...071-326-0501 Move as a Member of a Fire Team Conditions: In a designated position (other than team leader) in a moving fire team. Standards: React immediately to the fire team leader’s example. Perform the same actions as the fire team leader does in the designated position within the formation. Performance Steps 1. Fire team formations describe the relationship of the Soldiers in the fire team to each other. Standard fire team formations are the wedge (figure 071-326-0501-1), modified wedge (figure 071-326-0501-2), diamond (figure 071-326-0501-3), and file (figure 071-326-0501-4). a. Fire team wedge (figure 071-326-0501-1). This is the basic fire team formation which— (1) Is easy to control. (2) Is flexible. (3) Allows immediate fires in all directions. (4) Offers all round local security. Figure 071-326-0501-1. Fire team wedge STP 21-1-SMCT 18 June 2009 071-326-0501 3-165 Performance Steps b. Fire team modified wedge (figure 071-326-0501-2). When rough terrain, poor visibility, or other factors reduce control of the wedge formation, the sides are closed up to (almost) a single file. When moving in less rugged terrain and control becomes easier, resume your original positions. The modified wedge is also used for extended periods when traveling on roads or trails. The modified wedge— (1) Is easier to control in reduced visibility or rough terrain than are other formations. (2) Offers less flank security than a...
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...NRIC: 3087Z SECTION: G2 NRIC: 3087Z SECTION: G2 RESEARCH QUESTION The article focused on how consumers make decisions in remote and in-store environments. Remote environments are those where products cannot be physically examined and only descriptions (both visual and verbal) are available. An example of a remote environment is an online store such as Amazon.com. In-store environments are those where real products can be handled and touched. An example of an in-store environment is the local supermarket, Cold Storage. The rationale behind the research was that consumers’ preferences might differ based on the type of information they get in the two different environments. In remote environments, consumer must rely on information provided through the sense of vision. In in-store environments, consumers can use the sense of touch to add information to their purchase decisions. The research question addressed by this study was: What is the effect of the two different types of information – information gathered through the sense of touch and information gathered through vision – on consumer decision-making and preferences? INDEPENDENT, DEPENDENT AND MODERATING VARIABLES The independent variable in the main study was the type of environment: remote or in-store. The dependent variable in this study was the number of times a product was chosen. The moderating variable in this study was the product type. A typical product can be classified into 2 general categories: geometric...
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...Unit 6: Graph Theory - Assignment Total points for Assignment: 35 points. Assignments must be submitted as a Microsoft Word document and uploaded to the Dropbox for Unit 6. All Assignments are due by Tuesday at 11:59 PM ET of the assigned Unit. NOTE: Assignment problems should not be posted to the Discussion threads. Questions on the Assignment problems should be addressed to the instructor by sending an email or by attending office hours. You must show your work on all problems. If a problem is worth 2 points and you only show the answer, then you will receive only 1 point credit. If you use a calculator or online website, give the source and tell me exactly what you provided as input. For example, if you used Excel to compute 16 * 16, state “I typed =16*16 into Excel and got 256. You may type your answer right into this document. Part I. Basic Computations 1. (4 points) The plan for a four-room house is shown below. Draw a graph that models the connecting relationships between the areas in the floor plan. [Your graph does not [Your graph does not need to be fancy. You may use any drawing software such as Visio or Creatly.com] Answer: I used viso for graph[pic] 2. a. Identify all the vertices in the above graph with odd degree. Identify the degree of each of these vertices. (2 points) Answer: The odd number of edges is a odd degree vertices are D,E,F from the graph 3,1,3 it has a odd number. So D,E,F, is odd and the rest is...
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...Food Webs Report Week 5 MTH / 221 University of Phoenix Food Webs It may be difficult to know all the factors which determine an ecological niche, and some factors may be relatively unimportant. Hence it is useful to start with the concept of competition and try to find the minimum number of dimensions necessary for a phase space in which competition can be represented by niche overlap. One approach to this question is to consider the notion of the food web of an ecological community. Definition 1 A food web of an ecological community is a directed graph with a vertex for each species in the community and a directed edge from the vertex representing species A to the vertex representing species B if and only if A preys on B We can define competition using the food web. Two species compete if and only if they have a common prey. Thus, a Hawk and Wolf compete (since Rabbit is a common prey); Rabbit and Grasshopper compete, while Deer and Toad do not compete. We use this competition relation to define a graph called the competition graph. Definition 2 The competition graph of a food web is a simple graph with a vertex for each species. Two vertices are joined by an edge if and only if the species they represent have a common prey. Definition 3 A graph is an intersection graph for a family of sets if each vertex is assigned a set in such a way that two vertices are joined by an edge if and only if the corresponding sets have non-empty...
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...bandwidth and Channel capacity is not linear(proportional) relationship by examining the Shannon–Hartley theorem- , where capital C means Channel capacity and capital W means bandwidth. 2. Explain if ring topology is a good choice for WANs. ANS: WAN is a computer networking technology used to transmit data over long distances, and between different LANs, MANs. Thus, at first glance, we can say ring topology may NOT be a good choice for WAN because of the following reasons: The communication delay of ring topology is directly proportional to number of nodes in the network. It will result in serious delay if ring topology is being chosen to implement large networks such as WAN. Moving, adding and changing the devices (or even one malfunctioning device) in ring topology can affect the whole network. Especially in the large network such as WAN, there are hundreds of thousands of routers may be under maintenance. If one or more alternate devices are not being used to replace the original one, the whole network may be shut down. Bandwidth is shared between devices, which will result in low channel capacity if the ring topology is being used in case like WAN. However, there are ring protection methods such as line and path switched rings. These topologies evolved from telephony networks as an essential part of the SONET standard. Nodes on a switched ring monitor the health of optical transmission and in the event of a failure or cable break, switch a loop-back around...
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...ICT PROJECT ON TOPOLOGIES. STAR TOPOLOGY. In the star network topology, there is a central computer or server to which all the workstations are directly connected. Every workstation is indirectly connected to every other through the central computer. Star networks are one of the most common computer network topologies. ADVANTAGES :- * Better performance: star topology prevents the passing of data packets through an excessive number of nodes. At most, 3 devices and 2 links are involved in any communication between any two devices. * Easy to detect faults and to remove parts. * Installation and configuration is easy since every one device only requires a link and one input/output port to connect it to any other device(s). DISADVANTAGES :- * High dependence of the system on the functioning of the central hub. Failure of the central hub renders the network inoperable. * It can be expensive to purchase. * Star topology requires a large amount of cable to be connected. RING TOPOLOGY. In the ring network topology, the workstations are connected in a closed loop configuration. Adjacent pairs of workstations are directly connected. Other pairs of workstations are indirectly connected, the data passing through one or more intermediate nodes. Data travel from node to node, with each node along the way handling every packet. Because a ring topology provides only one pathway between any two nodes, ring networks may be disrupted by the failure of...
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...MTH 333 Writing Project Abdullah Aurko MTH 333 Writing Project Topic- Connectedness: A topological property I am working on concept of connectedness of a topological space as the topic for my writing project. I wish to proceed by giving a formal definition of the concept with proof and proving that connectedness is a topological property. This will be followed up with the proofs of several interesting and essential theorems involving connectedness of a topological space and how it enables us to distinguish between different topological spaces. A topological space X is disconnected if X=A B, where A and B are disjoint, nonempty, open subsets of X. (Roseman, 1999) Definition A topological space X is connected if it is not disconnected. Examples 1. A closed interval [a,b] is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets and whose union is [a,b]. Let X be the set equal to A or B and which does not contain the point b. Let s= sup X, the supremum of X. Since X does not contain b, s must be within the interval [a,b] and thus must be within either X or [a,b]\X. If s is within X, then by the definition of supremum there is an open set (s-ε, s+ε) within X. If s is not within X, then is within [a,b]\X, which is also open, and there is an open set (s-ε, s+ε) within [a,b]\X. Either case implies that s is not the supremum. 2. The topological space X=(0,1)\{1/2) is disconnected: A=(0,1/2), B=(1/2,1) (MathWorld) Lemma The following...
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...geometry and combinatorial geometry are branches of geometry that study combinatorial properties and constructive methods of discrete geometric objects. Most questions in discrete geometry involve finite or discrete sets of basic geometric objects, such as points, lines, planes, circles, spheres, polygons, and so forth. The subject focuses on the combinatorial properties of these objects, such as how they intersect one another, or how they may be arranged to cover a larger object. Discrete geometry has large overlap with convex geometry and computational geometry, and is closely related to subjects such as finite geometry, combinatorial optimization, digital geometry, discrete differential geometry, geometric graph theory,toric geometry, and combinatorial topology. Although polyhedra and tessellations have been studied for many years by people such as Kepler and Cauchy, modern discrete geometry has its origins in the late 19th century. Early topics studied were: the density of circle packings by Thue,projective configurations by Reye and Steinitz, the geometry of numbers by Minkowski, and map colourings by Tait, Heawood, and Hadwiger. László Fejes Tóth, H.S.M. Coxeter and Paul Erdős, laid the foundations of discrete geometry. "This is an introduction to the field of discrete geometry understood as the investigation of combinatorial properties of configurations of (usually finitely many) geometric objects … . The book is written in a lively and stimulating but very precise style and contains many...
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...Sample Exam 2 - MATH 321 Problem 1. Change the order of integration and evaluate. (a) (b) 2 0 1 0 1 (x y/2 + y)2 dxdy. + y 3 x) dxdy. 1 0 0 x 0 y 1 (x2 y 1/2 Problem 2. (a) Sketch the region for the integral f (x, y, z) dzdydx. (b) Write the integral with the integration order dxdydz. THE FUNCTION f IS NOT GIVEN, SO THAT NO EVALUATION IS REQUIRED. Problem 3. Evaluate e−x −y dxdy, where B consists of points B (x, y) satisfying x2 + y 2 ≤ 1 and y ≤ 0. − Problem 4. (a) Compute the integral of f along the path → if c − f (x, y, z) = x + y + yz and →(t) = (sin t, cos t, t), 0 ≤ t ≤ 2π. c → − → − → − (b) Find the work done by the force F (x, y) = (x2 − y 2 ) i + 2xy j in moving a particle counterclockwise around the square with corners (0, 0), (a, 0), (a, a), (0, a), a > 0. Problem 5. (a) Compute the integral of z 2 over the surface of the unit sphere. → → − − → − → − − F · d S , where F (x, y, z) = (x, y, −y) and S is → (b) Calculate S the cylindrical surface defined by x2 + y 2 = 1, 0 ≤ z ≤ 1, with normal pointing out of the cylinder. → − Problem 6. Let S be an oriented surface and C a closed curve → − bounding S . Verify the equality → − → − → → − − ( × F ) · dS = F ·ds − → → − if F is a gradient field. S C 2 2 1...
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...Exercises in Classical Real Analysis Themis Mitsis Contents Chapter 1. Numbers 5 Chapter 2. Sequences, Series and Limits 11 Chapter 3. Topology 23 Chapter 4. Measure and Integration 29 3 CHAPTER 1 Numbers E 1.1. Let a, b, c, d be rational numbers and x an irrational number such that cx + d 0. Prove that (ax + b)/(cx + d) is irrational if and only if ad bc. S. Suppose that (ax + b)/(cx + d) = p/q, where p, q ∈ Z. Then (aq − cp) x = d p − bq, and so we must have d p − bq = aq − cp = 0, since x is irrational. It follows that ad = bc. Conversely, if ad = bc then (ax + b)/(cx + d) = b/d ∈ Q. E 1.2. Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be real numbers. Prove that n i=1 ai n j= 1 b j ≤ n n ak bk k=1 and that equality obtains if and only if either a1 = an or b1 = bn . S. Since {ai }n=1 and {bi }n=1 are both increasing, we have i i n n 0≤ (ai − a j )(bi − b j ) = 2n ak bk − 2 ai 1≤i, j≤n k =1 i= 1 n j=1 b j . If we have equality then the above implies (ai − a j )(bi − b j ) = 0 for all i, j. In particular (a1 − an )(b1 − bn ) = 0, and so either a1 = an or b1 = bn . E 1.3. (a) If a1 , a2 , . . . , an are all positive, then n n 1 a ≥ n2 ...
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...Chapter 1 Overview of Statistics Chapter 2 Data Collection Assignment (32 points due by 11 pm September 30th) Note: You can team up with one of your classmates to complete the assignment (not more than two in a team); if you want to work on the assignment individually, that’s also fine. If you are working in teams, then only one submission is required per team; include both the team members’ last names as part of the assignment submission file name as well as in the assignment submission document. Please provide detailed solutions to the following problems/exercises (4 problems/exercises x 8 points each): 1) Which type of data (categorical, discrete numerical, continuous numerical) is each of the following variables? a) Number of spectators at a randomly chosen US Open tennis match. Continuous numerical. b) Water consumption (in liters) by a randomly chosen US Open player during a match. Discrete numerical. c) Gender of a randomly chosen tennis player in the US Open tennis tournament. Categorical 2) Which measurement level (nominal, ordinal, interval, ratio) is each of the following variables? a) Number of annual office visits by a particular Medicare subscriber. Ordinal b) Daily caffeine consumption by a six year old child. Interval c) Type of vehicle driven by a college student. Nominal 3) Would you use a sample or a census to measure each of the following? Why? a) The number of cans of Campbell’s soup on your local supermarket’s...
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...MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 Generally, a “solution” is something that would be acceptable if turned in in the form presented here, although the solutions given are often close to minimal in this respect. A “solution (sketch)” is too sketchy to be considered a complete solution if turned in; varying amounts of detail would need to be filled in. Problem 1.1: If r ∈ Q \ {0} and x ∈ R \ Q, prove that r + x, rx ∈ Q. Solution: We prove this by contradiction. Let r ∈ Q\{0}, and suppose that r +x ∈ Q. Then, using the field properties of both R and Q, we have x = (r + x) − r ∈ Q. Thus x ∈ Q implies r + x ∈ Q. Similarly, if rx ∈ Q, then x = (rx)/r ∈ Q. (Here, in addition to the field properties of R and Q, we use r = 0.) Thus x ∈ Q implies rx ∈ Q. Problem 1.2: Prove that there is no x ∈ Q such that x2 = 12. Solution: We prove this by contradiction. Suppose there is x ∈ Q such that x2 = 12. Write x = m in lowest terms. Then x2 = 12 implies that m2 = 12n2 . n Since 3 divides 12n2 , it follows that 3 divides m2 . Since 3 is prime (and by unique factorization in Z), it follows that 3 divides m. Therefore 32 divides m2 = 12n2 . Since 32 does not divide 12, using again unique factorization in Z and the fact that 3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and n, contradicting the assumption that the fraction m is in lowest terms. n Alternate solution (Sketch): If x ∈ Q satisfies x2 = 12, then x is in Q and satisfies 2 x 2 = 3. Now prove that...
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...Recall our main theorem about vector fields. Theorem. Let R be an open region in E2 and let F be a C1 vector field on R. The following statements about F are equivalent: (1) There is a differentiable function f : R → R such that ∇f = F. (2) If C is a piecewise C1 path in R, then C F · dx depends only on the endpoints of C. (3) C F · dx = 0 for every piecewise C1 simple, closed curve in R. Furthermore, statements (1)–(3) imply (4) curl F = 0, and (4) implies (1)–(3) when R is simply connected (so all four are equivalent when R is simply connected). The purpose of this handout is to explore what happens when R is not simply connected, that is, when there exist non-gradient vector fields with zero curl. An important example Consider the vector field F = −y x2 + y2 i + x x2 + y2 j defined on R = R2\{(0, 0)}, that is, on all of R2 except the origin. Letting P = −y/(x2+y2) and Q = x/(x2 + y2), it is a simple matter to show that ∂Q ∂x = ∂P ∂y , and so curlF = 0. We can show that (1)–(3) are false for F by finding a simple, closed curve C for which C F · dx = 0. Let C be the unit circle parameterized counterclockwise by x = cost, y = sin t, 0 ≤ t ≤ 2π. We have C F · dx = C −y x2 + y2 dx + x x2 + y2 dy = 2π 0 − sin t 1 (− sin t) dt + cost 1 cost dt = 2π 0 dt = 2π. It follows from the theorem that F is not the gradient of any...
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...Math 245C Homework 1 Yunbai Cao 904066974 Apr 11, 2014 Exercise 1.10.3 Proof. Since R equipped with the usual topology F is Hausdorff, (R, F ) is Haudorff as F is stronger than F. Given any x ∈ R, for any y ∈ R\{x}, there exists By open such that By ∩ Byx = ∅. Then R\{x} = is open. Let K = [0, 1]\Q and L = { 1 }. Then K c = (−∞, 0) ∪ (1, ∞) ∪ Q ∈ F , so K 2 is closed. And L is closed by previous paragraph. Let U ⊃ K, V ⊃ L be two open neighbourhoods of K, L. Claim: U ∩ V = ∅. As F is generated by open intervals of R and {Q}, there exists 1 2 y∈R\{x} By y open and Byx x is open. Therefore {x} ∈ V ⊂ V where V is a finite intersection of open intervals (thus an open interval) and {Q}. So L ⊂ I ∩ Q ⊂ V where I is an open interval. On the other hand, as K ∩ Q = ∅, we may assume U ∈ F ( since we can always find an open subset U of U such that K ⊂ U ∈ F). As I 1 2 is an open 1 2 interval, U ∩ I = ∅ since otherwise there will be some irrational number close to in (R, F). Thus U ∩ V = ∅ as claimed. Therefore (R, F ) is not normal. that is not contained in U . Therefore U ∩ I ∩ Q = ∅ as U ∩ I ∈ F and Q is dense Exercise 1.10.6 Proof. Since X is locally compact, for any x ∈ K there exists Ux that U x is compact. Then K ⊂ cover of K, we may write K ⊂ then O = n i=1 U i x∈K Ux . Since K is compact, there n i=1 Ui with Ui ∈ {Ux : x ∈ K}. Let x open such exists a finite O= n i=1 Ui , is compact since finite union of compact sets is compact (Exercise 1.8...
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...Sequence - 26 (Use First) Precision - 29 (Use First) Technical - 29 (Use First) Confluent - 20 (Use as Needed) Sequence: Sequence is my "use first" for learning, and I use this Learning Pattern first. I utilize this when it comes to my job, I use a very methodicaly system. I always get with the intel shop, prior to planning any missions. After the intel dump, I then go look to see what other teams have been to this venue. If they have, then I look at their routes and times they have used. Because one thing to avoid is using patterns, and same routes. Another part of my jobs is detecting other induviuals anomalies. Sequence is an important life factor for my life and is sometimes hard to implement. Precision: This very accurately precision...
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