...For the first round of the stimulation was random mating. After the 20 students drew out two new cards from starting genotype, the new genetic pool had five pairs of dominant homozygous, ten pairs of heterozygous, and five pairs of recessive homozygous. The new generation had the same number of the starting gene pool of the known allele frequency. Random selection frequency of alleles does not change over time, due to which all individuals have an equal chance of being selected. The Hardy-Weinberg equilibrium equation, p2 + 2pq + q2 = 1, is used to define a population in which to understand both allele and genotype frequencies. The equation only occurs when mating is at random on a large population, and the relative genotype and allele frequencies persist in being constant. In all possible combination defined by the Punnett square, 60% or 0.6 of the gene pool is the A allele, and 40% or 0.4 of the a allele, totaling in for 100%. When adding the probability, the results are 0.36 for AA, 0.24 for aA, 0.24 for Aa, and 0.16 aa. In total, will equal 1. The AA homozygotes will have the frequency of p2, and similarly, aa homozygotes will have the frequency of q2 when the population is in equilibrium. Then lastly, the Aa heterozygotes will have the relative genotype frequency of 2pq. The calculation would remain the same throughout generations, but only with random mating. The Hardy-Weinberg equilibrium will not equal 1 when it is disturbed by nonrandom mating, mutations, migration, and...
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...only a small amount of power. Step 3. Traffic density metric • Select the path that is lightly loaded from the three selected paths Algorithm 1.1 Algorithm for broadcasting RReq 1.2. Algorithm for node failures reduction Step1: Source broadcasts Route Request packets which are heard by neighbor nodes within the coverage area. Step 2: all node those get the RReq will calculate their remaining battery power by using: Rp = Einitial - Eci Step 3: Also all nodes will calculate their transition power needed to transmit the RReq packet to next-hop by using the following formula: Step 4: Check that they have enough power for transmitting the packet to next-hop. Step 5: The neighboring nodes re-broadcast the route request. Step 6: The destination will select all available path that has enough power 1.3. Algorithm for reducing traffic density Step 1: once all the path those have enough power have been selected the destination nodes can count the number of hop for all the available paths and order them based on the shortest path first. Step 2: The destination node will select the path that is lightly loaded from the ordered list. Step 3: The destination will send the RRep to the source node through the selected path. Step 4: The destination sends Route Reply only to the first received Route Request. ...
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...Adedamola Iyiola M.E. LAB: Lab View Section 06 Lab View Adedamola Iyiola Department of Mechanical and Aerospace Engineering Rutgers University, Piscataway, New Jersey 08854 A Data acquisition board, thermocouple and a BNC terminal block were used simultaneously with Lab view to obtain measurements of different waveforms with varying frequencies of 500, 100, and 3000 Hz. Additionally, we varied sampling rates, at an input of 1000 Hz, at 500 Hz and 2500 Hz (0.5x and 2.5x the input wave frequency). Furthermore for the second set of measurements, we obtained data from running a thermocouple data acquisition program. We acquired temperature measurements of a room for a period of 60 seconds and an individual’s finger for 50 seconds....
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...Probability of getting no more than 0 heads: 0.0625 Probability of getting no more than 1 head: 0.3125 Probability of getting no more than 2 heads: 0.6875 Probability of getting no more than 3 heads: 0.9375 Probability of getting no more than 4 heads: 1.000 4. True 5. > pbinom(1, size=3, prob=.5) [1] 0.5 6. > round(pbinom(3, size=5, prob=.5), 2) [1] 0.81 7. > lmb expectation expectation [1] 0.25 or 15 seconds 8. The answer is .06 for minutes or 225 for seconds. 9. > round(pexp(15.5/60, rate=4), 2) # using minutes [1] 0.64 10. > pexp(40.2/60, rate=4) - pexp(10.7/60, rate=4) [1] 0.421445 11. > round(qexp(.95, rate=4), 2) # minutes [1] 0.75 > round(qexp(.95, rate=4/60), 2) # seconds [1] 44.94 12. The R command pexp(1.2, rate=3) shows the estimated probability of randomly selecting a value less than or equal to 1.2 from an exponential distribution that has a rate of 3. 13. > round(1 - pnorm(9, mean=7, sd=3),...
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...Schematic drawing of the particle passed through SEC. The colored boxes represent the SEC modules with energy deposit and white boxes are the modules without energy deposit. The group of green and orange boxes are the reconstructed clusters created due to the particle The energy of a neutral track (y-axis) is plotted against the angle between two photon candidates (x-axis) in the laboratory frame. (e) represents the data at 1.50~GeV beam energy. (c), (d), (e) and (f) represent the Monte Carlo simulation for the reactions $pd$$\to$$^3$He$~\eta$($\eta$$\to$$\gamma\gamma$), $pd\to\text{$^{3}$He}~\pi^0\pi^0\pi^0$, $pd\to\text{$^{3}$He}~\pi^0\pi^0\pi^0$ and $pd\to\text{$^{3}$He}\omega(\omega\to\pi^0\gamma)$, respectively, at 1.50~GeV beam energy. The events coming from the split-offs can be seen as a small tail near 20$^{\circ}$. % (b) is the statistical significance estimated for 1.45~GeV and 1.50~GeV beam energy, respectively. The optimal cut chosen for two energies are shown as dashed lines. The solid black hyperbola ($E > \frac{1.52}{\angle(\gamma1,\gamma2)}$) in all plots illustrates the final optimal cut used to reject the split-off tail. %cut used to reject the events from the neutral split-offs. } \label{Nsplitoff} \end{figure} % \subsubsection{Neutral Split-offs}\label{pi0gdeteffect} %\subsubsection{Neutral Split-offs {\textcolor{red}{This section need to be re-written more scientifically :-(}}}\label{pi0gdeteffect} % Fig.~\ref{Nsplitoffclus} Apart from the kinematics...
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...files and their individual roles. 4. Opening a file and different modes using which we can open a file. 5. Programs to implement various operations on files. 15.1 Introduction Most programs written in C++ handle large volumes of data that needs to be stored permanently on secondary storage devices such as hard disk. The data resides in these devices in the structure of files. A file is basically a collection of bytes stored on a secondary storage device, such as disk. Programs are written to carry out the read/write operations on these files. Following are the types of data communication a program is typically involved in. These two data communication types may be used in individual or together. 1. Transfer of data among console unit and program. Figure 15.1 Program-file-console interaction 2. Transfer of data among program and a disk file. In this lesson, we are basically concerned with various methods provided by C++ for storing and retrieving the data from files. The I/O system of C++ is capable of managing the file operations just as managed by the console input and output. The file stream acts as an interface between the program and the file. Input stream provides data to the program and output stream gets data from the program. The concept is demonstrated in the figure given below: Figure 15.2 Program-file interaction The read operation is used to create input stream between the program and the input file. And, the write operation is used to create output...
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...CHAPTER-6 EXPERIMENT RESULT 6.1 System Model For experimental purposes we assume that total 5000 jobs will be scheduled on the grid consists of 7 clusters. In general, each resource/cluster contains 80 computing nodes (Machines), and each computing node contains 1 Processing Elements (PE). The processors of computing nodes in different resources have same processing power (i.e. 1GHz). Every computing node consist RAM of 53GB approx. Network speeds among the computing node are also assumed same for different resources. The characteristics of 7resources used in our experiment, shown in Table 6.1. Parameter Name Value Total Number of Jobs 5000 Number of Cluster 07 Number of CPU per Cluster 80 RAM 53 GB Baud Rate 10000 CPU Speed 1GHz...
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...Katie Carroll Mr. Miller APP1 November 15, 2015 1. Background a. Equations: i. Σ F = T-mhg = mha ii. fk=ukN 2. Materials a. Before assembling the materials, prepare a clean, dry workspace (away from food) with all necessary materials. i. Wooden board that can be adjusted to various angles ii. Spring scale iii. Wooden block (with tape on one side) iv. Pulley v. Paperclips (weights) vi. Safety goggles 3. Procedure a. (Part 1)-determine kinetic friction between the wooden block and the wooden board b. Read through the entire procedure and prepare to carefully collect your data before you begin. c. Before assembling the materials, prepare a clean, dry workspace (away from food) with all necessary materials. d. Weigh and record the mass of the wooden block. e. Make sure the surfaced of the wooden plank and the block are clean of dirt so that nothing will disrupt the experimental friction coefficient. f. Put the wooden block (with tape side down) on the end of the wooden board and attach the string to the block. Then attach the spring scale to the other end of the pulley so that it hangs over the far end of the wooden board. g. Put mass on the spring scale and start the block with a small push. i. If the wooden block speeds up, take some of the mass off. ii. If the wooden block slows down, add more mass. iii. Once the block moves across the whole wooden board with a constant speed, record the total hanging mass and its weight. h. (Part 2)- Determine the difference surface area...
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...The attractive force between the positive charge in the nucleus and the valence electrons decreases because these electrons are farther from the nucleus. 4. a) The elements that are found at the main peaks of my graph are Lithium, Sodium, Potassium, and Rubidium. These elements are all found in the same group. They are all Alkali Metals. b) The elements that are found at the main valleys of my graph are Neon, Argon, Krypton, and Helium. These elements are all found in the same group. They are all part of the Noble Gases. 5. I predict Rubidium would be the largest atom in the atomic table because it has an atomic radius of 248pm. I predict the smallest atom in the atomic table would be Hydrogen because it has an atomic radius of 32pm. 6. There is a significant jump in the size of the nucleus (protons + neutrons) each time you move from period to period down a group. Additionally, new energy levels of elections are added to the atom as you move from period to period down a group, making the each atom significantly more massive, both is mass and volume. This makes the atomic radii bigger. Part 2 – Ionization Energy 7. The ionization energy is the exact quantity of energy that it takes to remove the outermost electron from the atom. 9. a) The elements that are found at the main peaks of my graph are Helium, Neon, Argon, and Krypton. These elements are all found in the same group. They are all part of the Noble Gases. b) The elements that are found at the main valleys of my...
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...This results in a value of (1.4 ± 0.4)s. This period can then be used to find the undamped oscillation frequency through the formula for frequency. rad/s. Uncertainty for this can be found by the following steps plugging in the appropriate values in this equation: rad/s. = (4.4880 ± 1.2823) rad/s The value of damping time was (2.5698 ± 0.1404)s. This was calculated by taking the natural log of the each ratio, the dividing the period by this value and multiplying by -1, then averaging all the values found. Uncertainty was calculated with standard deviation, in which the difference between each data point and the average was found. Then the differences were all squared, added together and then divided by the number of data points (which was 6 in this case). The recorded measurement for the driven resonant frequency we found through adjusting Lissajous plots was (4.430 ± 0.031)Hz. This was found by finding the frequency, 0.705 ±0.031 Hz, in which the Lissajous plotted by the output voltage and angle was most symmetric and did not lean to either side. This value was multiplied by 2 pi and yielded 4.430 Hz. The uncertainty of the frequency was found by seeing how much of the frequency had to change before seeing a difference in the Lissajous plots, which was 0.005Hz. The uncertainty for the driven resonant frequency was found through this formula . This value of the driven resonant frequency and the damping time can be then used to calculate a measured value of Q by the following...
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...ITT Technical Institute IS3340 Windows Security Onsite Course SYLLABUS Credit hours: 4.5 Contact/Instructional hours: 60 (30 Theory Hours, 30 Lab Hours) Prerequisite(s) and/or Corequisite(s): Prerequisite: NT2580 Introduction to Information Security or equivalent Course Description: This course examines security implementations for a variety of Windows platforms and applications. Areas of study include analysis of the security architecture of Windows systems. Students will identify and examine security risks and apply tools and methods to address security issues in the Windows environment. Windows Security Syllabus Where Does This Course Belong? This course is required for the Bachelor of Science in Information Systems Security program. This program covers the following core areas: Foundational Courses Technical Courses BSISS Project The following diagram demonstrates how this course fits in the program: IS4799 NT2799 IS4670 ISC Capstone Project Capstone ProjectCybercrime Forensics NSA NT2580 NT2670 Introduction to Information Security IS4680 IS4560 NT2580 NT2670 Email and Web Services Hacking and Introduction to Security Auditing for Compliance Countermeasures Information Security Email and Web Services NT1230 NT1330 Client-Server Client-Server Networking I Networking II IS3230 IS3350 NT1230 NT1330 Issues Client-Server Client-Server SecurityContext in Legal Access Security Networking I Networking II NT1110...
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