the I-function was damaged, because it may very well be located elsewhere- connections may have simply been lost. A person with Arnold-Chiari can still think and have a sense of self, but somehow can not connect with the various body parts that can be affected. Some uses and pathways of the I-function can be understood, but the exact location of it remains vague.

Gainesboro Machine Tools Corporation †Essay Essay

Executive Summary Gainesboro Corporation was a company who designed and manufactured a number of machinery parts, including metal presses, dies, and molds. The company was found in 1923 in Concord, New Hampshire, by two mechanical engineers, James Gaines and David Scarboro. The two men had gone to school together and were disenchanted with their prospects as mechanics at a farm equipment manufacturer. In the 1940’s Gainesboro produced armored-vehicle and tank parts and miscellaneous equipment for the war effort. And then in the early 1980’s, they focused on manufacturing machinery parts, war equipment, and now entered new field of computer aided design and computer aided manufacturing (CAD/CAM). Objective Ashley Swenson, chief financial officer (CFO) in mid-September 2005 needed to submit recommendation to Gainesboro’s board of directors regarding the company’s dividend policy. The Gainesboro’s stock also fallen 18%to $22.15 due to post impact of the Hurricane Katrina. Now, Ashley Swenson’s dividend decision problem was compounded by the dilemma of whether to use company funds to pay shareholder dividends or to buy back stock. Analysis >>Buy-back Stock Stock Price per share = $22.15 Net income in year 2005 = $18,018,000 Number of shares = 18,600,000 shares (assumed number in year 2004 is still the same with year 2005) Earnings per share = $0.98 Price to earnings ratio ( P/E Ratio)=(Price per share)/EPS P⠁„E Ratio=22.15/0.98=22.6 Number of retired shares=(Net income)/(Price per share) Number of retired shares=18,018,000/22,15=813,453.72≈813,454 Therefore, number of shares outstanding =18,600,000-813,454=17,786,546 shares Then we can calculate the new EPS after repurchase stock, Earnings per Share (EPS) =(Net income)/(Number of shares) EPS =$18,018,000/17,786,546=$1,013 Thus, the new market price is =EPS x PE Ratio=1.013 x 22.6=$22.89 It can be seen that by buying back the stock, the market price can increase for 3.34%. >Pay shareholders dividend a. Zero dividend payout Policy This policy required the company will not pay dividend from 2005 to 2011.In the year 2005, The company expenditure was about $63.3 million dollars but the amount of the total sources was only $40 million, so in order to balanced the company financial condition, the company borrowed $22.7 million. The same thing was also happened in 2006, the company borrowed $7.3 million (total expenditure $72.8 million –total source $65.5 million). From 2007 to 2011, the company excess cash are positive ($4.2, $11.5, $29.4, $27.2, $77.6) million, these situation happened because the total expenditure remained lower than the company total source, so the company did not have to borrowing needs. So, by sum all of the excess cash and the borrowed money data from 2005 to 2011, we can calculate that the company total excess cash is $120 million. This kind of policy has the best impact on company’s financial condition because of the absence of dividend that will reduce the company’s retained earnings. Retained earning posses a greater role to make sure the company runs smoothly in the future by using minimum portion of debt required on a project, reflected in the industrial zero-dividend payout ratio. b. 40% dividend Payout From data in exhibit 8, 40% dividend payout means that the company will pay dividend 40% from net income from year 2005 to 2011. This results and the total excess cash for borrowing needs from 2005 to 2011 is ($95.1) million. The company will do borrowing from year 2005 to 2010. Amount of money borrowed respectively, ($29.9), ($23.3), ($18.8), (17.6), ($7.2), and ($12.0). All of the value comes from deduction of the total expenditures to  the total sources. Year 2011 the company will get $13.6 million excess cash ($212.5 million – $134.9 million). $134.9 million is from the total expenditures (capital expense + change in working capital). And $212.5 million comes from the total sources (net income + depreciation). By sum up all of values (excess cash and borrowed money) from year 2005 to 2011 we get the total cash flow of ($95.1) million. By raise dividend payout from 31.4% in 2004, 140,784(Net income)/0.25(dividend per share) to 40% company need excess cash 95.1 million; only in 2011 the company gain profit. The following is the calculation table: c. Residual-payout Dividend The following is the calculation for the residual-dividend payout: By applying residual payout policy, at the total of excess cash from year 2005 to year 2011, Gainesboro still experiences negative cash. It means they will still have to borrow extra cash to pay the dividend. Conclusion and Recommendation Based on the market price value, EPS, and P/E Ratio calculation, the company’s stock will have higher market price if they buy back the stock. Therefore, it’s recommended to buy back stock instead of paying dividend. It is also supported by the comparison between zero payout dividend, 40% payout ratio, and residual-payout. The best ending cash the company has is when they do zero payout ratio, which means they don’t give dividend at certain years. Since, to pay the dividend they will have borrowing need forcing them to increase the debt level. Meanwhile, they current debt level is already higher than the maximum level management expect which is 40%. The year 2005 debt to equity ratio is 140%. Also, without paying dividend, the company still can attract investors. It is shown from the P/E ratio that is in average if compared to other similar companies.

Psy/315 Week 1 Worksheet - 923 Words

University of Phoenix Material- Instructor: Mesha Mathis Week 1 Practice Worksheet Prepare a written response to the following questions. Chapter 1 1. Explain and give an example for each of the following types of variables: a. Nominal: Measurement where a number is assigned to represent something or someone else. An example of nominal could be credit card numbers, social security numbers, or zip codes. b. Ordinal: Measurement that shows the order or rank of items. An example of ordinal could be ranking places in a contest, or test scores. c. Interval: Measurements that do not have a true zero and are composed of equal units. An example of interval could be temperature as†¦show more content†¦|15-19 | |Forms of Bullying |N |% | |Electronic victims |41 |48.8 | | |Text-message victim |27 |32.1 | | |Internet victim (websites, chatrooms) |13 |15.5 | | |Picture-phone victim |8 |9.5 | |Traditional Victims |60 |71.4 | | |Physical victim |38 |45.2 | | |Teasing victim |50 |59.5 | | |Rumors victim |32 |38.6 | | |Exclusion victim |30 |50 | |Electronic Bullies |18 |21.4 | | |Text-message bully |18 |21.4 | | |Internet bully |11 |13.1Show MoreRelatedPsy/315 Week 1 Worksheet Essay1080 Words   |  5 PagesUniversity of Phoenix Material Week 1 Practice Worksheet Prepare a written response to the following questions. Chapter 1 1. Explain and give an example for each of the following types of variables: a. Nominal: This is a measurement that has a number assigned to show something or someone else, an example of this would be one’s social security number. b. Ordinal: This is a measurement that represent the order of a particular stat. A good example of this would the placement

Empirical Formula Definition and Examples

The empirical formula of a compound is defined as the  formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols. Also Known As: The empirical formula is also known as the  simplest formula  because the subscripts are the smallest whole numbers that indicate the ratio of elements. Empirical Formula Examples Glucose has a molecular formula of C6H12O6. It contains 2 moles of hydrogen for every mole of carbon and oxygen. The empirical formula for glucose is CH2O. The molecular formula of ribose is C5H10O5, which can be reduced to the empirical formula CH2O. How to Determine Empirical Formula Begin with the number of grams of each element, which you usually find in an experiment or have given in a problem.To make the calculation easier, assume the total mass of a sample is 100 grams, so you can work with simple percentages. In other words, set the mass of each element equal to the percent. The total should be 100 percent.Use the molar mass you get by adding up the atomic weight of the elements from the periodic table to convert the mass of each element into moles.Divide each mole value by the small number of moles you obtained from your calculation.Round each number you get to the nearest whole number. The whole numbers are the mole ratio of elements in the compound, which are the subscript numbers that follow the element symbol in the chemical formula. Sometimes determining the whole number ratio is tricky and youll need to use trial and error to get the correct value. For values close to x.5, youll multiply each value by the same factor to obtain the smallest whole number multiple. For example, if you get 1.5 for a solution, multiply each number in the problem by 2 to make the 1.5 into 3. If you get a value of 1.25, multiply each value by 4 to turn the 1.25 into 5. Using Empirical Formula to Find Molecular Formula You can use the empirical formula to find the molecular formula if you know the molar mass of the compound. To do this, calculate the empirical formula mass and then divide the compound molar mass by the empirical formula mass. This gives you the ratio between the molecular and empirical formulas. Multiply all of the subscripts in the empirical formula by this ratio to get the subscripts for the molecular formula. Empirical Formula Example Calculation A compound is analyzed and calculated to consist of 13.5 g Ca, 10.8 g O, and 0.675 g H. Find the empirical formula of the compound. Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H. 13.5 g Ca x (1 mol Ca / 40.1 g Ca) 0.337 mol Ca 10.8 g O x (1 mol O / 16.0 g O) 0.675 mol O 0.675 g H x (1 mol H / 1.01 g H) 0.668 mol H Next, divide each mole amount by the smallest number or moles (which is 0.337 for calcium) and round to the nearest whole number: 0.337 mol Ca / 0.337   1.00 mol Ca 0.675 mol O / 0.337 2.00 mol O 0.668 mol H / 0.337 1.98 mol H which rounds up to 2.00 Now you have the subscripts for the atoms in the empirical formula: CaO2H2 Finally, apply the rules of writing formulas to present the formula correctly. The cation of the compound is written first, followed by the anion. The empirical formula is properly written as Ca(OH)2

Causes And Effects Of Revolutions - 1237 Words

The Causes and Effects of Revolutions Revolutions have occurred throughout history and continue to arise even in the present day. Many revolution happens because of the miscommunication between the government and its citizens. In the end, the outcome of revolutions are usually good but, sometimes it might not be what people want. In the following paragraphs, examples of the general causes and effects that creates a revolution will be discussed in order to illustrate how revolutions are mainly caused by the dispute between the regime and its citizen. When a revolution begins, there is always a cause behind it and most of time it s the government s unfair treatment to its citizens. Inequality can come in many different types, such as†¦show more content†¦For example, in the French revolution, the bread prices rise due to the poor harvest of wheat before and during the French revolution and the deregulation of the grain market. Even though bread was expensive, people would still have to purchase and consume it in order to survive because it was the staple food in France. As a result, with the increased price of bread and the unchanged rate of taxation and wages, this caused a famine in France where many people from the lower class died from it. Therefore, it has prompted a mass of people to revolt. Hence, this leads to social inequality. Social inequality is when people in the upper class owns a greater amount of wealth than those who are in the middle and lower class. For instance, in the Chinese revolution, the regime ke pt increasing the taxes on its citizens, causing a disparity between the middle and upper classes. As a result, no matter how much the government increase in taxes, the upper class would always have the money to pay for it since they own a significant amount of wealth, while those who are in the middle class and lower class doesn t have the ability to pay for the large amount of taxes. Therefore, this causes the middle and lower classes to revolt. Despite that the regimes are treating its citizens unfairly in the past, this type of issue still occurs in the present day which causes the many different types of revolutions worldwide. In every

Poem Analysis “Out, Out” Free Essays

Poem Analysis: â€Å"Out, Out-† In the poem, â€Å"Out, Out-â€Å", author Robert Frost starts off his poem by giving an inanimate object, the buzzsaw, a sense of life. Using the literary device, Personification, the buzz saw is being written with characteristics a curious and rather playful child. The buzzsaw acts like once hears the young man’s mother call for supper time, that it wants to eat, so eats the young man’s hand. We will write a custom essay sample on Poem Analysis â€Å"Out, Out† or any similar topic only for you Order Now The buzzsaw takes (Cuts Off) the hand in a rather subtle demeanor, but in truth, it would be a very graphic to behold. Throughout the poem, everything is written in a peaceful and quite tone, even during the violent and gruesome ones to. To add to the fact of the buzzsaw is being personified in the story, the buzzsaw seems to only attack when the mother calls all for supper. The buzzsaw acts like it knows what the meaning of supper time is. Another literary device used in this poem is the process of along with the use of otomotapia(s). Using repetition first to deliver emphasis to the reader of the sounds that buzzsaw would be making, and then the actual sound being written/sounded out in an otomotapia base. The otomotapia in the story would be the grinding sounds made the buzzsaw ripping the through the poor young man’s arm. This quote from the story pretty much sums all that I describe above; â€Å"The saw snarled and rattled, snarled and rattled†¦Ã¢â‚¬  and it continues about three more times over and over. The use of the sound effects gives the once playful buzzsaw a more animalistic approach, making it seem like it is hungry after hearing the key word â€Å"supper†. To conclude, the literary devices used in Robert Frost’s poem are mostly to emphasis and give life to once lifeless piece of machinery. The story, rather bloody and saddening, is a well written example of poetry and depth behind each and every letter/word. The analysis is still to be assessed, but this all gives basic understanding as to what meant behind his more obvious literary devices. How to cite Poem Analysis â€Å"Out, Out†, Essay examples

Optical Network Design and Planning

Question: Discuss about the Optical Network Design and Planning. Answer: Network Diagram Figure: 1 Network Diagram The above network diagram portrays that a total of seven personal computer are connected to the network via wired line whereas, one laptop and one tablet computer are accessing the network through the wireless connection. In the above-designed network, diagram has total two routers and two switches. Router 1 is connected to the internet and the second router is connected to router one. Two routers are connected with two switches respectively. Switch one is connected with the router one and switch two is connected to the second router. Three PCs are connected to the switch one, and other three PCs are connected to the second switches which are subnet B and subnet A. The main network administration system is connected with the switch one. A wireless access point also connected with the switch one. A tablet computer and a laptop computer access, the network via access point which is subnet C. The subnet C network, is connected to the internet via switch and router. In the Subnet 3 netwo rk, it has wireless access to the internet. Subnet A System IP Address Subnet Mask Gateway Address PC1 172.22.0.2 255.255.254.0 172.22.0.1 PC2 172.22.0.3 255.255.254.0 172.22.0.1 PC3 172.22.0.4 255.255.254.0 172.22.0.1 Subnet B System IP Address Subnet Mask Gateway Address PC1 172.22.2.3 255.255.255.248 172.22.2.1 PC2 172.22.2.4 255.255.255.248 172.22.2.1 PC3 172.22.2.5 255.255.255.248 172.22.2.1 Subnet C System IP Address Subnet Mask Gateway Address Tablet computer 172.22.2.2 255.255.255.128 172.22.2.7 Laptop PC 172.22.2.6 255.255.255.128 172.22.2.7 The network diagram shows the various entities used in forming the network. The various other devices that need to be added to the network for ensuring the basic functionality are hub, bridge, repeater, central office server, firewall and many more. In order to understand the significant of connecting various devices, the architecture of the network need to be robust. An essential element required in the network is the Hub. Being a non-intelligent device the hub is required in the network to broadcast the data from one computer connected o the network to another device. The hub is added to the OSI model's physical layer. Thus, it has no acknowledgement of the different MAC address of the devices connected to the network. Repeater used in the network will amplify the signal while discarding the noise. Repeaters are required for broadcasting over a long distance to prevent the distorted signal. The firewall system is used for security purpose. A firewall system can be placed or connected the router, switch, or in between the internet and router. A firewall has two type of functions, one is packet filter firewall, and another one is proxy filter firewall. Packet filter firewall is used to filter the network packets, and proxy filter firewall is used to bypass the access system. A proxy server is used in proxy filter firewall to check the authorized access. A repeater can be used to in this network. The chief hardware recommendation of the network for basic functionality is a wireless router, wires based connection, and wireless NIC. The wireless router is required for the providing internet access to the components connected to the network. The wireless router acts as the hub of the wired network. Furthermore, a wired base connection or Ethernet connection is required to provide standard networking function in the network. The most vital part of the wireless network required is the Network Information Card (NIC). Usually, the laptops come with inbuilt NIC card, but in order to connect any personal computer on the desktop with the wireless network, the system must have installed NICs. To implement a wireless network few things must be considered, those are The environment creates major influence in a wireless network system. Before place any wireless access point, the environment should be checked. Application support must be enabled in a wireless network system. The application can be simple office application, like an email application, file transfer application, browsing application. It could be remote patient observing in a doctor's facility or voice telephony in a storeroom. The application prerequisites empower the network designer to indicate the material throughput, innovations and items when outlining the framework. Coverage area describes the place where users access the wireless network. The users might need only connectivity in their application. In addition, deliberately consider whether coverage is required in stairwells, lifts, and parking lots. These are hard-to-cover the area, and it can increase the wireless network device cost. By indicating the appropriate coverage, it can be avoid the extra expense for wireless network devices. Dynamic IP- Make sure to recognize whether the users are stationary or mobile, which gives evidence to incorporating upgraded roaming in the configuration. The user of mobile will move about the facility and possible roaming crosswise over IP spaces, making need to oversee IP addresses dynamically. Few users, in any case, might be stationary, for example, remote desktops. Security system describes the wireless network information sensitivity which will navigate the wireless system. The wireless network designer needs to put an authentication in the wireless network system. A robust firewall system needs to be installed in the network to provide security for communication through the network. Bibliography Simmons, J. M. (2014). Optical network design and planning. Springer. Dziubiski, M., Goyal, S. (2013). Network design and defence. Games and Economic Behavior, 79, 30-43. Ramezani, M., Bashiri, M., Tavakkoli-Moghaddam, R. (2013). A new multi-objective stochastic model for a forward/reverse logistic network design with responsiveness and quality level. Applied Mathematical Modelling, 37(1), 328-344. Quigley, T., MacInnis, A. G., Behzad, A. R., Karaoguz, J., Walley, J., Buer, M. (2015). U.S. Patent No. 9,198,096. Washington, DC: U.S. Patent and Trademark Office. Mayoral, A., Lpez, V., Gerstel, O., Palkopoulou, E., de Dios, . G., Fernndez-Palacios, J. P. (2014, March). Minimizing resource protection in IP over WDM networks: Multi-layer shared a backup router. In Optical Fiber Communication Conf.(OFC) (pp. M3B-1). Andrews, P. E., Harris, R., Plum, D. L. (2013). U.S. Patent No. 8,495,190. Washington, DC: U.S. Patent and Trademark Office. Li, Y., He, P., Hu, Y., Chen, C., Nie, J., Liang, Y. (2015). U.S. Patent Application No. 14/879,950. Tongxing, M. A., Pan, L., Yang, R., Jianguo, D. A. N. G. (2014). U.S. Patent Application No. 14/338,409. Scherzer, T., Scherzer, S. (2013). U.S. Patent No. 8,358,638. Washington, DC: U.S. Patent and Trademark Office