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# Physics Kinetic Theory and Thermodynamcis ## Describe the Fundamental Postulates kinetic Theory ideal Gas Describe the Fundamental Postulates kinetic Theory ideal Gas Describe the Fundamental Postulates kinetic Theory ideal Gas:-The Laws of Thermodynamics: The Zeroth law, various indicator diagrams, work done by and on the system, first law of thermodynamics. internal energy as a state function and other applications. Reversible and irreversible changes. Carnot cycle and its efficiency, Carnot […] ## What is the Entropy Physics Notes What is the Entropy Physics Notes What is the Entropy Physics Notes:-Real Gas: vander Waal’s gas, equation of state, nature of vander Waal’s forces, comparison with experimental P-V curves. The critical constants, gas and vapour, Joule expansion of ideal gas and of a vander Waal’s gas, Joule coefficient, estimates of J-T cooling.   प्रश्न 50. ## BSc Physics Real Gas and Vapour Notes BSc Physics Real Gas and Vapour Notes BSc Physics Real Gas and Vapour Notes :-Real Gas: vander Waal’s gas, equation of state, nature of vander Waal’s forces, comparison with experimental P-V curves. The critical constants, gas and vapour, Joule expansion of ideal gas and of a vander Waal’s gas, Joule coefficient, estimates of J-T cooling. ## BSc Physics Equation of State Notes BSc Physics Equation of State Notes BSc Physics Equation of State Notes:-Real Gas: vander Waal’s gas, equation of state, nature of vander Waal’s forces, comparison with experimental P-V curves. The critical constants, gas and vapour, Joule expansion of ideal gas and of a vander Waal’s gas, Joule coefficient, estimates of J-T cooling.Ideal Gas: Kinetic model, ## BSc Physics Very Important Question Answer BSc Physics Very Important Question Answer BSc Physics Very Important Question Answer:-Real Gas: vander Waal’s gas, equation of state, nature of vander Waal’s forces, comparison with experimental P-V curves. The critical constants, gas and vapour, Joule expansion of ideal gas and of a vander Waal’s gas, Joule coefficient, estimates of J-T cooling.     प्रश्न ## BSC Physics Short Question Answer BSC Physics Short Question Answer BSC Physics Short Question Answer:-Ideal Gas: Kinetic model, deduction of Boyle’s law, interpretation of temperature, estimation of rms speeds of molecules. Brownian motion estimate of the Avogadro number. Equipartition of energy, specific heat of monoatomic gas, extension to di- and tri-atomic gases, behaviour at low temperatures. Adiabatic expansion of an ## BSc Kinetic Theory and Thermodynamics Physics Notes BSc Kinetic Theory and Thermodynamics Physics Notes BSc Kinetic Theory and Thermodynamics Physics Notes:-Ideal Gas: Kinetic model, deduction of Boyle’s law, interpretation of temperature, estimation of rms speeds of molecules. Brownian motion estimate of the Avogadro number. Equipartition of energy, specific heat of monoatomic gas, extension to di- and tri-atomic gases, behaviour at low temperatures. Scroll to Top
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{[ promptMessage ]} Bookmark it {[ promptMessage ]} rate_change_2_sol # rate_change_2_sol - so that M = O'd M 3 ~-=-k dr-J nX~... This preview shows page 1. Sign up to view the full content. Free calculus worksheets from www.analyzemath.com Calculus Worksheet: Rate of Chanae (11 A block of ice start melting at t = 0 where t is the time. The rate at which its mass M decreases is proportional to the cube root of the mass. If the initial (at t = 0) mass M is equal to 5 Kilograms and it is equal to 3 Kilograms at t = 1 hour, find t This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: so that M = O. 'd . M- 3 . ~-=-k dr-J nX~ ~ \\.-~J.('~ LJ _ y_ t-L,: ~ . d r~ S k ~t-d-i--v ~ '-/3 k\-+ c( C( 2. )-L. ) (~'\ V e~",- ) M (\;) "' J~ k r-+U1S')~'--~~ He,)-= ,I (~ k ( \) d1:1, f ~~ \ \Ie ~ k' k-" { C~~-\h( J . M (t-) " J~'F'-'R) h ;Jti'J'" 'V\ It-l:: cvvJ ~ lJe 1uv t to ob~-... View Full Document {[ snackBarMessage ]} Ask a homework question - tutors are online
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# PTX - loan question Registered Posts: 452 Can anyone help with this question please? On 1 November 11, Roweena had been given £15,000 loan on which she paid her employer 1% interest. On 1 July 2013 , Reweena repaid £3000 of the loan, but shemade no other repayment were made How do you work out the taxable benefit in kind arising from the loan please? • Moderator, MAAT, AAT Licensed Accountant Posts: 1,369 You may use 2 methods (assuming 4% HMRC rate) 1) The average method uses the opening balance of the loan plus the ending balance of the loan divided by 2 i.e. at 5th April 2013 15,000 at 5th April 2014 12,000 (3,000 repaid) average 13,500 @ 4% = 540 2) The strict method 15,000 x 3/12 (April to July) x 4% + 12,000 x 9/12 x 4% =150 + 360 = 510 The strict method is more beneficial. Then deduct the interests paid by the employee Contributions = (15,000 x 3/12 + 12,000 x 9/12) x 1% = 127.50 Benefit in kind arising from the loan : 510 - 127.50 = 382.50 Hope this helps • Registered Posts: 452 You may use 2 methods (assuming 4% HMRC rate) 1) The average method uses the opening balance of the loan plus the ending balance of the loan divided by 2 i.e. at 5th April 2013 15,000 at 5th April 2014 12,000 (3,000 repaid) average 13,500 @ 4% = 540 2) The strict method 15,000 x 3/12 (April to July) x 4% + 12,000 x 9/12 x 4% =150 + 360 = 510 The strict method is more beneficial. Then deduct the interests paid by the employee Contributions = (15,000 x 3/12 + 12,000 x 9/12) x 1% = 127.50 Benefit in kind arising from the loan : 510 - 127.50 = 382.50 Hope this helps Thank you for your help it really is appreciated, especially on a Saturday night! I did the average method as you had done but then wasnt sure where to go from the £540 as the AAT answer is £405? so by power of deduction the interest paid by the employee has too be £135 but i have no idea where this is coming from, any ideas ? • Registered Posts: 214 ? ? ? You may use 2 methods (assuming 4% HMRC rate) 1) The average method uses the opening balance of the loan plus the ending balance of the loan divided by 2 i.e. at 5th April 2013 15,000 at 5th April 2014 12,000 (3,000 repaid) average 13,500 @ 4% = 540 You need to multiply 13,500 (the average balance) and mulitply by 3% (because HMRC rate is 4% and she paid 1% of it so the benefited from the remaining 3%) 13500 * 3% = 405 Hope this helps • Registered Posts: 214 ? ? ? And to answer to your last post 1% * 13500 (the average balance) = 135 as you deducted • Registered Posts: 452 And to answer to your last post 1% * 13500 (the average balance) = 135 as you deducted thank you very much ! all of the puzzle has finally fit together now :001_smile:
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11 Replies Latest reply: Jul 15, 2012 8:17 AM by Emmanuel Tenk # Empty Cells in Report Currently Being Moderated Hi Experts, After executing the query, I am getting  some cells which are empty without any zero / any value.... Is there any setting to display those cells with zeros ........ Regards, Nagamani • ###### Re: Empty Cells in Report Currently Being Moderated Hi, Please check the infoprovider to see if there is anything displayed for that particular selection. • ###### Re: Empty Cells in Report Currently Being Moderated Hi! Those blank cells are not actually equal to 0.  They have a different value.  What you can do is hide the current key figure and replace it with a calculated key figure which is equal to your hidden KF + 0.  This will then display those blank cells as 0. Hope this helps! R, • ###### Re: Empty Cells in Report Currently Being Moderated Actually I am getting the Empty cells for the Calculated KF  it self, I have followed the given suggesition , but still getting the Empty cells Actually my CKF was , Basic KF           ZFORECAST   (Hide) Calculate KF     (NODIM(ZFORECAST)/ 1000)+0 I have used the above one, Could you give me the alternative Solution for this ... Regards, Nagamani • ###### Re: Empty Cells in Report Currently Being Moderated Hi Nagmani, After calculation if ur getting few of ur cells empty even the desired is 0 then in this case u have to create 2 more Calculated key figure and to write something like this in ur Figure: Suppose the blank ciolumn name is KF1 Then create a KF2 and type below formula: in KF2 write : KF1 + 1. Now create another KF3 in that use the below formula: (KF2 == 1) * 0 + KF1. KF3 wil be ur final column then, As the blank space is not zero so directly adding 0 to that will not serve to ur purpose u need to create two more KF as suggested above. Its a common problem faced as such there is no settings for this so need to do little +, - with Zero so that u can get the 0 Thanks Dipika • ###### Re: Empty Cells in Report Currently Being Moderated Hi Dipika, Could you clearly explain the below KF2 write : KF1 + 1. if we have added 1 to the KF the value will changed ....and that to I am getting error while adding value '0' to the Calculated KF .... Regards, Nagamani • ###### Re: Empty Cells in Report Currently Being Moderated Hi, KF2=(KF1+0) KF3=((KF2==0)*0(KF20)) Where KF1=Original KF KF2,KF3 -->Cal.KF Thanks, cheers, shana • ###### Re: Empty Cells in Report Currently Being Moderated Hi All, I got the required thing , by setting the Query property Suppression Active All values = 0 Effect pon Rows and Columns Regards, Nagamani • ###### Re: Empty Cells in Report Currently Being Moderated Tried several alternatives of this approach; it does not work for BW 3.x. This persistent problem got to do with the difference between null values (blanks/empty cells) and zeros set up internally in the SAP system within the Infoproviders. Interestingly, zeros are more expensive (they take more disk space in the database), bigger in size than blank/empty cells. The solution could be to upgrade to a newer version or use a macro to do the conversion in your report. • ###### Re: Empty Cells in Report Currently Being Moderated Hi! Kindly try Calculate KF     (NODIM(ZFORECAST+0)/ 1000) R, • ###### Re: Empty Cells in Report Currently Being Moderated I have tried the below one , but  getting error like .. Formula not defined correctly, I have checked the Basic KF Values in the Infoprovider level, there I have zero values ... and tried to create a Formula variable  with default value '0' and added to the Calculated KF , then also   I am getting the empty cells, Please let me know any possible solution Regards, Nagamani • ###### Re: Empty Cells in Report Currently Being Moderated Use NODIV( ) in place of NODIM( ) and see.
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# Profit and Loss (4) 1. If a material is sold for Rs.34.80, there is a loss of 25%. Find out the cost price of the material? 1. Rs.46.40 2. Rs.44 3. Rs.42 4. Rs.47.20 Explanation :- SP=(100-25)=75% of CP 34.50=(75/100)*CP CP=(34.50*100)/75 CP=46.40 2. A fruit seller sells apples at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%? 1. 11.32 2. 11 3. 12 4. 11.81 Explanation :- SP=9 Loss=20% CP=(9*100)/80=45/4 Profit=(45/4)*(105/100)=11.81 3. A man sells two houses at the rate of Rs.1.995 lakhs each. On one he gains 5% and on the other, he loses 5%. What is his gain or loss percent in the whole transaction? 1. 0.25%. 2. .3% 3. .4% 4. .5% Explanation :- If a person sells two items at the same price; one at a gain of x % and another at a loss of x %, then the seller always incurs a loss expressed as Loss %age=((Common profit or loss %age)/10)^2 Therefore Loss %age = (5/10)^2 = 0.25% OR CP of item on which there is 5% loss = (1.995/95)×100 = 2.1 CP of item on which there is 5% Profit = (1.995/105)×100 = 1.9 Total cost price of both items = 2.1 + 1.9 = 4                            And total SP of both items is 1.995 + 1.995 = 3.99 so there is loss of (4 – 3.99) = .01                             Profit % = (.01/4)×100 = .25% 4. John purchased a machine for Rs. 80,000. After spending Rs.5000 on repair and Rs.1000 on transport he sold it with 25% profit. What price did he sell the machine? 1. Rs.107000. 2. Rs.107500. 3. Rs.108500. 4. None of these Explanation :- CP=80000+5000+1000 = 86000 SP= (86000/100)*125=107500 5. By selling an item for Rs.15, a trader loses one sixteenth of what it costs him. The cost price of the item is 1. Rs.14 2. Rs.15 3. Rs.16 4. Rs.17
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# Dividend Yield and Common Equity Pages: 3 (724 words) Published: February 5, 2013 otivationFM-II Assignment -1 Spring 2013 Cost of Capital Q1: Percy Motors has a target capital structure of 40% debt and 60% equity. The yield to maturity on the company’s outstanding bonds is 9% and the company tax rate is 40%. Percy’s CFO has calculated the company’s WACC as 9.96%. What is the company’s cost of common equity? Q2: Tunney Industries issued preferred stock at a price of \$47.50 a share. The issue is expected to pay a constant annual dividend of \$3.80 a share. What is the company’s cost of preferred stock, Kp? Q3: Javit & Son’s common stock is currently trading at \$30 a share. The stock is expected to pay a dividend of \$3.00 a share at the end of the year (D1=\$3.00), and dividend is expected to grow at a constant rate of 5% a year. If the company were to issue external equity, it would incur a 10% floatation cost. What are the costs of internal and external equity? Q4: The Patrick company’s cost of common equity is 16%. Its before tax cost of debt is 13% and its marginal tax rate is 40%. The stock sells at book value. Using the following balance sheet, calculate Patrick’s after tax weighted average cost of capital: Q5: Hook industries has a capital structure that consists solely of debt and common equity. The company can issue debt at 11%. Its stock currently pays a \$2 dividend per share (D0 =\$2), and the stock’s price is currently \$24.75. The company’s dividend is expected to grow at a constant rate of 7% per year; its tax rate is 35%; and the company estimates that its WACC is 13.95%. What percentage of the company’s capital structure consists of debt financing? Q6: Assume that there is an increase in the risk free rat. What impact would this increase have on the cost of debt? What impact would it have on cost of equity? Q7: Sidman products’ stock is currently selling for \$60 a share. The firm is expected to earn\$5.40 per share this year and to pay a year-end dividend of \$3.60. a. If investors require a 9% return, what rate of...
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A project log for DeepPlankter - Autonomous Drone Boat An open-source wave propelled / wing propelled drone boat that can sail in the ocean for months and travel thousands of miles. Jianjia Ma 02/23/2022 at 22:140 Comments Navigation is a challenge for a small boat like DeepPlankter. • The ship might equip with a free-rotated wing. Headwind and tailwind will generate little-to-no force. The boat needs to offset to the wind direction. • The propulsion (from wing or wave) is small and unstable. It can even be pushed back by wind or sea current. The navigation should be able to drive the boat back in this situation. • The boat cruising speed is very low. ## Simulator The source code is available in the dedicated navi-sim repo To ensure the navigation algorithm works as expected, I actually build a 2-D simulator out of Processing 3. Processing 3 is basically a java graphical lib with some customized interfaces. This simulator is extremely helpful for me to develop a navigation algorithm. Up until today, it has the features below: • Ture earth coordination • Dynamic wind simulation [direction, wind guest, wind speed] • Dynamic sea current [direction, speed] • Drag • Boat physical model • Interactive waypoint • Time wrapping Although most of them are very simple, they still can cover the most extreme situations such as strong wind, guest wind and strong current. Different drone navigations algorithm have been widely implemented on many open source flight controllers. Actually, most of the opensource flight controllers use the so-called L1 navigation algorithm (original paper) or its variances. It is robust and has already been validated by a lot of experienced users. The most significant principle for L1 is it uses the term "the acceleration back to track" (a_s_cmd) as the key parameter to control the course correction. Also, in many L1 implementations, they assume that η is a very small number so sine can be replaced by linear functions. L1 give drones a very good track following method no matter the track is a line or a curve. It is also very simple to tune because there is only one parameter called L1, which define many things. Such as the track width, the damper for the stay on track acceleration, the length of line-of-sign to track, and others. Here is the L1 principle (figure copyright belong to the paper author). The question for me is whether I should implement L1 or develop a specific algorithm for the boat. There are some considerations: • Compared to air drones or rovers, the boat I built is aimed to low speed ( < 2knots) but also at very large scale waypoints (10-100km). The "stay in track" capability is not very helpful since the track can be very wide(100m or ~km). • The navigation should also implement the zigzagging algorithm when the boat is heading into the wind direction or away from the wind direction. Otherwise, the air wing give us little to no propulsion or even drag. • The boat will also likely be in extreme situations such as strong wind and current in a storm. It is not clear whether the L1 can handle that. (not sure my algorithm can either) • L1 use only one parameter to define many terms that are used in the algorithm. This is not very feasible for this boat navigation situation. To sum up, using L1 here will not be ideal, since we are not taking advantage of the following track or the simplified parameter setting. Instead, we still need to have finer tuning parameters such as track width and others. We are more interested in large scale direction, for example, the path width can be 100 metres or ~km when in the ocean. Compared to these distances, the boat speed or accelerations seem very small. It is unnecessary to wake up the actuators to do a small correction. We only care about whether the boat is heading toward the right direction in a long time scale (minutes or hours.) ## Track following However, track following is still needed, as the distance between the waypoints can be too long, simply heading to the target waypoint will have very limited resistance to interference such as wind and current. What we don't want is not try too hard to get back to the track, which consumes our limited battery power. For the track following, unlike L1, I use a P controller on the cross track distance to get the course correction back to track. This correction is also limited by a fixed threshold, called max_offset_angle, this is normally in the range of 45 deg to 70 deg. This is a very simple P controller, but the P setting will be very low so I am not too worried about oscillation. In the actuator output implementation, there are also heavy filtering and lazy responding implementation, so a small correction will not trigger the servo action (servo will be power down most of the time to save power). ## Out of track detection and secondary track Out of track detection calculate the track distance to the primary track. If the distance is larger than n=3 times of track width. The boat will no longer sail follows the track to the primary target. Instead, it creates a secondary target waypoint on the primary track. Then it starts to follow the secondary track to get back into the primary track. This seems unnecessary because the track following always drive the boat back to track. However, it is very useful for wing sailing which we will discuss in the next section. The insertion point (the secondary target) between the secondary track and the primary track is defined by the boat's current location and the max_off_angle, the angle is set to < 90deg. At least, it will not go backward. Simulation : To be noticed that we use heading instead of course for this calculation for simplicity. It is expected strong current will push the boat out of track quite often. But as I mention, we are not aiming for an extreme track following. Also, the parameters we use do not contain derivative components (speed or acceleration), but we assume they are very small values compared to the waypoints and distance which hopefully make the parameters tuning cover most of the case. So, we don't need to do filtering because derivatives can be very noisy. We know that sailboat cannot sail directly into the wind, there is an angle threshold where the wind propulsion drop dramatically, when the wind direction changes from side to the headwind. Sailors call this no-sail zone. Many researches have been done with wing sailing, such as Design of a free-rotating wing sail for an autonomous sailboat by CLAES TRETOW. Below is the polar diagram from the thesis. This graph represents the wind direction (relative) vs boat speed. It is symmetric for the left and right sides. Here we only discuss the wind from the right. We can see that the boat start to struggle when the wind direction is over 150 deg (tailwind) or less than 30 deg (headwind). I assume our boat has a similar diagram, so our navigation has to be able to void sailing into these "no-sail" zone. When the course to the target waypoint is close to the wind direction, we need to sail in a zigzag pattern to approach while avoiding sailing into the zone. I considered the below ways to implement the zigzag path. • Add temporary waypoints to form zigzag path + strict track following (such as L1). • Allow some free sailing zone within a large track. For the first method, adding temporary waypoints requires extra calculation and waypoint management. Also, it is not responsive when the wind direction changes during the temporary path. When generating a temporary waypoint, require stable wind direction measurement. This method requires many measurements which will increase the complexity. For the second method, as I mentioned in the above sections, our navigation method has a large track width. So as long as the boat is on the track, it is free for the zigzag method to decide where to head. The zigzag method can decide the heading freely according to the current wind direction and wing angle. This method decoupled the zigzag algorism with navigation algorism. which make it easier to design. The below diagram shows the second method. When the boat is within the track, it can sail freely in any direction, and of course, the zigzag algorism will navi the boat following the edge of the "no-sail" zone. When the boat finally reaches the boundary of the track, the track following algorism will override the zigzag output and turn the boat back into the track immediately. There is a path shrinking angle that will shrink the path near the waypoint. This angle is normally set to arctan(2). In this zigzag mode, the "no-sail" zone is dynamically updated, depending on the wind angle measurement (which is relatively reliable). It is very good because the "wind direction" that we use is the relative direction, which combines the boat motion and the real wind motion. So it can continually update the heading to archive good wing efficiency. ## Zigzag algorithm The zigzag algorithm is very simple. When inside the track, the zone is free for zigzag to decide where to head. We first use the current wing angle to see if it is in the no sail zone. Then compare it to the course to target waypoint. If we will turn into the no sail zone, then we cancel the angle out so the boat follows the edge of the zone instead. It will continue to sail this until it reaches the other side track border, where it will reconsider which zone edge to follow. ## Code implementation Please refer to the firmware repo or the simulator repo for the implementation. ## Simulation in an extreme situation Finally, this is the simulation in frequently changed wind direction. • added 20deg P-P wind direction noise (every cycle) See the top-right corner arrow which indicates the average wind direction (the actual change of wind direction is much quicker). See also the "wing effi" statistics on the right-hand list, which records the percentage of time that when the wing is outside of "no-sail" zone (means the wings is effectively providing propulsion). The route that boat is changing very frequently to allow the wing in high-efficiency angles. Sometimes it even sails the boat backward. Overall, this navigation method can drive the boat to the destination even in an extreme scenario. You can run the simulator yourself by downloading the code from the simulator repo. The environment can be changed through the source code.
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# Proton Velocity Calculator ## About Proton Velocity Calculator (Formula) The Proton Velocity Calculator is a specialized tool used in the realm of particle physics and nuclear research to estimate the velocity of protons, one of the fundamental particles that make up the atomic nucleus. Protons play a pivotal role in understanding the structure of matter and the behavior of particles within atomic nuclei. Calculating the velocity of protons is essential for various scientific experiments and observations. This calculator relies on a specific formula designed to estimate the velocity of protons, providing scientists and researchers with crucial data for their studies. The formula for estimating the velocity of protons is based on fundamental physics principles and includes key parameters: 1. Momentum (p): Momentum represents the product of an object’s mass (m) and its velocity (v). In the case of protons, this mass is their rest mass, which is approximately 1.6726219 x 10^-27 kilograms. 2. Kinetic Energy (KE): Kinetic energy is the energy an object possesses due to its motion. For particles like protons, it’s calculated as (1/2)mv^2, where ‘m’ is the mass and ‘v’ is the velocity. 3. Relativistic Effects: When dealing with particles approaching the speed of light, relativistic effects come into play. The velocity of protons is often close to the speed of light (c), which is approximately 299,792,458 meters per second. The formula for estimating the velocity of protons incorporates these factors and can be expressed as: v = c * √[1 – (1 / (1 + (KE / (m * c^2))))] In this formula: • v: Velocity of protons (in meters per second, m/s). • c: Speed of light (approximately 299,792,458 m/s). • KE: Kinetic energy of protons (in joules, J). • m: Rest mass of protons (approximately 1.6726219 x 10^-27 kg). The Proton Velocity Calculator applies this formula, allowing users to input the kinetic energy of protons they are interested in studying. The calculator then computes the estimated velocity of these protons, taking into account relativistic effects as they approach the speed of light. This calculator serves a critical role in the field of particle physics and nuclear research: 1. Experimental Particle Physics: Researchers use it to estimate the velocity of protons in particle accelerators, helping plan and analyze experiments involving high-energy collisions. 2. Nuclear Astrophysics: In the study of nuclear reactions in astrophysical environments, understanding proton velocities is essential for modeling processes in stars and cosmic events. 3. Medical Applications: In proton therapy, a form of radiation therapy used to treat cancer, precise knowledge of proton velocities is crucial for treatment planning. 4. Fundamental Research: Scientists employ it in fundamental research to explore the behavior of particles and the structure of atomic nuclei. To use the Proton Velocity Calculator, users input the kinetic energy of the protons they are interested in, and the calculator provides the estimated velocity in meters per second. In conclusion, the Proton Velocity Calculator, driven by its specialized formula, is an indispensable tool for physicists, researchers, and scientists exploring the fascinating world of subatomic particles. It simplifies the process of estimating proton velocities, accounting for relativistic effects and providing vital data for experiments, studies, and applications in various scientific fields.
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# Chemistry posted by on . Heyy, okay I have this question about chemistry and its driving me nuts. It feels like I am so close but am just missing something. The question is: A vessel of 6.84 L in volume contains 3.61 L of pure water at 25°C. A partial pressure of 3.67 atm of CO2 is quickly injected into the space above the water. Calculate the partial pressure of carbon dioxide remaining once the solution has become saturated with the gas. Henry's constant for CO2 at this temperature is 0.0350 M atm-1 Okay so I know basically this...we dicussed with some other people and rearrganed equations and basically got this: (mol initial - mol final)/V = concentration x pressure. so the initial moles you get using PV=NRT, so N=PV/RT and same for moles final (diff volumes used of course) but I think that's where I'm messing up. Because after that you divide by a volume, then by the concentration. But I think I'm messing up what volumes go where. I know what the answer ius supposed to be and I cant get it for the life of me. I'm thinking that for inital moles the volume would be 6.84 cause that's the whole thing. And then I'm thinking for final moles it would be 3.61 cause that's where the liquid is. But I don't know what I would divde by...the leftover 3.23? anyways I've tried many combinations and I cant get it. ANy help would be greatly apprecaited. I could be totally on the wrong track. Thanks no youarent just doit and you see I already did it and it doesnt work, that's why I'm asking. Do you have an answer? The answer for the question is supposed to be 1.88 Its supposed to be 1.88 but I can't get it to that at all. I don't know waht I'm doing wrong or if I'm even going in the right direction. I get like 200 something I'm starting to think that's not the way to do it at all anymore....any other suggestions? ### Answer This Question First Name: School Subject: Answer: ### Related Questions More Related Questions Post a New Question
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Some sail area math questions... - SailNet Community Junior Member Join Date: Jun 2009 Posts: 7 Thanks: 0 Thanked 0 Times in 0 Posts Rep Power: 0 Some sail area math questions... I have been struggling with the math on this, and I am sure there is a simple equation for it, but I have yet to figure it out. I was hoping someone here could give me the answers I am looking for. I am currently sailing a Redwing 30 with a 155% Genoa on a roller furling setup. I want to make some marks on the foot for roller reefing the Genoa down to a 135% & 100% I need to know the distance from the tack or the clue to place the marks on the foot to equal these respective positions. I am looking for something like "2.3 feet from the tack for 100%" etc.. if my question makes any sense? Much thanks in advanced, sincerely confused mathematician Alstare... -Brian 1970 Hinterhoeller Redwing 30' #128 Alstare is offline Junior Member Join Date: Jun 2009 Posts: 7 Thanks: 0 Thanked 0 Times in 0 Posts Rep Power: 0 Being a new member I can't post links, but I have a link to the sail plan data fore the boat... -Brian 1970 Hinterhoeller Redwing 30' #128 Alstare is offline Junior Member Join Date: Jun 2009 Posts: 7 Thanks: 0 Thanked 0 Times in 0 Posts Rep Power: 0 3rd post should allow me to post links now... Here is the sailplan info for the boat: Redwing 30 C&C REDWING 30 Sailboat SailPlan Data and Sail Quoting System -Brian 1970 Hinterhoeller Redwing 30' #128 Alstare is offline Old 06-14-2011 Just another Moderator Join Date: Sep 2005 Location: New Westminster, BC Posts: 19,139 Thanks: 153 Thanked 545 Times in 519 Posts Rep Power: 10 Lay the sail out on the floor and measure from the luff to the clew, perpendicular to the luff. That dimension will be your "LP" (should be nearly 18 feet [17.85]) which is 155% of your J measurement (ie forestay to mast along the deck - 11.5 ft) So the 100% will be rolled up til there's still 11.5 feet of sail between the rolled portion and the clew (again, perpendicular to the luff). Just do the ratios for any other position you might want. Mark the position on that 'line' you measured, and extend it to the foot parallel to the luff. Ron 1984 Fast/Nicholson 345 "FastForward" ".. there is much you could do at sea with common sense.. and very little you could do without it.." Capt G E Ericson (from "The Cruel Sea" by Nicholas Monsarrat) Last edited by Faster; 06-14-2011 at 11:14 PM. Faster is online now Old 06-14-2011 Senior Member Join Date: Jul 2000 Location: Pennsylvania Posts: 4,565 Thanks: 42 Thanked 259 Times in 245 Posts Rep Power: 18 1. Measure the foot of the 155% sail = X 2. X times 135 divided by 155 = length of foot at 135% 2a. X times 100 divided by 155 = length of foot 100% The 'typical' maximum reduction you can get on a reefing-furler and still have a 'reasonable' sail shape is 30% reduction. 155 X .7 = 108 or 110% 3. X times 110 divided by 155 = length of foot at 108 - 110% :-) RichH is online now Junior Member Join Date: Jun 2009 Posts: 7 Thanks: 0 Thanked 0 Times in 0 Posts Rep Power: 0 Wow... Thanks for the quick and extremely informative replies guys! I will try out both solutions to verify my numbers. I know with the rolled furler I am not gonna get exact performance of the properly reduce hanked on jib sizes. But I think it will make for a good reference for setting the boat up quickly for different wind conditions as apposed to my "pull some in" method I am currently using! I got the beers if you guys ever sail down this way! -Cheers! -Brian 1970 Hinterhoeller Redwing 30' #128 Alstare is offline Old 06-14-2011 Senior Member Join Date: Aug 2008 Posts: 384 Thanks: 3 Thanked 7 Times in 6 Posts Rep Power: 9 Well, most of these suggestions are partly right. First determine the LP. Measure the length of the luff from the tack to the head ring. Mark the mid point of the luff. Then measure from the center of the clew ring to the midpoint you marked. That is the actual LP. The LP dimension is different from the foot dimension. You probably know your J dimension (look up in Mauripro rig database on their site if you don't). A 155% genoa should have an LP that is 155% x the J dimension. Calculate the the 135 and 110% dimensions. Measure from the midpoint of the luff to to a spot on the imaginary line from the clew to the tack which corresponds to135% and 110% dimension. If you measure to the foot it will be inaccurate because there usually a foot roach involved. Sanduskysailor is offline Old 06-15-2011 Senior Member Join Date: Jul 2000 Location: Pennsylvania Posts: 4,565 Thanks: 42 Thanked 259 Times in 245 Posts Rep Power: 18 There is absolutely NO need for the complication of find a the "LP" of a genoa to determine the percentages you want for the foot dimensions for a roll-up to 135 or 100%, etc. LP is a simple direct ratio of foot length. A genoa for this sized boat will NOT have roach of any significant measure ... its not a symmetrical spinnaker its a genoa. The 'roach', or as a sailmaker would call it an apron, will only extend several 'inches' and therefore will only add less than 1% of the foot length (as a tangent ratio). Just measure the exiting foot length and ratio that value by what you want .... (X)135/155 or (X)100/155 and youll be within 1±% of the 'actual' or 'spherical' dimension. ... and the rest of the world wonders why the USA is in 40th place out of ~195 countries in Math education ..... RichH is online now Old 06-15-2011 Just another Moderator Join Date: Sep 2005 Location: New Westminster, BC Posts: 19,139 Thanks: 153 Thanked 545 Times in 519 Posts Rep Power: 10 Quote: Originally Posted by Sanduskysailor View Post .... Measure the length of the luff from the tack to the head ring. Mark the mid point of the luff. Then measure from the center of the clew ring to the midpoint you marked. That is the actual LP. .... Not sure you want to measure from mid luff.... LP is for Luff Perpendicular, ie a line at right angles to the luff through the clew.. It would only be mid luff on a very high clewed sail... (as I understand it, anyhow) Ron 1984 Fast/Nicholson 345 "FastForward" ".. there is much you could do at sea with common sense.. and very little you could do without it.." Capt G E Ericson (from "The Cruel Sea" by Nicholas Monsarrat) Faster is online now Old 06-15-2011 Senior Member Join Date: Jul 2000 Location: Pennsylvania Posts: 4,565 Thanks: 42 Thanked 259 Times in 245 Posts Rep Power: 18 LP (length perpendicular) is determined by establishing a line at exactly 90°, perpendicular, to the luff* running through the clew, ... has nothing to do with the head (ring). What he's describing is to one segment of the 'centroid of area' .... finding the exact midpoint along an 'edge' (luff, leech or foot) and drawing a line from the opposite 'corner' TO the midpoint along the opposite edge ... the intersection of all three lines establishes the centroid of area - aka: center of effort or CE. * in actuality this is an 'imaginary' luff, as with jibs/genoas there is usually some material that is cut away in a smooth curve to compensate for forestay, etc. sag ... and with mainsails some smooth curve added to match the expected mast 'pre-bend'. So to calculate LP, you construct an imaginary line running between the tack and the head (an imaginary luff), then establish a line at 90° to the 'imaginary line' that bisects the clew. ;-) RichH is online now Message: Options By choosing to post the reply above you agree to the rules you agreed to when joining Sailnet. ## Register Now In order to be able to post messages on the SailNet Community forums, you must first register. Please note: After entering 3 characters a list of Usernames already in use will appear and the list will disappear once a valid Username is entered. User Name: OR ## Log-in Human Verification In order to verify that you are a human and not a spam bot, please enter the answer into the following box below based on the instructions contained in the graphic. Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
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# Crystal Reports 11 - Master-Sub report Hi, I have some questions on the Auto-Schedule feature in the Crystal. I found a report with that feature and I want to replicate that in to mine. Below are the codes used in those reports, I kinda understand whats going on, but not in detail.. Master Report ------------------- This has 4 formulas and 3 parameters: Formulas: 1. reporting_date Local numbervar m:=month(CurrentDate) + 1; Local numbervar y:=year(CurrentDate); If m>12 then (m:=m-12; y:=y + 1); Date(y,m,1) 2. begin_dt IF {?AutoSchedule} = 'N' THEN {?begin_date} ELSE IF {?AutoSchedule} = 'Y' AND Month({@reporting date}) in 1 to 3 else if {?AutoSchedule} = 'Y' AND Month({@reporting date}) in 4 to 12 3. end_dt IF {?AutoSchedule} = 'N' THEN {?end_date} ELSE IF {?AutoSchedule} = 'Y' THEN DateValue(DateAdd('m',-1,{@reporting date}))-1 4. qtr_display IF {?AutoSchedule} = "Y" THEN &'/' ELSE IF {?AutoSchedule} = "N" THEN 'Q'&Totext ( DatePart ( 'q' , {@end_dt}) , '0') & '/' & Totext (Year ({@end_dt}), '0000') Parameters: 1.AutoSchedule -- This has 2 values --  Y, N 2. BeginDate -- Empty parameter 3. EndDate  -- Empty parameter -------------------------------------------------------------------------- SUB-Report -------------- This has 3 parameters -- begin_date, end_date, qtr_display Can anyone explain me about this auto-schedule feature and whats going on in each and every part of the code above. Please also let me know if there is any website which explains this step by step with examples. Thanks@ ###### Who is Participating? I wear a lot of hats... "The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S. Commented: 1. reporting_date : You are moving the date to 1st of the next month. For example : if your are on dec 15th 2009 ,  then m:=month(CurrentDate) + 1;  ===> 12+1 =13 If m>12 then (m:=m-12; y:=y + 1);  ====> month will be 13-12 =1 (Jan) and year 2010 =====> it implies it is moving to ist of the next month. 2. begin_dt {? } ==> implies parameter {@ } ==> formula IF {?AutoSchedule} = 'N' THEN {?begin_date} =============== means if you enter  N  in parameter {?AutoSchedule}  then this formula gives what you selected from {?begin_date} ============ ELSE IF {?AutoSchedule} = 'Y' AND Month({@reporting date}) in 1 to 3 ============== means if you enter  Y  in parameter {?AutoSchedule}  and month of {@reporting date} formula is in 1 to 3 (1 st quarter )  then this formula(begin date) moving year back to 2 years back of the year of {@reporting date} and to month and date to  jan 1st ============== else if {?AutoSchedule} = 'Y' AND Month({@reporting date}) in 4 to 12 ========= means if you enter  Y  in parameter {?AutoSchedule}  and month of {@reporting date} formula is in 2nd quarter then this formula(begin date) moving year back to 1 year back of the year of {@reporting date} and to month and date to  jan 1st 0 Experts Exchange Solution brought to you by Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle. Commented: in the above not second quarter (but 4 to 12 months) 3. end_dt IF {?AutoSchedule} = 'N' THEN {?end_date} ============== means if you enter  N  in parameter {?AutoSchedule}  then this formula gives hat you selected from {?end_date} parameter ========= ELSE IF {?AutoSchedule} = 'Y' THEN DateValue(DateAdd('m',-1,{@reporting date}))-1 =================== means if you enter  Y  in parameter {?AutoSchedule}  this formula(end date) moving  month  of {@reporting date} to last day of before previous month. (for example :@reporting date => 1 st day of march , so this formula gives you last day of jan In another way : {@reporting date} gives you 1 st day of next month of currentdate so this formula gives you last day of previous month 0 Commented: 4. qtr_display: IF {?AutoSchedule} = "Y" THEN &'/' ========= means if you enter  Y  in parameter {?AutoSchedule} then this formula gives you : last quarter for next month-1 /  year of  previous currentdate month ============= ELSE IF {?AutoSchedule} = "N" THEN 'Q'&Totext ( DatePart ( 'q' , {@end_dt}) , '0') & '/' & Totext (Year ({@end_dt}), '0000') ============ For N current quarter of end date/ year of endate ======== check crystal help or click F1 for checking the functions like : 0 Author Commented: Thanks that was very informative... Just curious.. Is there something like a dual table (Oracle), in crystal to check these functions? what would be my begin_date and end_date parameter, if it is year to date, meaning, begin_date should always be the beginning of the year and the end_date should be previous months last day... This is a quartely report... For eg: It runs 4 times a year.. Apr 10, Jul 10, Oct 10 and Jan 10 For Apr 10 the reporting period should be from Jan 1 to Mar 31 For Jul 10 -- Jan 1 to Jun 30 For Oct 10 --  Jan 1 to sep 30 For Jan 10 -- Jan 1 to Dec 31 -- This one is kinda tricky because, even though the report is ran the next year it needs information for prev year... Lets say the above ones are 2010 and the Jan 10 2011 should pull records of Jan 1 2010 to Dec 31 2010... Please assist me in writing the formula.. Thanks! 0 Commented: To answer your question about the AutoSchedule feature, there is no such feature.  The report you are looking at uses a parameter by that name and the value is N or Y depending on how it is called.  I assume this report is scheduled through a tool like Crystal Server and when run that way the value passed is Y so the report knows to report appropriately.  If it is run manually then the value will be N and the report act accordingly. mlmcc 0 ###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today Crystal Reports From novice to tech pro — start learning today.
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Code covered by the BSD License # Surface Fitting using gridfit ### John D'Errico (view profile) 11 Nov 2005 (Updated ) Model 2-d surfaces from scattered data ### Editor's Notes: This file was a File Exchange Select file. Select files are submissions that have been peer-reviewed and approved as meeting a high standard of utility and quality. test_main.m ```% Gridfit test script % This script file is designed to be used in cell mode % from the matlab editor, or best of all, use the publish % to HTML feature from the matlab editor. Older versions % of matlab can copy and paste entire blocks of code into % the Matlab command window. %% Topographic data x=bluff_data(:,1); y=bluff_data(:,2); z=bluff_data(:,3); % Two ravines on a hillside. Scanned from a % topographic map of an area in upstate New York. plot3(x,y,z,'.') %% % Turn the scanned point data into a surface gx=0:4:264; gy=0:4:400; g=gridfit(x,y,z,gx,gy); figure colormap(hot(256)); surf(gx,gy,g); camlight right; lighting phong; line(x,y,z,'marker','.','markersize',4,'linestyle','none'); title 'Use topographic contours to recreate a surface' %% Fitting a trigonometric surface clear n1 = 15; n2 = 15; theta = rand(n1,1)*pi/2; r = rand(1,n2); x = cos(theta)*r; y = sin(theta)*r; x=x(:); y=y(:); x = [[0 0 1 1]';x;x;1-x;1-x]; y = [[0 1 0 1]';y;1-y;y;1-y]; figure plot(x,y,'.') title 'Data locations in the x-y plane' %% z = sin(4*x+5*y).*cos(7*(x-y))+exp(x+y); xi = linspace(0,1,51); [xg,yg]=meshgrid(xi,xi); zgd = griddata(x,y,z,xg,yg); figure surf(xi,xi,zgd) colormap(hot(256)) camlight right lighting phong title 'Griddata on trig surface' % Note the wing-like artifacts along the edges, due % to the use of a Delaunay triangulation in griddata. %% zgrid = gridfit(x,y,z,xi,xi); figure surf(xi,xi,zgrid) colormap(hot(256)) camlight right lighting phong title('Gridfit to trig surface') %% The trig surface with highly different scalings on the x and y axes xs = x/100; xis = xi/100; ys = y*100; yis = xi*100; % griddata has problems with badly scaled data [xg,yg]=meshgrid(xis,yis); zgd = griddata(xs,ys,z,xg,yg); figure surf(xg,yg,zgd) colormap(hot(256)) camlight right lighting phong title 'Serious problems for griddata on badly scaled trig surface' %% % autoscaling off zgrid = gridfit(xs,ys,z,xis,yis,'autoscale','off'); % autoscaling on (the default) zgrids = gridfit(xs,ys,z,xis,yis,'autoscale','on'); % plot the unscaled result figure surf(xis,yis,zgrid) colormap(hot(256)) camlight right lighting phong title 'Gridfit (unscaled) on trig surface' % plot the unscaled result figure surf(xis,yis,zgrids) colormap(hot(256)) camlight right lighting phong title 'Default gridfit (automatically scaled) on trig surface' %% Comparison of extrapolation behaviors % Note: griddata will not extrapolate using most most of % its methods, and is very slow/memory intensive for the % 'v4' method clear n = 50000; x = min(2,max(-2,randn(n,1))); y = min(2,max(-2,randn(n,1))); z = exp(x+y); nodes = -3:.1:3; zl = gridfit(x,y,z,nodes,nodes,'reg','laplacian','sm',.01); zs = gridfit(x,y,z,nodes,nodes,'reg','springs','sm',.01); % plot the results figure surf(nodes,nodes,zg) colormap(bone(256)) camlight right lighting phong hold on plot3(x,y,z,'r.') hold off title 'Extrapolation with gridfit (gradient regularization)' figure surf(nodes,nodes,zl) colormap(bone(256)) camlight right lighting phong hold on plot3(x,y,z,'r.') hold off title 'Extrapolation with gridfit (Laplacian regularization)' figure surf(nodes,nodes,zs) colormap(bone(256)) camlight right lighting phong hold on plot3(x,y,z,'r.') hold off title 'Extrapolation with gridfit (gradient regularization)' %% Fitting the "peaks" surface with fairly sparse data clear n = 100; x = (rand(n,1)-.5)*6; y = (rand(n,1)-.5)*6; z = peaks(x,y); xi = linspace(-3,3,101); zpgf = gridfit(x,y,z,xi,xi); [xg,yg]=meshgrid(xi,xi); zpgd = griddata(x,y,z,xg,yg,'cubic'); figure surf(xi,xi,zpgd) colormap(jet(256)) camlight right lighting phong title 'Griddata (method == cubic) on peaks surface' figure surf(xi,xi,zpgf) colormap(hsv(256)) camlight right lighting phong title('Gridfit on peaks surface') %% Fitting the seamount surface clear xg = linspace(min(x),max(x),50); yg = linspace(min(y),max(y),50); zsgf = gridfit(x,y,z,xg,yg); figure surf(xg,yg,zsgf) colormap(hot(256)) camlight right lighting phong title('Gridfit applied to seamount data') % Note that gridfit will extrapolate smoothly to the % corners of the grid, whereas griddata cannot, unless % the 'v4' option is chosen. But 'v4' is slow on even % moderately large problems. [xg,yg]=meshgrid(xg,yg); zsgd = griddata(x,y,z,xg,yg); figure surf(xg,yg,zsgd) colormap(pink(256)) camlight right lighting phong title 'Griddata' %% Using tiles in gridfit % Users of gridfit who have really huge problems now have % an option. I'll generate a large amount of data, % and hope to model a fairly large grid - 800 x 800. This % would normally require gridfit to solve a system of % equations with 640,000 unknowns. It might be a solveable % (though very slowly so) problem for my computer, were I % to use gridfit on the full problem. Gridfit allows you to % break the problem into smaller tiles if you choose. In % this case each tile is 120x120, with a 25% (30 element) % overlap between tiles. % Relax, this demo may take a couple of minutes to run!!!! n = 100000; x = rand(n,1); y = rand(n,1); z = x+y+sin((x.^2+y.^2)*10); xnodes = 0:.002:1; ynodes = xnodes; [zg,xg,yg] = gridfit(x,y,z,xnodes,ynodes,'tilesize',120,'overlap',0.25); surf(xg,yg,zg)
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Mathematics # 14 Liters ÷ 9500 millilters #### Jennn 1 year ago Convert litres into ml 14 litres=14000ml =14000/9500 =140/95 =28/19=1.47 #### AlleenHysom 1 year ago 14 l = 14000 mls so... 14000 divided by 9500 equals 1.47368421
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## Info CSB_aood = 0.0277[(4.762)(25/841)]a125 x (0.481/841)a1(500/8.66)a5 = 0.0947 m/s Alternatively, applying the Fair correlation: The flow parameter Flg = 0.021 [Eq. (14-89)]. From Fig. 14-31, Csbf = 0.095 m/s. Then, based on the net area, CSB = 0.095(25/20)02 = 0.0993 m/s about 5 percent higher than the answer obtained from Kister and Haas. For the design condition, the C-factor based on the net area is 25,500 3600(0.481X4.66) 0 0
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# Probit model Short description: Statistical regression where the dependent variable can take only two values In statistics, a probit model is a type of regression where the dependent variable can take only two values, for example married or not married. The word is a portmanteau, coming from probability + unit.[1] The purpose of the model is to estimate the probability that an observation with particular characteristics will fall into a specific one of the categories; moreover, classifying observations based on their predicted probabilities is a type of binary classification model. A probit model is a popular specification for a binary response model. As such it treats the same set of problems as does logistic regression using similar techniques. When viewed in the generalized linear model framework, the probit model employs a probit link function.[2] It is most often estimated using the maximum likelihood procedure,[3] such an estimation being called a probit regression. ## Conceptual framework Suppose a response variable Y is binary, that is it can have only two possible outcomes which we will denote as 1 and 0. For example, Y may represent presence/absence of a certain condition, success/failure of some device, answer yes/no on a survey, etc. We also have a vector of regressors X, which are assumed to influence the outcome Y. Specifically, we assume that the model takes the form $\displaystyle{ P(Y=1 \mid X) = \Phi(X^\operatorname{T}\beta), }$ where P is the probability and $\displaystyle{ \Phi }$ is the cumulative distribution function (CDF) of the standard normal distribution. The parameters β are typically estimated by maximum likelihood. It is possible to motivate the probit model as a latent variable model. Suppose there exists an auxiliary random variable $\displaystyle{ Y^\ast = X^T\beta + \varepsilon, }$ where ε ~ N(0, 1). Then Y can be viewed as an indicator for whether this latent variable is positive: $\displaystyle{ Y = \left.\begin{cases} 1 & Y^* \gt 0 \\ 0 &\text{otherwise} \end{cases} \right\} = \left.\begin{cases} 1 & X^\operatorname{T}\beta + \varepsilon \gt 0 \\ 0 &\text{otherwise} \end{cases} \right\} }$ The use of the standard normal distribution causes no loss of generality compared with the use of a normal distribution with an arbitrary mean and standard deviation, because adding a fixed amount to the mean can be compensated by subtracting the same amount from the intercept, and multiplying the standard deviation by a fixed amount can be compensated by multiplying the weights by the same amount. To see that the two models are equivalent, note that \displaystyle{ \begin{align} P(Y = 1 \mid X) &= P(Y^\ast \gt 0) \\ &= P(X^\operatorname{T}\beta + \varepsilon \gt 0) \\ &= P(\varepsilon \gt -X^\operatorname{T}\beta) \\ &= P(\varepsilon \lt X^\operatorname{T}\beta) & \text{by symmetry of the normal distribution}\\ &= \Phi(X^\operatorname{T}\beta) \end{align} } ## Model estimation ### Maximum likelihood estimation Suppose data set $\displaystyle{ \{y_i,x_i\}_{i=1}^n }$ contains n independent statistical units corresponding to the model above. For the single observation, conditional on the vector of inputs of that observation, we have: $\displaystyle{ P(y_i=1|x_i)= \Phi(x_i'\beta) }$[clarification needed] $\displaystyle{ P(y_i=0|x_i)= 1-\Phi(x_i'\beta) }$ where $\displaystyle{ x_i }$ is a vector of $\displaystyle{ K \times 1 }$ inputs, and $\displaystyle{ \beta }$ is a $\displaystyle{ K \times 1 }$ vector of coefficients. The likelihood of a single observation $\displaystyle{ (y_i, x_i) }$ is then $\displaystyle{ \mathcal{L}(\beta; y_i, x_i) = \Phi(x_i^\operatorname{T}\beta)^{y_i} [1-\Phi(x_i^\operatorname{T}\beta)]^{(1-y_i)} }$ In fact, if $\displaystyle{ y_i=1 }$, then $\displaystyle{ \mathcal{L}(\beta; y_i, x_i) = \Phi(x_i^\operatorname{T}\beta) }$, and if $\displaystyle{ y_i=0 }$, then $\displaystyle{ \mathcal{L}(\beta; y_i, x_i) = 1-\Phi(x_i^\operatorname{T}\beta) }$. Since the observations are independent and identically distributed, then the likelihood of the entire sample, or the joint likelihood, will be equal to the product of the likelihoods of the single observations: $\displaystyle{ \mathcal{L}(\beta; Y, X) = \prod_{i=1}^n \left( \Phi(x_i^\operatorname{T}\beta)^{y_i} [1-\Phi(x_i^\operatorname{T}\beta)]^{(1-y_i)} \right) }$ The joint log-likelihood function is thus $\displaystyle{ \ln\mathcal{L}(\beta; Y, X) = \sum_{i=1}^n \bigg( y_i\ln\Phi(x_i^\operatorname{T}\beta) + (1-y_i)\ln\!\big(1-\Phi(x_i^\operatorname{T}\beta)\big) \bigg) }$ The estimator $\displaystyle{ \hat\beta }$ which maximizes this function will be consistent, asymptotically normal and efficient provided that $\displaystyle{ \operatorname{E}[XX^\operatorname{T}] }$ exists and is not singular. It can be shown that this log-likelihood function is globally concave in $\displaystyle{ \beta }$, and therefore standard numerical algorithms for optimization will converge rapidly to the unique maximum. Asymptotic distribution for $\displaystyle{ \hat\beta }$ is given by $\displaystyle{ \sqrt{n}(\hat\beta - \beta)\ \xrightarrow{d}\ \mathcal{N}(0,\,\Omega^{-1}), }$ where $\displaystyle{ \Omega = \operatorname{E}\bigg[ \frac{\varphi^2(X^\operatorname{T}\beta)}{\Phi(X^\operatorname{T}\beta)(1-\Phi(X^\operatorname{T}\beta))}XX^\operatorname{T} \bigg], \qquad \hat\Omega = \frac{1}{n}\sum_{i=1}^n \frac{\varphi^2(x^\operatorname{T}_i\hat\beta)}{\Phi(x^\operatorname{T}_i\hat\beta)(1-\Phi(x^\operatorname{T}_i\hat\beta))}x_ix^\operatorname{T}_i, and \lt math\gt \varphi=\Phi' }$ is the Probability Density Function (PDF) of standard normal distribution. Semi-parametric and non-parametric maximum likelihood methods for probit-type and other related models are also available.[4] ### Berkson's minimum chi-square method Main page: Minimum chi-square estimation This method can be applied only when there are many observations of response variable $\displaystyle{ y_i }$ having the same value of the vector of regressors $\displaystyle{ x_i }$ (such situation may be referred to as "many observations per cell"). More specifically, the model can be formulated as follows. Suppose among n observations $\displaystyle{ \{y_i,x_i\}_{i=1}^n }$ there are only T distinct values of the regressors, which can be denoted as $\displaystyle{ \{x_{(1)},\ldots,x_{(T)}\} }$. Let $\displaystyle{ n_t }$ be the number of observations with $\displaystyle{ x_i=x_{(t)}, }$ and $\displaystyle{ r_t }$ the number of such observations with $\displaystyle{ y_i=1 }$. We assume that there are indeed "many" observations per each "cell": for each $\displaystyle{ t, \lim_{n \rightarrow \infty} n_t/n = c_t \gt 0 }$. Denote $\displaystyle{ \hat{p}_t = r_t/n_t }$ $\displaystyle{ \hat\sigma_t^2 = \frac{1}{n_t} \frac{\hat{p}_t(1-\hat{p}_t)}{\varphi^2\big(\Phi^{-1}(\hat{p}_t)\big)} }$ Then Berkson's minimum chi-square estimator is a generalized least squares estimator in a regression of $\displaystyle{ \Phi^{-1}(\hat{p}_t) }$ on $\displaystyle{ x_{(t)} }$ with weights $\displaystyle{ \hat\sigma_t^{-2} }$: $\displaystyle{ \hat\beta = \Bigg( \sum_{t=1}^T \hat\sigma_t^{-2}x_{(t)}x^\operatorname{T}_{(t)} \Bigg)^{-1} \sum_{t=1}^T \hat\sigma_t^{-2}x_{(t)}\Phi^{-1}(\hat{p}_t) }$ It can be shown that this estimator is consistent (as n→∞ and T fixed), asymptotically normal and efficient.[citation needed] Its advantage is the presence of a closed-form formula for the estimator. However, it is only meaningful to carry out this analysis when individual observations are not available, only their aggregated counts $\displaystyle{ r_t }$, $\displaystyle{ n_t }$, and $\displaystyle{ x_{(t)} }$ (for example in the analysis of voting behavior). ### Gibbs sampling Gibbs sampling of a probit model is possible because regression models typically use normal prior distributions over the weights, and this distribution is conjugate with the normal distribution of the errors (and hence of the latent variables Y*). The model can be described as \displaystyle{ \begin{align} \boldsymbol\beta & \sim \mathcal{N}(\mathbf{b}_0, \mathbf{B}_0) \\[3pt] y_i^\ast\mid\mathbf{x}_i,\boldsymbol\beta & \sim \mathcal{N}(\mathbf{x}^\operatorname{T}_i\boldsymbol\beta, 1) \\[3pt] y_i & = \begin{cases} 1 & \text{if } y_i^\ast \gt 0 \\ 0 & \text{otherwise} \end{cases} \end{align} } From this, we can determine the full conditional densities needed: \displaystyle{ \begin{align} \mathbf{B} &= (\mathbf{B}_0^{-1} + \mathbf{X}^\operatorname{T}\mathbf{X})^{-1} \\[3pt] \boldsymbol\beta\mid\mathbf{y}^\ast &\sim \mathcal{N}(\mathbf{B}(\mathbf{B}_0^{-1}\mathbf{b}_0 + \mathbf{X}^\operatorname{T}\mathbf{y}^\ast), \mathbf{B}) \\[3pt] y_i^\ast\mid y_i=0,\mathbf{x}_i,\boldsymbol\beta &\sim \mathcal{N}(\mathbf{x}^\operatorname{T}_i\boldsymbol\beta, 1)[y_i^\ast \lt 0] \\[3pt] y_i^\ast\mid y_i=1,\mathbf{x}_i,\boldsymbol\beta &\sim \mathcal{N}(\mathbf{x}^\operatorname{T}_i\boldsymbol\beta, 1)[y_i^\ast \ge 0] \end{align} } The result for $\displaystyle{ \boldsymbol\beta }$ is given in the article on Bayesian linear regression, although specified with different notation. The only trickiness is in the last two equations. The notation $\displaystyle{ [y_i^\ast \lt 0] }$ is the Iverson bracket, sometimes written $\displaystyle{ \mathcal{I}(y_i^\ast \lt 0) }$ or similar. It indicates that the distribution must be truncated within the given range, and rescaled appropriately. In this particular case, a truncated normal distribution arises. Sampling from this distribution depends on how much is truncated. If a large fraction of the original mass remains, sampling can be easily done with rejection sampling—simply sample a number from the non-truncated distribution, and reject it if it falls outside the restriction imposed by the truncation. If sampling from only a small fraction of the original mass, however (e.g. if sampling from one of the tails of the normal distribution—for example if $\displaystyle{ \mathbf{x}^\operatorname{T}_i\boldsymbol\beta }$ is around 3 or more, and a negative sample is desired), then this will be inefficient and it becomes necessary to fall back on other sampling algorithms. General sampling from the truncated normal can be achieved using approximations to the normal CDF and the probit function, and R has a function rtnorm() for generating truncated-normal samples. ## Model evaluation The suitability of an estimated binary model can be evaluated by counting the number of true observations equaling 1, and the number equaling zero, for which the model assigns a correct predicted classification by treating any estimated probability above 1/2 (or, below 1/2), as an assignment of a prediction of 1 (or, of 0). See Logistic regression § Model for details. ## Performance under misspecification Consider the latent variable model formulation of the probit model. When the variance of $\displaystyle{ \varepsilon }$ conditional on $\displaystyle{ x }$ is not constant but dependent on $\displaystyle{ x }$, then the heteroscedasticity issue arises. For example, suppose $\displaystyle{ y^*= \beta_0+B_1 x_1+\varepsilon }$ and $\displaystyle{ \varepsilon\mid x \sim N (0,x^2_1) }$ where $\displaystyle{ x_1 }$ is a continuous positive explanatory variable. Under heteroskedasticity, the probit estimator for $\displaystyle{ \beta }$ is usually inconsistent, and most of the tests about the coefficients are invalid. More importantly, the estimator for $\displaystyle{ P (y=1\mid x) }$ becomes inconsistent, too. To deal with this problem, the original model needs to be transformed to be homoskedastic. For instance, in the same example, $\displaystyle{ 1[\beta_0+\beta_1 x_1+\varepsilon\gt 0] }$ can be rewritten as $\displaystyle{ 1[\beta_0/x_1+\beta_1+\varepsilon/x_1\gt 0] }$, where $\displaystyle{ \varepsilon/x_1\mid x\sim N(0,1) }$. Therefore, $\displaystyle{ P(y=1\mid x) = \Phi (\beta_1 + \beta_0/x_1) }$ and running probit on $\displaystyle{ (1, 1/x_1) }$ generates a consistent estimator for the conditional probability $\displaystyle{ P(y=1\mid x). }$ When the assumption that $\displaystyle{ \varepsilon }$ is normally distributed fails to hold, then a functional form misspecification issue arises: if the model is still estimated as a probit model, the estimators of the coefficients $\displaystyle{ \beta }$ are inconsistent. For instance, if $\displaystyle{ \varepsilon }$ follows a logistic distribution in the true model, but the model is estimated by probit, the estimates will be generally smaller than the true value. However, the inconsistency of the coefficient estimates is practically irrelevant because the estimates for the partial effects, $\displaystyle{ \partial P(y=1\mid x)/\partial x_{i'} }$, will be close to the estimates given by the true logit model.[5] To avoid the issue of distribution misspecification, one may adopt a general distribution assumption for the error term, such that many different types of distribution can be included in the model. The cost is heavier computation and lower accuracy for the increase of the number of parameter.[6] In most of the cases in practice where the distribution form is misspecified, the estimators for the coefficients are inconsistent, but estimators for the conditional probability and the partial effects are still very good.[citation needed] One can also take semi-parametric or non-parametric approaches, e.g., via local-likelihood or nonparametric quasi-likelihood methods, which avoid assumptions on a parametric form for the index function and is robust to the choice of the link function (e.g., probit or logit).[4] ## History The probit model is usually credited to Chester Bliss, who coined the term "probit" in 1934,[7] and to John Gaddum (1933), who systematized earlier work.[8] However, the basic model dates to the Weber–Fechner law by Gustav Fechner, published in (Fechner 1860), and was repeatedly rediscovered until the 1930s; see (Finney 1971) and (Aitchison Brown).[8] A fast method for computing maximum likelihood estimates for the probit model was proposed by Ronald Fisher as an appendix to Bliss' work in 1935.[9] ## References 1. Oxford English Dictionary, 3rd ed. s.v. probit (article dated June 2007): Bliss, C. I. (1934). "The Method of Probits". Science 79 (2037): 38–39. doi:10.1126/science.79.2037.38. PMID 17813446. Bibcode1934Sci....79...38B. "These arbitrary probability units have been called ‘probits’.". 2. Agresti, Alan (2015). Foundations of Linear and Generalized Linear Models. New York: Wiley. pp. 183–186. ISBN 978-1-118-73003-4. 3. Aldrich, John H.; Nelson, Forrest D.; Adler, E. Scott (1984). Linear Probability, Logit, and Probit Models. Sage. pp. 48–65. ISBN 0-8039-2133-0. 4. Park, Byeong U.; Simar, Léopold; Zelenyuk, Valentin (2017). "Nonparametric estimation of dynamic discrete choice models for time series data". Computational Statistics & Data Analysis 108: 97–120. doi:10.1016/j.csda.2016.10.024. 5. Greene, W. H. (2003), Econometric Analysis , Prentice Hall, Upper Saddle River, NJ. 6. For more details, refer to: Cappé, O., Moulines, E. and Ryden, T. (2005): "Inference in Hidden Markov Models", Springer-Verlag New York, Chapter 2. 7. Bliss, C. I. (1934). "The Method of Probits". Science 79 (2037): 38–39. doi:10.1126/science.79.2037.38. PMID 17813446. Bibcode1934Sci....79...38B. 8. Cramer 2002, p. 7. 9. Fisher, R. A. (1935). "The Case of Zero Survivors in Probit Assays". Annals of Applied Biology 22: 164–165. doi:10.1111/j.1744-7348.1935.tb07713.x.
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# Need help on designing plot 1 vue (au cours des 30 derniers jours) KSK le 27 Mai 2015 Commenté : KSK le 27 Mai 2015 Dear all, Can anyone help me with the below? Appreciate if anyone can help. Thank you! Design an x-y plot for the following where the values of x are from 0 to 20π with a spacing of π/100. a. y = x.sin(x) b. z = x.cos(x) c. Convert the above 2D into a 3D plot using z as the third axis Regards, Martin ##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens Connectez-vous pour commenter. ### Réponse acceptée Youssef Khmou le 27 Mai 2015 Modifié(e) : Youssef Khmou le 27 Mai 2015 defining the variable is the first step : x=0:pi/100:20*pi; using element wise operation : y=x.*sin(x); z=x.*cos(x); Three dimensional graph is given by : plo3t(x,y,z) ##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens KSK le 27 Mai 2015 Youssef Khmou Thank you very much! its work! i got two more question i posted today. hope you can help me. =) Really appreciated! Connectez-vous pour commenter. ### Catégories En savoir plus sur 2-D and 3-D Plots dans Help Center et File Exchange ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by
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### Challenge Problem Showdown – January 14, 2013 We invite you to test your GMAT knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GMAT Prep item. Tell your friends to get out their scrap paper and start solving! Here is this week’s problem: For all non-negative integers x and n such that 0 ≤ x ≤ n, the function fn(x) is defined by the equation fn(x) = xn“x. The smallest value of n for which the maximum of fn(x) occurs when x = 4 is To see the answer choices, and to submit your answer, visit our Challenge Problem Showdown page on our site. Discuss this week’s problem with like-minded GMAT takers on our Facebook page. The weekly winner, drawn from among all the correct submissions, will receive One Year of Access to our Challenge Problem Archive, AND the GMAT Navigator, AND Our Six Computer Adaptive Tests (\$92 value). 1. Ziya January 16, 2013 at 2:16 pm Yeah you are absolutely wright ! 2. Rohan January 15, 2013 at 12:11 am I too agree with salman, since x<=n and both are +ve, Fn(4) = 4^(n-4) will keep increasing as n increases or in other word.. MAX(Fn(x)) is not constrained. Are we missing something? 3. Salman January 14, 2013 at 2:52 pm I don’t follow the question. What do u mean by “The smallest value of n for which the maximum of f (x) occurs when x = 4”. If x and n are both positive, and n is larger than x (which is 4), then as n increases, f(x) will also increase. So there isn’t any max value. Am I misreading the question?
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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  19.41vvv Unicode version Theorem 19.41vvv 1846 Description: Theorem 19.41 of [Margaris] p. 90 with 3 quantifiers. (Contributed by NM, 30-Apr-1995.) Assertion Ref Expression 19.41vvv Distinct variable groups:   ,   ,   , Allowed substitution hints:   (,,) Proof of Theorem 19.41vvv StepHypRef Expression 1 19.41vv 1845 . . 3 21exbii 1569 . 2 3 19.41v 1844 . 2 42, 3bitri 240 1 Colors of variables: wff set class Syntax hints:   wb 176   wa 358  wex 1528 This theorem is referenced by:  19.41vvvv  1847  eloprabga  5896  dftpos3  6214  eeeeanv  24354 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1636  ax-8 1644  ax-6 1704  ax-11 1716 This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1529  df-nf 1532 Copyright terms: Public domain W3C validator
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Time:2020-10-21 # preface Hello, this is tongge. In the last section, we learned everything about hashes, especially the evolution of hashes. I believe that through the learning in the previous section, you can tell the interviewer how the hash table developed to this stage from the beginning to the end. However, can the ultimate form of HashMap be realized only in the form of “array + linked list + red black tree”? Is there an alternative? Why doesn’t Java use this alternative? In this section, we’ll learn another data structure — skip table. I’ll divide it into two sections. The first section introduces the evolution process of jump table, the second section code implements jump table, and rewrites HashMap. # Ordered array As we all know, arrays can support random access, that is, they can quickly locate elements through subscripts, and the time complexity is O (1). So what is the use of this random access feature in addition to finding elements by subscript? Imagine, if an array is ordered, I’m looking for someSpecified elementHow to find out the most quickly? Simply, traverse the entire array from the beginning, and return when it encounters the element to be searched. For example, it takes 6 times to find the element 8, and 8 times to find the element 10. Therefore, the time complexity of searching elements in this way is O (n). Fast method, because the array itself is ordered, we can use binary search. First, search from the middle. If the specified element is smaller than the middle element, then search on the left half. If the specified element is larger than the middle element, search in the right half and proceed in turn until the specified element is found. For example, to find the element 8, first locate it in the middle (7 / 2 = 3). In the next search, add 1 to the left pointer, take position 4 as the left pointer, and change the middle position to the position of (4 + (7-4) / 2 = 5). It only takes two times to find the element 8. Using binary search, the efficiency is improved by more than a little bit. Even in the worst case, it only needs the time complexity of log (n). # Ordered list Above, we introduced the quick search of ordered arrays. Next, let’s take a look at the ordered linked list. The above is an ordered linked list. At this time, when I want to find the element 8, I can only search from the chain header until I encounter 8. The time complexity is O (n). It seems that there is no better way. Let’s consider the differences between ordered arrays and ordered linked lists. The reason why ordered arrays can be directly located to intermediate elements is that they can be accessed quickly through indexes (subscripts). Then, can we implement similar functions by adding indexes to ordered lists? # Skip Watch The first question is: how to give an ordered list a reference? Here, we need to add a concept of “layer”. Assuming that the level of the original linked list is 0, then select some elements to extend upward to form the level 1 index. Similarly, on the basis of the level 1 index, select some elements to extend upward to form the level 2 index until you feel that the number of layers of the index is almost the same. Yes, the skip table is so arbitrary that you can fill it That’s good^^ Suppose, for the ordered linked list above, I add the following indexes: The second question: where to access the hop table? 6? 3? 1? 9? It doesn’t seem to work. Therefore, we need to add a special node, the head node, in front of element 0. For example, after the header node is added to the hop table above, it looks like this: At this point, as long as we start from the node H2, we can quickly find any element in the hop table. For example, to find the element 8, H2 first looks to the right. Eh, it is 6, smaller than 8. Jump to the position of 6, and then look to the right. Ah, it is 9, which is larger than 8. Therefore, you can’t jump over. Jump down to the position of level 1, 6, and look right again. You can’t jump past it. Jump down one step to 6 on level 0. Since, to level 0, you can only follow the chain The table traverses backward in turn until 8 is encountered. The whole process is as follows: As you can see, the whole process is jumping and jumping, so it is named “Tiao watch”. The number of elements here is relatively small, and you may not see much advantage. If there are many elements, every two elements form an index upward, and every two indexes form an index upward. Finally, it is similar to a balanced binary tree As you can see, each search can reduce the search range by half. Therefore, the query time complexity of hop table is O (log n). However, it is impossible to use this completely balanced skip table in practice, because if we want to keep the balanced feature, we must do rebalancing when inserting or deleting elements, which greatly reduces the efficiency. Therefore, generally, we use random to determine whether an element or index should produce an index. The third question: when will the index be produced? The best time is to insert an element, because the next step is to use the index. Why? No matter insert, delete or query, in fact, you must first query to find the element before you can proceed to the next step. To put it bluntly, it means that no matter what the operation is, the query must go through the index. If you want to go through the index, you must first build the index. If you want to build the index, you should insert the element. OK, I’ll take you through the whole process of creating a skip table by using a step-by-step method: 1. In the initial state, there is only one header node H0 (no, there is also a watermark of tongge reading source code, naughty ^ ^). 2. Insert an element 4, put it after H0, and randomly decide whether to index upward. The result is no index. 3. Insert an element 3 and search from H0. The next element of H0 is 4, which is larger than 3. Therefore, 3 is placed between H0 and 4, and then asked whether to form an index. At this time, 3 forms an index upward. At the same time, H0 also forms an index H1 upward. The results are as follows: 4. Insert an element 9, search from H1, and then go through H1 > 3 > 3 > 4, but no position is found. Finally, insert it after 4 and ask whether to form an index. At random, they say that I want to form an index, and I want to form a two-level index (at most one more than the current level), and then it turns out to be like this: 5. Then, elements 1 and 7 are inserted, which are neither surprised nor pleased, nor indexed: 6. Insert element 6. According to the index, the search route is H2 > H1 > 3 > 3 > 4. Eh, it is found that the next 4 is 7. Therefore, 6 is placed between 4 and 7. Then, decide whether to form an index or not. At random, I want to form an index, and I also want to form a two-level index. At this time, it is very troublesome. When the index of element 6 is formed, the line 3 – > 9 and H2 – > need to be modified 9, the result is as follows: 7. At the end of the paper, elements 8 and 10 are inserted, both of which are risk-free and have no index. Therefore, the final result is as follows: As you can see, the skip table is a very random data structure. Even if the elements are re inserted in the same order, the generated skip table may be completely different and capricious. Therefore, I like the data structure of jump table very much. Fourth question: the process of inserting elements is described above. What is the deletion process like? In the deletion process, we should first find the elements. However, there are some small differences, very small differences, which are difficult to describe. For example, to delete the element 6, can I come from the path H2 – > 6 – > 6? No, because from this path, after deleting index 6 of the first layer, the line 3 – > 9 cannot be repaired. Therefore, when deleting an element, you can only take the path H2 – > H1 – > 3 – > 3 – > 4 – > 6, and remember the last index of each layer on the way, so that the index of each layer can be correctly repaired after deleting the element 6. Delete 6 as follows: Well, at this point, I can’t help but think of a small optimization item in the Java skipping table concurrentskiplistmap. In the concurrentskiplistmap, whether search, insert, or delete, they all follow the same search path as delete. In fact, we can simply optimize it, and we can take another path when inserting and searching. The source code analysis of my classmate skipplist can be analyzed Well, that’s all we have to say about the theory of skip watch. # Postscript In this section, we show the whole process of searching, inserting, and deleting elements in the skip table completely and clearly by means of step-by-step diagram. Did you get it? Can you hang up the interviewer? However, many students may say “talk is leap, show me the code”, OK. In the next section, I will show you the details of jump table implementation in code, and rewrite HashMap and next part with jump table. Pay attention to the princess “tongge read the source code” to unlock more source code, foundation and architecture knowledge. ## MVC and Vue MVC and Vue This article was written on July 27, 2020 The first question is: is Vue an MVC or an MVVM framework? Wikipedia tells us: MVVM is a variant of PM, and PM is a variant of MVC. So to a certain extent, whether Vue is MVC or MVVM or not, its ideological direction […]
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{[ promptMessage ]} Bookmark it {[ promptMessage ]} Lecture_5 # Lecture_5 - NewtonsLawsofMotion TOPICS FROM PREVIOUS... This preview shows pages 1–5. Sign up to view the full content. P123 Lecture 5 1 October 2010 Newton’s Laws of Motion TOPICS FROM PREVIOUS LECTURE (motion in 2-D) Components of Equation of Motion v x = v o x + a x tv y = v o y + a y t = + +1/2 a 2 = + a 2 x x o + v o x t + 1/2 x t y y o + v o y t + 1/2 y t etc., Projectile Motion Superposition of two motions Superposition of two motions: Constant v x and free-fall v y Then solve for : y = f ( x ) or trajectory: g v h o o 2 sin 2 2 y g v R o o 2 sin 2 h x R o v oy v ox v Relative Velocity: (e.g., cars on highway, airplane in wind, boat on river, …) Fixed Frame: A 1 Moving Frame: B Moving Object: P A B B P A P v v v / / / This preview has intentionally blurred sections. Sign up to view the full version. View Full Document i Clicker y h x R o v oy v o v 2 ox a x =0 ; a y = -g NEWTON’S LAWS OF MOTION So Far: Follow motion in 1 D and 2 D based solely on Follow motion in 1 D and 2 D based solely on definitions of v and a in terms of position as a function of time. (KINEMATICS) NOW Determine motion based on the underlying physica NOW: Determine motion based on the underlying physical cause of that motion. (DYNAMICS) “Laws of Motion” formulated by Newton over 300 years ago Forces and interactions Common experience: Force is a “push” or a “pull” Force is a vector pull push Different types of forces: Contact Force (kick soccer ball, push book, tow car, … ) Long range forces, act through “empty” space (electric force, magnetic force, gravitational force, … ) Unit of force in SI units: Newton [N] 3 We will focus on contact forces and gravity (near the earth’s surface ) This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Superposition of forces Forces are vectors add to give resultant. 2 F 1 F R x R y R R NET FORCE: ... 3 2 1 F F F F R F F F F R F F F F R ... 3 2 1 x x x x x ... 3 2 1 y y y y y 2 2 2 z y x R R R R EQUILIBRIUM: Net force = 0 3 F y F 3 ? 0 F 2 F 1 F ! This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 14 Lecture_5 - NewtonsLawsofMotion TOPICS FROM PREVIOUS... This preview shows document pages 1 - 5. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 24 May 2019, 19:03 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History GMAT Focus Score Range Accuracy? Author Message Intern Joined: 11 Dec 2016 Posts: 4 Location: United States GMAT 1: 680 Q42 V41 GPA: 3.42 GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 14 Aug 2017, 18:53 I have taken all but two CATs in the GMATPrep Exam Packs and usually score between 41-44 on those. In my two attempts at the GMAT Focus tests my score range has been 46-51 and 45-49. My question is, in the experience/opinion among GMAT takers, how accurate a predictor of test performance is the GMAT Focus scaled score range? CEO Joined: 15 Jul 2015 Posts: 2639 Location: India GMAT 1: 780 Q50 V51 GRE 1: Q170 V169 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 14 Aug 2017, 20:20 cananalyst wrote: I have taken all but two CATs in the GMATPrep Exam Packs and usually score between 41-44 on those. In my two attempts at the GMAT Focus tests my score range has been 46-51 and 45-49. My question is, in the experience/opinion among GMAT takers, how accurate a predictor of test performance is the GMAT Focus scaled score range? I haven't taken the Focus tests recently, but here's what I remember: (a) a lot (or all) repeat questions (from OGs), (b) a really wide score range (47-51), and (c) expensive for the number of questions provided. Maybe someone can confirm whether the question pool has been updated, but I'd recommend that you consider the GMATPrep scores as being far better indicators of your quant score. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14198 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 15 Aug 2017, 16:44 Hi cananalyst, Score 'ranges' (especially at higher levels) can be rather vague - the advice that I would offer to someone scoring Q46 would differ significantly from the advice I would offer to someone scoring Q50. This is meant to say that if you want an accurate assessment of your GMAT skills, then you should take a new, FULL-LENGTH CAT and do so in a realistic fashion (take the FULL CAT - with the Essay and IR sections, take it away from your home, at the same time of day as when you'll take the Official GMAT, etc.). While the 6 Official GMAC CATs are best, any of the CATs from Kaplan, MGMAT or Veritas are 'close enough' to the real thing to provide a realistic assessment (assuming that you take the CAT in the proper fashion). Studies: 1) How long have you studied? 2) What study materials have you used? 3) How have you scored on each of your CATs (including the Quant and Verbal Scaled Scores for each)? Goals: 4) What is your goal score? 5) When are you planning to take the GMAT? 6) When are you planning to apply to Business School? 7) What Schools are you planning to apply to? GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ GMAT Tutor Joined: 24 Jun 2008 Posts: 1527 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 15 Aug 2017, 17:20 GMATFocus uses real GMAT questions and the real algorithm, so the scoring is 'accurate', and will be more accurate than any company test. As Ajitesh points out above, some of the Focus questions have been published elsewhere (unless they revised the question bank recently), so if you knew some questions in advance, then your score might be inflated. One issue with the Focus tests is that they're shorter than the real GMAT, which is why they give score ranges. The margin of error on any score the Focus tests produce is too large for them to honestly report a pinpoint score, because the tests are a bit too short. Don't always take the midpoint of the score range as your score; you could very well be at the top of the range, or at the bottom, depending, and you should combine the information from a Focus test with your scores on other official tests to get a better idea of your level. If you've done some study since taking the GMATPrep tests, and didn't recognize any Focus questions, it seems like you've improved to at least a Q46 level. If you did know some questions, that might not be the case though. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Intern Joined: 11 Dec 2016 Posts: 4 Location: United States GMAT 1: 680 Q42 V41 GPA: 3.42 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 15 Aug 2017, 17:39 EMPOWERgmatRichC wrote: Hi cananalyst, Score 'ranges' (especially at higher levels) can be rather vague - the advice that I would offer to someone scoring Q46 would differ significantly from the advice I would offer to someone scoring Q50. This is meant to say that if you want an accurate assessment of your GMAT skills, then you should take a new, FULL-LENGTH CAT and do so in a realistic fashion (take the FULL CAT - with the Essay and IR sections, take it away from your home, at the same time of day as when you'll take the Official GMAT, etc.). While the 6 Official GMAC CATs are best, any of the CATs from Kaplan, MGMAT or Veritas are 'close enough' to the real thing to provide a realistic assessment (assuming that you take the CAT in the proper fashion). Studies: 1) How long have you studied? 2) What study materials have you used? 3) How have you scored on each of your CATs (including the Quant and Verbal Scaled Scores for each)? Goals: 4) What is your goal score? 5) When are you planning to take the GMAT? 6) When are you planning to apply to Business School? 7) What Schools are you planning to apply to? GMAT assassins aren't born, they're made, Rich Thanks Rich, I have done both realistic settings and home settings where I only practiced one session. With regard to your questions: Studies: 1) How long have you studied? 5 months 2) What study materials have you used? Target Test Prep, OG Guide, Manhattan SC Guide, Kaplan 800, GMATPrep, GMATFocus 3) How have you scored on each of your CATs (including the Quant and Verbal Scaled Scores for each)? 45/44, 42/38, 32/41, 40/42, 35/41, 38/44 Goals: 4) What is your goal score? Dream Score: 44/44 Happy with anything above 700 though 5) When are you planning to take the GMAT? Well, later this week but still time to re-take before Round 1 apps 6) When are you planning to apply to Business School? Round 1 (Sept 14-Oct. 4) 7) What Schools are you planning to apply to? Top Tier: Yale, CBS Second Tier: NYU & Darden Third Tier: McDonough, McCombs, Owen EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14198 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 15 Aug 2017, 21:54 1 Hi cananalyst, If you're taking the GMAT sometime in the next few days, then you really shouldn't try to do too much additional studying before then. A bit of light study and/or review would be fine, but NO 'cramming' (and you should not plan to take any more practice CATs). You would be better served by getting some extra rest and going into Test Day calm and ready to work. It could be that you'll be all set after your Exam, but you should plan to post back here with the results regardless. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Manager Joined: 27 Jan 2013 Posts: 239 GMAT 1: 780 Q49 V51 GMAT 2: 770 Q49 V47 GMAT 3: 760 Q47 V48 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 16 Aug 2017, 08:01 1 Hi Cananalyst, Good question! There's a lot of mystery surrounding the GMAT Focus quizzes. I find the score ranges accurate. As Ian pointed out, GMAT Focus uses the official scoring algorithm. Question overlap with the official guide used to be a problem. A great majority of the overlap has been eliminated. So for people using a relatively recent version of the guides (2017+) you should see very few if any repeats (if using older guides you could just avoid the repeats. There's a thread somewhere pointing them out). GF is expensive on a question per \$ basis but in my mind the GMAT Focus quizzes are worth it. They have a whole host of tough questions and keep you honest on the timing side of things. They're great for the last couple of weeks of preparation or for a re-take. Shoot over any follow up questions! Happy Studies, A. _________________ "It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land GMAT vs GRE Comparison If you found my post useful KUDOS are much appreciated. Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html Intern Joined: 11 Dec 2016 Posts: 4 Location: United States GMAT 1: 680 Q42 V41 GPA: 3.42 GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 16 Aug 2017, 17:16 Thanks everyone for your insight, here's how the real deal went today: 680 (42Q/41V) ......7 IR for those interested. ----Quant felt like quant. I didn't see any concepts I felt blindsided by, I may have over-invested in a few but only toward the end where I felt like I had banked some time. I budgeted four outright guesses and used them. ----For Verbal I started getting a little fatigued and may have fallen into relying on my ear for SC and maybe rushed a little on some CR questions. Finished with ample time. ----I chose to tackle the test in the QUANT>VERBAL>IR>AWA order. I didn't cancel my score and was fine "banking" a 680 for now but would like to give it a few weeks and try again. I can't absorb anymore math concepts. I have finished the Target Test Prep curriculum and feel like I have a strong enough understanding of the math concepts I need to change the way I am approaching the quant questions more broadly. Any advice on the best way to spend the next few weeks would be greatly appreciated! CEO Joined: 15 Jul 2015 Posts: 2639 Location: India GMAT 1: 780 Q50 V51 GRE 1: Q170 V169 Re: GMAT Focus Score Range Accuracy?  [#permalink] Show Tags 17 Aug 2017, 02:14 cananalyst wrote: Thanks everyone for your insight, here's how the real deal went today: 680 (42Q/41V) ......7 IR for those interested. ----Quant felt like quant. I didn't see any concepts I felt blindsided by, I may have over-invested in a few but only toward the end where I felt like I had banked some time. I budgeted four outright guesses and used them. ----For Verbal I started getting a little fatigued and may have fallen into relying on my ear for SC and maybe rushed a little on some CR questions. Finished with ample time. ----I chose to tackle the test in the QUANT>VERBAL>IR>AWA order. I didn't cancel my score and was fine "banking" a 680 for now but would like to give it a few weeks and try again. I can't absorb anymore math concepts. I have finished the Target Test Prep curriculum and feel like I have a strong enough understanding of the math concepts I need to change the way I am approaching the quant questions more broadly. Any advice on the best way to spend the next few weeks would be greatly appreciated! 680 is a very good score, and you should keep it. I'm also very impressed by the fact that you got a 41 on verbal even though you felt tired. Make sure that you're well rested for the next attempt, because you shouldn't normally be tired if you're not going with the traditional order. You might need to continue working on your quant concepts though. I don't know enough about you to rule out any other reasons for your 42, but a 42 on quant usually means that the test taker is not truly comfortable with the concepts being tested. Go ahead and make those changes in approach that you mentioned, but I think it will be good if you keep working on the basics as well. _________________ Re: GMAT Focus Score Range Accuracy?   [#permalink] 17 Aug 2017, 02:14 Display posts from previous: Sort by GMAT Focus Score Range Accuracy? Moderator: HKD1710 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
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## Frequent Practice of Basic Computation Skills Helping children learn the basic facts is an important goal in the Everyday Mathematics curriculum. Most children should have developed an automatic recall of the basic addition and subtraction facts by the end of the second grade. They should also know most of their 1, 2, 5, and 10 multiplication facts by this time. By the end of the third grade most students should have an automatic recall of all the basic multiplication facts and be familiar with the basic division facts. Multiplication and division facts are reinforced at the beginning of fourth grade. ### Developing Basic Fact Power The Everyday Mathematics curriculum employs a variety of techniques and contexts to help children develop their "fact power", or basic number-fact reflexes. ### Practice through Games The curriculum has a wide variety of fact-practice games. Children find these games much more engaging than standard drill exercises, so they are willing to spend more time practicing their basic facts. For more information on games, please visit our sample games page. ### 50-Facts Multiplication Tests Beginning in fourth grade, students take timed tests on multiplication facts about every three weeks. Students calculate their percentage correct and track their progress with line graphs. ### Choral Drills and Mental Math Exercises Beginning in first grade, short oral drills are suggested for fact review. In third through sixth grades, basic fact power is reinforced in routine mental math exercises called Mental Math and Reflexes. ### Fact Extension Practice Fact extensions are calculations made with larger numbers using knowledge of basic facts. If children know that 3 + 4 = 7, then they also know that 30 + 40 = 70, and 300 + 400 = 700. Children are introduced to fact extensions in first grade and are encouraged to practice them throughout the program. ### Fact Triangles Fact Triangles are Everyday Mathematics' version of flash cards. But Fact Triangles are more effective than flash cards because they help children learn fact families rather than isolated facts. Partner practice with Addition/Subtraction Fact Triangles begins in first grade. Multiplication/Division Fact Triangles are introduced in second grade. ### Frames and Arrows Diagrams These diagrams are visual representations of rule-based sequences of numbers. Variations of these diagrams are used routinely from first through third grades. The challenge of filling in the blank frames involves lots of practice with basic facts. Home Link homework assignments are included with every lesson and have many opportunities for basic fact practice built-into the suggested activities. ### What's My Rule? Function Machines Variations of these function machines are used routinely from kindergarten through sixth grade, and provide another avenue for basic fact practice. Home Link homework assignments are included with every lesson and have many opportunities for basic fact practice built-into the suggested activities. ### Teaching Everyday Mathematics Access guides to assessment, computation, differentiation, pacing, and other aspects of Everyday Mathematics instruction. ### Everyday Mathematics Virtual Learning Community Join the Virtual Learning Community to access EM lesson videos from real classrooms, share EM resources, discuss EM topics with other educators, and more. ### Professional Development The UChicago STEM Education offers strategic planning services for schools that want to strengthen their Pre-K–6 mathematics programs. ### On the Publisher's Site McGraw-Hill Education's website features supplemental materials, games, assessment and planning tools, technical support, and more.
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# Video: Solving Logarithmic Equations by Factorization What is the solution set of the equation log_(π‘₯) (9π‘₯ βˆ’ 18) = 2? 02:10 ### Video Transcript What is the solution set of the equation log to the base π‘₯ of nine π‘₯ minus 18 equals two? So to help us solve this, what we can use is some of our relationships for logarithms. So, if we had log to the base 𝑏 of π‘₯ equals π‘˜, well, then what we can say is that π‘₯ is equal to 𝑏 to the power of π‘˜. So what we’re gonna do now is take a look at our equation and see how we can change it into exponent form. So first of all, what we’re gonna do is take a look at the corresponding parts from our equation to the relationship we looked at. So we’ve got the 𝑏 in the relationship is gonna be our π‘₯. The π‘₯ is gonna be our nine π‘₯ minus 18. And the π‘˜ is gonna be our two. And we can do this because our equation is in that form, log to the base 𝑏 of π‘₯ equals π‘˜. So then what we can do is actually convert this into our exponent form. So, what we’re gonna get is nine π‘₯ minus 18 is equal to π‘₯ squared. Then what we can do is rearrange this to make it a quadratic that we know how to solve. So, if we subtract nine π‘₯ and add 18 to both sides of the equation, we’re gonna get zero is equal to π‘₯ squared minus nine π‘₯ plus 18. And this is actually a nice, straightforward quadratic for us to solve because what we can do is use factoring to solve it. So now, in order to actually factor our quadratic, what we need to do is find two values whose product is positive 18 and whose sum is negative nine. Well, these are negative three and negative six. So we get zero is equal to π‘₯ minus three multiplied by π‘₯ minus six. And then to check why we’ve got that, we do negative three multiplied by negative six, which is positive 18, and negative three add negative six gives us negative nine. So therefore, our π‘₯-values are gonna be three or six. And that’s cause these are the values that make our parentheses equal to zero. So, for instance, if we have three minus three, this is zero, and zero multiplied by anything is just zero. We want that on the right-hand side because the left-hand side is zero. So therefore, we can say that the solution set of the equation log to the base π‘₯ of nine π‘₯ minus 18 equals two is three and six.
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# Devices for Computation • Mika Hirvensalo Part of the Natural Computing Series book series (NCS) ## Abstract To study computational processes, we have to fix a computational device first. In this chapter, we study Turing machines and circuits as models of computation. We use the standard notations of formal language theory and represent these notations briefly now. An alphabet is any set A. The elements of an alphabet A are called letters. The concatenation of sets A and B is a set AB consisting of strings formed of any element of A followed by any element of B. Especially, A k is the set of strings of length k over A. These strings are also called words. The concatenation w 1 w 2 of words w 1 and w 2 is just the word w 1 followed by w 2. The length of word w is denoted by |w| or (w) and defined as the number of the letters that constitute w. We also define A 0 as the set that contains only the empty word that has no letters, and A* = A 0A lA 2 ∪... is the set of all words over A. Mathematically speaking, A* is the free monoid generated by the elements of A, having the concatenation as the monoid operation and as the unit element. ## Keywords Turing Machine Quantum Circuit Quantum Gate Toffoli Gate Boolean Circuit These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
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# When should we get into limits in introductory calculus courses? All of the calculus textbooks I've used (teaching at community colleges) start with the first chapter covering limits. (Perhaps after a review chapter.) I think this order is wrong. Historically, Newton and Leibniz thought in terms of fluxions or infinitesimals. It took mathematicians 150 years to develop the logical (epsilon-delta) machinery of limits that puts calculus on a sound logical foundation. (And it turns out that this is not the only way to do so. Non-standard analysis uses infinitesimals in a logically rigorous way.) The big ideas of calculus are derivatives for rate of change, and integration for areas and volumes. I think the course makes more sense for students if we start with "slopes of curvy lines" and velocity. So I'm bringing in outside material to help them explore the basic concept of the derivative from many perspectives. (Our textbook only has two sections that do this. Most only have one. I spend three weeks on it. Much of my material comes from Boelkins' Active Calculus.) I give the limit definition of the derivative, but I say that for now we'll think of the limit as meaning that we get infinitely close. After we are done with simple derivatives, I go back to chapter one's limit exercises before doing trig derivatives, because we hit some unusual limits there (e.g. sinx/x). I'd like to know why textbooks cover limits first. Is it just because the mathematicians writing them are stuck at a higher level, and want to first establish the validity of what will be done later? I don't think that's good pedagogy. I think calculus is beautiful and powerful, and that I can convey this better by waiting to introduce limits after students see why they need them. Why cover limits first? • On a whim, I thought I might google the phrase Calculus without limits. Indeed, a PDF of potential interest shows up: hawkeyecollege.edu/webres/File/employees/faculty-directory/… Commented Sep 5, 2014 at 16:33 • It's brilliant! I've seen the textbook using infinitesimals, but it didn't seem any easier for students than the limit treatment, and it's not the standard they'll see elsewhere. But this is great! Short enough for interested students to read. And a great way to show that you can "get away with" thinking about infinitely small bits. (There's a great proof of (sinx)' = cosx that uses infinitesimals. Takes it from my four-page explanation to about one page: thephysicsvirtuosi.com/posts/trigonometric-derivatives.html) Commented Sep 5, 2014 at 17:13 • It's another case of the "question answered before it is asked" approach to education that plagues our system. For those who think learning is just remembering everything you're told, this makes some sense. For the rest of us who think the students actually have to be engaged in meeting and addressing problems, it does not. Commented Sep 6, 2014 at 13:05 • @rschwieb, I'm not sure what 'it' refers to here. My approach, or the text's? Commented Sep 6, 2014 at 22:04 • @SueVanHattum I was referring to most run-of-the-mill texts for students at that level and lower. Commented Sep 6, 2014 at 23:29 I'd like to know why textbooks cover limits first. I don't think there's any big mystery as to why commercial textbooks tend to be similar. It's a market mechanism known as the network effect, the same mechanism that makes Microsoft Windows so popular. Once people start to see something as a standard, anything different becomes non-viable in the marketplace. Stephen J. Gould wrote a nice essay about this in the field of biology; he uses the example of how biology texts propagate the meme of Lamarckian and Darwinian explanations of the giraffe's neck. To see that there is no objective reason for doing limits first, we just have to look at some older textbooks. Hutton 1807 uses fluxions. Davies 1836 uses limits, but also connects with infinitesimals. Granville 1904 uses limits. Thompson 1910 uses infinitesimals. Different authors did different things, which is how it should be. There are also some modern texts that don't start with limits and then move on to derivatives: Marsden 1981, Keisler 1976. At the risk of seeming self-promoting, here's a link to a book that I'm a coauthor of. There may have been a time ca. 1850-1965 when limits were believed to be the only possible rigorous foundation for calculus. That era ended when NSA essentially vindicated Leibnizian infinitesimals.[Blaszczyk 2012] Other than the network effect, a second and more cynical explanation would be that a course in calculus is used today as a filter in order to get rid of some of the students seeking a valuable credential. This is the phenomenon of credential creep. For example, I teach quite a few students who want to be physical therapists, and many DPT programs require them to take a year of calculus. Calculus is fundamentally an easy subject, but it can be made much harder by emphasizing epsilontics and by requiring these folks to learn a bag of tricks for integration. I'm not a cynic at heart, but I do have a very hard time coming up with any other explanation for why we require future physical therapists to learn how to do integrals using trig substitutions. Blaszczyk, Katz, and Sherry, "Ten Misconceptions from the History of Analysis and Their Debunking," 2012, http://arxiv.org/abs/1202.4153 Charles Davies, Elements of the Differential and Integral Calculus (1836), https://archive.org/details/elementsdiffere03davigoog Granville, Elements of the Differential and Integral Calculus (1904), https://archive.org/details/elementsdiffere01smitgoog Charles Hutton, A Course of Mathematics: In Two Volumes. For the Use of Academies as Well as Private Tuition (1807), https://archive.org/details/acoursemathemat02huttgoog Keisler, Elementary Calculus: An Infinitesimal Approach, 1976, http://www.math.wisc.edu/~keisler/calc.html Marsden, Calculus Unlimited, 1981, http://www.cds.caltech.edu/~marsden/books/Calculus_Unlimited.html Thompson, Calculus Made Easy, 1910, http://www.gutenberg.org/ebooks/33283 • I'll look at your text. I love Boelkins for my first unit. And I mostly use our official text (Anton) for the rest, though I put things in a different order. I'd love to put together my own textbook, though I don't know how many others would use it. Commented Sep 6, 2014 at 5:59 • I think your approach is very... american. I was always told that we study for ourselves and not for our job: it is clearly a cultural difference between US and old Europe :-) Commented Sep 6, 2014 at 10:07 • @Siminore: If a student takes a year of calculus simply because he likes math, that's great, but requiring it is a different matter. Similarly, it's wonderful if a future physical therapist wants to learn Homeric Greek -- but I don't think it should be required. – user507 Commented Sep 6, 2014 at 23:13 • @BenCrowell Actually the university system in my country used to be less focused on future job perspectives than it is in your country. Commented Sep 7, 2014 at 8:23 • How is assuming calculus is used for weeding out students cynical? That seems like a perfectly sensible use for it. Even with the (extremely miniscule amounts) of epsilon-delta definitions of limits it is, as taught in the US, a fairly easy subject for anyone willing to think a little and do some work. If you do more work you don't really even have to think. Having barrier subjects is quite usual I believe to make sure you don't spend time and money on students who aren't willing to do the work. – DRF Commented May 4, 2017 at 7:43 My currently preferred approach is to start the course with an introductory lecture explaining the difference between average velocity over a time interval (something we can always in principle measure using a stopwatch) and instantaneous velocity at a given instant of time (which we do not have a means to measure). This motivates the question of what exactly instantaneous velocity is, using the intuition that most of us have that the notion makes sense. (Of course, that in and of itself is a philosophical issue and one that I want to avoid in the course for the obvious reason of limited time!) Then I explain that one can approximate instantaneous velocity by measuring the average velocity over smaller and smaller intervals of time centered around the given instant of time. (In fact, I never actually use a centered time interval, but rather use the formulation with the instant of time as one endpoint of the time interval, so that the formulas I write down look like what the definition of derivative will be when we define it officially later on.) Then I say that the instantaneous velocity is the number that we would get in the limit where the time interval became closer and closer to $0$. And I emphasize that, of course, we can never actually make a measurement with a time interval of duration $0$ (both physically because we can't start and stop our stopwatch at the exact same instant of time and mathematically because using the same starting and ending time in the formula for average velocity yields the undefined expression "$0/0$".) This serves to motivate the introduction of limits, and in fact, in the courses I teach, typically this is more or less the definition of a limit. I use the heuristic and imprecise version of $\lim\limits_{x \rightarrow a} f(x) = L$ means that as $x$ gets closer and closer to $a$ without actually reaching $a$, the values $f(x)$ get closer and closer to $L$. For me, this seems to strike the right balance of motivating limits and discussing how to compute them in the types of cases that will come up throughout the semester, while still being able to quickly get to derivatives so that most of the semester is spent on the two fundamental notions of derivative and integral and, ultimately, the link between the two provided by the Fundamental Theorem of Calculus. I strongly prefer to follow pretty closely to the textbook's development in a course such as introductory calculus, where I feel that students are often not sufficiently mature to handle the discrepancy between the textbook presentation and the presentation given in lecture. And they are not equipped to handle the inevitable situation where the textbook has referred to concept X in Section A which appears before Section B and I want to cover the ideas in Section B before covering concept X, but too much of the presentation and choice of problems refer to or are based on understanding concept X. As for why I prefer to discuss limits before derivatives and integrals (neglecting the obvious cultural bias that that is the way I learned the subject), since I cannot speak for textbook publishers, to me there are two big pictures to take away from calculus. One is the study of change (derivatives) and accumulation (integrals) and the fact that the two problems are intimately related (Fund. Thm. of Calc.). The other is that derivatives and integrals, key ideas for understanding our dynamic natural world, are best understood computationally as the limiting process of simpler calculations that only require basic arithmetic (difference quotients for derivatives, Riemann sums for integrals). It should be noted that the subjective value judgment indicated by the use of the word "best" is not universally agreed upon. In my opinion, these two big pictures should both be presented and stressed to the students. I find it easiest, logically and pedagogically, to present derivatives and then later integrals to model independent processes and stress the point of view that these are both limiting processes. As such, I need to have discussed what a limit is. After defining integrals as limits of Riemann sums, I point out that the fact that integration, like differentiation, is a limiting process is one of two reasons why the two topics are both covered in a common course. And then I foreshadow what is to come by alluding to the fact that there is a deeper and more important link between these two concepts, which I explain a few lectures later when we are ready to discuss the Fundamental Theorem of Calculus. • I tend to agree with your post here. I think the case can be made for the necessity of limits by discussing the speedometer on a car. By discussing how we want to measure rates of change over smaller and smaller durations of time it naturally leads to concept of a limit. Moreover, the fact that it will have the form $0/0$ is manifest. I then tell them that is why we are going to study limits with a focus on tricky ones which have this indeterminant form. I also try to connect it to geometry as many have physics-phobia... Commented Sep 7, 2014 at 2:35 • I agree very much that the derivative should motivate the definition of the limit, and not vice versa, but I would go farther: in an introductory calculus course, why introduce the limit as a separate concept at all? (I don't think 'continuity' is a satisfactory answer; no book I've seen gives any interesting discussion of continuity beyond what can be 'seen' from the intuitive description. Better to differentiate x^2sin(1/x) than to test x sin(1/x) for continuity!) Commented Sep 7, 2014 at 6:33 • @L Spice I think one good reason to do limits first and discuss continuity is to give the first example of linearity, products, quotients, composites. The inheritance for new functions from old. These are interesting for continuous functions. Also, just to have a language to ubiquitously say when the function is continuous you can just plug in the limit point. It's a baby example of structure and definitions. Is it entertaining? Probably not. Commented Sep 7, 2014 at 17:59 • @JamesS.Cook, differentiable functions also exhibit interesting behaviour under the operations you mention. Basically, I think that continuous but differentiable functions are not interesting objects to study in their own right in a calculus course. Saying that continuous functions are ones for which you can plug in the limit point is unconvincing for someone (like me) who argues that limits as independent objects also do not belong in an introductory calculus course! Commented Sep 7, 2014 at 23:50 • @LSpice maybe you should write a question which presents your view of what calculus I should cover, perhaps with a sample plan and pointed details of how definitions are altered as to avoid limits etc. It might be productive to understand our difference of opinion. Commented Sep 8, 2014 at 2:52 It has become almost a dogma that the math curriculum should teach technical prerequisites to what will be covered later. The consequence is that zillions of high-school students learn algorithms for doing partial-fractions decompositions and will never take later courses in which that is used. In effect the broad public learns that mathematics consists of learning to apply meaningless algorithms. High-school teachers don't know how to alter the curriculum to something that makes more sense, and math professors don't want to do things for students who are so weak that they think mathematics consists of learning to apply meaningless algorithms, which is exactly the lesson that the high-school curriculum designed by math professors taught them. The students know (and, as this applies to most students, they are right) that if they just learn the meaningless algorithms they will get good grades, but if they heed any hints that there something else to math besides that, that won't help their grades and takes time away from things that would. The professors fail to realize that what they perceive as stupidity among the students is in fact deliberate strategic stupidity; it is in fact a good strategy for what the students want to do. So both the students and the professors are deliberately stupid. That is why following conventional curricula merely because they are conventional is contemptible. Stewart's textbook covers $\qquad$(1) $\displaystyle\lim_x \frac{f(x)}{g(x)}$ where $f(x),g(x)\text{ (both)} \to0$ and $\qquad$(2) $\displaystyle \lim_x \frac{f(x)}{g(x)}$ where $g(x)\to\ell\ne0$ and $\qquad$(3) $\displaystyle\lim_x \frac{f(x)}{g(x)}$ where $f(x),g(x) \text{ (both)}\to\infty$ and $\qquad$(4) $\displaystyle\lim_x (f(x) - g(x))$ where $f(x),g(x) \text{ (both)} \to\infty$ and $\qquad$(5) one sided limits of artificial piecewise defined functions and $\qquad$(6) infinite limits and $\qquad$(7) various ways in which limits fail to exist and a lot of other possibilities. Guess what students learn from this? They do not learn that case (1) above plays a central role in differential calculus. They write on the final exam that in a particular problem the limit does not exist because the numerator and denominator both approach $0$, and the fact that the whole of differential calculus could not exist if that were true does not occur to them; after all point (1) above; they do not learn that limits justify formulas about derivatives, since as far as they know, math is dogmatic rather than having justifications. The practice of teaching technical prerequisites first is hardly, if at all, distinguishable from teaching that mathematics is dogmatic. It is wrong. Probably at least $99.99\%$ of those who would tell you to cover limits first have no reason for saying that except that they've never given the question an instant's thought after observing that the definition of "derivative" in its most usually seen form relies on a concept of limits. Regarding your parenthetical comment "And it turns out that limits are not the only way to do so. Non-standard analysis uses infinitesimals in a logically rigorous way" I would like to comment that the opposition limits vs infinitesimals implied here is not entirely accurate. Limits are present in both approaches; the true opposition is between epsilon-delta methods in the context of an Archimedean continuum, and infinitesimal methods in the context of a Bernoullian (i.e., infinitesimal-enriched, as practiced by Bernoulli) continuum. In Robinson's framework, the limit is defined in terms of the standard part function. Thus, $\lim_{x\to 0} f(x)$ is the standard part of $f(\epsilon)$ when $\epsilon$ is a nonzero infinitesimal. • I like your precision of language. If you can propose a minor edit to fix this in my post, I'd be grateful. I do want to set up an opposition between the standard (epsilon-delta) limits material and infinitesimals. Mainly because I love this proof: thephysicsvirtuosi.com/posts/trigonometric-derivatives.html Commented Apr 24, 2017 at 19:28 • @SueVanHattum, thanks. I will try to make a precise edit to the question. It will have to be approved because I am still below 1000. Commented Apr 25, 2017 at 7:31 • I wanted to make less distinction than you do, so I made a smaller edit. Thanks. Commented Apr 26, 2017 at 13:42 • @SueVanHattum, no problem. Commented Apr 26, 2017 at 13:46 I think this book hasn't been mentioned yet: John C. Sparks: Calculus without Limit - Almost [Disclaimer: I have it on my bookshelf but have to admit that I haven't read it yet, so I can't really comment on whether it's good or not.] I am not advocating any order but one reason to start with limits first is that the concept of continuity is extremely important and useful and limits are the natural way to introduce it. • Why is continuity important at the beginning of the course? Commented Sep 6, 2014 at 22:02 • After asking SueVanHattum's I think very apposite question, I would ask another: who says that limits are the natural way—or, more to the point, more natural for whom? If for students, then I disagree; but if for the instructor, then who cares? Commented Sep 7, 2014 at 6:35 • @SueVanHattum : it is important at the beginning if you think the concepts of continuity/discontinuity are more important than rate of change (derivatives) or areas (integrals). To apply mathematics to real life (in physics, biology or economics) we have to make lots of simplifications and use approximations. Continuity is crucial to guarantee the approximations still "work" but not all phenomena are continuous. I would like my students to have a critical eye and not assume all functions are continuous. Commented Sep 7, 2014 at 14:24 • @SergioParreiras: If not using limits how would you introduce/teach continuity? The notion of continuity, in some form, predates the notion of a limit by 2000 years, so certainly there are other ways to do it. The traditional informal description is that a function is continuous if you can draw it without picking up your pen. At a more formal level (which is irrelevant to the vast majority of students), there are also multiple ways of defining continuity. See Keisler, math.wisc.edu/~keisler/calc.html , p. 125, for a freshman-level example. In SIA it's an axiom. – user507 Commented Sep 7, 2014 at 15:56 • You write that limits are the natural way to introduce continuity but what evidence do you have for such a hypothesis? The historical evidence points to the contrary conclusion; namely, Cauchy who invented continuity defined it without reference to limits, by requiring that every infinitesimal change in the $x$-variable should always produce an infinitesimal change in the $y$-variable. Commented Apr 23, 2017 at 11:53 You ask why to cover limits prior the derivatives when it would be easier to cover derivatives first and limits later; to show why are limits good to know. Why shall we cover derivatives prior without [anything that uses derivatives]? I think it is easier to define derivatives when the students are "familiar" with limits. Same for integrals with the knowledge of derivatives. What really matters is HOW does the teacher present the part that is taught. One shall show the students an idea where the course is about to go and why. I accidentaly passed all the exams in a highschool and thought math is one of the most boring and useless subjects. Man, how wrong I was. At the university I was lucky to have the teacher I had. He followed the syllabus in terms: 1. Setting up the "vocabulary" (lemma, proof, set,...) and "tools" (proof,...) 2. Operations on sets and basics of boolean logic. 3. Sequence. 4. Limit of a infinite sequence. (What happens at the end of the universe?) 5. Series. 6. Sum of infinite series. (What happens when we try to sum infinite ammount of numbers?) 7. Function. 8. Limits of a function. (What happens at the end? And what happens rally closeto some point?) 9. Derivatives. 10. Integrals. He taught us the way that it was interesting and fun. I cannot forget his "mutated definitions" that were even in tests. He randomly changed quantificators and relation in a definition and ask us to find a sequence/seires/function that fits in. He forced us to think out of the box and use any tool available. Once you have students' passion, you can either lead them trough blind phase (limits prior derivatives) and they will follow you or directly to the finish (derivatives) and explain all the tools behind later.
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# second day of week Learn about second day of week, we have the largest and most updated second day of week information on alibabacloud.com Related Tags: ### Calculate the dayof the week How to calculate the day of the week?Slowtiger published on 2009-11-19 08:42:00How to calculate the day of the week?--Caille (Zeller) formulaWhat is the day of the week in history? What day of the future is the week? With regard to this problem, ### C language detailed Jiezzaile (Zeller) formula calculates the dayof the week is extremely convenient --Caille (Zeller) formulaWhat is the day of the week in history? What day of the future is the week? With regard to this problem, there are many computational formulae (two general formulas and some piecewise formulas), the most notable of which is ### SQL gets the current date, year, month, day, week, time, minute, second Select GETDATE() as 'Current Date', Datename( Year,GetDate()) as 'years', Datename(Month,GetDate()) as 'Month', Datename( Day,GetDate()) as 'Day', Datename(DW,GetDate()) as 'Week', Datename(Week,GetDate()) as 'Week number', ### MySQL Query day, query one week, query one months of data "go" Transferred from: http://www.cnblogs.com/likwo/archive/2010/04/16/1713282.htmlQuery Day:SELECT * FROM table where to_days (column_time) = To_days (now ());SELECT * FROM table where date (column_time) = Curdate ();Query one week:SELECT * FROM table ### Oracle Date minute solution and week start end time calculation 1 ORACLE mid-week related knowledge description 1.1 Date formatting functions To_char (X [,FORMAT]) : Press X toFORMATThe format is converted into a string. X is a date,FORMATis a format string that specifies the format in which X is Trending Keywords: ### mysql-Query Day, week, January, year, and MySQL basic date function Query Day:Select * from Table where = To_days (now ()); Select * from Table where =Query one week:Select * from Table where 7 Day Date (column_time);Query one months:Select * from Table where 1 MONTH Date (column_time);MySQL date and time ### Enter the first letter of the dayof the week to determine the dayof the week. If the first letter is the same, continue to judge the second letter. Def juge (Num, week_list): W = input ('enter the % s letter' % num) W = W. lower () RES = [] State = 0 for week in week_list: If W = week [0 + num-1: 1 + num-1]: State + = 1 res. append (week) If State = 1: Print ('% s' % res [0]) Elif State> 1: ### ASP now function, ASP time function detailed, ASP time function Daquan, ASP week function The date () function gets the dates, The time () function gets the times.The now () function can get the current date plus time. Now () gets the current system date and time, the ASP output can be written like this: Years (now ()) Gets the year, ASP ### SQL Query Day, this month, this week's record SELECT * FROM table WHERE convert (Nvarchar, DateAndTime, 111) = CONVERT (Nvarchar, GETDATE (), 111) ORDER by DateAndTime DESCRecord of the Month SELECT * FROM table WHERE DateDiff (Month,[dateadd],getdate ()) =0 Week ### MySQL gets a variety of SQL statements for day, week, January time data To create a table:The code is as follows:CREATE table if not exists t(ID int,Addtime datetime default ' 0000-00-00 00:00:00′)Add two initial data:Insert T VALUES (1, ' 2012-07-12 21:00:00′);Insert T VALUES (2, ' 2012-07-22 21:00:00′);Data inserted Related Keywords: Total Pages: 15 1 2 3 4 5 .... 15 Go to: Go The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email. If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days. ## A Free Trial That Lets You Build Big! Start building with 50+ products and up to 12 months usage for Elastic Compute Service • #### Sales Support 1 on 1 presale consultation • #### After-Sales Support 24/7 Technical Support 6 Free Tickets per Quarter Faster Response • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.
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# Understanding rnn_forward in utils.py of W1 A2 sampling dinosaur names This is the code provided by the exercise, but I don’t understand it at all. What’s the difference between the X and x? What is len(X), is it the number of time steps or the length of features, i.e. the possible number of characters in this case 27?? ``````def rnn_forward(X, Y, a0, parameters, vocab_size=27): # Initialize x, a and y_hat as empty dictionaries x, a, y_hat = {}, {}, {} a[-1] = np.copy(a0) # initialize your loss to 0 loss = 0 for t in range(len(X)): # Set x[t] to be the one-hot vector representation of the t'th character in X. # if X[t] == None, we just have x[t]=0. This is used to set the input for the first timestep to the zero vector. x[t] = np.zeros((vocab_size, 1)) if (X[t] != None): x[t][X[t]] = 1 # Run one step forward of the RNN a[t], y_hat[t] = rnn_step_forward(parameters, a[t - 1], x[t]) # Update the loss by substracting the cross-entropy term of this time-step from it. loss -= np.log(y_hat[t][Y[t], 0]) cache = (y_hat, a, x) return loss, cache `````` We wrote code very similar to this in the RNN Step by Step exercise that was the first assignment in Week 1 of Sequence Models, right? It might be worth comparing this code to what you wrote earlier. Yes, len(X) is the number of “timesteps” or elements in the sequence, not the number of features. That is vocab_size, right? The difference between x and X is explained in the comments: x is the “one hot” version of X initially. Thanks for your reply Paul. Yes, I did compare with the function I wrote, but I still don’t understand this one. So does it mean that the X here is not yet one hot coded? In the previous exercise it was already one hot coded. X has the dimension of (n_x, m, T_x), right? n_x here is vocab_size. Since in the comment, it says “x[t] to be the one-hot vector representation of the t’th character in X.”, so it looks like t is the number of training examples m. So X in this case has dimension of (m, T_x), so X[t] has one dimension (1, T_x). x[t] has dimension of (n_x, 1). But I’m lost again here: X[t] != None. how can you judge if a vector is None or not? Yes, they tell you that in the comments: X is not one-hot encoded. They are literally writing out the logic for you to create x as the one-hot encoded version of X one timestep at a time. You can print the shapes of the inputs if you want to confirm what shapes they are. The business about testing X[t] for None is also explained in the comments: the first element is apparently not set for timestep 0. Thanks so much Paul, I think I understand now. So we only consider one training example here, X is a single dimension vector containing the positions (0-26) of each of its character, could be something like [2, 6, 25, 16]. The length of X is the number of timesteps of the RNN. x is a dict with each key being each timestep of the RNN, and value being one-hot coded version of that timestep, 2 would become [0, 0, 1, 0, …, 0], a list of 27 length. But why in the first exercise the input has a dimension of (n_x, m, T_x), which m is the number of training examples in the mini-batch, and in this exercise we only use one example at a time? They specifically say in the comments that they are using Stochastic Gradient Descent here, so that means you only need to handle one sample at a time. Maybe they did that for simplification? Or maybe they have a priori knowledge that it works better than Minibatch for this particular case? I don’t know why they did it that way here. Of course the previous exercise was for the fully general case …
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showing 1-24 of 467 results Are your students starting to add fraction and subtract fractions with like denominators?This Add and Subtract Fractions and Simplify Bundle will give your students practice adding fractions with like denominators with and without simplifying (reducing fractions). Save 20% by buying this bundle! A Subjects: Types: CCSS: \$16.00 \$12.80 17 Ratings 4.0 Bundle This is a scavenger hunt that can be used as a review for a test or an enrichment activity. All problems cover either adding or subtracting fractions with like and unlike denominators. Many answers are improper, which forces the kids to simplify. There are 10 questions included. Post the questions Subjects: Types: \$3.25 36 Ratings 4.0 PDF (1010.11 KB) These task cards are unique because they TEACH and REVIEW how to subtract fractions that have different denominators. This is great for state assessment test prep, review and individualized instruction. ********Special Note: These are similar to the Common Core version of the fifth grade 5.NF.A.1 Subjects: Types: \$7.00 \$5.00 6 Ratings 4.0 PDF (1.11 MB) Fractions! Are fractions a mystery to your students? Need some extra practice? This fractions activity includes 40 task cards that deal with adding and subtracting fractions with unlike denominators. Students add or subtract, then determine if the fraction is in lowest terms, and if it isn't, they Subjects: Types: \$2.50 49 Ratings 4.0 PDF (2.11 MB) With these task/scoot cards students can practice adding and subtracting fractions with like denominators. There is no simplifying necessary. 7 word problem cards are included. These task cards can be used in a number of ways. Your students can use these cards for a board game, as a math center, Subjects: Types: \$2.50 27 Ratings 4.0 PDF (126.11 MB) What better way for students to review adding and subtracting fractions and mixed numbers with LIKE denominators at the end of the unit or before testing season than with an interactive lesson that allows students to teach one another? Find A Buddy is designed for students to teach their classmates Subjects: Types: CCSS: \$1.50 32 Ratings 4.0 PDF (1.76 MB) Practice adding and subtracting fractions with like denominators with this cut and paste activity. Great for centers! Students work the problems then cut and paste the answers in the appropriate box. This activity include the following skills: 1. Add fractions with like denominators. 2. Subtract Subjects: Types: \$3.60 28 Ratings 4.0 DOCX (33.73 KB) Mrs. Claus is having a Christmas Party before her husband leaves to deliver presents to children all over the world. She is getting her house ready for the party, but the elves keep playing tricks on her and hiding her decorations! They have left math problems for Mrs. Claus to solve so that she ca Subjects: \$1.50 22 Ratings 4.0 PDF (4.63 MB) This is a Mix n' Mingle Pack that includes: -Fraction Review Mix n' Mingle Review Activity (Skills Included: Adding and Subtracting Fractions, Mixed/Improper Fractions, Simplifying Fractions, Equivalent Fractions, Word Problems) -Fraction Quick-Check Individual Practice Sheet This activity works w Subjects: Types: CCSS: \$1.50 25 Ratings 4.0 PDF (1.34 MB) Fractions with Unlike Denominators - Addition and Subtraction Color by AnswerA fun way to practice fraction addition and subtraction! The answers must be simplified and/or changed from improper fractions to mixed numbers. This is a great practice for several different fraction skills!Students color Subjects: Types: Also included in: Fraction Games Bundle - Engaging Activities That Make Fractions Fun! \$1.75 24 Ratings 4.0 PDF (2.34 MB) With these task/scoot cards students can practice adding and subtracting fractions with unlike denominators. Simplifying is required in most problems. 7 word problem cards are included. These task cards can be used in a number of ways. Your students can use these cards for a board game, as a m Subjects: Types: \$2.50 14 Ratings 4.0 PDF (3.14 MB) This Christmas themed powerpoint game an adding and subtracting fractions game. Students will be in teams (up to 6 groups/players). All groups solve the problem on the screen. If they get it correct, they can choose an ornament. Sfter selecting the ornament, it will disapear and the number of po Subjects: Types: \$0.99 12 Ratings 4.0 PPTX (2.56 MB) Adding and Subtracting Fractions and Mixed Numbers assessment aligned with Math Common Core. This test will challenge students understanding of adding and subtracting fractions with like and unlike denominators, subtracting mixed numbers with borrowing, and simplifying fractions and improper fracti Subjects: Types: \$0.99 5 Ratings 3.9 DOCX (157.31 KB) Solve & Check cards POP with Color! This set features adding and subtracting fractions with unlike denominators. In addition, the final answer must be simplified in order to be counted as correct. Mixed number answers are included. A Solve & Check Math Skills set contains: • 32 vibrant Subjects: Types: \$3.50 5 Ratings 4.0 PDF (12.58 MB) 8 Domino Games that include both of my packets for Adding Fractions and Subtracting Fractions Dominoes Games Adding Fractions Includes: Adding fractions with like denominators and no simplifying, Adding fractions with like denominators simplifying, Adding fractions with unlike denominators, and Add Subjects: Types: \$5.00 \$3.75 3 Ratings 4.0 PDF (508.9 KB) ¬ This year, go green and let Google's powerful engine do the grading while breathing new life into your teaching of 5.NF.1 (Add & Subtract Fractions with Unlike Denominators). We cordially invite you to take a look at the PREVIEW above by clicking on the green button above this description. Subjects: Types: \$3.00 3 Ratings 3.7 PDF (13.32 MB) These Add and Subtract Fractions with Unlike Denominators Using Equivalent Fractions Organizers 1-3 allow students to practice the following common core standard: CCSS.Math.Content. 5. NF.A.1: Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fraction Subjects: Types: \$1.50 3 Ratings 4.0 DOC (60.5 KB) ¬ This year, go green and let Google's powerful engine do the grading while breathing new life into your teaching of CCSS 4.NF.3d (Adding and Subtracting Fractions Using Models). We cordially invite you to take a look at the PREVIEW above by clicking on the green button above this description.⧂⚪⧂⚪⧂ Subjects: Types: \$3.00 2 Ratings 4.0 PDF (15.15 MB) This product is a great review of finding common denominators, adding and subtracting fractions, and simplifying fractions! Includes examples of adding and subtracting unlike fractions. Great for a math station when reviewing fractions! Subjects: Types: CCSS: \$2.00 2 Ratings 4.0 PDF (1.69 MB) This assessment tests students' knowledge of finding least common denominators, adding and subtracting fractions (no mixed numbers), simplifying fractions, and real-world applications of fractions. Subjects: CCSS: \$1.50 2 Ratings 4.0 PDF (629.08 KB) Solve & Check cards POP with Color! This set features adding and subtracting fractions with unlike denominators. In addition, the student will need to simplify their answers in order for their answer to be correct. Some answers will be mixed numbers. A Solve & Check Math Skills set cont Subjects: Types: \$3.00 1 Rating 4.0 PDF (11.85 MB) Want an interactive and online way to teach your students how to add and subtract fractions with common denominators? This tour book is perfect for guiding your students step by step, using concepts and problem-solving. I have added links to online videos and activities to appeal to a variety of lea Subjects: Types: CCSS: \$5.00 not yet rated N/A PDF (1.66 MB) A visually appealing and interactive way to teach adding and subtracting fractions and mixed numbers. The notes begin by identifying key vocabulary and proceed to showcase three different ways of adding and subtracting fractions, including the use of pictures, decimals and common denominators. More Subjects: CCSS: \$5.00 not yet rated N/A PDF (2.55 MB) Students practice adding and subtracting fractions with common denominators on number lines in this card sort activity.They must match seven problems with seven number lines showing addition and subtraction. Students complete a number line to show the final problem.All fractions in this exercise hav Subjects: Also included in: Fraction Number Sense Bundle, Basics with Models and Number Lines \$2.00 not yet rated N/A PDF (363.77 KB) showing 1-24 of 467 results Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
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The volume of a sphere is V = 4/3(pi)r^3 If you differentiate with respect to r you get dV/dr = 4(pi)r^2 This is the surface area of the sphere. How come this works so well for the sphere? It doesn't work for the cube... volume of cube is V = s^3 differentiate with respect to s dV/ds = 3s^2 this is only half the true surface area 6s^2 Is there something special about the sphere, within the calculus, that makes it work out perfectly or is it just a coincidence? If you differentiate the surface area of the sphere with respect to r d(SA)/dr = 8(pi)r you get 4 times the circumference of a great circle on the sphere. It would be nice if you got 2(pi)r but you don't, you get 8(pi)r. On the other hand, differentiate the area of a circle with respect to r you get the circumference. If you differentiate the area of the square you get half the perimeter, or twice the side, either way, not as nice as the circle. You get half the surface area for the cube, you get half the perimeter for the square...coincidence? Are there other geometrical objects that work out nicely by applying differentiation or are the sphere, circle the only ones? I found that this nice relationship between volume and surface are also works for hyperspheres. http://en.m.wikipedia.org/wiki/N-sphere [end edit] 2. Re: Curiosity about surface are The procedure does work for a cube, if you apply it in the right way. In the case of a sphere, you measure the volume and surface area in terms of the radius, which is the distance from a point on the sphere to the centre. You need to do the same thing for the cube. More precisely, if you have a cube with side s, then the distance x from the centre of the cube to a midpoint of a face of the cube is given by $\displaystyle x=\tfrac12 s$, or $\displaystyle s=2x.$ The volume of the cube is $\displaystyle s^3 = 8x^3.$ Differentiate that with respect to x and you get $\displaystyle 24x^2 = 6s^2,$ which correctly tells you the surface area. The same thing works for area and perimeter in two dimensions. If you use the half-side, rather than the side, as the variable, you find that the derivative of the area of a square gives you the perimeter. , , , , , , What is the derivative of 1/8 pi r^3 Click on a term to search for related topics.
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# Sort array based on frequency ## Question: Given the array, sort the array based on the frequency in descending order. If two elements have the same frequency, then sort based on their values. Note: 3 occurs two times in array then frequency(3) = 2. Sample 1: Input : [5, 3, 2, 2, 1, 7, 9, 5, 3, 2, 1, 3] Output: [ 2, 2, 2, 3, 3, 3, 1, 1, 5, 5, 7, 9 ] ## Logic: • Create Hash-Map with elements as a key and their frequency as value • Then insert elements into the Hash-Map by calculating their corresponding frequency • Create Array List which contains the Hash-Map Values • Sort the array list based on frequency using Collections • Print the sorted value ## Java: ```import java.util.ArrayList; import java.util.Collections; import java.util.Comparator; import java.util.HashMap; import java.util.Map; import java.util.Map.Entry; public class SortArrayByFrequency { private static void sortArrayByFrequency(int[] array) { //Create HashMap with elements as key and their frequency as value Map<Integer, Integer> element = new HashMap<>(); int frequency; //Insert the array Elements into the HashMap for (int i = 0; i < array.length; i++) { //Check the key is already present in the HashMap if (element.containsKey(array[i])) { //Increment the Value if it is already present element.put(array[i], element.get(array[i])+1); } else { //Insert the Key and value as 1 element.put(array[i], 1); } } //Create ArrayList with HashMap Values ArrayList<Entry<Integer, Integer>> list = new ArrayList<>(element.entrySet()); Collections.sort(list, new Comparator<Entry<Integer, Integer>>() { @Override public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) { //Check the frequency is equal and Swap the values if(o2.getValue() == o1.getValue()) { if(o2.getKey() < o1.getKey()) { return 1; } else return -1; } return o2.getValue().compareTo(o1.getValue()); } }); //Print the Vales based on frequnency for (Entry<Integer, Integer> entry : list) { frequency = entry.getValue(); for(int count = 0; count < frequency; count++) { System.out.print(entry.getKey()+" "); } } } public static void main(String[] args) { sortArrayByFrequency(new int[] {5, 3, 2, 2, 1, 7, 9, 5, 3, 2, 1, 3}); } } ``` ## Approach 2: • Create a structure with key and value parameters • Calculate the frequency of each element and insert into the struct array of key and value • Sort the array based on frequency, if frequency is equal then sort based on key element • Print the sorted array value ## C: ```#include<stdio.h> struct map { int key; int value; }; void swap(struct map **value1,struct map **value2) { struct map *temp = *value1; *value1 = *value2; *value2 = temp; } void main() { int arr[] = {5,3,2,2,1,7,9,5,3,2,1,3}; int length = sizeof(arr)/sizeof(arr[0]); struct map mapList[length]; int i,j,index = -1; int temp; //find the frequency of each element for(i = 0; i < length; i++) { if(arr[i] != -1) { index++; mapList[index].key = arr[i]; mapList[index].value = 1; } else { continue; } for(j = i + 1; j< length; j++) { if(arr[i] == arr[j]) { mapList[index].value++; arr[j] = -1; } } } for(i = 0;i<=index;i++) printf("value:%d frequency:%d\n",mapList[i].key,mapList[i].value); //sort the array based on frequency for(i = 0; i <= index; i++) { for(j = i + 1; j <= index; j++) { if(mapList[i].value == mapList[j].value) { if(mapList[i].key > mapList[j].key) { swap(&mapList[i],&mapList[j]); } } else if(mapList[i].value < mapList[j].value) { swap(&mapList[i],&mapList[j]); } } } //print the value for(i = 0;i<=index;i++) { for(j = 0; j < mapList[i].value; j++) { printf("%d ",mapList[i].key); } } } ``` ## HashMap Java ### Vignesh A Computer Science graduate who likes to make things simpler. When he’s not working, you can find him surfing the web, learning facts, tricks and life hacks. He also enjoys movies in his leisure time. 5 1 vote Article Rating Subscribe Notify of
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# UPSEE: Important Questions and Preparation Tips - Complex Numbers Sep 12, 2017 16:38 IST UPSEE 2018 This article contains all important concepts, formulae and some important solved questions of chapter Complex Numbers. This will help all UPSEE aspirants to score good marks in the coming UPSEE examination. About 2-4 questions are asked from this chapter in UPSEE examination. This can be used as a quick revision material for UPSEE entrance examination. You will have to do a lot of practice to master this topic as this is not an easy topic. Uttar Pradesh State Entrance Examination (UPSEE) is a state level engineering entrance examination for degree level engineering institutions and other professional colleges. It is regulated by Dr. A.P.J. Abdul Kalam Technical University (AKTU) formerly known as Uttar Pradesh Technical University (UPTU). Definition:  A complex number is a number that can be written in form x + iy, where x and y are real numbers and i is an imaginary part. IIT JAM 2018: Online registration begins Some important concepts and formulae are given below: Top Engineering Colleges Other than IITs & NITs Some important solved questions are given below: Question 1: Solution 1: Hence, the correct option is (d). Mathematics Online Test Complex Numbers Question 2: Solution 2: Hence, the correct option is (b). Question 3: Solution 3: Hence, the correct option is (c). Question 4: Solution 4: Hence, the correct option is (b). Question 5: Solution 5: Hence, the correct option is (c). What parameters to check before selecting Engineering College WBJEE Physics & Chemistry Question Paper: 2015 DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article. ## Register to get FREE updates All Fields Mandatory • (Ex:9123456789)
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# Forums Join The Community Technical Battery Discussion So I am an engineer and really want to understand the battery. It is my understanding that there are 6831 NCR18650 Panasonic batteries in the "Battery pack". 11 moduals in series each with 9 "bricks" in series and each brick with 69 18650's in parallel. So 69 * 9 * 11 = 6,831. Are you with me so far? Now the hard part. On Panasonic's website i get specs for the NCR18650 showing a nominal voltage and capacity of 3.6 VDC and 2.9 AH, respectively. Lets do the voltage first. 9 bricks X 3.6 volts X 11 moduals = 356.4 volts. But the Tesla specs say the battery is 375 volts. Backing into the nominal voltage needed to get 375 volts each battery has to have a nominal of 3.78 volts. If someone knows the answer to this puzzle I would greatly appreciate an explanation. Thanks in advance. Now lets do kWH. 69 cells X 2.9 AH = 200.1 AH/brick X 9 bricks X 11 moduals X 3.78 volts /1000 = 74.9 kWH. So how do I match this up to the 56 kWH claimed by Tesla? At 75% the number is 56kWH. Does that mean that only 75% of the batteries nominal capacity is "usable"? Is this because there is about a 25% loss between input energy (from the HPC) and actual usable stored and and delivered energy from the battery to the inverter/motor? Finally, the new 18650 battery is supposed to be 4 AH instead of 2.9 AH in 2012. The 2.9 AH battery weighs 44 g while the new 4 AH battery weighs 54 g. 6831 * 44 g = 300.564 kg or 663 lbf. Since the entire battery weighs 990 lbf, the battery enclosure, coolant and electronics must weigh 990 - 663 = 327 lbf. Therefore if the new batteries weigh 6831 * 54 g = 368.874 kg or 814 lbf, then a new battery pack for my Roadster with the new batteries should weigh 814 + 327 = 1,141 lbf. For the extra 151 lbf of weight, one should get an increased nominal range of 245 * 4.0/2.9 = 338 miles minus a little for hauling around the extra weight. Do I have it right on all counts? As for weight, you have to consider that there is more in the Tesla battery than just the Panasonic NCR18659 in a pile. At the very least, there is a system for pumping coolant through the assembly, and the assembly itself probably weighs a substantial amount. In other words, a new battery pack made with new batteries will probably weigh more, but they could easily change more than just the Panasonic part when revising the pack. You have wrong battery. The one used in Roadster is 2.1Ah battery, not 2.9. The one used in Model S is probably 3.1Ah battery. Also that 4Ah battery weights only 46g not 54g. That is unless you have some better and newer information than I do. Those that go to Tesla are also modified somehow, which probably does something to the weight, not sure what though. Thanks Timo. The 2.1 AH clears up the kWH. I found the 54g on a website somewhere. 48g is a lot better news. Looks like all of them are the same physical dimensions (2.1, 2.9, 3.1 and 4.0). Do they all have the same discharge ability? Do you know what the xC rate is? Wouldn't the 3.1's give 47% more range (3.1/2.1 = 1.47)? In any event, it is exciting to think that much more range and or lower weight for the same range is just over the horizon. Can't wait. Any clarification on the nominal voltage? Is the 2.1 AH battery nominal 3.78 volts? My model airplane LiPo's are all 3.7 v. rsdio, If you look at the calculations again the batteries themselves made up 663# of the 990# weight so I assumed the balance of the weight was due to the enclosure, cooling system and electronics (about 327# worth).With the 48 gram weight for the 4 AH I get 1050# total pack weight (663*48/44 = 723 + 327 = 1050# The roadster batteries are 2200mah,3.7V and 44g. Model S 300 mile pack batteries are supposedly the 3100mah, 3.6v and weigh 44.5g. Thx to all. Calculations all work now (accept for voltage). I will be watching the battery technology and feel certain that someday we will all be able to get new batteries that either weigh a lot less for the same range or have considerably more range for the wieght when the time comes for replacement. 6 weeks left for #1364. we will all be able to get new batteries that either weigh a lot less for the same range or have considerably more range for the wieght when the time comes for replacement. I'd say it's a safe bet that the batteries will be imporved with respec to power/weight ratios, but far more important will be improved costs.Frankly, right now, with 300 miles of range, and a 45 minute recharge, they are good enough to be fully competitive with gas powered jobs. The Roadster Innovations / Battery page has a typographical error. It says that a brick contains sixty-nine cells, a sheet contains ninety-nine bricks, and a pack contains 11 sheets. That would be 75,141 cells in all, but we know it's only 6,831. Based upon the opening message of this thread, I'm assuming that the "Ninety-nine bricks" should read "Nine bricks" - but maybe they mean that the 99 bricks in series make 11 sheets. It's basically a little confusing the way it's worded. 45 minute recharge? Even with the 90A charger it takes about 3 hours. The 45 min recharge is referring to some kind of level 3 charging (such as DC fast charging), not the level 2 charging the Roadster can use. At the 70A level 2 charging level of the Roadster, the 300 mile pack would probably take around 5 hours to fully recharge. the way you calculate the energy available gives the correct number but its not representing the correct electrochemical event of discharge. even though you have 6831 cells, only the capacity (Q) of 69 of them (those arranged in parallel) is used for discharge at the nominal voltage of 375V (given by the arrangement in series of the remainder 99). E = Q x V so E = (69 x 2.2)A x 375V = 56.9 kWh. the capacity coming from the cells arranged in series is only there for "balancing". If they were "empty" then the capacity of the 69 arranged in parallel would have to be averaged over the entire number of cells (6831). as you can see an increase in capacity is most wanted as it allows for both an increase in energy (driving range) and power (acceleration, etc) EVEN if the individual cell operates at a much lower voltage. For example, if you had double the capacity in a single cell you would only need 35 of them in parallel for the same energy (driving range) and 195 in series which would now give a much higher nominal voltage. V = 195 x 3.5 = 683V for a 3.5V cell or 585 for a 3.0V cell, etc. so lets hope there is interest in post lithium ion batteries such that the chemistries of such batteries are ironed out in due time :). Does anyone know how much litium the battery contains? So I have calculated that tesla will be able to manufacture 367 million cars with the 11.000.000 tonnes of lithium in the world. Do you concur? I think we are safe on the lithium supply until we we are making a billion plus cars. See the very insightful article from Nick Butcher about battery constraints. http://seekingalpha.com/article/654441-ev-myths-and-realities-part-1-the... Yes, I miscalculated: 1,6 billion! i see lots of talk about the specs but nothing on cycles, does anybody know how well these things will do in the long run? the amount of lithium in the battery pack can be estimated by math. Lithium only exists in the LiCoO2 cathode and the LIPF6 electrolyte. The electrolyte is typically 1 mol/L so there is 1 mol of Li in 1L of electrolyte. Based on the total capacity of the unit cell (18650) of ~3000 mAh and the reversible specific capacity of LiCoO2 of 140 mAh/g, we can estimate 21.5g of reversible LiCoO2 in one cell. however, the total capacity of LiCoO2 is ~280 mAhg, but only ~50% of is reversible, so we have to multiply 21.5 x 2 = 43g LiCoO2 to obtain total amount in the unit cell. now the molecular weight of LiCoO2 is 97.87 g/mol and that of Li is 6.941 so Li weighs 7.1% of total active material in cathode. Then 43g x 0.071 = 3.05g total Li in the cathode of one 18650 cell with ~3 Ah capacity. I dont know how much electrolyte one 18650 cell contains but it must be as much as the free volume (~50%) of its celgard separator which must be the same geometric area as the cathode. The geometric area of the cathode in an 18650 cell is 265.65 cm2. So if i assume ~280 cm2 geometric area for the celgard separator with 0.1 mm thickness and ~50% porosity, i get ~ 280 cm2 x 0.01 cm = 2.8 cm3 / 2 = 1.4 cm3 or 1.4 ml of electrolyte. at a a typical concentration of 1M LiPF6, a volume of 1.4 ml corresponds to 0.269 g LiPF6 (mw of LiPF6 is 191.905 g/mol). there is 3.6% Li in LiPF6 by weight so 0.269 g x 0.036 = 0.0097 g Li. to add everything up we have 0.0097g + 3.05g = 3.06g Li in one 18650 cell. Model S has 7000 cells so a total of 21.4 kg Li per battery pack or just about 47 lbs of lithium. i hope i made only a few mistakes in my math. if anyone can pitch in how much that costs...
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# American Institute of Mathematical Sciences December  2020, 9(4): 1089-1114. doi: 10.3934/eect.2020047 ## Fully history-dependent evolution hemivariational inequalities with constraints 1 College of Sciences, Beibu Gulf University, Qinzhou, Guangxi 535000, China 2 Jagiellonian University in Krakow, Chair of Optimization and Control, ul. Lojasiewicza 6, 30348 Krakow, Poland 3 School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan 611731, China 4 College of Sciences, Beibu Gulf University, Qinzhou, Guangxi 535000, China Dedicated to Professor Meir Shillor on the occasion of his 70th birthday Received  October 2019 Published  March 2020 In this paper we study a new class of abstract evolution first order hemivariational inequalities which involves constraints and history-dependent operators. First, we prove the existence and uniqueness of solution by using a mixed equilibrium formulation with suitable selected bifunctions combined with a fixed-point principle for history-dependent operators. Next, we deduce existence, uniqueness and regularity results for some special subclasses of problems which include a constrained history-dependent variational–hemivariational inequality, an evolution quasi-variational inequality with constraints, and an evolution second order hemivariational inequality with constraints. Then, we provide an application of the results to a dynamic unilateral viscoelastic frictional contact problem and show its unique weak solvability. Citation: Stanisław Migórski, Yi-bin Xiao, Jing Zhao. Fully history-dependent evolution hemivariational inequalities with constraints. Evolution Equations & Control Theory, 2020, 9 (4) : 1089-1114. doi: 10.3934/eect.2020047 ##### References: show all references ##### References: [1] Yifan Chen, Thomas Y. Hou. Function approximation via the subsampled Poincaré inequality. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 169-199. doi: 10.3934/dcds.2020296 [2] Andrew D. Lewis. Erratum for "nonholonomic and constrained variational mechanics". Journal of Geometric Mechanics, 2020, 12 (4) : 671-675. doi: 10.3934/jgm.2020033 [3] Shipra Singh, Aviv Gibali, Xiaolong Qin. Cooperation in traffic network problems via evolutionary split variational inequalities. Journal of Industrial & Management Optimization, 2020  doi: 10.3934/jimo.2020170 [4] Lingfeng Li, Shousheng Luo, Xue-Cheng Tai, Jiang Yang. A new variational approach based on level-set function for convex hull problem with outliers. Inverse Problems & Imaging, , () : -. doi: 10.3934/ipi.2020070 [5] Lorenzo Zambotti. A brief and personal history of stochastic partial differential equations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 471-487. doi: 10.3934/dcds.2020264 [6] Mostafa Mbekhta. Representation and approximation of the polar factor of an operator on a Hilbert space. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020463 [7] Jie Zhang, Yuping Duan, Yue Lu, Michael K. Ng, Huibin Chang. Bilinear constraint based ADMM for mixed Poisson-Gaussian noise removal. Inverse Problems & Imaging, , () : -. doi: 10.3934/ipi.2020071 [8] Adel M. Al-Mahdi, Mohammad M. Al-Gharabli, Salim A. Messaoudi. New general decay result for a system of viscoelastic wave equations with past history. Communications on Pure & Applied Analysis, 2021, 20 (1) : 389-404. doi: 10.3934/cpaa.2020273 [9] Jiaquan Liu, Xiangqing Liu, Zhi-Qiang Wang. Sign-changing solutions for a parameter-dependent quasilinear equation. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020454 [10] Bahaaeldin Abdalla, Thabet Abdeljawad. Oscillation criteria for kernel function dependent fractional dynamic equations. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020443 [11] Guido Cavallaro, Roberto Garra, Carlo Marchioro. Long time localization of modified surface quasi-geostrophic equations. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020336 [12] Zedong Yang, Guotao Wang, Ravi P. Agarwal, Haiyong Xu. Existence and nonexistence of entire positive radial solutions for a class of Schrödinger elliptic systems involving a nonlinear operator. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020436 [13] Soniya Singh, Sumit Arora, Manil T. Mohan, Jaydev Dabas. Approximate controllability of second order impulsive systems with state-dependent delay in Banach spaces. Evolution Equations & Control Theory, 2020  doi: 10.3934/eect.2020103 [14] Marion Darbas, Jérémy Heleine, Stephanie Lohrengel. Numerical resolution by the quasi-reversibility method of a data completion problem for Maxwell's equations. Inverse Problems & Imaging, 2020, 14 (6) : 1107-1133. doi: 10.3934/ipi.2020056 [15] Wenmeng Geng, Kai Tao. Large deviation theorems for dirichlet determinants of analytic quasi-periodic jacobi operators with Brjuno-Rüssmann frequency. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5305-5335. doi: 10.3934/cpaa.2020240 [16] Zuliang Lu, Fei Huang, Xiankui Wu, Lin Li, Shang Liu. Convergence and quasi-optimality of $L^2-$norms based an adaptive finite element method for nonlinear optimal control problems. Electronic Research Archive, 2020, 28 (4) : 1459-1486. doi: 10.3934/era.2020077
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## Day 14: Going to the Rats by As I hinted at back in the in the Day 1 post, Perl 6 has rational numbers. They are created in the most straightforward fashion, by dividing an integer with another integer. But it can be a bit hard to see that there is anything unusual about the result: ```> say (3/7).WHAT Rat() > say 3/7 0.428571428571429 ``` When you convert a Rat to a Str (for example, to “say” it), it converts to a decimal representation. This is based on the principle of least surprise: people generally expect 1/4 to equal 0.25. But the precision of the Rat is exact, rather than the approximation you’d get from a floating point number like a Num: ```> say (3/7).Num + (2/7).Num + (2/7).Num - 1; -1.11022302462516e-16 > say 3/7 + 2/7 + 2/7 - 1 0 ``` The most straightforward way to see what is going on inside the Rat is to use the `.perl` method. `.perl` is a standard Perl 6 method which returns a human-readable string which, when eval’d, recreates the original object as closely as is possible: ```> say (3/7).perl 3/7 ``` You can also pick at the components of the Rat: ```> say (3/7).numerator 3 > say (3/7).denominator 7 > say (3/7).nude.perl [3, 7] ``` All the standard numeric operators and operations work on Rats. The basic arithmetic operators will generate a result which is also a Rat if that is possible; the rest will generate Nums: ```> my \$a = 1/60000 + 1/60000; say \$a.WHAT; say \$a; say \$a.perl Rat() 3.33333333333333e-05 1/30000 > my \$a = 1/60000 + 1/60001; say \$a.WHAT; say \$a; say \$a.perl Num() 3.33330555601851e-05 3.33330555601851e-05 > my \$a = cos(1/60000); say \$a.WHAT; say \$a; say \$a.perl Num() 0.999999999861111 0.999999999861111 ``` (Note that the `1/60000 + 1/60000` didn’t work in the last official release of Rakudo, but is fixed in the Rakudo github repository.) There also is a nifty method on Num which creates a Rat within a given tolerance of the Num (default is 1e-6): ```> say 3.14.Rat.perl 157/50 > say pi.Rat.perl 355/113 > say pi.Rat(1e-10).perl 312689/99532 ``` One interesting development which has not made it into the main Rakudo build yet is decimal numbers in the source are now spec’d to be Rats. Luckily this is implemented in the ng branch, so it is possible to demo how it will work once it is in mainstream Rakudo: ```> say 1.75.WHAT Rat() > say 1.75.perl 7/4 > say 1.752.perl 219/125 ``` One last thing: in Rakudo, the Rat class is entirely implemented in Perl 6. The source code is thus a pretty good example of how to implement a numeric class in Perl 6. ### 8 Responses to “Day 14: Going to the Rats” 1. Joe Z Says: It wasn’t immediately clear to me why “1 / 60000 + 1 / 60001″ shouldn’t also be a rational. Is it that Rats are built on Ints, and 60000 * 60001 doesn’t fit in an Int? If so, then that makes sense. “Int” doesn’t appear anywhere in the post, though. • colomon Says: It’s kind of wonky, which is why I think I skirted around the answer. Most directly, it’s because (in current Rakudo), 60000 * 60001 is a Num, not an Int, because it trips an Int overflow check after the multiplication. What’s weird about this is it trips that 32-bit overflow flag even if Rakudo is compiled with 64-bit Ints and thus is perfectly capable of storing 60000 * 60001 in an Int. (On my 32-bit OS X, (60000 * 60001).Int == -2147483648; on my 64-bit Linux machine, (60000 * 60001).Int == 3600060000. On both of them (60000 * 60001).WHAT is Num.) In theory, an Int should be a BigNum and able to handle 60000 * 60001 with ease, but that’s not the case in Rakudo. But then, in theory, the denominator of a Rat should be an Int64, a 64-bit Int, but Rakudo doesn’t have such a type yet. Once these types are in place, the gist of the example still is correct, but you’ll need larger numbers that 60001 to see the effect in practice.) 2. bened Says: Will there be automatic big numbers in Rakudo in future releases? That’s a feature I was always impressed by in Common Lisp. • colomon Says: They are in the spec, but probably won’t make it into Rakudo until the second half of 2010. • Roie Says: Given that automatic bignums are in the spec, shouldn’t the 1/60000+1/60001 example always work, even for huge numbers? I.e., shouldn’t (1/\$x+1/\$y).WHAT always be Rat, regardless of the size of \$x and \$y? • colomon Says: In the spec, Rats are Int over uint64 (that is, a unlimited size big num integer over an unsigned 64-bit int), so these issues do apply. If you need unlimited precision regardless of efficiency, the spec provides the FatRat type, which is an Int over an another Int. There’s been no attempt to implement this, as Rakudo does not have bignums yet. 3. Ricardo Signes Says: That was great — probably one of the most fun (for me) articles so far. I have no darn idea what I’d ever use them for, but the overall interface and simplicity of use seems, from this post, really nice.
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+0 # Algebra 0 102 1 In November's Math Meet, the Little Monsters of Beast Academy solved 1/3 of the problems. In March's Math Meet, they solved 2/7 of the problems. (There might be different numbers of problems asked in November and March.) In total, the Little Monsters solved 15 of 28 problems over both Math Meets. How many problems were asked at March's Math Meet? Sep 7, 2022 #1 0 Let the number of problems in November ==N Let the number of problems in March ==M 1/3N  +  2/7M ==15........................(1) N  +  M == 28.................................(2), solve for N, M N==147     and    M== - 119  !!!!!!  Do these numbers make sense to you ?? I think there is a major mistake in your question. Check it carefully. Sep 9, 2022
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Change language to: Français - 日本語 - Português - Русский Please note that the recommended version of Scilab is 2024.1.0. This page might be outdated. See the recommended documentation of this function # cdfbet cumulative distribution function Beta distribution ### Syntax ```[P,Q]=cdfbet("PQ",X,Y,A,B) [X,Y]=cdfbet("XY",A,B,P,Q) [A]=cdfbet("A",B,P,Q,X,Y) [B]=cdfbet("B",P,Q,X,Y,A)``` ### Arguments P,Q,X,Y,A,B five real vectors of the same size. P,Q (Q=1-P) The integral from 0 to X of the beta distribution (Input range: [0, 1].) Q 1-P X,Y (Y=1-X) Upper limit of integration of beta density (Input range: [0,1], Search range: [0,1]) A,B : The two parameters of the beta density (input range: (0, +infinity), Search range: [1D-300,1D300] ) ### Description Calculates any one parameter of the beta distribution given values for the others (The beta density is proportional to `t^(A-1) * (1-t)^(B-1)`. Cumulative distribution function (P) is calculated directly by code associated with the following reference. DiDinato, A. R. and Morris, A. H. Algorithm 708: Significant Digit Computation of the Incomplete Beta Function Ratios. ACM Trans. Math. Softw. 18 (1993), 360-373. Computation of other parameters involve a search for a value that produces the desired value of P. The search relies on the monotonicity of P with the other parameter. From DCDFLIB: Library of Fortran Routines for Cumulative Distribution Functions, Inverses, and Other Parameters (February, 1994) Barry W. Brown, James Lovato and Kathy Russell. The University of Texas. ### Examples ```x = 0:0.1:1; y = 1-x; A = 2*ones(x); B = 3*ones(x); [p,q]=cdfbet('PQ',x,y,A,B); plot2d2("gnn",[0:10]',p,5,"111","Repartition",[0,0,10,1])``` ### See also • cdfbin — cumulative distribution function Binomial distribution • cdfchi — cumulative distribution function chi-square distribution • cdfchn — cumulative distribution function non-central chi-square distribution • cdff — cumulative distribution function Fisher distribution • cdffnc — cumulative distribution function non-central f-distribution • cdfgam — cumulative distribution function gamma distribution • cdfnbn — cumulative distribution function negative binomial distribution • cdfnor — cumulative distribution function normal distribution • cdfpoi — cumulative distribution function poisson distribution • cdft — cumulative distribution function Student's T distribution Report an issue << binomial Cumulated Distribution Functions cdfbin >> Copyright (c) 2022-2023 (Dassault Systèmes)Copyright (c) 2017-2022 (ESI Group)Copyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Tue Feb 14 15:02:48 CET 2017
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: ## Symbolic integration over a region I want to set up a symbolic integration procedure over an region. I will have expressions such as z = c1 + c2*x + c3*y + c4*x*y .. but the number of terms in z can vary and should be determined automatically by the procedure. c1, c2 .. are arbitrary factors. For example, a possible z could be z = a + b/c*x + d/e*x*y^3 + f*y^5. The integral of z should be returned from the procedure as zint = c1*INT(0,0) + c2*INT(1,0) + c3*INT(0,1) + c4*INT(1,1) .. where INT(i,j) represents the integral over the region of x^i*y^j. These integrals will be calculated numerically by another procedure. Does anyone know how to do this? Thanks. John 
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## ››Convert rod [international] to parsec rod [international] parsec Did you mean to convert rod [international] rod [survey] to parsec How many rod [international] in 1 parsec? The answer is 6.1355237413505E+15. We assume you are converting between rod [international] and parsec. You can view more details on each measurement unit: rod [international] or parsec The SI base unit for length is the metre. 1 metre is equal to 0.19883878151595 rod [international], or 3.2407792700054E-17 parsec. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between rods and parsecs. Type in your own numbers in the form to convert the units! ## ››Want other units? You can do the reverse unit conversion from parsec to rod [international], or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Rod A rod is a unit of length, equal to 11 cubits, 5.0292 metres or 16.5 feet. A rod is the same length as a perch[1] and a pole. The lengths of the perch (one rod) and chain (four rods) were standardized in 1607 by Edmund Gunter. The length is equal to the standardized length of the ox goad used by medieval English ploughmen; fields were measured in acres which were one chain (four rods) by one furlong (in the United Kingdom, ten chains). ## ››Definition: Parsec The parsec (symbol pc) is a unit of length used in astronomy. It stands for "parallax of one arc second", and is approximately 19,131,554,073,600 (19 trillion) miles. ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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or # The determinant |a2\ a^2-(b-c)^2b c b^2b^2-(c-a)^2c a c^2c^2-(a-b)^2a b| is divisible by- a.a+b+c b. (a+b)(b+c)(c+a) c.a^2+b^2+c^2 d. (a-b)(b-c)(c-a) Question from  Class 12  Chapter Determinants Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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Question Details Urgent \$ 40.00 • From Mathematics, Statistics • Due on 12 Mar, 2016 05:04:00 • Asked On 12 Mar, 2016 04:12:38 • Due Date has already passed, but you can still Post Solutions. Question posted by It is for both Statistics and Probability. the deadline is today so I need help. Attachment Available solutions \$ 30.00 Statistics and Probability • This solution has not purchased yet. • Submitted On 12 Mar, 2016 04:33:56 Solution posted by Week One Homework Assignment - Statistics and Probability The head of the chemistry department at a local university has tasked a student teaching assistant to prepare a variety of statistics pertaining to the final course grades earned by students who have completed the university’s CHEM 103 course during the past ten years. These statistics will be used as a basis for predicting the potential performance of future students who enroll in the course. A total of 1,125 students have completed the course during the past ten years. Rather than trying to have the teaching assistant collect and analyze data for all 1,125 students, the department head has agreed to allow the teaching assistant to prepare the necessary statistics using a random sample consisting of 25 students who completed the course during the past ten years. The following table summarizes the final course grades earned by the 25 randomly selected students. Student No. Course Grade 1 83.17 2 82.31 3 90.18 4 93.63 5 90.14 6 96.06 7 83.75 8 85.17 9 86.67 10 93.81 11 83.11 12 82.16 13 89.48 14 86.76 15 83.76 16 86.89 17 88.57 18 96.93 19 86.48 20 95.82 21 92.32 22 78.70 23 92.67 24 74.56 25 78.48 For the purposes of this assignment, assume that: (1) the sample consisting of 25 students is truly representative of the population of 1,125 students from which it was drawn; and (2) the 1,125 students who have completed the course during the past ten years constitute a truly representative sample of all future students who will eventually enroll in the course. Based upon these assumptions, use the sample data provided in the preceding table to answer the following questions. 1. What is the predicted range for the mean grade for an average future student enrolling in the CHEM 103 course? o 27.65 o 22.37 o 19.87 o 34.68 Solution: Range = Maximum – Minimum Range = 96.93 – 74.56 Range = 22.37 2. What is the predicted mean grade for an average future student enrolling in the CHEM 103 course? o 76.31 o 83.26 o 87.26 o 90.89 Solution: Mean = ∑X/n Mean = 2181.58/25 Mean = 87.26 3. What is the predicted median grade for an average future student enrolling in the CHEM 103 course? o 86.76 o 91.94 o 79.16 o 90.91 Solution: Median = (n +1)/2th observation Median = (25 + 1)/2th observation Median = 13th observation Median = 86.76 4. What is the predicted standard error of the mean grade for an average future student enrolling in the CHEM 103 course? o 1.58 o 2.29 o 2.11 o 1.18 Solution: Standard deviation = √∑(X – mean)²/ (n – 1) Standard deviation = √832.44/24 Standard deviation = 5.89 Standard error = s/√n Standard error = 5.89/√25 Standard error = 1.18 5. Assuming the level of confidence for the interval estimate is not specified, what is the predicted int... Buy now to view full solution. Attachment \$ 60.00 Week One Homework Assignment - Statistics and Probability A+ use as a guide • This solution has not purchased yet. • Submitted On 12 Mar, 2016 07:17:05 Solution posted by For the purposes of this assignment, assume that: (1) the sample consisting of 25 students is truly representative of the population of 1,125 students from which it was drawn; and (2) the 1,125 students who have completed the course during the past ten years constitute a truly representative sample of all future students who will eventually enroll in the course. Based upon these assumptions, use the sample data provided in the preceding table to answer the following questions. 1. What is the predicted range for the mean grade for an average future student enrolling in the CHEM 103 course? o 27.65 o 22.37 o 19.87 o 34.68 Solution: Range = Maximum – Minimum Range = 96.93 – 74.56 Range = 22.37 2. What is the predicted mean grade for an average future student enrolling in the CHEM 103 course? o 76.31 o 83.26 o 87.26 o 90.89 Solution: Mean = ∑X/n Mean = 2181.58/25 Mean = 87.26 3. What is the predicted median grade for an average future student enrolling in the CHEM 103 course? o 86.76 o 91.94 o 79.16 o 90.91 Solution: Median = (n +1)/2th observation Median = (25 + 1)/2th observation Median = 13th observation Median = 86.76 4. What is the predicted standard error of the mean grade for an average future student enrolling in the CHEM 103 course? o 1.58 o 2.29 o 2.11 o 1.18 Solution: Standard deviation = √∑(X – mean)²/ (n – 1) Standard deviati... Buy now to view full solution. \$ 40.00 Statistics Homework • This Solution has been Purchased 1 time • Average Rating for this solution is A+ • Submitted On 12 Mar, 2016 08:07:37 Solution posted by
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# Converse of Spectral Theorem for Compact Self-Adjoint Operators If I have a bounded linear operator $A$ on a Hilbert space $H$ whose eigenvectors form an orthonormal basis for $H$ and whose corresponding eigenvalues go to $0$ then is $A$ compact and self-adjoint? I ask because I want to prove that $A$ defined on an orthonormal basis $\{e_k\}$ as $Ae_k=e_k/(k^2+1)$ is compact and self-adjoint. I know that it is, but I'm just wondering if appealing to the spectral theorem is valid. Thanks! • Yes, it's valid. In particular, we can define $A$ as the limit of finite-rank operators (with respect to the operator norm). $A$ inherits self-adjointness from the sequence since $A \mapsto A^*$ is operator-norm continuous. – Omnomnomnom May 22 '17 at 3:20 • Thanks! Just to be clear, this holds for all $A$ with eigenvectors that form an onb and eigenvalues that go to 0 and not just for the explicit operator $A$ I gave, correct? – Robert B. Marshall May 22 '17 at 3:37 • Yep!${}{}{}{}{}{}{}{}{}$ – Omnomnomnom May 22 '17 at 3:39 Yes, $A$ is compact. By considering truncations of $A$ that are zero on $e_k$ for $k\geq n$, you can write $A$ as a limit of finite-rank operators.
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1. ## trig integral integral dx/[(sin x)(cos x)] how do i approach such an example 2. Originally Posted by alex83 integral dx/[(sin x)(cos x)] how do i approach such an example $\displaystyle \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \csc 2x$ and $\displaystyle 2 \int \csc 2x\, dx$ is a standard integral (if $\displaystyle u = 2x$). 3. alex83 - Once again I'm going to ask that you show some effort before making another thread on trig integral problems. It seems you are going down the list of your homework. We want you to learn how to do these on your own and it doesn't seem you know how to. 4. Originally Posted by alex83 integral dx/[(sin x)(cos x)] how do i approach such an example sin x cos x can be re-written as 0.5 sin (2x). So your job is to integrate a cosec function. See here: http://www.mathhelpforum.com/math-he...barrassed.html You will not get any further help on this question until you show your working and state where you get stuck. You may not think so, but this is actually for your own academic benefit. Maths is not a spectator sport. 5. Originally Posted by Jameson alex83 - Once again I'm going to ask that you show some effort before making another thread on trig integral problems. It seems you are going down the list of your homework. We want you to learn how to do these on your own and it doesn't seem you know how to. actually im studying for my finals and i have my solution manual with me but when the book shows an answer differnet than mine and i dont understand the way they got to a certain step i seek help from this helpful forum which i learned from more than my class.. thank you anyway but pls. dont get me wrong 6. Originally Posted by alex83 actually im studying for my finals and i have my solution manual with me but when the book shows an answer differnet than mine and i dont understand the way they got to a certain step i seek help from this helpful forum which i learned from more than my class.. thank you anyway but pls. dont get me wrong If that happens you should post the solution you've been given and say what part of it you don't understand.
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# 投稿 2021.09.26 ## Lll Spread Betting Explained For Sports And Esports +++ Betting on the favorite with the point spread requires that team to win the contest by a certain amount of points. Betting on the underdog with the point spread will allow that team to lose the contest, as long as it’s not by more than the posted number. Hockey and baseball also have point spreads, but they work a little different. Implied probability of 75% would return Decimal Odds of 1.33. Knowing how to convert one form of odds to another can be helpful, especially if you have come into possession of a large amount of betting data that has odds formatted in an unfamiliar way. Multiplied then by 100, we get the implied probability percentage of 35.7%. Multiplied then by 100, we get the implied probability percentage of 54.5%. New York Jets to win against the New England Patriots at odds -120. Multiplied then by 100 to express as an implied probability percentage of 60.6%. ## Here Are Some Basics About Sports Betting: When it comes http://www.bee-vivid.co.jp/archives/19030 to sports betting, the more wagering options you have, the better. The typical bets of spreads, moneylines, totals, and parlays are available on almost all of the sites. Teasers, props, and futures will all be present in a large number of them. ## The Top Sports Spread Betting Companies In More Detail The offerings can vary, but you should see at least a few choices. For the MLB alternate run lines, the odds will vary based on the choice, but we can still have a general expectation based on the situation at hand. For example, a situation in which the alternate run line for our example is a complete switch, the odds might look like this. For a case where the run line is 0.5 runs, the odds could tighten even further. When the run line rises, you may find more favorable odds on the favored side while those on the underdog side get worse. Yes, bettors can generally combine point spread bets with other types of bets in a parlay – a multi-part that requires all components of the bet to hit. In fact, parlaying spread bets is very popular among many bettors. This means bettors must win more than 50% of the time to break even. Two-team parlays usually pay 2.6-to-1, three-team parlays pay 6-to-1, four-team parlays pay 10-to-1, etc. Though we refer to these as “two-team parlays,” parlays can consist of picks on the point totals and sometimes even props. Of course, if you are new to NBA betting, or sports betting in general, you will first want to understand exactly what you are looking at in terms of NBA odds, moneylines and point totals. A few of our favorite NBA betting tips won’t hurt you, either. Tennessee-based sportsbook app Action 24/7 will make its debut before the end of 2020. There are no regulations and laws in Macau that regulate various forms of remote gambling, including online gambling. However, it is mentioned in the law that interactive and online gaming concessions are detached from the operation of brick-and-mortar casino concessions. Six companies, namely Galaxy Entertainment, SJM Melco, Wynn Macau, MGM China, and Sands castle, can now run casinos in Macau. The team expected to win gives or lays, points to the team expected to lose for betting purposes. Do you know that how to read soccer betting odds and NFL Football odds are completely different? The two major outcomes to an NFL Football game is winning or losing unlike soccer that has home win, both teams to draw, and away team to win. You can also visit some sports news website to view the daily lines. At the end of the game, if things are tight and your team is winning narrowly, the opposing team will likely try to foul and get the ball back. The emergence of in-play betting has been seen across many sports including basketball. Handicap basket lines will go in play, and these will change depending on how the game is going. Again, this means shopping around as both the lines and odds will be different depending on your bookmaker. In the NBA, there is no reason to win big, so teams lack the motivation to run up a big score, and often they are playing very soon after, so they want to conserve energy.
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The "Teacher’s Corner" of The American Statistician enables statisticians to discuss topics that are relevant to teaching and learning statistics. Sometimes, the articles have practical relevance, too. Andersson (2023) "The Wald Confidence Interval for a Binomial p as an Illuminating 'Bad' Example," is intended for professors and masters-level students in mathematical statistics. However, the topic has practical importance. The introduction to the article states, "the standard two-sided Wald interval for the binomial [proportion]is ... known to be of low quality with respect to actual coverage, and it ought to be stricken from all courses" in favor of alternative intervals such as the Wilson (score). If Andersson wants the method "stricken from all courses," then presumably he also advocates that the Wald CI should not be used in practice. That is a strong statement with practical importance. The Wald confidence interval (CI) is the default CI for PROC FREQ in SAS, although the documentation states "the Wilson interval has ... better performance than the Wald interval." Let's investigate this issue. This article presents a simulation study that compares the coverage probability of the two-sided CIs for the (uncorrected) Wald method and for the Wilson (score) method. The simulation demonstrates that Andersson makes an excellent point: In practice, the Wald method is an inferior choice for the CI. You can use the Wilson CI to obtain a better confidence interval. In PROC FREQ, you can use the BINOMIAL(CL=WILSON) option on the TABLES statement to obtain the Wilson CI. ### The Wald confidence interval for a binomial proportion As Andersson (2023) points out, "generations" of students have memorized the formula for the Wald CI, "and probably some of them still remember it." If a sample has n observations and n1 are an event of interest, then a point-estimate for the proportion of events is $\hat{p} = n_1 / n$. This is called "the binomial proportion" because the distribution of n1 is Binomial(p, n). The Wald confidence interval for the proportion is given by the formula $\hat{p} \pm z \sqrt { \hat{p} (1 - \hat{p} ) / n }$ where z is the (1-α/2)th quantile of the standard normal distribution. The formula for the Wilson interval is given in the documentation of PROC FREQ. I won't repeat Andersson's mathematical arguments about why the Wald interval is deficient, but a few issues are: • The point estimator and its standard error (the quantity under the square root sign) are correlated. • The Wald CI approximates the discrete sampling distribution of the Wald statistic by using a continuous normal distribution. This leads to a mismatch between the quantiles of the actual sampling distribution and the approximate quantiles. • The Wald statistic is biased, skewed, and exhibits excess kurtosis. (See Table 1 in Andersson (2023).) • The Wald interval has lower-than-expected coverage. For example, a 95% confidence interval might contain the proportion parameter for only 93% of random samples. ### Coverage probability for the Wald confidence interval Andersson states that the Wald interval is "low quality with respect to actual coverage." The coverage probability of a confidence interval is the probability that the interval (which is a statistic) contains the value of the distribution parameter. For example, a 95% confidence interval should contain the distribution parameter in 95% of all random samples. For the binomial proportion, you can run a simulation to estimate the coverage probability of the Wald interval and Wilson intervals, as follows: • Choose a proportion parameter, p. • Simulate n independent Bernoulli trials with probability p. This results in n1 events. Equivalently, yhat you can simulate the n1 values directly from the Binomial(p, n) distribution. • Calculate the Wald and Wilson intervals. Assess whether the parameter is inside each interval. • Repeat the simulation B times, where B is a large number. Estimate the coverage probability as k/B, where k is the number of times that the parameter is in the confidence interval. The following SAS program simulates B=10,000 samples of size n=50 from the Binomial(p, n) distribution for p=0.2. You can use PROC FREQ to estimate the proportion and confidence intervals from each sample. The OUTPUT statement writes the statistics to a SAS data set. The variables L_Bin and U_Bin are the lower and upper confidence limits, respectively, for the Wald interval. The variables L_W_Bin and U_W_Bin are the corresponding limits for the Wilson interval. A second DATA step creates binary variables that indicate whether each interval contains the parameter, p. A second call to PROC FREQ calculates the proportion of samples for which the interval contains the parameter, which estimates the coverage probability. /* simulation of Wald interval vs Wilson interval for binomial proportion */ %let B = 10000; /* number of simulations in study */ %let N = 50; /* sample size */ data BinomSim; call streaminit(12345); N = &N; do p = 0.2; do ID = 1 to &B; Event = 1; Count = rand("Binom", p, N); /* number of events in N trials */ output; Event = 0; Count = N - Count; /* number of non-events */ output; end; end; run;   /* for each sample, estimate proportion and CIs (Wald and Wilson) */ proc freq data=BinomSim noprint; by p ID; weight Count / zeros; tables Event / nocum binomial(level='1' CL=Wald CL=Wilson); output out=Est binomial; run;   /* create binary (0/1) variables that indicate whether p is in the CI */ data Coverage; set Est; label _BIN_="Estimate (pHat)" L_Bin="Lower Wald CL" U_Bin="Upper Wald CL" L_W_Bin="Lower Wilson CL" U_W_Bin="Upper Wilson CL"; WaldInCL = (L_Bin <= p & p <= U_Bin); /* is p in Wald CI? */ WilsonInCL = (L_W_Bin <= p & p <= U_W_Bin); /* is p in Wilson CI? */ keep p ID _Bin_ L_Bin U_Bin L_W_Bin U_W_Bin WaldInCL WilsonInCL; run;   /* estimate the empirical coverage probability for the nominal 95% CIs */ proc freq data=Coverage; by p; tables WaldInCL WilsonInCl / nocum; run; The output for p=0.2 is shown. We wanted 95% confidence intervals, but the simulation p indicates that the empirical coverage probability for the Wald confidence interval is about 93.89%, which is substantially lower than 95%. In contrast, the empirical coverage probability for the Wald confidence interval is about 95.11% when p=0.2. ### Repeat the simulation for a range of parameters Andersson's article taught me something new. Namely, that the coverage probability for an interval is not a smooth (or even continuous) function of the proportion parameter. I knew that interval estimates are better when p ≈ 1/2 and can be bad when p is near 0 or 1, but I did not know that the coverage probability can jump dramatically if you change p by a small amount. Andersson shows this in Figure 1, which is reproduced on the right. Andersson states that "the ragged performance" is "due to the lattice structure of the binomial distribution." This phenomenon is explained in Brown, Cai, and DasGupta (2001), which contains several similar plots. The phenomenon is not some artifact of the Wald interval. It is a real phenomenon that affects all confidence intervals, including the Wilson interval. Let's extend the simulation so that it estimates the Wald and Wilson coverage probability for a range of proportion parameters. Most textbooks caution that you should not use the Wald interval unless np(1-p) > 5, which means that when n=50 we should consider only p > 0.11. You can convert the previous simulation to a series of simulations by changing one line in the first DATA step. You should replace the line that sets p=0.2 with a DO loop that iterates over a range of values. /* simulation of Wald interval vs Wilson interval for binomial proportion */ %let B = 10000; /* number of simulations in study */ %let N = 50; /* sample size */ /* Simulate random samples of size N where Pr(Event=1)=p */ data BinomSim; call streaminit(12345); N = &N; /* n*p*(1-p) > 5 ==> p > 0.1127 */ do p = 0.11 to 0.5 by 0.005; /* <== iterate over a range of proportion parameters */ do ID = 1 to &B; Event = 1; Count = rand("Binom", p, N); /* number of events in N trials */ output; Event = 0; Count = N - Count; /* number of non-events */ output; end; end; run;   /* construct Wald and Wilson CIs for estimate of proportion, \hat{p} */ proc freq data=BinomSim noprint; by p ID; weight Count / zeros; tables Event / nocum binomial(level='1' CL=Wald CL=Wilson); output out=Est binomial; run;   /* create binary variable to indicate whether interval contains p */ data Coverage; set Est; label _BIN_="Estimate (pHat)" L_Bin="Lower Wald CL" U_Bin="Upper Wald CL" L_W_Bin="Lower Wilson CL" U_W_Bin="Upper Wilson CL"; WaldInCL = (L_Bin <= p & p <= U_Bin); /* is p in Wald CI? */ WilsonInCL = (L_W_Bin <= p & p <= U_W_Bin); /* is p in Wilson CI? */ keep p ID _Bin_ L_Bin U_Bin L_W_Bin U_W_Bin WaldInCL WilsonInCL; run; Instead of looking at tables of the empirical coverage probabilities, the following program statements graph the estimated coverage versus the parameter for both the Wald and Wilson intervals: /* plot empirical coverage probability of the Wald CI for range of p */ proc freq data=Coverage noprint; by p; tables WaldInCL / nocum binomial(level='1' CL=Wilson); output out=CoverageWard binomial; run; title "Coverage Probability of Ward CI"; proc sgplot data=CoverageWard; label _BIN_="Ward Coverage" p="Binomial Probability"; scatter y=p x=_BIN_ / xerrorlower=L_W_Bin xerrorupper=U_W_Bin; refline 0.95 / axis=x noclip; xaxis min=0.88 max=0.98; run;   /* plot empirical coverage probability of the Wilson CI for range of p */ proc freq data=Coverage noprint; by p; tables WilsonInCL / nocum binomial(level='1' CL=Wilson); output out=CoverageWilson binomial; run; title "Coverage Probability of Wilson CI"; proc sgplot data=CoverageWilson; label _BIN_="Wilson Coverage" p="Binomial Probability"; scatter y=p x=_BIN_ / xerrorlower=L_W_Bin xerrorupper=U_W_Bin; refline 0.95 / axis=x noclip; xaxis min=0.88 max=0.98; run; I've aligned the graphs for comparison. From the graphs, you can notice the following: • The coverage probability "jumps around" in both graphs as the parameter, p, changes. This is a real phenomenon and is not due to Monte Carlo variation. • In general, the Wald confidence interval exhibits less than 95% coverage. For proportion parameters less than 0.2, the coverage is especially bad. • Sometimes the Wilson confidence interval is less than 95% coverage, but other times it is more. The Wilson intervals appear to be roughly symmetric around the 95% reference line. This is a good property to have, as opposed to the systematic downward bias of the Wald CI. These graphs show why the Wilson interval is preferred over the Wald interval. The Wald interval has actual coverage that is less than its nominal coverage. If you do not know the proportion parameter (and you never do!) then the Wilson interval is a more reliable and dependable interval. If you pardon the pun, you can have more "confidence" in the Wilson interval. ### Summary Andersson (2023) states, "the standard two-sided Wald interval for the binomial [proportion]is ... known to be of low quality with respect to actual coverage." The documentation for PROC FREQ in SAS agrees by stating, "the Wilson interval has ... better performance than the Wald interval." This article runs a simulation study that provides evidence for those statements. Andersson suggests that the Wald CI should not be used in practice. In PROC FREQ in SAS, you can use the BINOMIAL(CL=WILSON) option on the TABLES statement to obtain the Wilson CI. With this small change to your programs, you will obtain a more reliable interval estimate for a binomial proportion. The post Should you use the Wald confidence interval for a binomial proportion? appeared first on The DO Loop. A previous article describes the metalog distribution (Keelin, 2016). The metalog distribution is a flexible family of distributions that can model a wide range of shapes for data distributions. The metalog system can model bounded, semibounded, and unbounded continuous distributions. This article shows how to use the metalog distribution in SAS by downloading a library of SAS IML functions from a GitHub site. The article contains the following sections: • An overview of the metalog library of SAS IML functions • How to download the functions from GitHub and include the functions in a program • How to use a metalog model to fit an unbounded model to data • A discussion of infeasible models For a complete discussion of the metalog functions in SAS, see "A SAS IML Library for the Metalog Distribution" (Wicklin, 2023). ### The metalog library of functions in SAS The metalog library is a collection of SAS IML modules. The modules are object-oriented, which means that you must first construct a "metalog object," usually from raw data. The object contains information about the model. All subsequent operations are performed by passing the metalog object as an argument to functions. You can classify the functions into four categories: • Constructors create a metalog object from data. By default, the constructors fit a five-term unbounded metalog model. • Query functions enable you to query properties of the model. • Computational functions enable you to compute the PDF and quantiles of the distribution and to generate random variates. • Plotting routines create graphs of the CDF and PDF of the metalog distribution. The plotting routines are available when you use PROC IML. Before you create a metalog object, you must choose how to model the data. There are four types of models: unbounded, semibounded with a lower bound, semibounded with an upper bound, or bounded. Then, you must decide the number of terms (k) to use in the metalog model. A small number of terms (3 ≤ k ≤ 6) results in a smooth model. A larger number of terms (k > 6) can fit multimodal data distributions, but be careful not to overfit the data by choosing k too large. The most common way to create a metalog object is to fit a model to the empirical quantiles of the data, which estimates the k parameters in the model. For an unbounded metalog model, the quantile function is $Q(p) = a_1 + a_2 L(p) + a_3 c(p) L(p) + a_4 c(p) + a_5 c(p)^2 + a_6 c(p)^2 L(p) + a_7 c(p)^3 + \ldots$ where L(p) = Logit(p) is the quantile function of the logistic distribution and c(p) = p – 1/2 is a linear function. You use ordinary least squares regression to fit the parameters ($a_1, a_2, \ldots, a_k$) in the model. The semibounded and bounded versions are metalog models of log-transformed variables. • If X > L, fit a metalog model to Z = log(X – L) • If L < X < U, fit a metalog model to Z = log((X – L) / (U – X)) The process of downloading SAS code from a GitHub site is explained in the article, "How to use Git to share SAS programs." The process is also explained in Appendix A in the metalog documentation. The following SAS program downloads the metalog functions to the WORK libref. You could use a permanent libref, if you prefer: /* download metalog package from GitHub */ options dlcreatedir; %let repoPath = %sysfunc(getoption(WORK))/sas-iml-packages; /* clone repo to WORK, or use permanent libref */   /* clone repository; if repository exists, skip download */ data _null_; if fileexist("&repoPath.") then put 'Repository already exists; skipping the clone operation'; else do; put "Cloning repository 'sas-iml-packages'"; rc = gitfn_clone("https://github.com/sassoftware/sas-iml-packages/", "&repoPath." ); end; run;   /* Use %INCLUDE to read source code and STORE functions to current storage library */ proc iml; %include "&repoPath./Metalog/ML_proc.sas"; /* each file ends with a STORE statement */ %include "&repoPath./Metalog/ML_define.sas"; quit; Each file ends with a STORE statement, so running the above program results in storing the metalog functions in the current storage library. The first file (ML_PROC.sas) should be included only when you are running the program in PROC IML. It contains the plotting routines. The second file (ML_define.sas) should be included in both PROC IML and the iml action. ### Fit an unbounded metalog model to data The Sashelp.heart data set contains data on patients in a heart study. The data set contains a variable (Systolic) for the systolic blood pressure of 5,209 patients. The following call to PROC SGPLOT creates a histogram of the data and overlays a kernel density estimate. title 'Systolic Blood Pressure'; proc sgplot data=sashelp.heart; histogram Systolic; density Systolic / type=kernel; keylegend /location=inside; run; The kernel density estimate (KDE) is a nonparametric model. The histogram and the KDE indicate that the data are not normally distributed, and it is not clear whether the data can be modeled by using one of the common "named" probability distributions such as gamma or lognormal. In situations such as this, the metalog distribution can be an effective tool for modeling and simulation. The following SAS IML statements load the metalog library of functions, then fit a five-term unbounded metalog distribution to the data. After you fit the model, it is a good idea to run the ML_Summary subroutine, which displays a summary of the model and displays the parameter estimates. proc iml; load module=_all_; /* load the metalog library */ use sashelp.heart; read all var "Systolic"; close;   MLObj = ML_CreateFromData(Systolic, 5); /* 5-term unbounded model */ run ML_Summary(MLObj); The ML_Summary subroutine displays two tables. The first table tells you that the model is a 5-term unbounded model, and it is feasible. Not every model is feasible, which is why it is important to examine the summary of the model. (Feasibility is discussed in a subsequent section.) The second table provides the parameter estimates for the model. The MLObj variable is an object that encapsulates relevant information about the metalog model. After you create a metalog object, you can pass it to other functions to query, compute, or plot the model. For example, in PROC IML, you can use the ML_PlotECDF subroutine to show the agreement between the cumulative distribution of the model and the empirical cumulative distribution of the data, as follows: title '5-Term Metalog Model for Systolic Blood Pressure'; run ML_PlotECDF(MLObj); The model is fit to 5,209 observations. For this blog post, I drew the CDF of the model in red so that it would stand out. The model seems to fit the data well. If you prefer to view the density function, you can run the statement run ML_PlotPDF(MLObj); Again, note that the MLObj variable is passed as the first argument to the function. One way to assess the goodness of fit is to compare the sample quantiles to the quantiles predicted by the model. You can use the ML_Quantile function to obtain the predicted quantiles, as follows: /* estimates of quantiles */ Prob = {0.01, 0.05, 0.25, 0.5, 0.75, 0.95, 0.99}; call qntl(SampleEst, Systolic, Prob); /* sample quantiles */ ModelEst = ML_Quantile(MLObj, Prob); /* model quantiles */ print Prob SampleEst ModelEst; The model quantiles agree well with the sample quantiles. Note that the median estimate from the model (the 0.5 quantile) is equal to the a1 parameter estimate. This is always the case. An advantage of the metalog model is that it has an explicit quantile function. As a consequence, it is easy to simulate from the model by using the inverse CDF method. You can use the ML_Rand function to simulate from the model, as follows: /* simulate 500 new blood pressure values from the model */ call randseed(1234); title 'Simulated Systolic Values'; BP = ML_Rand(MLObj, 500); call histogram(BP); Notice that the shape of this histogram is similar to the histogram of the original data. The simulated data seem to have similar properties as the original data. The metalog documentation provides the syntax and documentation for every public method in the library. ### Infeasible metalog functions As discussed in a previous article, not every metalog model is a valid distribution. An infeasible model is one for which the quantile function is not monotone increasing. This can happen if you have a small data set that has extreme outliers or gaps, and you attempt to fit a metalog model that has many parameters. The model will overfit the data, which results in an infeasible model. I was unable to create an infeasible metalog model for the blood-pressure data, which has thousands of observations and no extreme observations. However, if I use only the first 20 observations, I can demonstrate an infeasible model by fitting a six-term metalog model. The following SAS IML statements create a new data vector that contains only 20 observations. A six-term metalog model is fit to those data. /* What if we use a much smaller data set? Set N = 20. */ S = Systolic[1:20]; /* N = 20 */ MLObj = ML_CreateFromData(S, 6); /* 6-term model */ run ML_Summary(MLObj); As shown by the output of the ML_Summary call, the six-term metalog model is not feasible. The following graph shows the quantile function of the model overlaid on the empirical quantiles. Notice that the quantile function of the model is not monotone increasing, which explains why the model is infeasible. If you encounter this situation in practice, Keelin (2016) suggests reducing the number of terms in the model. ### Summary This article describes how to use the metalog distribution in SAS. The metalog distribution is a flexible family that can model a wide range of distributional shapes. You can download the Metalog package from the SAS Software GitHub site. The package contains documentation and a collection of SAS IML modules. You can call the modules in a SAS IML program to fit a metalog model. After fitting the model, you can compute properties of the model and simulate data from the model. More information and examples are available in the documentation of the Metalog package. The post Use the metalog distribution in SAS appeared first on The DO Loop. A SAS programmer asked for help to simulate data from a distribution that has certain properties. The distribution must be supported on the interval [a, b] and have a specified mean, μ, where a < μ < b. It turns out that there are infinitely many distributions that satisfy these conditions. This article describes the shapes for a family of beta distributions that solve this problem. ### Common bounded distributions There are three common distributions that are used to model data on a bounded interval: • The triangular distribution has a peak (mode) that is easy to specify. The PDF looks like a triangle, so this distribution might not be a good model for real data. • The PERT distribution also has a mode that is easy to specify. The PERT distribution is a particular example of a beta distribution that is used in decision analysis. • The two-parameter beta distribution is a flexible family that can model a wide range of distributional shapes. An interesting fact about the two-parameter beta distribution is that you can model many different shapes. The parameters for the beta distribution enable you to model distributions for which the PDF is decreasing, increasing, U-shaped, and has either positive or negative skewness. If Y is a beta-distributed random variable on [0,1] that has mean p, then X = (ba)Y + a is a random variable on [a, b] that has mean μ = (ba)p + a. Thus, we can simulate beta-distributed data, and then scale and translate the data to any other bounded interval. ### Beta distributions that have a common mean Let's examine the shapes of some beta distributions that all have the same mean, p, in [0,1]. The mean of the Beta(α, β) distribution is p = α/(α+β). Thus, for any specified mean, there is a one-parameter family of beta distributions, each with a different shape, that all have the same mean. For any value of the β parameter, choose α = p / (1 – p) β to ensure that the Beta(α, β) distribution has mean p. Let's compute the PDF for a few members of the family to see what they look like. In the following program, I specify that I want a beta distribution that has mean value p = 2/3, which forces α = 2 β. I then plot the PDF for several values of β to visualize the different shapes: /* show PDFs for a sample of (alpha, beta) values such that the Beta(alpha, beta) distribution has mean=2/3 */ data BetaPDF; keep alpha beta y pdf; p = 2/3; /* mean of Y ~ Beta(alpha, beta) distribution */ do beta = 0.2, 0.8, 2, 6; alpha = p/(1-p) * beta; /* choose alpha so that distrib has mean p */ do y = 0.01 to 0.99 by 0.01; PDF = pdf("beta", y, alpha, beta); output; end; end; run;   title "A Family of Beta Distributions for Mean = 2/3"; proc sgplot data=BetaPDF; series x=y y=PDF / group=beta lineattrs=(thickness=2); yaxis min=0 max=4 label="Density"; run; Notice the shapes of the resulting beta distributions: • The PDF for β=0.2 is U-shaped. • The PDF for β=0.8 is monotonic increasing. • The PDF for β=2 has a mode at 0.75. • The PDF for β=6 has a mode at 0.6875. It appears to be approximately bell-shaped. All these distributions have the same mean, which is p = 2/3. As β increases, the distribution becomes nearly normal, and the mode approaches the mean. ### Simulate data from a bounded distribution with a specified mean The PDF of the distributions is easier to visualize than a random sample. But you can modify the program to generate random variates instead of a PDF. To obtain a random sample on [a, b] that has mean μ, you can transform the problem: use the beta distribution to simulate a sample on [0, 1], then transform the data into the interval [a, b]. For example, suppose you want a random sample from a distribution that has mean 20 and is bounded on the interval [10, 25]. Because 20 is two-thirds of the way between 10 and 25, you can simulate from a beta distribution on [0, 1] that has mean p = 2/3. If Y is a beta-distributed random variable on [0, 1], then X = (25-10)*Y + 10 is a random variable on [10, 25]. The following SAS DATA step demonstrates this technique. Because the problem does not have a unique solution, the program generates six random samples, each with N=200 observations. Each sample has a different shape, but they are all generated from a distribution whose mean is 20. /* Define interval [a,b] and mean, mu */ %let a = 10; %let b = 25; %let mu = 20; /* note that mu is 2/3 of the way from a to b */ /* if X is r.v. on [a,b] with mean mu, then Y = (X-a)/(b-a) is r.v. on [0,1] with mean p=a + (b-a)*mu */ data BetaSim; call streaminit(1234); keep alpha beta x y; a = &a; b = &b; mu = &mu; p = (mu - a)/(b-a); /* mean of Y ~ Beta in [0, 1] */ do beta = 0.2, 0.5, 0.8, 1, 2, 6; alpha = p/(1-p) * beta; /* choose alpha so that distrib has mean p */ do i = 1 to 200; /* N = 200 for this example */ y = rand("beta", alpha, beta); /* Y ~ Beta(alpha, beta) on [0,1] */ x = (b-a)*y + a; /* transform values into [a,b] */ output; end; end; run;   proc sgpanel data=BetaSim; panelby alpha beta / columns=3; histogram x; colaxis grid; run; The panel shows six different samples. Each sample is drawn from a distribution that has mean 20. Four of the samples are generated from a (rescaled) distribution that was shown in the previous section. As you can see, the shape of the distributions vary. Some are U-shaped, some are nearly linear, and some are bell-shaped. If you want a unique solution to this problem, you must add an additional constraint. A common choice is to match not just the mean of some sample data, but also the variance. These beta distributions all have different variances, so adding a constraint on the variance ensures a unique beta distribution. ### Summary This article shows how to simulate data from a distribution on the interval [a, b] that has a specified mean, μ. There are infinitely many distributions that satisfy these constraints. This article visualizes the shapes for a family of beta distributions that you can use to solve this problem. To get a unique solution, you can specify an additional requirement, such as a value for the variance. The post Simulate from a bounded distribution that has a specified mean appeared first on The DO Loop. SAS programmers love to make special graphs for Valentine's Day. In fact, there is a long history of heart-shaped graphs and love-inspired programs written in SAS! Last year, I added to the collection by showing how a ball bounces on a heart-shaped billiards table. This year, I create a similar image, but one that is stochastic instead of deterministic. This year's image is a random walk in a heart-shaped region. In the figure to the right, the red lines trace the path of a random walk that starts in the middle of the heart. For each step, the new location is determined by a random bivariate normal vector. That is, from the position (x,y), the next location is (x+dx, y+dy) where (dx, dy) is a random bivariate normal vector. If the next location would be outside of the heart-shaped region, then shorten the step and land on the boundary. The figure shows the path of a constrained random walk for 2000 steps. Perhaps this image is a visual metaphor for randomly exploring love? Some readers will be content to view this image and reflect on the love in their lives. Others will want to learn more about how the image is created in SAS. If you are in the second group, read on! ### An unconstrained random walk Before I show how to constrain the random walk to stay inside the heart, let's create and visualize an unconstrained random walk. There are many kinds of random walks, but this one takes a step according to a random bivariate normal vector. Starting from a point (x,y), you generate a vector (dx,dy) ~ MVN(0, I(2)) and move to the new point (x+dx, y+dy). You repeat this process many time, then draw the path that results. The following SAS DATA step generates the random walk by using 0.2 as the standard deviation of each variate: /* random 2-D walk where each step is a vector v, which is bivariate normal v ~ MVN(0, I(2)) */ data RandomWalk; N = 2000; call streaminit(12345); x = 0; y = 0; do i = 1 to N; dx = rand("Normal", 0, 0.2); dy = rand("Normal", 0, 0.2); x = x + dx; y = y + dy; output; end; keep x y; run; ods graphics / reset width=480px height=480px; title "Random Walk with Bivariate Normal Steps"; proc sgplot data=RandomWalk; series x=x y=y / lineattrs=(color=lightred); xaxis grid; yaxis grid; run; This graph displays a random walk with 2,000 steps, starting from the origin. Most of the time, the two consecutive points are close to each other. However, sometimes the next point is far away from the previous point. ### A constrained random walk Now imagine a variation on the previous random walk. The initial point is inside a compact closed region. Instead of always moving from an initial point (x,y) to a final point (x+dx, y+dy), you compute whether the final point is inside the region. If so, move to the point. If not, truncate the step so that you land on the boundary of the region. Repeat this many times to obtain a constrained random walk that is always inside the region. For Valentine's Day, I will choose the region to be heart-shaped. It is easiest to perform these computations in the SAS IML language. The following program defines a few helper functions and generates 2,000 points for the constrained random walk. The points are written to a SAS data set and overlaid on a graph of the heart-shaped region: /* Instead of an unconstrained walk, at each step check whether the new location is outside of a region. If so, truncate the step to land on the boundary of the region. */ proc iml; /* define the heart-shaped region */ start H(xy); x = xy[,1]; y = xy[,2]; return ( (x**2 + y**2 - 1)**3 - x**2 * y**3 ); finish;   /* return 1 if (x,y) is inside the region */ start InRegion(xy, tol=1e-14); return (H(xy) <= tol); finish;   /* given a point, x, and a vector, v, this function returns the function f(t) = H(x + t*v), t in [0,1]. If f has a root, then the line segment from x to x+v intersects the boundary of the reqion. This function is used by FROOT to find points on the boundary of the region. */ start OnBoundary(t) global(G_x, G_v); return ( H( G_x + t*G_v ) ); finish;   /* Start at x. Try to step to x+v. If x+v is in the region, take the step. Otherwise, see if we can take part of the step to land on the boundary. */ start StepInRegion(x, v) global(G_x, G_v); if InRegion(x + v) then return (x + v); if InRegion(x) then do; G_x = x; G_v = v; /* does the line from x to x+v cross the boundary? */ t = froot("OnBoundary", {0 1}); /* find intersection of line and region */ if t > 0 then return (x + t*v); /* step onto the boundary */ else return (x); /* cannot step in this direction */ end; /* something is wrong: x is not in the region */ return ( {. .} ); finish;   N = 2000; call randseed(12345); vel = randfun(N//2, "Normal", 0, 0.2);   x = j(1, 2, 0); create walk from x [c={x y}]; do i = 1 to N; x = StepInRegion(x, vel[i,]); append from x; end; close; QUIT;   title "Random Walk Inside a Heart-Shaped Region"; proc sgplot data=Walk; series x=x y=y / lineattrs=(color=lightred); run; The trajectory of the random walk is shown for 2,000 random steps inside a heart-shaped region. You can carry out this computation for any compact region, provided that you can represent the region as the level set of some continuous function H(x,y)=0 that divides the plane into an inside region (where H(x,y) < 0) and an outside region (where H(x,y) > 0). To visualize the heart-shaped region, you can overlay the trajectory on the region, as done at the top of this article. The images below visualize how the random walk evolves and gradually fills the region. For details about how these images were created, see the SAS program that creates these images. ### Summary It is easy to simulate an unconstrained random walk. This article shows how to use SAS to generate a random walk in which the points are contained in a compact region. During the simulation, you can monitor whether the next step will occur outside the region. If so, you can decrease the step so that the point is on the boundary of the region. This is illustrated by using a heart-shaped region, but the same algorithm can constrain the random walk to any compact region of the plane. The post A random walk inside a heart appeared first on The DO Loop. A previous article shows that you can use the Intercept parameter to control the ratio of events to nonevents in a simulation of data from a logistic regression model. If you decrease the intercept parameter, the probability of the event decreases; if you increase the intercept parameter, the probability of the event increases. The probability of Y=1 also depends on the slope parameters (coefficients of the regressor effects) and on the distribution of the explanatory variables. This article shows how to visualize the probability of an event as a function of the regression parameters. I show the visualization for two one-variable logistic models. For the first model, the distribution of the explanatory variable is standard normal. In the second model, the distribution is exponential. ### Visualize the probability as the Intercept varies In a previous article, I simulated data from a one-variable binary logistic model. I simulated the explanatory variable, X, from a standard normal distribution. The linear predictor is given by η = β0 + β1 X. The probability of the event Y=1 is given by μ = logistic(η) = 1 / (1 + exp(-η)). In the previous article, I looked at various combinations of Intercept (β0) and Slope (β1) parameters. Now, let's look systematically at how the probability of Y=1 depends on the Intercept for a fixed value of Slope > 0. Most of the following program is repeated and explained in my previous post. For your convenience, it is repeated here. First, the program simulates data (N=1000) for a variable, X, from a standard normal distribution. Then it simulates a binary variable for a series of binary logistic models as the intercept parameter varies on the interval [-3, 3]. (To get better estimates of the probability of Y=1, the program simulates 100 data sets for each value of Intercept.) Lastly, the variable calls PROC FREQ to compute the proportion of simulated events (Y=1), and it plots the proportion versus the Intercept value. /* simulate X ~ N(0,1) */ data Explanatory; call streaminit(12345); do i = 1 to 1000; x = rand("Normal", 0, 1); /* ~ N(0,1) */ output; end; drop i; run;   /* as Intercept changes, how does P(Y=1) change when Slope=2? */ %let Slope = 2; data SimLogistic; call streaminit(54321); set Explanatory; do Intercept = -3 to 3 by 0.2; do nSim = 1 to 100; /* Optional: param estimate is better for large samples */ eta = Intercept + &Slope*x; /* eta = linear predictor */ mu = logistic(eta); /* mu = Prob(Y=1) */ Y = rand("Bernoulli", mu); /* simulate binary response */ output; end; end; run;   proc sort data=SimLogistic; by Intercept; run; ods select none; proc freq data=SimLogistic; by Intercept; tables Y / nocum; ods output OneWayFreqs=FreqOut; run; ods select all;   title "Percent of Y=1 in Logistic Model"; title2 "Slope=&Slope"; footnote J=L "X ~ N(0,1)"; proc sgplot data=FreqOut(where=(Y=1)); series x=Intercept y=percent; xaxis grid; yaxis grid label="Percent of Y=1"; run; In the simulation, the slope parameter is set to Slope=2, and the Intercept parameter varies systematically between -3 and 3. The graph visualizes the probability (as a percentage) that Y=1, given various values for the Intercept parameter. This graph depends on the value of the slope parameter and on the distribution of the data. It also has random variation, due to the call to generate Y as a random Bernoulli variate. For these data, the graph shows the estimated probability of the event as a function of the Intercept parameter. When Intercept = –2, Pr(Y=1) = 22%; when Intercept = 0, Pr(Y=1) = 50%; and when Intercept = 2, Pr(Y=1) = 77%. The statement that Pr(Y=1) = 50% when Intercept=0 should be approximately true when the data is from a symmetric distribution with zero mean. ### Vary the slope and intercept together In the previous section, the slope parameter is fixed at Slope=2. What if we allow the slope to vary over a range of values? We can restrict our attention to positive slopes because logistic(η) = 1 – logistic(-η). The following program varies the Intercept parameter in the range [-3,3] and the Slope parameter in the range [0, 4]. /* now do two-parameter simulation study where slope and intercept are varied */ data SimLogistic2; call streaminit(54321); set Explanatory; do Slope = 0 to 4 by 0.2; do Intercept = -3 to 3 by 0.2; do nSim = 1 to 50; /* optional: the parameter estimate are better for larger samples */ eta = Intercept + Slope*x; /* eta = linear predictor */ mu = logistic(eta); /* transform by inverse logit */ Y = rand("Bernoulli", mu); /* simulate binary response */ output; end; end; end; run;   /* Monte Carlo estimate of Pr(Y=1) for each (Int,Slope) pair */ proc sort data=SimLogistic2; by Intercept Slope; run; ods select none; proc freq data=SimLogistic2; by Intercept Slope; tables Y / nocum; ods output OneWayFreqs=FreqOut2; run; ods select all; The output data set (FreqOut2) contains Monte Carlo estimates of the probability that Y=1 for each pair of (Intercept, Slope) parameters, given the distribution of the explanatory variable. You can use a contour plot to visualize the probability of the event for each combination of slope and intercept: /* Create template for a contour plot https://blogs.sas.com/content/iml/2012/07/02/create-a-contour-plot-in-sas.html */ proc template; define statgraph ContourPlotParm; dynamic _X _Y _Z _TITLE _FOOTNOTE; begingraph; entrytitle _TITLE; entryfootnote halign=left _FOOTNOTE; layout overlay; contourplotparm x=_X y=_Y z=_Z / contourtype=fill nhint=12 colormodel=twocolorramp name="Contour"; continuouslegend "Contour" / title=_Z; endlayout; endgraph; end; run; /* render the Monte Carlo estimates as a contour plot */ proc sgrender data=Freqout2 template=ContourPlotParm; where Y=1; dynamic _TITLE="Percent of Y=1 in a One-Variable Logistic Model" _FOOTNOTE="X ~ N(0, 1)" _X="Intercept" _Y="Slope" _Z="Percent"; run; As mentioned earlier, because X is approximately symmetric, the contours of the graph have reflective symmetry. Notice that the probability that Y=1 is 50% whenever Intercept=0 for these data. Furthermore, if the points (β0, β1) are on the contour for Pr(Y=1)=α, then the contour for Pr(Y=1)=1-α contains points close to (-β0, β1). The previous line plot is equivalent to slicing the contour plot along the horizontal line Slope=2. You can use a graph like this to simulate data that have a specified probability of Y=1. For example, if you want approximately 70% of the cases to be Y=1, you can choose any pair of (Intercept, Slope) values along that contour, such as (0.8, 0), (1, 1), (2, 3.4), and (2.4, 4). If you want to see all parameter values for which Pr(Y=1) is close to a desired value, you can use the WHERE statement in PROC PRINT. For example, the following call to PROC PRINT displays all parameter values for which the Pr(Y=1) is approximately 0.7 or 70%: %let Target = 70; proc print data=FreqOut2; where Y=1 and %sysevalf(&Target-1) <= Percent <= %sysevalf(&Target+1); var Intercept Slope Y Percent; run; ### Probabilities for nonsymmetric data The symmetries in the previous graphs are a consequence of the symmetry in the data for the explanatory variable. To demonstrate how the graphs change for a nonsymmetric data distribution, you can run the same simulation study, but use data that are exponentially distributed. To eliminate possible effects due to a different mean and variance in the data, the following program standardizes the explanatory variable so that it has zero mean and unit variance. data Expo; call streaminit(12345); do i = 1 to 1000; x = rand("Expon", 1.5); /* ~ Exp(1.5) */ output; end; drop i; run;   proc stdize data=Expo method=Std out=Explanatory; var x; run; If you rerun the simulation study by using the new distribution of the explanatory variable, you obtain the following contour plot of the probability as a function of the Intercept and Slope parameters: If you compare this new contour plot to the previous one, you will see that they are very similar for small values of the slope parameter. However, they are different for larger values such as Slope=2. The contours in the new plot do not show reflective symmetry about the vertical line Intercept=0. Pairs of parameters for which Pr(Y=1) is approximately 0.7 include (0.8, 0) and (1, 1), which are the same as for the previous plot, and (2.2, 3.4), and (2.6, 4), which are different from the previous example. ### Summary This article presents a Monte Carlo simulation study in SAS to compute and visualize how the Intercept and Slope parameters of a one-variable logistic model affect the probability of the event. A previous article notes that decreasing the Intercept decreases the probability. This article shows that the probability depends on both the Intercept and Slope parameters. Furthermore, the probability depends on the distribution of the explanatory variables. You can use the results of the simulation to control the proportion of events to nonevents in the simulated data. The ideas in this article generalize to logistic regression models that contain multiple explanatory variables. For multivariate models, the effect of the Intercept parameter is similar. However, the effect of the slope parameters is more complicated, especially when the variables are correlated with each other. This article shows that you can use the intercept parameter to control the probability of the event in a simulation study that involves a binary logistic regression model. For simplicity, I will simulate data from a logistic regression model that involves only one explanatory variable, but the main idea applies to multivariate regression models. I also show mathematically why the intercept parameter controls the probability of the event. ### The problem: Unbalance events when simulating regression data Usually, simulation studies are based on real data. From a set of data, you run a statistical model that estimates parameters. Assuming the model fits the data, you can simulate new data by using the estimates as if they were the true parameters for the model. When you start with real data, you obtain simulated data that often have similar statistical properties. However, sometimes you might want to simulate data for an example, and you don't have any real data to model. This is often the case on online discussion forums such as the SAS Support Community. Often, a user posts a question but cannot post the data because it is proprietary. In that case, I often simulate data to demonstrate how to use SAS to answer the question. The simulation is usually straightforward for univariate data or for data from an ordinary least squares regression model. However, the simulation can become complicated for generalized linear regression models and nonlinear models. Why? Because not every choice of parameters leads to a reasonable data set. For example, if you choose parameters "out of thin air" in an arbitrary manner, you might end up with a model for which the probability of the event (Pr(Y=1)) is close to 1 for all values of the explanatory variables. This model is not very useful. It turns out that you can sometimes "fix" your model by decreasing the value of the intercept parameter in the model. This reduces the value of Pr(Y=1), which can lead to a more equitable and balanced distribution of the response values. ### Simulate and visualize a one-variable binary logistic regression model Let's construct an example. For a general discussion and a multivariate example, see the article, "Simulating data for a logistic regression model." For this article, let's keep it simple and use a one-variable model. Let X be an independent variable that is distributed as a standardized normal variate: X ~ N(0,1). Then choose an intercept parameter, β0, and a slope parameter, β1, and form the linear predictor η = β0 + β1 X. A binary logistic model uses a logistic transformation to transform the linear predictor to a probability: μ = logistic(η), where logistic(η) = 1 / (1 + exp(-η)). The graph of the S-shaped logistic function is shown to the right. You can then simulate a response as a Bernoulli random variable Y ~ Bern(μ). In SAS, the following DATA step simulates data from a standard normal variable: data Explanatory; call streaminit(12345); do i = 1 to 1000; x = rand("Normal", 0, 1); /* ~ N(0,1) */ output; end; drop i; run; Regardless of the distribution of X, the following DATA step simulates data from a logistic model that has a specified set of values for the regression parameters. I embed the program in a SAS macro so that I can run several models with different values of the Intercept and Slope parameters. The macro also runs PROC FREQ to determine the relative proportions of the binary response Y=0 and Y=1. Lastly, the macro calls PROC SGPLOT to display a graph that shows the logistic regression model and the "observed" responses as a function of the linear predictor values, /* Simulate data from a one-variable logistic model. X ~ N(0,1) and Y ~ Bern(eta) where eta = logistic(Intercept + Slope*X). Display frequency table for Y and a graph of Y and Prob(Y=1) vs eta */ %macro SimLogistic(Intercept, Slope); title "Logistic Model for Simulated Data"; title2 "Intercept=&Intercept; Slope=&Slope"; data Sim1; call streaminit(54321); set Explanatory; eta = &Intercept + &Slope*x; /* eta = linear predictor */ mu = logistic(eta); /* mu = Prob(Y=1) */ Y = rand("Bernoulli", mu); /* simulate binary response */ run;   proc freq data=Sim1; tables Y / nocum; run; proc sort data=Sim1; by eta; run;   proc sgplot data=Sim1; label Y="Observed" mu="P(Y=1)" eta="Linear Predictor"; scatter x=eta y=Y / transparency=0.8 markerattrs=(symbol=CircleFilled); series x=eta y=mu; xaxis grid values = (-16 to 16) valueshint; yaxis grid label="Probability"; keylegend / location=inside position=E opaque; run; %mend; ### How does the probability of the event depend on the intercept? Let's run the program. The following call uses an arbitrary choice of values for the regression parameters: Intercept=5 and Slope=2: /* call the macro with Intercept=5 and Slope=2 */ %SimLogistic(5, 2); The output shows what can happen if you aren't careful when choosing the regression parameters. For this simulation, 96.5% of the response values are Y=1 and only a few are Y=0. This could be a satisfactory model for a rare event, but not for situations in which the ratio of events to nonevents is more balanced, such as 75-25, 60-40, or 50-50 ratios. The graph shows the problem: the linear predictor (in matrix form, η = X*β) is in the range [-2, 12], and about 95% of the linear prediction are greater than 1.5. Because logistic(1.5) > 0.8, the probability of Y=1 is very high for most observations. Since you have complete control of the parameters in the simulation, you can adjust the parameters. If you shift η to the left, the corresponding probabilities will decrease. For example, if you subtract 4 from η, then the range of η becomes [-6, 8], which should provide a more balanced distribution of the response category. The following statement implements this choice: /* call the macro with Intercept=1 and Slope=2 */ %SimLogistic(1, 2); The output shows the results of the simulation after decreasing the intercept parameter. When the Intercept=1 (instead of 5), the proportion of simulated responses is more balanced. In this case, 63% of the responses are Y=1 and 37% are Y=0. If you want to further decrease the proportion of Y=1, you can decrease the Intercept parameter even more. For example, if you run %SimLogistic(-1, 2), then only 35.9% of the simulated responses are Y=1. ### The mathematics I have demonstrated that decreasing η (the linear predictor) also decreases μ (the probability of Y=1). This is really a simple mathematical consequence of the formula for the logistic function. Suppose that η1 < η2. Then -η1 > -η2 ⇒    1 + exp(-η1) > 1 + exp(-η2) because EXP is a monotonic increasing function ⇒    1 / (1 + exp(-η1)) < 1 / (1 + exp(-η2)) because the reciprocal function is monotonic decreasing ⇒    μ1 < μ2 by definition. Therefore, decreasing η results in decreasing μ For some reason, visualizing the simulation makes more of an impact than the math. ### Summary This article shows that by adjusting the value of the Intercept parameter in a simulation study, you can adjust the ratio of events to nonevents. Decrease the intercept parameter to decrease the number of events; increase the intercept parameter to increase the number of events. I demonstrated this fact for a one-variable model, but the main ideas apply to multivariate models for which the linear predictor (η) is a linear combination of multiple variables. The probability of Y=1 depends continuously on the regression parameters because the linear predictor and the logistic function are both continuous. In my next article, I show how to visualize the probability of an event as a function of the regression parameters. A probabilistic card trick is a trick that succeeds with high probability and does not require any skill from the person performing the trick. I have seen a certain trick mentioned several times on social media. I call it "ladders" or the "ladders game" because it reminds me of the childhood board game Chutes and Ladders (previously known as Snakes and Ladders in India and the UK). In Chutes and Ladders, some squares on the board tell the player to move forward (go up a ladder) or to move backward (go down a chute). In the "ladders" game, you only move forward. I have been told that Martin Gardner wrote about this game. He referred to mathematical card tricks as those that "do not require prestidigitation," or sleight of hand. The rules of the game and a Python simulation are presented in a blog post by Bob Carpenter, who saw the trick performed at a children's museum. The performer shuffles an ordinary deck of 52 playing cards and places them face up in a row on a table. The performer then describes the game of ladders: • A player starts by pointing to any card in the first five positions. • The value of the card indicates the number of cards to move forward. If the card is 2, 3, ..., 9, you move forward that many cards. If the card is a 10, jack, queen, king, or ace, you move forward one space. • From the new position, apply the rule again. Repeat this process until you cannot advance any more. The performer selects two volunteers from the audience and tells them each to point randomly to any card from among the first five cards. Each player does so. The performer then announces that both players will end up on a certain "magical" card and he tells everyone the name of the magical card. The players then follow the game rules. Amazingly, each person's path terminates on the magical card that was previously announced. The trick works about 98.2% of the time. To illustrate the ladders game, first consider a simpler situation of a deck that contains only 13 cards, one for each card value (for example, use only spades). One random shuffle of the 13 cards is shown to the right. For this arrangement of cards, the performer announces that both volunteers will end up on the 7 card (the magical card for this shuffle), which is the 11th card in the row of cards. Suppose the first volunteer points to the ace in the first position as her starting card. The rules of the game dictate the following sequence of moves: • From the A, the player moves forward one card and lands on the J. • From the J, the player moves forward one card and lands on the 2. • From the 2, the player moves forward two cards and lands on the 6. • From the 6, the player moves forward six cards and lands on the 7. The player cannot move forward seven spaces, so the 7 is her final card, as predicted in advance. Thus, the path for the first player is the sequence of positions (1, 2, 3, 5, 11). What happens for the other player? Well, if the second volunteer chooses his initial card at positions 2, 3, or 5, then his path will instantly coincide with the path of the first player, so his path, too, terminates at the 7 card in the 11th position. If the second volunteer chooses his initial card at position 4, then his path hits positions 4, 8, and 11. This path also terminates at the 7 card in the 11th position. Thus, for this set of cards and this shuffle, all initial positions follow a path that eventually lands on the 7. How does the performer know the name of the final magical card? As he deals, he secretly counts the path determined by the first card in the deck. ### Simulate the ladders game in SAS You can use Monte Carlo simulation to estimate the probability that the ladders card trick "works." To keep it simple, let's say the trick "works" when the paths of the two players converge. That is, the players initially start on two random positions (1-5), but they end up on the same final card after they each follow the path dictated by their initial position. The simulation for the ladders card trick has the following major steps: 1. Set up the card deck. For each card, assign the number of steps to move forward. 2. Create a function that computes the final position, given the initial position and the sequence of moves determined by the shuffled deck. 3. Run the Monte Carlo simulation. This consists of several sub-steps: • Shuffle the deck (that is, shuffle the array of moves). • Use sampling without replacement to choose two initial positions. • Compute the final card for each initial position. • Record whether the two paths converged. 4. Estimate the probability of the trick working as the proportion of times that the paths converged in the simulation. As is often the case, this simulation is easiest to perform in SAS by using the SAS/IML language. You can use the RANPERM function to shuffle the deck. You can use the SAMPLE function to sample without replacement from the values 1-5. You can write a function that computes the final position based on the initial position. The following program is one way to implement a Monte Carlo simulation for the ladders card trick: /* The Ladders card trick: https://statmodeling.stat.columbia.edu/2022/10/07/mira-magic-a-probabilistic-card-trick/ Lay out a shuffled deck of cards. Choose one of the first five cards as a starting point. Move down the deck as follows: If the current card is 2-9, move that many cards forward. For other cards, move one space forward. */ proc iml; /* 1. set up the card deck and assign values. Note: We don't actually need the deck, only the instructions about how to move. */ values = {A,'2','3','4','5','6','7','8','9','10', J, Q, K}; advance= {1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1}; deck = repeat(values, 4); /* full deck of 52 cards */ moves = repeat(advance, 4); /* corresponding moves */ nCards = nrow(moves);   /* 2. Find the final position, given the first position and the vector of moves for the current shuffled deck. startPos : a scalar value 1-5 Move : a vector of values. If your current card is in position i, the next card is position (i + Move[i]) */ start EndPos(startPos, Move); Pos = startPos; do while(Pos + Move[Pos] < nrow(Move)); Pos = Pos + Move[Pos]; end; return( Pos ); finish;   /* 3. Run the Monte Carlo simulation: A. Shuffle the deck (vector of moves) B. Use sampling w/o replacement to choose two initial positions C. Compute the final card for each initial position D. Record whether the two paths converged */ call randseed(1234); NSim = 1E5; match = j(NSim,1); do j = 1 to nSim; k = ranperm(nCards); /* shuffle the deck */ Move = moves[k]; /* choose two different starting positions in 1-5 */ startPos = sample(1:5, 2, "NoReplace"); EndPos1 = EndPos(startPos[1], Move); EndPos2 = EndPos(startPos[2], Move); match[j] = (EndPos1=EndPos2); end;   /* 4. Estimate probability as the proportion of matches (1s) in a binary vector. Optional: Compute a Wald CL for the proportion */ ProbMatch = mean(match); StdErr = sqrt( ProbMatch*(1-ProbMatch)/ NSim ); CL = ProbMatch + {-1 1}*1.96*StdErr; print NSim[F=comma9.] ProbMatch StdErr CL[c={'LCL' 'UCL'}]; The simulation shows that the two players end up on the same card more than 98% of the time. ### Summary This article describes a probabilistic card trick, which I call the game of ladders. Two players choose random cards as an initial condition, then follow the rules of the game. For about 98.2% of the deals and initial positions, the paths of the two players will converge to the same card. As the performer is laying out the card, he can predict in advance that the paths will converge and, with high probability, announce the name of the final card. Bob Carpenter's blog post mentions the connection between this card trick and Markov chains. He also presents several variations, such as using more cards in the deck. The post Ladders: A probabilistic card trick appeared first on The DO Loop. A SAS programmer was trying to simulate poker hands. He was having difficulty because the sampling scheme for simulating card games requires that you sample without replacement for each hand. In statistics, this is called "simple random sampling." If done properly, it is straightforward to simulate poker hands in SAS. This article shows how to generate a standard deck of 52 cards. It then shows how to generate many card hands for a game that involves several players. This can be done in two ways: The SAMPLE function in the SAS/IML language and PROC SURVEYSELECT in SAS/STAT software. ### Encode a deck of cards The first step is to generate a standard deck of 52 playing cards. There are many ways to do this, but the following DATA step iterates over 13 values (A, 2, 3, ..., Q, K) and over four suits (clubs, diamonds, hearts, and spades). The step creates a data set that has 52 cards. To make the output more compact, I concatenate the value of the card with the first letter of the suit to obtain a two-character string for each card. For example, "5C" is the "five of clubs" and "AS" is the "ace of spades." So that each card can be represented by using two characters, I use "T" instead of "10." For example, "TH" is the "ten of hearts." The data Cards; length Value $2 Suit$8 Card $3; array Values[13]$ _temporary_ ('A','2','3','4','5','6','7','8','9','T','J','Q','K'); array Suits[4] \$ _temporary_ ('Clubs', 'Diamonds', 'Hearts', 'Spades'); do v = 1 to 13; do s = 1 to 4; Value = Values[v]; Suit = Suits[s]; Card = cats(Value,substr(Suit,1,1)); output; end; end; run;   proc print data=Cards(obs=8); var v s Value Suit Card; run; TABLE The cards are generated in order. The output shows the first eight cards in the deck, which are the aces and twos. For some types of poker hands (such as determining pairs and straights), the V variable can be useful, and for other analyses (such as determining flushes), the S variable can be useful. ### Deal cards by using SAS/IML When humans deal cards, we use a two-step method: shuffle the deck and then deal the top cards to players in a round-robin fashion. A computer simulation can simluate dealing cards by using a one-step method: deal random cards (without replacement) to every player. SAS provides built-in sampling methods to make this process easy to program. In SAS/IML, the default sampling method for the SAMPLE function is sampling without replacement. If there are P players and each player is to receive C cards, then a deal consists of P*C cards, drawn without replacement from the deck. The deck does not need to be shuffled because the SAMPLE function outputs the cards in random order. Because the cards are in random order, it does not matter how you assign the cards to the players. The following SAS/IML program uses P=3 players and generates C=5 cards for each player. The program simulates one deal by assigning the first C cards to the first player, the second C cards to the second player, and so on: %let nCards = 5; /* cards per player */ %let nPlayers = 3; /* number of players */ %let nDeals = 10; /* number of deals to simulate */   /* simulate dealing many card hands in SAS/IML */ proc iml; call randseed(1234); /* set random number seed */ use Cards; read all var "Card"; close; /* read the deck of cards into a vector */   nCardsPerDeal = &nCards * &nPlayers; /* each row is a complete deal */ D = sample(Card, nCardsPerDeal, "WOR" ); /* sample without replacement from the deck */ Deal = shape(D, &nPlayers, &nCards); /* reshape into nPlayers x nCards matrix */   /* print one deal */ cardNames = "C1":"C&nCards"; playerNames = "P1":"P&nPlayers"; print Deal[c=cardNames r=playerNames]; TABLE For this deal, the first player has two tens, two fives, and a king. The second player has a four, a five, a nine, an ace, and a two. You could sort the cards in each row, but I have left each player's cards unsorted so that you can see the random nature of the assignment. ### Simulate multiple deals The previous program simulates one deal. What if you want to simulate multiple deals? One way is to loop over the number of deals, but there is a more efficient method. The second argument to the SAMPLE function can be a two-element vector. The first element specifies the number of cards for each deal, and the second element specifies the number of deals. Thus, for example, if you want to simulate 10 deals, you can use the following statement to generate a matrix that has 10 rows. Each row is a sample (without replacement) from a 52-card deck: /* use one call to simulate many deals */ D = sample(Card, nCardsPerDeal//&nDeals, "WOR" ); /* simulate many deals */   /* Ex: The 8th row contains the 8th deal. Display it. */ Deal8 = shape(D[8,], &nPlayers, &nCards); print Deal8[c=cardNames r=playerNames]; TABLE To demonstrate that each row is a deal, I have extracted, reshaped, and displayed the eighth row. The eighth deal is unusual because each player is dealt a "good" hand. The first player has a pair of eights, the second player has two pairs (threes and sixes), and the third player has a pair of sevens. If you want to print (or analyze) the 10 deals, you can loop over the rows, as follows: do i = 1 to &nDeals; Deal = shape(D[i,], &nPlayers, &nCards); /* analyze or print the i_th deal */ print i, Deal[c=cardNames r=playerNames]; end; ### Deal random hands by using PROC SURVEYSELECT You can also simulate card deals by using the SURVEYSELECT procedure. You should use the following options: • Use the METHOD=SRS option to specify sampling without replacement for each deal. To obtain the cards in random order, use the OUTRANDOM option. • Use the N= option to specify the number of cards in each deal. • Use the REPS= option to specify the total number of deals. An example is shown in the following call to PROC SURVEYSELECT: /* similar approach for PROC SURVEYSELECT */ proc surveyselect data=Cards seed=123 NOPRINT method=SRS /* sample without replacement */ outrandom /* randomize the order of the cards */ N=%eval(&nCards * &nPlayers) /* num cards per deal */ reps=&nDeals /* total number of deals */ out=AllCards; run; The output data set (AllCards) contains a variable named Replicate, which identifies the cards in each deal. Because the cards for each deal are in a random order, you can arbitrarily assign the cards to players. You can use a round-robin method or assign the first C cards to the first player, the next C cards to the second player, and so on. This is done by using the following DATA step: /* assign the cards to players */ data AllDeals; set AllCards; by Replicate; if first.Replicate then do; Cnt = 0; end; Player = 1 + floor(Cnt/&nCards); Cnt + 1; drop Cnt; run; OUTPUT The simulated data are in long form. For some analyses, it might be more convenient to convert it into wide form. You can do this by using the DATA step or by using PROC TRANSPOSE, as below: /* OPTIONAL: Convert from long form to wide form */ proc transpose data=AllDeals out=Deals(rename=(col1-col&nCards=C1-C&nCards) drop=_NAME_); var Card; by Replicate Player; run;   proc print data=Deals; where Replicate=8; var Replicate Player C:; run; OUTPUT To demonstrate the results, I display the result of the eighth deal (Replicate=8). For this deal, the first player has several face cards but no pairs, the second player has a pair of threes, and the third player has a pair of fours. ### Summary SAS provides many tools for simulation studies. For simulations of card games, this article shows how to create a data set that represents a standard deck of 52 playing cards. From the cards, you can use the SAMPLE function in SAS/IML software or the SURVEYSELECT procedure to simulate many deals, where each deal is a sample without replacement from the deck. For each deal, you can assign the cards to the players in a systematic manner. The post Simulate poker hands in SAS appeared first on The DO Loop. I recently blogged about how to compute the area of the convex hull of a set of planar points. This article discusses the expected value of the area of the convex hull for n random uniform points in the unit square. The article introduces an exact formula (due to Buchta, 1984) and uses Monte Carlo simulation to approximate the sampling distribution of the area of a convex hull. A simulation like this is useful when testing data against the null hypothesis that the points are distributed uniformly at random. ### The expected value of a convex hull Assume that you sample n points uniformly at random in the unit square, n ≥ 3. What is the expected area of the convex hull that contains the points? Christian Buchta ("Zufallspolygone in konvexen Vielecken," 1984) proved a formula for the expected value of the area of n random points in the unit square. (The unit square is sufficient: For any other rectangle, simply multiply the results for the unit square by the area of the rectangle.) Unfortunately, Buchta's result is published in German and is behind a paywall, but the formula is given as an answer to a question on MathOverflow by 'user9072'. Let $s = \sum_{k=1}^{n+1} \frac{1}{k} (1 - \frac{1}{2^k})$. Then the expected area of the convex hull of n points in [0,1] x [0,1] is $E[A(n)] = 1 -\frac{8}{3(n+1)} \bigl[ s - \frac{1}{(n+1)2^{n+1}} \bigr]$ The following SAS DATA step evaluates Buchta's formula for random samples of n ≤ 100. A few values are printed. You can use PROC SGPLOT to graph the expected area against n: /* Expected area of the convex hull of a random sample of size n in the unit square. Result by Buchta (1984, in German), as reported by https://mathoverflow.net/questions/93099/area-enclosed-by-the-convex-hull-of-a-set-of-random-points */ data EArea; do n = 3 to 100; s = 0; do k = 1 to n+1; s + (1 - 1/2**k) / k; end; EArea = 1 - 8/3 /(n+1) * (s - 2**(-n-1) / (n+1)); output; end; keep n EArea; run;   proc print data=EArea noobs; where n in (3,4,6,8,10,12,13) or mod(n,10)=0; run;   title "Expected Area of Convex Hull"; proc sgplot data=EArea; series x=n y=EArea; xaxis grid label="Number of Random Points in Unit Square"; yaxis grid label="Expected Area" min=0 max=1; run; The computation shows that when n is small, the expected area, E[A(n)], is very small. The first few values are E[A(3)]=11/144, E[A(4)]=11/72, E[A(5)] = 79/360, and E[A(6)] = 199/720. For small n, the expected area increases quickly as n increases. When n = 12, the expected area is about 0.5 (half the area of the bounding square). As n gets larger, the expected area asymptotically approaches 1 very slowly. For n = 50, the expected area is 0.8. For n = 100, the expected area is 0.88. ### The distribution of the expected area of a convex hull Buchta's result gives the expected value, but the expected value is only one number. You can use Monte Carlo simulation to compute and visualize the distribution of the area statistics for random samples. The distribution enables you to estimate other quantities, such as the median area and an interval that predicts the range of the areas for 95% of random samples. The following SAS/IML program performs a Monte Carlo simulation to approximate the sampling distribution of the area of the convex hull of a random sample in the unit square of size n = 4. The program does the following: 1. Load the PolyArea function, which was defined in a previous article that shows how to compute the area of a convex hull. Or you can copy/paste from that article. 2. Define a function (CHArea) that computes the convex hull of a set of points and returns the area. 3. Let B be the number of Monte Carlo samples, such as B = 10,000. Use the RANDFUN function to generate B*n random points in the unit square. This is an efficient way to generate B samples of size n. 4. Loop over the samples. For each sample, extract the points in the sample and call the CHArea function to compute the area. 5. The Monte Carlo simulation computes B sample areas. The distribution of the areas is the Monte Carlo approximation of the sampling distribution. You can compute descriptive statistics such as mean, median, and percentiles to characterize the shape of the distribution. proc iml; /* load the PolyArea function or copy it from https://blogs.sas.com/content/iml/2022/11/02/area-perimeter-convex-hull.html */ load module=(PolyArea);   /* compute the area of the convex hull of n pts in 2-D */ start CHArea(Pts); indices = cvexhull(Pts); /* compute convex hull of the N points */ hullIdx = indices[loc(indices>0)]; /* the positive indices are on the CH */ P = Pts[hullIdx, ]; /* extract rows to form CH polygon */ return PolyArea(P); /* return the area */ finish;   call randseed(1234); NSim = 1E4; /* number of Monte Carlo samples */ n = 4; X = randfun((NSim*n)//2, "Uniform"); /* matrix of uniform variates in [0,1]x[0,1] */ Area = j(NSim, 1, .); /* allocate array for areas */ do i = 1 to NSim; beginIdx = (i-1)*n + 1; /* first obs for this sample */ ptIdx = beginIdx:(beginIdx+n-1); /* range of obs for this sample */ Pts = X[ptIdx, ]; /* use these n random points */ Area[i] = CHArea(Pts); /* find the area */ end;   meanArea = mean(Area); medianArea = median(Area); call qntl(CL95, Area, {0.025 0.975}); print n meanArea medianArea (CL95)[c={LCL UCL}]; For samples of size n = 4, the computation shows that the Monte Carlo estimate of the expected area is 0.154, which is very close to the expected value (0.153). In addition, the Monte Carlo simulation enables you to estimate other quantities, such as the median area (0.139) and a 95% interval estimate for the predicted area ([0.022, 0.369]). You can visualize the distribution of the areas by using a histogram. The following SAS/IML statements create a histogram and overlay a vertical reference line at the expected value. title "Distribution of Areas of Convex Hull"; title2 "N=4; NSim = 10,000"; refStmt = "refline 0.15278 / axis=x lineattrs=(color=red) label='E(Area)';"; call histogram(Area) other=refStmt; The histogram shows that the distribution of areas is skewed to the right for n = 4. For this distribution, a better measure of centrality might be the median, which is smaller than the expected value (the mean). Or you can describe the distribution by reporting a percentile range such as the inter-quartile range or the range for 95% of the areas. ### Graphs of the distribution of the area of a convex hull You can repeat the previous Monte Carlo simulation for other values of n. For example, if you use n = 50, you get the histogram shown to the right. The distribution for n = 50 is skewed to the left. A prediction interval for 95% of areas is [0.69, 0.89]. Whether or not you know about Buchta's theoretical result, you can use Monte Carlo simulation to estimate the expected value for any sample size. The following SAS/IML program simulates 10,000 random samples of size n for n in the range [3, 100]. For each simulation, the program writes the estimate of the mean and a prediction interval to a SAS data set. You can use the BAND statement in PROC SGPLOT to overlay the interval estimates and the estimate of the expected value for each value of n: /* repeat Monte Carlo simulation for many values of N. Write results to a data set and visualize */ create ConvexHull var {'n' MeanArea LCL UCL};   numPts = 3:100; NSim = 1E4; /* number of Monte Carlo samples */ do j = 1 to ncol(numPts); /* number of random pts for each convex hull */ N = numPts[j]; X = randfun((NSim*N)//2, "Uniform"); /* matrix of uniform variates in [0,1]x[0,1] */ Area = j(NSim, 1, .); /* allocate array for areas */ LCL = j(NSim, 1, .); /* allocate arrays for 95% lower/upper CL */ UCL = j(NSim, 1, .); do i = 1 to NSim; beginIdx = (i-1)*N + 1; /* first obs for this sample */ ptIdx = beginIdx:(beginIdx+N-1); /* range of obs for this sample */ Pts = X[ptIdx, ]; /* use these N random points */ Area[i] = CHArea(Pts); end;   meanArea = mean(Area); call qntl(CL95, Area, {0.025 0.975}); LCL = CL95[1]; UCL = CL95[2]; append; end; close ConvexHull; QUIT;   title "Monte Carlo Estimate of Expected Area of Convex Hull"; proc sgplot data=ConvexHull noautolegend; band x=n lower=LCL upper=UCL / legendlabel="95% Prediction"; series x=n y=meanArea / legendlabel="Expected Area"; keylegend / location=inside position=E opaque across=1; xaxis grid label="Number of Random Points in Unit Square"; yaxis grid label="Expected Area" min=0 max=1; run; From this graph, you can determine the expected value and a 95% prediction range for any sample size, 3 ≤ n ≤ 100. Notice that the solid line is the Monte Carlo estimate of the expected value. This estimate is very close to the true expected value, which we know from Buchta's formula. ### Summary Some researchers use the area of a convex hull to estimate the area in which an event occurs. The researcher might want to test whether the area is significantly different from the area of the convex hull of a random point pattern. This article presents several properties of the area of the convex hull of a random sample in the unit square. The expected value of the area is known theoretically (Buchta, 1984). Other properties can be estimated by using a Monte Carlo simulation. The post The area of the convex hull of random points appeared first on The DO Loop. One of the benefits of social media is the opportunity to learn new things. Recently, I saw a post on Twitter that intrigued me. The tweet said that the expected volume of a random tetrahedron in the unit cube (in 3-D) is E[Volume] = 0.0138427757.... This number seems surprisingly small! After all, the cube has unit volume, and the volume of the largest tetrahedron in the cube is 1/3. So, the result says that the average volume of a random tetrahedron is 24 times smaller than the largest tetrahedron! The result is attributed to Zinani, A. (2003) "The Expected Volume of a Tetrahedron whose Vertices are Chosen at Random in the Interior of a Cube," which is behind a paywall. The abstract says that the analytical result is exactly E[Volume] = 3977 / 216000 - π2 / 2160 and that this number is "in convincing agreement with a simulation." This article shows how to run a Monte Carlo simulation in SAS to simulate random tetrahedra and compute their volumes. We use the simulation to confirm Zinani's result and to visualize the distribution of the volume. ### The volume of a tetrahedron In multivariable calculus, students are introduced to the scalar triple product of three vectors. Given three vectors u, v, and w, the scalar triple product is the quantity (u x v)⋅w, where (u x v) is the cross product of the vectors, and zw = zT*w is the scalar dot product. The scalar triple product is the (signed) volume of the parallelepiped formed by the vectors. Recall that the volume of a tetrahedron with vertices (A,B,C,D) is 1/6 of the volume of the parallelepiped formed by the vectors between vertices. Consequently, the volume of a tetrahedron is V = 1/6 |(AB x AC)⋅AD| where AB is the vector from A to B, AC is the vector from A to C, and AD is the vector from A to D. ### Compute the volume of a tetrahedron in SAS Since the volume formula includes vectors and vector operations, it is natural to use the SAS/IML language to compute the volume. The following functions implement the definition of the vector cross-product. To test the function, the program computes the volume of the largest tetrahedron that can be imbedded in the unit cube. One such tetrahedron is shown to the right. The tetrahedron has vertices A(0,0,0), B(1,1,0), C(0,1,1), and D(1,0,1). The volume of this tetrahedron is 1/3, as shown by the following program. /* volume of tetrahedron in 3D */ proc iml; /* vector cross product (3D vectors only) https://blogs.sas.com/content/iml/2013/05/20/matri-multiplication.html */ start CrossProd(u, v); i231 = {2 3 1}; i312 = {3 1 2}; return( u[i231]#v[i312] - v[i231]#u[i312] ); finish;   start VolumeTetra(A,B,C,D); z = CrossProd(B-A, C-A); /* z = AB x AC */ w = colvec(D - A); /* w = AD */ volume = abs(z*w) / 6; /* V = 1/6 | z*w | */ return( volume ); finish;   /* test the formula: compute the volume of the largest tetrahedron inside the unit cube */ A = {0 0 0}; B = {1 1 0}; C = {0 1 1}; D = {1 0 1}; maxVol = VolumeTetra(A,B,C,D); print maxVol; ### A Monte Carlo simulation of random tetrahedron It is not hard to write a SAS/IML program that generates four random vertices for a tetrahedron inside the unit cube in 3-D. To generate the four vertices, you can generate 12 uniform variates independently in the interval [0,1]. Assign the first three numbers to be the (x,y,z) Euclidean coordinates of the first vertex, the next three numbers to be the coordinates of the second vertex, and so forth. From the vertices, you can call the VolumeTetra function to produce the volume. The following program generates 100,000 random tetrahedra and computes their volumes. The program then calls the MEAN function computes the average volume, which is the Monte Carlo estimate of the expected value. call randseed(1234); N = 1E5; X = randfun(N//12, "Uniform"); /* Z x 12 matrix of uniform variates */ Vol = j(N, 1, .); /* allocate array for volumes */ do i = 1 to N; A = X[i, 1:3]; B = X[i, 4:6]; /* form four vertices from the 12 random numbers */ C = X[i, 7:9]; D = X[i, 10:12]; Vol[i] = VolumeTetra(A,B,C,D); end;   avgVol = mean(Vol); /* Monte Carlo estimate of mean */ expectedValue = 3977/216000 - constant('pi')**2 / 2160; /* Zinani's exact result for E[Volume] */ print avgVol expectedValue (expectedValue-avgVol)[L="Diff"]; The output shows that the Monte Carlo estimate is very close to the expected value. This confirms Zinani's result: the average volume of a random tetrahedron is smaller than you might think. ### Visualize the distribution of volumes The expected value for the distribution of tetrahedral volumes is small. Since we know that at least one tetrahedron has a relatively large value (1/3), we might conjecture that most of the tetrahedra have very small volumes and only a few have larger volumes. A way to test that conjecture is to visualize the distribution by drawing a histogram of the volumes for the 100,000 random tetrahedra. The following SAS/IML statements write the expected value to a macro variable and write the volumes of the random tetrahedra to a data set. PROC UNIVARIATE then computes descriptive statistics for the volumes and plots a histogram of the distribution of volumes: /* E[Volume] = 3977 / 216000 - pi**2 / 2160 = 0.01384277574... */ call symputx("EV",expectedValue); /* write E[Volume] to a macro variable */ create Vol var "Vol"; append; close; /* write volumes to a data set */ QUIT;   title "Volume of a Random Tetrahedron"; proc univariate data=Vol; histogram Vol / endpoints href=&EV odstitle=title; run; The graph indicates that a random tetrahedron is likely to have a volume that is near 0. The vertical line is the expected value of the distribution. Notice that only a small proportion of the tetrahedra have volumes that exceed 0.1. In this collection of 100,000 random tetrahedra, the largest volume was 0.1777, which is about 50% of the volume of the largest possible tetrahedron. The distribution has a long but thin tail. If you look carefully at the table of descriptive statistics (the "Moments" table), you will see that the mean and the standard deviation are nearly the same. This property is characteristic of an exponential distribution. An exponential distribution has a skewness of 2 and an excess kurtosis of 6. However, the sample skewness and sample kurtosis are biased statistics, and the estimates shown here (1.8 and 5.1, respectively) are typical for a large sample from an exponential distribution. This suggests that the tetrahedral volume is distributed as V ~ Exp(λ=E[Volume]). Let's use PROC UNIVARIATE again and overlay the density of the Exp(λ=0.0139) distribution. Furthermore, you can use a Q-Q plot to visualize how well the exponential model fits the simulated volumes. proc univariate data=Vol; histogram Vol / exponential(theta=0 sigma=&EV) endpoints; qqplot Vol / exponential(theta=0 sigma=&EV); run;` The fit is excellent! Not only does the density curve overlay perfectly on the histogram, but the quantiles of the data (not shown) are in excellent agreement with the expected quantiles of the exponential distribution. If you look at the Q-Q plot, you again see excellent agreement between the data and the exponential model. The only shortcoming is that the simulated volumes have fewer extreme values than predicted by the model. The model enables you to compute the probability of certain events. For example, the probability that a random tetrahedron has volume greater than 0.1 is 1-cdf("Expo", 0.1, &EV), which is 0.00073. ### Summary A result by Zinani (2003) gives the expected volume of a tetrahedron whose vertices are chosen uniformly at random in the unit cube. Zinani's result is E[Volume] = 0.0138427757.... Although this number seems surprisingly small, the expected value is confirmed by using a Monte Carlo simulation in SAS. The simulation suggests that the volumes are exponentially distributed. For an alternative mathematical proof of the result, see Philip, J. (2007) "The Expected Volume of a Random Tetrahedron in a Cube", which is free to view and is very well written. The post The expected volume of a random tetrahedron in a cube appeared first on The DO Loop.
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# How Much Power Does a 40 Amp Battery Charger Use? When it comes to charging batteries, understanding the power requirements of the charger is essential. In this article, we will answer the question of how much power a 40 amp DC-DC charger uses. We will provide an explanation of power consumption and offer recommendations to ensure safe and efficient charging. ## Understanding Power Consumption Power consumption is measured in watts (W) and is calculated by multiplying the voltage (V) by the current (A). In the case of a 40A battery charger, the power consumption can be determined by multiplying 40 amps by the voltage of the charger. ## Power required for a 12V 40A charger A 40 amp battery charger typically operates at a voltage of 12 volts. Therefore, the power consumption can be calculated as follows: Power (W) = Voltage (V) x Current (A) Power (W) = 12 V x 40 A Power (W) = 480 W Therefore, a 40A battery charger consumes approximately 480 watts of power during operation. ## Tips for Battery Charger Usage ### 1. Power Source Ensure that the power source you are using to supply electricity to the charger can handle the power consumption. Check the electrical circuit's capacity and use a dedicated outlet to avoid overloading. ### 2. Safety Precautions When using a high-power charger like a 40A 12V battery charger, it is crucial to follow safety guidelines. Use appropriate protective gear, such as gloves and safety glasses, and ensure proper ventilation in the charging area to prevent overheating. ### 3. Charging Time Consider the charging time required for your batteries. A higher amp charger can charge batteries faster, but it may also generate more heat. Monitor the charging process and avoid leaving the batteries unattended. ### 4. Battery Compatibility Ensure that the charger is compatible with the type and capacity of the batteries you are charging. Follow the manufacturer's recommendations regarding the charger's compatibility with different battery chemistries and sizes. A 40 amp battery charger consumes approximately 480 watts of power during operation. Understanding the power consumption of your charger is crucial for selecting an appropriate power source and ensuring safe and efficient charging. Follow the recommendations provided to optimize your charging experience and maintain the longevity of your batteries. ## 1 comentario Power output is not equal to power consumption. You forget inefficiencies and losses to heat. Your 40A charge probably consumes 600+ watts, not 480. Maybe more. ## Shop now Using the most advanced technology, we can provide customers with efficient, reliable, and energy-saving power conversion solutions.
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# Inertial oscillations A drifting buoy set in motion by strong westerly winds in the Baltic Sea in July 1969. When the wind has decreased the uppermost water layers of the oceans tend to follow approximately inertia circles due to the Coriolis effect. This is reflected in the motions of drifting buoys. In the case there are steady ocean currents the trajectories will become cycloides. The inertia circles are not eddies; a set of buoys close to each other would be co-moving, rather than revolve around each other ## Rotating planet Picture 2. Animation Schematic representation of inertial oscillation. The pattern of motion of the buoy that is being tracked is due to a pattern of dynamics that is present only in the case of a rotating planet. Animation 2 shows a preview. I will first discuss a simpler case, and then I will proceed to how it works for the Earth as a whole. Also available: the java applet inertial oscillation, a 3D simulation of the physics of inertial oscillation. Understanding of the physics of inertial oscillation is the key to understanding how the rotation of the Earth affects the oceanic and atmospheric dynamics. Inertial oscillation is the simplest, purest case. Usually the motion of air mass is affected by both pressure gradient and Coriolis effect. Inertial oscillation shows how water mass and air mass move when there is no pressure gradient to begin with, and no buildup of pressure gradient in the course of the motion. ## Simpler case as physical model: parabolic dish The dynamics of terrestrial inertial oscillations can be understood by noting the parallels with a simpler case: motion over the surface of a dish with a parabolic cross section. In the fluid dynamics lab of the MIT Earth, Atmosphere and Planetary sciences faculty students can set up various demonstrations, and one of them is the construction of a parabolic turntable. A platform with a diameter of one meter, with a rim of a couple of centimeters, rotating at a constant angular velocity of about 30 revolutions per minute, is filled with a resin that takes several hours to set. Because the platform is rotating, the resin gets redistributed in such a way that in the final state the surface has a parabolic shape; the center is a centimeter or so deeper than the perimeter. The equilibrium state of the rotating fluid is called 'solid body rotation'. The formal name of the solid that is formed is 'paraboloid of revolution'. After the resin has set the surface is sanded to a smooth finish. For demonstrations a small disk of dry ice is placed on the parabolic dish. The evaporating carbondioxide formes an air cushion, resulting in very low friction. Also, in demonstrations the parabolic dish must be rotating at exactly the same angular velocity as when it was manufactured. Picture 3. Image The blue arrow represents the force of gravity. The red arrow represents the normal force. The green arrow represents the resultant force. When it was still liquid the resin was in solid body rotation, hence at every distance to the center of rotation the inclination of the surface is such that the resultant force of gravity and the normal force provides the amount of centripetal force that is necessary to co-rotate with the dish. The strength of this centripetal force is exactly proportional to the distance to the axis of rotation. If the puck is released in such a way that it has a small velocity relative to the dish then it will follow (to a first approximation) the trajectory that is shown on the left side of animation 4. The shape of the trajectory is an ellipse, and the motion is rather like an orbit. The right side depicts the motion as seen from a point of view that is co-rotating with the dish. Picture 4. Animation Frictionless motion along the surface of a parabolic dish. The arrow represents the centripetal force due to the inclination of the surface. ### Orbit Normally, the word 'orbit' is used for the motion of a satellite in space, being subject to the inverse square law of gravity. There are in fact two force laws that give rise to periodic orbits. One is the inverse square law of gravity, and the other one is this case: a proportional force. The shallower the parabolic dish the better the validity of approximating the trajectories as ellipses. The MIT rotating dish demonstration shows a picture. Picture 5. Image This motion allong an ellipse-shaped orbit can be thought of as a linear combination of two (perpendicular) harmonic oscillations. The following parametric equation of the position as a function of time describes the motion of the puck over the surface of the dish. The parametric equation provides a complete description: it describes the shape of the orbit and the velocity at each point in time. a half the length of the major axis b half the length of the minor axis Ω 360° divided by the duration of one revolution In the articles on this site I use the greek capital Ω (Omega) to refer to a constant factor; when the small greek letter ω (omega) is used it refers to the instantaneous angular velocity of some object. The instantaneous angular velocity of some object may fluctuate, depending on the circumstances. In the case of a harmonic force the trajectory of the object can be expressed as a function of a constant factor Ω. Usually the orbit will be an eccentric orbit, and then the actual angular velocity oscillates, together with an oscillation in distance to the center of rotation. ### Circle and epi-circle Interestingly, the parametric equation can be rearranged into the following two components: Picture 6. Animation Animation 6 represents this rearrangement geometrically. It shows that the ellipse-shaped orbit has some remarkable symmetries: it can be thought of as a combination of two circular motions, both with uniform angular velocity Ω. The epi-circle corresponds with the degree of eccentricity of the overall orbit as compared to perfectly circular motion. When motion along this particular ellipse-shaped orbit is mapped in a rotating coordinate system the eccentricity of the ellipse-shaped trajectory shows up as motion along a small circle. In the case of modeling atmospheric dynamics on a parabolic dish this small circle is called an inertia circle. The factor 2 in the above parametric equation is especially noteworthy. The frequency of the motion along the inertia circle is twice the frequency of the overall rotation. ### Separating the coordinate transformation In this discussion I separate the task of taking coordinate transformation into account from the discussion of the physics. In this article, the coordinate transformations are taken care of by the animations, allowing me to focus on the physics taking place. ### Java applet When motion is mapped in a rotating coordinate system the rotation is taken into account by adding the centrifugal term and the coriolis term. The centrifugal term and the Coriolis term are depicted in the following interactive animation: Coriolis effect. ### Careful arrangement When the parabolic dish was manufactured the dish was rotating at a certain angular velocity, and the resin assumed a shape that matched the rotation rate. In other words, the manufacturing process tuned the parabolic dish to a certain rotation rate. The period of an orbit over the surface of that parabolic dish coincides with the dish's rotation rate. In demonstrations the instruction is to use precisely the same rate of rotation as when the parabolic dish was manufactured, and to use that angular velocity for the rotating coordinate system. The outcome of that careful arrangement is that the Coriolis term in the equation of motion matches the motion along the epi-circle precisely. This correspondence is visualized in the above mentioned interactive animation: Coriolis effect ## Energy conversions: doing work Picture 7. Animation Harmonic oscillation; the restoring force is proportional to the distance to the center. The case of inertial oscillation on a shallow parabolic dish and the rotational-vibrational coupling discussed in the previous article are perfectly analogous. In the case of inertial oscillations the potential energy reaches its maximum at the extremal points. As the object is being drawn closer to the center of rotation the centripetal force is doing work, converting potential energy to kinetic energy. The kinetic energy reaches its maximum at the points where the object is at its closest to the center of rotation. When the object moves away from the center of rotation again the centripetal force is doing negative work, decreasing the kinetic energy of the object. Picture 8. Animation Final situation after friction has dissipated the energy that was associated with the eccentricity of the trajectory. #### Dissipation of energy I will discuss dissipation of energy now, for that will reveal some interesting aspects. The natural process is: as long as a system can dissipate energy, it will. Consider a situation where the only way to dissipate energy is friction between the surface and the object that is supported by the surface. In the case of inertial oscillation there is dissipation of energy as long as there is a velocity with respect to the rotating parabolic dish. Friction reduces the eccentricity of the orbit. When the orbit has become circular there is still a lot of potential and kinetic energy, but as there is no friction there is no opportunity for energy dissipation. Animation 8 depicts the state that the system will evolve to. Interestingly, it is an example of equipartition of energy. It is a common property of many kinds of dynamical systems that when the system has reached a final state (no more opportunity for energy dissipation) then the system's total energy will be divided evenly over the available degrees of freedom. Here, the available degrees of freedom are gravitational potential energy and rotational kinetic energy. The final state can be understood as energized degrees of freedom in a state of equilibrium with each other. In the final state the ratio of rotational kinetic energy to gravitational potential energy is 1:1 at every distance to the center of rotation. During manufacture of the parabolic dish, this is the state of equilibrium that the still liquid resin ended in when it reached a state of solid body rotation. ## Analogy between parabolic dish and oblate spheroid Picture 9. Image The blue arrow represents the force of gravity. The red arrow represents the normal force. The green arrow represents the resultant force. Picture 10. Image Forces in the case of an oblate spheroid. The resultant force provides the required centripetal force. Picture 11. Animation Inertial oscillation over the surface of an oblate speroid. Picture 12. Diagram The theoretically predicted period of inertial oscillations at various latitudes. The shape of the Earth and the shape of a parabolic dish manufactured as described above are analogous cases. The solid Earth is ductile; while the Earth's rock is brittle under impact, over periods of millions of years it behaves as a very viscous fluid. (For a demonstration of a very viscous fluid see the Pitch drop experiment.) The Earth's shape develops automatically towards an equilibrium shape. The equilibrium state is one of hydrostatic equilibrium. When the solar system was formed the Earth started out as a protoplanetary disk, and over time this distribution of space debris contracted and compacted to a final rotation rate and shape: an oblate spheroid. The current oblateness is tuned to the current rotation rate, just as the parabolic dish's manufacturing process tuned it to a particular rotation rate. On the poles and on the equator the force of gravity and the normal force are exactly opposite in direction; on all other latitudes the two are not exactly opposite, giving rise to a resultant force, that provides the centripetal force that is necessary for buoyant objects to remain on the same latitude. Without this centripetal force objects would slide towards the equator. The centripetal force that is depicted in Image 10 can be decomposed in a component parallel to the local surface and a component perpendicular to the local surface. The component that is parallel to the local surface is called the poleward force. ### Inertial oscillation over the surface of an oblate spheroid In effect each hemisphere is a bowl, with the poles as points of lowest gravitational potential. Animation 11 represents the motion of a buoy that is co-moving with a mass of seawater that is in inertial oscillation as depicted in the image at the start of the article. The motion pattern is the same as the motion pattern depicted in animation 4. If you would be far in space, exactly above the north pole, but thousends of kilometers above it, then the ellipse-shape of the trajectory would be plain to see. The depiction in animation 11 is highly schematized. Actual inertial oscillations of surface level seawater have amplitudes in the order of several kilometers, in the animation the oscillations have an amplitude that is a considerable fraction of the Earth's diameter. As in the analogous case of motion over the surface of parabolic dish the motion pattern is determined by a centripetal force. There is an oscillation between the poleward force doing positive work and doing negative work. When the buoy is furthest away from the poles it is circumnavigating the Earth's axis slower than the Earth, and consequently the poleward force pulls it closer to the Earth's axis, and in pulling it closer the angular velocity of the buoy increases again. There is a very interesting inertial oscillation visualizer available on the oceanmotion.org website. Try for example the following settings: · Starting direction: Southward. · Starting velocity 20 m/s. · Latitude 15 degrees. The Oceanmotion.org inertial oscillation visualizer also incorporates the beta-effect. Closer to the equator the Coriolis effect is weaker, so closer to the equator the deflection is less strong. As a consequence the inertion oscillation motion as a whole will travel westward. Picture 13. Image At 30 degrees latitude, when the inertial oscillation has an amplitude of L, the motion towards and away from the Earth's axis is 1/2 L #### Dependency on latitude The period of revolution of inertial oscillation is different at different latitudes. Diagram 12 shows the theoretically predicted period of inertial oscillation at various latitudes. The closer to the equator, the longer the period. #### Work Image 13 illustrates why the inertial oscillations have longer periods the further away from the poles. The amount of work that the centripetal force does is proportional to how much distance is travelled towards and away from the central axis of rotation. The inertial mass of the object in inertial oscillation is always the same, but the amount of work that the centripetal force does is not the same at every latitude; the closer to the equator, the smaller the amount of work that the centripetal force can do. #### Buoyancy and the poleward force Picture 14 . Image The poleward force. In Image 14 the blue arrow represents the poleward force. It's important to note that the poleward force applies if and only if the object is supported; it applies only when a normal force sustains the object's height with respect to the Earth's surface. Without the normal force as shown in the image there is no poleward force. An airship (also known as a zeppelin) is an interesting example of a buoyant object. It is tempting to think of an airship as weightless, having released itself from effects of gravity altogether, and accordingly it's tempting to expect an airship to move in the same way as a satellite does, only slower. In fact an airship is subject to a normal force, as the airship displaces a large volume of air. Gravity and the normal force do not cancel each other, there is a small angle between them, that provides required centripetal force. A satellite on the other hand is does not experience a normal force; a satellite is subject to gravity only. To denote the sense of an object's weight being supported, with little to no friction to move parallel to the supporting surface, I use the expression 'buoyant'. All of the air mass of the atmosphere is buoyant, and hence subject to the poleward force. In the next article Comparison of ballistics and inertial oscillations I discuss the contrast between inertial oscillation and ballistic trajectories. ### Presence of other influences Inertial oscillation is the case where any pressure gradient is negligably small, the only influence is the terrestrial Coriolis effect. In the actual atmosphere pressure gradients are affecting the direction of air mass all the time. In the article formation of cyclonic flow. I discuss how the terrestrial Coriolis effect and pressure gradient force are both involved. ### Why the oscillations are called inertial oscillations Picture 15. Animation Inertial oscillation over the surface of an oblate speroid. Generally, inertial motion is asscociated with uniform motion along a straight line, and the motion depicted in the animation is far from uniform: it's not along a straight line and the velocity isn't constant. But there is another sense of 'inertial motion' that does apply in the case of inertial oscillation. In general, motion is referred to as 'inertial motion' when an accelerometer doesn't register any acceleration. Accelerometers onboard a satelite in orbit do not register acceleration since both the accelerometer casing and the movable parts inside are identically influenced by the Earth's gravitation. Orbital motion is free fall motion, the satellites are falling towards the Earth all the time - but they don't get there because they also have sufficient tangential velocity to overshoot the Earth all the time. In the case of a buoy or a weather balloon that is moving along with inertial oscillation an onboard accelerometer would not register acceleration in the direction parallel to the local surface. Of course the accelerometer will register the gravitational acceleration that is associated with the local vertical component of the gravitational force; the buoy is supported by its buoyancy, the buoyancy support prevents it from "going with the flow" of the local vertical component of the gravitational acceleration. But the accelerometer won't register an acceleration parallel to the surface. The motion of the object parallel to the surface is motion that is "going with the flow" of gravitational acceleration, and then an accelerometer doesn't register acceleration. This illustrates again that it's useful to think of inertial oscillation as a form of orbital motion. Next article: Comparison of ballistics and inertial oscillations. Sources of information that were instrumental in developing this article: Rigorous mathematical treatment of inertial oscillation: History of understanding of the Trade Winds:
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# 2xsquared + 6x - 14 = 0 2 by Crissamil 2016-07-19T21:24:05+08:00 A=2 b=6 c=-14 x=-b+and- • Brainly User 2016-07-20T14:19:34+08:00
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Upcoming SlideShare × # Module 3 lesson 17 583 views Published on Published in: Education, Technology • Full Name Comment goes here. Are you sure you want to Yes No Your message goes here • Be the first to comment • Be the first to like this ### Module 3 lesson 17 1. 1. Module 3 Lesson 17.notebook February 06, 2014 Module 3, Lesson 17 02/6/14 The Area of a Circle Homework: Problem Set 17 #2, 3, 4, 8 Quiz Friday 2/7/14 Do Now: Do Lesson 16 Exit 1 2. 2. Module 3 Lesson 17.notebook February 06, 2014 2 3. 3. Module 3 Lesson 17.notebook February 06, 2014 Solve and graph the inequality. 3x – 12 < ­21 3 4. 4. Module 3 Lesson 17.notebook February 06, 2014 Solve and graph the inequality. ­5x > 25 4 5. 5. Module 3 Lesson 17.notebook February 06, 2014 Solve and graph the inequality. x -3 + 7 < -2 5 6. 6. Module 3 Lesson 17.notebook February 06, 2014 You have \$30 to spend at the bowling alley. It costs \$4 per game and \$3 to rent shoes. What is the maximum number of games you can play. Write, solve, and graph an inequality. 6 7. 7. Module 3 Lesson 17.notebook February 06, 2014 What is the measure of the missing angle? 2x 40 7 8. 8. Module 3 Lesson 17.notebook February 06, 2014 What is the measure of the missing angle? 3x 45 235 8 9. 9. Module 3 Lesson 17.notebook February 06, 2014 80 2x + 10 9 10. 10. Module 3 Lesson 17.notebook February 06, 2014 Homework Answers 10 11. 11. Module 3 Lesson 17.notebook February 06, 2014 Homework Answers 11 12. 12. Module 3 Lesson 17.notebook February 06, 2014 Homework Answers 12 13. 13. Module 3 Lesson 17.notebook February 06, 2014 S.94 13 14. 14. Module 3 Lesson 17.notebook February 06, 2014 14 15. 15. Module 3 Lesson 17.notebook February 06, 2014 15 16. 16. Module 3 Lesson 17.notebook February 06, 2014 16 17. 17. Module 3 Lesson 17.notebook February 06, 2014 17 18. 18. Module 3 Lesson 17.notebook February 06, 2014 18 19. 19. Module 3 Lesson 17.notebook February 06, 2014 19 20. 20. Module 3 Lesson 17.notebook February 06, 2014 20 21. 21. Module 3 Lesson 17.notebook February 06, 2014 21 22. 22. Module 3 Lesson 17.notebook February 06, 2014 22 23. 23. Module 3 Lesson 17.notebook February 06, 2014 Problem Set 23 24. 24. Module 3 Lesson 17.notebook February 06, 2014 24 25. 25. Module 3 Lesson 17.notebook February 06, 2014 Click me! 25 26. 26. Module 3 Lesson 17.notebook February 06, 2014 26
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Question Thu March 25, 2010 # Polynomials Sun March 28, 2010 Let α, β and γ be the roots, comparing x3 +3px2 +3qx + r with, ax3 +bx2 +cx + d α + β + γ = -b/a = -3p αβ+ βγ + αγ = c/a = 3q αβγ = -d/a = -r, Now as α, β, γ are required to be in AP. Let α = a-d,  β = a, γ = a+d then, α + β + γ = -3p a-d + a + a + d  = -3p 3a = -3p, a = -p and αβ+ βγ + αγ = 3q (a-d)a + a(a+d) + (a-d)(a+d) = 3q a2 - ad + a2 + ad + a2 - d2 = 3q 3a2 - d2 = 3q 3p2 - d2 = 3q d2 =3p2 -  3q αβγ = -r (a-d)(a)(a+d) = -r a(a2-d2) = -r -p(p2-d2) = -r (p2- 3p2 + 3q) = r/p (3q - 2p2) = r/p The required condition. Regards, Team, TopperLearning. Related Questions Mon November 20, 2017 # FROM P = 5K I HAVE NOT UNDER STAND GIVE SOME DTAILED CONCEPT Mon November 20, 2017
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# Different types of craps bets complexity of craps Pass Line and Don’t Pass Line bets The easiest bet to roll is the Pass Line bet. As a rule, casinos will require that you place at least the minimum bet on the pass line if you are the shooter. It is optional for the other players, but we recommend beginners to start with this bet. You will also notice that most experienced players are very fond of this bet. There is a large ring that goes around the table that says Pass Line. If you suspect that you are placing your bet here, you are a smart guy. Here you place your bet and the shooter then fires the dice for the come out roll. A few things can happen depending on what he rolls: If a 7 or 11 is thrown, you will automatically win this bet. If a 3 or 12 is rolled, you will automatically lose that bet and the game will start over. If the shooter rolls ANY other number, that number becomes the point and your bet remains on the table. If a 7 is rolled afterwards, you will lose that bet. If the point is thrown, you will win this bet. The reverse bet on the Pass Line is of course the Don’t Pass Line bet. This bet is commonly known as betting with the house or several other curses that we do not share here. Basically, this bet wins when most of the other players who play pass line bets and other point bets lose. This is what happens with this come out roll bet: If a 2 or 3 is thrown, you will automatically win this bet. If a 7 or 11 is rolled, you will automatically lose that bet. If the shooter rolls a 12, the bet is a push. If the shooter rolls ANY other number, that number becomes the point and your bet remains on the table. If a 7 is rolled afterwards, you will win this bet. If the point is thrown, you will lose that bet. Come and don’t come bets These bets are pretty similar to Pass Line and Don’t Pass bets, except that they are made AFTER the come out roll. You can complete these at any time between throws in the corresponding areas where a point is set. This happens on a come bet based on what the shooter rolls: If the first roll after you make your come bet is 7 or 11, you win your bet. If a 2, 3 or 12 is thrown, you lose your bet. If ANY other number is rolled then your bet remains on the table and that number is the corresponding “point” of the bet. If a 7 is rolled afterwards, you will lose that bet. If this “point” is rolled, you win the bet. The don’t come bet is basically the opposite of the come bet. You can complete these at any time between throws where a point is set. This happens on a don’t come bet based on what the shooter rolls: If the first roll after making your don’t come bet is a 7 or 11, you lose your bet. If a 2, 3 or 3 is thrown, you win that bet. If ANY other number is rolled then your bet remains on the table and that number is the corresponding “point” of the bet. If a 7 is rolled afterwards, you will win this bet. If this “point” is rolled, you lose the bet. Odds betting Do you remember the bets that have no house edge? You found her. These are bets that are made after the point has been set and are essentially extensions of the four bets mentioned above (pass line, don’t pass line, come bet, don’t come bet). These bets follow the same rules as the bets that accompany them. For example, if you place a odds bet on your Pass Line bet, your odds bet will win if your Pass Line wins and lose if it loses. The same applies to the other three bets. real odds 4.7 Star App Store Review! Cpl.dev***uke The Communities are great you rarely see anyone get in to an argument :) king***ing Love Love LOVE
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Categories Flash Sparkler Just a quick post to show off one of the new effects that was in my recent FlashBelt session on particles. [kml_flashembed movie=”/wp-content/uploads/manual/2007/sparkler.swf” width=”445″ height=”280″ FVERSION=”9″ /] I’m really pleased how this one came out, it’s a combination of a little spark particle and a smoke effect, although the smoke looks a little more like retinal burn. Which is fine by me 🙂 It looks like there’s a kind of motion blur as the particles fly out, but all I’m doing is actually rotating the lozenge shaped spark particle to match the direction the particle is moving in. The way we calculate the angle of the direction of movement uses pythagorus’ theorum – it’s the same way that we calculate the angle of a right-angled triangle. ```clip.rotation = Math.atan2(velocityY, velocityX); ``` Except of course that this will return the value in radians and we need degrees. And we do that by multiplying by 180/PI: ```clip.rotation = Math.atan2(velocityY, velocityX) * 180 / Math.PI; ``` Source is in my earlier FlashBelt post. 12 replies on “Flash Sparkler” Angel Wallenbergsays: Very useful information was found here, thank you for your work. Is there any way you could easily explain to an AS3 illiterate person how to modify particles52.fla to obtain this sparkler effect you have? I love that thing…thanks! sebsays: I’ve decided I’m going to try harder to reply to comments on my blog 🙂 Mawkus – there are some AS2 examples here from the original particles session here //www.sebleedelisle.com/?p=4 The only difference with the sparkler effect is that the particle image is rotated in the direction it’s travelling. Look at the AS3 particle class and see if you can copy the maths bit into AS2. You may find that you don’t get the performance that you need in AS2 though. Good luck! Seb Mawkussays: Hey Seb, Is there anyway you could supply me with the source files for this same exact sparkler effect? I have CS3 so I have AS3 as well. I would be more than happy to give you credit for the effect. It’s for an intro animation for a website. Let me know what you think. Thanks! Mawkussays: Hey Seb, I found the exact sparkler effect I was looking for in the files you provided. I have one more question. I am trying to move this effect as an external .swf in an empty movie clip in another .fla file. So far it loads fine but the effect stays in one position. I noticed in your code that the coordinates are defined to one spot. Is there anyway to remove/edit that coding to where I could control its movement? Thanks! sebsays: hey mawkus, yeah you can just change the x and y position to anything you want! It’s in the constructor for the particles i think. Good luck! Seb Mawkussays: Thanks for the response again Seb. But I don’t just want to move the position of the effect. I actually want the sparkler effect to move on a path, say over 20 frames in the timeline from left to right on the stage. Can I do omit the AS that has the x/y coordinates in the sparkler file so that I can move the effect in an empty movie clip in another flash file? Thanks. sebsays: hey mawkus I’m not sure what you mean when you say that you want the effect in “another flash file”. There are many ways to move the position of the sparkler over 20 frames, the easiest way would be to store an x position in a variable, increment it every frame, and use that value for the x position of the sparkler. I hope this helps. kindest Seb Mawkussays: When I say “another flash file” I mean I am trying to move the sparkler.swf with an empty movie clip in a new Flash (.fla) file. Is there an easy way to move the sparkler over 20 frames with the movie clip method? I am afraid I don’t know Flash and AS enough to store an x position in a variable to move the sparkler, which you suggested in the previous post. If it’s easy and you have time, could you explain further in detail? Thanks Seb! -Mawkus sebsays: Hi Mawkus, yeah I think it may be a little complicated to teach you all of that in the comments on my blog 🙂 I would definitely recommend Keith Peter’s book, Actionscript 3 animation, which is a great introduction to creating cool effects in code. Good luck! Seb […] just for FlashBelt, I’d rewritten all of my particle experiments into AS3 and realised I could get at least 10 times as many particles as in the AS2 […]
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# PCalc binomial functions Over the weekend, I read Federico Viticci’s excellent post on using iOS 12 Shortcuts to access PCalc. I haven’t installed iOS 12 yet (I usually wait a few days to let Apple’s servers cool down), but I expect to use Federico’s advice to make a few PCalc shortcuts when I do. In the meantime, his post inspired me to clean up and finish off a PCalc function that I’d half-written some time ago. The function calculates a probability from the binomial distribution. Imagine there are $N$ independent trials of some event. There are only two possible outcomes of each event, which we’ll call success and failure, although you could give them any names you like. For each trial, the probability of success is $p$. The probability of $n$ successes in those $N$ trials is $(Nn)pn(1−p)N−n\binom{N}{n}\; p^n\; (1 - p)^{N-n}$ The first term in the formula is the binomial coefficient, which is defined as $(Nn)=N!n!(N−n)!\binom{N}{n} = \frac{N!}{n!\;(N - n)!}$ The binomial probability, then, is a function of three inputs, $n$, $N$, and $p$, that yields one output. The PCalc function I wrote to implement the function starts with three numbers on the stack,1 say $n=5$, $N=10$, and $p=\mathrm{.5}$, Any entries on the stack “above” the three inputs will still be there after the function runs. The function is called “Binomial PMF.” The PMF stands for probability mass function, which is standard terminology for the function that returns the probability of a random variable equaling a discrete value. The function is defined this way, which is a lot of steps to enter. You’d probably prefer to download it. If you look through the function definition, you’ll notice that I’m using up a lot of registers—more than is truly necessary to run out the calculation. I do this for a couple of reasons: 1. The registers are there to be used, and by keeping all the intermediate calculations in separate registers, I’m able to make the logic of the function easier to follow. 2. I have further plans for this function. I want to use it as the starting point for a more complicated function, the cumulative distribution function, which will need to keep track of some of those intermediate calculations. We’ll look at the Binomial CDF function in the next post. 1. Yes, it’s an RPN-only function. RPN is the natural way to handle this many inputs, and because I don’t use PCalc’s algebraic mode, I didn’t see any reason to write the function to work in that mode.
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# How do you simplify sqrt10 (3- 2sqrt6 )? $3 \sqrt{10} - 4 \sqrt{15}$ $\sqrt{10} \left(3 - 2 \sqrt{6}\right) = 3 \sqrt{10} - 2 \sqrt{60} = 3 \sqrt{10} - 2 \sqrt{4 \cdot 15} = 3 \sqrt{10} - 2 \cdot 2 \sqrt{15} = 3 \sqrt{10} - 4 \sqrt{15}$
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• Create Account FREE SOFTWARE GIVEAWAY We have 4 x Pro Licences (valued at \$59 each) for 2d modular animation software Spriter to give away in this Thursday's GDNet Direct email newsletter. Read more in this forum topic or make sure you're signed up (from the right-hand sidebar on the homepage) and read Thursday's newsletter to get in the running! Posted 16 January 2013 - 11:41 PM I've always heard and known of the conditional operator in C++ though for some reason my brain would always forget exactly how it's read.  I decided to take a quick look at it again as I do see some places in my code that is a lot longer than it should be just because I always used a full if/else statement to do one quick test. I just wanted to ask to make sure I am reading these right. lets say I have the following: // assume x has already been declared and defined earlier in the program. int a = (x > 100) ? 0 : 1; This would read and be the same as: // again assume x is defined and declared earlier in the program if(x > 100) a = 0; else a = 1; I just want to make sure I am right in reading this.  Even though I do feel dumb that I have always forgot about this operator. So in other words really if the first expression in true, then the second expression is executed.  If the first expression is false then the third expression is executed.  Right? Posted 16 January 2013 - 11:39 PM I've always heard and known of the conditional operator in C++ though for some reason my brain would always forget exactly how it's read.  I decided to take a quick look at it again as I do see some places in my code that is a lot longer than it should be just because I always used a full if/else statement to do one quick test. I just wanted to ask to make sure I am reading these right. lets say I have the following: // assume x has already been declared and defined earlier in the program. int a = (x > 100) ? 0 : 1; This would read and be the same as: // again assume x is defined and declared earlier in the program if(x > 100) a = 0; else a = 1; I just want to make sure I am right in reading this.  Even though I do feel dumb that I have always forgot about this operator. PARTNERS
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# Egyptian Triangle By Paper Folding III What is known as Haga's First Theorem is a two fold construction of the 3-4-5 ("Egyptian") triangle from a square piece of paper. The construction I believe beats hands down tying knots on a rope (a procedure assumed to be used by the ancient Egyptian and which gave them the moniker of "rope-stretchers".) The first fold in Haga's construction is to mark the midpoint of a side (the top side in the diagram below). The second fold places one of the lower corners at the midpoint of the top edge: This simple procedure generates a 3-4-5 triangle (three of them naturally, as there are similar triangles.) Assume the side of the stare is $8$ units of length. Let the vertical side of the right triangle in the top right corner be $a.$ Then its hypotenuse equals $8-a$ (which is the remaining part of the vertical side of the square.) The Pythagorean theorem then gives an equation $a^{2}+4^{2}=(8-a)^2.$ This simplifies to $16=64-16a,$ from which $a=3,$ as shown in the diagram. [Haga, p.6] draws attention to the fact that the same folding procedure reveals several fractions, e.g. $2/3$ and $1/8.$ The triangle in the upper left corner has the same proportions: $3:4:5$ but its smallest side is now $4,$ implying that the vertical leg equals $4\cdot 4/3=8\cdot 2/3$ so that the vertical leg is $2/3$ of the side of the square. The hypotenuse of that triangle equals $4\cdot 5/3=8\cdot 5/6,$ i.e. $5/6$ of the side of the square. Thus the long leg of the small protruding triangle at the left edge of the square is $1/6$ of the side of the square, making the short side $1/8$ and the hypotenuse $5/24.$ It follows that the second fold marks exactly $8\cdot 1/8=1$ unit up from the bottom edge of the square. ### References 1. Kazuo Haga. et al, Origamics: Mathematical Explorations Through Paper Folding, World Scientific Publishing Company, 2008 Paper Folding Geometry
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# SPSS Review CENTRAL TENDENCY & DISPERSION ## Presentation on theme: "SPSS Review CENTRAL TENDENCY & DISPERSION"— Presentation transcript: SPSS Review CENTRAL TENDENCY & DISPERSION SOC 3155 SPSS Review CENTRAL TENDENCY & DISPERSION Survey Items Nominal Measurement Good Not as good What college (e.g., CLA, CHESP) are you enrolled in? Not as good Are you a freshman, sophomore, junior, or senior? What is your college major? Political affiliation/Religion Survey Items Ordinal Interval/Ratio Mostly Likert-type Freshman, sophomore…. How far do you live from campus (0-5 miles…) Interval/Ratio GPA Hours study per week Number of credits carried Review Class Exercise: Professor Numbscull surveys a random sample of Duluth residents. He wants to know if age predicts the support for banning so-called “synthetic marijuana” products (very supportive, somewhat supportive, not supportive). What are the IV and DV—what is their level of measurement? Ratios/Proportions/Percents What percent of people enjoy fishing? What is the ratio of Ford F-150 owners to Toyota Prius Owners? What proportion of those who do not enjoy fishing drive a Honda Accord? Type of Vehicle Enjoy Fishing Do not enjoy Fishing TOTAL Honda Accord 11 14 25 Ford Escort 33 10 43 Toyota Prius 3 8 Ford F-150 Truck 41 2 TOTALS 88 34 122 SPSS CODING ALWAYS do “recode into different variable” INPUT MISSING DATA CODES Variable labels Check results with original variable Useful to have both numbers and variable labels on tables EDITOPTIONSOUTPUTPIVOT TABLES Variable values in label shown as values and labels Class Exercise Recode a variable from GSS Create a variable called “married_r” where: 0 = not married 1 = married Be sure to label your new variable! Run a frequency distribution for the original married variable and “married_r” Distribution of Scores All the observations for any particular sample or population Examples: Distribution (GSS Histogram) Exercise Create your own histogram from the GSS data using the variable “TVHOURS” Concept of “skew” Positive vs. negative Skew is only possible for ordered data, and typically relevant for interval/ratio data Measures of Central Tendency Purpose is to describe a distribution’s typical case – do not say “average” case Mode Median Mean (Average) Measures of Central Tendency Mode Value of the distribution that occurs most frequently (i.e., largest category) Only measure that can be used with nominal-level variables Limitations: Some distributions don’t have a mode Most common score doesn’t necessarily mean “typical” Often better off using proportions or percentages Measures of Central Tendency A more ambiguous mode Measures of Central Tendency 2. Median value of the variable in the “middle” of the distribution same as the 50th percentile When N is odd #, median is middle case: N=5: median=6 When N is even #, median is the score between the middle 2 cases: N=6: median=(5+9)/2 = 7 MEDIAN: EQUAL NUMBER OF CASES ON EACH SIDE Measures of Central Tendency 3. Mean The arithmetic average Amount each individual would get if the total were divided among all the individuals in a distribution Symbolized as X Formula: X = (Xi ) N Measures of Central Tendency Characteristics of the Mean: It is the point around which all of the scores (Xi) cancel out. Example: X (Xi – X) 3 3 – 7 -4 6 6 – 7 -1 9 9 – 7 2 11 11- 7 4 X = 35 (Xi – X) = Mean as the “Balancing Point” Think of the values in the previous slide as weights Think of the mean as the point at which the weights would be balanced. X Measures of Central Tendency Characteristics of the Mean: 2. Every score in a distribution affects the value of the mean As a result, the mean is always pulled in the direction of extreme scores Example of why it’s better to use MEDIAN family income POSITIVELY SKEWED NEGATIVELY SKEWED Measures of Central Tendency In-class exercise: Find the mode, median & mean of the following numbers: 2 Does this distribution have a positive or negative skew? Measures of Central Tendency Levels of Measurement Nominal Mode only (categories defy ranking) Often, percent or proportion better Ordinal Mode or Median (typically, median preferred) Interval/Ratio Mode, Median, or Mean Mean if skew/outlier not a big problem (judgment call) Measures of Dispersion provide information about the amount of variety or heterogeneity within a distribution of scores Necessary to include them w/measures of central tendency when describing a distribution Measures of Dispersion Range (R) The scale distance between the highest and lowest score R = (high score-low score) Simplest and most straightforward measure of dispersion Limitation: even one extreme score can throw off our understanding of dispersion Measures of Dispersion 2. Interquartile Range (Q) The distance from the third quartile to the first quartile (the middle 50% of cases in a distribution) Q = Q3 – Q1 Q3 = 75% quartile Q1 = 25% quartile Example: Prison Rates (per 100k), 2001: R = 795 (Louisiana) – 126 (Maine) = 669 Q = 478 (Arizona) – 281 (New Mexico) = 197 25% 50% 75% 281 478 126 366 795 MEASURES OF DISPERSION Problem with both R & Q: Calculated based on only 2 scores MEASURES OF DISPERSION Standard deviation Uses every score in the distribution Measures the standard or typical distance from the mean Deviation score = Xi - X Example: with Mean= 50 and Xi = 53, the deviation score is = 3 The Problem with Summing Deviations From Mean 2 parts to a deviation score: the sign and the number Deviation scores add up to zero Because sum of deviations is always 0, it can’t be used as a measure of dispersion X Xi - X 8 +5 1 -2 3 0 0 -3 12 0 Mean = 3 Average Deviation (using absolute value of deviations) Works OK, but… AD =  |Xi – X| N X |Xi – X| AD = 10 / 4 = 2.5 X = 3 Absolute Value to get rid of negative values (otherwise it would add to zero) Variance & Standard Deviation Purpose: Both indicate “spread” of scores in a distribution Calculated using deviation scores Difference between the mean & each individual score in distribution To avoid getting a sum of zero, deviation scores are squared before they are added up. Variance (s2)=sum of squared deviations / N Standard deviation Square root of the variance Xi (Xi – X) (Xi - X)2 5 1 2 -2 4 6  = 20  = 0  = 14 Terminology “Sum of Squares” = Sum of Squared Deviations from the Mean =  (Xi - X)2 Variance = sum of squares divided by sample size =  (Xi - X)2 = s2 N Standard Deviation = the square root of the variance = s Calculation Exercise Number of classes a sample of 5 students is taking: Calculate the mean, variance & standard deviation mean = 20 / 5 = 4 s2 (variance)= 14/5 = 2.8 s= 2.8 =1.67 Xi (Xi – X) (Xi - X)2 5 1 2 -2 4 6  = 20 14 Calculating Variance, Then Standard Deviation Number of credits a sample of 8 students is are taking: Calculate the mean, variance & standard deviation Xi (Xi – X) (Xi - X)2 10 -4 16 9 -5 25 13 -1 1 17 3 15 2 4 14 18  = 112 72 AGAIN, A SMALL NUMBER OF CASES. IN THIS CASE, N=8 Calculate the variance & standard deviation of these 8 scores. MEAN = 14 (112 / 8) VARIANCE = SUM OF SQUARED DEVIATIONS/# OF SCORES 72 / 8 = 9 Standard deviation = sq. rt. of variance Summary Points about the Standard Deviation Uses all the scores in the distribution Provides a measure of the typical, or standard, distance from the mean Increases in value as the distribution becomes more heterogeneous Useful for making comparisons of variation between distributions Becomes very important when we discuss the normal curve (Chapter 5, next) 1 – Like the mean in this sense 2 – Good index of variation 3 – Another form of standardization (putting things into a common metric, just like %s or rates) Becomes very important when we discuss the normal curve (Chapter 5, next) 4 – Read Chapter 5 for next time Mean & Standard Deviation Together Tell us a lot about the typical score & how the scores spread around that score Useful for comparisons of distributions: Example: Class A: mean GPA 2.8, s = 0.3 Class B: mean GPA 3.3, s = 0.6 Mean & Standard Deviation Applet
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## Commutative residuated lattice-ordered semigroups Abbreviation: CRLSgrp ### Definition A commutative residuated lattice-ordered semigroup is a residuated lattice-ordered semigroup $\mathbf{A}=\langle A, \vee, \wedge, \cdot, \to\rangle$ such that $\cdot$ is commutative: $xy=yx$ Remark: This is a template. If you know something about this class, click on the Edit text of this page'' link at the bottom and fill out this page. It is not unusual to give several (equivalent) definitions. Ideally, one of the definitions would give an irredundant axiomatization that does not refer to other classes. ##### Morphisms Let $\mathbf{A}$ and $\mathbf{B}$ be commutative residuated lattice-ordered semigroups. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a homomorphism: $h(x \vee y)=h(x) \vee h(y)$, $h(x \wedge y)=h(x) \wedge h(y)$, $h(x \cdot y)=h(x) \cdot h(y)$, and $h(x \to y)=h(x) \to h(y)$. ### Definition A is a structure $\mathbf{A}=\langle A,...\rangle$ of type $\langle ...\rangle$ such that $...$ is …: $axiom$ $...$ is …: $axiom$ Example 1: ### Properties Feel free to add or delete properties from this list. The list below may contain properties that are not relevant to the class that is being described. Classtype variety yes yes ### Finite members $\begin{array}{lr} f(1)= &1\\ f(2)= &\\ f(3)= &\\ f(4)= &\\ f(5)= &\\ \end{array}$ $\begin{array}{lr} f(6)= &\\ f(7)= &\\ f(8)= &\\ f(9)= &\\ f(10)= &\\ \end{array}$ ### Subclasses [[Commutative distributive residuated lattice-ordered semigroups]] subvariety [[Commutative residuated lattices]] expansion ### Superclasses [[Residuated lattice-ordered semigroups]] supervariety [[Commutative lattice-ordered semigroups]] subreduct
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# Thread: Complex polygon perimeter/area equation 1. ## Complex polygon perimeter/area equation Hi. What I'm looking to do is probably impossible with only the given data, but I've come here to make certain before giving up! I have a non-regular polygon with several concave and convex characteristics. I know the number of sides, the lengths of each side (therefore the total perimeter as well), and the area of the polygon. I do no, however, know the locations of each side, nor what the polygon actually looks like. Is it possible to calculate the new area of the polygon if its perimeter is offset outward by a certain factor. (i.e., each side would be shifted perpendicularly to its length outward by a specified amount.) I want to say no, based on the sheer number of unknowns, but there's this nagging feeling that says there must be a general correlation between area, perimeter, and # of sides (the polygon's complexity). Any help is appreciated! Thanks 2. ## Re: Complex polygon perimeter/area equation Yes. If the sides of the polygon are scaled by a constant $k$, the perimeter is multiplied by $k$ and the area is multiplied by $k^2$. 3. ## Re: Complex polygon perimeter/area equation richard1234, thanks for the reply. If I were scaling the polygon as a whole, or if it were a regular polygon, that'd be great. Unfortunately, the scenario I'm describing offsets each side by the specified distance. If you're familiar with any CAD software, this would be an offset command, not a scale command. For example, in concave conditions (such as a U or C shape), the offset sides of the polygon would be moving toward each other. So, although the sides are being offset perpendicularly by the same distance, some sides might actually decrease in length.
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# Chi-Squared Probability Function (χ2) (Redirected from chi-squared distribution) A Chi-Squared Probability Function (χ2) is the Gamma probability function from a Chi-squared distribution family (based on a sum of squares of $k$ independent standard normal random variables). • AKA: $\chi^2$. • Context: • Example(s): • $\chi^2(k=2,x=0.5) = \frac{1}{2^{0.5} \times (0.5 - 1)!) \times (0.5^{(0.5 - 1)}) \times (e^{(((-1) * 0.5) / 2)}} = 0.43939129...$ • $\chi^2(k=2,x=1) =$ (1 / ((2^0.5) * ((0.5 - 1) !)) * (1.0^(0.5 - 1)) * (e^(((-1) * 1.0) / 2)) = 0.24197072... • $\chi^2(k=2,x=2) =$ (1 / ((2^0.5) * ((0.5 - 1) !)) * (2.0^(0.5 - 1)) * (e^(((-1) * 2.0) / 2)) = 0.10377687... • Counter-Example(s): • See: Chi-Squared Test. ## References ### 1999 • (Lane, 1999) ⇒ D. Lane. (1999). “HyperStat Online Textbook". Chapter 16: Chi Square.
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# Accounting (26th Edition) Edit edition Problem 18E from Chapter 26: Internal rate of return method—two projectsMunch N’ Crunch S... We have solutions for your book! Chapter: Problem: Internal rate of return method—two projects Munch N’ Crunch Snack Company is considering two possible investments: a delivery truck or a bagging machine. The delivery truck would cost \$43,056 and could be used to deliver an additional 95,000 bags of pretzels per year. Each bag of pretzels can be sold for a contribution margin of \$0.45. The delivery truck operating expenses, excluding depreciation, are \$1.35 per mile for 24,000 miles per year. The bagging machine would replace an old bagging machine, and its net investment cost would be \$61,614. The new machine would require three fewer hours of direct labor per day. Direct labor is \$18 per hour. There are 250 operating days in the year. Both the truck and the bagging machine are estimated to have seven-year lives. The minimum rate of return is 13%. However, Munch N’ Crunch has funds to invest in only one of the projects. a. Compute the internal rate of return for each investment. Use the present value of an annuity of \$1 table appearing in this chapter (Exhibit 5). b. Provide a memo to management, with a recommendation. Step-by-step solution: Chapter: Problem: • Step 1 of 3 a) Internal rate of return (IRR): IRR for an investment proposal is the discount rate that equates the present value of the expected net cash flows with the initial cash out flow. The IRR considers the time value of money, the initial cash investment and all cash investment and all cash from the investment. But unlike the net present value method The IRR method does not use the desired rate of return but estimates the discount rate that makes the present value of subsequent net cash flows equal to the initial investment. This discount rate is calling IRR. It is based on the project cost and project benefit. In ignores the fact how the project is founded. IRR calculated as follows: For TRUCK: Step: 1Calculation of annual net cash inflows: Particulars Amount Sale proceeds of truck \$42750 Less: Operating cost (32,400) Annual net cash inflows \$10,350 Step: 2 Calculated the Payback period =4.16 Step: 3Ascertain the present value annuity factor In PVAF table looking for a discount rate that gives PVAF equals to PBP (step-1) in the year equal to life of the project. PVAF (15%, 7years) = 4.16 Step: 4 calculate the NPV and find out the IRR If NPV is’ 0’, that itself is IRR If NPV is ‘Positive’, try higher rate If NPV is ‘Negative’, try Lower rate =\$0 On the basis of above calculation NPV is ‘0’. Hence IRR is 15%. For BAGGING MACHINE: Step: 1Calculation of annual net cash inflows: Sale proceeds of truck =\$13,500 Step: 2 Calculated the Payback period =4.564 Step: 3Ascertain the present value annuity factor In PVAF table looking for a discount rate that gives PVAF equals to PBP (step-1) in the year equal to life of the project. PVAF (12%, 7years) = 4.564 Step: 4 calculate the NPV and find out the IRR If NPV is’ 0’, that itself is IRR If NPV is ‘Positive’, try higher rate If NPV is ‘Negative’, try Lower rate =\$0 On the basis of above calculation NPV is ‘0’. Hence IRR is 12%. • Chapter , Problem is solved. Corresponding Textbook Accounting | 26th Edition 9781305088405ISBN-13: 1305088409ISBN: Jim ReeveAuthors:
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CSE 2320 Section 501/571 Fall 1999 Homework 5 Solution 1. TOPOLOGICAL-SORT 2. STRONGLY-CONNECTED-COMPONENTS 3. MST-PRIM using vertex a as the root vertex. 4. Execution of the Edmonds-Karp algorithm. 5. Compute the Knuth-Morris-Pratt prefix function for the following patterns when the alphabet is . (a) abccbabbcaabcbabccbabcba i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 P[i] a b c c b a b b c a a b c b a b c c b a b c b a 0 0 0 0 0 1 2 0 0 1 1 2 3 0 1 2 3 4 5 6 7 3 0 1 (b) aaaaaaaabbbbbbbbaaaaaaaa i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 P[i] a a a a a a a a b b b b b b b b a a a a a a a a 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 6. Determine a possible pattern (if any) consistent with the following KMP prefix functions when the alphabet is . (a) i 1 2 3 4 5 6 7 8 0 1 2 3 0 1 0 1 P[i] a a a a b a b a The general form of the pattern has characters 1-4, 6 and 8 the same, and characters 5 and 7 different from the others. (b) i 1 2 3 4 5 6 7 8 0 1 2 1 2 1 2 1 No pattern would generate this prefix function, because the first three characters must be the same to generate a prefix of 012. And means that P[4]=P[1]=a. But then P[1..4] are the same, and therefore, would be 3, not 1. (c) i 1 2 3 4 5 6 7 8 0 0 0 1 1 0 0 0 P[i] a b b a a c c c The general form of the pattern is that characters 1, 4 and 5 must be the same. Characters 2, 3, 6, 7 and 8 must be different from P[1], and characters 2 and 6 must be different from eachother.
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Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: fraction and exponent calculators Related topics: learn algebra 1 | algebra square root calculator | free 3nd grade math printouts | graphing 3 variables on ti 83 | algebra for dummies online free | real life questions in trigonometry | balancing chemical equations online | difficult algebra problems | long division involving x cubed | fractions for dummies Author Message ocla Registered: 12.04.2007 From: Posted: Saturday 30th of Dec 20:51 I am in a real problem . Somebody save me please. I am getting a lot of dilemma with absolute values, least common measure and adding fractions and especially with fraction and exponent calculators. I have to show some snappy growth in my math. I heard there are many Applications available online which can assist you in algebra. I can pay some cash too for an effective and inexpensive tool which helps me with my studies. Any reference is greatly appreciated. Thanks. Jahm Xjardx Registered: 07.08.2005 From: Odense, Denmark, EU Posted: Monday 01st of Jan 14:48 Hey. I believe I can assist . Can you elucidate some more on what your troubles are? What precisely are your problems with fraction and exponent calculators? Getting a first rate teacher would have been the greatest thing. But do not fret . I think there is a way out . I have come across a number of math programs. I have tried them out myself. They are pretty smart and first rate . These might just be what you need. They also do not cost a lot. I feel what you need is Algebrator. Why not try this out? It could be just be the answer for your problems . DoniilT Registered: 27.08.2002 From: Posted: Wednesday 03rd of Jan 07:44 It would really be nice if you could tell us about a resource that can offer both. If you could get us a home tutoring software that would offer a step-by-step solution to our assignment , it would really be good . Please let us know the genuine links from where we can get the software. Jrahan Registered: 19.03.2002 From: UK Posted: Wednesday 03rd of Jan 20:34 I remember having often faced problems with factoring, linear algebra and geometry. A truly great piece of math program is Algebrator software. By simply typing in a problem homework a step by step solution would appear by a click on Solve. I have used it through many algebra classes – College Algebra, Pre Algebra and College Algebra. I greatly recommend the program. cxrecthotsme Registered: 16.01.2005 From: St. Albans, UK Posted: Thursday 04th of Jan 18:53 Great! Dudes . Only one more issue . Where can I obtain the magical program ? Matdhejs Registered: 08.12.2001 From: The Netherlands Posted: Friday 05th of Jan 10:40 Visit https://linear-equation.com/syllabus-for-differential-equations-and-linear-alg.html and hopefully all your problems will be resolved .
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Subscribe to our Youtube Channel - https://you.tube/teachoo 1. Chapter 5 Class 10 Arithmetic Progressions 2. Concept wise 3. Checking if AP or not and finding a, d Transcript Ex 5.1 ,4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (xi) a, a2, a3, a4 a, a2 , a3 , a4 Difference between second and first term = a2 a = a (a 1) Difference between third and second term = a3 a2 = a2 (a 1) Since, difference is not same. a (a 1) a2 (a 1) Hence it is not an AP Ex 5.1 ,4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (xii) 2, 8, 18, 32 2 , 8 , 18 , 32 Difference between second and first term = 8 2 = (2 2 2) 2 = (2 22) 2 = 2 2 2 = 2 Difference between third and second term = 18 8 = (3 3 2) (2 2 2) = (32 2) (22 2) = 3 2 2 2 = 2 Difference between fourth and third term = 32 18 = (4 4 2) (2 3 3) = (42 2) (32 2) = 4 2 3 2 = 2 Since difference is same, it is an AP Common difference = d = 2 We have to find next 3 terms, We are given 4 terms So, we need to find 5th, 6th and 7th term Fifth term = fourth term + common difference = 32 + 2 = (4 4 2) + 2 = (42 2) + 2 = 4 2 + 2 = 5 2 Sixth term = Fifth term + Common difference = 5 2 + 2 = 6 2 Seventh term = Sixth term + common difference = 6 2 + 2 = 7 2 Hence, 5th, 6th and 7th terms are 5 2 , 6 2, 7 2 Ex 5.1 ,4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (xiii) 3, 6, 9, 12 3 , 6 , 9 , 12 Difference between second and first term = 6 3 = (3 2) 3 = 3 2 3 = 3 2 3 1 = 3 ( 2 1) Difference between third and second term = 9 6 = (3 3) (3 2) = 3 3 3 2 = 3 ( 3 2) The difference is not common So, it is not A.P. Ex 5.1 ,4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms (xiv) 12, 32, 52, 72 12, 32, 52, 72 Difference between second and first term = 32 12 = 9 1 = 8 Difference between third and second term = 52 32 = 25 9 = 16 Difference is not same So, it is not A.P Ex 5.1 ,4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms (xv) 12, 52, 72, 73 12, 52, 72, 73 Difference between second and first term = 52 12 = 25 1 = 24 Difference between third and second term = 72 52 = 49 25 = 24 Difference between fourth and third term = 73 49 = 24 Since difference is same, it is an AP. Common difference = d = 24 We have to find next 3 terms, We are given 4 terms So, we need to find 5th, 6th and 7th term Fifth term = fourth term + common difference = 73 + 24 = 97 Sixth term = fifth term + common difference = 97 + 24 = 121 Seventh term = sixth term + common difference = 121 + 24 = 145 So, next 3 terms of AP are 97, 121 and 145 Checking if AP or not and finding a, d
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Artificial Intelligence 2012-02-20 • Weak AI - solving hard problems. State of art. • AI - a machine can understand problems. (Classical problem of Chinese room) How to cope with uncertainty? • Act as in believe state. • We eliminated some states by performing actions. • Big problem of large number of n states. • Trying can explore actions by going to $$2^n$$ (I can use 2 actions from one state) • Apriorni pravděpodobnost (?aprior probability?) – unconditional probability • Příklad: $$P(Cavity \mid toothache) = \alpha$$ • Nezavislost nam umoznuje ukladat mensi tabulky pro vypocet marginalni pravdepodobnosti. • Podminena nezavislost je castejsi a take setri $$P(X \mid Y,Z) = P(X \mid Y)*P(X \mid Z)$$ vypocet
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###### a few weeks ago I had done a simple calculation that had shown Boeings flying a distance of ~10,000 kms on earth lose a time of the range of micro-second. I had done it without effects of gravity. A Boeing flies ~10 kms above earth surface for a good part of the flight so it would measure a smaller time effected by a smaller gravitational potential of it’s height. An airplane is synchronized by the GPS clocks. Two airports 10,000 kms apart if not synchronized properly by this GPS system would lose a big time even in an hr. In a 15 hr trip it could reflect really well. (one can do the calculation) SO for the OPERA baseline which is not even 1000 kms we can not allow as big an error as 60 nano-secs since we are covering that distance at speed of light. That is we are allowing ourselves ~2.5 milli-secs only for the entire flight. A 60 nano-sec error in 2.5 milli-secs becomes 1.3 seconds in 15 hr flight. (not to mention what error we accumulate for satelites which govern our daily routines such as a favorite TV show, two places such as Delhi and Bhubaneswar would see the show at very different times, eg people in one city may lose few minutes of the show because the satelites warned us much later as they are so far above us) Also note that a 2 second error in a day is not something even a dead-battery would do (joke). You can do this experiment with your digital watch if your watch is not synchronized by a GPS satelite, if it says 2 seconds either way after you have fixed all other factors you would know that there is something wrong with GPS system, we do not need very precise experiments to claim GPS precision if we are to lack such in the first place … Before you start at Narita airport, you ask your friends in the destination country to show you a watch over video conference. You fix your watch time. Then you fly and keep precise note of what time it is when you landed the airport according to airport clocks and your clock after you made your time correction as per airport announcement. (You know how much you adjusted) Then you go to your friend’s place and he would broadcast his time again which will be noted as per another friend in Japan who did not fly. You are done. basically you have more than one clock, all fixed differently as per your need. (that is reference clocks at different points and experimental clocks) To get honest clocks you have to be careful, you may install pre-arranged clocks from the lab with airports permission or someone may just be present with such a clock .. SO this can also be done to test the actual syncronization of clocks at both ends of OPERA. If it gives you synch below 1 sec for a corresponding time-of-flight you have set an upper limit for the actual synch offset thereby making a very confident measurement when neutrinos are involved . OPERA calibrated it’s clocks in 3 methods. 1. GPS clock, 2. Cs clocks on lab (these are world wide synched by GPS clocks hence only reflect GPS precision) 3. fiber optics cables It can have electric cables (they must already have) and send electric currents or something like that which travel much slower so a much longer TOF than a second is available. Once that is the case any errors of second level will be determined .. Two computer servers pre-synched at one place and placed at two different labs will note the timing of these two clocks on their servers and in a 24 hrs they will show a 2 second difference. I think this they already did, suggestion: transport them by bullock carts since the relativists think 4 km/s will bring in a time dilation or even a flight transportation, then the debate will go onto oh we transported them via Grey Hound buses and these buses offer a bumpy ride hence that may also be the cause for time dilation.
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# Column calculations and formula 2 months 4 days ago #30648 by Janinem Column calculations and formula was created by Janinem Hi, I am hoping that someone can help me with coding for the following. I am waiting to be able to do the following. within section A there are columns (I am going to say column a,b,c,d,e,f,g,h) I need it to be that if columns a-h are 0 then the total is 0 and If the columns have a value in them, then we need to add the columns up, x2 add 4. I hope that it makes sense. janine ##### Attachments: Please Log in or Create an account to join the conversation. 1 month 3 weeks ago #30651 by bert Replied by bert on topic Column calculations and formula Hi janine! Sorry for the delay, I missed your post If you explain with a bit more detail what to do with those values once they appear I'll prepare the formula Please Log in or Create an account to join the conversation. 1 month 3 weeks ago #30652 by Janinem Replied by Janinem on topic Column calculations and formula Hi Bert So basically this is what I want.  There are 8 sub columns within the A folder, Lets name them 1,2,3,4,5,6. - if all the values in each column are 0 then the total needs to be zero. (must not take into account the next formula) - however if there is a value in each column (the value can only be 1) then we would need to add all the columns up to get a total. - this total would then need to be multipled by 2 and add 4 - this will give me a total out of 20 I hope this makes sense, Janine Please Log in or Create an account to join the conversation. 1 month 3 weeks ago #30653 by Munay Replied by Munay on topic Column calculations and formula Hello, I believe this formula should work for what you want to do. I attach a screenshot of the formula and one of the result in the gradebook. In your A folder change the calculation to formula and enter your grade over 20 or a total points of 20. In the formula just replace the HÜ columns (my columns) for your 8 columns and recreate the rest. Let me know how it does. ##### Attachments: Please Log in or Create an account to join the conversation. 1 month 3 weeks ago #30657 by Janinem Replied by Janinem on topic Column calculations and formula That you so much, that does exactly what I want. Janine Please Log in or Create an account to join the conversation.
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ALGEBRA 1 EOC REVIEW 4 EXPONENTS, FOIL, FACTORING, SYSTEMS 1 / 35 # ALGEBRA 1 EOC REVIEW 4 EXPONENTS, FOIL, FACTORING, SYSTEMS - PowerPoint PPT Presentation ALGEBRA 1 EOC REVIEW 4 EXPONENTS, FOIL, FACTORING, SYSTEMS. SIMPLIFY:. 7 x 2 – 5 x – 3 + 2 x. 1). A) 7 x 2 + 3 x – 3. B) x 4. C) – 3. D) 7 x 2 – 3 x – 3. SIMPLIFY:. (- 3 c 3 d 4 )(5 c 5 d 2 ). 2). A) - 15 c 15 d 8. B) - 15 c 8 d. C) - 15 c 8 d 6. D) - 8 c 8 d 8. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'ALGEBRA 1 EOC REVIEW 4 EXPONENTS, FOIL, FACTORING, SYSTEMS' - ivory-wilkinson Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ALGEBRA 1 EOC REVIEW 4 EXPONENTS, FOIL, FACTORING, SYSTEMS SIMPLIFY: 7 x2 – 5 x – 3 + 2 x 1) A) 7 x2 + 3 x – 3 B) x4 C) – 3 D) 7 x2 – 3 x – 3 SIMPLIFY: (- 3 c3 d4)(5 c5 d2) 2) A) - 15 c15 d8 B) - 15 c8 d C) - 15 c8 d6 D) - 8 c8 d8 SIMPLIFY: 3) B) A) - 15 x2 y3 z3 D) C) SIMPLIFY: (6 b2 c3)2 4) A) 12 b4 c6 B) 36 b4 c6 C) 12 b4 c5 D) 36 b4 c5 SIMPLIFY: 5) B) A) D) C) SIMPLIFY: 6) B) A) D) C) SIMPLIFY: 7) B) A) D) C) 8) Find the perimeter of a triangle whose sides are (3 x2 + 5); (5 x – 2); and (6 x2 + 5 x). A) 9 x2 + 10 x + 3 B) 19 x2 + 3 C) 9 x4 + 10 x2 – 3 D) 7 x4 + 10 x – 3 9) Simplify: (2 x + 5) (2 x – 3) A) 4 x2 – 15 B) 4 x2 – 4 x – 15 C) 4 x2 + 4 x – 15 D) 8 x – 15 10) Simplify: (3 x2 + 5 x + 1) – (7 x2 – 2) A) - 4 x2 + 5 x + 3 B) - 4 x2 + 5 x – 1 C) x2 – 1 D) x2 + 3 11) Simplify: (x – 2) (3 x2 – x + 4) A) 3 x3 – 7 x2 + 6 x – 8 B) 3 x3 – 6 x2 + 6 x + 4 C) 3 x3 + 7 x2 – 6 x – 8 D) 2 x3 – 8 12) Find the perimeter of a rectangle if the width is (2 x – 4) and the length is (5 x + 1). A) 7 x – 3 B) 7 x + 3 C) 14 x – 6 D) 14 x + 6 13) Find the area of a triangle if the base is (2 x – 4) and the height is (x + 6). A) x2 + 4 x – 12 B) 2 x2 + 8 x – 24 C) 2 x2 – 8 x – 24 D) 3 x + 2 SIMPLIFY: 14) A) 2 x y – 3 + 4 x B) 2 y – 3 + 4 x C) 2 y – 3 + 4 y D) 2 x y – 3 + 4 x2 SIMPLIFY: 15) A) B) C) D) SIMPLIFY: 16) A) B) C) D) 17) The area of a rectangle is given by the expression: x2 – 5 x – 6. The length and width have only integral coefficients. Which of the following could represent the length of the rectangle? A) x – 6 B) x – 2 D) x – 1 C) x – 3 Given: 2 x – 3 y = 12 6 x + 2 y = 42 18) What is x + y? 2(2 x – 3 y = 12) 4 x – 6 y = 24 3(6 x + 2 y = 42) 18 x + 6 y = 126 22 x = 150 x = 150/22 = 75/11 2(75/11) – 3 y = 12 x + y = 75/11 + 6/11 y = 6/11 150/11 – 3 y = 12 – 3 y = - 18/11 x + y = 81/11 = 7.4 19) A restaurant received 270 hamburger patties and 350 hotdogs on Monday for \$ 450. On Friday the restaurant received 550 hamburger patties and 425 hotdogs for \$ 630. a) How much did each hamburger cost? x = cost of a hamburger cost of a hamburger = \$ .38 y = cost of a hotdog cost of a hotdog = \$ .995 270 x + 350 y = 450 550 x + 425 y = 630 b) How much will 25 hamburgers and 50 hotdogs be? 25(\$.38) + 50(\$.995) = \$ 59.25 20) A local pet store has triple the amount of fish as birds and has a total of 250 fish and birds. Write a system of equations represents the number of fish and birds using the variables F and B. F = number of fish B = number of birds F = 3(62.5) = 187.5 F = 3 B F + B = 250 No solution since you cannot have a fraction of a bird or of a fish. 3 B + B = 250 4 B = 250 B = 62.5 Given: 4 x + 3 y = 60 x – y = 10 21) x = y + 10 What is the value of x ? 4(y + 10) + 3 y = 60 4 y + 40 + 3 y = 60 7 y + 40 = 60 7 y = 20 y = 20/7 = 2 6/7 = 2.86 x = 2 6/7 + 10 = 12 6/7 = 12.86 Given: 2 x + y = 15 5 x – 6 y = - 22 22) y = - 2 x + 15 What is the value of x – y ? 5 x – 6(- 2 x + 15) = - 22 5 x + 12 x – 90 = - 22 17 x – 90 = - 22 y = - 2(4) + 15 y = 7 17 x = 68 y = - 8 + 15 x = 4 x – y = 4 – 7 = - 3 A) 11 B) 2 D) - 3 C) 3 Given: w = 1 – v 2 v + w = 4 23) What is the value of w ? 2 v + 1 – v = 4 w = 1 – v = 1 – 3 = - 2 v + 1 = 4 v = 3 A) 3 B) 2 D) - 2 C) 1 A limosine company charges a flat-fee of \$ 80 plus \$.05 per mile. A shuttle van company charges a flat-fee of \$ 60 plus \$.50 per mile. Approximately what mileage will yield the same fare for both? 24) limo: y = .05 x + 80 shuttle: y = .5 x + 60 .05 x + 80 = .5 x + 60 2000 = 45 x 5 x + 8000 = 50 x + 6000 44.4 = x A) 24 miles B) 34 miles D) 54 miles C) 44 miles The price of six sodas and four candy bars is \$ 18.50. The price of two candy bars and eight sodas is \$ 20.50. What is the price of a candy bar? 25) x = number of sodas y = number of candy bars 6 x + 4 y = 18.50 x = \$ 2.25 y = \$ 1.25 8 x + 2 y = 20.50 A) \$ 1.25 B) \$ 2.25 D) \$ 2.15 C) \$ 1.65 The area of a rectangle is given by the expression: x2 – 5 x – 6. The length and width only have integral coefficients. Which of the following could represent the length of the rectangle? 26) (x – 6)(x + 1) A) x – 6 B) x – 3 D) x – 1 C) x – 2 27) Factor: x2 + 9 x + 18 (x + 6) (x + 3) 28) Factor: x2 – 13 x y – 30 y2 (x – 3 y)(x – 10 y) 29) Factor: w2 + 2 w – 15 (w + 5) (w – 3) 30) Factor: x3 + 5 x2 + 6 x x(x2 + 5 x + 6) x(x + 2)(x + 3) A restaurant makes at least 50 pizzas a night, but no more than 250 pizzas. The restaurant makes at least 20 salads but no more than 90 salads. A total of no less than 325 pizzas and salads are made each night. Each pizza makes a profit of \$ 3.00. Each salad makes a profit of \$ 2.25. What is the maximum profit the restaurant can make in a night? 31) Constraints: P(x, y) = 3 p + 2.25 s 50 ≤ p ≤ 250 20 ≤ s ≤ 90 p + s ≥ 325 A) \$ 998.25 B) \$ 881.25 D) \$ 952.50 C) \$ 907.50 31) x + y ≥ 325 P3(250, 90) Corner points: P1(235, 90) y ≤ 90 P2(250, 75) x > 50 y ≥ 50 x ≤ 250 P(235, 90) = 3(235) + 2.25(90) = \$ 672.50 P(250, 75) = 3(250) + 2.25(75) = \$ 918.75 HIGHEST PROFIT P(250, 90) = 3(250) + 2.25(90) = \$ 952.50 SIMPLIFY: 32) A) B) C) D) 34) Solve the following system of inequalities: 2 x + y < 3 x – 2 y ≤ 8 y < - 2 x + 3 y ≥ ½ x – 4 y = - 2 x + 3 y = ½ x – 4 34) Solve the following system of inequalities: 3 y ≥ 6 x – 9 3 x + 4 y < 12 y ≥ 2 x – 3 y < - ¾ x + 3 y = - ¾ x + 3 y = 2 x – 3
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# tf.keras.metrics.Hinge Computes the hinge metric between `y_true` and `y_pred`. ``````tf.keras.metrics.Hinge( name='hinge', dtype=None ) `````` `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. For example, if `y_true` is [-1., 1., 1.], and `y_pred` is [0.6, -0.7, -0.5] the hinge metric value is 1.6. #### Usage: ``````m = tf.keras.metrics.Hinge() m.update_state([-1., 1., 1.], [0.6, -0.7, -0.5]) # result = max(0, 1-y_true * y_pred) = [1.6 + 1.7 + 1.5] / 3 print('Final result: ', m.result().numpy()) # Final result: 1.6 `````` Usage with tf.keras API: ``````model = tf.keras.Model(inputs, outputs) model.compile('sgd', metrics=[tf.keras.metrics.Hinge()]) `````` #### Args: • `fn`: The metric function to wrap, with signature `fn(y_true, y_pred, **kwargs)`. • `name`: (Optional) string name of the metric instance. • `dtype`: (Optional) data type of the metric result. • `**kwargs`: The keyword arguments that are passed on to `fn`. ## Methods ### `reset_states` View source ``````reset_states() `````` Resets all of the metric state variables. This function is called between epochs/steps, when a metric is evaluated during training. ### `result` View source ``````result() `````` Computes and returns the metric value tensor. Result computation is an idempotent operation that simply calculates the metric value using the state variables. ### `update_state` View source ``````update_state( y_true, y_pred, sample_weight=None ) `````` Accumulates metric statistics. `y_true` and `y_pred` should have the same shape. #### Args: • `y_true`: The ground truth values. • `y_pred`: The predicted values. • `sample_weight`: Optional weighting of each example. Defaults to 1. Can be a `Tensor` whose rank is either 0, or the same rank as `y_true`, and must be broadcastable to `y_true`. Update op.
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# Counting connected manifolds Apparently, there is the following fact: The set of homeomorphism classes of connected manifolds has the same cardinality $c$ as that of $\mathbb R$. I find it to be interesting; but would be happier to see a proof, and would be grateful for a reference somewhere. - Already in dimension 3 there are continuously many non-homeomorphic manifolds (in fact even contractible open subsets of $\mathbf{R}^3$, see my answer here mathoverflow.net/questions/4155/…). – algori May 17 '10 at 2:05 Upper bound (assuming the manifolds are second countable): every manifold admits a complete metric, and the "set" of isometry classes of complete separable metric spaces is of cardinality continuum. Indeed, every such a space is a completion of a countable metric space, and there is only continuum of metrics on a countable set. Lower bound: there is a continuum of non-isomorphic fundamental groups of manifolds. E.g. any countable sum of cyclic groups is a first homology group of a connected manifold: just take a connected sum of an appropriate countable collection of lens spaces. - To be more explicit in the second part: You can encode an infinite binary string as a subset $S$ of the prime numbers. Then you can take a connected sum of the lens spaces $L(p,1)$ for each $p \in S$. You can then use the fundamental group or first homology as an invariant to recover $S$, or even Milnor's theorem that 3-manifolds uniquely factor with respect to connected sum. Technically speaking Milnor's theorem is in the closed category, but it applies to this case. I think that there are only countably many compact manifolds. – Greg Kuperberg May 16 '10 at 22:20 Greg, your last statement is correct. Any compact manifold can be triangulated with finitely many simplexes, and there are only countably many ways to glue these guys together. – Kevin Ventullo May 17 '10 at 1:54 Kevin, are you talking about the smooth case? Aren't there non-triangulable topological manifolds? – Sergei Ivanov May 17 '10 at 8:40 There are countably many compact topological manifolds, as was shown by Cheeger-Kister in "Counting topological manifolds", Topology 9 1970 149--151. The proof depends on results of Edwards-Kirby of deformation of homeomorphisms. A sketch of the proof is here. Incidentally, a good discussion of counting non-metrizable manifolds is in Gauld's paper, section 3. - I am worried that this may not be true in TOP (the topological category). Let us work in PL (piece-wise linear) instead. Every PL manifold can be expressed as a locally finite simplical complex, with at most a countable number of vertices. Thus there are at most $2^N$ of these. (Here $N$ is the natural numbers.) I think I will leave the lower bound as an exercise -- I have a way of doing this by encoding binary sequences using generalized Heegaard splittings of three-manifolds, but that is a hack. I am sure that there is a more beautiful way to give the lower bound just using non-compact surfaces. I looked, but could not find a reference. EDIT: Just in case the above is too brief here is an "easier" exercise. The number of simple locally finite graphs on at most a countable number of vertices is again $2^N$. [Hint: consider the adjacency matrix.] [In fact, the locally finite hypothesis is not necessary in this case.] EDIT2: After thinking a bit more about my three-manifold examples, I realized the statement can be simplified a bit. You can embedd the set of binary sequences into the set of homeomorphism classes of submanifolds of $S^3$ as follows: Choose your two favorite distinct hyperbolic links in $S^3$, each with two components. Lets call these links $L_0$ and $L_1$. Given a binary sequence $s \colon N \to \{0,1\}$ form a three manifold $M_s$ by gluing copies of $L_0$ and $L_1$ as instructed by $s$. Note that $M_s$ embeds in the three-sphere; the complement looks a bit like Antoine's necklace. Finally, $M_s$ determines $s$ by the uniqueness of the JSJ decomposition. As a remark - it is also possible to do this with hyperbolic manifolds (thus having trivial JSJ decomposition) again embedding in the three-sphere. Ah - following the link that algori provides reminds me that there are (up to homeomorphism) uncountably many Whitehead manifolds - open, contractable submanifolds of $S^3$. The point of all of these examples is that it is "easy" to encode information in the end of a three-manifold. I'll leave my examples here, as they are a useful warm-up to understanding the Whitehead manifolds. -
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# 2023/24 Taught Postgraduate Module Catalogue ## MATH5802M Time Series and Spectral Analysis ### 15 creditsClass Size: 88 Module manager: Professor Charles Taylor Email: C.C.Taylor@leeds.ac.uk Taught: Semester 1 (Sep to Jan) View Timetable Year running 2023/24 ### Pre-requisite qualifications MATH2715 or MATH2735. ### This module is mutually exclusive with MATH3802 Time Series This module is approved as an Elective ### Module summary In time series, measurements are made at a succession of times, and it is the dependence between measurements taken at different times which is important. The module will concentrate on techniques for model identification, parameter estimation, diagnostic checking and forecasting within the autoregressive moving average family of models and their extensions. Extensions will include transformations and differencing to deal with non-stationarity, the incorporation of seasonal dependence into the model to deal, for example with monthly series. ### Objectives On completion of this module, students should be able to: a) assess graphically the stationarity of a time series, including the calculation and use of a sample autocorrelation function; b) evaluate the autocorrelation function and partial autocorrelation function for AR, MA and ARMA models; c) use the autocorrelation and partial autocorrelation functions and other diagnostics to formulate, test and modify suitable hypotheses about time series models; d) forecast future values of a time series; e) develop the frequency representation of a stationary time series; f) use the periodogram to carry out harmonic analyses; g) use a statistical package with real data to facilitate the analysis of time series data and write a report giving and interpreting the results. ### Syllabus 1. Overview. Stationarity, outline of Box-Jenkins approach through identification of model, fitting, diagnostic checking, and forecasting. Mean, autocorrelation function, partial autocorrelation function. 2. Models. Autoregressive (AR) models, moving average (MA) models, ARMA models, their autocorrelation functions, and partial autocorrelation functions. Transformations and differencing to achieve stationarity, ARIMA models. 3. Estimation and diagnostics. Identifying possible models using autocorrelation function, and partial autocorrelation function. Estimation, outline of maximum likelihood, conditional and unconditional least squares approaches. Diagnostic checking, methods and suggestions of possible model modification. 4. Forecasting. Minimum mean square error forecast and forecast error variance, confidence intervals for forecasts, updating forecasts, other forecasting procedures. 5. Seasonality, time series regression. 6.The frequency representation of a stationary time series. 7. The use of a periodiogram to carry out harmonic analysis. ### Teaching methods Delivery type Number Length hours Student hours Lecture 33 1.00 33.00 Practical 1 2.00 2.00 Private study hours 115.00 Total Contact hours 35.00 Total hours (100hr per 10 credits) 150.00 ### Private study Studying and revising of course material. Completing of assignments and assessments. ### Opportunities for Formative Feedback Regular example sheets and practicals. ### Methods of assessment Coursework Assessment type Notes % of formal assessment Assignment . 20.00 Total percentage (Assessment Coursework) 20.00 There is no resit available for the coursework component of this module. If the module is failed, the coursework mark will be carried forward and added to the resit exam mark with the same weighting as listed above. Exams Exam type Exam duration % of formal assessment Standard exam (closed essays, MCQs etc) 2 hr 30 mins 80.00 Total percentage (Assessment Exams) 80.00 Normally resits will be assessed by the same methodology as the first attempt, unless otherwise stated
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# Convert exajoule/second to terajoule/second ## How to Convert exajoule/second to terajoule/second To convert exajoule/second to terajoule/second , the formula is used, $\mathrm{TJ/s}=\mathrm{EJ/s}×1,000,000$ where the EJ,s to TJ,s value is substituted to get the answer from Power Converter. 1 EJ,s = 10000e+2 TJ,s 1 TJ,s = 0 EJ,s Example: convert 15 EJ,s to TJ,s: 15 EJ,s = 15 x 10000e+2 TJ,s = 15000e+3 TJ,s ## exajoule/second to terajoule/second Conversion Table exajoule/second (EJ,s) terajoule/second (TJ,s) 0.01 EJ,s 10000 TJ,s 0.1 EJ,s 100000 TJ,s 1 EJ,s 1000000 TJ,s 2 EJ,s 2000000 TJ,s 3 EJ,s 3000000 TJ,s 5 EJ,s 5000000 TJ,s 10 EJ,s 10000000 TJ,s 20 EJ,s 20000000 TJ,s 50 EJ,s 50000000 TJ,s 100 EJ,s 100000000 TJ,s 1000 EJ,s 1000000000 TJ,s ### Popular Unit Conversions Power The most used and popular units of power conversions are presented for quick and free access.
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# How Do You Find A Residual Example? ## How do you find the residual error? The residual is the error that is not explained by the regression equation: e i = y i – y^ i. homoscedastic, which means “same stretch”: the spread of the residuals should be the same in any thin vertical strip. The residuals are heteroscedastic if they are not homoscedastic.. ## What is a residual in statistics? In statistical models, a residual is the difference between the observed value and the mean value that the model predicts for that observation. Residual values are especially useful in regression and ANOVA procedures because they indicate the extent to which a model accounts for the variation in the observed data. ## How do you find the residual in multiple regression? A studentized residual is calculated by dividing the residual by an estimate of its standard deviation. The standard deviation for each residual is computed with the observation excluded. For this reason, studentized residuals are sometimes referred to as externally studentized residuals. ## What is the formula for residual value? The formula to figure residual value follows: Residual Value = The percent of the cost you are able to recover from the sale of an item x The original cost of the item. For example, if you purchased a \$1,000 item and you were able to recover 10 percent of its cost when you sold it, the residual value is \$100. ## What is a good residual value? So when you’re shopping for a lease, the first rule of thumb is to look for cars that hold their value better — the ones that have high residual values. Residual percentages for 36-month leases tend to hover around 50 percent but can dip into the low 40s or be as high as the mid-60s. ## How do you record residual value? In depreciation the residual value is the estimated scrap or salvage value at the end of the asset’s useful life. In the accounting equation, owner’s equity is considered to be the residual of assets minus liabilities. In investment evaluations, the residual value is the profit minus the cost of capital. ## Who determines residual value? This estimate comes from the bank that will hold your lease contract. Let’s say you leased a vehicle with a manufacturer’s suggested retail price (MSRP) of \$20,000 for a three-year lease term, and this vehicle had a 60% residual value. ## What is residual value example? When it comes to the residual value of a leased car, for example, it equals the estimated value of the car at the end of the lease. … If, for example, a bank believes that a \$32,000 car has a residual value of \$15,000 at the end of the lease term, the lessee would need to pay the \$17,000 difference. ## How do you find the predicted and residual value? So, to find the residual I would subtract the predicted value from the measured value so for x-value 1 the residual would be 2 – 2.6 = -0.6. Mentor: That is right! The residual of the independent variable x=1 is -0.6. ## What does the residual tell you? A residual is the difference between the observed y-value (from scatter plot) and the predicted y-value (from regression equation line). It is the vertical distance from the actual plotted point to the point on the regression line. … The plot will help you to decide on whether a linear model is appropriate for your data.
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This is an archived post. You won't be able to vote or comment. [–] 39 points40 points  (21 children) This paper strikes me as just trying to take potshots at economic theory without doing some of the more interesting practical analysis. I don't think anyone will argue that markets fully match the weak efficiency model. What I would like to hear is how comparing it to computing complexity reveals new characteristics of the markets, and where the divergence between model and reality is particularly large. [–] 33 points34 points  (15 children) Simple algorithmic traders (and some humans) don't buy or sell stock based on news about a company's profitability. Instead, they only look at the curve of the stock's recent values, and hope that the information contained in it encodes all the other relevant information which attentive humans have access to. In fact, if that is true, then that's what it means for the market to be weakly efficient. I read this paper a while ago, but from what I can remember the idea is that they assume that the next value of the stock is completely deterministic, and is given by an unknown function f(x1, x2, ..., x_n) of the stock's recent values. Next, they show that an algorithm having access to historic data could not recover f from this data in polynomial time. Indeed if f(x1, x2, ..., x_n) is arbitrary, then we will need to examine all the potential input sequences in order to recover the output value for each case. In the simplest case where each x_i is just a boolean and not a floating point number, there are already 2n inputs to check, so of course recovering f is exponential in the number of inputs f has. Now, where do real world markets differ from this model? I know much less about finance than about computer science, but there are two things which sound wrong to me about this model. • f is not completely arbitrary. For example I expect that similar curves would yield similar results (i.e., f is continuous) • why is n the number of parameters and not the length of the historical data? [–] 1 point2 points  (0 children) The word 'hope' is the important one. Algo's might expect prices to move in certain ways (i.e. they take on some random distribution) but that does NOT mean that they think the next value is deterministic. The process is less like finding a deterministic function, and more like updating probabilistic knowledge. [–] 0 points1 point  (13 children) I'd say that the unpredictable mood swings in the market brought about by high frequency algorithmic trading are a perfect example of why this not only does not work long term, but is damaging to the whole point of having a market to begin with. If your trading volume is small, then the margin of error generated by your actions are likely to be unnoticeable, but the aggregate of all those errors are creating more and more problems for the real world. [–] 1 point2 points  (12 children) How does trading more frequently increase the margin of error? You can do as much damage by submitting a few big bad trades, if not more (e.g. London Whale), which is the opposite of HFT activity. [–] -2 points-1 points  (11 children) Well, it's like this: when you throw all the numbers into a machine and tell it to figure out how to make YOUR numbers go UP you are likely not thinking at all about the long term consequences of the decisions you are making. Add to that: you have multiple large systems fighting against each other focusing only on making their numbers go up, and working on imperfect models of each others behavior and you are bound to have unexplainable negative side effects in the systems those numbers represent: namely, the real world, and the people whose labor are supposedly the source of all those numbers to begin with. When are we going to get back to what the markets were supposed to be about in the first place? That being, distributing the risk of large, costly projects such that we ensure the continual progress of the human race? [–] 1 point2 points  (10 children) I don't think people in the rest of the market, whether they are McDonalds franchisees or traders or petroleum engineers, make decisions on the basis of the long term consequences of their actions, so I don't see how or why this suddenly becomes a problem with respect to financial markets. I can certainly see cases where people might make poor decisions without coordination, but this what does this have to do with trading more quickly or more often? [–] 0 points1 point  (9 children) Beacuse a third of the world's wealth is now tied up in pure number games rather than financial decisions that involve planning beyond a week or so in advance. [–] 0 points1 point  (8 children) Okay, this is extremely confusing to me. What does it mean the world's wealth is tied up? And what is a financial decision but a number game where you try to maximize profit while minimizing risk, etc.? Are we supposed to want our financial decisions to be made non-quantitatively? [–] 0 points1 point  (7 children) When you decouple the true value of things (lets say, the social effects of one family having a roof over their head vs. being homeless) from their monetary value (the dollar value of their mortgage in the derivatives market), you make decisions based on the dollars that make up that financial construct alone, and you ignore the future consequences of having one more family that will be able to contribute less to society in the future because you were playing financial games with something that has worth beyond its monetary value. Is it really that difficult for you to understand? I feel like it's kind of obvious. [–] -1 points0 points  (6 children) If something have 'true value' to someone, it is going to be reflected in their understanding of its 'monetary value'. People trade because they see items priced at 'monetary values' that differ from their 'true value' - and work towards bringing the monetary values back in line. Furthermore, is it really traders and investors' jobs to be deciding what values we should have as a society? It seems much more reasonable to leave those kinds of decisions to civic institutions like governments, churches, etc. [–][S] 10 points11 points  (4 children) where the divergence between model and reality is particularly large. We're running into a problem of self-reference (e.g. reality as portrayed in an Escher drawing diverges the most near the Droste bits) because financial markets are supposed to be the national subsystems that most efficiently analogize/model/abstractly compare stuff (that's also why some of the most cutting-edge AI is on Wall St.). When you have the means to invest in supercomputing clusters that essentially just take advantage of discretization error you end up creating demand for anything that increases the magnitude of the change of the market in the short term (effectively making a profit off instability): Financial black swans driven by ultrafast machine ecology Society's drive toward ever faster socio-technical systems, means that there is an urgent need to understand the threat from 'black swan' extreme events that might emerge. On 6 May 2010, it took just five minutes for a spontaneous mix of human and machine interactions in the global trading cyberspace to generate an unprecedented system-wide Flash Crash. However, little is known about what lies ahead in the crucial sub-second regime where humans become unable to respond or intervene sufficiently quickly. Here we analyze a set of 18,520 ultrafast black swan events that we have uncovered in stock-price movements between 2006 and 2011. We provide empirical evidence for, and an accompanying theory of, an abrupt system-wide transition from a mixed human-machine phase to a new all-machine phase characterized by frequent black swan events with ultrafast durations (<650ms for crashes, <950ms for spikes). Our theory quantifies the systemic fluctuations in these two distinct phases in terms of the diversity of the system's internal ecology and the amount of global information being processed. Our finding that the ten most susceptible entities are major international banks, hints at a hidden relationship between these ultrafast 'fractures' and the slow 'breaking' of the global financial system post-2006. More generally, our work provides tools to help predict and mitigate the systemic risk developing in any complex socio-technical system that attempts to operate at, or beyond, the limits of human response times. [–] 5 points6 points  (1 child) Why do I feel like I'm reading a cyberpunk novel? [–][S] 10 points11 points  (0 children) Good scifi is supposed to mimic reality. [–] 0 points1 point  (1 child) Pardon, dumb question here, but you seem to have a combination of (a) understanding of this space, and (b) demonstrate a great degree of patience :-) [–] 1 point2 points  (0 children) The identity of people behind electronic exchanged-based trades and orders are anonymous, so you can't per se tell who the other traders are. People do try to figure how who's trading against them depending on the situation, e.g. if you're consistently being picked off by an algo, you might try to figure out what they're doing. [–] 14 points15 points  (6 children) Fortunately, when searching for a solution, one is often happy to find one that is close the best possible solution, such as within a constant factor. So is there any reason to believe that the markets are not approximately efficient? I know that not all NP problems can be approximated in P time so hopefully someone here with more knowledge than me can provide an answer to this question based on the constructions used in the paper. :-) [–] 5 points6 points  (2 children) Part of the problem with your question is defining "approximately" appropriately. Without any test datasets or exact solutions, its hard to tell what, if anything, the heuristic constant might be. [–] 6 points7 points  (1 child) If I remember correctly from my C.S. classes, when a problem cannot be approximated efficiently there exists no such heuristic constant, so the question is whether any such constant exists. [–] 1 point2 points  (0 children) You're talking about the APX complexity class I presume? [–] 1 point2 points  (2 children) If its as close to be a constant factor then the polynomial time hierarchy would collapse. There is a good informal overview of why heuristics will never be good enough, here (warning: pdf). [–] 0 points1 point  (1 child) Well constant factor in correctness, not running time, so the polynomial time hierarchy remains intact. For example, the MST heuristic gives a 2-approximation for TSP. [–] 0 points1 point  (0 children) The article I linked is about correctness, their argument is that most of the answer sets that have real world use value are sparse so heuristics look good but cannot approximate the whole set with the same error rate without collapsing the polynomial time hierarchy. [–] 3 points4 points  (0 children) Couple reactions: • Arbitrage opportunities reflect current public information, not past. So they are not an example of enduring profits from past information. • Many people who endorse the weak EMH do so only in the short run. In the long run, things like mean reversion come into play reliably. • The paper links to a lot of good stuff, but the text itself just comes across as gimmicky. I don't think anyone other than Fama maybe really believes that people go through literally every possible strategy to test if it's profitable. • EMH can be motivated by the idea that short run changes in prices are random, rather than the fact that all opportunities have already been taken. [–] 5 points6 points  (37 children) showing how we can “program” the market to solve NP-complete problems. It's not a matter of solving them, it's a matter of solving them quickly. Furthermore, "quickly" is defined in terms of the algorithm used, not the absolute time it takes to compute something. If the algorithm involves a black box (such as "the market"), its computational complexity is impossible to calculate. This claim, even if absolutely true, is irrelevant to P=NP. [–][S] 0 points1 point  (27 children) Furthermore, "quickly" is defined in terms of the algorithm used, not the absolute time it takes to compute something. If the algorithm involves a black box (such as "the market"), its computational complexity is impossible to calculate. Except this is physical reality. There aren't any black boxes (just ask schrodinger's cat). [–] 1 point2 points  (26 children) P vs NP is not a problem of physics. It's a math problem. Unless the market's mechanism is well-understood, we cannot introduce it to a rigorous mathematical proof. [–][S] -1 points0 points  (25 children) P vs NP is not a problem of physics. It's a math problem. Actually it's a theoretical computer science problem that overlaps with core components of both of those academic fields. [–] 1 point2 points  (24 children) IMHO "theoretical computer science" is math. [–][S] -5 points-4 points  (23 children) Only someone who doesn't know any physics would think that. Edit: Because the hive sees fit to downboat me here I should clarify that I'm being hyperbolic, and many people who have degrees in physics are completely unaware of this connection. [–] 2 points3 points  (12 children) [–][S] 0 points1 point  (11 children) [–] 1 point2 points  (2 children) Yeah, that page doesn't say anything about physics . . . [–][S] -1 points0 points  (1 child) I take it you don't know what the word "hamiltonian" means then? [–] 1 point2 points  (7 children) Perhaps you could articulate an actual argument, instead of just linking to random Wikipedia pages? P vs. NP has nothing to do with quantum computation. [–][S] 0 points1 point  (6 children) QMA, which stands for Quantum Merlin Arthur, is the quantum analog of the deterministic complexity class NP or the probabilistic complexity class MA. It is related to BQP in the same way NP is related to P, or MA is related to BPP. [–] 0 points1 point  (9 children) Care to elaborate on that? Sounds like a personal slur to me. [–][S] 0 points1 point  (8 children) Sounds like a perfectly reasonable observation about quantum information and computability. [–] 0 points1 point  (7 children) My background's in finance - give me the layman's rundown. [–][S] 0 points1 point  (6 children) We have been assuming markets are efficient, and markets constantly rely on the production of efficiently confirmed solutions to hard optimization problems (see: cryptocurrency, or The Matter, Forme and Power of a Common Wealth Ecclesiasticall and Civil). In the long-term, markets have to be regulated by non-market external entities. This means treating systems as if they are reversible when in fact they are not (unless they are of a certain class of quantum computer, which is why the D-Wave has financial applications). [–] -1 points0 points  (8 children) It's not a matter of solving them, it's a matter of solving them quickly. For some definition of "quickly". There are very reasonable real-life datasets that could take longer than the lifetime of the universe to compute using an exponential time algorithm. Consider a travelling salesman problem with every city in the United States. [–] 2 points3 points  (6 children) There are very reasonable real-life datasets that could take longer than the lifetime of the universe to compute using an exponential time algorithm. This includes assumptions about available computational power, based on the (estimated) total number and speed of all the computers we've built. While it's not totally unreasonable to extend such assumptions to cover "the market," it does seem to lack a rigorous foundation. How much computing power does "the market" have? How does that compare to a modern supercomputer? What about all the humans involved in "the market" -- do they count? For how much? [–] 0 points1 point  (4 children) Before asking what kind of computational power we have, we need to figure out what the approximate solution to the 'profiting strategy search' problem would look like. EDIT: Looks like i actually just posted responding to the wrong comment XD [–] 0 points1 point  (3 children) Sorry, are you referring to something defined in the paper? I only bothered reading the abstract. Or maybe I Just Don't Get It (TM). [–] 1 point2 points  (2 children) So the paper sets up the problem as "Does there exist a strategy that statistically significantly (after accounting for possible data mining) makes money?" (it's also stated more formally) Then it claims that a potential solution to this problem is verifiable in polynomial time, but an algorithm for solving this problem could be mapped onto the knapsack problem which is NP-complete. So what is an approximate answer to this problem? Also, what is a 'black box' and why is it a problem? (and maybe I don't get it either - my sole experience is a UG Algo class so take what I say with a grain of salt :D ) [–] 1 point2 points  (1 child) Also, what is a 'black box' and why is it a problem? A "black box" is a component of the system that is poorly understood, or not understood at all. Even if the market is efficient, we don't know how it achieves this. In order to show that P=NP under that assumption, we'd first need to prove that the market's solution to the profiting strategy search problem is a polynomial-time algorithm. But since we don't even know what that algorithm is, such a proof is a lost cause. We could try to informally demonstrate it based on the total computational power of the market vs. the speed at which it is efficient, but IMHO the "total computational power of the market" is ill-defined given all the human brains involved. [–] 1 point2 points  (0 children) Agreed, I think this paper is flawed in a number of ways. Weak EMH holding does not necessarily imply the existence of solution(s) to such a profit finding algorithm. [–] -1 points0 points  (0 children) Assuming 1. The current monetary value of computing resources on Wall St is non-decreasing, 2. Computing power scales with the number of transistors on a die, and 3. Moore's law, we can show that Wall St's computing power increases exponentially. Given an algorithm that runs in exponential-time (in the CS sense) then, the "real-world" runtime may scale only linearly. Say Wall St can perform 1 operation per unit time, and a market-efficiency algorithm requires O(2n ) operations to complete. On input of size n, our linear-real-time algorithm could be as simple as 1. Wait n*18 months, 2. Run the algorithm on the new computers. (TWAJS) [–] 1 point2 points  (0 children) TSP has been done for Euclidian data sets with 100,000 items, and other NP-hard problems have been solved with data sets of over 10,000,000 items. An algorithm being worst-case exponential does not mean you'll often see that behaviour in practice or on real world data sets. Start with this: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.97.3555 [–] 1 point2 points  (9 children) If this paper is right, Keynes was right. As a computer scientist who believes in P <> NP, this is kind of a fact. And the implications... The implications are HUGE. Specially for modern capitalism. [–] 0 points1 point  (8 children) What does it show Keynes was right about? [–] 0 points1 point  (7 children) Keynes(more or less) said that free markets aren't perfect. So, you need a gov agency to leverage the inefficiencies. [–] 0 points1 point  (6 children) Modern capitalism already involves (and indeed requires) laws and government. I doubt there are many economists who disagree on that point. What are you implications you see? [–] 0 points1 point  (4 children) I answered this question before, but domehow the abswer didn't appear. :-/ There is many. Our system, today, is defined to be the "best" and in the US, specially, market capitalism is defined like the top of the possible economic paradises, that only requires a few mote patches... With this, the paper states: "you'll never reach that state. You'll be patching forever" This is tremendous cuz financials & economies are growing into monstrous huge worldwide spiderwebs. So, crises are going to persist and continue to grow and correlating wars. I know sounds like too much. I only ask you to analize current trends in five years or ten. This paper means, in a mathematical way that crises will have no end, neither poverty, there is many more... [–] 0 points1 point  (3 children) There's no reason that not being able to reach perfection means we can't have improvements - poverty can be alleviated without achieving complete equality. [–] 0 points1 point  (2 children) Problem is crisis cycles. [–] 0 points1 point  (1 child) If we can decrease the severity of those cycles, though, they become less problematic even if we cannot eliminate them. [–] 0 points1 point  (0 children) Every crisis is worse than the one left behind. This is due to the fact that the system is everytime more and more coupled. And what this paper states is: there is no way to avoid this, IF THERE IS, then P=NP. As a CompSci pro I told you this: VERY unlikely. @hernanemartinez [–] 0 points1 point  (0 children) ... Many more implications. Comunism, proven faulty system is the alternative. So where you go when there is no place to go? [–] -5 points-4 points  (0 children) Well, that's a unique way of looking at the problem. [–] -2 points-1 points  (4 children) I did not read the entire article but I confess I was put off in the intro where it stated "Since P probably does not equal NP" ... That's a very unscientific way of putting it. There's no probability involved. [–] 0 points1 point  (3 children) probably != probability [–] 0 points1 point  (2 children) That's sort of my point: "probably" shouldn't be used in that way. Imagine reading a paper from the 1980s that said that since a majority of mathematicians polled said that Fermat's last theorem is true, it probably is. You wouldn't be taken aback? That's the kind of BS I'd expect from a sociology paper. [–] 0 points1 point  (0 children) It works fine if you consider probability to be a measure of belief.
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Question If $${x}^{4}+\cfrac{1}{{x}^{4}}=194$$, find $${x}^{3}+\cfrac{1}{{x}^{3}}$$, $${x}^{2}+\cfrac{1}{{x}^{2}}$$ and $$x+\cfrac{1}{x}$$ Solution Given,$$\Rightarrow x^4+\dfrac{1}{x^4}=194$$$$\Rightarrow x^4+\dfrac{1}{x^4}+2=194+2$$$$\Rightarrow \left ( x^2+\dfrac{1}{x^2} \right )^2=196$$$$\therefore x^2+\dfrac{1}{x^2}=14$$.......$$(ii)$$$$\Rightarrow x^2+\dfrac{1}{x^2}+2=14+2$$$$\left ( x+\dfrac{1}{x} \right )^2=16$$$$\therefore x+\dfrac{1}{x}=4$$.......$$(iii)$$$$\Rightarrow \left ( x+\dfrac{1}{x} \right )^3=x^3+\dfrac{1}{x^3}+3\left ( x+\dfrac{1}{x} \right )$$$$\Rightarrow 4^3=x^3+\dfrac{1}{x^3}+3(4)$$$$\therefore x^3+\dfrac{1}{x^3} = 52$$........$$(i)$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More
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## 留言板 引用本文: 孙强,彭东升,张义明,董庆辉,岳前进,吴峰,阎军,钟万勰. 悬链式单点系泊浮筒主尺度可行域研究 [J]. 应用数学和力学,2022,43(3):281-289 SUN Qiang, PENG Dongsheng, ZHANG Yiming, DONG Qinghui, YUE Qianjin, WU Feng, YAN Jun, ZHONG Wanxie. Feasible Region Study on Main Dimensions of CALM Buoys[J]. Applied Mathematics and Mechanics, 2022, 43(3): 281-289. doi: 10.21656/1000-0887.420115 Citation: SUN Qiang, PENG Dongsheng, ZHANG Yiming, DONG Qinghui, YUE Qianjin, WU Feng, YAN Jun, ZHONG Wanxie. Feasible Region Study on Main Dimensions of CALM Buoys[J]. Applied Mathematics and Mechanics, 2022, 43(3): 281-289. • 中图分类号: TE53 ## Feasible Region Study on Main Dimensions of CALM Buoys • 摘要: 为了能够设计合理的单点浮筒主尺度,归纳提出了悬链式单点浮筒5条主要设计原则。基于实际工程经验给出了浮筒重量的母型预估方法,探讨储备浮力,提出等干舷方案集,基于静力学原理和水动力分析探讨了浮筒的自由漂浮稳性及浮筒的运动响应,梳理了垂荡及摇摆随浮筒直径的变化规律。以25 m和100 m水深的浮筒直径上限和下限为边界建立可行域,分别以三个实际项目采纳的浮筒直径和45 m水深下单点浮筒直径的可行范围为参考,开展浮筒直径可行域的验证。研究表明:环境条件等不同因素也会引起相同水深下浮筒直径可行域的变化,因此在可行域的实际应用中应进一步考虑一定的偏差裕量。该文的可行域研究对于悬链式单点浮筒的主尺度确定具有积极的指导意义。 • 图  1  悬链式单点系泊系统 Figure  1.  Catenary anchor leg mooring system 图  2  浮筒干舷 Figure  2.  The CALM buoy freeboard 图  3  单点浮筒稳性 Figure  3.  The CALM buoy stability 图  4  单点浮筒垂荡 Figure  4.  The CALM buoy heave 图  5  单点浮筒摇摆 Figure  5.  The CALM buoy roll & pitch 图  6  装置浮筒水动力模型 Figure  6.  The CALM buoy hydrodynamic model 图  7  浮筒直径的可行域 Figure  7.  The CALM buoy diameter feasible field • [1] HARING R E, ADAMS R B, BEAZLEY R A, et al. Application of probability to design of single-point moorings[J]. Society of Petroleum Engineers Journal, 1969, 10(3): 219-228. [2] BERNITSAS M M, PAPOULIAS F A. Stability of single point mooring systems[J]. Applied Ocean Research, 1986, 8(1): 49-58. [3] HWANG Y L. Dynamic analysis for the design of CALM system in shallow and deep waters[J]. Journal of Offshore Mechanics and Arctic Engineering-Transactions of the ASME, 1997, 119(3): 151-157. [4] SCHELLIN T E. Mooring load of a ship single-point moored in a steady current[J]. Marine Structures, 2003, 16(2): 135-148. [5] AKYUZ E, CELIK E, A modified human reliability analysis for cargo operation in single point mooring (SPM) off-shore units[J]. Applied Ocean Research, 2016, 58: 11-20. [6] BROWN A, GORTER W, PAALVAST M, et al. Design approach for CALM buoy moored vessel in squall conditions[C]//35th International Conference on Ocean, Offshore and Arctic Engineering. American Society of Mechanical Engineers, 2016. [7] GU H Y, CHEN H C, ZHAO L Y. Coupled mooring analysis of a CALM buoy by a CFD approach[C]//The 27th International Ocean and Polar Engineering Conference. OnePetro, 2017. [8] 黄国樑, 藤野正隆. 关于风和潮流作用下单点系泊船体的鱼尾状摆动的研究[J]. 海洋工程, 1987, 5(3): 1-13. (HUANG Guoliang, FUJINO M. Fishtailing oscillation of a ship moored to single point mooring systems in wind and current[J]. The Ocean Engineering, 1987, 5(3): 1-13.(in Chinese) [9] 季春群, 王磊, 彭涛. 单点系泊的系泊稳定性分析[J]. 中国海洋平台, 1998, 13(5/6): 12-14. (JI Chunqun, WANG Lei, PENG Tao. The moored stability of single point mooring system[J]. China Offshore Platform, 1998, 13(5/6): 12-14.(in Chinese) [10] 杜度, 张纬康, 单点系泊船舶运动的仿真研究[J]. 计算机仿真, 2003, 20(6): 96-99, 103.DU Du, ZHANG Weikang, Simulation study on motion of sigle point mooring vessel[J]. Computer Simulation, 2003, 20(6): 96-99, 103. (in Chinese) [11] 孙强, 彭贵胜, 董庆辉. 悬链式单点系泊系统设计与分析方法研究[C]//第十六届中国科协年会(分8): 绿色造船与安全航运论坛论文集. 2014: 126-131.SUN Qiang, PENG Guisheng, DONG Qinghui. Design and analysis method research on CALM type single point mooring system[C]//16th Annual Meeting of China Association for Science and Technology (Part Eight): Forum on Green Shipbuilding and Safe Shipping. 2014: 126-131. (in Chinese) [12] 周楠, 刘旭平, 李俊汲. CALM单点系泊系统设计综述[J]. 海洋工程装备与技术, 2017, 4(2): 102-104. (ZHOU Nan, LIU Xuping, LI Junji. Design method summary of CALM single point mooring[J]. Ocean Engineering Equipment and Technology, 2017, 4(2): 102-104.(in Chinese) [13] American Bureau of Shipping. Single point moorings[S]. Act of Legislature of the State of New York, 2019. [14] BERNARD M. 海洋工程水动力学[M]. 北京: 国防工业出版社, 2012: 151-154.BERNARD M. Hydrodynamique des Structures Offshore[M]. Beijing: National Defense Industry Press, 2012: 151-154. (in Chinese) [15] 芦苇, 赵冬, 李东波, 等. 基于期望函数的土遗址锚固参数组合优化方法[J]. 应用数学和力学, 2019, 40(6): 650-662.LU Wei, ZHAO Dong, LI Dongbo, et al. A composite optimization method for anchorage parameters of rammed earth sites based on desirability functions[J] Applied Mathematics and Mechanics, 2019, 40(6): 650-662. (in Chinese) ##### 计量 • 文章访问数:  657 • HTML全文浏览量:  241 • PDF下载量:  42 • 被引次数: 0 ##### 出版历程 • 收稿日期:  2021-04-28 • 录用日期:  2021-04-08 • 修回日期:  2022-01-07 • 网络出版日期:  2022-02-16 • 刊出日期:  2022-03-08 / • 分享 • 用微信扫码二维码 分享至好友和朋友圈
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# beta distribution cdf calculator This calculator will compute the cumulative distribution function (CDF) for the beta distribution (i.e., the area under the beta distribution from 0 to x), given values of the shape parameters, and the point at which to evaluate the function. The beta distribution plays a critical role is calculating many common statistics, and is hence important to a wide variety of analytics-related tasks. To improve this 'Beta distribution Calculator', please fill in questionnaire. The beta distribution plays a critical role is calculating many common statistics, and is hence important to a wide variety of analytics-related tasks.Please provide the necessary values, and then click 'Calculate'. Beta Distribution CDF and Quantile Calculator An implementation of the Beta Distribution CDF and Quantile function Calculator occurs below. Male or Female ? All rights reserved. Compute the cumulative distribution function (CDF) for the beta distribution, given the upper limit of integration x, and values of the shape parameters. For shape parameters and the Beta density function is:- Note that the domain of the Beta distribution function is restricted to the range 0 to 1. Click Calculate! All rights reserved. The Cumulative Distribution Function of a Beta random variable is defined by: where Ix(α, β) is the regularized incomplete beta function. Please enter the necessary parameter values, and then click 'Calculate'. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student The Analytics Calculators index now contains 106 free analytics calculators! {\displaystyle {\begin{aligned}f(x;\alpha ,\beta )&=\mathrm {constant} \cdot x^{\alpha -1}(1-x)^{\beta -1}\\[3pt]&={\frac {x^{\alpha -1}(1-x)^{\beta -1}}{\displaystyle \int _{0}^{1}u^{\alpha -1}(1-u)^{\beta -1}\,du}}\\[6pt]&={\frac {\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}}\,x^{\alpha -1}(1-x)^{\beta -1}\\[6pt]&={\frac {1}{\mathrm {B} (\alpha ,\beta )}}x^{\alpha -1}(1-x)^{\beta -1}\end{al… Cumulative Distribution Function (CDF) for the Beta Distribution Related Calculators Below you will find descriptions and links to 11 different statistics calculators that are related to the free cumulative distribution function (cdf) calculator for the beta distribution. Copyright © 2015 - 2020 by Dr. Daniel Soper. Home / Probability Function / Beta distribution; Calculates a table of the probability density function, or lower or upper cumulative distribution function of the beta distribution, and draws the chart. Beta distribution (chart) Calculator . Compute the cumulative distribution function (CDF) for the beta distribution, given the upper limit of integration x, and values of the shape parameters. This calculator will compute the cumulative distribution function (CDF) for the beta distribution (i.e., the area under the beta distribution from 0 to x), given values of the shape parameters, and the point at which to evaluate the function. Define the Beta variable by setting the shape (α) and the shape (β) in the fields below. Beta Distribution Cumulative Distribution Function (CDF) Calculator Compute the cumulative distribution function (CDF) for the beta distribution, given the upper limit of integration x, and values of the shape parameters. better insights and decisions, one calculation at a time! The Free Statistics Calculators index now contains 106 free statistics calculators! This calculator will compute the cumulative distribution function (CDF) for the beta distribution (i.e., the area under the beta distribution from 0 to x), given values of the shape parameters, and the point at which to evaluate the function.Please enter the necessary parameter values, and then click 'Calculate'. select function: probability density f lower cumulative distribution P Choose the parameter you want … Beta Distribution Cumulative Distribution Function (CDF) Calculator. ©2020 Matt Bognar Department of Statistics and Actuarial Science University of Iowa and find out the value at x in [0,1] of the cumulative distribution function for that Beta variable. cited in more than 3,000 scientific papers! Calculator: Cumulative Distribution Function (CDF) for the Beta Distribution, Cumulative Distribution Function (CDF) for the Beta Distribution Calculator, Cumulative Distribution Function (CDF) Calculator for the Beta Distribution. Copyright © 2006 - 2020 by Dr. Daniel Soper. Parameters Calculator - Beta Distribution - Define the Beta variable by setting the shape (α) and the shape (β) in the fields below. 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# Math-o-Fun In Maths-o-Fun Quiz, the team of Swapnil and Manish needed to answer the last question to win the quiz. As soon as the last question appeared on the screen, both of them worked on the question and got the answer. They pressed the buzzer immediately and hence they won the quiz. Can you guess what was their answer? The question was: Find the sum of the remainders when $${ 3 }^{ 15 }$$ is divided by 2 and when $${ 3 }^{ 15 }$$ is divided by 4. ×
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## If the volume of the pyramid shown is 12 centimeters cubed, what is its height? A rectangular pyramid with a base of 3 centimeters by 2 cent Question If the volume of the pyramid shown is 12 centimeters cubed, what is its height? A rectangular pyramid with a base of 3 centimeters by 2 centimeters and a height of h. 1 cm 2 cm 6 cm 7 cm Mark this and return in progress 0 6 hours 2021-09-14T22:13:47+00:00 1 Answer 0
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# Pev june 2008 Options Registered Posts: 20 New contributor 🐸 Section 2.1 gearing ratio can someone please explain gearing ratio i worked the question by using long term loan /net assets i got 1.43 but they seems to get 58 percentage need explaination thanks for you help • Registered Posts: 585 Epic contributor 🐘 Options nosaj wrote: » Section 2.1 gearing ratio can someone please explain gearing ratio i worked the question by using long term loan /net assets i got 1.43 but they seems to get 58 percentage need explaination thanks for you help gearing ratio = Long term liability(LTL) divide by (net assets plus LTL) therefore gearing = 12000 /(8400+12000) x100 = 58.82% • Registered Posts: 32 Regular contributor ⭐ Options Hi, I think the equation is long term loan/net assets+long term loan 12000/20400 = 0.588 multiply by 100 = 58.8% • Registered Posts: 20 New contributor 🐸 Options Thanks much i was not adding back the long term loan • Registered Posts: 122 Dedicated contributor 🦉 Options we've been taught debt over debt plus equity or: Long term liabilities / net assets plus long term liabilities. I'm just doing that paper now. Had completely forgotten about interest cover and oncorrectly took net profit (after interest) instead of operating profit (PBIT). Finding the comments of the Financial position question bit tough. • Registered Posts: 122 Dedicated contributor 🦉 Options So stuck on 2.2.c. Can anyone help? I said do contract put because the Romanian cost per turbine for 10,000 units would be £650 plus £105 unavoidable fixed production costs, as opposed to £800 for Voltair. Answer doesn't factor in these FPC to contract out. Why? I said don't contract out for 14000 units as economies of scale allow prod cost of £714.29 per turbine as opposed to £650 plus £75 contracted out. HELP??? • Registered Posts: 122 Dedicated contributor 🦉 Options DOH! Just scrolled down the answer page! Time for a break I think.
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Purchase Solution # Kinematics Problems Not what you're looking for? Here, we provide a brief summary of some common kinematics, and solve three common kinematics problems, specifically from College Physics, by J. Serway, et al. We specifically find: For a car traveling east at a given velocity, the final velocity for a constant acceleration in two different directions. For an arrow shot straight up in the air, the maximum height of the arrow and the time the arrow is in the air. For a mailbag released from a descending helicopter, the speed of the mailbag and its distance below the helicopter. We focus on the strategy for solving kinematic equations efficiently. The full solution/explanation is included in both PDF and PS format. ##### Solution Summary In this solution, we provide a brief summary of some common kinematics, and solve three common kinematics problems, specifically from College Physics, by J. Serway, et al. ##### Solution Preview ** Please see the attached PDF or PS file for the complete solution ** Here, we follow a summarized introduction to kinematics along the lines of College Physics, written by Serway et al., and solve a few common kinematic problems from the second chapter of that text. Displacement in the x direction is given by: ∆x ≡ x_f − x_i, (1) where x_f is the ... ##### Intro to the Physics Waves Some short-answer questions involving the basic vocabulary of string, sound, and water waves. ##### Introduction to Nanotechnology/Nanomaterials This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.
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# Re: The Giza Plateau Grid Disocvered October 30, 2009 08:33PM Registered: 5 years ago Posts: 4,374 Hi Don i feel measuring astronomical ratios and planet diameters are subject to change in the future as more accuracy is achieved, thats why i feel a fixed number system that has its begining in the distant past and may come from an advanced race who relised that the Earth , Mars the moon and even the sun are the product of a race of beings so far evolved that we would say they are gods, they may even exist outside of this univers in a timeless univers and are watching every move we make as we play on the planet that they made for themselves , so" are we human , or are we dancers, " "The Killers", i belive that this lost civilisation had a measuring system based on number and earth commensurate measure , but with the belief that this planet is the property of a devine race of beings, and so they revered those numbers and one of those numbers could be a fraction of 315/8 = 39.375 inches , take 1152 x 39.375 = 4536 the half base side of G1s south side, so number must be a very important aspect of the system that would relate to the solar system , take that 1152 number for an example , The sun has a symbolic number of 8640/75 = 1152 , the Earths symbolic number is 7920/6.875 = 1152 , then the moon,s number is 2160/1.875 = 1152, finaly Mars has a symbolic number of 4224/11 x 3 = 1152 , and so if there is a solar system commensurate cubit it would only relate to these four bodies The Speed of Light = 186282398 mps / 510000 = 365.259604 Subject Author Views Posted #### • The Giza Plateau Grid Disocvered Ahatmose 671 October 21, 2009 02:46PM #### • Re: The Giza Plateau Grid Disocvered Sterling 329 October 21, 2009 03:02PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 299 October 21, 2009 03:12PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 409 October 21, 2009 11:09PM #### • Re: The Giza Plateau Grid Disocvered Modeh 304 October 22, 2009 12:03AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 322 October 22, 2009 03:33AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 306 October 22, 2009 03:55AM #### • Re: The Giza Plateau Grid Disocvered Modeh 334 October 22, 2009 05:06AM #### • More interesting stuff from The Giza Template Ahatmose 319 October 22, 2009 05:59AM #### • Re: More interesting stuff from The Giza Template Modeh 320 October 22, 2009 07:29AM #### • The Giza Grid Unit Squares The Circle Ahatmose 348 October 22, 2009 12:47PM #### • Even More interesting stuff from The Giza Template Ahatmose 321 October 22, 2009 11:42PM #### • Re: More interesting stuff from The Giza Template Ahatmose 308 October 23, 2009 12:16AM #### • Re: More interesting stuff from The Giza Template Ahatmose 346 October 23, 2009 12:23AM #### • Re: More interesting stuff from The Giza Template Ahatmose 310 October 23, 2009 12:37AM #### • Re: More interesting stuff from The Giza Template Ahatmose 303 October 23, 2009 12:51AM #### • Re: More interesting stuff from The Giza Template Ahatmose 277 October 23, 2009 01:58AM #### • Re: More interesting stuff from The Giza Template Ahatmose 311 October 23, 2009 04:02AM #### • Re: The Giza Plateau Grid Disocvered DUNE 327 October 22, 2009 10:15PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 328 October 22, 2009 11:39PM #### • Re: The Giza Plateau Grid Disocvered DUNE 327 October 23, 2009 07:09AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 356 October 25, 2009 12:34AM #### • Re: The Giza Plateau Grid Disocvered debraregypt 328 October 25, 2009 01:50PM #### • The Giza Plateau Grid Discovered: Why 39.37 ? Ahatmose 314 October 25, 2009 11:32AM #### • Re: The Giza Plateau Grid Discovered: Why 39.37 ? Ahatmose 251 October 25, 2009 12:22PM #### • Re: The Giza Plateau Grid Discovered: Why 39.37 ? Ahatmose 324 October 25, 2009 01:28PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 338 October 29, 2009 03:21AM #### • Re: The Giza Plateau Grid Disocvered DUNE 382 October 29, 2009 08:25PM #### • Re: The Giza Plateau Grid Disocvered Chris Partridge 326 October 30, 2009 12:42AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 337 October 30, 2009 02:51AM #### • Reconnecting to the Source Nick L 286 October 31, 2009 05:50AM #### • Re: The Giza Plateau Grid Disocvered DUNE 305 October 30, 2009 07:33AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 306 October 30, 2009 11:37AM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 330 October 31, 2009 12:09PM #### • Re: The Giza Plateau Grid Disocvered Nick L 311 October 31, 2009 03:11PM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 236 October 31, 2009 04:11PM #### • Re: The Giza Plateau Grid Disocvered Nick L 343 October 31, 2009 04:51PM #### » Re: The Giza Plateau Grid Disocvered DUNE 323 October 30, 2009 08:33PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 358 October 31, 2009 02:28AM #### • Re: The Giza Plateau Grid Disocvered DUNE 355 October 31, 2009 01:04PM #### • Re: The Giza Plateau Grid Disocvered DUNE 355 October 31, 2009 04:10PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 290 October 31, 2009 10:44PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 315 October 31, 2009 11:23PM #### • Re: The Giza Plateau Grid Disocvered Jon B 332 November 01, 2009 09:41AM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 319 November 01, 2009 11:00AM #### • Re: The Giza Plateau Grid Disocvered Jon B 335 November 01, 2009 11:10AM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 304 November 01, 2009 12:18PM #### • Re: The Giza Plateau Grid Disocvered Jon B 317 November 01, 2009 12:23PM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 313 November 01, 2009 01:33PM #### • Re: The Giza Plateau Grid Disocvered Jon B 327 November 01, 2009 01:45PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 297 November 01, 2009 02:33PM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 321 November 01, 2009 02:43PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 334 November 01, 2009 03:30PM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 253 November 02, 2009 05:12AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 332 November 01, 2009 03:34PM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 333 November 01, 2009 02:36PM #### • Re: The Giza Plateau Grid Disocvered DUNE 333 November 01, 2009 11:13AM #### • Re: The Giza Plateau Grid Disocvered DUNE 388 November 03, 2009 08:36PM #### • Re: The Giza Plateau Grid Disocvered Scott Creighton 356 November 03, 2009 09:29PM #### • Re: The Giza Plateau Grid Disocvered Sirfiroth 320 November 03, 2009 09:41PM #### • Re: The Giza Plateau Grid Disocvered DUNE 334 November 03, 2009 10:07PM #### • Re: The Giza Plateau Grid Disocvered DUNE 386 November 03, 2009 10:10PM #### • Re: The Giza Plateau Grid Disocvered DUNE 287 November 04, 2009 06:04PM #### • Re: The Giza Plateau Grid Disocvered DUNE 329 November 04, 2009 06:30PM #### • Re: The Giza Plateau Grid Disocvered Chris Partridge 323 November 05, 2009 04:50AM #### • Re: The Giza Plateau Grid Disocvered Ahatmose 287 November 05, 2009 11:42AM #### • Re: The Giza Plateau Grid Disocvered Chris Partridge 299 November 06, 2009 03:40AM #### • Re: The Giza Plateau Grid Disocvered DUNE 366 November 06, 2009 08:03PM #### • Re: The Giza Plateau Grid Disocvered DUNE 330 November 09, 2009 08:22PM #### • Re: The Giza Plateau Grid Disocvered DUNE 316 November 12, 2009 07:51PM #### • Re: The Giza Plateau Grid Disocvered DUNE 348 November 14, 2009 03:35PM #### • Re: The Giza Plateau Grid Disocvered DUNE 342 November 20, 2009 02:50PM #### • Re: The Giza Plateau Grid Disocvered DUNE 346 November 23, 2009 08:51PM Sorry, only registered users may post in this forum.
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## Welcome to the Treehouse Community Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here. ### Looking to learn something new? Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today. # What number is represented by the byte 0000011? I don't get how to get that answer? What number is represented by the byte 0000011? I don't get how to get that answer? Here's one way to decode binary numbers. Make a column for each digit, and starting from the right, put a "1" and then as you move left put double the value of the column next to it. Remember, right-to-left ( THIS WAY ). You'll get something like this, which is a horizontal version of the helper table shown in the quiz: ```128 64 32 16 8 4 2 1 <-- 8 columns for 8 digits ``` Then, put your binary number on the next line, spread into the columns. Now below that, multiply each top number by the binary digit and put the result below. It's easy, since binary digits are either 1 or 0, so you either put the number on top again or zero. Then finally, add up all the numbers on the bottom row and that's your answer. Like this: ```128 64 32 16 8 4 2 1 <-- starting columns 0 0 0 0 0 0 1 1 <-- multiply by your binary digits --- --- --- --- --- --- --- --- 0 0 0 0 0 0 2 1 <-- add these up: 2 + 1 = 3 ``` Also, since zeros in front don't count you can skip them. So in this case we really only needed 2 columns.
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We previously studied about current. But how to determine the diameter of wire for carrying a specific amount of current. Wire gauge is the measurement of wire diameter. Wire gauge actually explains how much electrical current a wire can safely carry. American Wire Gauge or AWG is a wire table which is the primary system used for selecting wires. Let's have a look at it: Here a few points for using the table. 1. The table is for solid wires. Practically, the wires are stranded (Made of many strands instead of a single solid conductor. 2. For a change of gauge number by 3 the area is doubled (approx). That is for AWG 37 the area is 19.83 whereas for 34 it is 39.76 which is approximately double in number.
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# stealth game AI solver I'm making an AI for solving a simple stealth game. The objective is to go to the goal point without being caught by enemies(with flashlights). I already implemented a pathfinding algorithm(A*) but without the logic for avoiding or waiting at a certain spot before moving again. There are no other elements on the map. There's just the player, the guards, and the goal. The only idea I have right now is, when the AI has already constructed the path, the coordinates in the path that would be spotted will be excluded and the AI constantly reconstructs the path. But the enemies are constantly moving(dynamic obstruction) so I think it's impossible to reach the goal with only this. Any other ideas? - Assuming you've got enough memory to play with and the guard movements are known/deterministic, you can model time as another dimension (e.g. your 2D map becomes a 3D space-time arena). Then you can do A* through time..... the enemies with moving searchlights will be "static" in space-time. - Thanks. I actually thought of something like that, that at a certain time the guards won't spot you if you go that path, but I didn't know how to implement it. –  Zik Feb 24 '13 at 10:22 Don't tell me anymore how to implement that space-time, I'll just think about it. :) At least I got an idea that there would be another dimension. –  Zik Feb 24 '13 at 10:31 Constructing a good heuristic is the key. Your heuristic should consider the probability that a guard is likely to move toward your path and spot you, and weight moves accordingly. This will naturally tend toward paths that avoid guards. Note that with chance elements like random moving guards, there's no way to guarantee a goal. - Thanks. The guards have a fixed pattern. –  Zik Feb 24 '13 at 10:19 If the guards have a fixed pattern, it's relatively easy to have your heuristic take that pattern into account, and then you will have a deterministic A* search that can find an optimal solution every time. –  Ian McMahon Feb 24 '13 at 10:20 Ok, I will think about that heuristic. :) –  Zik Feb 24 '13 at 10:27
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# Teaching Treasures NOTE: You must start the timer in order to check your answers 7 × 5 = 6 × 6 = 49 ÷ 7 = 32 ÷ 8 = 9 × 4 = 8 × 7 = 3245 - 2116 = 4560 + 2441 = 36 ÷ 9 = 9 × 5 = Thank you for using Teaching Treasures online learning interactive activities. Copyright. Home | Site Map | Terms of Use | Privacy Policy | Contact Us | Links | Copyright | © Teaching Treasures™ Publications
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# Central Net Force Model Worksheet 2 Central Net Force Model Worksheet 2 - Web central net force model worksheet 2. A woman flying aerobatics executes a maneuver as illustrated below. (c) analyze and describe the accelerated movement in two dimensions using equations, including. Web central net force model worksheet 2: Web read online central net force model worksheet 2 radial net force answers pdf. A car travels through a valley at constant. A woman flying aerobatics executes a maneuver as illustrated below. Sum of forces = net force = ma. Web physics central net force model test flashcards | quizlet. A car travels through a valley at constant speed, though not at constant. (c) analyze and describe the accelerated movement in two dimensions using equations, including. Web central net force model worksheet 2. Web name date pd central net force model worksheet 2:radial net force 1. A woman flying aerobatics executes a maneuver as illustrated below. We as pay for hundreds of the books. A car travels through a valley at constant speed, though not at constant. Web radial net forces and circular motion 1. Central Net Force Model Worksheet 2 - A bowling ball rolls down the hallway. A woman flying aerobatics executes a maneuver as illustrated below. Web name date pd central net force model worksheet 2:radial net force 1. Web central net force model worksheet 2: At the top of the first hill of the rollercoaster, point “a,” a. Web physics central net force model test flashcards | quizlet. Study with quizlet and memorize flashcards containing terms. A woman flying aerobatics executes a maneuver as illustrated below. The total amount of force on an object. Web central net force model worksheet 3: Web read online central net force model worksheet 2 radial net force answers pdf. A car travels through a valley at constant. Web central net force model worksheet 1: Web gravitational field strength on earth's moon is 1.6 m/s2, or 1.6 n/kg.) when the satellite is in orbit, how big will the centripetal force. Central net force particle model: ## A Woman Flying Aerobatics Executes A Maneuver As Illustrated Below. Central net force particle model: A car enters a circular turn. Study with quizlet and memorize flashcards containing terms. A woman flying aerobatics executes a maneuver as illustrated below. ## Draw A Force Diagram (Side View) For. Web read online central net force model worksheet 2 radial net force answers pdf. Web central net force model worksheet 3: Web central net force model worksheet 2: Web physics central net force model test flashcards | quizlet. ## Web Central Net Force Model Worksheet 1: Web central net force model worksheet 3: Web central net force model worksheet 1: Web radial net forces and circular motion 1. A car travels through a valley at constant speed, though not at constant. ## At The Top Of The First Hill Of The Rollercoaster, Point “A,” A. Web central net force model worksheet 2. Web central net force model. (c) analyze and describe the accelerated movement in two dimensions using equations, including. The total amount of force on an object.
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Cody # Problem 411. Back to basics 21 - Matrix replicating Solution 1312452 Submitted on 24 Oct 2017 by Yaakoub Ben Romdhane This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass x = [1]; y_correct = [1 1;1 1]; assert(isequal(matrix_replication(x),y_correct)) 2   Pass x = [1 2;3 4]; y_correct = [1 2 1 2; 3 4 3 4; 1 2 1 2; 3 4 3 4]; assert(isequal(matrix_replication(x),y_correct)) 3   Pass x = [1 2]; y_correct = [1 2 1 2; 1 2 1 2]; assert(isequal(matrix_replication(x),y_correct))
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# Learning transistor based circuits simulate this circuit – Schematic created using CircuitLab In one of our electronic circuit tests, this question was asked: What is the output voltage of given circuit? This circuit looks like a common emitter amplifier circuit to me, but when I tried to calculate the output, I am getting 0.047 V. My questions are: • Why is this circuit not amplifying, is it required to add capacitors? • How to select suitable resistor values of Rb, Rc and Re in any transistor based circuits? I know some of the formulas but I don't know how to implement: $$I_b = V_{in} - \frac{V_{be}}{R_b}$$ $$V_{be} = 0.7 \mathrm{\;assumed}$$ $$I_b = 0.372\mathrm{\,mA}$$ $$I_c = \beta I_b$$ Considered β = 100, so: $$I_c = 0.037\mathrm{\,A}$$ $$\mathrm{Gain} = \frac{V_{out}}{V_{in}}$$ • $47mV$ seems quite reasonable to me. Keep in mind that this is an inverting amplifier. – Dzarda Aug 27 '14 at 10:03 • EDIT: 47mV is of course complete bogus, since this is a BJT. – Dzarda Aug 27 '14 at 12:45 • I formatted the formulas using MathJax, please let me know if anything is incorrect. – JYelton Aug 27 '14 at 16:25 • If you ground Q1_c the I_R1 is V/R = 15V/10k = 1.5 mA. | /I_R2 ~= V2/R2 = 10V/25k =~~ 0.4 mA. (Slightly less as Olin says due to Vbe <>0. | For Beta = 100. Ic = Ib x Beta = 0.4 mA x 100 = 40 mA. BUT we saw above that the absolute max R1 can carry is 1.5 mA when the transistor is a hard short to ground. So the transistor has enough base currentto support Ic = 40 mA BUT 1.5 mA in R1 is enough to take Q1_c to ground so the transistor is turned on hard in saturation. As Olin said. As Olin said (again) what exactly Vce is in saturation varies with specific transistor. .... – Russell McMahon Sep 15 '14 at 8:40 • .... If you drive the base VERY hard Vce may fall to very close to zero. Long ago I drove a transistor with Ib ~= 10 x Ic !!!. This turned it on VERY hard so Vce was a few millivolts. I was used to switch a divider string on and off so this made sense to do. (Nowadays I'd probably use a low Rdson MOSFET instead). – Russell McMahon Sep 15 '14 at 8:41 You should be able to see immediately from inspection (without doing any calculations) that the transistor is saturated. Generally you figure the C-E voltage of a saturated transistor is 200 mV or less unless the current is unusually large, which in this case it's obviously not due to the size of R1 and R3. The answer for any real electrical engineering purpose is therefore "200 mV or less", which we can see immediately from inspection. Part of designing good circuits is to make sure that this 200 mV uncertainty doesn't matter. If your professor wants a more accurate answer, then he's being academic and unrealistic, and you can tell him I said so. Now let's do the math to back up what we already know from inspection is happening. Let's say the B-E drop is 700 mV, so there is 9.3 V accross R2, which means the base current is 370 µA. R1, R3, and V1 form a Thevenin source of 7.5 V and 5 kΩ. That means the collector current can't be more than 1.5 mA (C-E drop 0 which can't happen, but is useful to get the guaranteed not to exceed current). (1.5 mA)/(370 µA) = 4, which is the gain required for the transistor to saturate. You can easily rely on a 2N3904 to have well more gain than that at 1.5 mA collector current. The transistor is clearly well into saturation, which is why it was easy to see from inspection without actually running the numbers. So now the question is what will the C-E drop be for the transistor at 1.5 mA collector current and well into saturation. Again, the basic electrical enginnering answer is "less than 200 mV". If you need a more accurate answer then two things are going on. First, the robustness of your overall circuit design is suspect if this really matters. Second, you have to look in the datasheet to see what details they tell you about this particular transistor. This is where things get a little tricky since there are various variants of the 2N3904 out there, and different manufacturers will spec it a little differently. I just grabbed the Fairchild datasheet to use as a example. I wouldn't assume that this level of detail applied to 2N3904 from other manufacturers without checking. This is yet another reason it is better if your circuit works with anything in the 0-200 mV range. You don't have to worry about which variant from which manufacturer you are using, and purchasing can get the cheapest generic 2N3904 they can find that week. On page 2, there is actually a spec for Collector-Emitter Saturation Voltage, which is a maximum of 200 mV for IC at 10 mA and 300 mV for IC at 50 mA. This is typical of datasheets in that they don't explicitly tell you what the part will do at all possible operating points. However, since our collector current is well below 10 mA, we can safely count on the C-E drop to be 200 mV or less. Note that's what we already knew from 3 seconds of inspection in the first place. With this level of information in the datasheet, "200 mV or less" is actually the only correct answer. This is all the manufacturer promises the transistor will do. Now we know that almost certainly in our particular case the C-E voltage will be lower, but the absolute worst case spec is 200 mV. Claiming anything lower is actually wrong, and I would argue strongly with anyone that accepted a specific lower answer as correct in a engineering course. Second guessing the datasheet is irresponsible engineering. So if I was grading the test, I'd mark any answer that gave a specific number below 200 mV as wrong. Even if you built the circuit and measured the C-E drop to be 89.3 mV, for example, I still wouldn't accept that as a valid answer to the question because it can't be counted on accross part variations within a batch, and accross parts from various manufacturers. This is a area where theory and engineering differ, and is something engineering students need to learn. If your professor disagrees with this, and this is in a engineering course, then he's just plain wrong and needs to get out into the real world. And yes, you can (and should) tell him I said so. • Here i did not understand, how did you find Vce? I am a beginner so taking allot of time to understand each and every steps explained,comparing with the theoretical calculation,some where i am going wrong with the formulas getting totally different answers. please help me to understand proper working of this circuit,and when i googled i found similar circuit as common emitter amplifier,so what exactly difference with the common emitter amplifier from this circuit?? how can i analyse circuit?? – yasmi Aug 28 '14 at 5:04 • I remember an old professor in our Elec Eng dept (I'm a Mech) that used to go nuts if a) like you say, students returned values like 89.3mV to something like Vce and b) students returned answers with a precision greater than that asked. He'd get "40.1249576V" as an answer to a question with 5% resistors. I want a University friend to start a course called "Real World Electronics", where caps are -20+80% and hfe is "in that region somewhere". :-) – carveone Aug 28 '14 at 10:03 • @yasmi: Too many questions at once. Pick what you really want to know. The basic answer to where 200 mV or less came from is from the datasheet. Take a look at the datasheet I mentioned. – Olin Lathrop Aug 28 '14 at 12:38 • @carveone: Exactly. All freshman engineering students at RPI in 1974 were required to take a course called Elementary Engineering. There was a lot of estimating, how close "close" really needed to be, etc. It was a real eye-opener for me. In one problem we had to decide the wire guage to use for something, and the numbers deliberately came out to a little smaller than one available guage, so we all naturally picked that one. Those were all marked wrong due to not leaving any margin for stuff to happen outside the problem statement. Great course. – Olin Lathrop Aug 28 '14 at 12:43 This isn't an amplifier as such. The transistor is acting as a switch. Let's have a quick look at the numbers using rules of thumb. (I'm assuming those voltages are DC with no AC component) 10V - 0.6V (Vbe) = 9.4V at the base. Divide by 25k = 0.37mA roughly base current. If we multiply that by (hand wave) a hfe of 100, we get 37mA into the collector before the transistor will drop out of saturation. I personally go for 10 times that, so I'm positive the transistor will be fully saturated at 4mA into the collector. So, without doing anything too hard like getting the output load involved, imagine the transistor is saturated. That means the collector is nearly at 0V. So 15V / 10k = 1.5mA. That's well below 4mA so the output will be close to 0V. Under 0.2 anyway. • How do you calculate output voltage ?? and if i want to make it as common emitter amplifier what modifications i need to make in this circuit?? – yasmi Aug 27 '14 at 10:35 • Add an emitter resistor, for instance 1 KOhm for an amplification of 10. Note that you can't amplify 10V much when you have only a 15V supply, and that the output will be inverted. – Wouter van Ooijen Aug 27 '14 at 10:37 • Wouter gets there first! Add emitter resistor. If you can find a book called The Art of Electronics, a lot of this stuff is explained quite well there (in my opinion of course)! I realise I answered the circuit not really the questions you had below. – carveone Aug 27 '14 at 10:42 • I understood this circuit working,can you tell me,if i want to make it as amplifier circuit,how i have to make changes?? what all the parameters i need to consider to design transistor based circuits?? @carveone – yasmi Aug 28 '14 at 5:08 • That's kind of a big question for a comment box! I personally used the Art of Electronics book back in the day. These days there's lot of online resources but I don't know how many of them are good. I see allaboutcircuits.com which looks ok. Bit mathsy. This one looks a bit more fun and less mathsy: talkingelectronics.com/projects/TheTransistorAmplifier/…. Yes, it's tricky to get started and I find a lot of university stuff is a bit too theory based for me. – carveone Aug 28 '14 at 10:18
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Math Parser, Part 4: Tests November 10, 2008 / category: Parsing / 8 comments Writing test code is a worthwhile practice and building a parser is a good example to prove this claim. We have seen previously that the Parser class consists of many methods and each method is a part of a chain in top-down analysis. It won't be a good idea to start writing the parser from scratch, add all methods we think necessary and then run the code for the first time. Instead of tempting fate, we should write tests for every line of code we create. The parser code was showed in the previous part at once. The fact is that I wrote it step by step, adding a new functionality and test code every time. I run the test often and fix any errors as soon as they appeared, before proceeding to the next functionality. Perhaps illustrating this process would be very instructive for those not familiar with test driven development, but I meant to focus on the parser itself. Anyway, here is the test code I wrote: My eBook: “Memoirs of a Software Team Leader” ``````require 'parser' describe Parser do before(:each) do @parser = Parser.new end it 'should compute 5 when given 2 + 3' do @parser.parse('2 + 3').should == 5 end it 'should compute 6 when given 2 * 3' do @parser.parse('2 * 3').should == 6 end it 'should compute 89 when given 89' do @parser.parse('89').should == 89 end it 'should raise an error when input is empty' do lambda {@parser.parse('')}.should raise_error() end it 'should omit white spaces' do @parser.parse(' 12 - 8 ').should == 4 @parser.parse('142 -9 ').should == 133 @parser.parse('72+ 15').should == 87 @parser.parse(' 12* 4').should == 48 @parser.parse(' 50/10').should == 5 end it 'should treat dot separated floating point numbers as a valid input' do @parser.parse('2.5').should == 2.5 @parser.parse('4*2.5 + 8.5+1.5 / 3.0').should == 19 @parser.parse('5.0005 + 0.0095').should be_close(5.01, 0.01) end it 'should handle tight expressions' do @parser.parse('67+2').should == 69 @parser.parse(' 2-7').should == -5 @parser.parse('5*7 ').should == 35 @parser.parse('8/4').should == 2 end it 'should calculate long additive expressions from left to right' do @parser.parse('2 -4 +6 -1 -1- 0 +8').should == 10 @parser.parse('1 -1 + 2 - 2 + 4 - 4 + 6').should == 6 end it 'should calculate long multiplicative expressions from left to right' do @parser.parse('2 -4 +6 -1 -1- 0 +8').should == 10 @parser.parse('1 -1 + 2 - 2 + 4 - 4 + 6').should == 6 end it 'should calculate long, mixed additive and multiplicative expressions from left to right' do @parser.parse(' 2*3 - 4*5 + 6/3 ').should == -12 @parser.parse('2*3*4/8 - 5/2*4 + 6 + 0/3 ').should == -1 end it 'should return float pointing numbers when division result is not an integer' do @parser.parse('10/4').should == 2.5 @parser.parse('5/3').should be_close(1.66, 0.01) @parser.parse('3 + 8/5 -1 -2*5').should be_close(-6.4, 0.01) end it 'should raise an error on wrong token' do lambda {@parser.parse(' 6 + c')}.should raise_error() lambda {@parser.parse(' 7 &amp; 2')}.should raise_error() lambda {@parser.parse(' %')}.should raise_error() end it 'should raise an error on syntax error' do lambda {@parser.parse(' 5 + + 6')}.should raise_error() lambda {@parser.parse(' -5 + 2')}.should raise_error() end it 'should return Infinity when attempt to divide by zero occurs' do @parser.parse('5/0').should be_infinite @parser.parse(' 2 - 1 + 14/0 + 7').should be_infinite end it 'should compute 2 when given (2)' do @parser.parse('(2)').should == 2 end it 'should compute complex expressions enclosed in parenthesis' do @parser.parse('(5 + 2*3 - 1 + 7 * 8)').should == 66 @parser.parse('(67 + 2 * 3 - 67 + 2/1 - 7)').should == 1 end it 'should compute expressions with many subexpressions enclosed in parenthesis' do @parser.parse('(2) + (17*2-30) * (5)+2 - (8/2)*4').should == 8 @parser.parse('(5*7/5) + (23) - 5 * (98-4)/(6*7-42)').should be_infinite end it 'should handle nested parenthesis' do @parser.parse('(((((5)))))').should == 5 @parser.parse('(( ((2)) + 4))*((5))').should == 30 end it 'should raise an error on unbalanced parenthesis' do lambda {@parser.parse('2 + (5 * 2')}.should raise_error() lambda {@parser.parse('(((((4))))')}.should raise_error() lambda {@parser.parse('((2)) * ((3')}.should raise_error() lambda {@parser.parse('((9)) * ((1)')}.should raise_error() end end `````` Test code is written in the format accepted by the RSpec framework (any other test framework could be used as well). Each test covers the other part of functionality: those parts that come directly from the parser specification (like "it should compute 6 when given 2 * 3") and those part that might go wrong, like "it should calculate long additive expressions from left to right". Test code has to be stored in a file, e.g. parser_spec.rb. To run the spec, type this at the command line: ``````spec parser_spec.rb --format specdoc `````` The "--format specdoc" can be shortened to: ``````spec parser_spec.rb -f s `````` Anyway, spec output should be similar to this: ``````Parser - should compute 5 when given 2 + 3 - should compute 6 when given 2 * 3 - should compute 89 when given 89 - should raise an error when input is empty - should omit white spaces - should treat dot separated floating point numbers as a valid input - should handle tight expressions - should calculate long additive expressions from left to right - should calculate long multiplicative expressions from left to right - should calculate long, mixed additive and multiplicative expressions from left to right - should return float pointing numbers when division result is not an integer - should raise an error on wrong token - should raise an error on syntax error - should return Infinity when attempt to divide by zero occurs - should compute 2 when given (2) - should compute complex expressions enclosed in parenthesis - should compute expressions with many subexpressions enclosed in parenthesis - should handle nested parenthesis - should raise an error on unbalanced parenthesis Finished in 0.018552 seconds 19 examples, 0 failures `````` We end up with a properly working and quite functional parser. It can be further developed or create a base for a completely different program, like some simple XML or YAML parser. I hope you enjoyed this short series of articles. Thanks for reading! Andy June 04, 2009 01:23 PM Thanks a lot. Very useful series. Bill December 04, 2009 05:49 AM Why not try writing your math parser in either c, c++, or java, since those languages are the most widely used in most computer science courses? Lukasz Wrobel December 05, 2009 07:24 AM Because I've been writing parsers in C/C++ etc. during my studies, and now I'm doing this in Ruby just for fun. Anyway, I'm pretty sure there are many C/C++ parsers available on the net, if this is what you were searching for. Theory remains the same across all programming languages, it's just a matter of implementation. dragoljub February 27, 2010 01:05 AM it 'should treat sign before parentheses as a valid input' do ``````@parser.parse('-(2)').should == -2 `````` end right? :) Lukasz Wrobel March 01, 2010 09:03 PM Well, it depends. Notice that there is no way to derive your expression from the grammar. However, a quick patch can be applied to the "number" method: ``````def number token = @lexer.get_next_token reverse = 1 signs = [Token::Minus, Token::Plus] if signs.include?(token.kind) if token.kind == Token::Minus reverse = -1 end token = @lexer.get_next_token end if token.kind == Token::LParen value = expression expected_rparen = @lexer.get_next_token raise 'Unbalanced parenthesis' unless expected_rparen.kind == Token::RParen elsif token.kind == Token::Number value = token.value else raise 'Not a number' end value * reverse end `````` By the way, when trying to investigate the problem you described, I've found an incosistency with the grammar. What's more, even the tests reflect it. Do you know where it is? dragoljub March 02, 2010 12:41 AM I'm not sure exactly which "inconsistency" you were referring to. What I've found is quite a few inconsistencies throughout the whole article. For example, the number production in the grammar was mysteriously changed to exclude sign handling. You were changing how productions work in the code. You mixed up your code so it does not reflect your grammar (there is no component method, factor works as component, number works as factor). All in all, you do have a "teacher-like" style and approach, but your article is far from consistent. I wouldn't say it's bad, because it is good, it can be very helpful for newcomers. It's short, it goes straight to the point, it covers all essential parts of developing a parser. But the inconsistencies may easily mislead and confuse your readers. Also, in my opinion, Ruby is not such a good teaching tool. I hope you'd take my opinion as not offensive, but rather friendly, and, please, do take some time to edit your article. Lukasz Wrobel March 03, 2010 07:53 PM First, I'd like to thank you for the time you've spent on reading the articles. I'm also far from treating your comments as offensive. This short article series was meant - just as you've said - for newcomers. I realise that there are some incosistencies, but some of them were intentional and pretty harmless. On the contrary, disappearance of the sign handling is a good example of what I didn't plan. The articles were meant for people unfamiliar with the topic and I think they do more good than harm. I wasn't trying to write a fully-fledged dissertation and I realise my articles are far from being proffesional. In fact, "proffesional style" is what I was trying to avoid (it doesn't justify unintentional misstatements, anyway). And is Ruby a good teaching tool? I think it's not significantly better or worse than other programming languages. What makes it slightly easier to read is the abstraction of technical details, like pointers manipulation in C. Unfortunately, I can't promise that I'll find time soon to update the article. But I'd like to thank you for enunciating your opinions. I can only be proud of having such attentive readers! SK March 05, 2012 12:44 PM Thanks for a nice explanation. `*emphasis*` emphasis `**strong**` strong ``inline code`` `inline code` `[My blog](http://lukaszwrobel.pl)` ```# use 4 spaces to indent # a block of code def my_method(x) x = x + 1 end``` ``````def my_method(x) x = x + 1 end`````` ```* First. * Second.``` • First. • Second. ```> This is a citation. > Even more citation. I don't agree with you.``` This is a citation. Even more citation. I don't agree with you.
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# English to Tamil Meaning :: mile Mile : மைல் மைல்மைல் #### Show English Meaning (↑) Noun(1) a unit of length equal to 1,760 yards or 5,280 feet; exactly 1609.344 meters(2) a unit of length used in navigation; exactly 1,852 meters; historically based on the distance spanned by one minute of arc in latitude(3) a large distance(4) a former British unit of length once used in navigation; equivalent to 6,000 feet (1828.8 meters(5) a former British unit of length equivalent to 6,080 feet (1,853.184 meters(6) an ancient Roman unit of length equivalent to 1620 yards(7) a Swedish unit of length equivalent to 10 km(8) a footrace extending one mile #### Show Examples (↑) (1) It seems to me like if it takes more kilometers to make a mile , then it should take more kilograms to make a pound.(2) this is my favourite film by a mile(3) he rode the fastest mile of his entire career in 1914(4) Every five seconds counted is equal to approximately one mile between you and the storm.(5) The three-year-old colt had won each of his five starts this year, all Group I races at a mile .(6) The earth is approximately 93 million miles / 150 million kilometers from the sun.(7) In one area some 10 square miles [25 square kilometers] of the city was completely flattened.(8) Come daybreak, the atoll was about three miles (five kilometers) away and had rough water.(9) Each village is considered to own three miles into the forest in every direction.(10) Apart from The West Wing, it's the best thing on television by miles and miles .(11) And it wasn't a close win - it was a win by miles , so that was nice.(12) Etched into the stone are the Roman numerals LIII, the distance in Roman miles to Carlisle.(13) The US has the highest rates of incarceration in the civilized world, and I mean we hold the record by miles .(14) The deer taught her how to run, and keep running for miles at a steady pace.(15) A hunt can last from a few seconds to several minutes and cover up to two miles (three kilometers).(16) And for those who rarely venture South of the river - this beats every bar in West London by miles . Related Words (1) square mile :: சதுர மைல் (2) nautical mile :: கடல் மைல் (3) half a mile :: அரை மைல் (4) last mile :: கடந்த மைல் (5) go the extra mile :: கூடுதல் மைல் சென்று (6) stand out a mile :: ஒரு மைல் வெளியே நிற்க Synonyms Noun 1. statute mile :: சட்ட மைல் 2. sea mile :: கடல் மைல் 3. mil :: மில் கடற்படை மைல் 5. nautical mile :: கடல் மைல் Antonyms 1. hair :: முடி 2. inch :: அங்குலம் 3. step :: படி Different Forms mile, miles Word Example from TV Shows The best way to learn proper English is to read news report, and watch news on TV. Watching TV shows is a great way to learn casual English, slang words, understand culture reference and humor. If you have already watched these shows then you may recall the words used in the following dialogs. Used to be you couldn't find a tree within a MILE of the Wall. #### Game of Thrones Season 3, Episode 6 They're a MILE north. #### Game of Thrones Season 1, Episode 9 he's never been within a MILE of a real battle. #### Game of Thrones Season 3, Episode 4 You'll find a weirwood a MILE north of the Wall. #### Game of Thrones Season 1, Episode 7 Spare me the homilies. I can smell a fraud from a MILE away. #### Game of Thrones Season 5, Episode 7 English to Tamil Dictionary: mile Meaning and definitions of mile, translation in Tamil language for mile with similar and opposite words. Also find spoken pronunciation of mile in Tamil and in English language. Tags for the entry 'mile' What mile means in Tamil, mile meaning in Tamil, mile definition, examples and pronunciation of mile in Tamil language. ## English-Tamil.Net | English to Tamil Dictionary This is not just an ordinary English to Tamil dictionary & Tamil to English dictionary. This dictionary has the largest database for word meaning. It does not only give you English toTamil and Tamil to English word meaning, it provides English to English word meaning along with Antonyms, Synonyms, Examples, Related words and Examples from your favorite TV Shows. This dictionary helps you to search quickly for Tamil to English translation, English to Tamil translation. It has more than 500,000 word meaning and is still growing. This English to Tamil dictionary also provides you an Android application for your offline use. The dictionary has mainly three features : translate English words to Tamil translate Tamil words to English, copy & paste any paragraph in the Reat Text box then tap on any word to get instant word meaning. This website also provides you English Grammar, TOEFL and most common words.
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## Mother Nature and Game Theory: Evaluating Hurricane Risks It seems that the natural disaster of choice for the end of Summer 2017 has been hurricanes. Harvey devastated Southern Texas with torrential flooding a mere two weeks ago, and Irma’s danger looms as it approaches Florida. The damages caused by these natural disasters may be inescapable, but the option to evacuate flood-prone areas is a choice which residents have the ability to make. Utilizing game theoretic thinking, we can analyze how people evaluate hurricane risk. In order to analyze how residents evaluate hurricane risks via game theory, we must first acknowledge the details and the drawbacks of this comparison. In this scenario, one player would be the resident (whose strategies are to either evacuate or stay), while the other player would be the weather. This scenario fits with the most basic definition of a game since the outcome depends not only on what strategy the resident chooses, but also on the choices of what they are interacting with. Of course, Mother Nature as the second player generates problems because you are not exactly sure what it is going to do (after all, how often are meteorologists actually accurate?). In addition, nature has no “payoffs,” it has nothing to gain or lose. Thus, for our simplified game, a more accurate description of the second player would be how we believe Mother Nature will affect us. In our simplified version, Nature’s strategies are either worst-case or best-case: for example, either your house gets flooded, destroying much of your livelihood, or it just rains pretty heavily for a few hours and then it’s over. Since only residents receive payoffs, we can interpret this payoff as the negative amount of money that they would have to spend. If evacuation costs \$1000, worst-case scenario with evacuation: \$5000, and the worst-case scenario without evacuation: \$9000, we can generate the following payoff matrix: In this matrix, if the resident believes that the storm will not affect them greatly, the resident’s best response would be to stay. If the resident believes the storm will be damaging, the resident’s best response would be to evacuate beforehand. However, the article link showcases that people’s abilities to access risks in these situations are rather irrational. The effects “associated with the last hurricane to make landfall in their county was the most powerful predictor of how coastal residents” anticipated the risks of the next big storm to be (Vox). This logic fails residents as it does not anticipate the unforeseen consequences of these unprecedented storms: what if the damage that occurs is detrimentally worse than the imagined “worst case scenario”. Unfortunately, this is what happened in Houston. As a born and raised Houstonian, I watched from afar as my city drowned. However, this was not a completely new experience for Southern Texas: Houston has faced two other major flooding events in the past three years. Both of these caused city-wide damage and school cancellations, and Houstonians began to think of these events as the normal “worst-case scenario.” After all, they were damaging, but survivable. When the threat of Hurricane Harvey began, it was easy to assume that Harvey would be a similar experience to those previous floods. In the end, it was much worse. This bias affected what Houstonians believed Mother Nature’s “strategies” were and how they played this “game”. There is another logistical aspect of evacuations which must be considered. Houston is a city of over six million, and evacuation is a logistical nightmare. In 2005, my family attempted to evacuate for Hurricane Rita. We spent four hours in evacuation route traffic only to move a mere five miles. After turning around, we stayed in our home throughout the storm and although there were two days of power outage, everything was fine. Experiences like these showcase that there is also an incentive for resident’s to not evacuate, for the payoff decreases when other residents decide to also evacuate. Of course, this piece is a retrospective analysis on hurricane preparation and is subject to much hindsight bias. Still, utilizing game theory in this manner should make one wonder how we can better prepare for our next natural disaster. Further, we should consider whether it is always better to be safe rather than sorry as Southern Texas, the Caribbean, and Florida begin to heal.
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# associativity of operators in C language Discussion in 'Engineering Concepts' started by hitesh_best, Aug 3, 2007. Not open for further replies. 1. ### hitesh_bestNew Member Joined: Aug 3, 2007 Messages: 4 0 Trophy Points: 0 Hello friends, I am following Let us C, 6th edition, by Yashavant Kanetkar, I have a problem understanding the concept of associativity of operators, I tried to find some good material on internet by searching in google but could not find much. Kanetkar writes, "Left to Right associativity means that the left operand must be unambiguous. Unambiguous in what sense? It must not be involved in evaluation of any other sub-expression. Similarly in case of Right to Left associativity the right operator must be unambiguous." I could not understand what it means by "being unambiguous". There is a expression, Code: ``` g = 10/5/2/1 ``` Now we have to find the order in whcih the calculations will take place In his solution book, he gives the solution. Code: ``` operator left Right Remark / 10 5 or 5/2/1 Left operand is unambiguous, Right is not / 10/5 or 5 2 or 2/1 Left operand is unambiguous, Right is not / 10/5/2 or 2 1 Right operand is unambiguous, Left is not ``` Now in first row and right column why he writes only two operands that are possible: 5 or 5/2/1 why also not 5/2 In the second row in the remark column why he says that left operand is unambiguous when there are two operands that are possible i.e. 10/5 or 5. Thanks a lot 2. ### hitesh_bestNew Member Joined: Aug 3, 2007 Messages: 4 0 Trophy Points: 0 Friends, admitting my mistake I have started the same thread in C/C++ forum . I think this thread should be there. Please forgive me for this. Joined: Jul 12, 2004 Messages: 15,326
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Select Page Your Perfect Assignment is Just a Click Away Starting at \$8.00 per Page 100% Original, Plagiarism Free, Customized to Your instructions! 1, ## Branches: Leap Year A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: 1) The year must be divisible by 4 2) If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400 Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines whether that year is a leap year. Ex: If the input is: ``1712`` the output is: ``1712 is a leap year.`` Ex: If the input is: ``1913`` the output is: ``1913 is not a leap year.`` `2, `Branches: Remove gray from RGB Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color’s value can range from 0 to 255. Thus (255, 0, 0) is bright red, (130, 0, 130) is a medium purple, (0, 0, 0) is black, (255, 255, 255) is white, and (40, 40, 40) is a dark gray. (130, 50, 130) is a faded purple, due to the (50, 50, 50) gray part. (In other words, equal amounts of red, green, blue yield gray). Given values for red, green, and blue, remove the gray part. Ex: If the input is: ``130 50 130`` the output is: ``80 0 80`` Hint: Find the smallest value, and then subtract it from all three values, thus removing the gray 3,  Branches: Largest number Write a program that reads three integers as inputs, and outputs the largest of the three values. Ex: If the input is: ``7 15 3`` the output is: ``15`` `3,`Variables/Assignments: Divide by x Write a program using integers userNum and x as input, and output userNum divided by x four times. Ex: If the input is: ``2000 2`` the output is: ``1000 500 250 125`` Note: In Coral, integer division discards fractions. Ex: 6 / 4 is 1 (the 0.5 is discarded). 4,  Variables/Assignments: Driving costs Driving is expensive. Write a program with a car’s miles/gallon and gas dollars/gallon (both floats) as input, and output the gas cost for 10 miles, 50 miles, and 400 miles. Output the gas cost (gasCost) with two digits after the decimal point, which can be achieved as follows: `Put gasCost to output with 2 decimal places` Ex: If the input is: ``20.0 3.1599`` the output is: ``1.58 7.90 63.20`` Note: Small expression differences can yield small floating-point output differences due to computer rounding. Ex: (a + b)/3.0 is the same as a/3.0 + b/3.0 but output may differ slightly. Because our system tests programs by comparing output, please obey the following when writing your expression for this problem. First use the dollars/gallon and miles/gallon values to calculate the dollars/mile. Then use the dollars/mile value to determine the cost per 10, 50, and 400 miles. Note: Real per-mile cost would also include maintenance and depreciation. 5,  Variables/Assignments: Using math functions Given three floating-point numbers x, y, and z, output x to the power of y, x to the power of (y to the power of z), the absolute value of x, and the square root of (x * y to the power of z). Output all results with five digits after the decimal point, which can be achieved as follows: `Put result to output with 5 decimal places` Ex: If the input is: ``5.0 2.5 1.5`` the output is: ``55.90170 579.32402 5.00000 6.64787`` Hint: Coral has built-in math functions (discussed elsewhere) that may be used. 6,  Variables/Assignments: Using math functions Given three floating-point numbers x, y, and z, output x to the power of y, x to the power of (y to the power of z), the absolute value of x, and the square root of (x * y to the power of z). Output all results with five digits after the decimal point, which can be achieved as follows: `Put result to output with 5 decimal places` Ex: If the input is: ``5.0 2.5 1.5`` the output is: ``55.90170 579.32402 5.00000 6.64787`` Hint: Coral has built-in math functions (discussed elsewhere) that may be used. 7,  Arrays: Middle item Given a sorted list of integers, output the middle integer. A negative number indicates the end of the input (the negative number is not a part of the sorted list). Assume the number of integers is always odd. Ex: If the input is: ``2 3 4 8 11 -1 `` the output is: ``4`` The maximum number of inputs for any test case should not exceed 9 positive values. If exceeded, output “Too many inputs”. Hint: Use an array of size 9. First, read the data into an array. Then, based on the number of items, find the middle item. 8,Arrays: Output values below an amount Write a program that first reads a list of 5 integers from input. Then, read another value from the input, and output all integers less than or equal to that last value. Ex: If the input is: ``50 60 140 200 75 100`` the output is: ``50 60 75`` For coding simplicity, follow every output value by a space, including the last one. Then, output a newline. Such functionality is common on sites like Amazon, where a user can filter results. "Place your order now for a similar assignment and have exceptional work written by our team of experts, guaranteeing you A results." ## Our Service Charter 1. Professional & Expert Writers: Eminence Papers only hires the best. Our writers are specially selected and recruited, after which they undergo further training to perfect their skills for specialization purposes. Moreover, our writers are holders of masters and Ph.D. degrees. They have impressive academic records, besides being native English speakers. 2. Top Quality Papers: Our customers are always guaranteed of papers that exceed their expectations. 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"this class of problems lie"s in **RE**, so its name is "**RE**". <br><br><br> $\operatorname{Prob}(\hspace{.02 in}M \text{ accepts}) \; = \; \operatorname{Prob}\hspace{-0.04 in}\big(\hspace{-0.02 in}(\exists n)(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})\big)$ <br>$=$ $\displaystyle\sum_n \operatorname{Prob}(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})$ $=$<br> $\displaystyle\lim_N \left(\displaystyle\sum_{n\leq \hspace{.02 in}N} \operatorname{Prob}(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})\hspace{-0.04 in}\right)$ $=$ $\displaystyle\lim_N \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } N \text{ steps}) \;\;\; = \;\;\; \displaystyle\lim_n \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})$ <br><br> Since $\: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps}) \:$ is non-decreasing in $n$, <br><br><br> $\frac12 \;\; < \;\; \operatorname{Prob}(\hspace{.02 in}M \text{ accepts})$ $\iff$ $\frac12 \;\; < \;\; \displaystyle\lim_n \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})$ $\iff$ $(\exists n)\left(\frac12 < \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})\hspace{-0.02 in}\right)$<br><br>
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## Duotrigintatrecentilliquattuorsexagintaquingentillion A Duotrigintatrecentilliquattuorsexagintaquingentillion (1 Duotrigintatrecentilliquattuorsexagintaquingentillion) is 10 to the power of 997695 (10^997695). This is a stupendously astronomical number! ## How many zeros in a Duotrigintatrecentilliquattuorsexagintaquingentillion? There are 997,695 zeros in a Duotrigintatrecentilliquattuorsexagintaquingentillion. ## What's before Duotrigintatrecentilliquattuorsexagintaquingentillion? A Duotrigintatrecentillitresexagintaquingentillion is smaller than a Duotrigintatrecentilliquattuorsexagintaquingentillion. ## What's after Duotrigintatrecentilliquattuorsexagintaquingentillion? A Duotrigintatrecentilliquinsexagintaquingentillion is larger than a Duotrigintatrecentilliquattuorsexagintaquingentillion. ## Duotrigintatrecentilliquattuorsexagintaquingentillionaire A Duotrigintatrecentilliquattuorsexagintaquingentillionaire is someone whos assets, net worth or wealth is 1 or more Duotrigintatrecentilliquattuorsexagintaquingentillion. It is unlikely anyone will ever be a true Duotrigintatrecentilliquattuorsexagintaquingentillionaire. If you want to be a Duotrigintatrecentilliquattuorsexagintaquingentillionaire, play Tap Tales! ## Is Duotrigintatrecentilliquattuorsexagintaquingentillion the largest number? Duotrigintatrecentilliquattuorsexagintaquingentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is. ## Duotrigintatrecentilliquattuorsexagintaquingentillion written out Duotrigintatrecentilliquattuorsexagintaquingentillion is written out as: 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# Isomorphism of polynomial rings implying isomorphism of the coefficient rings [duplicate] Let $R$ and $S$ be commutative rings. Let $x, y$ be indeterminates, and assume that one has an isomorphism $R[x] \rightarrow S[y]$ (not necessarily mapping $x$ to $y$ of course). Does this imply $R \cong S$? If not, what is a counterexample? This may seem like a homework problem, but is not.
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Paul's Online Notes Home / Calculus I / Limits / Computing Limits Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. Section 2.5 : Computing Limits 9. Evaluate $$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{x}{{3 - \sqrt {x + 9} }}$$, if it exists. Show Solution There is not really a lot to this problem. Simply recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Recall that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. Simply factoring will not do us much good here so in this case it looks like we’ll need to rationalize the denominator. \begin{align*}\mathop {\lim }\limits_{x \to 0} \frac{x}{{3 - \sqrt {x + 9} }} & = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\left( {3 - \sqrt {x + 9} } \right)}}\frac{{\left( {3 + \sqrt {x + 9} } \right)}}{{\left( {3 + \sqrt {x + 9} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {3 + \sqrt {x + 9} } \right)}}{{9 - \left( {x + 9} \right)}}\\ & = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {3 + \sqrt {x + 9} } \right)}}{{ - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3 + \sqrt {x + 9} }}{{ - 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 6}}\end{align*}
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## Encyclopedia > Triangular number Article Content # Triangular number A triangular number is a number that can be arranged in the shape of an equilateral triangle (by convention, the first triangular number is 1): 1: + x 3: x x + + x x 6: x x x x x x + + + x x x 10: x x x x x x x x x x x x + + + + x x x x 15: x x x x x x x x x x x x x x x x x x x x + + + + + x x x x x 21: x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x + + + + + + x x x x x x The formula for the nth triangular number is ½n(n+1) or (1+2+3+...+ n-2 + n-1 + n). It is the binomial coefficient ${n+1 \choose 2}$ It can also be shown that for any n-dimensional simplex with sides of length x, the formula $\frac {(x)(x+1)...(x+(n-1))} {n!}$ will accurately show the number of that simplex. For example, a tetrahedron with sides of length 2 has a number of $\frac {(2)(2+1)(2+2)} {6}$, or 4. (Note: A tetrahedron can be created by taking a number, getting the triangle of that number, and then adding to it all the triangles of the numbers before it, so a tetrahedron of 2 would have 2 triangled=3 plus 1 triangled=1 =4.) One of the most famous triangular numbers is 666, also known as the Number of the Beast. Every perfect number is triangular. All Wikipedia text is available under the terms of the GNU Free Documentation License Search Encyclopedia Search over one million articles, find something about almost anything! Featured Article Dynabee ... will cause one end of the axis to push against the upper rim of the groove, while the other end of the axis pushes against the lower rim of the groove. While the ax ...
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# Weierstrass’ criterion of uniform convergence ###### Theorem. Let the real functions $f_{1}(x)$, $f_{2}(x)$, … be defined in the interval $[a,b]$.  If they all the condition $|f_{n}(x)|\leqq M_{n}\quad\forall\,x\in[a,b],$ with $\sum_{n=1}^{\infty}M_{n}$ a convergent series of , then the function series $f_{1}(x)\!+\!f_{2}(x)\!+\!\cdots$ converges uniformly (http://planetmath.org/SumFunctionOfSeries) on the interval $[a,b]$. The theorem is valid also for the series with complex function terms, when one replaces the interval with a subset of $\mathbb{C}$. Title Weierstrass’ criterion of uniform convergence WeierstrassCriterionOfUniformConvergence 2013-03-22 14:38:21 2013-03-22 14:38:21 pahio (2872) pahio (2872) 9 pahio (2872) Theorem msc 26A15 msc 40A30 Weierstrass’ M-test
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472,356 Members | 1,890 Online # Need Help Sorting a JavaScript Array ... ???? Hello All, whilst I know how to sort an array I am having problems with my situation, what I would ideally like is something like a 'sort keys' action with Perl hashes. I am reading in data and building an array entry that will be of the form:- array[0] - horse1,2,4,6,1,3,6 = 22 ... array[6] - horse6,2,3,6,1,9,8 = 29 etc etc. I know want to sort these array entries based on the totals (i.e 22, 29 etc), I figured you cannot do this in one array so maybe 2 arrays, one array would be as follows:- array1[0] - horse1,2,4,6,1,3,6 = ... array1[6] - horse6,2,3,6,1,9,8 = and the second array:- array2[0] - 22 ... array2[6] - 29 I could then sort on the 2nd array, the trouble is that I loose the relationship between the 2 arrays. I thought about creating an object array to sort upon but am unsure if this would work. Has anyone a solution to this and if so some example code to help me along. thanx, Mark .... Jul 20 '05 #1 3 1392 In article <24**************************@posting.google.com >, ma***********@talk21.com (Tommo) writes: Hello All, whilst I know how to sort an array I am having problems with my situation, what I would ideally like is something like a 'sort keys' action with Perl hashes. I am reading in data and building an array entry that will be of the form:- array[0] - horse1,2,4,6,1,3,6 = 22 ... array[6] - horse6,2,3,6,1,9,8 = 29 etc etc. Making the assumption, based on your above data, that array[X] will not always correspond to horseX since your array[6] is horse6 but array[0] is not horse0. Anyway, its not relevant for this approach. is the array created as a string, or as seperate entries? myArray = new Array() myArray[0] = new Array('horse0',2,4,6,1,3,6) //more data myArray[6] = new Array('horse6',2,3,6,1,9,8) //more efficient to use bracket notation: myArray = new Array() myArray[0] = ['horse0',2,4,6,1,3,6] //more data myArray[6] = ['horse6',2,3,6,1,9,8] //although I personally never use this approach, its too confusing sometimes. myArray2 = new Array() for (i=0;i<myArray.length;i++) { myArray2[i] = new Array() //lets save the reference myArray2[i][1] = myArray[i][0] k = 0; //lets total the rest of it for (j=0;j<myArray[i].length;j++) { k = k + (+myArray[i][j]); } myArray2[i][0] = k; } myArray2.sort() //In my personal experience, when you sort an array that is an array of arrays, it sorts on the first element in the secondary array Untested. How you would go about it depends a lot on how your original array is constructed though. If the original array is constructed as a plain string, then you would have to split it on the delimiter. It might be more efficient to construct it as a hash array and go from there. Needs testing. Its 7:30AM here and no time to test it. Will test it further later today. -- Randy Jul 20 '05 #2 thanks for your assistance, I will also do some testing and post back results .. cheers, Mark .. Jul 20 '05 #3 JRS: In article <24**************************@posting.google.com >, seen in news:comp.lang.javascript, Tommo <ma***********@talk21.com> posted at Mon, 15 Dec 2003 02:12:27 :- whilst I know how to sort an array I am having problems with my situation, what I would ideally like is something like a 'sort keys' action with Perl hashes. I am reading in data and building an array entry that will be of the form:- array[0] - horse1,2,4,6,1,3,6 = 22 ... array[6] - horse6,2,3,6,1,9,8 = 29 etc etc. You can use the sort method of the outer array by providing it with a suitable compare function. function Compare(X, Y) { var SX = SY = 0 for (J=1; J<X.length; J++) SX += X[J] // form score for (J=1; J<Y.length; J++) SY += Y[J] // form score return SY - SX } That is probably about right; it should return the difference between the scores of the two horses. If X.length = Y.length always, the loops can merge; otherwise, the job seems unfair. For efficiency, though, the sums should be calculated once per horse; and, as someone said, if they are placed at the start of the data structure the default sort could suffice. To use sort keys in general, just provide a compare function that compares them appropriately, returning <0, 0, >0 accordingly. Consider representing each animal as Nag = {Score:22, Name:"Shergar", Data:[2,4,6,1,3,6]} -- © John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 IE 4 © <URL:http://jibbering.com/faq/> Jim Ley's FAQ for news:comp.lang.javascript <URL:http://www.merlyn.demon.co.uk/js-index.htm> Jsc maths, dates, sources. <URL:http://www.merlyn.demon.co.uk/> TP/BP/Delphi/Jsc/&c, FAQ topics, links. Jul 20 '05 #4 This thread has been closed and replies have been disabled. Please start a new discussion. ### Similar topics 3 by: Paul Kirby | last post by: Hello All I am trying to update me code to use arrays to store a group of information and I have come up with a problem sorting the multiple array :( Array trying to sort: (7 arrays put into... 4 by: Derek | last post by: Hi, I've built a rather large CGI that dumps a lot of data and a fairly complex javascript app out to the client's browser. 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