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https://apps.dtic.mil/sti/citations/ADA158555	[ "# Abstract:\n\nThis document shows that a sum of dependent random variables is approximately compound Poisson when the variables are rarely nonzero and, given they are nonzero, their conditional distributions are nearly identical. It give several upper bounds on the total-variation distance between the distribution of such a sum and a compund Poisson distribution. Included is an example for Markovian occurrences of a rare event. The bounds are consistent with those that are known for Poisson approximations for sums of uniformly small random variables. Author\n\n# Subject Categories:\n\n• Statistics and Probability" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8694123,"math_prob":0.96298885,"size":891,"snap":"2021-43-2021-49","text_gpt3_token_len":181,"char_repetition_ratio":0.10935739,"word_repetition_ratio":0.0,"special_character_ratio":0.18630752,"punctuation_ratio":0.11486486,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9771549,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T14:49:30Z\",\"WARC-Record-ID\":\"<urn:uuid:d92eda3f-3e5f-4e90-973a-91edbc5c10b9>\",\"Content-Length\":\"33451\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d505ee3-7e7b-493b-a8de-c884864e6bfa>\",\"WARC-Concurrent-To\":\"<urn:uuid:660b6f45-50ee-4983-b347-20d378f16118>\",\"WARC-IP-Address\":\"140.18.188.11\",\"WARC-Target-URI\":\"https://apps.dtic.mil/sti/citations/ADA158555\",\"WARC-Payload-Digest\":\"sha1:WFNK47US5REGT3526QZZLJ4PQWFSAFLN\",\"WARC-Block-Digest\":\"sha1:KE5YDI5GZSYPQ7VHTEY6Q4DCDAAKZHK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584886.5_warc_CC-MAIN-20211016135542-20211016165542-00350.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/1401.3305/	[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Open Quantum Walks on Graphs\n\nS. Attal Université de Lyon, Université de Lyon 1, C.N.R.S., Institut Camille Jordan, 21 av Claude Bernard, 69622 Villeubanne cedex, France F. Petruccione Quantum Research Group, School of Chemistry and Physics and National Institute for Theoretical Physics, University of KwaZulu-Natal, Durban, 4001, South Africa I. Sinayskiy Quantum Research Group, School of Chemistry and Physics and National Institute for Theoretical Physics, University of KwaZulu-Natal, Durban, 4001, South Africa\n###### Abstract\n\nOpen quantum walks (OQW) are formulated as quantum Markov chains on graphs. It is shown that OQWs are a very useful tool for the formulation of dissipative quantum computing algorithms and for dissipative quantum state preparation. In particular, single qubit gates and the CNOT-gate are implemented as OQWs on fully connected graphs. Also, dissipative quantum state preparation of arbitrary single qubit states and of all two-qubit Bell-states is demonstrated. Finally, the discrete time version of dissipative quantum computing is shown to be more efficient if formulated in the language of OQWs.\n\n###### keywords:\nOpen quantum walk, dissipative quantum computing, dissipative state engineering\njournal: Physics Letters A\n\n## 1 Introduction\n\nRecently, the experimental realization of a quantum computer has been the focus of extensive research QCR . One of the main problems of the physical implementation of well known quantum algorithms QA is the creation and manipulation of entanglement between qubits. Any physical system is subject to interaction with the environment, which inevitably leads to decoherence and dissipation toqs . One way to compensate for this destructive environmental influence in the unitary implementations of the quantum algorithms is to introduce error-correcting codes QECC . However, this approach treats the interaction with the environment as an effect the influence of which needs to be minimized.\n\nA paradigm shift in looking for alternative strategies to realize quantum computers was induced with the theoretical prediction that dissipation can be used to create complex entangled states Diehl and to perform universal quantum computation Vers . This fundamental change is based on the assumption that one can manipulate the coupling of a system to an environment in such a way that the system is driven towards a thermal state, which is the solution of a particular quantum computing task Vers or a target state in quantum state engineering Diehl ; Vers . The feasibility of this strategy was demonstrated by implementing dissipative quantum state engineering with ensembles of atoms Polzik and trapped ions Blatt .\n\nQuantum algorithms for universal quantum computing are conveniently formulated in the language of quantum walks uqwqc . For example, a discrete time quantum walk implementation of the search algorithm on complex graphs has been shown to be more efficient than other known implementations of this quantum algorithm QS . As for any other unitary implementation of quantum computing the efficiency of the quantum walk based realization decreases due to interaction with the environment. In view of the appeal of dissipative quantum computing it seems natural to formulate a dissipative version of the quantum walk so that algorithms for dissipative quantum computing and quantum state engineering are implemented efficiently. In other words, if dissipative quantum computing can make use of the interaction with the environment for performing universal quantum computation, can one introduce a framework which will use dissipative rather than unitary dynamics as a ”driving force” of the quantum walk?\n\nDuring the last few years attempts were made to take into account dissipation and decoherence in the description of quantum walks ken1 . In these approaches decoherence is treated as an extra modification of the unitary quantum walk scheme, the effect of which needs to be minimized and eliminated. In fact, the general framework of quantum stochastic walks was proposed qsw , which incorporates unitary and non-unitary effects of the quantum Markovian dynamics. In particular, by adding extra decoherence in experimental realizations of quantum walks, the transition from quantum to classical random walks was observed exp .\n\nRecently, a formalism for discrete time open quantum walks (OQW), which is exclusively based on the non-unitary dynamics induced by the environment was introduced APSS . OQWs are formulated in the language of quantum Markov chains gudder and rest upon the implementation of appropriate completely positive maps toqs ; kraus .\n\n## 2 Formalism\n\nTo review briefly the formalism of OQWs, we consider a random walk on a set of nodes or vertices with oriented edges as illustrated in Fig. 1(a). The number of nodes is considered to be finite or countable infinite. The space of states corresponding to the dynamics on the graph specified by the set of nodes will be denoted by and has an orthonormal basis . The internal degrees of freedom of the quantum walker, e.g. the spin or -energy levels, will be described by a separable Hilbert space attached to each node of the graph. More concisely, any state of the quantum walker will be described on the direct product of the Hilbert spaces .", null, "Figure 1: Schematic illustration of the formalism of the Open Quantum Walk. (a) The walk is realized on a graph with a set of vertices denoted by i,j,k∈V. The operators Bji describe transitions in the internal degree of freedom of the “walker” jumping from node (i) to node (j). (b) The simplest non-trivial example of the OQW on the finite graph is a walk on a two-node network. In this case the walk is performed using four operators Mji(i,j=1,2). In particular, the transitions between node 1 and node 2 are induced by the operators M21=B1⊗\|1⟩⟨2\| and M12=B2⊗\|2⟩⟨1\|; the operators describing changes in internal degrees of freedom of the ”walker”, if the ”walker” does not jump, are Mii=Ci⊗\|i⟩⟨i\|,(i=1,2).\n\nTo describe the dynamics of the quantum walker, for each edge we introduce a bounded operator . This operator describes the change in the internal degree of freedom of the walker due to the shift from node to node . By imposing for each that,\n\n ∑iBij†Bij=I, (1)\n\nwe make sure, that for each vertex of the graph there is a corresponding completely positive map on the positive operators of : Since the operators act only on and do not perform transitions from node to node , an operator can be introduced in the following form It is clear that, if the condition expressed in Eq. (1) is satisfied, then . This latter condition defines a completely positive map for density matrices on , i.e.,\n\n M(ρ)=∑i∑jMijρMij†. (2)\n\nThe above map defines the discrete time open quantum walk (OQW). For an arbitrary initial state the density matrix will take a diagonal form after just one step of the open quantum walk. By the direct insertion of an arbitrary initial condition in Eq. (2) we get\n\n M(∑k,mρk,m⊗\|k⟩⟨m\|) = ∑i,j,k,mBij⊗\|i⟩⟨j\|(ρk,m⊗\|k⟩⟨m\|)Bij†⊗\|j⟩⟨i\| (3) = ∑i,j,k,mBijρk,mBij†⊗\|i⟩⟨i\|δj,kδj,m = ∑i(∑jBijρj,jBij†)⊗\|i⟩⟨i\|.\n\nHence, we will assume that the initial state of the system has the form , with . It is straightforward to give an explicit formula for the iteration of the OQW from step to step : , where . This iteration formula gives a clear physical meaning to the mapping : the state of the system on site is determined by the conditional shift from all connected sites , which are defined by the explicit form of the operators . Also, one can see that . Generic properties of OQWs have been discussed in APSS .\n\nAs a first illustration of the application of the formalism of open quantum walks we consider the walk on a 2-node graph (see Fig. 1b). To be specific, the transition operators are and and the operators describing changes in internal degrees of freedom of the ”walker”, if the ”walker” does not jump, are . In this case for each node we have:\n\n B†iBi+C†iCi=I. (4)\n\nThe state of the walker after steps is given by,\n\n ρ[n]=ρ[n]1⊗\|1⟩⟨1\|+ρ[n]2⊗\|2⟩⟨2\|, (5)\n\nwhere the particular form of the is found by recursion,\n\n ρ[n]1=C1ρ[n−1]1C†1+B2ρ[n−1]2B†2, (6) ρ[n]2=C2ρ[n−1]1C†2+B1ρ[n−1]2B†1.", null, "Figure 2: Open quantum walk on Z. (a) A schematic representation of the OQW on Z: all transitions to the right are induced by the operator Bi+1i≡B, while all transitions to the left are induced by the operator Bi−1i≡C (see Eq. (7)); Figures (b)-(e) show the occupation probability distribution for the “walker” with the initial state I2/2⊗\|0⟩⟨0\| and transition operators given by Eq. (7) after 10, 20, 50 and 100 steps, respectively. Two distinctive behaviors of the walker are observed: a Gaussian wave-packet moving slowly to the left (dots) and a deterministic trapped state propagating to the right at a speed of 1 in units of cells per time step (cross).\n\nNext we consider an open quantum walk on the line. The situation is depicted in Fig. 2 (a). For this simple one-dimensional walk, the only non-zero transitions out of cell are given by operators of the form and . Obviously, the operators and satisfy the condition , as imposed in Eq. (1). Assuming the initial state of the system to be localized on site , i.e., , after one step the system will jump to sites so that the new density matrix will be . The procedure can easily be iterated. In Figs. 2(b)-(e) we show the probability to find a “walker” on a particular lattice site for different numbers of steps. For this simulation we have chosen the transition matrices and as follows,\n\n B=sinθ\|−⟩⟨−\|+\|+⟩⟨+\|,C=cosθ\|−⟩⟨−\|, (7)\n\nand . One can see that already after 10 steps, there are two distinctive behaviors of the “walker”. The first is a Gaussian wave-packet moving slowly to the left and the second one is a completely deterministic trapped state propagating to the right at a speed of 1 in units of cells per time step. Interestingly, the state of the “walker” in the “soliton like” part is given by , while in the other, Gaussian part is given by the , where, of course and states are defined as and is the probability to find the “walker” on the site . Even this simple example demonstrates a remarkable dynamical richness of OQW. Further examples of OQW on which show a behavior distinct from the classical random walk and the unitary quantum walk can be found in APSS .\n\n## 3 OQW for dissipative quantum computing and state engineering", null, "Figure 3: OQW preparation of the Bell-states of two qubits. A generic OQW walk on a 4-node graph is decomposed in two walks on two-nodes graphs: “left-right” and “down-up” walks. For an initial unpolarized state of two-qubits on any nodes ρ0=14I4⊗\|i⟩⟨i\| the corresponding Bell-pairs are obtained by measuring the position of the walkers after performing the OQW.\n\nTo motivate the potential of the suggested approach for the formulation of quantum algorithms for dissipative quantum computing and quantum state engineering we consider the example of an OQW on the 2-node fully connected graph in Fig. 1 (b). This example, will show that it is possible to implement all single-qubit gates and the CNOT-gate in the language of the suggested formalism. To be specific, in order to realize an X-gate with OQWs we prepare the system in some initial state in node 1 and we will read the result of the computation in the node 2 (See Fig. 1(b)). If we choose , , and , where and are positive constants such that , then the stationary state of this walk will have the following form . Therefore, the OQW on this graph realizes the X-gate with probability upon read-out of the presence of the “walker” in node 2. In a similar way, in order to implement the CNOT-gate we initially place two “walkers” in node 1. We choose , , and and if the presence of both “walkers” is measured in the node 2 then the OQW realized the CNOT-gate with probability .\n\nOn the same 2-node network we can also implement dissipative state preparation (see Fig. 1(b)). To this end, we consider trivial transition matrices on the node 1, i.e., and non-unitary transition matrices on the node 2, i.e., and , where and are positive constants such that , and . With this choice an arbitrary initial state will converge to a unique steady state, namely . The probability of detecting the system in the steady state after steps of the walk is given by , where are the elements of the initial density matrix of the system, .\n\nOQWs on more complex graphs allow the dissipative preparation of entangled multi-qubit states. With two “walkers” on a 4-node network (see Fig. 3) we can prepare all two-qubits Bell-states. In this particular case, the OQW can be decomposed in a combination of two walks on two independent 2-node networks. The first “walker” moves up and down, while the second one moves left and right (see Fig.3). We choose the transition operators to be , , and , where and denotes Pauli matrices acting on the corresponding qubit (). Starting with “walkers” initially in an unpolarized state in any node, i.e. , the OQW will converge to a state . This means that measuring the position of the “walkers” will determine corresponding Bell-state of their internal degrees of freedom note .", null, "Figure 4: Efficient transport with Open Quantum Walk. Fig. (a): a scheme of the chain of the N nodes with neighbor-neighbor interaction. Fig. (b): occupation probability distribution as a function of time and lattice sites. The initial state of the walker is localized in the first node and given by ρ0=12I2⊗\|1⟩⟨1\|.\n\nAnother application of the OQWs is a description of a dissipatively driven quantum bus between computational quantum registers. To this end we consider a chain of nodes (see Fig. 4(a)). Initially, the first node of the chain is in the exited state, so that . To be specific, we chose transition operators and as follows, and . It is very interesting to note that in this case the state propagates through the chain with velocity almost equal to 1 (in units of cells per time step): the initial excitation in node (1) is completely transferred to the last node (N) in N+2 steps. In Fig. 4 (b) we consider a 100 node chain with , , and show that the initial excitation reaches the final node (100) in 102 steps. The high performance of transport of excitations in the OQW formalism opens up new avenues of research into the understanding of quantum efficiency in open systems.\n\nOQWs include the discrete time version of the dissipative quantum computing (DQC) introduced by Verstraete et al. Vers . In the original setup a linear chain of time registers is considered and the initial state is prepared in the time register . A quantum computation is performed by the dissipative evolution of the system into its unique steady state Vers . The result of the quantum computation can be read-out by measuring the state of the system in the last time register which is given by , where is an appropriate sequence of unitary operators Vers . The probability of a successful read-out is . A discrete time version of DQC can be realized as an OQW on a linear chain of time registers (Fig. 5) by choosing the transition operators as it is shown in Fig. 5 and constants . However, with the same number of steps in the OQW formulation of DQC the probability of successful read-out can be increased arbitrarily close to one. In order to understand this dramatic improvement in efficiency we recall that in the original DQC formulation the probability of read-out of the final state is determined by the form of “jumping” operators between time registers, i.e., . The probability to ”jump” forward and backward in the time register is the same. In the OQW formulation of the DQC we have the freedom of choosing which induces the steady state of the OQW with probability of read-out of the final state between and .", null, "Figure 5: OQW formulation of dissipative quantum computing (DQC). The initial state is prepared in the time-register 0. After performing the OQW, the results of the algorithm can be readout from the time-register N. The internal state of the ”walker” in the register T will be given by \|ψT⟩=UT…U1\|ψ0⟩. The positive constants ω and λ satisfy ω+λ=1.\n\n## 4 Conclusion\n\nIn conclusion, we have shown that the recently introduced formalism of OQWs is a very useful tool for the formulation of dissipative quantum computing algorithms and for dissipative quantum state preparation. OQWs are to dissipative quantum computing what Hadamard quantum walks are to circuit based quantum computing. In particular, we have shown OQW implementation of circuit and dissipative models of quantum computing. Remarkably, the OQW discretisation of dissipative quantum computing increases the probability of successful implementation of the quantum algorithm with respect to the original formulation. It is to be expected, that OQWs will lead to the optimal formulation of certain classes of quantum algorithms. Furthermore, we have indicated that OQW can be used to explain non-trivial highly efficient transport phenomena not only in linear but also in more complex topologies of the underlying graphs. This implies that OQW as quantum walks which are driven by dissipation and decoherence are one of the candidates for understanding the remarkable transport efficiency in photosynthetic complexes qbt . We expect the potential of this framework to be soon revealed in the realms of quantum computing and quantum biology.\n\n## Acknowledgements\n\nThis work is based upon research supported by the South African Research Chair Initiative of the Department of Science and Technology and National Research Foundation. Work supported by ANR project “HAM-MARK” N ANR-09-BLAN-0098-01" ]	[ null, "https://media.arxiv-vanity.com/render-output/2952892/x1.png", null, "https://media.arxiv-vanity.com/render-output/2952892/x2.png", null, "https://media.arxiv-vanity.com/render-output/2952892/x3.png", null, "https://media.arxiv-vanity.com/render-output/2952892/x4.png", null, "https://media.arxiv-vanity.com/render-output/2952892/x5.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8771803,"math_prob":0.9727628,"size":17464,"snap":"2020-24-2020-29","text_gpt3_token_len":4194,"char_repetition_ratio":0.15595648,"word_repetition_ratio":0.03565365,"special_character_ratio":0.24633531,"punctuation_ratio":0.16107768,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9807754,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T09:14:08Z\",\"WARC-Record-ID\":\"<urn:uuid:2e3e02e5-4a59-4746-a0a9-2679160350be>\",\"Content-Length\":\"434096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d461fdb7-6f13-460b-a0bc-75498a6b091c>\",\"WARC-Concurrent-To\":\"<urn:uuid:dabb8f8c-365e-4aba-830e-b3e4755eef24>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1401.3305/\",\"WARC-Payload-Digest\":\"sha1:REIBTW7CMD4OTQDJYBT3NRUSTSJH2NNB\",\"WARC-Block-Digest\":\"sha1:HJXHW4O5TZ7GPIAJHRTHVPWQWZJWZUV6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881984.34_warc_CC-MAIN-20200703091148-20200703121148-00151.warc.gz\"}"}
https://jp.maplesoft.com/support/help/maple/view.aspx?path=geometry/CrossRatio&L=J	[ "", null, "geometry - Maple Programming Help\n\nHome : Support : Online Help : Mathematics : Geometry : 2-D Euclidean : Point Functions : geometry/CrossRatio\n\ngeometry\n\n CrossRatio\n compute the cross ratio (or anharmonic ratio, or double ratio) of four points\n\n Calling Sequence CrossRatio(A, B, C, F)\n\nParameters\n\n A, B, C, F - four points\n\nDescription\n\n • Given four points A, B, C, F, the routine CrossRatio(A, B, C, F) computes the cross ratio of A, B, C, F taken in this order.\n • If A, B, C, F are four distinct points on an ordinary line, the cross ratio of A, B, C, F taken in this order is defined as:\n\n$\\frac{\\frac{\\mathrm{SensedMagnitude}\\left(A,C\\right)}{\\mathrm{SensedMagnitude}\\left(C,B\\right)}}{\\frac{\\mathrm{SensedMagnitude}\\left(A,F\\right)}{\\mathrm{SensedMagnitude}\\left(F,B\\right)}}$\n\n • The command with(geometry,CrossRatio) allows the use of the abbreviated form of this command.\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{geometry}\\right):$\n > $\\mathrm{point}\\left(A,0,0\\right),\\mathrm{point}\\left(B,3,3\\right),\\mathrm{point}\\left(C,7,7\\right),\\mathrm{point}\\left(F,\\frac{21}{11},\\frac{21}{11}\\right):$\n > $\\mathrm{CrossRatio}\\left(A,B,C,F\\right)$\n ${-}\\frac{\\sqrt{{98}}{}\\sqrt{{32}}{}\\sqrt{{882}}{}\\sqrt{{288}}}{{28224}}$ (1)\n > $\\mathrm{CrossRatio}\\left(A,C,B,F\\right)$\n $\\frac{\\sqrt{{18}}{}\\sqrt{{32}}{}\\sqrt{{882}}{}\\sqrt{{6272}}}{{28224}}$ (2)" ]	[ null, "https://bat.bing.com/action/0", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7019268,"math_prob":0.9998772,"size":899,"snap":"2020-34-2020-40","text_gpt3_token_len":298,"char_repetition_ratio":0.16424581,"word_repetition_ratio":0.118644066,"special_character_ratio":0.26028922,"punctuation_ratio":0.23963134,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998449,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T12:09:58Z\",\"WARC-Record-ID\":\"<urn:uuid:a18d56ed-bb5b-41e1-b178-69df0d891d97>\",\"Content-Length\":\"161386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4db4c5ae-dab9-4757-b78e-f93b61c9496f>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e0e8268-8a2d-490d-8faf-10a12ac88295>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://jp.maplesoft.com/support/help/maple/view.aspx?path=geometry/CrossRatio&L=J\",\"WARC-Payload-Digest\":\"sha1:RT7FMYNSMNDTSIQFYQ72QNRCHDNMVYJC\",\"WARC-Block-Digest\":\"sha1:OHXFBHO3FWIPGCOE6TPRFI4E6THA3PAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737645.2_warc_CC-MAIN-20200808110257-20200808140257-00162.warc.gz\"}"}
https://corporatefinanceinstitute.com/resources/excel/functions/choose-function/	[ "# CHOOSE Function\n\nReturns a value from an array corresponding to the index number provided\n\n## What is the CHOOSE Function?\n\nThe CHOOSE function is categorized under Excel Lookup and Reference functions. It will return a value from an array corresponding to the index number provided. The function will return the nth entry in a given list.\n\nAs a financial analyst, the CHOOSE function is very useful when creating scenarios in financial models. By using the CHOOSE formula, an analyst is able to select between 5 different scenarios (for example) that can flow through the entire model. Scenario analysis is an important part of building a robust financial model.", null, "### Formula\n\n=CHOOSE(index_num, value1, [value2], …)\n\nThe formula uses the following arguments:\n\n1. Index_num (required argument) – This is an integer that specifies which value argument is selected. Index_num must be a number between 1 and 254, or a formula or reference to a cell containing a number between 1 and 254.\n2. Value1, Value2 – Value1 is a required option but the rest are optional. It is a list of one or more values that we want to return a value from.\n\n#### Notes:\n\n1. If index_num is 1, CHOOSE returns value1; if it is 2, CHOOSE returns value2; and so on.\n2. Value1, value2 must be entered as individual values (or references to individual cells containing values).\n3. If the argument index_num is a fraction, it is truncated to the lowest integer before being used.\n4. If the argument index_num is an array, every value is evaluated when CHOOSE is evaluated.\n5. The value arguments can be range references, as well as single values.\n\n### How to use the CHOOSE Function in Excel?\n\nTo understand the uses of the CHOOSE function, let’s consider an example:\n\n#### Example 1\n\nSuppose we are given the following dates:", null, "We can calculate the fiscal quarter from the dates given above. The fiscal quarters start in a month other than January.\n\nThe formula to use is:", null, "The formula returns a number from the array 1-4, which corresponds to a quarter system that begins in April and ends in March.\n\nWe get the results below:", null, "### A few notes about the CHOOSE Function\n\n1. VALUE! error – Occurs when:\n• The given index_num is less than 1 or is greater than the given number of values.\n• The given index_num argument is non-numeric.\n1. #NAME? error – Occurs when the value arguments are text values that are not enclosed in quotes and are not valid cell references.\n\nThanks for reading CFI’s guide to important Excel functions! By taking the time to learn and master these functions, you’ll significantly speed up your financial modeling and analysis. To learn more, check out these additional CFI resources:\n\n• Excel for Finance" ]	[ null, "https://cdn.corporatefinanceinstitute.com/assets/scenarios-1024x534.png", null, "https://cdn.corporatefinanceinstitute.com/assets/choose-function01.png", null, "https://cdn.corporatefinanceinstitute.com/assets/choose-function02.png", null, "https://cdn.corporatefinanceinstitute.com/assets/choose-function03.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8378321,"math_prob":0.962921,"size":2662,"snap":"2019-51-2020-05","text_gpt3_token_len":579,"char_repetition_ratio":0.12979683,"word_repetition_ratio":0.013274336,"special_character_ratio":0.21675432,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934081,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T21:59:18Z\",\"WARC-Record-ID\":\"<urn:uuid:fc1b86de-9a42-4bda-8a33-ff6902ce9cd5>\",\"Content-Length\":\"54136\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ed8e2ed-844b-4417-8263-2a67869fa272>\",\"WARC-Concurrent-To\":\"<urn:uuid:89f88661-65d9-4f61-86b4-ef1cb6ab0b17>\",\"WARC-IP-Address\":\"104.26.4.129\",\"WARC-Target-URI\":\"https://corporatefinanceinstitute.com/resources/excel/functions/choose-function/\",\"WARC-Payload-Digest\":\"sha1:DUHH63TLPTEXHH63W3WUMRQ7B2XYPNK5\",\"WARC-Block-Digest\":\"sha1:UPL2FKCK2X3SXDPFBZ4KTQ2XROIOXU2G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540523790.58_warc_CC-MAIN-20191209201914-20191209225914-00451.warc.gz\"}"}
https://www.freesteel.co.uk/wpblog/2007/06/11/point-inside-triangle-correct/	[ "## Point inside triangle correct\n\nMonday, June 11th, 2007 at 11:02 am", null, "Following from the incorrect method for finding which triangle of a mesh a given point is inside, here is the right answer.\n\nThe trick is to know that the only lines you can reliably tell sidedness along are those parallel to the X or Y axes, and the special cases of the lines x = y and x = -y. All other lines, defined by, say, a + tv, are going to prove somewhat noisy close in when attempting to determine which side of them a given point lies if you use the standard relation:\n\n(p - a) . vT > 0\n\nIn particular, if you have three lines passing through the same point and that point is not the a point in all of them, they won’t, and you’ll get a little unexpected area inside the triangle where they don’t all meet.\n\nSo, the scan line method works along the Y-axis. The triangle here has, relative to this direction, ay < cy < by. All the intervals are half-open, which means the point p can hit the triangle only when ay <= py < cy. The triangle also separates into two parts, the lower where py < cy and the upper where py >= cy.\n\nIn this diagram p is in the upper part, so we find the intersections of the line Y = py with the lines ab and cb to get points xab and xcb (not labeled). It doesn’t matter what order because they can be sorted and then treated as a half-open interval for comparison to the value px.\n\nActually, it is a good question whether the condition xab < xcb will remain the same for all values of py. I don’t know the answer to this one, and it probably depends on how carefully you calculate it. Ideally, if it’s determined that c is to the right of the line ab, then it’s consistent for all Y.\n\nOne has to assume the 2D triangle meshes are very nasty because the problem we’re actually trying to solve is the intersection of a 3D line through a polyhedral mesh representing the surface of a solid volume. The first step is always to project the entire model onto the plane perpendicular to the line (so that the line becomes a point). Even though you began with a well-behaved triangular surface, there are projections where the triangles turn into slivers and shards.\n\n• 1. SJ replies at 12th June 2007, 5:52 am :\n\nBogus logic. If you start with line segments represented in terms of their endpoints, and triangles represented in terms of line segments with common endpoints, then you can solve the problem robustly using only that single ‘sign of the determinant’ function. But if you choose to throw away\ninformation (and accuracy) by converting the line representation to a ‘standard’ parametric one, you are introducing the noise yourself. Frankly, you are big-headed and unbearable…\n\n• 2. Julian replies at 21st November 2008, 4:18 pm :\n\nYou needed to follow the link given to know what it is correcting:\n\nhttp://www.freesteel.co.uk/wpblog/2007/05/finding-points-in-triangles/\n\nThe standard method is fine if you only have one triangle. The reason for this non-standard method is it guarentees that the point will be found in exactly one triangle of a mesh covering of the plane.\n\nBut obviously if you don’t know this, the method looks kind of stupid. Depending on how quick-witted you want to be, you can leap to the conclusion that you are right and I am wrong, if that’s how you’d like it to be.\n\n• 3. anderswallin.net ›&hellip replies at 17th December 2011, 11:39 am :\n\n[…] check if the CC point lies within the triangle, but that will have to wait until the next post… (Mr. Todd has some thoughts on this) This was written by admin. Posted on Monday, June 25, 2007, at 21:06. Filed under Drop-Cutter. […]" ]	[ null, "http://www.freesteel.co.uk/images/blog/insidetriangscan.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9349541,"math_prob":0.93663365,"size":3662,"snap":"2020-45-2020-50","text_gpt3_token_len":862,"char_repetition_ratio":0.12575178,"word_repetition_ratio":0.0,"special_character_ratio":0.24003276,"punctuation_ratio":0.105882354,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9624731,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T22:21:02Z\",\"WARC-Record-ID\":\"<urn:uuid:5a825e32-083d-4b97-a0c2-dc69c2034afe>\",\"Content-Length\":\"29647\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb8ca30b-ec71-43e6-b753-3aa76d4ab70b>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf44efe7-3b44-4693-ac17-662dde7df718>\",\"WARC-IP-Address\":\"93.93.131.30\",\"WARC-Target-URI\":\"https://www.freesteel.co.uk/wpblog/2007/06/11/point-inside-triangle-correct/\",\"WARC-Payload-Digest\":\"sha1:4UTJI5BALE6FFC52P2FGJXHVEMW5A6O4\",\"WARC-Block-Digest\":\"sha1:NWXLLGGUHQRB3O3JLOWIKONNATU3BPUA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878633.8_warc_CC-MAIN-20201021205955-20201021235955-00239.warc.gz\"}"}
https://oemz-online.at/exportword?pageId=34373859	[ "Message-ID: <27643314.26951.1607169619488.JavaMail.confluence@omzconfluence> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary=\"----=_Part_26950_716305635.1607169619488\" ------=_Part_26950_716305635.1607169619488 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html 2020/4/3 (EN): Mader\n\n=20\n=20\n=20\n=20\n\n# =E2=80=9ESoldier dri= ll=E2=80=9C in the k. (u.) k. Army\n\n=20\n=20\n=20\n=20\n=20\n=20\n\n=20\n=20\n=20\n=20\n=20\n=20\n\nHugo Kerchnawe, latter Austrian ma= jor general and military historian, authored the book =E2=80=9EMilitary and= Civilian=E2=80=9C as =E2=80=9EAnonymous=E2=80=9D in 1904. In this publication, Kerchnawe demonstrated ar= ticulately his critical distance towards the army and its practice of train= ing soldiers. In the introduction he writes that he =E2=80=9Edoes not try to hide his sympathy for the ar= my at all=E2=80=9D, but that he, on the other hand, =E2=80=9Ehad not been rendered blind=E2=80=9D for its= weaknesses and been hindered from discussing and =E2=80=9Evituperate=E2=80=9D them. Naturally, he knew t= hat this courage was resented and criticized by many in the army. Kerchnawe= proved convinced that the course taken by him was the only correct one and= would contribute to the best of the army. The hush-up system customary in = the country did not help the military much. It only strengthened its (numer= ous) opponents and achieved the opposite of what was to be purposed. Nevert= heless, there were some =E2=80=9Eshort-sighted=E2=80=9D circles in the army who suspected animosity or at= least inappropriate deprecation in such positive criticism. He himself wis= hed that these =E2=80=9Eshort-= sighted=E2=80=9D people understood that it certainly was not true exaltatio= n to deny the mistakes of one stands for, and not to mend them. In his book= , Kerchnawe also deals with the topic =E2=80=9Emaltreatment of soldiers=E2=80=9D. Apart from the reproach= of =E2=80=9Emilitarism=E2=80= =9D (military dominance), nothing was so often brought forward against the = military as these maltreatments of soldiers. There was not a single parliam= entary session without such accusations, which were discussed in the most <= span style=3D\"color: rgb(51,51,51);\">=E2=80=9Espiteful=E2=80=9D way.= An objective treatment of this problem did not exist in the Austrian press= - nearly the entire press =E2=80=9E<= /span>stood under the banner of the ideals of 1848 without exception=E2=80= =9D, opposing the military, bringing forward not a single word in its favou= r. Thus, Kerchnawe continued, one ought to advise the military authorities = emphatically to investigate every suspicion of maltreatment by a superior. = After an accurate investigation of the circumstances, the guilty must be pu= nished without lenience, thus stopping the spiteful palaver against the mil= itary. As Kerchnawe states, nothing was worse than the operated (popular) h= ush-up system. In the =E2=80=9Edarkness of keeping secrets=E2=80=9C and =E2=80=9Ein the uncertain twiligh= t of sugar-coated presentments=E2=80=9D, the maltreatments of soldiers look= ed even uglier than they were in reality. For this reason, the public were = permanently given the levelheadedness and verisimilitude they had the right= to receive. The military would =E2= =80=9Eprofit=E2=80=9D from that. Despite his understanding for super= iors sometimes losing their temper, Kerchnawe did not support =E2=80=9Eviolent pedagogy=E2=80=9D at all= , meaning that he depreciated brutality. As Kerchnawe pointed out, there we= re indirect maltreatments in the military as well. The originators were off= icers, often even with higher ranks, and their motives were recklessness an= d indifference for the well-being of their subordinates. Often together wit= h the ambition to =E2=80=9Eshi= ne with extraordinary military strains=E2=80=9D. These often resulted in (s= tartling) high numbers of sick soldiers, numerous sunstroke and heatstroke = cases, and even fatalities. So, this maltreatment of the personnel manifest= ed in =E2=80=9Eunreasonable an= d not seldom barbarous exigencies=E2=80=9D. Thus, Kerchnawe (stridently) cr= iticized the evils in the k. u. k. Army without questioning the entire syst= em. Military drill aims at forming up the drives and emotions of single sol= diers appropriately. That means that the training (drill) ought to evoke an= d consolidate military obedience with the soldiers (especially in the turmo= il of battle). According to that, the total character of the military is ba= sed on a strict hierarchy, as well as on certain assumptions concerning the= social behaviour both of the individual and the troops in the extraordinar= y situations of combat. The ultimate purpose of achieving military discipli= ne thus was eliminating of uncertainties in the face of the unpredictable c= ircumstances of war. The traditional appearance of the soldier of the 1860i= es was re-analysed due to different reasons, but pure humaneness was (mostl= y) not the impetus. On the contrary, many people realized that with soldier= s who had been =E2=80=9Eunlear= ned=E2=80=9D any independence one could not win wars. On the other hand, ho= wever, a deep hiatus arose between theory and practice as far as the =E2=80=9Edrill=E2=80=9D of soldiers = was concerned. Obviously, the (positive) theoretical considerations<= /p>\n\n=20\n=20\n=20\n=20\n=20\n=20\n\n=20\n=20\n=20\n=20\n\n=20\n=20\n=20\n=20\n\n=20\n=20\n=20" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9454082,"math_prob":0.88262755,"size":5282,"snap":"2020-45-2020-50","text_gpt3_token_len":1408,"char_repetition_ratio":0.14115195,"word_repetition_ratio":0.0,"special_character_ratio":0.25691026,"punctuation_ratio":0.099585064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995915,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T12:00:19Z\",\"WARC-Record-ID\":\"<urn:uuid:aafe3845-620f-4778-a1f7-685caca4d30c>\",\"Content-Length\":\"85807\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60d9e0e6-ac00-41c6-8f64-c3e8e6686124>\",\"WARC-Concurrent-To\":\"<urn:uuid:449f650f-8e0f-49f3-bf9e-6fcf30aedca6>\",\"WARC-IP-Address\":\"188.165.63.123\",\"WARC-Target-URI\":\"https://oemz-online.at/exportword?pageId=34373859\",\"WARC-Payload-Digest\":\"sha1:JP42YSUF6TGZM6MECT4M2VK43XRPIXOP\",\"WARC-Block-Digest\":\"sha1:XBDAGUV4NNWNOPEDJ3RG3BWJ5RNHACDW\",\"WARC-Identified-Payload-Type\":\"message/rfc822\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141747774.97_warc_CC-MAIN-20201205104937-20201205134937-00619.warc.gz\"}"}
https://convertoctopus.com/30-3-hours-to-minutes	[ "## Conversion formula\n\nThe conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:\n\n1 hr = 60 min\n\nTo convert 30.3 hours into minutes we have to multiply 30.3 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:\n\n1 hr → 60 min\n\n30.3 hr → T(min)\n\nSolve the above proportion to obtain the time T in minutes:\n\nT(min) = 30.3 hr × 60 min\n\nT(min) = 1818 min\n\nThe final result is:\n\n30.3 hr → 1818 min\n\nWe conclude that 30.3 hours is equivalent to 1818 minutes:\n\n30.3 hours = 1818 minutes\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 0.00055005500550055 × 30.3 hours.\n\nAnother way is saying that 30.3 hours is equal to 1 ÷ 0.00055005500550055 minutes.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that thirty point three hours is approximately one thousand eight hundred eighteen minutes:\n\n30.3 hr ≅ 1818 min\n\nAn alternative is also that one minute is approximately zero point zero zero one times thirty point three hours.\n\n## Conversion table\n\n### hours to minutes chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from hours to minutes\n\nhours (hr) minutes (min)\n31.3 hours 1878 minutes\n32.3 hours 1938 minutes\n33.3 hours 1998 minutes\n34.3 hours 2058 minutes\n35.3 hours 2118 minutes\n36.3 hours 2178 minutes\n37.3 hours 2238 minutes\n38.3 hours 2298 minutes\n39.3 hours 2358 minutes\n40.3 hours 2418 minutes" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80221975,"math_prob":0.9852338,"size":1633,"snap":"2022-05-2022-21","text_gpt3_token_len":458,"char_repetition_ratio":0.21853898,"word_repetition_ratio":0.0,"special_character_ratio":0.32639313,"punctuation_ratio":0.10746269,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99169713,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T11:27:11Z\",\"WARC-Record-ID\":\"<urn:uuid:36f653c3-6225-4263-a54a-d1102686d1ef>\",\"Content-Length\":\"30729\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee41df68-bf92-4a81-a7fd-23426d7174e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:d991164a-e7b4-424f-a1be-e0d4d49e4947>\",\"WARC-IP-Address\":\"172.67.171.60\",\"WARC-Target-URI\":\"https://convertoctopus.com/30-3-hours-to-minutes\",\"WARC-Payload-Digest\":\"sha1:S7VVL2L7JWCPLX7D4KT4JDHLWETMUNVV\",\"WARC-Block-Digest\":\"sha1:QAMF6FBUBOQNZKTWM2A5MQIAX7LWEF5M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016373.86_warc_CC-MAIN-20220528093113-20220528123113-00725.warc.gz\"}"}
https://www.amansmathsblogs.com/s-chand-icse-maths-solutions-class-10-chapter-5-quadratic-equations-exercise-5e/	[ "Thursday, August 11, 2022\nHome > ICSE Solutions for Class 10 Maths > S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5E\n\n# S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5E\n\nHi students, Welcome to Amans Maths Blogs (AMB). In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5E. This is the chapter of introduction of ‘Quadratic Equations‘ included in ICSE 2020 Class 10 Maths syllabus.\n\nIn S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5E, word problems are given. First, we need to make a quadratic equation according to the given question and then we need to solve them.\n\n## S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5E\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 1\n\nThe sum of the squares of two consecutive positive even integers is 100. Find the integers.\n\nS Chand ICSE Maths Solutions:\n\nLet two consecutive positive even integers are x and (x + 2). Then, we have\n\nx2 + (x + 2)2 = 100\n\n⇒ x2 + x2 + 4x + 4 = 100\n\n⇒ 2x2 + 4x – 96 = 0\n\n⇒ x2 + 2x – 48 = 0\n\n⇒ x2 + 8x – 6x – 48 = 0\n\n⇒ x(x + 8) – 6(x + 8) = 0\n\n⇒ (x – 6)(x + 8) = 0\n\n⇒ x = 6 or -8\n\nSince x is positive even integer, then x = 6 and hence other positive even integer is (x + 2) = 8.\n\nThus, the required integers are 6 and 8.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 2\n\nFind two rational numbers which differ by 3 and the sum of whose squares is 117.\n\nS Chand ICSE Maths Solutions:\n\nLet two rational numbers are x and (x + 3). Then, we have\n\nx2 + (x + 3)2 = 117\n\n⇒ x2 + x2 + 6x + 9 = 117\n\n⇒ 2x2 + 6x + 9 = 117\n\n⇒ 2x2 + 6x – 108 = 0\n\n⇒ x2 + 3x – 54 = 0\n\n⇒ x2 + 9x – 6x – 54 = 0\n\n⇒ x(x + 9) – 6(x + 9) = 0\n\n⇒ (x – 6)(x + 9) = 0\n\n⇒ x = 6 or -9\n\nIf x = 6, then (x + 3) = 9\n\nIf x = -9, then (x + 3) = -6\n\nThus, the required rational numbers are 6 and 9 or -9 and -6.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 3\n\nWhat number increased by its reciprocal equals 65/8?\n\nS Chand ICSE Maths Solutions:\n\nLet the number is x. Then, we have\n\nx + 1/x = 65/8\n\n⇒ (x2 + 1)/x = 65/8\n\n⇒ 8(x2 + 1) = 65x\n\n⇒ 8x2 – 65x + 8 = 0\n\n⇒ 8x2 – 64x – x + 8 = 0\n\n⇒ 8x(x – 8) – 1(x – 8) = 0\n\n⇒ (8x – 1)(x – 8) = 0\n\n⇒ x = 1/8 or 8\n\nThus, the required number is 8 or 1/8.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 4\n\nThe sum of the numerator and denominator of a certain fraction is 8. If 2 is added to the numerator and to the denominator, the value of the fraction is increased by 4/35. Find the fraction.\n\nS Chand ICSE Maths Solutions:\n\nLet the fraction is x / (8 – x). Then, we have\n\n(x + 2)/(8 – x + 2) = x/(8 – x) + 4/35\n\n⇒ (x + 2)/(10 – x) – x/(8 – x) = 4/35\n\n⇒ [(x + 2)(8 – x) – x(10 – x)]/(10 – x)(8 – x) = 4/35\n\n⇒ [8x – x2 + 16 – 2x – 10x + x2]/(80 – 10x – 8x + x2) = 4/35\n\n⇒ 35[16 – 4x] = 4(x2 – 18x + 80)\n\n⇒ 35[4 – x] = (x2 – 18x + 80)\n\n⇒ x2 + 17x – 60 = 0\n\n⇒ x2 + 20x – 3x – 60 = 0\n\n⇒ x(x + 20) – 3(x + 20) = 0\n\n⇒ (x – 3)(x + 20) = 0\n\n⇒ x = 3 or -20\n\nThus, the required fraction is x/(8 – x) = 3/5.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 5\n\nThere are three consecutive integers such that the square of the first increased by the product of the other two gives 154. Find the integers.\n\nS Chand ICSE Maths Solutions:\n\nLet three consecutive integers are x, (x + 1) and (x + 2). Then, we have\n\nx2 + (x + 1)(x + 2) = 154\n\n⇒ x2 + x2 + 3x + 2 = 154\n\n⇒ 2x2 + 3x – 152 = 0\n\n⇒ 2x2 – 16x + 19x – 152 = 0\n\n⇒ 2x(x – 8) + 19(x – 8) = 0\n\n⇒ (2x + 19)(x – 8) = 0\n\n⇒ x = -19/2 or 8.\n\nThus, the required three consecutive integers are 8, 9, 10.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 6\n\nThe sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.\n\nS Chand ICSE Maths Solutions:\n\nLet two numbers are x and (8 – x). Then, we have\n\n15[1/x + 1/(8 – x)] = 8\n\n⇒ 15(8 – x + x)/x(8 – x) = 8\n\n⇒ 15(8) = 8x(8 – x)\n\n⇒ 15 = x(8 – x)\n\n⇒ x2 – 8x + 15 = 0\n\n⇒ x2 – 3x – 5x + 15 = 0\n\n⇒ x(x – 3) – 5(x – 3) = 0\n\n⇒ (x – 5)(x – 3) = 0\n\n⇒ x = 3 or 5\n\nIf x = 3, then (8 – x) = 5\n\nIf x = 5, then (8 – x) = 3\n\nThus, the required numbers are 3 and 5.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 7\n\nA two digit number is such that the product of the digits is 12. When 36 is added to this number, the digits interchange their places. Find the number.\n\nS Chand ICSE Maths Solutions:\n\nLet the unit digit of the number is x, then its tens digit is 12/x. Then, we have\n\n10(12/x) + x + 36 = 10x + 12/x\n\n⇒ 9x – 9(12/x) = 36\n\n⇒ x – (12/x) = 4\n\n⇒ (x2 – 12)/x = 4\n\n⇒ (x2 – 12) = 4x\n\n⇒ x2 – 4x – 12 = 0\n\n⇒ x2 – 6x + 2x – 12 = 0\n\n⇒ x(x – 6) + 2(x – 6) = 0\n\n⇒ (x + 2)(x – 6) = 0\n\n⇒ x = -2 or 6\n\nSince x is a digit, then x = 6.\n\nThus, the unit digit is x = 6 and tens digit 12/x = 2 and hence the number is 26\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 8\n\nFive times a certain whole number is equal to three less than twice the square of the number. Find the number.\n\nS Chand ICSE Maths Solutions:\n\nLet the whole number is x. Then, we have\n\n5x = 2x2 – 3\n\n⇒ 2x2 – 5x – 3 = 0\n\n⇒ 2x2 – 6x + x – 3 = 0\n\n⇒ 2x(x – 3) + 1(x – 3) = 0\n\n⇒ (2x + 1)(x – 3) = 0\n\n⇒ x = -1/2 or 3\n\nSince x is a whole number, then the required number is x = 3.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 9\n\nThree consecutive numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence, find the three numbers.\n\nS Chand ICSE Maths Solutions:\n\nLet the three number are (x – 1), x, (x + 1). Then, we have\n\nx2 = (x + 1)2 – (x – 1)2 + 60\n\n⇒ x2 = x2 + 2x + 1 – x2 + 2x – 1 + 60\n\n⇒ x2 – 4x – 60 = 0\n\n⇒ x2 – 10x + 6x – 60 = 0\n\n⇒ x(x – 10) + 6(x – 10) = 0\n\n⇒ (x + 6)(x – 10) = 0\n\n⇒ x = -6 or 10\n\nThus, the three numbers are 9, 10 and 11.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 10\n\nThe sum of the ages (in years) of a son and his father is 35 and the product of their ages is 150. Find their ages.\n\nS Chand ICSE Maths Solutions:\n\nLet the father’s age is x years then his son’s age is (35 – x) years. Then, we have\n\nx(35 – x) = 150\n\n⇒ 35x – x2 = 150\n\n⇒ x2 – 35x + 150 = 0\n\n⇒ x2 – 30x – 5x + 150 = 0\n\n⇒ x(x – 30) – 5(x – 30) = 0\n\n⇒ (x – 5)(x – 30) = 0\n\n⇒ x = 5 or 30\n\nThus, the father’s age is x = 30 years then his son’s age is (35 – x) = 5 years\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 11\n\nAnjali was born in 1985 A.D. In the year x2 AD, she was (x – 5) years old. Find the value of x.\n\nS Chand ICSE Maths Solutions:\n\nIn the year x2 AD, Anjali was (x – 5) years old. Then, we have\n\nx2 – 1985 = (x – 5)\n\n⇒ x2 – x – 1980 = 0\n\n⇒ x2 – 45x + 44x – 1980 = 0\n\n⇒ x(x – 45) + 44(x – 45) = 0\n\n⇒ (x + 44)(x – 45) = 0\n\n⇒ x = -44 or 45\n\nSince x is year, x = 45.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 12\n\nThe difference of mother’s age and her daughter’s age is 21 years and the twelfth part of the product of their ages is less than the mother’s age by 18 years. Find their ages.\n\nS Chand ICSE Maths Solutions:\n\nLet the daughter’s age is x years then her mother’s age is (x + 21) years. Then, we have\n\nx(x + 21) / 12 = (x + 21) – 18\n\n⇒ x2 + 21x = 12(x + 3)\n\n⇒ x2 + 21x = 12x + 36\n\n⇒ x2 + 9x – 36 = 0\n\n⇒ x2 + 12x – 3x – 36 = 0\n\n⇒ (x + 12) – 3(x + 12) = 0\n\n⇒ (x – 3)(x + 12) = 0\n\n⇒ x = 3 or -12\n\nSince x is an age, x = 3\n\nThus, daughter’s age is x = 3 years and mother’s age is (x + 21) = 24 years.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 13\n\nReena is x year old, while her father Mr. Sunil is x2 years old. 5 years hence, Mr Sunil will be three times as old as Reena. Find their present ages.\n\nS Chand ICSE Maths Solutions:\n\nGiven that Reena is x year old, while her father Mr. Sunil is x2 years old.\n\nAfter 5 years, Mr Sunil will be three times as old as Reena.\n\nThen, we have\n\nx2 + 5 = 3(x + 5)\n\n⇒ x2 – 3x – 10 = 0\n\n⇒ x2 – 5x + 2x – 10 = 0\n\n⇒ x(x – 5) + 2(x – 5) = 0\n\n⇒ (x + 2)(x – 5) = 0\n\n⇒ x = -2 or 5\n\nSince x is an age, x = 5.\n\nThus, Reena is x = 5 year old and her father Mr. Sunil is x2 = 25 years old.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 14\n\nForm a quadratic equation from the following information, taking as width x ∈ N.\n\n(i) The area of rectangle whose length is five more than twice its width is 75.\n\n(ii) Solve the equation and find its length.\n\nS Chand ICSE Maths Solutions:\n\nGiven that the width of rectangle is x, and hence the length of the rectangle is (2x + 5).\n\nThen, we have\n\nx(2x + 5) = 75\n\n⇒ 2x2 + 5x – 75 = 0 [This is required quadratic equation.]\n\n⇒ 2x2 + 15x – 10x – 75 = 0\n\n⇒ x(2x + 15) – 5(2x + 15) = 0\n\n⇒ (x – 5)(2x + 15) = 0\n\n⇒ x = 5 or -15/2\n\nThus, the width of the rectangle is x = 5.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 15\n\nThe sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the triangle be 60 cm2, calculate the lengths of the sides of triangle.\n\nS Chand ICSE Maths Solutions:\n\nGiven that the area of a right angled triangle, whose legs are 5x cm and (3x – 1) cm, is 60 cm2.\n\nThen, we have\n\n(1/2) × (5x) × (3x – 1) = 60\n\n⇒ 15x2 – 5x – 120 = 0\n\n⇒ 5(3x2 – x – 24) = 0\n\n⇒ 3x2 – x – 24 = 0\n\n⇒ 3x2 – 9x + 8x – 24 = 0\n\n⇒ 3x(x – 3) + 8(x – 3) = 0\n\n⇒ (3x + 8)(x – 3) = 0\n\n⇒ x = 3 or -8/3\n\nThus, the legs of the right angled triangle are 5x = 15 and (3x – 1) = 8 and hence its hypotenuse is √(82 + 152) = 17.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 16\n\nThe hypotenuse of a right angle triangle is 13 cm and the difference between the other two sides is 7 cm.\n\n(i) Taking x as the length of the shorter of the two sides, write an equation in x that represents the above statement.\n\n(ii) Solve the equation obtained in (i) above and hence find the two unknowns sides of the triangle.\n\nS Chand ICSE Maths Solutions:\n\nLet the length of legs of right angled triangle are x and (x + 7). Then, we have\n\nx2 + (x + 7)2 = 132\n\n⇒ x2 + x2 + 14x + 49 = 169\n\n⇒ 2x2 + 14x – 120 = 0\n\n⇒ x2 + 7x – 60 = 0 [This is required quadratic equation.]\n\n⇒ x2 + 12x – 5x – 60 = 0\n\n⇒ x(x + 12) – 5(x + 12) = 0\n\n⇒ (x – 5)(x + 12) = 0\n\n⇒ x = 5 or -12\n\nThus, the two legs of the right angled triangle are x = 5 and (x + 7) = 12.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 17\n\nThe length of a verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.\n\n(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement.\n\n(ii) Solve the equation in (i) above and hence find the dimension of the verandah.\n\nS Chand ICSE Maths Solutions:\n\nLet the dimension of the verandah are x and (x + 3). Then, we have\n\nx(x + 3) = 2(x + x + 3)\n\n⇒ x2 + 3x = 4x + 6\n\n⇒ x2 – x – 6 = 0 [This is required quadratic equation.]\n\n⇒ x2 – 3x + 2x – 6 = 0\n\n⇒ x(x – 3) + 2(x – 3) = 0\n\n⇒ (x + 2)(x – 3) = 0\n\n⇒ x = 3 or -2.\n\nThus, the dimension of the verandah are width as x = 3 and length as (x + 3) = 6.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 18\n\nTwo squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq cm. Express this as an algebraic equation in x and solve the equation to find sides of the squares.\n\nS Chand ICSE Maths Solutions:\n\nTwo squares have sides x cm and (x + 4) cm. Then, we have\n\nx2 + (x + 4)2 = 656\n\n⇒ x2 + x2 + 8x + 16 = 656\n\n⇒ 2x2 + 8x – 640 = 0\n\n⇒ x2 + 4x – 320 = 0 [This is required quadratic equation.]\n\n⇒ x2 – 16x + 20x – 320 = 0\n\n⇒ x(x – 16) + 20(x – 16) = 0\n\n⇒ (x + 20)(x – 16) = 0\n\n⇒ x = -20 or 16\n\nSince x represents the length of the side of square, then x = 16.\n\nThus, the sides of the squares are 16 cm and 20 cm.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 19\n\nThe perimeter of a rectangular plot is 180 m and its area is 1800 cm2. Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.\n\nS Chand ICSE Maths Solutions:\n\nGiven that the length of rectangular plot is x m. Then, we have\n\nNow, Area = 1800\n\nx(90 – x) = 1800\n\n90x – x2 = 1800\n\nx2 – 90x + 1800 = 0\n\nx2 – 60x – 30x + 1800 = 0\n\n⇒ x(x – 60) – 30(x – 60) = 0\n\n⇒ (x – 30)(x – 60) = 0\n\n⇒ x = 30 or 60.\n\nIf x = 30, then 90 – x = 60\n\nIf x = 60, then 90 – x = 30\n\nThus, length = 60 cm and width = 30 cm\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 20\n\nA rectangle has an area of 24 cm2. If its length is x cm, write down its breadth in terms of x. Given that its perimeter is 20 cm, form an equation in x and solve it.\n\nS Chand ICSE Maths Solutions:\n\nGiven that the length of rectangle is x m. Then, we have\n\nNow, Perimeter = 20\n\n2(x + 24/x) = 20\n\n(x + 24/x) = 10\n\nx2 + 24 = 10x\n\nx2 – 10x + 24 = 0\n\nx2 – 6x – 4x + 24 = 0\n\n⇒ x(x – 6) – 4(x – 6) = 0\n\n⇒ (x – 4)(x – 6) = 0\n\n⇒ x = 4 or 6\n\nSince x is length of the rectangle, then x = 6 and 24/x = 4.\n\nThus, length and breadth of the rectangle are 6 cm and 4 cm.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 21\n\nA rectangular garden 10 m by 16 m is to be surrounded by a concrete path of uniform width. Given that the area of the path is 120 squares meters, assuming the width of the path to be x, form an equation in x and solve it to find the value of x.\n\nS Chand ICSE Maths Solutions:", null, "Given that the width of the wall to be x. Then, we have\n\n(2x + 16)(2x + 10) – 10 × 16 = 120\n\n⇒ 4x2 + 52x + 160 – 160 = 120\n\n⇒ 4x2 + 52x – 120 = 0\n\nx2 + 13x – 30 = 0\n\nx2 + 15x – 2x – 30 = 0\n\n⇒ x(x + 15) – 2(x + 15) = 0\n\n⇒ (x – 2)(x + 15) = 0\n\n⇒ x = 2 or -15\n\nThus, the width of the path is x = 2 cm.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 22\n\nA man covers a distance of 200 km travelling with a uniform speed of x km per hour. The distance could have been covered in 2 hours less, had the speed been (x + 5)km/he. Calculate the value of x.\n\nS Chand ICSE Maths Solutions:\n\nAccording to question, we have\n\n200/x – 200/(x + 5) = 2\n\n⇒ 1/x – 1/(x + 5) = 1/100\n\n⇒ (x + 5 – x)/x(x + 5) = 1/100\n\n⇒ 5/x(x + 5) = 1/100\n\n⇒ x2 + 5x – 500 = 0\n\n⇒ x2 + 25x – 20x – 500 = 0\n\n⇒ x(x + 25) – 20(x + 25) = 0\n\n⇒ (x – 20)(x + 25) = 0\n\n⇒ x = 20 or -25.\n\nThus, required speed is x = 20 km/hr.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 23\n\nAn express train makes a run of 240 km at a certain speed. Another train, whose speed is 12 km/hr less than the first takes an hour longer to make the same trip. Find the speed of the express train to km/hr.\n\nS Chand ICSE Maths Solutions:\n\nLet the speed of the express train is x km/hr, we have\n\n240/(x – 12) – 240/x = 1\n\n⇒ 1/(x – 12) – 1/x = 1/240\n\n⇒ (x – x + 12)/x(x – 12) = 1/240\n\n⇒ 12/x(x – 12) = 1/240\n\n⇒ x2 – 12x – 2880 = 0\n\n⇒ x2 – 60x + 48x – 2880 = 0\n\n⇒ x(x – 60) + 48(x – 60) = 0\n\n⇒ (x + 48)(x – 60) = 0\n\n⇒ x = -48 or 60\n\nThus, the speed of the express train is 60 km/hr.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 24\n\nA train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.\n\nS Chand ICSE Maths Solutions:\n\nLet the original speed of the train is x km/hr, we have\n\n90/x – 90/(x + 15) = 1/2\n\n⇒ 1/x – 1/(x + 15) = 1/180\n\n⇒ (x + 15 – x)/x(x + 15) = 1/180\n\n⇒ 15/x(x + 15) = 1/180\n\n⇒ x2 + 15x – 2700 = 0\n\n⇒ x2 + 60x – 45x – 2700 = 0\n\n⇒ x(x + 60) – 45(x + 60) = 0\n\n⇒ (x – 45)(x + 60) = 0\n\n⇒ x = 45 or -60\n\nThus, the original speed of the train is x = 45 km/hr.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 25\n\nA plane left 30 minutes later than the scheduled time and in order to reach its distance, 1500 km away it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.\n\nS Chand ICSE Maths Solutions:\n\nLet the usual speed of the plain is x km/hr, we have\n\n1500/x – 1500/(x + 250) = 1/2\n\n⇒ 1/x – 1/(x + 250) = 1/3000\n\n⇒ (x + 250 – x)/x(x + 250) = 1/3000\n\n⇒ 250/x(x + 250) = 1/3000\n\n⇒ x2 + 250x – 750000 = 0\n\n⇒ x2 + 1000x – 750x – 750000 = 0\n\n⇒ x(x + 1000) – 750(x + 1000) = 0\n\n⇒ (x – 750)(x + 1000) = 0\n\n⇒ x = 750 or -1000.\n\nThus, the usual speed of the plain is 750 km/hr.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 26\n\nA boat takes 1 hour longer to go 36 km up a river than to return. If the river flows at 3 km/hr, find the rate at which the boat travels in still water.\n\nS Chand ICSE Maths Solutions:\n\nLet the speed of the boat in still water is x km/hr, we have\n\n36/(x – 3) – 36/(x + 3) = 1\n\n⇒ 1/(x – 3) – 1/(x + 3) = 1/36\n\n⇒ (x + 3 – x + 3)/(x + 3)(x – 3) = 1/36\n\n⇒ 6/(x + 3)(x – 3) = 1/36\n\n⇒ 1/(x2 – 9) = 1/216\n\n⇒ x2 = 225\n\n⇒ x = 15 or -15.\n\nThus, the speed of the boat in still water is x = 15 km/hr.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 27\n\nA man purchased some horses for Rs. 3000. Three of them died, and he sold the rest at Rs. 65 more than what he paid for each horse and thus gained 6% more his outlay. How many horses did he buy?\n\nS Chand ICSE Maths Solutions:\n\nLet the number of horses purchased is x. Then, the price of each horse is Rs. 3000/x.\n\nNow, CP = 3000 and SP = 3000(1 + 6/100) = 3180.\n\nAccording to question, we have\n\n3180/(x – 3) = 3000/x + 65\n\n⇒ 636/(x – 3) – 600/x = 13\n\n⇒ 636x – 600x + 1800 = 13x(x – 3)\n\n⇒ 36x + 1800 = 13x2 – 39x\n\n⇒ 13x2 – 75x – 1800 = 0\n\n⇒ 13x2 – 195x + 120x – 1800 = 0\n\n⇒ 13x(x – 15) + 120(x – 15) = 0\n\n⇒ (13x + 120)(x – 15) = 0\n\n⇒ x = 15 or -120/13.\n\nThus, the number of horses purchased is x = 15.\n\nS Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5E: Ques No 28\n\nA trader bought a number of articles for Rs. 1200. Ten were damaged and he sold each of the rest at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of article he bought as x, form an equation in x and solve.\n\nS Chand ICSE Maths Solutions:\n\nLet the number of article purchased is x. Then, the price of each article is Rs. 1200/x.\n\nNow, CP = 1200 and SP = 1200 + 60 = 1260.\n\nAccording to question, we have\n\n1260/(x – 10) = 1200/x + 2\n\n⇒ 625/(x – 10) – 600/x = 1\n\n⇒ 625/(x – 10) – 600/x = 1\n\n⇒ 625x – 600x + 6000 = x(x – 10)\n\n⇒ x2 – 40x – 6000 = 0\n\n⇒ x2 – 100x + 60x – 6000 = 0\n\n⇒ x(x – 100) + 60(x – 100) = 0\n\n⇒ (x + 60)(x – 100) = 0\n\n⇒ x = -60 or 100\n\nThus, the number of article purchased is x = 100.\n\nerror: Content is protected !!" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20300%20154'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8703506,"math_prob":1.0000018,"size":18630,"snap":"2022-27-2022-33","text_gpt3_token_len":7761,"char_repetition_ratio":0.21904865,"word_repetition_ratio":0.24619122,"special_character_ratio":0.44170693,"punctuation_ratio":0.080242425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999893,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T15:04:47Z\",\"WARC-Record-ID\":\"<urn:uuid:78435b2c-b75f-437a-a66e-0a2c3b2c876a>\",\"Content-Length\":\"194528\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6604a3d-d338-4f2f-a5b6-4755d7d66b48>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c782565-6b62-4f9a-8268-6a13ab628cdd>\",\"WARC-IP-Address\":\"104.21.82.96\",\"WARC-Target-URI\":\"https://www.amansmathsblogs.com/s-chand-icse-maths-solutions-class-10-chapter-5-quadratic-equations-exercise-5e/\",\"WARC-Payload-Digest\":\"sha1:OVC5RFPUTMR6UHXD5O3R2UT6YU2TAGBW\",\"WARC-Block-Digest\":\"sha1:UR6UQ6HXANBGKMZTOFUCHD7CUJYIGF5V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571472.69_warc_CC-MAIN-20220811133823-20220811163823-00504.warc.gz\"}"}
https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos.html	[ "geom_mlb_logos() and geom_mlb_scoreboard_logos() are used to plot MLB team and league logos instead of points in a ggplot. It requires x, y aesthetics as well as a valid MLB team abbreviation. The latter can be checked with valid_team_names() but is also cleaned before being plotted.\n\n## Usage\n\ngeom_mlb_logos(\nmapping = NULL,\ndata = NULL,\nstat = \"identity\",\nposition = \"identity\",\n...,\nnudge_x = 0,\nnudge_y = 0,\nna.rm = FALSE,\nshow.legend = FALSE,\ninherit.aes = TRUE\n)\n\ngeom_mlb_scoreboard_logos(\nmapping = NULL,\ndata = NULL,\nstat = \"identity\",\nposition = \"identity\",\n...,\nnudge_x = 0,\nnudge_y = 0,\nna.rm = FALSE,\nshow.legend = FALSE,\ninherit.aes = TRUE\n)\n\n## Arguments\n\nmapping\n\nSet of aesthetic mappings created by aes(). If specified and inherit.aes = TRUE (the default), it is combined with the default mapping at the top level of the plot. You must supply mapping if there is no plot mapping.\n\ndata\n\nThe data to be displayed in this layer. There are three options:\n\nIf NULL, the default, the data is inherited from the plot data as specified in the call to ggplot().\n\nA data.frame, or other object, will override the plot data. All objects will be fortified to produce a data frame. See fortify() for which variables will be created.\n\nA function will be called with a single argument, the plot data. The return value must be a data.frame, and will be used as the layer data. A function can be created from a formula (e.g. ~ head(.x, 10)).\n\nstat\n\nThe statistical transformation to use on the data for this layer, either as a ggproto Geom subclass or as a string naming the stat stripped of the stat_ prefix (e.g. \"count\" rather than \"stat_count\")\n\nposition\n\nPosition adjustment, either as a string naming the adjustment (e.g. \"jitter\" to use position_jitter), or the result of a call to a position adjustment function. Use the latter if you need to change the settings of the adjustment.\n\n...\n\nOther arguments passed on to ggplot2::layer(). These are often aesthetics, used to set an aesthetic to a fixed value. See the below section \"Aesthetics\" for a full list of possible arguments.\n\nnudge_x, nudge_y\n\nHorizontal and vertical adjustment to nudge labels by. Useful for offsetting text from points, particularly on discrete scales. Cannot be jointly specified with position.\n\nna.rm\n\nIf FALSE, the default, missing values are removed with a warning. If TRUE, missing values are silently removed.\n\nshow.legend\n\nlogical. Should this layer be included in the legends? NA, the default, includes if any aesthetics are mapped. FALSE never includes, and TRUE always includes. It can also be a named logical vector to finely select the aesthetics to display.\n\ninherit.aes\n\nIf FALSE, overrides the default aesthetics, rather than combining with them. This is most useful for helper functions that define both data and aesthetics and shouldn't inherit behaviour from the default plot specification, e.g. borders().\n\n## Value\n\nA ggplot2 layer (ggplot2::layer()) that can be added to a plot created with ggplot2::ggplot().\n\n## Aesthetics\n\ngeom_mlb_logos() and geom_mlb_scoreboard_logos() understand the following aesthetics (required aesthetics are in bold):\n\n• x - The x-coordinate.\n\n• y - The y-coordinate.\n\n• team_abbr - The team abbreviation. Need to use Savant's abbreviation.\n\n• alpha = NULL - The alpha channel, i.e. transparency level, as a numerical value between 0 and 1.\n\n• colour = NULL - The image will be colourized with this colour. Use the special character \"b/w\" to set it to black and white. For more information on valid colour names in ggplot2 see https://ggplot2.tidyverse.org/articles/ggplot2-specs.html?q=colour#colour-and-fill\n\n• angle = 0 - The angle of the image as a numerical value between 0° and 360°.\n\n• hjust = 0.5 - The horizontal adjustment relative to the given x coordinate. Must be a numerical value between 0 and 1.\n\n• vjust = 0.5 - The vertical adjustment relative to the given y coordinate. Must be a numerical value between 0 and 1.\n\n• width = 1.0 - The desired width of the image in npc (Normalised Parent Coordinates). The default value is set to 1.0 which is big but it is necessary because all used values are computed relative to the default. A typical size is width = 0.075 (see below examples).\n\n• height = 1.0 - The desired height of the image in npc (Normalised Parent Coordinates). The default value is set to 1.0 which is big but it is necessary because all used values are computed relative to the default. A typical size is height = 0.1 (see below examples).\n\n## Examples\n\n# \\donttest{\nlibrary(mlbplotR)\nlibrary(ggplot2)\n\nteam_abbr <- valid_team_names()\n# remove conference logos from this example\nteam_abbr <- team_abbr[!team_abbr %in% c(\"NL\", \"AL\", \"MLB\")]\n\ndf <- data.frame(\na = rep(1:6, 5),\nb = sort(rep(1:5, 6), decreasing = TRUE),\nteams = team_abbr\n)\n\n# keep alpha == 1 for all teams including an \"A\"\nmatches <- grepl(\"A\", team_abbr)\ndf$alpha <- ifelse(matches, 1, 0.2) # also set a custom fill colour for the non \"A\" teams df$colour <- ifelse(matches, NA, \"gray\")\n\n# scatterplot of all logos\nggplot(df, aes(x = a, y = b)) +\ngeom_mlb_logos(aes(team_abbr = teams), width = 0.075) +\ngeom_label(aes(label = teams), nudge_y = -0.35, alpha = 0.5) +\ntheme_void()", null, "# apply alpha via an aesthetic from inside the dataset df\n# please note that you have to add scale_alpha_identity() to use the alpha\nggplot(df, aes(x = a, y = b)) +\ngeom_mlb_scoreboard_logos(aes(team_abbr = teams, alpha = alpha), width = 0.075) +\ngeom_label(aes(label = teams), nudge_y = -0.35, alpha = 0.5) +\nscale_alpha_identity() +\ntheme_void()", null, "# apply alpha and colour via an aesthetic from inside the dataset df\n# please note that you have to add scale_alpha_identity() as well as\n# scale_colour_identity() to use the alpha and colour values in your dataset!\nggplot(df, aes(x = a, y = b)) +\ngeom_mlb_logos(aes(team_abbr = teams, alpha = alpha, colour = colour), width = 0.075) +\ngeom_label(aes(label = teams), nudge_y = -0.35, alpha = 0.5) +\nscale_alpha_identity() +\nscale_colour_identity() +\ntheme_void()", null, "# apply alpha as constant for all logos\nggplot(df, aes(x = a, y = b)) +\ngeom_mlb_scoreboard_logos(aes(team_abbr = teams), width = 0.075, alpha = 0.6) +\ngeom_label(aes(label = teams), nudge_y = -0.35, alpha = 0.5) +\ntheme_void()", null, "# it's also possible to plot league logos\nleague <- data.frame(a = 1:3, b = 0, teams = c(\"AL\", \"NL\", \"MLB\"))\nggplot(league, aes(x = a, y = b)) +\ngeom_mlb_logos(aes(team_abbr = teams), width = 0.3) +\ngeom_label(aes(label = teams), nudge_y = -0.4, alpha = 0.5) +\ncoord_cartesian(xlim = c(0.5,3.5), ylim = c(-0.75,.75)) +\ntheme_void()", null, "# }" ]	[ null, "https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos-1.png", null, "https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos-2.png", null, "https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos-3.png", null, "https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos-4.png", null, "https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos-5.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7478234,"math_prob":0.9858435,"size":6618,"snap":"2023-14-2023-23","text_gpt3_token_len":1818,"char_repetition_ratio":0.118234046,"word_repetition_ratio":0.24219489,"special_character_ratio":0.28180718,"punctuation_ratio":0.17720531,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9664613,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-10T13:25:29Z\",\"WARC-Record-ID\":\"<urn:uuid:32e73299-872a-49e1-974e-acc8e6a3572b>\",\"Content-Length\":\"34690\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97567aff-e121-4707-b67c-02693c860963>\",\"WARC-Concurrent-To\":\"<urn:uuid:0364dffc-348a-4600-b928-b20d64b6165b>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://camdenk.github.io/mlbplotR/reference/geom_mlb_logos.html\",\"WARC-Payload-Digest\":\"sha1:CJS63GAMBRXAHOMTXQALHCK547K6HG6B\",\"WARC-Block-Digest\":\"sha1:LRAJQHYTJZX7GTDBNMV464NYG2WXW6LU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224657720.82_warc_CC-MAIN-20230610131939-20230610161939-00602.warc.gz\"}"}
https://www.tensorflow.org/versions/r2.0/api_docs/python/tf/signal/linear_to_mel_weight_matrix	[ "Watch keynotes, product sessions, workshops, and more from Google I/O\n\n# tf.signal.linear_to_mel_weight_matrix\n\nReturns a matrix to warp linear scale spectrograms to the mel scale.\n\nReturns a weight matrix that can be used to re-weight a `Tensor` containing `num_spectrogram_bins` linearly sampled frequency information from `[0, sample_rate / 2]` into `num_mel_bins` frequency information from `[lower_edge_hertz, upper_edge_hertz]` on the mel scale.\n\nFor example, the returned matrix `A` can be used to right-multiply a spectrogram `S` of shape `[frames, num_spectrogram_bins]` of linear scale spectrum values (e.g. STFT magnitudes) to generate a \"mel spectrogram\" `M` of shape `[frames, num_mel_bins]`.\n\n``````# `S` has shape [frames, num_spectrogram_bins]\n# `M` has shape [frames, num_mel_bins]\nM = tf.matmul(S, A)\n``````\n\nThe matrix can be used with `tf.tensordot` to convert an arbitrary rank `Tensor` of linear-scale spectral bins into the mel scale.\n\n``````# S has shape [..., num_spectrogram_bins].\n# M has shape [..., num_mel_bins].\nM = tf.tensordot(S, A, 1)\n# tf.tensordot does not support shape inference for this case yet.\nM.set_shape(S.shape[:-1].concatenate(A.shape[-1:]))\n``````\n\n`num_mel_bins` Python int. How many bands in the resulting mel spectrum.\n`num_spectrogram_bins` An integer `Tensor`. How many bins there are in the source spectrogram data, which is understood to be `fft_size // 2 + 1`, i.e. the spectrogram only contains the nonredundant FFT bins.\n`sample_rate` Python float. Samples per second of the input signal used to create the spectrogram. We need this to figure out the actual frequencies for each spectrogram bin, which dictates how they are mapped into the mel scale.\n`lower_edge_hertz` Python float. Lower bound on the frequencies to be included in the mel spectrum. This corresponds to the lower edge of the lowest triangular band.\n`upper_edge_hertz` Python float. The desired top edge of the highest frequency band.\n`dtype` The `DType` of the result matrix. Must be a floating point type.\n`name` An optional name for the operation.\n\nA `Tensor` of shape `[num_spectrogram_bins, num_mel_bins]`.\n\n`ValueError` If `num_mel_bins`/`num_spectrogram_bins`/`sample_rate` are not positive, `lower_edge_hertz` is negative, frequency edges are incorrectly ordered, `upper_edge_hertz` is larger than the Nyquist frequency, or `sample_rate` is neither a Python float nor a constant Tensor.\n\n[{ \"type\": \"thumb-down\", \"id\": \"missingTheInformationINeed\", \"label\":\"Missing the information I need\" },{ \"type\": \"thumb-down\", \"id\": \"tooComplicatedTooManySteps\", \"label\":\"Too complicated / too many steps\" },{ \"type\": \"thumb-down\", \"id\": \"outOfDate\", \"label\":\"Out of date\" },{ \"type\": \"thumb-down\", \"id\": \"samplesCodeIssue\", \"label\":\"Samples / code issue\" },{ \"type\": \"thumb-down\", \"id\": \"otherDown\", \"label\":\"Other\" }]\n[{ \"type\": \"thumb-up\", \"id\": \"easyToUnderstand\", \"label\":\"Easy to understand\" },{ \"type\": \"thumb-up\", \"id\": \"solvedMyProblem\", \"label\":\"Solved my problem\" },{ \"type\": \"thumb-up\", \"id\": \"otherUp\", \"label\":\"Other\" }]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59827965,"math_prob":0.93309855,"size":2496,"snap":"2021-21-2021-25","text_gpt3_token_len":659,"char_repetition_ratio":0.15930979,"word_repetition_ratio":0.0,"special_character_ratio":0.24399039,"punctuation_ratio":0.1719457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98619604,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T20:23:53Z\",\"WARC-Record-ID\":\"<urn:uuid:2ba8a494-cfe3-4bfd-afaa-06c419434fb6>\",\"Content-Length\":\"720659\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7da71b11-74b8-41cf-a429-bc3f348eec4a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2aa8d88-6f77-4fab-917f-19f2d788e5e3>\",\"WARC-IP-Address\":\"172.217.2.110\",\"WARC-Target-URI\":\"https://www.tensorflow.org/versions/r2.0/api_docs/python/tf/signal/linear_to_mel_weight_matrix\",\"WARC-Payload-Digest\":\"sha1:LT52JSDQG3ZZHW2X3JYBFEJEK23WUNIA\",\"WARC-Block-Digest\":\"sha1:5JH4UXQ5GVSPQZEIZA7EWRF6ZWUPXDSU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487613453.9_warc_CC-MAIN-20210614201339-20210614231339-00001.warc.gz\"}"}
https://chemistrygod.com/hund-rule	[ "Search the World of Chemistry\n\n×\n\n# Hund's Rule of Maximum Multiplicity\n\n28th Feb 2020 @ 6 min read\n\nPhysical Chemistry\n\n## Statement\n\nThe Hund rule of maximum multiplicity states:\n\n1. When two or more orbitals of equal energy (or very close energy) are available, electrons will fill the orbitals singly before filling doubly.\n2. All the electrons in the orbitals will have the same spin to maximize the multiplicity.\n\nIn simple words, the rule says the lowest-energy electronic configuration is attained with the maximum number of parallel electron spins.\n\nThe rule was formulated by Friedrich Hund, a German physicist, in 1925. Hund's contributions to quantum mechanics are significant, particularly in the electronic structure of atoms. He is also known for quantum tunneling.\n\n## Explanation\n\nAlong with the aufbau principle and Pauli exclusion principle, the Hund is very helpful in understanding the electronic configuration of an atom. The aufbau principle is used to predict the electronic configuration when the energy difference between orbitals is considerably large, and the Pauli exclusion principle restricts the maximum number of electrons in an orbital to two. The Hund rule useful to predict the ground state of atoms or molecules where the equal energy orbitals are available. This is especially seen in the orbitals of the same subshell which are equal in energy. For example, in the p subshell, all its orbitals (px, py, and pz) have the same energy. Such equal energy orbitals are called degenerate orbitals.\n\nAs the rule says electrons occupy the same energy orbital singly before pairing. It means the pairing of electrons does not occur until each equal energy orbital has the occupation of one electron. Also, all of these unpaired electrons must have the same spin, i.e., either spin up (↑) or spin down (↓).\n\n### Filling of p subshell\n\nLet's consider the filling of 2p orbitals. There are three orbitals in every p subshell, which are equal in energy. The first electron enters either of three orbitals since they are degenerate. Even though the first-occupied orbital is not completely filled and it can take one more electron, the second electron will not occupy it. Instead, it will enter another orbital with the same spin as the first electron. Similarly, the third will occupy the remaining orbital. Thus, the first three electrons occupy all three orbitals with either spin up (shown in the diagram below) or spin down.\n\nAfter singly filling, the following three electrons will fill the remaining vacancies, which results in pairing.\n\n### Filling of d subshell\n\nThe same is true for the d subshell. There are five d orbitals; each has the same energy. So, the first five electrons fill each orbital with the same spin. And the next five will form pairs.\n\nThe Hund rule is not only seen for the orbitals of the same subshell but also for the orbitals of different subshells with a small difference in energy. Chromium is one such example; it is discussed later in the article.\n\n## What is multiplicity?\n\nIn quantum chemistry, the multiplicity (or spin multiplicity) is defined as the total number of spin orientations and is given by 2S + 1. Here, S is the total spin quantum number, and its value is the sum of all unpaired half spins. According to the Hund rule, the lowest energy configuration is attained when the multiplicity, i.e., 2S + 1, is maximum.\n\nThe maximum value of S is obtained only when all the spins are either up or down.\n\nConsider an example of 2p orbitals with three electrons. The possible number of arrangements are shown below.\n\nThe maximum value of S is obtained in the first and fourth case, and it is calculated as:", null, "From the total spin quantum number, we can find the multiplicity.", null, "## Examples\n\n### Carbon\n\nCarbon has six electrons. Its electronic configuration is 1s2 2s2 2p2. The last two electrons are in 2p subshell, and both of them occupy the different orbitals with the same spin.\n\nThe total spin quantum number (S) of carbon is", null, ". And the multiplicity is 2S + 1 = 3.\n\n### Nitrogen\n\nAdding one more electron to carbon, we get the electronic configuration of nitrogen: 1s2 2s2 2p3. In nitrogen, all 2p orbitals are half-filled.\n\nS is equal to", null, ". And the multiplicity is 2S + 1 = 4.\n\n### Oxygen\n\nOxygen follows nitrogen and has an extra electron in its electronic configuration: 1s2 2s2 2p4. The fourth electron in 2p subshell forms a pair.\n\nS = 1 and the multiplicity is 2S + 1 = 3\n\n### Chromium\n\nChromium, a d block element, has the atomic number of 24. Its electronic configuration is [Ar] 3d54s1. The last two orbitals are 3d and 4s. The difference between the energies of these two orbitals is very small. As we can see from the diagram below, both orbitals are half-filled and all electrons have the same spin as per the rule.\n\nThere are six unpaired electrons in chromium. So,", null, "and the multiplicity is 2S + 1 = 7.\n\n### Oxygen molecule\n\nThe Hund rule is also used in molecular orbitals. The figure below is the molecular orbital diagram of dioxygen.\n\nThe filling of electrons in the π antibonding orbitals is achieved as per the Hund rule. Both molecular orbitals (π\nx\nand π\ny\n) are equal in energy and accept one electron each.\n\n## Justification\n\nThere are several justifications proposed for the explanation of the Hund rule. One of them is the electron-electron repulsion. The stability of the electronic system increases by minimizing the repulsion among electrons. This can be achieved by maximizing the spatial space. In a singly filled orbital, electrons are far apart from one another. This minimizes the repulsion and increases the stability.\n\nThe second justification is the shielding effect. In an atom, the columbic attraction between the nucleus and outer electrons is shielded by the inner electrons. Consequently, the outer electrons experience a lesser nuclear charge than the inner electrons. It causes the expansion of outer orbitals and increases the energy of the system. The energy of the system is lowered by decreasing the shielding effect, which is possible by the symmetrical distribution of electrons. Thus, the singly filled orbital has relatively small shielding and the stability is further increases.\n\nThe third factor that affects the energy of the system is the exchange energy. Electrons of the same spin residing in degenerate orbitals exchanges their positions. During these exchanges, the energy is released called the exchange energy. The more the exchanges, the more the stability. When degenerate orbitals are filled with unpaired electrons of the same spin, the number of exchanges is the highest, and the system reaches stability.\n\n## Associated articles\n\nIf you appreciate our work, consider supporting us on ❤️ patreon.\n\nCopy Article Cite\n\nWrite a response", null, "Joseph\n14th Oct 2020\nI'm truly amazed by how this was explained I don't fully understand everything because it's not all my level yet... But it's a great start", null, "" ]	[ null, "https://chemistrygod.com/assets/media/placeholder.svg", null, "https://chemistrygod.com/assets/media/placeholder.svg", null, "https://chemistrygod.com/assets/media/placeholder.svg", null, "https://chemistrygod.com/assets/media/placeholder.svg", null, "https://chemistrygod.com/assets/media/placeholder.svg", null, "https://chemistrygod.com/assets/media/avatar.png", null, "https://chemistrygod.com/assets/media/en.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9199453,"math_prob":0.94077694,"size":6211,"snap":"2023-14-2023-23","text_gpt3_token_len":1357,"char_repetition_ratio":0.1760915,"word_repetition_ratio":0.032535885,"special_character_ratio":0.20640798,"punctuation_ratio":0.11177474,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9863715,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T10:11:27Z\",\"WARC-Record-ID\":\"<urn:uuid:e880c86b-fcf2-467a-ba97-c586004ae605>\",\"Content-Length\":\"25113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8225742f-0818-46cd-9633-7b9c85e45f58>\",\"WARC-Concurrent-To\":\"<urn:uuid:82828fd8-06cf-4154-98e1-d908b31d0c03>\",\"WARC-IP-Address\":\"143.198.77.105\",\"WARC-Target-URI\":\"https://chemistrygod.com/hund-rule\",\"WARC-Payload-Digest\":\"sha1:4IBRDQU4DDPJDAMACNE6G77TUYD4H3GT\",\"WARC-Block-Digest\":\"sha1:MUPAIE4PNTNJMD32RVUXUUZGEGYCB4QW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224651815.80_warc_CC-MAIN-20230605085657-20230605115657-00658.warc.gz\"}"}
https://percent-of.com/calculate/what-is-36-of-1542/	[ "# We use percentages in almost everything.\n\nPercentages are a very important part of our daily lives. They are used in Economics, Cooking, Health, Sports, Mathematics, Science, Jewellery, Geography, Medicine and many other areas.\n\n## Percent of Calculator\n\nCalculate percentage of X, quick & simple.\n\n%\n?\n\n36% of 1542 is:\n555.12\n\n## Percent of - Table For 1542\n\nPercent of Difference\n1% of 1542 is 15.42 1526.58\n2% of 1542 is 30.84 1511.16\n3% of 1542 is 46.26 1495.74\n4% of 1542 is 61.68 1480.32\n5% of 1542 is 77.1 1464.9\n6% of 1542 is 92.52 1449.48\n7% of 1542 is 107.94 1434.06\n8% of 1542 is 123.36 1418.64\n9% of 1542 is 138.78 1403.22\n10% of 1542 is 154.2 1387.8\n11% of 1542 is 169.62 1372.38\n12% of 1542 is 185.04 1356.96\n13% of 1542 is 200.46 1341.54\n14% of 1542 is 215.88 1326.12\n15% of 1542 is 231.3 1310.7\n16% of 1542 is 246.72 1295.28\n17% of 1542 is 262.14 1279.86\n18% of 1542 is 277.56 1264.44\n19% of 1542 is 292.98 1249.02\n20% of 1542 is 308.4 1233.6\n21% of 1542 is 323.82 1218.18\n22% of 1542 is 339.24 1202.76\n23% of 1542 is 354.66 1187.34\n24% of 1542 is 370.08 1171.92\n25% of 1542 is 385.5 1156.5\n26% of 1542 is 400.92 1141.08\n27% of 1542 is 416.34 1125.66\n28% of 1542 is 431.76 1110.24\n29% of 1542 is 447.18 1094.82\n30% of 1542 is 462.6 1079.4\n31% of 1542 is 478.02 1063.98\n32% of 1542 is 493.44 1048.56\n33% of 1542 is 508.86 1033.14\n34% of 1542 is 524.28 1017.72\n35% of 1542 is 539.7 1002.3\n36% of 1542 is 555.12 986.88\n37% of 1542 is 570.54 971.46\n38% of 1542 is 585.96 956.04\n39% of 1542 is 601.38 940.62\n40% of 1542 is 616.8 925.2\n41% of 1542 is 632.22 909.78\n42% of 1542 is 647.64 894.36\n43% of 1542 is 663.06 878.94\n44% of 1542 is 678.48 863.52\n45% of 1542 is 693.9 848.1\n46% of 1542 is 709.32 832.68\n47% of 1542 is 724.74 817.26\n48% of 1542 is 740.16 801.84\n49% of 1542 is 755.58 786.42\n50% of 1542 is 771 771\n51% of 1542 is 786.42 755.58\n52% of 1542 is 801.84 740.16\n53% of 1542 is 817.26 724.74\n54% of 1542 is 832.68 709.32\n55% of 1542 is 848.1 693.9\n56% of 1542 is 863.52 678.48\n57% of 1542 is 878.94 663.06\n58% of 1542 is 894.36 647.64\n59% of 1542 is 909.78 632.22\n60% of 1542 is 925.2 616.8\n61% of 1542 is 940.62 601.38\n62% of 1542 is 956.04 585.96\n63% of 1542 is 971.46 570.54\n64% of 1542 is 986.88 555.12\n65% of 1542 is 1002.3 539.7\n66% of 1542 is 1017.72 524.28\n67% of 1542 is 1033.14 508.86\n68% of 1542 is 1048.56 493.44\n69% of 1542 is 1063.98 478.02\n70% of 1542 is 1079.4 462.6\n71% of 1542 is 1094.82 447.18\n72% of 1542 is 1110.24 431.76\n73% of 1542 is 1125.66 416.34\n74% of 1542 is 1141.08 400.92\n75% of 1542 is 1156.5 385.5\n76% of 1542 is 1171.92 370.08\n77% of 1542 is 1187.34 354.66\n78% of 1542 is 1202.76 339.24\n79% of 1542 is 1218.18 323.82\n80% of 1542 is 1233.6 308.4\n81% of 1542 is 1249.02 292.98\n82% of 1542 is 1264.44 277.56\n83% of 1542 is 1279.86 262.14\n84% of 1542 is 1295.28 246.72\n85% of 1542 is 1310.7 231.3\n86% of 1542 is 1326.12 215.88\n87% of 1542 is 1341.54 200.46\n88% of 1542 is 1356.96 185.04\n89% of 1542 is 1372.38 169.62\n90% of 1542 is 1387.8 154.2\n91% of 1542 is 1403.22 138.78\n92% of 1542 is 1418.64 123.36\n93% of 1542 is 1434.06 107.94\n94% of 1542 is 1449.48 92.52\n95% of 1542 is 1464.9 77.1\n96% of 1542 is 1480.32 61.68\n97% of 1542 is 1495.74 46.26\n98% of 1542 is 1511.16 30.84\n99% of 1542 is 1526.58 15.42\n100% of 1542 is 1542 0\n\n### Here's How to Calculate 36% of 1542\n\nLet's take a quick example here:\n\nYou have a Target coupon of \\$1542 and you need to know how much will you save on your purchase if the discount is 36 percent.\n\nSolution:\n\nAmount Saved = Original Price x Discount in Percent / 100\n\nAmount Saved = (1542 x 36) / 100\n\nAmount Saved = 55512 / 100\n\nIn other words, a 36% discount for a purchase with an original price of \\$1542 equals \\$555.12 (Amount Saved), so you'll end up paying 986.88.\n\n### Calculating Percentages\n\nSimply click on the calculate button to get the results of percentage calculations. You will see the result on the next page. If there are errors in the input fields, the result page will be blank. The program allows you to calculate the difference between two numbers in percentages. You can also input a percentage of any number and get the numeric value. Although it is a simple calculator, it can be very useful in many scenarios. Our goal is to give you an easy to use percentage calculator that gives you results you want fast.\n\nPercentage in mathematics refers to fractions based in 100. It is usually represented by “%,” “pct,” or “percentage.” This web app allows a comma or dot as a decimal separator. So you can use both freely.\n\nWe have provided several examples for you to use. You can use the examples to feed in your own data correctly. We hope you will find this site useful for calculang percentages. You can even use it for crosschecking the accuracy of your assignment results.\n\nNB. Americans use “percent,” which the British prefer “per cent.”\n\n#### Examples\n\nExample one\n\nCalculate 20% of 200?\n20% of 200 =____\n(200/100) x 20 = _____\n2 x 20 = 40\n\nIt is quite easy. Just divide 200 by 100 to get one percent. The result is 2. Then multiply it by 20 ( 20% = 20 per hundred) = 20 x 2 = 40\n\nExample two\n\nWhat percentage of 125 is 50?\n\n50 = ---% of 125\n50 x (100/125) = 40%\n\nGet the value of one percent by dividing 100 by 125. After that, multiply the value by 50 to get the percentage value of 50 units, which is 40% That is how to calculate the percentage.\n\nExample three\n\nWhat is the percentage (%) change (increase or decrease) from 120 to 150?\n\n(150-120) x (100/120) = 36\n\nSince 150 represents 100%. One percent will be equal to 100/150. 150-120 is 30. Therefore, 30 units represents 30 x (100/150) = 36 % This is how to calculate the percentage increase.\n\nwe do not use a percentage at all times. There are scenarios where we simply want to show the ratio of numbers. For instance, what is 20% of 50? This can also be interpreted as 20 hundredths of 50. This equates to 20/100 x 50 = 10.\n\nYou can use a calculation trick here. Anyme you want to divide a number by 100, just move the decimal two places to the left. 20/100 x 50 calculated above can also be writen as (20 x 50)/100. Since 20x 50 =1000. You can simply divide 1000 by 100 by moving two decimal places to the left, which gives you 10.\n\nIn another scenario, you want to calculate the percentage increase or decrease. Supposing you have \\$10 and spend \\$2 to buy candy, then you have spent 20% of your money. So how much will be remaining? All the money you have is 100%, if you spend 20%, you will have 80% remaining. You can simply use the percentage reduction tool above to calculate this value.\n\n#### Origin\n\nThe word percent is derived from the Latin word percenter which means per hundred, and it is designated by %" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91458464,"math_prob":0.9864803,"size":6898,"snap":"2021-04-2021-17","text_gpt3_token_len":2618,"char_repetition_ratio":0.22497824,"word_repetition_ratio":0.039412674,"special_character_ratio":0.55552334,"punctuation_ratio":0.16309255,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998326,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T12:21:16Z\",\"WARC-Record-ID\":\"<urn:uuid:91ab27ce-65e8-4aa5-8afc-ccf21c67a2c5>\",\"Content-Length\":\"47886\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:272ffc00-e995-4b2f-a594-c5aa20d345b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:8af7a185-6f72-4e16-b721-9089e829998e>\",\"WARC-IP-Address\":\"209.42.195.149\",\"WARC-Target-URI\":\"https://percent-of.com/calculate/what-is-36-of-1542/\",\"WARC-Payload-Digest\":\"sha1:G5UH4MOAZXE7AOEA4QIGWXYGMHCYUAIU\",\"WARC-Block-Digest\":\"sha1:7XGBNFMSIH46XO3O5RK2VLYR35EWWHKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703524743.61_warc_CC-MAIN-20210121101406-20210121131406-00273.warc.gz\"}"}
http://eqworld.ipmnet.ru/eqarchive/view.php?id=51	[ "", null, "EqWorld The World of Mathematical Equations", null, "", null, "EqArchive: Add Equation/Solution > View Equation English only\n\n## View Equation\n\nThe database contains 327 equations (14 equations are awaiting activation).\n\nEquation data\nCategory:2. First-Order Partial Differential Equations\nSubcategory:2.3. Nonlinear Equations\nEquation(s):[can not create temporary dvi file]\nSolution(s),\nTransformation(s),\nIntegral(s)\n:\n[can not create temporary dvi file]\nRemarks:[can not create temporary dvi file]\nNovelty:Material has been fully published elsewhere\nReferences:Akulenko, L.D. (1987) Asymptotic Methods of Optimal Cotrol. Nauka, Moscow.\nAuthor/Contributor's Details\nLast name:Zhurov\nFirst name:Alexei\nCountry:Russia\nStatistic information\nSubmission date:Sun 17 Dec 2006 06:13\nEdits by author:0\n\nEdit (Only for author/contributor)\n\nThe EqWorld website presents extensive information on solutions to various classes of ordinary differential equations, partial differential equations, integral equations, functional equations, and other mathematical equations.\n\nCopyright © 2006-2011 Andrei D. Polyanin, Alexei I. Zhurov and Alexander L. Levitin" ]	[ null, "http://eqworld.ipmnet.ru/images/eqworld-logo-small.gif", null, "http://eqworld.ipmnet.ru/images/ipmlogo24b.gif", null, "http://counter.rambler.ru/top100.cnt", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72020036,"math_prob":0.8819326,"size":1051,"snap":"2021-43-2021-49","text_gpt3_token_len":269,"char_repetition_ratio":0.13180515,"word_repetition_ratio":0.046511628,"special_character_ratio":0.23977165,"punctuation_ratio":0.19428572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9874756,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T04:44:46Z\",\"WARC-Record-ID\":\"<urn:uuid:57439d1d-177b-4e8d-922e-24740e39e4bf>\",\"Content-Length\":\"7676\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18571ace-1873-4b53-bf4e-ff91139362df>\",\"WARC-Concurrent-To\":\"<urn:uuid:0ac19b63-41e1-4643-810d-2047cf6d30df>\",\"WARC-IP-Address\":\"188.44.34.72\",\"WARC-Target-URI\":\"http://eqworld.ipmnet.ru/eqarchive/view.php?id=51\",\"WARC-Payload-Digest\":\"sha1:7J2SYXUBMQ2QYXBHBRHHETH6SFZUHOQ4\",\"WARC-Block-Digest\":\"sha1:HTLYGAW3JTXLD54TDUA654KZHXF5A7Y5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358469.34_warc_CC-MAIN-20211128043743-20211128073743-00277.warc.gz\"}"}
https://stackoom.com/question/1zAkZ	[ "# 如果没有其他的话，Ocaml会嵌套Ocaml nested if without else\n\n``````let rec move_helper b sz r = match b with\n[] -> r\n\|(h :: t) ->\nif h = 0 then\nif h - 1 = sz then h - 1 ::r\nif h + 1 = sz then h + 1 ::r\nelse move_helper t sz r\n;;\n\nlet move_pos b =\nmove_helper b 3 r\n;;\n\nlet g = move_pos [0;8;7;6;5;4;3;2;1]\n``````\n\n## 2 个回复2\n\n### #1楼票数：4 已采纳\n\n``````let f x =\nif x land 1 <> 0 then print_string \"1\";\nif x land 2 <> 0 then print_string \"2\";\nif x land 4 <> 0 then print_string \"4\"\n``````\n\n### #2楼票数：1\n\n``````(if test then f else g) 1\n``````\n\n``````let rec move_helper b sz r = match b with\n\| [] -> r\n\| h :: t ->\nif h = 0 then\nif h - 1 = sz then (h - 1) :: r\nelse if h + 1 = sz then (h + 1) :: r\nelse (* What do you want to return here? *)\nelse move_helper t sz r\n``````", null, "3回复\n\n2回复\n\n1回复\n\n1回复\n\n1回复\n\n1回复\n\n1回复\n\n1回复" ]	[ null, "https://stackoom.com/img/dash.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.8604039,"math_prob":0.9986519,"size":312,"snap":"2021-43-2021-49","text_gpt3_token_len":176,"char_repetition_ratio":0.14935064,"word_repetition_ratio":0.03125,"special_character_ratio":0.42948717,"punctuation_ratio":0.225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996237,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T14:19:56Z\",\"WARC-Record-ID\":\"<urn:uuid:3777319c-612f-43d9-af83-26d29fff613a>\",\"Content-Length\":\"29086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b38a9e3a-f034-4289-9b1f-717fdf8cd186>\",\"WARC-Concurrent-To\":\"<urn:uuid:430a6861-95f2-4e9b-a964-d1617ddebf0e>\",\"WARC-IP-Address\":\"118.31.174.222\",\"WARC-Target-URI\":\"https://stackoom.com/question/1zAkZ\",\"WARC-Payload-Digest\":\"sha1:TMFKEGQ3I4H3TUN3MZBYVWMUHTJ3HZZW\",\"WARC-Block-Digest\":\"sha1:RWP2NKPKJSJGFXNPHBROF4GH7O64ELUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585696.21_warc_CC-MAIN-20211023130922-20211023160922-00069.warc.gz\"}"}
https://www.onlinemathlearning.com/statistics-lecture-2.html	[ "# Statistics Lectures: Variables\n\nA series of free Statistics Lectures with lessons, examples & solutions in videos.\n\nThis is the second page of the series of free video lessons, “Statistics Lectures”. These lectures introduce the concepts of variables, independent and dependent variables and discuss measurement scales.\n\nDescriptive statistics summarizes a sample, giving us some ideas about the mean and spread of the data, etc. Inferential statistics use a sample or samples to describe what the population is like; in other words, infering the characteristics of the population from the sample/s.\n\n### Statistics - Lecture 3: Types Of Variables\n\nA variable is a property that can take on many values.\nTwo kinds of variable:\n\n• Quantitative Variable\n• Qualitative/Categorical Variables.\n\nQuantitative variables are measured numerically. With measurements of quantitative variables you can do things like add, subtract, multiply and divide, and get a meaningful result.\nThere are two kinds of quantitative variables:\n\n• Discrete variables\n• Continuous Variables.\n\nA Discrete Variable is a quantitative variable with a finite number of values.\n\nA Continuous Variable is a quantitative variable with an infinite number of values.\n\nQualitative/Categorical variables allow for classification based on some characteristic.\n\n### Statistics - Lecture 4: Independent And Dependent Variables\n\nAn independent variable is any variable that is being manipulated.\nA dependent variable is any variable that is being measured.\n\n### Statistics - Lecture 5: Variable Measurement Scales\n\nThere are four data types of measured variables:\n\n• Nominal\n• Ordinal\n• Interval\n• Ratio\n\nNominal data (also known as qualitative/categorical data) is data that is split into categories.\nOrdinal data is data where order matters, but distance between values does not.\nInterval data is data where order matters, and distances between values are equal and meaningful, and a natural zero is not present. Ratio data is data where order matters, distances between values are equal and meaningful, and a natural zero is present.", null, "" ]	[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8731456,"math_prob":0.99247676,"size":2222,"snap":"2022-40-2023-06","text_gpt3_token_len":418,"char_repetition_ratio":0.1330027,"word_repetition_ratio":0.0952381,"special_character_ratio":0.17911792,"punctuation_ratio":0.11597938,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99765784,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T08:52:57Z\",\"WARC-Record-ID\":\"<urn:uuid:02859cf0-6579-48b2-8719-d226d8bf3b8d>\",\"Content-Length\":\"47850\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:be12ce58-160b-4d03-8e0c-354dedb2b46f>\",\"WARC-Concurrent-To\":\"<urn:uuid:3aa279da-01e9-4b10-bd35-038d6e38f5fe>\",\"WARC-IP-Address\":\"173.247.219.45\",\"WARC-Target-URI\":\"https://www.onlinemathlearning.com/statistics-lecture-2.html\",\"WARC-Payload-Digest\":\"sha1:UJRF63C26VJG5W55JQH2YZHOPP6V625X\",\"WARC-Block-Digest\":\"sha1:7JRYG5YGRVOFHR5QND6VY5FAU65QI5BM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337595.1_warc_CC-MAIN-20221005073953-20221005103953-00612.warc.gz\"}"}
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-12-exponential-functions-and-logarithmic-functions-12-3-logarithmic-functions-12-3-exercise-set-page-803/73	[ "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)\n\n$\\log_{z}6=m$\nSince $b^y=x$ is equivalent to $y=\\log_bx,$ the given equation, $z^m=6 ,$ is equivalent to \\begin{array}{l}\\require{cancel} \\log_{z}6=m .\\end{array}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81853193,"math_prob":0.99996364,"size":435,"snap":"2019-26-2019-30","text_gpt3_token_len":122,"char_repetition_ratio":0.09976798,"word_repetition_ratio":0.0,"special_character_ratio":0.26436782,"punctuation_ratio":0.08988764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999523,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T22:46:03Z\",\"WARC-Record-ID\":\"<urn:uuid:7badf8ee-0302-426d-a894-d468f0427a40>\",\"Content-Length\":\"129103\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:522166c5-4a87-49cf-ac63-8e6caf7c9001>\",\"WARC-Concurrent-To\":\"<urn:uuid:65b466c0-4c02-426b-ab4e-732e4400502b>\",\"WARC-IP-Address\":\"54.210.251.76\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-12-exponential-functions-and-logarithmic-functions-12-3-logarithmic-functions-12-3-exercise-set-page-803/73\",\"WARC-Payload-Digest\":\"sha1:2GSOS45W2W6ZVQVIT2RHNI253WAJ4SQ2\",\"WARC-Block-Digest\":\"sha1:2H4P6H4WUXJRDCQLICLIMCRICOKYCVYH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525414.52_warc_CC-MAIN-20190717221901-20190718003901-00157.warc.gz\"}"}
https://au.mathworks.com/help/simbio/ref/simdata.getdata.html	[ "# getdata\n\nGet simulation data from `SimData` object\n\n## Syntax\n\n``[t,x,names] = getdata(simdata)``\n``sdOut = getdata(simdata)``\n``___ = getdata(simdata,format)``\n\n## Description\n\nexample\n\n````[t,x,names] = getdata(simdata)` returns the simulation time points `t`, the simulation data `x`, and corresponding `names` for the data columns.```\n````sdOut = getdata(simdata)` returns the simulation results as a `SimData` object `sdOut`.```\n\nexample\n\n````___ = getdata(simdata,format)` returns the simulation data in the specified `format`.```\n\n## Examples\n\ncollapse all\n\n`sbioloadproject('gprotein.sbproj');`\n\nSimulate the model.\n\n```sdObj = sbiosimulate(m1); sbioplot(sdObj);```", null, "The plot shows all the states together. Plot each state separately on its own axes in a subplot.\n\nFirst, extract the simulation data from the `SimData` object.\n\n`[time,data,names] = getdata(sdObj);`\n\nCalculate the number of rows and columns needed for the subplot.\n\n```sqrtnames = sqrt(numel(names)); nrows = round(sqrtnames); ncolumns = ceil(sqrtnames);```\n\nCreate a subplot and plot each state on its own axes.\n\n```figure for(i = 1:numel(names)) subplot(nrows,ncolumns,i) plot(time,data(:,i)); title(names(i)); end```", null, "## Input Arguments\n\ncollapse all\n\nSimulation data, specified as a `SimData` object or array of `SimData` objects.\n\nSimulation data format, specified as a character vector or string. Some formats require you to specify only one output argument. The valid formats follow.\n\n• `'num'` — This format returns simulation time points and simulation data in numeric arrays and the names of quantities and sensitivities as a cell array. This format is the default when you run `getdata` with multiple output arguments.\n\n• `'nummetadata'` — This format returns a cell array of metadata structures instead of the names of quantities and sensitivities as the third output argument.\n\n• `'numqualnames'` — This format returns qualified names in the third output argument to resolve ambiguities.\n\nYou must specify only one output argument for the following formats.\n\n• `'simdata'` — This format returns data in a new `SimData` object or an array of `SimData` objects. This format is the default when you specify a single output argument.\n\n• `'struct'` — This format returns a structure or structure array that contains both data and metadata.\n\n• `'ts'` — This format returns data as a cell array.\n\n• If `simdata` is scalar, the cell array is an m-by-1 array, where each element is a `timeseries` object. m is the number of quantities and sensitivities logged during the simulation.\n\n• If `simdata` is not scalar, the cell array is k-by-1, where each element of the cell array is an m-by-1 cell array of `timeseries` objects. k is the size of `simdata`, and m is the number of quantities or sensitivities in each `SimData` object in `simdata`. In other words, the function returns an individual time series for each state or column and for each `SimData` object in `simdata`.\n\n• `'tslumped'` — This format returns the data as a cell array of `timeseries` objects, combining data from each `SimData` object into a single time series.\n\n## Output Arguments\n\ncollapse all\n\nSimulation time points, returned as a numeric vector or cell array. If `simdata` is scalar, `t` is an n-by-1 vector, where n is the number of time points. If `simdata` is an array of objects, `t` is a k-by-1 cell array, where k is the size of `simdata`.\n\nSimulation data, returned as a numeric matrix or cell array. If `simdata` is scalar, `x` is an n-by-m matrix, where n is the number of time points and m is the number of quantities and sensitivities logged during the simulation. If `simdata` is an array of objects, `x` is a k-by-1 cell array, where k is the size of `simdata`.\n\nNames of quantities and sensitivities logged during the simulation, returned as a cell array. If `simdata` is scalar, `names` is an m-by-1 cell array. If `simdata` is an array of objects, `names` is a k-by-1 cell array, where k is the size of `simdata`.\n\nSimulation results, returned as a `SimData` object.", null, "" ]	[ null, "https://au.mathworks.com/help/examples/simbio/win64/ExtractDataFromSimDataObjectExample_01.png", null, "https://au.mathworks.com/help/examples/simbio/win64/ExtractDataFromSimDataObjectExample_02.png", null, "https://au.mathworks.com/images/responsive/supporting/apps/doc_center/bg-trial-arrow.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7387123,"math_prob":0.98662835,"size":2898,"snap":"2020-34-2020-40","text_gpt3_token_len":677,"char_repetition_ratio":0.17657222,"word_repetition_ratio":0.22088353,"special_character_ratio":0.20979986,"punctuation_ratio":0.09189189,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967395,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T15:10:51Z\",\"WARC-Record-ID\":\"<urn:uuid:a1af9ed0-7ff2-475c-afb3-7c34538b8691>\",\"Content-Length\":\"86873\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e1f0f66-60ec-4dbc-9863-529e701cd551>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9182b4f-e753-4b4d-b1f6-80445e6b062f>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://au.mathworks.com/help/simbio/ref/simdata.getdata.html\",\"WARC-Payload-Digest\":\"sha1:4CUOKEXYFAPPOK4VXFSG4VR7KKEZIZJ5\",\"WARC-Block-Digest\":\"sha1:GUNU5SW3E7MGOKVGYPND6CGNERJX5UB7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737883.59_warc_CC-MAIN-20200808135620-20200808165620-00366.warc.gz\"}"}
https://easystats.github.io/performance/reference/r2_efron.html	[ "Calculates Efron's pseudo R2.\n\n## Usage\n\nr2_efron(model)\n\n## Arguments\n\nmodel\n\nGeneralized linear model.\n\nThe R2 value.\n\n## Details\n\nEfron's R2 is calculated by taking the sum of the squared model residuals, divided by the total variability in the dependent variable. This R2 equals the squared correlation between the predicted values and actual values, however, note that model residuals from generalized linear models are not generally comparable to those of OLS.\n\nEfron, B. (1978). Regression and ANOVA with zero-one data: Measures of residual variation. Journal of the American Statistical Association, 73, 113-121." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8558617,"math_prob":0.9826356,"size":578,"snap":"2023-40-2023-50","text_gpt3_token_len":139,"char_repetition_ratio":0.114982575,"word_repetition_ratio":0.0,"special_character_ratio":0.21972318,"punctuation_ratio":0.16190477,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934402,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T18:06:44Z\",\"WARC-Record-ID\":\"<urn:uuid:1d4f19f4-f7aa-4672-b705-960861e7c725>\",\"Content-Length\":\"13938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3541855a-6e3e-446f-8caa-7d8a5cf3dd3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d847eef8-5fe0-44c9-87c5-c1c0a95e8185>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://easystats.github.io/performance/reference/r2_efron.html\",\"WARC-Payload-Digest\":\"sha1:J46B3SOW7VN6UGBGJW6TX4JLEUQZZXXC\",\"WARC-Block-Digest\":\"sha1:D43HATUXQPZCARWKDHBZSD6OUO6EIADF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510319.87_warc_CC-MAIN-20230927171156-20230927201156-00122.warc.gz\"}"}
https://www.whirldatascience.com/2019/03/27/segnet-applicability-in-single-object-detection-with-low-resolution-images/	[ "We thank Madhusudhan B A, Ara Ramalingam and Muthupandian Thangarajan for their guidance and support towards making this paper a reality\n\nAbstract\n\nWe implemented the SegNet neural network architecture to detect strays of transparent and colored plastic on a conveyor belt containing paper from recycled garbage at a recycling / sorting facility in California and to determine whether this architecture was suitable for single object detection. We have also analyzed the performance of the model generated during the different stages of training on our test set to identify patterns if any.\n\nSegNet was designed to be an efficient architecture for pixel-wise semantic segmentation and it was our opinion that a pixel-wise semantic segmentation will aide in detecting the contours of plastic. FCN1 is also a popular architecture for pixel-wise segmentation but we chose SegNet as the authors of SegNet claim that their architecture was more efficient in terms of memory, accuracy, computation time compared to FCN and also used fewer trainable parameters.\n\nAn alternate was to try the faster R-CNN2. Since one of our initial requirements was to identify contours of the plastic, we did not consider Faster R-CNN as it would have only provided us with bounding boxes.\n\n1.Introduction", null, "Our aim was to do identify strays of transparent plastic on a moving conveyor belt containing mostly paper at a garbage sorting facility using SegNet and to determine if it is the right method to solve our problem. SegNet is a deep fully convolutional neural network for pixel-wise segmentation. The main challenge we faced was that, even visually, it was not possible to identify plastic in certain cases.\n\nFigure 1 and 2 serves as an example. Figure 1 can be clearly concluded as plastic while in Figure 2 it is unclear.\n\n2.Architecture\n\nThis SegNet architecture was derived from the deep FCN.", null, "It consists of two parts – an encoder and a decoder. The encoder is similar to the VGG163, but without the fully connected layer thus making the model smaller. The decoder consists of the deconvolutional network which is used for the upsampling process. This upsampling is done for accurate pixel-wise classication and to increase the resolution of the downsampled feature maps. Every decoder has a corresponding encoder.\n\nThe decoder makes use of the max-pooling indices which is obtained from the corresponding encoder to perform upsampling of the feature maps from the encoder.", null, "The reusing of max-pooling indices helps in reducing the training parameters and also improves the boundary delineation. We have made use of the SegNet – Basic architecture. This consists of four convolutional layers in encoder coupled with batch normalization and ReLu followed by a maxpooling layer. Decoder consists of the same but instead of the max-pooling layer an upsampling layer is present. The architecture of FCN and SegNet-Basic have been shown in Figure 3 and Figure 4 respectively.\n\n3.Dataset preparation\n\nStep 1: Dataset Collection & Preparation\n\nOur source data was a video shot through an entry level mobile phone camera. We collected the images by cutting frames from the video. The images were resized as our model takes an input of 480px of width and 360px of height. Then we took a total of 1,000 images such that 750 images contained plastic and 250 images were just paper without any plastic. Allotted a total of 600 images of plastic and 150 images without plastic to the training set and the rest to the validation set.\n\nStep 2: Dataset Ground Truth\n\nWe annotated the dataset using SegNet-annotate-tool4 . The annotation tool produces an output in a format which is not readily acceptable by the SegNet algorithm. We were able to suitably handengineer and automate the conversion process so that the output format was acceptable.\n\nStep 3: Weight Calculation\n\nThe weights5 have to be calculated using the train set. These weights are used to calculate the loss during the training process. According to the SegNet paper, when there is a large variation in the number of pixels in each class in the training set there is a requirement to weight the loss differently. This is known as class balancing.\n\n4.Training the model", null, "The process of training involves passing batches of the train images to the network using a concept of input pipeline. This input pipeline involves the use of two main queues, filename queue and example queue. Filename queue collects the filenames, which are passed through a decoder and enqueued to the example queue where the images are dequeued and passed to the network in batches and the training process starts. Figure 5 illustrates this process, this process is called input pipelining.\n\nWe used an input of 360 x 480 on an NVIDIA GeForce GTX 1050Ti with cuDNN6 v6 acceleration. There are two variants of the SegNet architecture – first the standard SegNet and second the SegNet Basic.\n\nThe rst variant being standard SegNet architecture makes use of 13 convolutional layers in both encoder and decoder networks. We tried this on our 4GB GTX 1050Ti GPU, but, encountered a CUDA7 out of memory error.\n\nSince all the images which we pass to the network in batches are stored in the GPU memory, the more layers we use the more memory gets utilized and thus we encountered this CUDA out of memory error. Then we tried reducing the layersby half to six layers and still encountered the same problem. Then we tried the second variant SegNet – Basic which makes use of four convolutional layers in both encoder and decoder networks as shown in Figure 4. This ran smoothly on our GPU and thus we made use of the egNet-Basic architecture. Figure 5 Input Pipeline to pass images to the network\n\nAt each step of the training process the mean Intersection over Union (mIoU), loss and accuracy is calculated for the the batch of training set.\n\n4.1.The mIoU is calculated by using the following formula :", null, "Where : True Positive[TP] is the number of pixels correctly classified as positive, False Positive[FP] is the number of pixels wrongly classified as positive, False Negative[FN] is the number of pixels wrongly classified as negative\n\n4.2.The accuracy value is calculated using the following formula:", null, "Where: True Negative [TN] is the number of pixels [n(px)] correctly classified as negative. The loss is calculated by using the the ground truth image as well as the SegNet networks output using the pixel values (These images are gray images). At every 100 steps the mean Intersection over Union, loss and accuracy is calculated for the validation set. The graphs for the above process is maintained in a TensorBoard8 and the values are maintained in a csv. At every 1,000 steps the models checkpoint file (which consists of model weights to test our model) will be saved.\n\n5.Hyper-parameter settings\n\nWe first set the maximum steps to 30,000. The validation loss started increasing approximately from the 6,000th step but when it reached the 15,000th step it started increasing at a faster rate thus we chose the maximum steps to be 15,000. Batch size is the number of images passed to the network at every step of the training process. The higher the batch size the more sooner the training would complete. We chose three values for our batch size and they are as follows:\n\n5.1.Training with batch size 6\n\nBatch-size = 6, Learning rate = 0.001, min-after-dequeue9 = 0.4 X (Size of training set), max-steps = 15,000 When we used batch size of 6 we encountered the CUDA out of memory error which we have mentioned about earlier.\n\n5.2.Training with batch size 5\n\nBatch-size = 5,\n\nLearning rate = 0.001,\n\nmin-after-dequeue = 0.4 X (Size of training set), max-steps = 15,000\n\nThis ran smoothly without any errors.\n\n5.3.Training with batch size 4\n\nBatch-size = 4,\n\nLearning rate = 0.001,\n\nmin-after-dequeue = 0.4 X (Size of training set), max-steps = 15000\n\nWe tried the same using batch size 4 to check if this hyperparameter(batch-size) has a significant impact on the performance and accuracy. Repeated the training five times for both batch size 4 and batch size 5.\n\n5.4.Steps and epochs\n\nAt every step, x number of images are passed to the network where x is the batch size. Thus one epoch is completed once all the images in our training set have been passed on to the network.", null, "Where ts is the training set and bs is the batch size.\n\nExample: Since we have used a batch size of 4, 1 epoch will be completed in approximately 188 steps. In that case 15000 steps would correspond to almost 80 epochs.\n\n6.Analysis of the training process\n\nThe TensorBoard results for validation accuracy and validation loss for both the batch sizes are shown below.\n\n6.1.Batch Size 4", null, "The validation loss in the case of batch 4 reached a minimum of 0.2450 at the 5,800th step and from there on it started increasing. The figures 6a and 6b illustrate this.\n\n6.2.Batch Size 5", null, "The same kind of behaviour was observed in the case of batch 5, only difference was that the increase in loss started at the 4,900th step with a loss of 0.2258 which is shown in Figure 7a and 7b. The training loss in both the cases of batch size were gradually decreasing. But there was a sudden increase in the validation loss while the training loss continued to decrease from which we concluded that the model was overfitting.\n\nNext we created a test set of 150 images with 101 of them having plastic and 49 without plastic. We used every checkpoint of the model to test on our test set. The reason for using all the checkpoints was to analyze the differences in the model’s result at different steps. Since we maintained a checkpoint le every 1,000 steps we had a total of 15 checkpoints. We found that the batch size 5 model performed slightly better than batch size 4 model.\n\n7. Results from the model\n\n7.1.Pass Cases", null, "", null, "", null, "7.2.Failed Cases\n\nThe results were not perfect and to analyze the results we created our own set of categories to classify the results to compare the performance between different checkpoint files.", null, "We first tried to obtain a single number metric to categorize the results but this was not possible as the predicted results were not confined to a fixed number of cases. Example: Some cases involved the predicted plastic to be correctly located but the size of the blob varied from the ground truth. Another case where the plastic was detected correctly but extra non existent plastics were also detected along with it. To overcome this, we came up with a total of 6 unique categories for the images with plastic as shown Table 1. Our test set contained 101 images with plastic and 49 without plastic. We maintained a separate table for both. The main reason we considered the columns B (Table 1) as true positive is because our aim was to make sure the model detects all plastic regardless of the paper detected along with it. If our aim were to detect plastic without detecting a single paper we would have to consider the columns B (Table 1) as False Negative. The images without plastic have two categories either True Negative or False Positive as shown in Table 2. Table 1 gives a brief explanation of each category, for Table 2 the True Negative are those cases where there is no plastic in the image and the model also predicted the same, whereas False Positive are those cases where there is no plastic 6 in image but the model has predicted some non existent plastic.\n\n7.3.An excerpt of the sheet we maintained:\n\nThe categories which are significant in defining the performance of our model are in columns A (Table 1)\n\nTable 1: Images with plastic", null, "1 – Location of the detected plastics and size of it are perfect matched with the visual appearance\n2 – Locations are precise, but difference in the size of the plastics\n3 – Plastic locations and size are correct, but predicted plastics at additional locations.\n4 – Plastic locations precise with different sizes, also predicted plastics at additional locations\n5 – Not detected any plastics but showing ones where it doesn’t exist\n6 – No plastic detected\n\nTable 2: Images without plastics", null, "A common pattern that we observed during our analysis of test set on different checkpoint files (for both batch size 4 and 5) was that the number of extra non-existent plastic decreased whereas the number of null predictions increased as the steps increased.\n\n8.Increasing the Size of the Dataset\n\nWe found that a common reason for overfitting is usually due to a small dataset where the model usually tends to memorize the training images along with its corresponding labels and the model settings would be very specific to the train set. Then we increased the dataset from 1000 images to 2000 images of which we allotted 1500 images to train set and the rest to validation set. The graph of validation loss was no different (gradual decrease with a sudden increase) expect for the fact that the validation loss reduced, and the increased datasets model performed much better and got more True positives compared to the one we trained with 750 image.\n\nTable 3: Comparison of true Positives between 1000 and 2000 datasets", null, "From the above Table 3 it is clear that, the number of True Positives (TP) are higher at every stage for the 2,000 images dataset. And then, we made use of the same 2,000 images and considered 1,750 images as the training set and 250 images as the validation set. The result of this training gave a significant decrease in the loss to 0.1848 at the 9,300th step and started increasing from there on. Thus from the above results, we concluded the size of the dataset has a significant impact on the accuracy and validation loss value.\n\n9.Dropout inclusion", null, "The most efficient way to avoid the problem of overfitting is to include a dropout layer, but, when we ran the code with the included dropout on our NVIDIA GeForce GTX 1050 Ti, we once again encountered the CUDA out of memory error. To solve this problem, we ran the same code on Google Colab10 on a 12GB GPU with the 2,000 images dataset and the result of the validation loss graph showed no sudden increase as in the previous runs. The dropout layer significantly reduced the problem of overfitting but the loss value did not go below 0.1840. The graphs have been shown below in Figure 14a and 14b.\n\n10.Challenges\n\nIt was very dificult to segregate the images with plastic and without plastic visually. We found it dificult to find a single valued metric for classifying the test results but came up with a fixed set of categories to categorize them as explained earlier.", null, "We did not get the required accuracy for the plastic dataset. We assumed that the dataset might have been too small for such a complex network like SegNet which involves the use of lot of parameters. To check, if this holds true we took a smaller dataset of 1,000 images of T-shirts compared to our 2,000 images of plastic. We had an idea to make the process of purchasing shirts more engaging and interactive by which users could select the t-shirt they like on a screen and this would be correctly placed on the body of the user, which can be viewed on the same screen below. To accomplish this task, we required the Tshirt which a person was wearing, to be correctly segmented out. SegNet was the perfect architecture as it was capable of delineating the boundaries by which the exact shape of the shirt could be extracted. For this we downloaded images of people wearing t-shirts from the internet. We took this new dataset to see, if the image quality also had a significant impact on the detection results.", null, "We took a total of 1,045 images allotted 900 as the training set and the rest as the validation set. We used an input of 360 X 480 on an NVIDIA GeForce GTX 1050Ti with cuDNN v6 acceleration. The labeling of these images were easy for us compared to the plastic labeling. The images were of high quality compared to the plastic dataset which was mediocre. We trained the model with the following hyperparameters: Batch-size=5, Learning rate=0.001, min-after-dequeue= 0.4 x (Size of training set), max-steps= 20000 Once the training of the t-shirts completed, we observed that there was a gradual decrease without any sudden increase in the validation loss value. The curve that we were expecting from our plastic dataset. The validation loss reached a minimum of 0.05 which was significantly better (Figure 15a & 15b).\n\n12.Sample results of the t-shirt dataset", null, "We succeeded in segmenting out the t-shirts a person is wearing, which is clear from the above results. From this additional process, we learnt that a small dataset was sufficient to obtain a good accuracy with minimal loss. We also found that increasing the dataset may not have a significant impact on accuracy and loss as proved by the plastic dataset, unless the image quality is good. Thus we identified that the required result could be achieved using a small dataset provided that we have a clear or unambiguous dataset.", null, "From this we could conclude that the dataset size is not a limiting factor for accuracy and the quality of images do play a vital role in obtaining a good accuracy.\n\n13.Quantitative comparison\n\nTable 4: 1000 Dataset-plastic (750-train, 250-validation)", null, "Table 5: 2000 Dataset-plastic (1500-train, 500-validation)", null, "BOOK A FREE CONSULTATION" ]	[ null, "https://www.whirldatascience.com/wp-content/uploads/2019/06/fig1fig2.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-3.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-4.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-5.jpg", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-formu-1.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-formu-2.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-formu-3.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-6.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-7.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-a.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-b.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-c.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/01/seg-fig-9.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-tab-1-1.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-tab-2.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-tab-3.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-10.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-11.png", null, "https://www.whirldatascience.com/2019/03/27/assets/images/seg-fig-12.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-13.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-fig-14.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-tab-4.png", null, "https://www.whirldatascience.com/wp-content/uploads/2019/04/seg-tab-5.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9466666,"math_prob":0.94668114,"size":16926,"snap":"2019-51-2020-05","text_gpt3_token_len":3611,"char_repetition_ratio":0.1396407,"word_repetition_ratio":0.03876774,"special_character_ratio":0.21357675,"punctuation_ratio":0.07970343,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9586108,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T02:01:11Z\",\"WARC-Record-ID\":\"<urn:uuid:ac20d19d-247a-4081-b26e-9f4adcbccc20>\",\"Content-Length\":\"72542\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf8c333b-4522-4b70-b1ad-f3e9b3e1585a>\",\"WARC-Concurrent-To\":\"<urn:uuid:40d5f968-138c-48d7-9661-ce20779a6138>\",\"WARC-IP-Address\":\"35.226.14.252\",\"WARC-Target-URI\":\"https://www.whirldatascience.com/2019/03/27/segnet-applicability-in-single-object-detection-with-low-resolution-images/\",\"WARC-Payload-Digest\":\"sha1:D42TUR5QGGTF7GTSCERWFY3HQQXTFRVX\",\"WARC-Block-Digest\":\"sha1:4WNKARAPIJ3F7TEQYVX6ZFIIHXZ7PGZP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606269.37_warc_CC-MAIN-20200122012204-20200122041204-00016.warc.gz\"}"}
https://www.systutorials.com/docs/linux/man/3-slassq.f/	[ "# slassq.f (3) - Linux Manuals\n\nslassq.f -\n\n## SYNOPSIS\n\n### Functions/Subroutines\n\nsubroutine slassq (N, X, INCX, SCALE, SUMSQ)\nSLASSQ updates a sum of squares represented in scaled form.\n\n## Function/Subroutine Documentation\n\n### subroutine slassq (integerN, real, dimension( * )X, integerINCX, realSCALE, realSUMSQ)\n\nSLASSQ updates a sum of squares represented in scaled form.\n\nPurpose:\n\n``` SLASSQ returns the values scl and smsq such that\n\n( scl*2 )smsq = x( 1 )2 +...+ x( n )2 + ( scale*2 )sumsq,\n\nwhere x( i ) = X( 1 + ( i - 1 )INCX ). The value of sumsq is\nassumed to be non-negative and scl returns the value\n\nscl = max( scale, abs( x( i ) ) ).\n\nscale and sumsq must be supplied in SCALE and SUMSQ and\nscl and smsq are overwritten on SCALE and SUMSQ respectively.\n\nThe routine makes only one pass through the vector x.\n```\n\nParameters:\n\nN\n\n``` N is INTEGER\nThe number of elements to be used from the vector X.\n```\n\nX\n\n``` X is REAL array, dimension (N)\nThe vector for which a scaled sum of squares is computed.\nx( i ) = X( 1 + ( i - 1 )INCX ), 1 <= i <= n.\n```\n\nINCX\n\n``` INCX is INTEGER\nThe increment between successive values of the vector X.\nINCX > 0.\n```\n\nSCALE\n\n``` SCALE is REAL\nOn entry, the value scale in the equation above.\nOn exit, SCALE is overwritten with scl , the scaling factor\nfor the sum of squares.\n```\n\nSUMSQ\n\n``` SUMSQ is REAL\nOn entry, the value sumsq in the equation above.\nOn exit, SUMSQ is overwritten with smsq , the basic sum of\nsquares from which scl has been factored out.\n```\n\nAuthor:\n\nUniv. of Tennessee\n\nUniv. of California Berkeley" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8147171,"math_prob":0.993763,"size":1675,"snap":"2023-14-2023-23","text_gpt3_token_len":471,"char_repetition_ratio":0.11310592,"word_repetition_ratio":0.116883114,"special_character_ratio":0.2710448,"punctuation_ratio":0.14836796,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989957,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T21:46:22Z\",\"WARC-Record-ID\":\"<urn:uuid:525ac60f-bf4f-42ed-9fed-2c55b561ed65>\",\"Content-Length\":\"24829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ced94a4-7767-4cd3-9a33-72c40205c9aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:12dceead-1210-452d-9fe1-7cf286d7027b>\",\"WARC-IP-Address\":\"104.21.34.36\",\"WARC-Target-URI\":\"https://www.systutorials.com/docs/linux/man/3-slassq.f/\",\"WARC-Payload-Digest\":\"sha1:Z2PBP4RLZGTNH5YZOOAB54MIZMLUZFLY\",\"WARC-Block-Digest\":\"sha1:5B6VYI2B66EW3EHGZSEVNAAJVYE2D5ZD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00445.warc.gz\"}"}
http://num.bubble.ro/d/49/4000/	[ "Division table for N = 49 / 3999÷4000\n\n49 / 3999 = 0.0123 [+]\n49 / 3999.01 = 0.0123 [+]\n49 / 3999.02 = 0.0123 [+]\n49 / 3999.03 = 0.0123 [+]\n49 / 3999.04 = 0.0123 [+]\n49 / 3999.05 = 0.0123 [+]\n49 / 3999.06 = 0.0123 [+]\n49 / 3999.07 = 0.0123 [+]\n49 / 3999.08 = 0.0123 [+]\n49 / 3999.09 = 0.0123 [+]\n49 / 3999.1 = 0.0123 [+]\n49 / 3999.11 = 0.0123 [+]\n49 / 3999.12 = 0.0123 [+]\n49 / 3999.13 = 0.0123 [+]\n49 / 3999.14 = 0.0123 [+]\n49 / 3999.15 = 0.0123 [+]\n49 / 3999.16 = 0.0123 [+]\n49 / 3999.17 = 0.0123 [+]\n49 / 3999.18 = 0.0123 [+]\n49 / 3999.19 = 0.0123 [+]\n49 / 3999.2 = 0.0123 [+]\n49 / 3999.21 = 0.0123 [+]\n49 / 3999.22 = 0.0123 [+]\n49 / 3999.23 = 0.0123 [+]\n49 / 3999.24 = 0.0123 [+]\n49 / 3999.25 = 0.0123 [+]\n49 / 3999.26 = 0.0123 [+]\n49 / 3999.27 = 0.0123 [+]\n49 / 3999.28 = 0.0123 [+]\n49 / 3999.29 = 0.0123 [+]\n49 / 3999.3 = 0.0123 [+]\n49 / 3999.31 = 0.0123 [+]\n49 / 3999.32 = 0.0123 [+]\n49 / 3999.33 = 0.0123 [+]\n49 / 3999.34 = 0.0123 [+]\n49 / 3999.35 = 0.0123 [+]\n49 / 3999.36 = 0.0123 [+]\n49 / 3999.37 = 0.0123 [+]\n49 / 3999.38 = 0.0123 [+]\n49 / 3999.39 = 0.0123 [+]\n49 / 3999.4 = 0.0123 [+]\n49 / 3999.41 = 0.0123 [+]\n49 / 3999.42 = 0.0123 [+]\n49 / 3999.43 = 0.0123 [+]\n49 / 3999.44 = 0.0123 [+]\n49 / 3999.45 = 0.0123 [+]\n49 / 3999.46 = 0.0123 [+]\n49 / 3999.47 = 0.0123 [+]\n49 / 3999.48 = 0.0123 [+]\n49 / 3999.49 = 0.0123 [+]\n49 / 3999.5 = 0.0123 [+]\n49 / 3999.51 = 0.0123 [+]\n49 / 3999.52 = 0.0123 [+]\n49 / 3999.53 = 0.0123 [+]\n49 / 3999.54 = 0.0123 [+]\n49 / 3999.55 = 0.0123 [+]\n49 / 3999.56 = 0.0123 [+]\n49 / 3999.57 = 0.0123 [+]\n49 / 3999.58 = 0.0123 [+]\n49 / 3999.59 = 0.0123 [+]\n49 / 3999.6 = 0.0123 [+]\n49 / 3999.61 = 0.0123 [+]\n49 / 3999.62 = 0.0123 [+]\n49 / 3999.63 = 0.0123 [+]\n49 / 3999.64 = 0.0123 [+]\n49 / 3999.65 = 0.0123 [+]\n49 / 3999.66 = 0.0123 [+]\n49 / 3999.67 = 0.0123 [+]\n49 / 3999.68 = 0.0123 [+]\n49 / 3999.69 = 0.0123 [+]\n49 / 3999.7 = 0.0123 [+]\n49 / 3999.71 = 0.0123 [+]\n49 / 3999.72 = 0.0123 [+]\n49 / 3999.73 = 0.0123 [+]\n49 / 3999.74 = 0.0123 [+]\n49 / 3999.75 = 0.0123 [+]\n49 / 3999.76 = 0.0123 [+]\n49 / 3999.77 = 0.0123 [+]\n49 / 3999.78 = 0.0123 [+]\n49 / 3999.79 = 0.0123 [+]\n49 / 3999.8 = 0.0123 [+]\n49 / 3999.81 = 0.0123 [+]\n49 / 3999.82 = 0.0123 [+]\n49 / 3999.83 = 0.0123 [+]\n49 / 3999.84 = 0.0123 [+]\n49 / 3999.85 = 0.0123 [+]\n49 / 3999.86 = 0.0123 [+]\n49 / 3999.87 = 0.0123 [+]\n49 / 3999.88 = 0.0123 [+]\n49 / 3999.89 = 0.0123 [+]\n49 / 3999.9 = 0.0123 [+]\n49 / 3999.91 = 0.0123 [+]\n49 / 3999.92 = 0.0123 [+]\n49 / 3999.93 = 0.0123 [+]\n49 / 3999.94 = 0.0123 [+]\n49 / 3999.95 = 0.0123 [+]\n49 / 3999.96 = 0.0123 [+]\n49 / 3999.97 = 0.0123 [+]\n49 / 3999.98 = 0.0123 [+]\nNavigation: Home \| Addition \| Substraction \| Multiplication \| Division Tables for 49: Addition \| Substraction \| Multiplication \| Division\n\nOperand: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 3991 3992 3993 3994 3995 3996 3997 3998 3999 4000 4001 4002 4003 4004 4005 4006 4007 4008 4009 5000 6000 7000 8000 9000\n\nDivision for: 1 2 3 4 5 6 7 8 9 10 20 30 40 41 42 43 44 45 46 47 48 49 50 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8368793,"math_prob":0.99997175,"size":13008,"snap":"2019-43-2019-47","text_gpt3_token_len":3481,"char_repetition_ratio":0.3032913,"word_repetition_ratio":0.6343319,"special_character_ratio":0.31396064,"punctuation_ratio":0.071687475,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999068,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T20:28:26Z\",\"WARC-Record-ID\":\"<urn:uuid:6746d279-6c00-4549-a792-5b4c1e57bd67>\",\"Content-Length\":\"45273\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aea3986a-76fc-478e-96ff-4ad475eb5a3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:652256f1-1cb9-413d-a582-11712bf4895d>\",\"WARC-IP-Address\":\"104.24.96.16\",\"WARC-Target-URI\":\"http://num.bubble.ro/d/49/4000/\",\"WARC-Payload-Digest\":\"sha1:5DXUYMK57FVWJLECJ262ZMSHUOJXLLK6\",\"WARC-Block-Digest\":\"sha1:VNJJOPNO6FYCAS2O7IYAYCHD3TTKSMYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986655310.17_warc_CC-MAIN-20191014200522-20191014224022-00445.warc.gz\"}"}
https://www.gkkv.com/113_113015/	[ "# 你是我的命中注定\n\n《你是我的命中注定》最新章节（提示：已启用缓存技术，最新章节可能会延时显示，登录书架即可实时查看。）\n\nChapter101. 你是我的命中注定\nChapter99.沈靖情况危急。\nChapter100. 危机重重。\nChapter97. 失望的心。\nChapter98. 沈靖病情暴露。\n《你是我的命中注定》正文\nChapter1. 其实，并不坚强。\nChapter2. 你若安好便是晴天\nChapter3. 假如时光倒流。\nChapter4. 天不遂人愿。\nChapter5. 扰乱了谁的心。\nChapter6. 一波未平一波又起。\nChapter7.直播秀风波四起。\nChapter8.泰宇出现了。\nChapter9. 遇见，是最美的年华。\nChapter10. 物是人非事事休。\nChapter11. 隐形的心动。\nChapter12. 误会加深。\nChapter13.交友不慎。\nChapter14. 沈若溪的报复计划。\nChapter15.为伊消得人憔悴\nChapter16. 自作孽不可活。\nChapter17. 谢谢你的出现。\nChapter18. 不期而遇。\nChapter19. 滚蛋吧，人渣。\nChapter20. 逝去的美好。\nChapter21.爱情是上瘾的毒药\nChapter22. 你真的能忘了我吗？\nChapter23. 看清自己的心。\nChapter24. 越在乎，越忐忑。\nChapter25. 我一定要见到你。\nChapter26. 原来，只是局外人。\nChapter27. 要的只是一个结果。\nChapter28. 感受你的温度。\nChapter29.升级矛盾。\nChapter30. 女人心海底针。\nChapter31. 等着你上钩。\nChapter32. 突然的告白。\nChapter33. 解开心结。\nChapter34.你是我的专属天使。\nChapter35. 困扰不断。\nChapter36. 沈若溪回沈家。\nChapter37.惊天秘密。\nChapter38.泰宇出手救沈若溪。\nChapter39.智闯沈氏武馆。\nChapter40.藏起来的秘密。\nChapter41.误中圈套。\nChapter42.姜成勋识破诡计。\nChapter43.游艇爆炸。\nChapter44. 命运的捉弄。\nChapter45. 遗失的美好。\nChapter46. 分我一半的眼泪。\nChapter47. 意外发生。\nChapter48. 针锋相对。\nChapter49. 斗志盎然。\nChapter50. 时光荏苒，岁月匆匆\nChapter51. 解梦。\nChapter52. 奋不顾身靠近你。\nChapter53. 唯恐天下不乱。\nChapter54.若溪，是道不明的情愫\nChapter55. 我的眼里只有你。\nChapter56. 请你离我越远越好。\nChapter57. 一场闹剧。\nChapter58. 背道而驰。\nChapter59. 关于秘密，太入人心\nChapter60. 离别意味重逢。\nChapter61. 毛里求斯的惊喜。\nChapter62.毛里求斯的惊喜2.\nChapter63. 惹恼沈若溪。\nChapter64. 隐藏的心。\nChapter65. 回到武馆。\nChapter66.带不走的思念。\nChapter67.任务空降。\nChapter68. 龙潭虎穴。\nChapter69.两人再次相遇。\nChapter70. 父爱如山。\nChapter71.令人窒息的见面。\nChapter72.沈若溪落入险境。\nChapter73.触不及防的吻。\nChapter74.任务成功。\nChapter75.暗生情愫。\nChapter76. 意外得知秘密。\nChapter77. 一尘不染的真心\nChapter78. 慢慢靠近彼此。\nChapter79. 如果爱下去。\nChapter80. 隐藏的感情。\nChapter81. 好戏上演。\nChapter82.自己挖的坑，自己跳。\nChapter83. 热泪盈眶。\nChapter84. 你是卧底？\nChapter85. 怦然心动。\nChapter86.沈若溪，服从命令。\nChapter87.爱情萌芽的初期。\nChapter88. 我承认，我喜欢你。\nChapter89. 忘不掉的悲伤。\nChapter90. 突然，想爱你。\nChapter91. 触动心的失望。\nChapter92.生不如死。\nChapter93.关于毛里求斯的秘密。\nChapter94．心有了暖的温度。\nChapter95. 泰宇避而不见。\nChapter96. 痛声大哭。\nChapter97. 失望的心。\nChapter98. 沈靖病情暴露。\nChapter99.沈靖情况危急。\nChapter100. 危机重重。\nChapter101. 你是我的命中注定" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.96996886,"math_prob":0.5159693,"size":205,"snap":"2021-31-2021-39","text_gpt3_token_len":226,"char_repetition_ratio":0.039800994,"word_repetition_ratio":0.0,"special_character_ratio":0.29268292,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9588911,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T17:22:58Z\",\"WARC-Record-ID\":\"<urn:uuid:b15f7c5b-6055-4c42-8d46-a7c12014c811>\",\"Content-Length\":\"20243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c27eeda5-3ff6-4024-9ae5-acf823ba5c39>\",\"WARC-Concurrent-To\":\"<urn:uuid:645c71f5-3357-430d-b2f3-61474040132a>\",\"WARC-IP-Address\":\"104.251.123.125\",\"WARC-Target-URI\":\"https://www.gkkv.com/113_113015/\",\"WARC-Payload-Digest\":\"sha1:XT6LIUMJ2RYWZUHMUDOJFHGY3VQJRU62\",\"WARC-Block-Digest\":\"sha1:G34PEISRO5WBUN3PMIT5WTASF7DC3JO2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154466.61_warc_CC-MAIN-20210803155731-20210803185731-00012.warc.gz\"}"}
https://edurev.in/course/quiz/attempt/17630_Olympiad-Test-Pattern-1-/f04f2b78-94b1-4939-8e3e-34ad7a2de09b	[ "# Olympiad Test: Pattern - 1\n\nTest Description\n\n## 15 Questions MCQ Test Science Olympiad Class 7 \| Olympiad Test: Pattern - 1\n\nOlympiad Test: Pattern - 1 for Class 7 2023 is part of Science Olympiad Class 7 preparation. The Olympiad Test: Pattern - 1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Pattern - 1 MCQs are made for Class 7 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Pattern - 1 below.\n 1 Crore+ students have signed up on EduRev. Have you?\nOlympiad Test: Pattern - 1 - Question 1\n\n### Directions: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 1\n\n7 × 5 + 6 = 41\n8 × 7 + 5 = 61\n6 × 7 + 9 = 51\n\nOlympiad Test: Pattern - 1 - Question 2\n\n### Directions: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 2\n\n7 × 4 + 3 × 5 = 28 + 15 = 43\n6 × 2 + 6 × 4 = 12 + 24 = 36\n9 × 3 + 5 × 7 = 27 + 35 = 62\n\nOlympiad Test: Pattern - 1 - Question 3\n\n### Directions: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 3\n\n84 ÷ 7 + 42 ÷ 2 = 12 + 21 = 33\n117 ÷ 13 + 96 ÷ 16 = 9 + 6 = 15\n153 ÷ 17 + 63 ÷ 3 = 9 + 21 = 30\n\nOlympiad Test: Pattern - 1 - Question 4\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 4\n\n5 × 2 – 1 = 9\n9 × 2 + 1 = 19\n19 × 2 – 1 = 37\n37 × 2 + 1 = 75\n75 × 2 – 1 = 149\n149 × 2 + 1 = 299\n299 × 2 – 1 = 597\n\nOlympiad Test: Pattern - 1 - Question 5\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 5\n\n3 × 7 + 6 × 9 = 21 + 54 = 75\n8 × 3 + 11 × 5 = 24 + 55 = 79\n9 × 4 + 8 × 7 = 36 + 56 = 92\n\nOlympiad Test: Pattern - 1 - Question 6\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 6\n\n(46 + 18) ÷ 8 = 64 ÷ 8 = 8\n(62 + 34) ÷ 8 = 96 ÷ 8 = 12\n(57 + 15) ÷ 8 = 72 ÷ 8 = 9\n\nOlympiad Test: Pattern - 1 - Question 7\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 7\n\n(13 – 5)2 + (7 – 3)2 = 64 + 16 = 80\n(23 – 17)2 + (8 – 1)2 = 36 + 49 = 85\n(25 – 12)2 + (5 – 2)2 = 169 + 9 = 178\n\nOlympiad Test: Pattern - 1 - Question 8\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 8\n\n(31 + 17) – (17 + 8) = 48 – 25 = 23\n(47 + 35) – (24 + 7) = 82 – 31 = 51\n(67 + 12) – (19 + 8) = 79 – 27 = 52\n\nOlympiad Test: Pattern - 1 - Question 9\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 9", null, "", null, "Olympiad Test: Pattern - 1 - Question 10\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 10", null, "Olympiad Test: Pattern - 1 - Question 11\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 11", null, "Olympiad Test: Pattern - 1 - Question 12\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 12\n\n(8 × 7) + (9 × 3) = 56 + 27 = 83\n(13 × 5) + (8 × 4) = 65 + 32 = 97\n(17 × 4) + (7 × 9) = 68 + 63 = 131\n\nOlympiad Test: Pattern - 1 - Question 13\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 13", null, "Olympiad Test: Pattern - 1 - Question 14\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 14", null, "Olympiad Test: Pattern - 1 - Question 15\n\nDirections: Choose the appropriate number which follow the given pattern.", null, "Detailed Solution for Olympiad Test: Pattern - 1 - Question 15\n\n(12 + 52 + 42 + 32) × 10 = 510\n(22 + 72 + 62 + 42) × 10 = 1050\n(52 + 32 + 82 + 92) × 10 = 1790" ]	[ null, "https://edurev.gumlet.io/ApplicationImages/Temp/4aa02965-e995-4a2e-890c-882949b7999b_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/c0613866-8338-4282-9816-e757193c396d_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/ddfc6472-7d37-40a1-8f7a-fedc73238937_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/8aa64788-9b3a-4880-9736-8b5a1ca3e5e0_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/9bf805fc-4ff2-40ed-83a5-3989aeb514bb_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/fca30473-b8d2-413a-a562-4e0276542198_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/7a6d5786-8d2c-4310-937b-21f28eb5c0d0_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/8899f7cc-f708-4c3d-acb6-3e78dda82433_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/984d9c27-7062-4b6c-96fd-0916a8068aa6_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/53df0c93-0c69-4ecb-a81c-4e9a10ff08da_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/044585b2-e664-401f-8549-75a8f4aea65d_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/be793eca-2b5e-42a0-af20-2b1c95f6e81e_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/0bce8a33-41ee-480f-acc3-7272ea6bfcdb_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/1f0a124a-c78d-4da3-8979-1e37ed4b2fca_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/fb9ac500-4a0b-40db-8436-0de291736e1d_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/09845034-f066-4e46-9e91-80cadba826f4_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/cdf033ed-f42b-42db-b2fc-7e56f0f8ce95_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/863d5bb8-c1fe-442e-ab9e-fccea09c4ae2_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/556b909a-7027-423c-bf04-8445f8604906_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/79c58b65-6962-43fa-a3b1-32a1bf1de513_lg.jpg", null, "https://edurev.gumlet.io/ApplicationImages/Temp/42af6198-a478-446c-b7f0-0ffebddbaffb_lg.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7068244,"math_prob":0.9764989,"size":2162,"snap":"2022-40-2023-06","text_gpt3_token_len":847,"char_repetition_ratio":0.21547729,"word_repetition_ratio":0.2244898,"special_character_ratio":0.52821463,"punctuation_ratio":0.0776699,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99904835,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42],"im_url_duplicate_count":[null,2,null,3,null,3,null,5,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T14:18:51Z\",\"WARC-Record-ID\":\"<urn:uuid:3c3a9bc3-538d-4b0d-a375-7f4e15682e58>\",\"Content-Length\":\"294322\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b77b5817-f288-44ec-9b9f-17ce267dc065>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff0c061a-7bd8-41fc-aaca-a6bac35a5010>\",\"WARC-IP-Address\":\"20.198.201.88\",\"WARC-Target-URI\":\"https://edurev.in/course/quiz/attempt/17630_Olympiad-Test-Pattern-1-/f04f2b78-94b1-4939-8e3e-34ad7a2de09b\",\"WARC-Payload-Digest\":\"sha1:ZQ3IA5KIY2RP5LZFBKYHM2EEHW4TKGEG\",\"WARC-Block-Digest\":\"sha1:EMHDC3PPNBZ46WCUMRB4VDTYRKM2PMMF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499819.32_warc_CC-MAIN-20230130133622-20230130163622-00331.warc.gz\"}"}
https://www.sanfoundry.com/electric-drives-questions-answers-three-phase-half-wave-ac-dc-conversion-resistive-loads/	[ "# Electric Drives Questions and Answers – Solid-State Switching Circuits – Three Phase, Half-Wave, AC/DC Conversion for Resistive Loads\n\nThis set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Solid-State Switching Circuits – Three Phase, Half-Wave, AC/DC Conversion for Resistive Loads”.\n\n1. Calculate the Vo(avg) for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 440 V.\na) 297.25 V\nb) 298.15 V\nc) 298.11 V\nd) 300.15 V\n\nExplanation: The Vo(avg) for the 3-Φ phase Half-wave uncontrolled rectifier is 3Vml÷2π. The value of output voltage is 3Vml÷2π=3×√2×440÷6.28=297.25 V.\n\n2. Calculate the pulse number if the supply frequency is 2π and the output frequency is 2π÷3.\na) 4\nb) 5\nc) 6\nd) 3\n\nExplanation: The pulse number can be calculated using the ratio of input frequency to the output frequency. The value of pulse number (P) is 2π÷(2π÷3)=3. It is a three-pulse converter or 3-Φ phase Half-wave uncontrolled rectifier.\n\n3. Calculate the average value of current for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 400 V and resistive load value is 5 Ω.\na) 54.04 A\nb) 57.26 A\nc) 51.64 A\nd) 58.15 A\n\nExplanation: The Io(avg) for the 3-Φ phase Half-wave uncontrolled rectifier is 3Vml÷2πR. The value of output voltage is 3Vml÷2πR=3×√2×440÷6.28×5=54.04 A.\n\n4. In 3-Φ Fully controlled rectifier calculate the average value of the voltage if the supply is 400 V and firing angle is 45°.\na) 381.15 V\nb) 382.16 V\nc) 383.19 V\nd) 384.25 V\n\nExplanation: In 3-Φ Fully controlled rectifier, the average value of the voltage is 3Vml(cos(∝))÷π=3×400×√2(cos(45°))÷3.14=382.16 V.\n\n5. Calculate the Vo(r.m.s) (Highly inductive load) for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 441 V.\na) 297.25 V\nb) 298.15 V\nc) 298.11 V\nd) 300.15 V\n\nExplanation: The Vo(r.m.s) for the 3-Φ phase Half-wave uncontrolled rectifier is 3Vml÷2π. The value of output voltage is 3Vml÷2π=3×√2×441÷6.28=297.25 V.\n\n6. Calculate the Vo(avg) for the 3-Φ phase Half-wave controlled rectifier if the supply value is 364 V and the value of firing angle is 2°.\na) 247.62 V\nb) 245.76 V\nc) 214.26 V\nd) 233.26 V\n\nExplanation: The Vo(avg) for the 3-Φ phase Half-wave controlled rectifier is 3Vmlcos(α)÷2π. The value of output voltage is 3Vmlcos(α)÷2π=3×√2×364×(.99)÷6.28=245.76 V.\n\n7. Calculate the value of the Input power factor for 3-Φ Fully controlled rectifier if the firing angle value is 16°.\na) .96\nb) .91\nc) .93\nd) .94\n\nExplanation: The value of the input power factor for 3-Φ Fully controlled rectifier is .95cos(16°)=.91. The input power factor is a product of distortion factor and displacement factor.\n\n8. In 3-Φ Semi-controlled rectifier calculate the average value of the voltage if the supply is 299 V and firing angle is 56.15°.\na) 311.26 V\nb) 304.26 V\nc) 314.51 V\nd) 312.45 V\n\nExplanation: In 3-Φ Semi-controlled rectifier, the average value of the voltage is 3Vml(1+cos(∝))÷2π=3×299×√2(1+cos(56.15°))÷6.28=314.51 V.\n\n9. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 13° and supply frequency is 49.5 Hz.\na) 730.2 msec\nb) 740.8 msec\nc) 754.5 msec\nd) 755.1 msec\n\nExplanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is (240°-α)÷ω. The value of circuit turn-off time for ∝ < 60° is (240°-13°)÷6.28×49.5=730.2 msec.\n\n10. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 170° and supply frequency is 50.5 Hz.\na) 32.4 msec\nb) 35.2 msec\nc) 39.3 msec\nd) 31.5 msec\n\nExplanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is (180°-α)÷Ω. The value of circuit turn-off time for ∝ ≥ 60° is (180°-170°)÷6.28×50.5=31.5 msec.\n\n11. Calculate the form factor if V(r.m.s)=45 V, V(avg)=15 V.\na) 2\nb) 5\nc) 3\nd) 8\n\nExplanation: The form factor is defined as the ratio of the Vr.m.s÷Vavg=45÷15=3. The r.m.s value is three times of average value Vavg.\n\n12. Calculate the Vo(avg) for the 3-Φ phase Half-wave controlled rectifier if the supply value is 480 V and the value of firing angle is 92°.\na) 264.02 V\nb) 487.26 V\nc) 858.26 V\nd) 248.25 V\n\nExplanation: The Vo(r.m.s) for the 3-Φ phase Half-wave controlled rectifier is 3Vm(1+cos(α))÷2π. The value of output voltage is 3Vm(1+cos(α+30))÷2π=3×480×√3×√2(1+cos(92))÷2π=264.02 V.\n\n13. Full form of SCR is ____________\na) Silicon controlled rectifier\nb) State-controlled rectifier\nc) State cover rectifier\nd) State-controlled reset\n\nExplanation: SCR stands for silicon controlled rectifier. It is a semi-controlled, bipolar, uni-directional switch. It is a high power rating than other power electronic devices.\n\n14. Full form of TRIAC is __________\na) Triode for Alternating current\nb) Tri for Alternating current\nc) Triode for Alternating counter\nd) Tri for Alternating counters\n\nExplanation: TRIAC stands for Triode for Alternating current. It is a bipolar, bidirectional switch. It is used in fan regulators to control the speed of fans.\n\n15. TRIAC is a unipolar switch.\na) True\nb) False\n\nExplanation: TRIAC stands for Triode for Alternating current. It is a bipolar, bidirectional switch. It is used in fan regulators to control the speed of fans.\n\nSanfoundry Global Education & Learning Series – Electric Drives.\n\nTo practice all areas of Electric Drives, here is complete set of 1000+ Multiple Choice Questions and Answers.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20150%20150%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.655024,"math_prob":0.9860603,"size":5362,"snap":"2023-40-2023-50","text_gpt3_token_len":1844,"char_repetition_ratio":0.1815976,"word_repetition_ratio":0.3264368,"special_character_ratio":0.34203655,"punctuation_ratio":0.13216561,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99604684,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T01:51:20Z\",\"WARC-Record-ID\":\"<urn:uuid:dab9270f-f222-4b59-ad4c-6c92b09900be>\",\"Content-Length\":\"152522\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1214feb6-6d7a-4403-a619-bda00583e3e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff5adb1d-4bc2-4647-8427-ee106a08cb1e>\",\"WARC-IP-Address\":\"172.67.82.182\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/electric-drives-questions-answers-three-phase-half-wave-ac-dc-conversion-resistive-loads/\",\"WARC-Payload-Digest\":\"sha1:GKRNHWFWE4F3QDGF37IE7BADQKH23TX6\",\"WARC-Block-Digest\":\"sha1:J7IUXVGV35DKDQJGANCM3OLKVDVM6VPW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100047.66_warc_CC-MAIN-20231129010302-20231129040302-00538.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/30-10-minus-8-6	[ "Solutions by everydaycalculation.com\n\n## Subtract 8/6 from 30/10\n\n1st number: 3 0/10, 2nd number: 1 2/6\n\n30/10 - 8/6 is 5/3.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 10 and 6 is 30\n2. For the 1st fraction, since 10 × 3 = 30,\n30/10 = 30 × 3/10 × 3 = 90/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n8/6 = 8 × 5/6 × 5 = 40/30\n4. Subtract the two fractions:\n90/30 - 40/30 = 90 - 40/30 = 50/30\n5. After reducing the fraction, the answer is 5/3\n6. In mixed form: 12/3\n\n#### Subtract Fractions Calculator\n\n-\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6332645,"math_prob":0.9985766,"size":363,"snap":"2019-35-2019-39","text_gpt3_token_len":171,"char_repetition_ratio":0.24791086,"word_repetition_ratio":0.0,"special_character_ratio":0.55647385,"punctuation_ratio":0.091836736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992885,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T17:22:29Z\",\"WARC-Record-ID\":\"<urn:uuid:a2da1715-37f6-49cd-9bd3-06fc3d047a0c>\",\"Content-Length\":\"8631\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6d671dd-077a-455b-8166-c04fb6d135b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7579103-71a9-4603-be3c-afdf5ac4a6b5>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/30-10-minus-8-6\",\"WARC-Payload-Digest\":\"sha1:NSC6OTRR5QZK6EMDM3IZUNQEXL5AT7MK\",\"WARC-Block-Digest\":\"sha1:WHGCHDFR6TNO674AQZHIE3GAENHMOOQ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514577478.95_warc_CC-MAIN-20190923172009-20190923194009-00204.warc.gz\"}"}
https://www.codetd.com/article/6804477	[ "## qbzt day2 晚上（竟然还有晚上）\n\nA\n\ng 当前点到根节点的深度\n\nh 当前点到终点理想的最优情况需要走几步\n\nf f=g+h\n\nA就是把所有的f从小到大排序\n\ng(x) : S x 的理论最近距离\n\ng(x) : S x 对于 g(x) 的估计\n\nf(x) : x T 的理论最近距离,F(x)=g(x)+h(x)\n\nf(x) : x T 对于 f(x) 的估计,F(x)=g(x)+h(x)", null, "BFS 几乎一样，只是每次都弹出当前局面中f(x)最小的那个局面进行扩展\n\n——故需要维护一个优先队列（小根堆）\n\n——使用系统的priority_queue<>即可\n\nf(x) = g(x) + h(x) h(x) = 0 即失去了启发式函数，则变为Breath First Search\n\nf(x) = g(x) + h(x) g(x) = 0 则变为 Best First Search", null, "", null, "", null, "IDA\n\ng：从根节点往下几步\n\nh：步数\n\ng+h>d return\n\nh(x)>h(x)?\n\nh(x)越接近h(x)跑得越快", null, "x 对一个数 p 有逆元当且仅当 (x, p) = 1\n\n……，等等，岂不是 ax == 1 (mod p)", null, "", null, "", null, "", null, "P = p1 * p2 * ... * pn\n\nPi = P / pi\n\nQi = Pi pi 的逆元\n\nx = sigma(ai * Pi * Qi)", null, "" ]	[ null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714202445784-1214273290.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203253675-452209959.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203257422-1282618944.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203304901-2039870549.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203404781-650514888.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203418139-2098593632.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203426359-1677604083.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203432261-542380459.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203436644-1582735826.png", null, "https://img2018.cnblogs.com/blog/1605984/201907/1605984-20190714203446497-1301643348.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.8690495,"math_prob":0.9999875,"size":1177,"snap":"2020-10-2020-16","text_gpt3_token_len":957,"char_repetition_ratio":0.08695652,"word_repetition_ratio":0.040201005,"special_character_ratio":0.37383178,"punctuation_ratio":0.059479553,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995494,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T19:44:58Z\",\"WARC-Record-ID\":\"<urn:uuid:f6bf3659-3014-457f-9934-54511accc796>\",\"Content-Length\":\"32669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:532c9525-248f-408f-86f7-4b824df3b800>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb53327b-1b8f-45c7-9c88-a23ffe01eb53>\",\"WARC-IP-Address\":\"104.27.188.244\",\"WARC-Target-URI\":\"https://www.codetd.com/article/6804477\",\"WARC-Payload-Digest\":\"sha1:HMDR2A754XPMXX2W5FB4QUPGGBTO5DJR\",\"WARC-Block-Digest\":\"sha1:LLYUNPPUCQXX2W7WTISBXH3EC46SEHQP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146485.15_warc_CC-MAIN-20200226181001-20200226211001-00253.warc.gz\"}"}
https://www.skyneel.com/sumif-and-sumifs-function	[ "Categories\n\n# What is Difference between SUMIF and SUMIFS Function\n\nSUMIF function allows you to conditionally sum the values which match the given single criteria, While SUMIF is used to conditionally sum the values which match the multiple criteria. Most of the person little bit confused during using these functions. Today we have to discuss what is the difference between the SUMIF and SUMIFS function in Excel.\n\n## Difference between SUMIF and SUMIFS Function\n\nSUMIF and SUMIFS are both important functions which are located in the Math and Trig category. SUMIF function is designed to work with only one criteria and return the result which match the given condition. SUMIFS function is used to get the conditional total which match the given more than one criteria in active worksheet.\n\n### SUMIF Function in Excel\n\nSUM is an important function used to get the total of selected range of cells. But sometime we need to sum selected range of cells which match the given condition. In that situation you have to use SUMIF function at the place of SUM function in Excel. SUMIF is an important function of Excel which is located in the Mathematical and Trig category. This function is used to conditionally sum the specified range values which match the given condition.", null, "SUMIF(range, criteria, [sum_range])\n\nrange: It the range of continuous cells that you want to evaluated by the given criteria.\n\ncriteria: It is a condition which specifies that which items you want to add that match the given criteria in the range.\n\nsum_range: It is an optional argument which specifies which cells to be added. If “sum_range” argument is omitted then “range” is used as a “sum_range”.\n\nStep 1: First you have to prepare the given table in new worksheet.\n\nStep 2: If you want to get the total units only for “North 1” region then you have to apply the given SUMIF function.\n\n=SUMIF(B27:B35, “North 1”, C27:C35)\n\nIf you want to get the total units for all those cells which date is greater than “31-12-2016” then you have to apply the given SUMIF function.\n\n=SUMIF(A27:A35,”>31-12-2016″,C27:C35)\n\n### SUMIFS Function in Excel\n\nSUMIFS is an important Excel function which is found in Mathematical and Trig category. This function allows you to add the continuous range of cells which is specified by a given set of conditions or criteria in active worksheet. If you have more than one conditions or criteria then you have to use SUMIFS function. If you have Excel 2007 or upper version then you have to use SUMIF function on active worksheet. But if you don’t Excel 2003 or lower version then you are not able to use this function in active worksheet.", null, "SUMIFS(sum_range, criteria_range1, criteria1, [criteria_range2, criteria2], …)\n\nsum_range – It is the range of cells which you want to add which match the given conditions. All blank and text values are ignored.\n\ncriteria_range1 – It is the first range which is evaluated by the first specified criteria.\n\ncriteria1 – It is the first condition that must be matched in the criteria_range 1.\n\ncriteria_range2, criteria2, … – These are additional ranges and criteria which you can associated with them as per your need. It is an optional part. You have to specified up to 127 range or criteria as per your requirement during using the SUMIFS function.\n\nStep 1: Create the following given table in new worksheet.\n\nStep 2: If you want to get the total units only for “North 1” region which date is greater than “31-12-2016” then you have to apply following given formula.\n\n=SUMIFS(C27:C35,B27:B35,”North 1″,A27:A35,”>31-12-2016″)\n\nIf you want to get the total units only for “North 2” region which date is greater than “31-12-2016” then you have to apply following given formula.\n\n=SUMIFS(C27:C35,B27:B35,”North 2″,A27:A35,”>31-12-2016″)\n\nIf you want to get the total units of “North 1” & “North 2” region which date is greater than “31-12-2016” then you have to apply following given formula.\n\n=SUM(SUMIFS(C27:C35,B27:B35,{“North 1″,”North 2″},A27:A35,”>31-12-2016″))\n\nI hope after reading this difference between SUMIF and SUMIFS Function article, you can understand what is the difference between the SUMIF and SUMIFS function in Excel. If you have any suggestion regarding this guide then please write us in the comment box. Thanks to all." ]	[ null, "https://www.skyneel.com/wp-content/uploads/2017/01/What-is-the-Difference-between-the-SUMIF-and-SUMIFS-Function-1.jpg", null, "https://www.skyneel.com/wp-content/uploads/2017/01/What-is-the-Difference-between-the-SUMIF-and-SUMIFS-Function.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.889316,"math_prob":0.85792196,"size":4353,"snap":"2020-34-2020-40","text_gpt3_token_len":1033,"char_repetition_ratio":0.17314325,"word_repetition_ratio":0.2083916,"special_character_ratio":0.24442913,"punctuation_ratio":0.103044495,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98292243,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T22:56:16Z\",\"WARC-Record-ID\":\"<urn:uuid:0f1f7ecf-46b5-4d9f-bc3e-5aed9f275cdc>\",\"Content-Length\":\"67959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6ce9c75-cf79-41d3-9a62-fcf9ec3f6a37>\",\"WARC-Concurrent-To\":\"<urn:uuid:f91fd485-5b32-4487-8165-c73dc349470c>\",\"WARC-IP-Address\":\"67.205.42.159\",\"WARC-Target-URI\":\"https://www.skyneel.com/sumif-and-sumifs-function\",\"WARC-Payload-Digest\":\"sha1:FQHMJ6UN6ZZPZTFKRFUIEFTX52CCQSPX\",\"WARC-Block-Digest\":\"sha1:AKBNIZF7ETBCQ7ZYXS5DA5ANAHRXXB3I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400212959.12_warc_CC-MAIN-20200923211300-20200924001300-00411.warc.gz\"}"}
https://www.yuheng.tech/page/37/	[ "原SDAU暑假训练第18天做题（2018816）\n\nHDU一直关着，很烦。做了几个洛谷的题，挑了几个题解发我新博客上了，点击查看\n\nCodeforces Round 48 (Rated for Div. 2)", null, "", null, "", null, "大意：输入字符串s和t、各自的长度和q个询问，询问由第l到r个字母构成的s的子串中，包含t的个数\n\n#include <iostream>\n#include <string>\n#include <cstdio>\nusing namespace std;\nint n,m,w,sum=0;\nstring s,t;\nint main()\n{\nint n,m,q;\nscanf(\"%d%d%d\",&n,&m,&q);//s长度,n长度，q次数\ns.resize(1200);\nt.resize(1200);\nscanf(\"%s%s\",&s,&t);//s串，t串\nint l,r,ans=0;\nwhile(q--)\n{\nans=0;\nscanf(\"%d%d\",&l,&r);\nif(r-l+1>=m)\n{\nfor(int i=l-1;i+m-1<r;i++)\n{\nstring temp=s.substr(i,m);\nif(s.substr(i,m)==t.substr(0,m)) ans++;\n}\n}\nprintf(\"%d\\n\",ans);\n\n}\n}\n\n后来采用了类似打表的思想过了。\n\n#include<iostream>\n#include<cstdio>\n#include<string>\nusing namespace std;\nint a;\nint main()\n{\nint n,m,k,q,l,r,ans;\nstring s,t;\ncin>>n>>m>>q;\ncin>>s>>t;\nfor(int i=0; i<=n-m; i++)\nif(s.substr(i,m)==t)\na[i]=1;\nwhile(q--)\n{\nans=0;\ncin>>r>>l;\nfor(int i=r-1; i<=l-m; i++)\nif(a[i])\nans++;\ncout<<ans<<endl;\n}\nreturn 0;\n}\n\n别人家的题解：\n\n#include<cstdio>\n#include<cstring>\n#include<string>\n#include<iostream>\n#include<algorithm>\nusing namespace std;\nint main()\n{\nint n,m,q,i,j,l,r,len;\nint counts;\nint vis;\nstring s1,s2;\ncin>>n>>m>>q;\ncin>>s1>>s2;\nlen=s2.size();\nmemset(vis,0,sizeof(vis));\nstring::size_type pos=0;\nwhile((pos=s1.find(s2,pos))!=string::npos)\n{\nvis[pos+1]=pos+1;\npos++;\n}\nfor(i=1;i<=q;i++)\n{\ncounts=0;\nscanf(\"%d%d\",&l,&r);\nfor(j=l;j<=r;j++)\n{\nif(vis[j]!=0&&vis[j]+len-1<=r)\n{\ncounts++;\n}\n}\nprintf(\"%d\\n\",counts);\n}\nreturn 0;\n}\n\n还有前缀数组的大佬：\n\n#include <bits/stdc++.h>\nusing namespace std;\nint n,m,q;\nstring s,t;\nint l,r;\nint hay,shay;\nint main()\n{\ncin>>n>>m>>q;\ncin>>s>>t;\ns=\" \"+s;\nt=\" \"+t;\nfor(int i=1;i<=n-m+1;i++)\n{\nhay[i]=1;\nfor(int j=0;j<m;j++)\n{\nif(s[i+j]!=t[1+j])\n{\nhay[i]=0;\nbreak;\n}\n}\nshay[i]=shay[i-1]+hay[i];\n}\n\nwhile(q--)\n{\ncin>>l>>r;\nr=r-m+1;\nif(r<l) cout<<\"0\"<<endl;\nelse cout<<shay[r]-shay[l-1]<<endl;\n}\nreturn 0;\n}\n\n原【模拟】NewBuildingforSIS", null, "", null, "题意：给你n个相邻的建筑（从左到右编号1到n），每个建筑有h层，每个建筑的a层到b层中任意一层c层可以花一秒通过连廊直接到隔壁建筑的c层。上下楼1层需要1秒。求q次查询(建筑a,层a)到(建筑b,层b)的最短时间。\n\n#include <iostream>\n#include <cmath>\n#include <cstdio>\nusing namespace std;\nint main()\n{\nint n,h,a,b,k;\nscanf(\"%d%d%d%d%d\",&n,&h,&a,&b,&k);\nwhile(k--)\n{\nint ta,fa,tb,fb;long long ans=0;\nscanf(\"%d%d%d%d\",&ta,&fa,&tb,&fb);\nif(ta==tb)\nans+=abs(fb-fa);//同楼层\nif(ta!=tb)\n{\nans+=abs(tb-ta);//计算横向移动时间\nif(fa>=a&&fa<=b)\n{\nans+=abs(fb-fa);//如果都处于连廊直接通过的楼层，附加纵向距离\n}\nelse\n{\nint temp1=0,temp2=0;\ntemp1+=abs(fa-a);//temp1代表如果从最下层连廊走\ntemp1+=abs(fb-a);\ntemp2+=abs(fa-b);//temp1代表如果从最上层连廊走\ntemp2+=abs(fb-b);\nans+=min(temp1,temp2);//取最短距离\n}\n}\ncout<<ans<<endl;\n}\nreturn 0;\n}" ]	[ null, "https://www.yuheng.tech/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.yuheng.tech/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.yuheng.tech/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.yuheng.tech/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.yuheng.tech/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.5018531,"math_prob":0.99156106,"size":3487,"snap":"2019-43-2019-47","text_gpt3_token_len":2253,"char_repetition_ratio":0.1073787,"word_repetition_ratio":0.04511278,"special_character_ratio":0.35847434,"punctuation_ratio":0.22692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969871,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T12:41:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ce39ecfe-6c1b-4f8a-9f0d-39c6dc0b8d68>\",\"Content-Length\":\"71032\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2bf2200-2c97-4a15-8ada-f097176624be>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ad08526-62f1-48b8-874a-dcd507ee1a53>\",\"WARC-IP-Address\":\"47.94.134.93\",\"WARC-Target-URI\":\"https://www.yuheng.tech/page/37/\",\"WARC-Payload-Digest\":\"sha1:LN75GCXT7CCQH3SY5Y4A55SWXYNNAEZH\",\"WARC-Block-Digest\":\"sha1:X2ESE5Z64QGP4OC7QOHDS4DRJNU2HQSO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987817685.87_warc_CC-MAIN-20191022104415-20191022131915-00239.warc.gz\"}"}
https://www.textbook.ds100.org/ch/23/classification_log_model.html	[ "import warnings\n# Ignore numpy dtype warnings. These warnings are caused by an interaction\n# between numpy and Cython and can be safely ignored.\n# Reference: https://stackoverflow.com/a/40846742\nwarnings.filterwarnings(\"ignore\", message=\"numpy.dtype size changed\")\nwarnings.filterwarnings(\"ignore\", message=\"numpy.ufunc size changed\")\n\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport pandas as pd\nimport seaborn as sns\n%matplotlib inline\nimport ipywidgets as widgets\nfrom ipywidgets import interact, interactive, fixed, interact_manual\nimport nbinteract as nbi\n\nsns.set()\nsns.set_context('talk')\nnp.set_printoptions(threshold=20, precision=2, suppress=True)\npd.options.display.max_rows = 7\npd.options.display.max_columns = 8\npd.set_option('precision', 2)\n# This option stops scientific notation for pandas\n# pd.set_option('display.float_format', '{:.2f}'.format)\n\ndef df_interact(df, nrows=7, ncols=7):\n'''\nOutputs sliders that show rows and columns of df\n'''\ndef peek(row=0, col=0):\nreturn df.iloc[row:row + nrows, col:col + ncols]\nif len(df.columns) <= ncols:\ninteract(peek, row=(0, len(df) - nrows, nrows), col=fixed(0))\nelse:\ninteract(peek,\nrow=(0, len(df) - nrows, nrows),\ncol=(0, len(df.columns) - ncols))\nprint('({} rows, {} columns) total'.format(df.shape, df.shape))\n\ndef jitter_df(df, x_col, y_col):\nx_jittered = df[x_col] + np.random.normal(scale=0, size=len(df))\ny_jittered = df[y_col] + np.random.normal(scale=0.05, size=len(df))\nreturn df.assign(*{x_col: x_jittered, y_col: y_jittered})\n\nlebron = pd.read_csv('lebron.csv')\n\n\n# 23.2. The Logistic Model¶\n\nIn this section, we introduce the logistic model, a regression model that we use to predict probabilities.\n\nRecall that fitting a model requires three components: a model that makes predictions, a loss function, and an optimization method. For the by-now familiar least squares linear regression, we select the model:\n\n\\begin{aligned} f_\\hat{\\boldsymbol{\\theta}} (\\textbf{x}) &= \\hat{\\boldsymbol{\\theta}} \\cdot \\textbf{x} \\end{aligned}\n\nAnd the loss function:\n\n\\begin{split} \\begin{aligned} L(\\boldsymbol{\\theta}, \\textbf{X}, \\textbf{y}) &= \\frac{1}{n} \\sum_{i}(y_i - f_\\boldsymbol{\\theta} (\\textbf{X}_i))^2\\\\ \\end{aligned} \\end{split}\n\nWe use gradient descent as our optimization method. In the definitions above, $$\\textbf{X}$$ represents the $$n \\times p$$ data matrix ($$n$$ is the number of data points and $$p$$ is the number of attributes), $$\\textbf{x}$$ represents a row of $$\\textbf{X}$$, and $$\\textbf{y}$$ is the vector of observed outcomes. The vector $$\\boldsymbol{\\hat{\\theta}}$$ contains the optimal model weights whereas $$\\boldsymbol{\\theta}$$ contains intermediate weight values generated during optimization.\n\n## 23.2.1. Real Numbers to Probabilities¶\n\nObserve that the model $$f_\\hat{\\boldsymbol{\\theta}} (\\textbf{x}) = \\hat{\\boldsymbol{\\theta}} \\cdot \\textbf{x}$$ can output any real number $$\\mathbb{R}$$ since it produces a linear combination of the values in $$\\textbf{x}$$, which itself can contain any value from $$\\mathbb{R}$$.\n\nWe can easily visualize this when $$x$$ is a scalar. If $$\\hat \\theta = 0.5$$, our model becomes $$f_\\hat{\\theta} (\\textbf{x}) = 0.5 x$$. Its predictions can take on any value from negative infinity to positive infinity:\n\nxs = np.linspace(-100, 100, 100)\nys = 0.5 xs\nplt.plot(xs, ys)\nplt.xlabel('$x$')\nplt.ylabel(r'$f_\\hat{\\theta}(x)$')\nplt.title(r'Model Predictions for $\\hat{\\theta} = 0.5$');", null, "For classification tasks, we want to constrain $$f_\\hat{\\boldsymbol{\\theta}}(\\textbf{x})$$ so that its output can be interpreted as a probability. This means that it may only output values in the range $$[0, 1]$$. In addition, we would like large values of $$f_\\hat{\\boldsymbol{\\theta}}(\\textbf{x})$$ to correspond to high probabilities and small values to low probabilities.\n\n## 23.2.2. The Logistic Function¶\n\nTo accomplish this, we introduce the logistic function, often called the sigmoid function:\n\n\\begin{aligned} \\sigma(t) = \\frac{1}{1 + e^{-t}} \\end{aligned}\n\nFor ease of reading, we often replace $$e^x$$ with $$\\text{exp}(x)$$ and write:\n\n\\begin{aligned} \\sigma (t) = \\frac{1}{1 + \\text{exp}(-t)} \\end{aligned}\n\nWe plot the sigmoid function for values of $$t \\in [-10, 10]$$ below.\n\nfrom scipy.special import expit\nxs = np.linspace(-10, 10, 100)\nys = expit(xs)\nplt.plot(xs, ys)\nplt.title(r'Sigmoid Function')\nplt.xlabel('$t$')\nplt.ylabel(r'$\\sigma(t)$');", null, "Observe that the sigmoid function $$\\sigma(t)$$ takes in any real number $$\\mathbb{R}$$ and outputs only numbers between 0 and 1. The function is monotonically increasing on its input $$t$$; large values of $$t$$ correspond to values closer to 1, as desired. This is not a coincidence—the sigmoid function may be derived from a log ratio of probabilities, although we omit the derivation for brevity.\n\n## 23.2.3. Logistic Model Definition¶\n\nWe may now take our linear model $$\\hat{\\boldsymbol{\\theta}} \\cdot \\textbf{x}$$ and use it as the input to the sigmoid function to create the logistic model:\n\n\\begin{aligned} f_\\hat{\\boldsymbol{\\theta}} (\\textbf{x}) = \\sigma(\\hat{\\boldsymbol{\\theta}} \\cdot \\textbf{x}) \\end{aligned}\n\nIn other words, we take the output of linear regression—any number in $$\\mathbb{R}$$— and use the sigmoid function to restrict the model’s final output to be a valid probability between zero and one.\n\nTo develop some intuition for how the logistic model behaves, we restrict $$x$$ to be a scalar and plot the logistic model’s output for several values of $$\\hat{\\theta}$$.\n\ndef flatten(li): return [item for sub in li for item in sub]\n\nthetas = [-2, -1, -0.5, 2, 1, 0.5]\nxs = np.linspace(-10, 10, 100)\n\nfig, axes = plt.subplots(2, 3, sharex=True, sharey=True, figsize=(10, 6))\nfor ax, theta in zip(flatten(axes), thetas):\nys = expit(theta * xs)\nax.plot(xs, ys)\nax.set_title(r'$\\hat{\\theta} =$' + str(theta))\n\n# add a big axes, hide frame\nfig.add_subplot(111, frameon=False)\n# hide tick and tick label of the big axes\nplt.tick_params(labelcolor='none', top='off', bottom='off',\nleft='off', right='off')\nplt.grid(False)\nplt.xlabel('$x$')\nplt.ylabel(r'$f_\\hat{\\theta}(x)$')\nplt.tight_layout()", null, "We see that changing the magnitude of $$\\hat{\\theta}$$ changes the sharpness of the curve; the further away from $$0$$, the sharper the curve. Flipping the sign of $$\\hat{\\theta}$$ while keeping magnitude constant is equivalent to reflecting the curve over the y-axis.\n\n## 23.2.4. Summary¶\n\nWe introduce the logistic model, a new prediction function that outputs probabilities. To construct the model, we use the output of linear regression as the input to the nonlinear logistic function." ]	[ null, "https://www.textbook.ds100.org/_images/classification_log_model_6_0.png", null, "https://www.textbook.ds100.org/_images/classification_log_model_10_0.png", null, "https://www.textbook.ds100.org/_images/classification_log_model_14_0.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61809087,"math_prob":0.9995722,"size":6077,"snap":"2021-04-2021-17","text_gpt3_token_len":1632,"char_repetition_ratio":0.11460563,"word_repetition_ratio":0.016091954,"special_character_ratio":0.29488236,"punctuation_ratio":0.17,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999906,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-21T13:52:02Z\",\"WARC-Record-ID\":\"<urn:uuid:1af980ee-06e9-4454-85cc-28098c1ee165>\",\"Content-Length\":\"57135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3164ea6b-02e5-42ae-ae64-13f3fe35c6c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:62c9d908-d953-4230-86cf-0d338377fa09>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://www.textbook.ds100.org/ch/23/classification_log_model.html\",\"WARC-Payload-Digest\":\"sha1:JC54YKL6YG4GYK6JD3PVZ3ZHGATFRRAX\",\"WARC-Block-Digest\":\"sha1:GVRRVRWAGYWAGOGL6JDY7BV6OZPQDRZ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039544239.84_warc_CC-MAIN-20210421130234-20210421160234-00600.warc.gz\"}"}
https://www.olympiadsuccess.com/courses/sof/international-mathematics-olympiad	[ "International Mathematics Olympiad(IMO) IMO Preparation Materials and Sample Papers for class 3,4,5,6,7,8,9,10\n\n#### Unified Council", null, "We are providing complete preparation guide for SOF IMO exams through multiple sample papers, practice papers & mock test papers for classes 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.\n\nBuy Level 1 & 2 Practice Papers\n\nOne of the popular Math Olympiad, conducted by SOF, the IMO is conducted at two levels\n\nLevel 1: The first level of the exam is organized in the respective schools of the participants during school hours only.\n\n• The Level 1 exam is an objective-type test having duration of 60 minutes comprising of 35 objective-type questions for classes 1 to 4 and 50 objective-type questions for classes 5 to 12.\n\n• The exam consists of four sections:\n\n• Section-1: Logical Reasoning\n\n• Section-2: Mathematical Reasoning\n\n• Section-3: Everyday Mathematics\n\n• Section-4: Achievers Section\n\n• There are separate question papers for each and every class.\n\n• The medium of the exam is English.\n\n• CBSE, ICSE/ISC and other State Board syllabus is followed for the setting of test papers.\n\n• The exam is conducted during school hours only\n\nLevel 2: The Level 2 is conducted for students of classes 3 to 12. The qualifiers to second round would include the following:\n\n• Top 5% of candidates class wise that appear for the 1st level exam. Due weightage to marks scored in different sections will be given. Each section is accorded with a separate weightage.\n• Zone wise top 25 rank holders class wise.\n• Class topper where at least 10 students from a class appear in the exam & have scored 50% qualifying marks.\n\nSOF IMO Syllabus and Marking Scheme\n\n Class Section No. of Questions Marks/Question Total Marks 1 to 4 Logical Reasoning 10 1 10 Mathematical Reasoning 10 1 10 Everyday Mathematics 10 1 10 Achievers Section 5 2 10 Grand Total 35 40 5 to 12 Logical Reasoning 15 1 15 Mathematical Reasoning 20 1 20 Everyday Mathematics 10 1 10 Achievers Section 5 3 15 Grand Total 50 60\n\nLearn How to Prepare for IMO\n\nSOF IMO Schedule for 2019-2020 examination\n\nThe exam will be held on 5th & 17th December 2019.\n\nBenefits\n\n• SOF IMO build some, but far from all, of the skills required to master at mathematical research.\n• SOF IMO make participants into very sharp-minded and so they are able to solve clever problems (which, perhaps, have some advantages in some contexts in research).\n\nCheck Class-wise Syllabus here:\n\n#### Class 1\n\nSection – 1 : Patterns, Odd One Out, Measuring Units, Geometrical Shapes, Spatial Understanding, Grouping, Analogy, Ranking Test, Problems based on Figures.\n\nSection – 2 : Numerals, Number names, Number Sense (2 digit numbers), Addition, Subtraction, Lengths, Weights and Comparisons, Time, Money, Geometrical Shapes and Solids, Patterns.\n\nSection – 3 : Syllabus as per Section–1 and 2.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section–1 and 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 1)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 2\n\nSection – 1 : Patterns, Odd One Out, Measuring Units, Geometrical Shapes, Analogy, Ranking Test, Grouping of Figures, Embedded Figures, Coding-Decoding.\n\nSection – 2 : Numerals, number names and Number Sense (3-Digit Numbers), Computation Operations, Length, Weight, Capacity, Time, Temperature, Money, Lines, Shapes and Solids, Pictographs, Patterns.\n\nSection – 3 : Syllabus same as Section 2.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 2)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 3\n\nSection – 1 : Patterns, Analogy and Classification, Alphabet Test, Coding-Decoding, Ranking Test, Grouping of Figures and Figure Matrix, Mirror Images, Geometrical Shapes, Embedded Figures, Days and Dates & Possible Combinations.\n\nSection – 2 : Numerals, Number names and Number Sense (4-digit numbers), Computation Operations, Fractions, Length, Weight, Capacity, Temperature, Time, Money, Geometry, Data Handling.\n\nSection – 3 : The Syllabus of this section will be based on the Syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 3)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 4\n\nSection – 1 : Patterns, Alphabet Test, Coding-Decoding, Ranking Test, Mirror Images, Geometrical Shapes and Solids, Embedded Figures, Direction Sense Test, Days and Dates & Possible Combinations, Analogy and Classification.\n\nSection – 2 : Numerals and Number Names, Number Sense (5-digit numbers), Computation Operations, Fractions, Length, Weight, Capacity, Time, Money, Geometry, Perimeter of Various Shapes, Symmetry, Conversions, Data Handling\n\nSection – 3 : The Syllabus of this section will be based on the Syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 4)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 5\n\nSection – 1 : Patterns, Analogy and Classification, Geometrical Shapes, Mirror and Water Images, Direction Sense Test, Ranking Test, Alphabet Test, Logical Sequence of Words, Puzzle Test, Coding-Decoding.\n\nSection – 2 : Numerals, Number names and Number Sense (7 and 8 digit numbers), Computation Operations, Fractions and Decimals, Measurement of Length, Weight, Capacity, Volume, Time, Temperature and Money, Conversions, Geometrical Shapes and Solids, Angles, Perimeter of Various Shapes & Area of Rectangle and Square, Symmetry, Data Handling\n\nSection – 3 : The Syllabus of this section will be based on the Syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 5)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 6\n\nSection – 1 : Verbal and Non-Verbal Reasoning.\n\nSection – 2 : Knowing our Numbers, Whole Numbers, Playing with Numbers, Basic Geometrical Ideas, Understanding Elementary Shapes, Integers, Fractions, Decimals, Data Handling, Mensuration, Algebra, Ratio And Proportion, Symmetry, Practical Geometry\n\nSection – 3 : The Syllabus of this section will be based on the syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 6)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 7\n\nSection – 1 : Verbal and Non-Verbal Reasoning.\n\nSection – 2 : Integers, Fractions and Decimals, Exponents and Powers, Algebraic Expressions, Simple Linear Equations, Lines and Angles, Comparing Quantities, The Triangle and its Properties, Symmetry, Congruence of Triangles, Rational Numbers, Perimeter and Area, Data Handling, Visualizing Solid Shapes, Practical Geometry.\n\nSection – 3 : The Syllabus of this section will be based on the syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 7)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 8\n\nSection – 1 : Verbal and Non-Verbal Reasoning.\n\nSection – 2 : Rational Numbers, Squares and Square Roots, Cubes and Cube Roots, Exponents and Powers, Comparing Quantities, Algebraic Expressions and Identities, Linear Equations in One Variable, Understanding Quadrilaterals, Constructions, Mensuration, Visualizing Solid Shapes, Data Handling, Direct and Inverse Variations, Factorization, Introduction to Graphs, Playing with Numbers.\n\nSection – 3 : The Syllabus of this section will be based on the syllabus of Mathematical Reasoning.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 8)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 9\n\nSection – 1 : Verbal and Non-Verbal Reasoning.\n\nSection – 2 : Number Systems, Polynomials, Coordinate Geometry, Linear Equations in Two Variables, Introduction to Euclid’s Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas of Parallelograms and Triangles, Circles, Constructions, Heron’s Formula, Surface Areas and Volumes, Statistics, Probability.\n\nSection – 3 : The syllabus of this section will be based on the syllabus of Mathematical Reasoning and Quantitative Aptitude.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 9)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber\n\n#### Class 10\n\nSection – 1 : Verbal and Non-Verbal Reasoning.\n\nSection – 2 : Real Numbers, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Coordinate Geometry, Introduction to Trigonometry, Some Applications of Trigonometry, Circles, Constructions, Areas Related to Circles, Surface Areas and Volumes, Statistics, Probability.\n\nSection – 3 : The Syllabus of this section will be based on the syllabus of Mathematical Reasoning and Quantitative Aptitude.\n\nSection – 4 : Higher Order Thinking Questions - Syllabus as per Section – 2.\n\nWhat we Offer?\n\n• Comprehensive Test Series: 60+ worksheets + 10+ mock test papers\n• Mock Test Series: 10+ mock test papers (IMO Mock Test Papers are based on Previous Year question papers for Class 10)\n\n* 20+ logical reasoning worksheets complimentary for every subscriber" ]	[ null, "https://assets.olympiadsuccess.com/designs/images/unicus.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7984936,"math_prob":0.6302945,"size":8352,"snap":"2019-43-2019-47","text_gpt3_token_len":2032,"char_repetition_ratio":0.16782463,"word_repetition_ratio":0.43065134,"special_character_ratio":0.23515326,"punctuation_ratio":0.20149761,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97679657,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T00:21:16Z\",\"WARC-Record-ID\":\"<urn:uuid:0a199241-8723-4d24-b44b-0da4c7d66519>\",\"Content-Length\":\"158999\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b4b26a7e-f97f-404a-8cf2-121e688a5101>\",\"WARC-Concurrent-To\":\"<urn:uuid:7579b827-2e6d-4ff1-aea2-21a6b8c4f9c8>\",\"WARC-IP-Address\":\"104.26.11.203\",\"WARC-Target-URI\":\"https://www.olympiadsuccess.com/courses/sof/international-mathematics-olympiad\",\"WARC-Payload-Digest\":\"sha1:3MCAD2DR7K6XT7B6DN5EQZC7U4N4X2WY\",\"WARC-Block-Digest\":\"sha1:4EB3S4R3Y62QODLPC7ABUH422VK3HN2S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668544.32_warc_CC-MAIN-20191114232502-20191115020502-00044.warc.gz\"}"}
http://www.softmath.com/math-com-calculator/radical-inequalities/lowest-common-denominator.html	[ "English \| Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/keras/losses/SquaredHinge?hl=he	[ "# tf.keras.losses.SquaredHinge\n\nComputes the squared hinge loss between `y_true` and `y_pred`.\n\n`loss = square(maximum(1 - y_true * y_pred, 0))`\n\n`y_true` values are expected to be -1 or 1. If binary (0 or 1) labels are provided we will convert them to -1 or 1.\n\n#### Usage:\n\n``````sh = tf.keras.losses.SquaredHinge()\nloss = sh([-1., 1., 1.], [0.6, -0.7, -0.5])\n\n# loss = (max(0, 1 - y_true * y_pred))^2 = [1.6^2 + 1.7^2 + 1.5^2] / 3\n\nprint('Loss: ', loss.numpy()) # Loss: 2.566666\n``````\n\nUsage with the `compile` API:\n\n``````model = tf.keras.Model(inputs, outputs)\nmodel.compile('sgd', loss=tf.keras.losses.SquaredHinge())\n``````\n\n## Methods\n\n### `from_config`\n\nView source\n\nInstantiates a `Loss` from its config (output of `get_config()`).\n\nArgs\n`config` Output of `get_config()`.\n\nReturns\nA `Loss` instance.\n\nView source\n\n### `__call__`\n\nView source\n\nInvokes the `Loss` instance.\n\nArgs\n`y_true` Ground truth values. shape = `[batch_size, d0, .. dN]`\n`y_pred` The predicted values. shape = `[batch_size, d0, .. dN]`\n`sample_weight` Optional `sample_weight` acts as a coefficient for the loss. If a scalar is provided, then the loss is simply scaled by the given value. If `sample_weight` is a tensor of size `[batch_size]`, then the total loss for each sample of the batch is rescaled by the corresponding element in the `sample_weight` vector. If the shape of `sample_weight` is `[batch_size, d0, .. dN-1]` (or can be broadcasted to this shape), then each loss element of `y_pred` is scaled by the corresponding value of `sample_weight`. (Note on`dN-1`: all loss functions reduce by 1 dimension, usually axis=-1.)\n\nReturns\nWeighted loss float `Tensor`. If `reduction` is `NONE`, this has shape `[batch_size, d0, .. dN-1]`; otherwise, it is scalar. (Note `dN-1` because all loss functions reduce by 1 dimension, usually axis=-1.)\n\nRaises\n`ValueError` If the shape of `sample_weight` is invalid.\n\n[{ \"type\": \"thumb-down\", \"id\": \"missingTheInformationINeed\", \"label\":\"Missing the information I need\" },{ \"type\": \"thumb-down\", \"id\": \"tooComplicatedTooManySteps\", \"label\":\"Too complicated / too many steps\" },{ \"type\": \"thumb-down\", \"id\": \"outOfDate\", \"label\":\"Out of date\" },{ \"type\": \"thumb-down\", \"id\": \"samplesCodeIssue\", \"label\":\"Samples / code issue\" },{ \"type\": \"thumb-down\", \"id\": \"otherDown\", \"label\":\"Other\" }]\n[{ \"type\": \"thumb-up\", \"id\": \"easyToUnderstand\", \"label\":\"Easy to understand\" },{ \"type\": \"thumb-up\", \"id\": \"solvedMyProblem\", \"label\":\"Solved my problem\" },{ \"type\": \"thumb-up\", \"id\": \"otherUp\", \"label\":\"Other\" }]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6359234,"math_prob":0.9776282,"size":2037,"snap":"2022-27-2022-33","text_gpt3_token_len":624,"char_repetition_ratio":0.12100344,"word_repetition_ratio":0.07073955,"special_character_ratio":0.3024055,"punctuation_ratio":0.21078432,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99357635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T13:26:17Z\",\"WARC-Record-ID\":\"<urn:uuid:aa0d5868-5a33-4582-88f3-503b2646f964>\",\"Content-Length\":\"758505\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0049bb3-9d30-42e4-953e-ee1dde4b1c16>\",\"WARC-Concurrent-To\":\"<urn:uuid:dd94050a-c0e3-4c6f-bd49-b0aba4a4b2a9>\",\"WARC-IP-Address\":\"142.251.16.139\",\"WARC-Target-URI\":\"https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/keras/losses/SquaredHinge?hl=he\",\"WARC-Payload-Digest\":\"sha1:ZD36ZBWEZF4L5OXMFYDUD72ZNBOZI7NC\",\"WARC-Block-Digest\":\"sha1:X4UGTR73R6RZTUOM4T5OYFOCMODRKO2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103941562.52_warc_CC-MAIN-20220701125452-20220701155452-00766.warc.gz\"}"}
https://math.answers.com/Q/What_is_the_volume_of_50_cm_25cm_and_16_cm	[ "", null, "", null, "", null, "", null, "0\n\n# What is the volume of 50 cm 25cm and 16 cm?\n\nUpdated: 9/26/2023", null, "Wiki User\n\n6y ago\n\nBe notified when an answer is posted", null, "Earn +20 pts\nQ: What is the volume of 50 cm 25cm and 16 cm?\nSubmit\nStill have questions?", null, "", null, "Related questions\n\n### What is the volume of a cube with 25 cm sides?\n\nVolume = lengthwidthheight, which for the 25cm cube are all 25 cm. Vol = 25cm * 25cm * 25cm = 15625 cubic cm = 15.625 cubic decimetres = 15.625 litres.\n\n### What is the volume of a block 50 cm long 25 cm high and 16 cm wide?\n\nVolume of block: 502516 = 20,000 cubic cm\n\n### What is the volume of a sphere with the diameter of 25 cm?\n\nThe volume of a sphere with a diameter of 25cm is about 8181.2cm3\n\n### Which is longer 25cm or 200mm?\n\n25cm is longer than 200mm which equals 20cm\n\n### How do you find the height of a box when the volume is 100 cm and the area of the bottom is 50 square cm?\n\nVolume = area x height volume is 100 cm cubed height = volume/area = 100/50 = 2 cm\n\n25cm\n\n15625 cubic cm\n\n### 0.5m plus 25cm equals cm?\n\nHalf a meter is equal to 50 centimeters. 50 centimeters plus 25 centimeters is equal to 75 centimeters.\n\n15625 cubic cm\n\n### How many cm in a 0.25 meters?\n\nOne meter is 100cm so a quarter of that is 25cm. There is 25cm in 0.25m.\n\n### What is the volume of a box that has 50 cm sq anda base of 35 cm?\n\nIf the box had a 50 cm square transverse section and the third dimension were 35 cm, the volume would be:50 × 50 × 35 = 87500 cubic centimeters or eighty seven and a half liters\n\n~35.36 cm" ]	[ null, "https://math.answers.com/icons/searchIcon.svg", null, "https://math.answers.com/icons/searchGlassWhiteIcon.svg", null, "https://math.answers.com/icons/notificationBellIcon.svg", null, "https://math.answers.com/icons/coinIcon.svg", null, "https://math.answers.com/images/avatars/default.png", null, "https://math.answers.com/icons/coinIcon.svg", null, "https://math.answers.com/icons/searchIcon.svg", null, "https://st.answers.com/html_test_assets/imp_-_pixel.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8472105,"math_prob":0.9981468,"size":1019,"snap":"2023-40-2023-50","text_gpt3_token_len":327,"char_repetition_ratio":0.1773399,"word_repetition_ratio":0.29015544,"special_character_ratio":0.3689892,"punctuation_ratio":0.09633028,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99429905,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T17:01:06Z\",\"WARC-Record-ID\":\"<urn:uuid:522107a1-2a98-43bd-8948-c5394379a192>\",\"Content-Length\":\"164461\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50cb2516-052a-41b2-99a2-d64232bb8d99>\",\"WARC-Concurrent-To\":\"<urn:uuid:541ea3d9-92f6-44d1-adaa-16e12433c815>\",\"WARC-IP-Address\":\"146.75.32.203\",\"WARC-Target-URI\":\"https://math.answers.com/Q/What_is_the_volume_of_50_cm_25cm_and_16_cm\",\"WARC-Payload-Digest\":\"sha1:7C4X6FAZTEAX4J2QNYNOT3TNREXBHDYP\",\"WARC-Block-Digest\":\"sha1:R5DNENHJ36NQFDTZZCOHKFVMDWBYLPSO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679515260.97_warc_CC-MAIN-20231211143258-20231211173258-00725.warc.gz\"}"}
https://stats.stackexchange.com/questions/566994/glrt-of-exponential-distribution-and-critical-region/567011	[ "# GLRT of exponential distribution and critical region\n\nFind the Generalized likelihood ratio test (GLRT) for $$H_0: \\lambda = \\lambda_0$$ when $$H_A: \\lambda \\ne \\lambda_0$$ for $$X_1 ... X_n$$ taken from $$X \\sim Exp(\\lambda;x)$$, with a test size of $$0.06$$, obtain the critical region.\n\nHere's what I have tried:\n\n$$L(\\lambda_0;x) = \\prod_{i=1}^nf(x_i;\\lambda_0) = \\prod_{i=1}^n\\lambda_0 e^{-\\lambda_0 x_i}=\\lambda_0^ne^{-\\lambda_0 \\sum_{i=1}^nx_i}$$\n\n$$\\mathbf{L}(\\lambda_0;x) = \\log L(\\lambda_0;x) = n\\log(\\lambda_0) - \\lambda_0\\sum_{i=1}^nx_i$$\n\n$$\\frac{\\partial\\mathbf{L}(\\lambda_0;x)}{\\partial\\lambda_0} = \\frac{n}{\\lambda_0}-\\sum_{i=1}^nx_i \\implies \\frac{n}{\\lambda_0}-\\sum_{i=1}^nx_i=0 \\implies \\lambda_0 = \\frac{1}{\\bar{x}}$$\n\n$$\\mathbf{L}(\\hat{\\lambda};x) = \\left(\\frac{1}{\\bar{x}}\\right)^ne^{-\\left(\\frac{1}{\\bar{x}} \\right) \\sum_{i=1}^nx_i}$$\n\nThen applying the likelihood ratio:\n\n$$\\Lambda :=\\frac{L(\\lambda_0;x)}{\\mathbf{L}(\\hat{\\lambda};x)} = \\frac{\\lambda_0 ^ne^{-\\lambda_0 \\sum_{i=1}^nx_i}}{\\left(\\frac{1}{\\bar{x}}\\right)^ne^{-\\left(\\frac{1}{\\bar{x}} \\right) \\sum_{i=1}^nx_i}}= \\left(\\bar{x}\\lambda_0 e^{1-\\lambda_0\\bar{x}} \\right)^n$$\n\nWe can find the critical region by:\n\n\\begin{align} C &= \\{x : \\lambda \\le k\\} \\\\ &= \\{x : \\left(\\bar{x}\\lambda_0 e^{1-\\lambda_0\\bar{x}} \\right)^n \\le k\\} \\\\ \\end{align}\n\nHowever, I am unsure on how to proceed from here to obtain the critical region/\n\nEdit:\n\nI have been doing further reading and it seems that the critical region can be found by the following, Wilks Theorem $$-2\\log(\\Lambda) \\approx \\chi^2_{1, 1-\\alpha}$$\n\nAnd as such: \\begin{align} -2\\log(\\Lambda) &= -2n\\log\\left(\\bar{x}\\lambda_0 e^{1-\\lambda_0\\bar{x}} \\right) \\\\ &= -2n \\left(\\log(\\bar{x})+\\log(\\lambda_0) + 1-\\lambda_0\\bar{x} \\right) \\le \\chi^2_{1, 0.94} = 0.2571744 \\\\ &= \\log(\\bar{x})+\\log(\\lambda_0) + 1-\\lambda_0\\bar{x} \\ge -\\frac{\\chi^2_{1, 0.94}}{2n} \\end{align}\n\n[Not enough reputation to comment, so I write in an answer]\n\nHint: consider the function $$g(y)=y\\lambda_0e^{1-\\lambda_0 y}$$ for $$y>0$$. Take the derivative to help draw the curve. You'll find a region of $$y$$ such that $$g(y)\\le k^{1/n}$$ by drawing a horizontal line intersecting the curve. After getting a form of the region based on $$y$$, you can forget about $$k$$, and determine the boundary (intersection points of the horizontal line and the curve) using the distribution of $$y$$. Here $$y$$ is your $$\\bar{x}$$.\n\nBy the way, $$\\lambda_0$$ is a specific value of $$\\lambda$$ and it's generally not recommended to write $$\\hat{\\lambda}_0$$, because there is nothing unknown to estimate. Instead, I'll write $$\\hat{\\lambda}$$ as the value that maximizes your log-likelihood function $$\\mathbf{L}(\\lambda;x)$$. The likelihood ratio has nothing to do with the parameter $$\\lambda$$ of the exponential distribution, so I'll suggest using another letter, say $$\\Lambda := \\frac{L(\\lambda_0;x)}{L(\\hat{\\lambda};x)}$$, where the denominator is the likelihood function evaluated at $$\\hat{\\lambda}$$, not $$\\mathbf{L}(\\hat{\\lambda};x)$$.\n\n• Thanks for the hint, I'll look into it some further. I have edited my post from a reading, not sure if you're familiar with it? I might have got it totally wrong though. Also, thanks for the tips on notation also. I'll make an update later today on your hint and will upload it to my post for progress. Mar 8 at 16:24\n• I have taking the first derivative and I get $e^{1-\\lambda_0 y}(2-\\lambda_0 y)$ by setting this equal to $0$ I find that I get $\\bar{x} = y = \\frac{2}{\\lambda_0}$ how do I proceed from here? Mar 8 at 16:36\n• @dollarbill Are you familiar with drawing the function curve with the help of derivatives? You got the first derivative, which contains some important information about the function. The first part, $e^{1-\\lambda_0y}$, is always positive, so the sign of $g'$ is solely determined by the second part, $2-\\lambda_0 y$, where $\\lambda_0>0$. Then you can see that $g'(y)>0$ for all $0<y<2/\\lambda_0$, and $g'(y)\\le 0$ otherwise. What do these tell you about the shape of $g$?\n– Min\nMar 9 at 7:01\n• @dollarbill About your update. That is the asymptotic distribution of $-2\\log\\Lambda$, which approximately holds when you have a finite sample. In this case, actually, you can derive an exact distribution, which is true for any finite sample size.\n– Min\nMar 9 at 7:04\n• @dollarbill In addition, you may confuse a distribution itself with critical values under a distribution. $-2\\log\\Lambda\\overset{approx.}{\\sim} \\chi^2_1$, where $\\sim$ means the random variable on the left side follows a distribution whose name is indicated on the right, and the critical region is the set of all viable values of $\\bar{x}$ such that $-2\\log\\Lambda\\ge \\chi^2_{1,\\alpha}$.\n– Min\nMar 9 at 7:15" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5025922,"math_prob":0.99985325,"size":1864,"snap":"2022-27-2022-33","text_gpt3_token_len":765,"char_repetition_ratio":0.21451613,"word_repetition_ratio":0.0,"special_character_ratio":0.430794,"punctuation_ratio":0.09668508,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999951,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T18:41:23Z\",\"WARC-Record-ID\":\"<urn:uuid:ca0a68d6-ce68-49b4-a38c-88f3db932513>\",\"Content-Length\":\"232132\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:baedd309-fabf-4093-a0e0-d3cac9abbfb2>\",\"WARC-Concurrent-To\":\"<urn:uuid:99057b99-e473-4503-a25d-5b38e35d4b64>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/566994/glrt-of-exponential-distribution-and-critical-region/567011\",\"WARC-Payload-Digest\":\"sha1:MVSCOHY4JMAYNOO4VW32HLWYKCDR5XWT\",\"WARC-Block-Digest\":\"sha1:7IICJJEJHEPWP5SMUMSA5EUJTPI4B6MU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572515.15_warc_CC-MAIN-20220816181215-20220816211215-00048.warc.gz\"}"}
https://drmf-beta.wmflabs.org/wiki/Formula:KLS:01.13:15	[ "# Formula:KLS:01.13:15\n\n${\\displaystyle{\\displaystyle{\\displaystyle\\qHyperrphis{2}{1}@@{q^{-n},b}{c}{q}% {z}=\\frac{\\left(b;q\\right)_{n}}{\\left(c;q\\right)_{n}}q^{-n-\\binomial{n}{2}}(-z% )^{n}\\ \\qHyperrphis{2}{1}@@{q^{-n},c^{-1}q^{1-n}}{b^{-1}q^{1-n}}{q}{\\frac{cq^{% n+1}}{bz}}}}}$", null, "## Proof\n\nWe ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space." ]	[ null, "https://drmf-beta.wmflabs.org/index.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68175274,"math_prob":0.9992934,"size":641,"snap":"2022-05-2022-21","text_gpt3_token_len":212,"char_repetition_ratio":0.1789639,"word_repetition_ratio":0.1388889,"special_character_ratio":0.2979719,"punctuation_ratio":0.27672955,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99762547,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T21:19:00Z\",\"WARC-Record-ID\":\"<urn:uuid:4a40f695-5965-4cc3-a597-b40df6aa9c69>\",\"Content-Length\":\"49983\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8c96cae-c907-4c95-a5e8-7091823118bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2f75820-0308-4dae-8658-c20a018ffbf4>\",\"WARC-IP-Address\":\"185.15.56.49\",\"WARC-Target-URI\":\"https://drmf-beta.wmflabs.org/wiki/Formula:KLS:01.13:15\",\"WARC-Payload-Digest\":\"sha1:XSQHMPXZG76NLBLFD2NWVL7IYFOOQLUX\",\"WARC-Block-Digest\":\"sha1:USSGBMKLNB2NIZDCIAP2LUFNGXRYOUUV\",\"WARC-Truncated\":\"disconnect\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662625600.87_warc_CC-MAIN-20220526193923-20220526223923-00319.warc.gz\"}"}
https://socratic.org/questions/find-the-function-values-of-cos-tan-sec-csc-and-cot#551189	[ "# Find the function values of cos θ, tan θ, sec θ, csc θ and cot θ ?\n\n## If sin θ = −(4)/(5) and cos θ < 0, in which quadrant is θ? Find the function values of cos θ, tan θ, sec θ, csc θ and cot θ.\n\nFeb 13, 2018\n\nGiven that, sin θ = (−4)/(5)\n\napplying the identity, ${\\sin}^{2} \\theta + {\\cos}^{2} \\theta = 1$\n\n${\\cos}^{2} \\theta = 1 - {\\sin}^{2} \\theta$\n\n${\\cos}^{2} \\theta = 1 - {\\left(\\frac{- 4}{5}\\right)}^{2}$\n\n${\\cos}^{2} \\theta = 1 - \\frac{16}{25} = \\frac{9}{25}$\n\n$\\cos \\theta = \\sqrt{\\frac{9}{25}} = \\pm \\frac{3}{5}$\n\nits given that, cos θ < 0 ; color(magenta)(costheta =(-3)/5\nSince, both $\\sin \\theta$ and $\\cos \\theta$ are negative, $\\theta$ lies in the $\\textcolor{red}{\\text{third quadrant}}$.\n\n$\\tan \\theta = \\sin \\frac{\\theta}{\\cos} \\theta$\n=(-4)/5 xx 5/(-3) => color(magenta)(tantheta= 4/3\n\nsec θ = 1/costheta => color(magenta)(sec theta=(-5)/3\n\ncsc θ = 1/sintheta => color(magenta)(csctheta = (-5)/4\n\ncottheta = costheta/sintheta = 1/tantheta => color(magenta)(cottheta=3/4" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56419885,"math_prob":1.0000068,"size":437,"snap":"2021-43-2021-49","text_gpt3_token_len":159,"char_repetition_ratio":0.12009238,"word_repetition_ratio":0.69148934,"special_character_ratio":0.32036614,"punctuation_ratio":0.17391305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T11:19:43Z\",\"WARC-Record-ID\":\"<urn:uuid:83d09f7a-518e-4394-9b00-e9be9062b575>\",\"Content-Length\":\"33314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:482d20ef-38df-4a9e-b3c1-8f518180dbb0>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9626e14-7b07-42a5-8617-f970286c9027>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/find-the-function-values-of-cos-tan-sec-csc-and-cot#551189\",\"WARC-Payload-Digest\":\"sha1:TVD7WKGFDLQJFL64U5BBGJYDJUY6HRQA\",\"WARC-Block-Digest\":\"sha1:J6UY75YGT2NNBUU6CRVXTOTLIGMIQ6BH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587659.72_warc_CC-MAIN-20211025092203-20211025122203-00365.warc.gz\"}"}
https://moam.info/biomedical-engineering-online-biomedsearch_5c5cae76097c478d4f8b4594.html	[ "## BioMedical Engineering OnLine - BioMedSearch\n\nOct 27, 2005 - Open Access. Research. Reliability of old and new ventricular fibrillation detection ... http://www.biomedical-engineering-online.com/content/4/1/60. Page 2 of 15 ...... \"BioMed Central will be the most significant development for.\n\nBioMedical Engineering OnLine\n\nBioMed Central\n\nOpen Access\n\nResearch\n\nReliability of old and new ventricular fibrillation detection algorithms for automated external defibrillators Anton Amann1, Robert Tratnig2 and Karl Unterkofler2 Address: 1Innsbruck Medical University, Department of Anesthesia and General Intensive Care, Anichstr. 35, A-6020 Innsbruck, Austria and Department of Environmental Sciences, ETH-Hönggerberg, CH-8093 Zürich, Switzerland and 2Research Center PPE, FH-Vorarlberg, Achstr. 1, A6850 Dornbirn, Austria Email: Anton Amann* - [email protected]; Robert Tratnig - [email protected]; Karl Unterkofler* - [email protected] * Corresponding authors\n\nPublished: 27 October 2005 BioMedical Engineering OnLine 2005, 4:60\n\ndoi:10.1186/1475-925X-4-60\n\nReceived: 06 May 2005 Accepted: 27 October 2005\n\nventricular fibrillation detectionautomated external defibrillator (AED)ventricular fibrillation (VF)sinus rhythm (SR)ECG analysis\n\nAbstract Background: A pivotal component in automated external defibrillators (AEDs) is the detection of ventricular fibrillation by means of appropriate detection algorithms. In scientific literature there exists a wide variety of methods and ideas for handling this task. These algorithms should have a high detection quality, be easily implementable, and work in real time in an AED. Testing of these algorithms should be done by using a large amount of annotated data under equal conditions. Methods: For our investigation we simulated a continuous analysis by selecting the data in steps of one second without any preselection. We used the complete BIH-MIT arrhythmia database, the CU database, and the files 7001 – 8210 of the AHA database. All algorithms were tested under equal conditions. Results: For 5 well-known standard and 5 new ventricular fibrillation detection algorithms we calculated the sensitivity, specificity, and the area under their receiver operating characteristic. In addition, two QRS detection algorithms were included. These results are based on approximately 330 000 decisions (per algorithm). Conclusion: Our values for sensitivity and specificity differ from earlier investigations since we used no preselection. The best algorithm is a new one, presented here for the first time.\n\nBackground Sudden cardiac arrest is a major public health problem and one of the leading causes of mortality in the western world. In most cases, the mechanism of onset is a ventricular tachycardia that rapidly progresses to ventricular fibrillation . Approximately one third of these patients\n\ncould survive with the timely employment of a defibrillator. Besides manual defibrillation by an emergency paramedic, bystander defibrillation with (semi-)automatic external defibrillators (AEDs) has also been recommended for resuscitation. These devices analyze the\n\nPage 1 of 15 (page number not for citation purposes)\n\nelectrocardiogram (ECG) of the patient and recognize whether a shock should be delivered or not, as, e.g., in case of ventricular fibrillation (VF). It is of vital importance that the ECG analysis system used by AEDs differentiates well between VF and a stable but fast sinus rhythm (SR). An AED should not deliver a shock to a collapsed patient not in cardiac arrest. On the other hand, a successfully defibrillated patient should not be defibrillated again. The basis of such ECG analysis systems of AEDs is one or several mathematical ECG analysis algorithms. The main purpose of this paper is to compare various old and new algorithms in a standardized way. To gain insight into the quality of an algorithm for ECG analysis, it is essential to test the algorithms under equal conditions with a large amount of data, which has already been commented on by qualified cardiologists. Commonly used annotated databases are Boston's Beth Israel Hospital and MIT arrhythmia database (BIH-MIT), the Creighton University ventricular tachyarrhythmia database (CU), and the American Heart Association database (AHA). We used the complete CU and BIH-MIT arrhythmia database, and the files 7001 – 8210 of the AHA database, , [3,4]. For each algorithm approximately 330 000 decisions had been calculated. No preselection of certain ECG episodes was made, which mimics the situation of a bystander more accurately. In this investigation we analyzed 5 well-known standard and 5 new ventricular fibrillation detection algorithms. In addition, two QRS detection algorithms were included. The results are expressed in the quality parameters Sensitivity and Specificity. In addition to these two parameters, we calculated the Positive Predictivity and Accuracy of the investigated algorithms. Furthermore, the calculation time in comparison to the duration of the real data was calculated for the different algorithms. The calculation times were obtained by analyzing data from the CU database only. The quality parameters were obtained by comparing the VF/no VF decisions suggested by the algorithm with the annotated decisions suggested by cardiologists. The cardiologists' decisions are considered to be correct. We distinguished only between ventricular fibrillation and no ventricular fibrillation, since the annotations do not include a differentiation between ventricular fibrillation and ventricular tachycardia. The closer the quality parameters are to 100%, the better the algorithm works. Since an AED has to differentiate between VF and no VF, the sensitivity and specificity are the appropriate parameters. To represent the quality of an algorithm by its sensitivity and specificity bears some problems. A special algorithm can\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\n100\n\n80 Sensitivity / per cent\n\nBioMedical Engineering OnLine 2005, 4:60\n\n60\n\n40\n\n20\n\n0\n\n0\n\n20\n\n40\n\n60\n\n80\n\n100\n\n(1−Specificity) / per cent\n\nFigureof Receiver plexity length measure\" 1operating 8s described characteristic in the for introduction, the algorithm for a\"comwindow Receiver operating characteristic for the algorithm \"complexity measure\" described in the introduction, for a window length of 8 s. The calculated value for the area under the curve, IROC, is 0.87.\n\nhave a high sensitivity, but a low specificity, or conversely. Which one is better? To arrive at a common and single quality parameter, we use the receiver operating characteristic (ROC). The sensitivity is plotted in dependence of (1 - specificity), where different points in the plot are obtained by varying the critical threshold parameter in the decision stage of the algorithm. By calculating the area under the ROC curve (we call this value \"integrated receiver operating characteristic\", and denote it by IROC), it is possible to compare different algorithms by one single value. Figure 1 shows a typical example of an ROC curve. Section 1 provides the necessary background for the algorithms under investigation. Section 2 describes the methods of evaluation and represents our results in Table 1, 2, 3 and Figure 5 and 6. A discussion of the results follows in Section 3. Appendix A recalls the basic definitions of the quality parameters Sensitivity, Specificity, Positive Predictivity, Accuracy, and ROC curve. In Appendix B we provide more details on one of the new algorithms.\n\n1 Methods The ventricular fibrillation detection algorithms considered here are partly taken from the scientific literature, five of them are new. Some of them have been evaluated in and . For all algorithms we used the same prefiltering process. First, a moving average filter of order 5 is applied to the\n\nPage 2 of 15 (page number not for citation purposes)\n\nBioMedical Engineering OnLine 2005, 4:60\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\nTable 1: >Quality of ventricular fibrillation detection algorithms (sensitivity (Sns), specificity (Spc), integrated receiver operating characteristic (IROC)) in per cent, rounded on 3 significant digits, wl = window length in seconds. () ... no appropriate parameter exists.\n\nData Source\n\nMIT DB\n\nParameter\n\nwl\n\nTCI TCI ACF95 ACF99 VF VF SPEC CPLX STE MEA SCA WVL1 WVL2 LI TOMP\n\n3 8 8 8 4 8 8 8 8 8 8 8 8 8 8\n\nSns. 82.5 74.5 33.2 59.4 36.7 29.4 23.1 6.3 54.5 62.9 72.4 28.7 81.1 3.1 68.5\n\nCU DB\n\nSpc. 78.1 83.9 45.9 30.1 100 100 100 92.4 83.4 80.8 98.0 99.9 89.0 95.1 40.6\n\nSns. 73.5 71.0 38.1 54.7 32.2 30.8 29.0 56.4 52.9 60.1 67.7 26.2 61.0 7.5 71.3\n\nAHA DB\n\nSpc. 62.6 70.5 58.9 49.3 99.5 99.5 99.3 86.6 66.6 87.5 94.9 99.4 72.1 94.8 48.4\n\nSns. 75.2 75.7 51.5 71.5 17.6 16.9 29.2 60.2 49.6 49.8 71.7 26.8 73.5 9.3 95.9\n\noverall result\n\nSpc. 78.3 86.9 52.2 40.3 99.9 100 99.8 91.9 81.0 88.6 99.7 99.5 89.6 92.0 39.7\n\nSns. 75.0 75.1 49.6 69.2 19.6 18.8 29.1 59.2 50.1 51.2 71.2 26.7 72.0 9.0 92.5\n\nSpc.\n\nIROC\n\n77.5 84.4 49.0 35.0 99.9 100 99.9 92.0 81.7 84.1 98.5 99.7 88.4 93.9 40.6\n\n82 82 49 49 85 87 89 87 67 82 92 80 () 58 67\n\nTable 2: Quality of ventricular fibrillation detection algorithms (positive predictivity (PP), accuracy (Ac), calculation time(ct)) for a window length of 8 seconds. Positive predictivity and accuracy in per cent, rounded on 3 digits; calculation time in per cent of the real time of the data, rounded on 2 digits, wl = window length in seconds.\n\nData Source\n\nMIT DB\n\nParameter\n\nwl\n\nTCI TCI ACF95 ACF99 VF VF SPEC CPLX STE MEA SCA WVL1 WVL2 LI TOMP\n\n3 8 8 8 4 8 8 8 8 8 8 8 8 8 8\n\nPP. 0.6 0.8 0.1 0.1 91.3 82.4 60.6 0.1 0.5 0.5 5.6 38.9 1.2 0.1 0.2\n\nCU DB\n\nAc. 78.1 83.9 45.9 30.2 99.9 99.9 99.8 92.3 83.4 80.8 97.9 99.8 88.9 94.9 40.6\n\nPP. 33.9 38.9 19.7 22.2 94.0 94.5 92.0 52.7 29.5 56.0 77.8 92.1 36.6 27.5 26.7\n\nAHA DB\n\nAc. 64.8 70.6 54.5 50.4 85.5 85.2 84.6 80.3 63.8 81.8 89.2 84.1 69.8 76.5 53.2\n\nsignal. (Note 1: Fixing the order of 5 seems to bear a problem. Data with different frequencies (AHA, CU ... 250 Hz, BIH-MIT ... 360 Hz) are filtered in a different way. But this is neglectable, when the Butterworth filter is applied afterwards.) This filter removes high frequency noise like inter-\n\nPP. 41.7 54.4 18.2 19.9 98.0 98.9 97.3 60.7 35.1 47.5 98.0 91.8 59.3 19.4 24.8\n\noverall result\n\nAc. 77.8 84.9 52.1 45.7 85.8 85.7 87.7 86.5 75.6 81.9 94.9 87.0 86.8 77.8 49.4\n\nPP. 23.7 31.1 8.3 9.1 97.0 97.7 96.1 40.8 20.4 23.2 81.6 90.5 36.8 12.1 12.7\n\nAc. 77.2 83.6 49.0 37.9 93.1 93.0 93.8 89.2 79.0 81.3 96.2 93.5 87.0 86.6 45.0\n\nct. 1.5 2.1 3.6 3.6 1.4 1.9 1.9 2.5 1.9 2.5 5.9 1.9 40 15 0.84\n\nspersions and muscle noise. Then, a drift suppression is applied to the resulting signal. This is done by a high pass filter with a cut off frequency of one Hz. Finally, a low pass Butterworth filter with a limiting frequency of 30 Hz is applied to the signal in order to suppress needless high-\n\nPage 3 of 15 (page number not for citation purposes)\n\nBioMedical Engineering OnLine 2005, 4:60\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\nTable 3: Sensitivity of ventricular fibrillation detection algorithms in per cent, wl = window length in seconds.\n\nSns. if Spc. = 95\n\nSns. if Spc. = 99\n\nTCI TCI ACF95 ACF99 VF VF SPEC CPLX STE MEA SCA WVL1 LI TOMP\n\n3 8 8 8 4 8 8 8 8 8 8 8 8 8\n\n15.0 25.3 3.0 3.0 71.0 73.4 69.8 38.8 29.4 7.0 79.0 56.7 7.3 9.1\n\n1.0 1.3 0.6 0.6 59.2 59.7 58.9 5.8 10.8 0.5 66.4 35.2 1.4 1.8\n\nfrequency information even more. This filtering process is carried out in a MATLAB routine. It uses functions from the \"Signal Processing Toolbox\". (Note 2: We used MATLAB R13 – R14 and \"Signal Processing Toolbox\" version 6.1 – 6.3 on a Power Mac G5, 2 GHz.) In order to obtain the ROC curve we have to change a parameter which we call \"critical threshold parameter\" below. TCI algorithm The threshold crossing intervals algorithm (TCI) operates in the time domain. Decisions are based on the number and position of signal crossings through a certain threshold.\n\nFirst, the digitized ECG signal is filtered by the procedure mentioned above. Then a binary signal is generated from the preprocessed ECG data according to the position of the signal above or below a given threshold. The threshold value is set to 20% of the maximum value within each one-second segment S and recalculated every second. Subsequent data analysis takes place over successive one-second stages. The ECG signal may cross the detection threshold one or more times, and the number of pulses is counted. For each stage, the threshold crossing interval TCI is the average interval between threshold crossings and is calculated as follows\n\nTCI =\n\n1000 (N − 1) +\n\nt2 t1 +t 2\n\n+\n\nt3 t3 +t4\n\n[ms].\n\nFigure 2 illustrates the situation.\n\n(1)\n\nACF95 STE MEA WVL1\n\n80 Sensitivity / per cent\n\nwl\n\nLI TOMP\n\n60\n\n40\n\n20\n\n0\n\n0\n\n20\n\n40\n\n60\n\n80\n\n100\n\n(1−Specificity) / per cent\n\nROC Figure TOMP curve 5 for the algorithms ACF, STE, MEA, WVL1, LI, ROC curve for the algorithms ACF, STE, MEA, WVL1, LI, TOMP.\n\n100 TCI 8s VF 8s SPEC CPLX SCA\n\n80 Sensitivity / per cent\n\nParameter\n\n100\n\n60\n\n40\n\n20\n\n0\n\n0\n\n20\n\n40\n\n60\n\n80\n\n100\n\n(1−Specificity) / per cent\n\nFigure ROC curve 6 for the algorithms TCI, VF, SPEC, CPLX, SCA ROC curve for the algorithms TCI, VF, SPEC, CPLX, SCA.\n\nHere, N is the number of pulses in S. t1 is the time interval from the beginning of S back to the falling edge of the preceding pulse. t2 is the time interval from the beginning of S to the start of the next pulse. t3 is the interval between the end of the last pulse and the end of S and t4 is the time from the end of S to the start of the next pulse. If TCI ≥ TCI0 = 400 ms, SR is diagnosed. Otherwise sequential hypothesis testing is used to separate ventricular tachycardia (VT) from VF. As stated above, the original algorithm works with single one-second time segments, (see , page 841). To achieve this the algorithm picks a 3-second episode. The first second and the third second are used to determine t1 and t4. The 2nd second yields the value for TCI. When picking an\n\nPage 4 of 15 (page number not for citation purposes)\n\nBioMedical Engineering OnLine 2005, 4:60\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\n@ I @\n\n\u0012 @ @\n\nN=2\n\n\u001b t1 -\u001b t2 -\n\n\u001b\n\nt3\n\n\u001b\n\nS (1 second interval)\n\nt4-\u001b -\n\nBinary Figuresignal 2 with 2 pulses in threshold crossing intervals algorithm Binary signal with 2 pulses in threshold crossing intervals algorithm.\n\n8-second episode we can hence evaluate 6 consecutive TCI values. Final SR decision is taken if diagnosed in four or more segments otherwise the signal is classified as VF. The critical threshold parameter to obtain the ROC is TCI0. ACF algorithm The autocorrelation algorithms ACF95 (Note 3: Probability of 95% in the Fisher distribution → α = 0.05 in F(α, k1, k2) with k1 = 1, k2 = 5) and ACF99 (Note 4: Probability of 99% in the Fisher distribution → α = 0.01 in F(α, k1, k2) with k1 = 1, k2 = 5) analyze the periodicities within the ECG. Given a discrete signal x(m), the short-term autocorrelation function (ACF) of x(m) with a rectangular window is calculated by\n\nR(k) =\n\nN −1−k\n\nm =0\n\nx(m)x(m + k), 0 ≤ k ≤ N − 1.\n\n(2)\n\nHere, this technique is used to separate VT and SR from VF. It is assumed that VF signals are more or less aperiodic and SR signals are approximately periodic. This assumption is however questionable since VF signals may have a cosine like shape. Compare this assumption with the assumption made by the algorithm in the next subsection.\n\nNote that the autocorrelation function of a function f is connected to the Power Spectrum of \| f \|2 by the \"WienerKhinchin Theorem\". The detection algorithm performs a linear regression analysis of ACF peaks. An order number i is given to each peak according to its amplitude. So, the highest peak is called P0, etc., ranged by decreasing amplitudes. In a SR signal, which is considered to be periodic or nearly periodic, a linear relationship should exist between the peaks lag and their index number i. No such relationship should exist in VF signals. The linear regression equation of the peak order and its corresponding lag of m peaks in the ACF is described as yi = a + bxi,\n\n(3)\n\nwhere xi is the peak number (from 0 to (m - 1)), and yi is the lag of Pi.\n\nx=\n\n1 m 1 m xi , y = ∑ i =1 yi , ∑ i = 1 m m\n\nb = ∑ i =1( xi − x )yi m\n\n(∑\n\nm (x i =1 i\n\n− x )2\n\n)\n\n−1\n\n.\n\n(4)\n\na = y − bx , In this study, m = 7. The variance ratio VR is defined by\n\nPage 5 of 15 (page number not for citation purposes)\n\nBioMedical Engineering OnLine 2005, 4:60\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\nThe critical threshold parameter to obtain the ROC is the leakage l0.\n\nm\n\nVR =\n\nb∑ ( xi − x )yi i =1\n\nR /(m − 2)\n\n,\n\n( 5)\n\nSpectral algorithm The spectral algorithm (SPEC) works in the frequency domain and analyses the energy content in different frequency bands by means of Fourier analysis.\n\nwhere m\n\nR = ∑ (yi − y − b( xi − x ))2 . i =1\n\nIf VR ≥ VR0 is greater than the Fisher statistics for degrees of freedom k1 = 1 and k2 = m - 2 = 5 with 95% (99%) probability, the rhythm is classified by ACF95 (ACF99) to be SR, otherwise it is VF. The critical threshold parameter to obtain the ROC is VR0, (VR0 ≈ 6.61(16.3) at 95% (99%)). VF filter algorithm The VF filter algorithm (VF) applies a narrow band elimination filter in the region of the mean frequency of the considered ECG signal.\n\nAfter preprocessing, a narrow band-stop filter is applied to the signal, with central frequency being equivalent to the mean signal frequency fm. Its calculated output is the \"VF filter leakage\". The VF signal is considered to be approximately of sinusoidal waveform.\n\nThe ECG of most normal heart rhythms is a broadband signal with major harmonics up to about 25 Hz. During VF, the ECG becomes concentrated in a band of frequencies between 3 and 10 Hz (cf. [11,12], with particularly low frequencies of undercooled victims). After preprocessing, each data segment is multiplied by a Hamming window and then the ECG signal is transformed into the frequency domain by fast Fourier transform (FFT). The amplitude is approximated in accordance with ref. by the sum of the absolute value of the real and imaginary parts of the complex coefficients. (Note 5: Normally one would take the modulus of complex amplitudes.) Let Ω be the frequency of the component with the largest amplitude (called the peak frequency) in the range 0.5 – 9 Hz. Then amplitudes whose value is less than 5% of the amplitude of Ω are set to zero. Four spectrum parameters are calculated, the normalized first spectral moment M\n\nThe number N of data points in an average half period N = T/2 = 1/(2fm) is given by −1    m  m  1  N = π  ∑\| Vi \|   ∑\| Vi − Vi −1 \|  +  ,     2   i =1     i =1\n\n(6)\n\nwhere Vi are the signal samples, m is the number of data points in one mean period, and Q...N denotes the floor function. The narrow band-stop filter is simulated by combining the ECG data with a copy of the data shifted by a half period. The VF-filter leakage l is computed as  m l =  ∑\| Vi + Vi − N   i =1\n\n  m \|   ∑ (\| Vi \| + \| Vi − N \|)      i =1 \n\n−1\n\n.\n\n(7)\n\nIn the original paper this algorithm is invoked only if no QRS complexes or beats are detected. This is done by other methods. Since we employ no prior QRS detection, we use the thresholds suggested by . If the signal is higher than a third of the amplitude of the last by the VFfilter detected QRS complex in a previous segment and the leakage is smaller than l0 = 0.406, VF is identified. Otherwise the leakage must be smaller than l0 = 0.625 in order to be classified as VF.\n\njmax\n\n1 M= Ω\n\naj w j\n\nj =1 jmax\n\nj =1\n\n,\n\n(8)\n\naj\n\njmax being the index of the highest investigated frequency, and A1, A2, A3. Here wj denotes the j-th frequency in the FFT between 0 Hz and the minimum of (20Ω, 100 Hz) and aj is the corresponding amplitude. A1 is the sum of amplitudes between 0.5 Hz and Ω/2, divided by the sum of amplitudes between 0.5 Hz and the minimum of (20Ω, 100 Hz). A2 is the sum of amplitudes between 0.7Ω and 1.4Ω divided by the sum of amplitudes between 0.5 Hz and the minimum of (20Ω, 100 Hz). A3 is the sum of amplitudes in 0.6 Hz bands around the second to eighth harmonics (2Ω – 8Ω), divided by the sum of amplitudes in the range of 0.5 Hz to the minimum of (20Ω, 100 Hz). VF is detected if M ≤ M0 = 1.55, A1 80% our current investigations indicate a promising good performance for new algorithms based on other methods coming from chaos theory which are currently under development. When finished these algorithms will be presented elsewhere.\n\nTable 2 shows the values for the positive predictivity, the accuracy and the calculation time of the investigated algorithms. Table 3 shows the values for the sensitivity of the investigated algorithms, if, due to an appropriate adaption of the threshold parameters, the specificity were 95% and 99%, respectively. Figure 5 and 6 show the ROC curves for all algorithms. For the computation of the ROC curves, we used 64 nodes. Since some critical threshold parameters are discrete the points of the ROC curve are not equidistant.\n\n3 Discussion and Conclusion In real applications of AEDs the specificity is more important than the sensitivity, since no patient should be defibrillated due to an analysis error which might cause cardiac arrest. Therefore, a low number of false positive decisions should be achieved, even if this process makes the number of false negative decisions higher. But one has to distinguish between our calculated values for specificity and sensitivity and the values in . Our values were determined for the basic mathematical algorithms, whereas this paper gives recommendations for whole ECG analysis systems. It also does not consider an analysis without preselection. Our results show that no algorithm achieves its proclaimed values for the sensitivity or specificity as described in the original papers or in and when applied to an arbitrary ECG episode. The main reason for this is the following: Whereas all other researchers made a preselection of signals, we simulated the situation of a bystander, who is supposed to use an AED, more accurately. Hence no preselection of ECG episodes were made. The best algorithm SCA, which yields the best value for the integrated receiver operating characteristic (IROC) is a new algorithm followed by the algorithms SPEC and VF. Studying the ROC curves in Figure 5 and Figure 6 we see that the relevant part of the ROC curves lies at the left side. The ROC curve also enables us to compare different algorithms given a specified specificity.\n\nAppendix A: Sensitivity, Specificity, Positive Predictivity, Accuracy, and ROC Sensitivity is the ability (probability) to detect ventricular fibrillation. It is given by the quotient\n\ndetected cases of VF TP = , all cases of VF TP + FN\n\n( 24 )\n\nwith TP being the number of true positive decisions, FN the number of false negative decisions. Specificity is the probability to identify \"no VF\" correctly. It is given by the quotient\n\ndetected cases of \"no VF\" TN = , all cases of \"no VF\" TN + FP\n\n( 25 )\n\nwhere TN is the number of true negative decisions, and FP is the number of false positive decisions. This means that if a defibrillator has a sensitivity of 90% and a specificity of 99%, it is able 90% of the time to detect a rhythm that should be defibrillated, and 99% of the time to recommend not shocking when defibrillation is not indicated. Remark: a trivial algorithm which classifies every ECG episode as \"no VF\" will reach a specificity of 100%, but will have sensitivity 0%. On the other hand, a trivial algorithm which classifies every ECG episode as VF will reach a sensitivity of 100%, but will have specificity 0%. The ROC curve (see below) describes this inherent tradeoff between sensitivity and specificity. Furthermore, we calculated the Positive Predictivity and the Accuracy of the investigated algorithms. Positive predictivity is defined by\n\nAll other algorithms yielded only mixed results in our simulations. We also conclude that algorithms developed for QRS detection, like LI and TOMP, are not suitable for VF detection even when the thresholds are suitably adapted.\n\ndetected cases of \"VF\" TP = . m as \"VF\" TP + FP all cases classified by the algorithm\n\n( 26 )\n\nPositive predictivity is the probability, that classified VF is truly VF:\n\nPage 12 of 15 (page number not for citation purposes)\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\n4\n\n4\n\n3\n\n3\n\n2\n\n2\n\nvoltage / a.u.\n\nvoltage / a.u.\n\nBioMedical Engineering OnLine 2005, 4:60\n\n1\n\n1\n\n0\n\n0\n\n−1\n\n−1\n\n−2 0\n\n2\n\n4\n\n6\n\n8\n\ntime / s\n\n2\n\n4\n\n6\n\n8\n\ntime / s\n\nFigure ECG applying signal 7step with 1 ofrelative Signal maxima Comparison (indicated Algorithm by stars (SCA) left) after ECG signal with relative maxima (indicated by stars left) after applying step 1 of Signal Comparison Algorithm (SCA). This ECG episode is annotated as no VF, a.u. ... arbitrary units.\n\nAccuracy is defined by\n\nall true decisions TP + TN = . all decisions TP + FP + TN + FN\n\n−2 0\n\n( 27 )\n\nAccuracy is the probability to obtain a correct decision. Specificity and sensitivity always depend on the chosen critical threshold parameters which depend, on the other hand, on the databases used for evaluation (see Note 7). To get rid of at least of the dependence on the chosen critical threshold parameter one uses the ROC curve. The sensitivity is plotted in dependence of (1 - specificity), where different points in the plot are obtained by varying the critical threshold parameter in the decision stage of an algorithm. The ROC curves enables us to compare different algorithms when choosing a specified specificity. For more information on ROC curves see [20,21].\n\nAppendix B: Algorithm details: Signal Comparison Algorithm (SCA) Here we describe the search for relative maxima and the appropriate choice among them, used in Section 1 in more detail.\n\nFigure ECG applying signal 8step with 2 ofrelative Signal maxima Comparison left (indicated Algorithmby(SCA) stars) after ECG signal with relative maxima left (indicated by stars) after applying step 2 of Signal Comparison Algorithm (SCA).\n\nAn offset is added to the ECG signal to make its mean value to zero. We construct a set Z containing the values aj and temporal positions tj of this new signal, i.e., Z = {(tj, aj)\|aj is the value of the ECG signal at time tj}. All further steps are executed both with the set Z and the set -Z, where -Z = {(tj, bj)\|bj = -aj is the value of the negative ECG signal at time tj} with the help of the reference signals rECGᐍ, ᐍ being VF, SR1, SR2 or SR3, or, equivalently, ᐍ = 0,1, 2, 3. Note, that the maxima of Z correspond to the minima of -Z. So we get 2 * 4 = 8 tests to find out whether a signal is VF or SR. If any of the 8 tests yields SR, the signal is considered to be SR. Step 1 All relative maxima aj of Z and their corresponding times tj are determined. The resulting set is called M0, i.e., M0 = {(tj, aj)\|aj is a local maximum}, so M0 ⊂ Z. All aj in M0, that are smaller than A, where A = ∆·max(aj) and ∆ is a threshold, are deleted. The threshold ∆ is set to 0.1 for the VF reference signal and to 0.2 for the SR reference signals. We call the reduced set M1. In Figure 7 we see an ECG episode from the CU database (cu21, from t = 148 s until t = 156 s) together with its selected relative maxima according to the status after processing step 1.\n\nNow, we introduce an index l and set it to l = 1.\n\nPage 13 of 15 (page number not for citation purposes)\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\n4\n\n4\n\n3\n\n3\n\n2\n\n2\n\nvoltage / a.u.\n\nvoltage / a.u.\n\nBioMedical Engineering OnLine 2005, 4:60\n\n1\n\n1\n\n0\n\n0\n\n−1\n\n−1\n\n−2 0\n\n2\n\n4\n\n6\n\n8\n\n−2 0\n\n2\n\n4\n\ntime / s\n\nFigure ECG applying signal 9step with 4 ofrelative Signal Comparison maxima (indicated Algorithm by stars) (SCA) after ECG signal with relative maxima (indicated by stars) after applying step 4 of Signal Comparison Algorithm (SCA).\n\n8\n\nFigure ECG applying signal 11 step with 6 ofrelative Signal maxima Comparison and SR Algorithm reference(SCA) signal after ECG signal with relative maxima and SR reference signal after applying step 6 of Signal Comparison Algorithm (SCA).\n\ninterval are equal or smaller than amax and larger than 0.2amax. All pairs (aj, tj) except (amax, tmax) in Ml, that are referred to the found interval Il, are deleted. We get a set that we call Ml+1. This procedure is repeated with all untreated aj in Ml, until every aj has been considered and afterwards either been deleted or kept. After each step, l is increased by 1. This means, first we consider M1, then M2 = M1\\I1, then M3 = M1\\{I1 ∪ I2} and so on, until we reach a highest l, called lmax. In the end, we get a set that we call\n\n4 3 2 voltage / a.u.\n\n6\n\ntime / s\n\n1 0 −1\n\nl\n\nM, with M = M1 \\{∪ jmax =1\n\n−1\n\nI j} .\n\n−2 −3 −4 0\n\n2\n\n4\n\n6\n\n8\n\ntime / s\n\nFigure ECG applying signal 10 step with 6 ofrelative Signal maxima Comparison and VF Algorithm reference(SCA) signal after ECG signal with relative maxima and VF reference signal after applying step 6 of Signal Comparison Algorithm (SCA).\n\nIn the end, the aj in M are the relative maxima in Z, that are higher than A and are the only ones in certain subintervals of Z. Two different aj in M can only be neighbors in Z, if they are separated by a valley that is deeper than 20% of the higher peak of the two. In Figure 8 we again see the ECG episode, together with its newly selected relative maxima according to the status after processing step 2. Step 3 A value Ω is calculated from M. The frequency Ω of \"peaks\" is given by\n\nStep 2 Ml is reduced further: The maximum aj in Ml is searched. Here, we call it amax. amax has a corresponding temporal position tmax. Then, the largest possible temporal interval Il in Z around tmax is searched, so that all values aj in this\n\nΩ=\n\n60N M t max − t min\n\n[min−1],\n\n( 28 )\n\nwhere NM is the number of points in M and tmax - tmin is the maximum temporal range of the elements in M.\n\nPage 14 of 15 (page number not for citation purposes)\n\nBioMedical Engineering OnLine 2005, 4:60\n\nhttp://www.biomedical-engineering-online.com/content/4/1/60\n\nStep 4 Now, if two different elements (ai, ti) and (aj, tj) of M are\n\n12.\n\n24 , Ω the element with the smaller a is deleted from M. This final set is called X. In Figure 9 we again see the ECG episode as in the figures above and together with its newly selected relative maxima according to the status after processing step 4.\n\n13.\n\nseparated by a temporal distance \|ti - tj\| smaller than\n\nStep 5 Ω is recalculated by Equation (28) with the help of the recalculated set X. If Ω > 280, r is set to 2, if Ω < 180, r is set to 0.9, else r is set to 1. Step 6 The decision is calculated by Equation (16). VRF is calculated for the ventricular fibrillation reference signal, VRS for the sinus rhythm reference signal.\n\n14. 15. 16. 17. 18. 19.\n\n20. 21.\n\nMurray A, Campbell R, Julian D: Characteristics of the ventricular fibrillation waveform. In Proc. Computers in Cardiology Washington, DC: IEEE Computer Society Press; 1985:275-278. Zhang X, Zhu Y, Thakor N, Wang Z: Detecting ventricular tachycardia and fibrillation by complexity measure. IEEE Trans Biomed Eng 1999, 46(5):548-55. Lempel A, Ziv J: On the complexity of finite sequences. IEEE Trans Inform Theory 1976, 22:75-81. Louis A, Maaß P, Rieder A: Wavelets Stuttgart: Teubner; 1998. Li C, Zheng C, Tai C: Detection of ECG characteristic points using wavelet transforms. IEEE Trans Biomed Eng 1995, 42:21-8. Hamilton P, Tompkins W: Quantitative investigation of QRS detection rules using the MIT/BIH arrhythmia database. IEEE Trans Biomed Eng 1986, 33(12):1157-65. Pan J, Tompkins W: A real-time QRS detection algorithm. IEEE Trans Biomed Eng 1985, 32(3):230-6. Kerber R, Becker L, Bourland J, Cummins R, Hallstrom A, Michos M, Nichol G, Ornato J, Thies W, White R, Zuckerman B: Automatic External Deflbrillators for Public Access Defibrillation: Recommendations for Specifying and Reporting Arrhythmia Analysis Algorithm Performance, Incorporating New Waveforms, and Enhancing Safety. Circulation 1997, 95:1677-1682. Swets J, Pickett R: Evaluation of diagnostic systems: Methods from signal detection theory New York: Academic Press; 1992. The magnificent ROC [http://www.anaesthetist.com/mnm/stats/ roc/]\n\nIn Figure 10 we see the ECG episode together with the corresponding VF reference signal. In Figure 11 we see the ECG episode together with the first corresponding SR reference signal. c1 is set to 2/r, c2 is set to 1.\n\nAcknowledgements We are indebted to the referees for numerous helpful suggestions. A. A. gratefully appreciates continuous support by the Bernhard-Lang Research Association.\n\nReferences 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.\n\n11.\n\nZheng Z, Croft J, Giles W, Mensah G: Sudden cardiac death in the United States, 1989 to 1998. Circulation 2001, 104(18):2158-63. American Heart Association, AHA database [http:// www.americanheart.org] Massachusetts Institute of Technology, MIT-BIH arrhythmia database [http://www.physionet.org/physiobank/database/mitdb] Massachusetts Institute of Technology, CU database [http:// www.physionet.org/physiobank/database/cudb] Clayton R, Murray A, Campbell R: Comparison of four techniques for recognition of ventricular fibrillation from the surface ECG. Med Biol Eng Comput 1993, 31(2):111-7. Jekova I: Comparison of five algorithms for the detection of ventricular fibrillation from the surface ECG. Physiol Meas 2000, 21(4):429-39. Thakor N, Zhu Y, Pan K: Ventricular tachycardia and fibrillation detection by a sequential hypothesis testing algorithm. IEEE Trans Biomed Eng 1990, 37(9):837-43. Chen S, Thakor N, Mower M: Ventricular fibrillation detection by a regression test on the autocorrelation function. Med Biol Eng Comput 1987, 25(3):241-9. Kuo S, Dillman R: Computer Detection of Ventricular Fibrillation. Computers in Cardiology, IEEE Computer Society 1978:347-349. Barro S, Ruiz R, Cabello D, Mira J: Algorithmic sequential decision-making in the frequency domain for life threatening ventricular arrhythmias and imitative artefacts: a diagnostic system. J Biomed Eng 1989, 11(4):320-8. Clayton R, Murray A, Campbell R: Changes in the surface ECG frequency spectrum during the onset of ventricular fibrillation. In Proc. Computers in Cardiology 1990 Los Alamitos, CA: IEEE Computer Society Press; 1991:515-518.\n\nPublish with Bio Med Central and every scientist can read your work free of charge \"BioMed Central will be the most significant development for disseminating the results of biomedical researc h in our lifetime.\" Sir Paul Nurse, Cancer Research UK\n\nYour research papers will be: available free of charge to the entire biomedical community peer reviewed and published immediately upon acceptance cited in PubMed and archived on PubMed Central yours — you keep the copyright\n\nBioMedcentral" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8561295,"math_prob":0.9575144,"size":34839,"snap":"2022-27-2022-33","text_gpt3_token_len":9697,"char_repetition_ratio":0.13233817,"word_repetition_ratio":0.10963956,"special_character_ratio":0.27954304,"punctuation_ratio":0.14677289,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.96734834,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T17:34:51Z\",\"WARC-Record-ID\":\"<urn:uuid:cd8f7712-35fe-4437-9e14-a701d345f7a8>\",\"Content-Length\":\"94720\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfd4b5e0-9405-4d13-8bed-24d609063d5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c3a81e53-5937-4679-a313-22fa617d2106>\",\"WARC-IP-Address\":\"104.21.83.11\",\"WARC-Target-URI\":\"https://moam.info/biomedical-engineering-online-biomedsearch_5c5cae76097c478d4f8b4594.html\",\"WARC-Payload-Digest\":\"sha1:IHMWHBOME7ZUIEHWWG7X66HZLITXYXW6\",\"WARC-Block-Digest\":\"sha1:GT6TAH6FM6ARBVHTFAYNEVGF4S2XKDOW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571483.70_warc_CC-MAIN-20220811164257-20220811194257-00472.warc.gz\"}"}
https://www.victorvaquero.me/publication/vvaquero2017ecmr/	[ "", null, "# Deconvolutional Networks for Point-Cloud Vehicle Detection and Tracking in Driving Scenarios\n\n### Abstract\n\nVehicle detection and tracking is a core ingredient for developing autonomous driving applications in urban scenarios. Recent image-based Deep Learning (DL) techniques are obtaining breakthrough results in these perceptive tasks. However, DL research has not yet advanced much towards processing 3D point clouds from lidar range-finders. These sensors are very common in autonomous vehicles since, despite not providing as semantically rich information as images, their performance is more robust under harsh weather conditions than vision sensors. In this paper we present a full vehicle detection and tracking system that works with 3D lidar information only. Our detection step uses a Convolutional Neural Network (CNN) that receives as input a featured representation of the 3D information provided by a Velodyne HDL-64 sensor and returns a per-point classification of whether it belongs to a vehicle or not. The segmented point cloud is then geometrically processed to generate observations for a multi-object tracking system implemented via a number of Multi-Hypothesis Extended Kalman Filters (MH-EKF) that estimate the position and velocity of the surrounding vehicles. The system is thoroughly evaluated on the KITTI tracking dataset, and we show the performance boost provided by our CNN-based detector over the standard geometric approach. Our lidar-based approach uses about 4% a fraction of the data needed for an image-based detector with similarly competitive results\n\nType\nPublication\nEuropean Conference on Mobile Robotics\nDate", null, "## Paper Summary", null, "We introduce a novel CNN-based vehicle detector on 3D range data. The proposed model is fed with an encoded representation of the point cloud and computes for each 3D point its probability of belonging to a vehicle. The classified points are then clustered generating trustworthy observations that are fed to our MH-EKF based tracker. Note: Bottom left RGB image is shown here only for visualization purposes.\n\n### Lidar Data Representation", null, "To obtain a useful input for our CNN-based vehicle detector, we project the 3D point cloud raw data to a featured image-like representation containing ranges and reflectivity information by means of $G(\\cdot)$. In this way, we first project the 3D Cartesian point cloud to spherical coordinates $sph(p_i)~=~{\\phi_i,\\theta_i, \\rho_i}$. According to the Velodyne HDL-64 specifications, elevation angles $\\theta$ are represented as a $H \\in \\mathbb{R}^{64}$ vector with a resolution $\\Delta\\theta$ of 13 degrees for the upper laser rays and 12 degrees the lower half respectively. Moreover, $G(\\cdot)$ needs to restrict the azimuth field of view, $\\phi \\in [-40.5,40.5]$ to avoid the presence of unlabelled vehicle points, as the Kitti tracking benchmark has labels only for the front camera viewed elements. The azimuth resolution was set to a value of $\\Delta\\phi=0.18$ degrees according to the manufacturer, and hence lying in $W \\in \\mathbb{R}^{451}$. Each $H,W$ pair encodes the range ($\\rho$) and reflectivity of each projected point, so finally our input data representation lies in an image-like space $G(\\mathcal{P}) \\in \\mathbb{R}^{64 \\times 451 \\times 2}$.\n\nGround-truth for learning the proposed classification task is obtained by first projecting the image-based Kitti tracklets over the 3D Velodyne information, and then applying again $G(\\cdot)$ over the selected points.\n\n### Network Used", null, "Our network encompasses only convolutional and deconvolutional blocks followed by Batch Normalization (BN) and ReLu (RL) non-linearities. The first three blocks conduct the feature extraction step controlling, according to our vehicle detection objective, the size of the receptive fields and the feature maps generated. The next three deconvolutional blocks expanse the information enabling the point-wise classification. After each deconvolution, feature maps from the lower part of the network are concatenated (CAT) before applying the normalization and non-linearities, providing richer information and better performance.\n\nDuring training, three losses are calculated at different network points, as shown in the bottom part of the graph:\n\n$\\mathcal{L}(\\hat{\\mathcal{Y}}, \\mathcal{Y})$ = $\\sum_{r=1}^{3}$ $\\lambda_r \\mathcal{L}_r (\\hat{\\mathcal{Y}_r}, {\\mathcal{Y}_r})$ ,\n\nwhere $r$ represents the intermediate loss-control positions, $\\lambda_{r}$ are regularization weights for the loss at each resolution, and $\\hat{\\mathcal{Y}}, \\mathcal{Y}$ are respectively the predictions and ground-truth classes at those resolutions.\n\nIn our approach, $\\mathcal{L}_r$ is a multi-class Weighted Cross Entropy loss (WCE) defined as:\n\n$\\mathcal{L}_r^{WCE}$ = $-\\sum_{i,j,k}^{H_r,W_r,K_r}$ $\\omega(\\mathcal{Y}_{i,j})$ $\\textit{Id}_{[\\mathcal{Y}_{i,j}]}$ $\\text{log}(\\hat{\\mathcal{Y}}_{i,j,k})$ ,\n\nwhere $\\textit{Id}_{[x’]}(x)$ is an index function that selects the probability associated to the expected ground truth class and $\\omega(k)$ is the previously mentioned class-imbalance regularization weight computed from the training set statistics.\n\n### Results", null, "", null, "Acknowledgements: This work has been supported by the Spanish Ministry of Economy and Competitiveness projects ROBINSTRUCT (TIN2014-58178-R) and COL- ROBTRANSP (DPI2016-78957-R), by the Spanish Ministry of Education FPU grant (FPU15/04446), the Spanish State Research Agency through the Marı́a de Maeztu Seal of Excellence to IRI (MDM-2016-0656) and by the EU H2020 project LOGIMATIC (H2020-Galileo-2015-1-687534). The authors also thank Nvidia for hardware donation under the GPU grant program.", null, "" ]	[ null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/featured_hu21f3d1329605c8ee2cdbdc2bc3ddf02f_551899_1600x400_fill_q90_box_smart1_2.png", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ECMR_17_DeepLidar.jpg", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ecmr_17_fig1.png", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ecmr_17_fig2.png", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ecmr_17_fig3.png", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ecmr_17_finalFigure.jpg", null, "https://www.victorvaquero.me/publication/vvaquero2017ecmr/ecmr_17_results.png", null, "https://www.victorvaquero.me/img/vvaquero_truck.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8629175,"math_prob":0.96386653,"size":5409,"snap":"2023-14-2023-23","text_gpt3_token_len":1251,"char_repetition_ratio":0.111933395,"word_repetition_ratio":0.0026954177,"special_character_ratio":0.22665927,"punctuation_ratio":0.08038585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9899282,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T11:45:55Z\",\"WARC-Record-ID\":\"<urn:uuid:366baede-3a8a-4133-88a4-15a41185448d>\",\"Content-Length\":\"33337\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c21b2af-08b2-4f92-af00-15b94c2b4e2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0ba7a6a-3df0-44cf-8e42-f2aa0ee014c5>\",\"WARC-IP-Address\":\"34.73.83.172\",\"WARC-Target-URI\":\"https://www.victorvaquero.me/publication/vvaquero2017ecmr/\",\"WARC-Payload-Digest\":\"sha1:CXFVZRDJJWZ7LX4IKFIHDRYM6EV253WR\",\"WARC-Block-Digest\":\"sha1:ZTZAVOMA3APITZP6AOMBGCO6DQQHRLMZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943809.76_warc_CC-MAIN-20230322114226-20230322144226-00282.warc.gz\"}"}
https://wordpress.stackexchange.com/questions/342409/update-existing-post-times-to-random-times	[ "# Update existing post times to random times?\n\nI have found a code here that will randomize the date of all posts to a random date. Here is the code that someone has posted:\n\n``````<?php\n/*\n Plugin Name: WPSE 259750 Random Dates\n* Description: On activation, change the dates of all posts to random dates\n/\n\n// We want to do this only once, so add hook to plugin activation\nregister_activation_hook( __FILE__ , 'wpse_259750_activation' );\nfunction wpse_259750_activation() {\n\n//* Get all the posts\n\\$posts = get_posts( array( 'numberposts' => -1, 'post_status' => 'any' ) );\nforeach( \\$posts as \\$post ) {\n\n//* Generate a random date between January 1st, 2015 and now\n\\$random_date = mt_rand( strtotime( '1 January 2015' ), time() );\n\\$date_format = 'Y-m-d H:i:s';\n\n//* Format the date that WordPress likes\n\\$post_date = date( \\$date_format, \\$random_date );\n\n//* We only want to update the post date\n\\$update = array(\n'ID' => \\$post->ID,\n'post_date' => \\$post_date,\n'post_date_gmt' => null,\n);\n\n//* Update the post\nwp_update_post( \\$update );\n}\n}\n``````\n\nHow would you do the same, but only randomize the time without randomizing the date of the post. So, the posts should keep the same date they currently have, but only randomize the time of the day they were posted.\n\nI tried changing the post_date to post_time and changing the random_date to time() only. Then of course changing the date_format to 'H:i:s' however it did nothing.\n\nThis is the part that is responsible for random date:\n\n``````//* Generate a random date between January 1st, 2015 and now\n\\$random_date = mt_rand( strtotime( '1 January 2015' ), time() );\n``````\n\nSo you need to change it. And if you want to randomize only the time part, then this is one way to achieve that:\n\n``````\\$random_date = strtotime( date( 'Y-m-d 00:00:00', strtotime(\\$post->post_date) ) ) + mt_rand(0, 246060);\n``````\n\nThis line will:\n\n• take the date of post\n• get only day part of it\n• concert it to time\n• add random number of seconds to it\n\nSo it will randomize the time of publication.\n\n• This code does exactly what I've asked. Thank you so much for your prompt answer. Jul 9, 2019 at 14:15\n\nhow can I set up a cronjob for this? My website has 10k post and I want to setup a cronjob to random date for 10k post each day?\n\n• This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. To get notified when this question gets new answers, you can follow this question. Once you have enough reputation, you can also add a bounty to draw more attention to this question. - From Review Aug 15 at 21:50" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8149933,"math_prob":0.798765,"size":1358,"snap":"2022-40-2023-06","text_gpt3_token_len":353,"char_repetition_ratio":0.16100442,"word_repetition_ratio":0.008888889,"special_character_ratio":0.31222385,"punctuation_ratio":0.14042553,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96961755,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T02:14:06Z\",\"WARC-Record-ID\":\"<urn:uuid:73451eed-1e18-4096-8c39-53c9f48702ca>\",\"Content-Length\":\"155020\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bfeb4f5-8b7e-4676-a9c1-59f91155bf1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f49ea685-93a7-452a-840d-7c0b23d72d72>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://wordpress.stackexchange.com/questions/342409/update-existing-post-times-to-random-times\",\"WARC-Payload-Digest\":\"sha1:EHVCJO77V3SQJEKSUE3U4TLZ6VD6NWYB\",\"WARC-Block-Digest\":\"sha1:6OC6LIZUUWOPUO3VPDYY4FUGM5GQI7RA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337531.3_warc_CC-MAIN-20221005011205-20221005041205-00739.warc.gz\"}"}
https://trading-market.org/what-is-pip-value/	[ "The meaning of pip value can vary between currencies, but as most major currency pairs are priced to four decimal places, a pip is usually equal to the fourth figure after the decimal point. Calculating the value of pips, especially for cross currency pairs that do not include the U.S. dollar as one of the currencies, can be time consuming and difficult. For that reason, there are pip value calculators that will do the job for you. With a pip value calculator, you simply input your account base currency and the calculator will let you know the value of a pip for the standard lot , mini lot and micro lot . This can save you a lot of time and since it can update in real time you will always easily know what your potential profit or loss is. In addition, since forex transactions are typically leveraged, the pip value of positions gets multiplied by the amount of leverage used. By knowing the pip value of a currency pair, you can use money management techniques to calculate the ideal position size for any trade within the limits of the size of your account and your risk tolerance.", null, "The value of a pip is calculated by multiplying the amount of the trade in lots by one pip in decimal form, and then dividing it by the current exchange rate of the quote currency in your pair. However, we often find market forward points to be slightly different to the theoretical implied forward points. In this example, the current market tradeable forward point is 86 pips. The difference is called cross currency basis and it is driven by relative supply and demand between currencies. To do so they purchase USD in the spot market while selling USD in the forward market to return the USD to their native currency. This dynamic – excess demand to buy USD today but also to sell USD in the future – pushes USD forward rates down below levels implied from interest rate differentials.\n\n## Try A Demo Account\n\nThis one pip movement would equal a shift in value of \\$0.10 on a micro lot of 1,000 euros, \\$1 on a mini lot of 10,000 euros and \\$10 for a full lot of 100,000 euros. Those would be your pip values when trading in a U.S. dollar denominated account. If you trade in an account denominated in a specific currency, the pip value for currency pairs that do not contain your accounting currency are subject to an additional exchange rate. This is due to the fact that you need to convert pip value into your accounting currency to compare it with the pip value of your other positions. If we keep the EUR, in a year’s time with interest rate being -0.5%, we will have EUR 995,000. If we keep the money in USD, with interest rate 0.2%, it will become USD 1,202,400. Dividing USD 1,202,400 by EUR 995,000 gives us an implied forward rate of 1.2084.\n\nMost currency pairs are priced out to four decimal places and the pip change is the last decimal point. Most other currency pairs have the U.S. dollar as the base currency, such as USD/JPY and USD/CAD, for example, and they have different pip values. To calculate the pip value where the USD is the base currency when trading in a U.S. dollar-denominated account, you need to divide the position size by the exchange rate. Currencies must be exchanged to facilitate international trade and business.\n\n## Fractional Pip\n\nThe effect that a one-pip change has on the dollar amount, or pip value, depends on the number of euros purchased. If an investor buys 10,000 euros with U.S. dollars, the price paid will be US\\$12,908.22 ([1/0.7747] x 10,000). If the exchange rate for this pair experiences a one-pip increase, the price paid would be \\$12,906.56 ([1/0.7748] x 10,000). Forex pairs are used to disseminate exchange quotes through bid and ask quotes that are accurate to four decimal places.\n\nFor pairs in which the euro isn’t the quote currency, you would divide the usual pip value by the exchange rate between the euro and the quote currency. When trading in the foreign exchange market, it’s hard to underestimate the importance of pips. A pip, which stands for either “percentage in point” or “price interest point,” represents the basic movement a currency pair can make in the market.\n\n## What Is A Pip?\n\nThe currency you used to open your forex trading account will determine the pip value of many currency pairs. Those pip values would change only if the value of the U.S. dollar rose or fell significantly—by more than 10%. Most forex brokers allow a very high leverage ratio, or, to put it differently, have very low margin requirements.\n\n• Those pip values would change only if the value of the U.S. dollar rose or fell significantly—by more than 10%.\n• The pip value is defined by the currency pair being traded, the size of the trade and the exchange rate of the currency pair.\n• Then, multiply that figure by your lot size, which is the number of base units that you are trading.\n• To calculate pip value, divide one pip (usually 0.0001) by the current market value of the forex pair.\n\nFor most currency pairs—including, for example, the British pound/U.S. dollar (GBP/USD)—a pip is equal to 1/100 of a percentage point, or one basis point, and pips are counted in the fourth place after the decimal in price quotes. For currency pairs involving the Japanese yen, a pip is one percentage point, and pips are counted in the second place after the decimal in price quotes. Electronic trading platforms have brought greater price transparency and price competition to the foreign exchange markets. Several trading platforms have extended the quote precision for most of the major currency pairs by an additional decimal point; the rates are displayed in 1/10 pip. Pip is an acronym for “percentage in point” or” price interest point.” A pip is the smallest price move that an exchange rate can make based on forex market convention.\n\nThis is why profits and losses vary greatly in forex trading even though currency prices do not change all that much — certainly not like stocks. Stocks can double or triple in price, or fall to zero; currency never does.\n\nThe pip value is defined by the currency pair being traded, the size of the trade and the exchange rate of the currency pair. To calculate pip value, divide one pip (usually 0.0001) by the current market value of the forex pair. Then, multiply that figure by your lot size, which is the number of base units that you are trading. This means that the value of a pip will be different between currency pairs, due to the variations in exchange rates. However, when the quote currency is the US dollar, the value of a pip is always the same – if the lot size is 100,000, the pip will equal \\$10.\n\n## Example Of Pip Value\n\nThe forex market is where such transactions happen—along with bets made by speculators who hope to make money off of price moves in pairs of currencies. Pips are used in calculating the rates participants in the forex market pay when carrying out currency trades. Since most currency pairs are quoted to a maximum of four decimal places, the smallest change for these pairs is 1 pip. The value of a pip can be calculated by dividing 1/10,000 or 0.0001 by the exchange rate. In addition to determining costs, tracking the change of pips is important for traders in determining the potential profit, or loss, that might be made on a trade. As exchange rates vary throughout trading, traders can make or lose money depending on whether the bid and ask prices change enough, and in the right direction, to offset any costs imposed by the spread. Basically, the movement of a currency pair such as EUR/USD from 1.2000 to 1.2001 would represent a one pip rise in the exchange rate, so the pip size in EUR/USD is 0.0001.\n\nBecause currency prices do not vary substantially, much lower margin requirements are less risky than it would be for stocks. Note, however, that there is considerable risk in forex trading, so you may be subject to margin calls when currency exchange rates change rapidly. If your account is funded with a currency other than the U.S. dollar, the same pip value amounts apply when that currency is the quote currency. For example, for a euro-denominated account, the pip value will be 10 euros for a standard lot, 1 euro for a mini lot, and 0.10 euro for a micro lot when the euro is the second currency in the pair.\n\n## Performance Improvement Project (pip) Library\n\nThe implied forward points based on interest rate differential are 0.0084 . A pip refers to the last decimal place quoted for a currency pair. Most currency pairs are priced to 4 decimal places, so a pip will be the same as 0.01%, or a basis point, in those cases. The exact value of a pip depends on the currency pair and the quoting convention. For example, a pip in USD-JPY denotes the second decimal place, whereas a pip in EUR-CZK denotes the third decimal place. The pip value is the price attributed to a one-pip move in a forex trade – it is often used when referencing a position’s losses or gains.\n\nWithout this knowledge, you might wind up taking either too much or too little risk on a trade. If your account is funded with U.S. dollars and the dollar isn’t the quote currency, you would divide the usual pip value by the exchange rate between the dollar and the quote currency. For example, if the U.S. dollar/Canadian dollar (USD/CAD) exchange rate is 1.33119, the pip value for a standard lot is \\$7.51 (\\$10 / 1.3319)." ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20395%20296'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9229827,"math_prob":0.9736965,"size":11158,"snap":"2021-31-2021-39","text_gpt3_token_len":2491,"char_repetition_ratio":0.17016317,"word_repetition_ratio":0.09727822,"special_character_ratio":0.23032801,"punctuation_ratio":0.10732759,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96668017,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T17:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:1c3a78ba-f71d-4463-bac3-f1b1361f86c1>\",\"Content-Length\":\"46406\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7649068f-0630-4876-b5d6-56c98fe07a83>\",\"WARC-Concurrent-To\":\"<urn:uuid:dad8f261-d272-4a3e-94a9-0970dfb1ac1f>\",\"WARC-IP-Address\":\"104.21.7.157\",\"WARC-Target-URI\":\"https://trading-market.org/what-is-pip-value/\",\"WARC-Payload-Digest\":\"sha1:WGJUAXKAX75MUQECHSEQDM65MAIXPS3W\",\"WARC-Block-Digest\":\"sha1:LZJ3WCCBQIYJ2QXSRC72HAAGCAXNBZDT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.63_warc_CC-MAIN-20210801154943-20210801184943-00639.warc.gz\"}"}