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http://www.physicspages.com/2017/01/09/infinite-square-well-force-to-decrease-well-width/ | 1,501,029,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00061.warc.gz | 533,143,673 | 14,902 | # Infinite square well – force to decrease well width
References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.2, Exercise 5.2.4.
One way of comparing the classical and quantum pictures of a particle in an infinite square well is to calculate the force exerted on the walls by the particle. If a particle is in state ${\left|n\right\rangle }$, its energy is
$\displaystyle E_{n}=\frac{\left(n\pi\hbar\right)^{2}}{2mL^{2}} \ \ \ \ \ (1)$
If the particle remains in this state as the walls are slowly pushed in, so that ${L}$ slowly decreases, then its energy ${E_{n}}$ will increase, meaning that work is done on the system. The force is the change in energy per unit distance, so the force required is
$\displaystyle F=-\frac{\partial E_{n}}{\partial L}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (2)$
If we treat the system classically, then a particle with energy ${E_{n}}$ between the walls is effectively a free particle in this region (since the potential ${V=0}$ there), so all its energy is kinetic. That is
$\displaystyle E_{n}$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}\ \ \ \ \ (3)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2E_{n}}{m}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n\pi\hbar}{mL} \ \ \ \ \ (5)$
The classical particle bounces elastically between the two walls, which means its velocity is exactly reversed at each collision. The momentum transfer in such a collision is
$\displaystyle \Delta p=2mv=\frac{2n\pi\hbar}{L} \ \ \ \ \ (6)$
The time between successive collisions on the same wall is
$\displaystyle \Delta t=\frac{2L}{v}=\frac{2mL^{2}}{n\pi\hbar} \ \ \ \ \ (7)$
Thus the average force exerted on one wall is
$\displaystyle \bar{F}=\frac{\Delta p}{\Delta t}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (8)$
Comparing with 2, we see that the quantum and classical forces in this case are the same. | 589 | 1,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-30 | latest | en | 0.806517 |
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Dec4 accepted Applying several patterns to Split Dec4 comment Applying several patterns to Split I like your approach better from the perspective of comprehension. However, LeonidShifrins approach is faster. Dec4 asked Applying several patterns to Split Nov29 revised Restricted accumulation of values added 5 characters in body Nov29 accepted How to define a large number of assumptions? Nov29 accepted Why is an integer not considered as a real? Nov29 asked How to define a large number of assumptions? Nov27 comment Determining the week of a year from a given date @Heike stevenvh is right. Is there a way to avoid these two problems (week 0 and week 53)? Nov27 comment Determining the week of a year from a given date Why would 2010-01-01 return week 53? Doesn't one year only contain 52 weeks? Nov27 comment Determining the week of a year from a given date I was not aware of using Java within Mathematica. What you wrote sounds interesting. Can you recommend any website or so where I can learn more from implementing java-code into Mathematica and also on available java-scripts? Nov26 comment Usage of plot results @SimonWoods Two more things I did not mentioned before: (1) It would be nice to have the button to appear above the line and (2) if the button touches one of the PlotMarkers Clipboard should only copy the corresponding data-point. Nov26 accepted Usage of plot results Nov23 comment Usage of plot results @SimonWoods The behaviour of the button is a bit strange. It does not change its colour immediately when I touch another line. (Maybe my Mac is reacting to slow?) Nov22 asked Usage of plot results Nov9 comment Leveling peaks in list Almost. With {a___, b_, c_, d_, e___} instead of {a__, b_, c_, d_, e__} it works. (i.e. BlankNullSequence instead of BlankSequence) Nov9 comment Leveling peaks in list The following works but maybe there is a better approach: first = {left_, right_, rest___} /; left > right :> {left/2, left/2, right, rest}; sec = {rest___, left_, right_} /; right > left :> {rest, Mean[{left, right}], Mean[{left, right}]}; third = {a___, b_, c_, d_, e___} /; b < c > d :> {a, Mean[{b, c}], Mean[{b, c}], d, e}; (# //. first //. sec //. third) & @ data Nov8 comment Leveling peaks in list An extreme value is always to be levelled if its neighbour is lower (e.g. dataNoLevelling={1,5,5,5,1} -> (*resNoLevelling={1,5,5,5,1}*)and dataLevelling={100,5,5,5,100} -> (*resLevelling={50,50,5,5,52.5,52,5}*) Nov8 comment Leveling peaks in list I just discovered that there is one problem with your approach. Lets say you have data={10,0,0,0,10}. Then both, the first and the last value, are extrem values which I would like to level as described. I did not pay attention to this case when I asked my question. The problem I see, is that the approach must extend data to the left. (desired result: res={5,5,0,0,5,5}. Note: Length@res-Length@data=1) Nov5 accepted Difference between AbsoluteTiming and Timing Nov5 comment Restricted accumulation of values @J.M. What about the If-clause? Do have an idea how one could substitute this part of my code with another expression? | 822 | 3,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-18 | longest | en | 0.916038 |
http://lofi.forum.physorg.com/please-help%2C-inelastic-collision_13833.html | 1,369,478,275,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705936437/warc/CC-MAIN-20130516120536-00062-ip-10-60-113-184.ec2.internal.warc.gz | 156,492,912 | 4,556 | fluvly
Forgive the probable incorrect use of words of physics, I'm not used to writing them in English Hope you will understand anyway.
Object number 1 is moving horizontally (left to right) with Uniform rectilinear motion at the speed of 10.5 m/s. Its mass is 7.5 Kg. The coefficient of dynamic friction is 0.15. A drawing force drags object number one from left to right (with an angle of 45) and its vertical component(projection) has the same value of its horizontal component, which is 9,6N.
Evidently the frictional force is -9,6N, otherwise the motion would not be uniform rectilinear.
Now Object number 2 (same mass of number 1, 7.5 Kg) falls on object number 1 vertically, causing a perfectly inelastic collision and it stays attached to object number 1. The drawing force stays constant (it doesn't change from 9.6N vertically and 9.6N horizontally).
what is the speed right after the collision?
what distance does the object number 1+2 cover before it stops?
Thanks for the help.
mr_homm
Here is the simple version of the solution (the one that textbooks and teachers usually want):
Because the collision takes very little time to complete, the drawing force and frictional force do not have time to transfer a significant amount of momentum to the system. Therefore, momentum is conserved during the moment of the collision, and so the final momentum = (2m)v_final = mv_initial = initial momentum. The velocity just after the collision is therefore exactly 1/2 of the velocity before the collision, 5.25 m/s.
Here is the more careful solution (which is actually true):
At the moment of impact, there will be a very large downward force on the bottom mass, as the top mass transfers all of its vertical momentum to the bottom mass. Because of this, the normal force between the bottom block and the ground will temporarily be very large, as the downward momentum of the top block passes through the bottom block and is absorbed by the ground. Therefore, the friction will also be very large during this brief time. Therefore, you CANNOT ignore the change of momentum due to friction during the collision.
The total momentum removed from the system by friction will be the average value of the friction force multiplied by the time of the collision. Since the original friction force is canceled by the horizontal part of the drawing force, the loss of momentum will be due only to the excess friction, which is caused by the impact. This is easy to calculate: If the second mass has initially a downward velocity v_y, it must lose all of its vertical momentum mv_y during the collision time t. Therefore, the average net upward force on the second block is mv_y/t. Since F_net = N - mg, it follows that N = mv_y/t + mg. The "mg" term is not really part of the collision, but is part of the new steady normal force which the first mass will feel after the collision. Therefore, the collision itself exerts, by Newton's third law, an average force mv_y/t driving the first mass downward during the collision, and so it generates an average friction of 0.15m(v_y/t).
During the time of collision, this friction will remove a quantity of momentum which is 0.15m(v_y/t)*t = 0.15mv_y. Therefore the momentum of the system immediately after the collision will be the initial momentum minus this change, i.e. (2m)v_final = mv_initial - 0.15mv_y. Thus the velocity just after the collision will be v_final = 0.5v_initial - 0.075v_y.
Since the problem did not specify v_y, this more careful solution cannot lead to a numerical answer. This is an example of poor teaching in my opinion, because the problem is constructed so that a student who does NOT see deeply into the problem is rewarded with a correct mark, while a student who DOES see deeply into the problem becomes lost because the necessary information is not present. The important thing to see is that the simple answer (which your textbook will say is correct) is ACTUALLY WRONG. To make the simple answer be the correct one, it is necessary to specify that the second mass is dropped onto the first mass from a height very close to zero, so that it does not have a vertical velocity when it strikes the first mass. Only if v_y = 0 is the simple solution correct, and the problem fails to state this assumption.
It may or may not be wise to inform your teacher of this fact. Some teachers are glad to see that students are really thinking, while other teachers hate to be corrected by a student. I do not know your teacher, so I cannot predict the reaction.
Now for the second part of the problem, assuming that the simple solution is correct, you must find the net force which is slowing the blocks. This is fairly easy. The normal force on the bottom of the first block is now 2mg - 9.6, and so the friction force is now 0.15*(2mg - 9.6) = 0.15*(2*7.5*9.81 - 9.6) = 20.6325 Newtons. Since the drawing force is still pulling forward with 9.6 Newtons, the net force slowing the blocks is 9.6 - 20.6325 = -11.0325 Newtons. Therefore, its acceleration is -11.0325/(2m) = -11.0325/15 = -0.7355 m/s^2.
Since its initial velocity is 5.25 m/s, its final velocity is 0, and its acceleration is -0.7355 m/s^2, you can use the kinematic equation (v_final)^2 - (v_initial)^2 = 2*a*(x_final - x_initial). Solving this gives x_final - x_initial = (0 - 5.25^2)/(2*-0.7355) = 18.737 meters. You can check this by doing the kinematic calculation another way: The time to stop must be (v_final - v_initial)/a = 7.138 seconds. Since the average speed during constant accleration is always (v_initial + v_final)/2, you have in this problem an average speed of (0 + 5.25)/2 = 2.625 m/s. Therefore, the distance traveled should be average speed * time = 2.625*7.138 = 18.737 meters, the same answer as you get by the other method.
Hope this helps!
--Stuart Anderson
fluvly
Thank you Stuart, I really appreciated the detailed explanation. It seemed to me a bit banal that the speed after collision was just half the initial speed. | 1,441 | 5,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2013-20 | latest | en | 0.933469 |
http://stackoverflow.com/questions/14423565/excel-search-through-a-column-to-find-if-text-matches-and-count-string | 1,448,869,876,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461113.77/warc/CC-MAIN-20151124205421-00241-ip-10-71-132-137.ec2.internal.warc.gz | 217,340,369 | 19,314 | # Excel - Search through a column to find if text matches and count string
I want to add values from 1 worksheet to another.
The first worksheet ("November 2011") simply contains name & status, names are however duplicated and a percentage of a string needs to be added on the second worksheet ("Sales").
e.g. "November 2011"
`````` A B
1 Name Status
2 McDonalds Completed
3 McDonalds Won
4 Burger King Won
5 Burger King Won
6 Wendys Completed
7 iHop Completed
8 iHop Completed
``````
The second worksheet ("Sales") contains 3 columns Name, Completed & Won. However, Name has had duplicates removed.
e.g. "Sales"
`````` A B C
1 Name Completed Won
2 McDonalds
3 Burger King
4 Wendys
5 iHop
``````
I want the "Sales" worksheet to look like this however:
`````` A B C
1 Name Completed Won
2 McDonalds 50% 50%
3 Burger King 0% 100%
4 Wendys 100% 0%
5 iHop 100% 0%
``````
Just wondering what formula to use? It needs to check against the name.
-
I'd use a pivot table. – lc. Jan 20 '13 at 10:26
unable to, unfortunately – Accolade Jan 20 '13 at 10:35
What version are you using? – glh Jan 20 '13 at 10:47
I'd use a `countif` function to find the number of sales and `countifs` to find the number of wins and completed.
``````=COUNTIFS('November 2011'!\$A:\$A, A2, 'November 2011'!\$B:\$B, "won")
/ COUNTIF('November 2011'!\$A:\$A,A2)
``````
This would give you the ratio or percent if formatted of the won.
-
Doesn't seem to work? – Accolade Jan 20 '13 at 10:59
The syntax may be out I can't test as I'm on my phone. There are some links in my answer. – glh Jan 20 '13 at 11:02
=COUNTIFS('November 2011'!\$A:\$A, A2, 'November 2011'!\$B:\$B, "won") / COUNTIF('November 2011'!\$A:\$A,A2) – Accolade Jan 20 '13 at 11:03
Thanks I'll edit the answer – glh Jan 20 '13 at 11:03
Another alternative is to use an "array formula" with AVERAGE function, i.e. this formula in B2
`=AVERAGE(IF(November 2011'!\$A\$2:\$A\$1000=\$A2,IF('November 2011'!\$B\$2:\$B\$1000=B\$1,1,0)))`
confirmed with CTRL+SHIFT+ENTER and copied across and down
-
This answer is definitely needed for pre-2007 Excel. Weldone. – glh Jan 21 '13 at 10:04 | 749 | 2,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2015-48 | longest | en | 0.882054 |
https://rochaksafar.com/how-far-are-15-miles/ | 1,686,019,818,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652207.81/warc/CC-MAIN-20230606013819-20230606043819-00479.warc.gz | 557,895,722 | 73,126 | 0
123
# How Far Are 15 Miles?
### What is the mile?
The mile refers to a measurement unit of length, widely utilized in the United States and other countries that adopted the Imperial measurement system. One mile equals 1.609 km or 1760 yards. It determines the distance in various situations, including road signage, race distances, and sports events.
## What is the distance of 15 miles?
If you’re thinking about the distance of 15 miles, it depends on the situation. For instance, 15 miles could be long or short if you travel by vehicle, depending on the conditions. On the other hand, if you face massive roads or traffic, it may take around 15 minutes to travel 15 miles over a smooth road without traffic or longer.
Running or walking at 15 miles is an extensive distance that can be difficult to finish without proper preparation and training. For example, a seasoned athlete could take many hours to complete the 15 miles, whereas an average walker can require a whole day.
Compared to other measures, 15 miles corresponds to approximately 24.14 km or 13.04 nautical miles. These conversions, like navigating the sea, could be crucial in specific international travel contexts.
## How Long Is A Fifteen Miles Drive?
### What is an inch?
The mile refers to a measurement unit of length, widely utilized in the United States and other countries using the Imperial measurement system. One mile is equal to 1.609 km or 1760 yards. It determines the distance in many scenarios, such as road signals, race distances, and sports events.
### What is the average time for 15 miles of driving?
The amount of time needed to travel 15 miles depends on many factors, such as traffic conditions, speed limits, and road conditions. Generally, driving 15 miles on a smooth road with no traffic can take 15 minutes at 60 mph. But, if you experience high traffic or construction, it may take a lot longer.
It is important to remember that driving times can differ significantly based on your location and the type of road you’re traveling on. For instance, a 15-mile drive on a freeway or highway could be quicker than a 15-mile drive along an unpaved country road with many intersections and stop signs.
### Factors that could impact the speed of your car
There are a variety of factors that influence the amount of time required to travel 15 miles. A few of these are:
• Conditions for traffic: Traffic jams, accidents, road closures, and accidents can all reduce your travel time and cause major delays to your trip.
• Speed limitations: If you’re traveling on roads with low-speed limits, it’ll take you longer to travel the same distance as if you were on roads with greater speed limits.
• Road conditions: Road conditions that are not in good shape, like potholes, roads in construction, or uneven surfaces, could slow down your time on the road and make driving more difficult.
• The type the vehicle: The kind of vehicle you’re driving may also affect the speed of your travel. For instance, a sports car could travel 15 miles quicker than a bigger, heavier vehicle.
### Tips for estimating the duration of your drive
If you’re making plans for a vacation and need to figure out the time to drive 15 miles, there are various methods you can employ to get an accurate estimate. These suggestions include:
• Verify traffic conditions: Utilize an app or website to determine whether there are any road delays or closures on your route.
• Be aware of how fast you can travel: Check out the speed limit of the roads you’ll be driving on, and then factor this into your estimate.
• Take note of stoppages: If you have to make stops for fuel, for example, food, be sure to add them into your estimation.
• Be honest: Keep in mind that unexpected delays are possible. Therefore, it’s always recommended to be on the safe side and allow ample time to get to the destination you want to be.
## How Far Are 15 Miles In Just 15 Minutes?
A mile is a measurement unit of length that is widely utilized throughout the United States and other countries using the Imperial measuring system. It’s equivalent in length to 1.609 miles or 1760 yards. Miles are used to quantifying distance in various instances, like road markers, race distances, and sports events.
### What is the distance between 15 miles in a minute?
The time required to travel 15 miles depends on many variables, including the mode of transport and the speed at which you’re traveling. For example, if you’re driving in a car and traveling 15 miles, it could be between 15 and 20 minutes with a speed average of around 60mph over a smooth road without traffic. But, if there’s lots of roadway construction or heavy traffic, you might be much longer.
If you’re walking you need to cover 15 miles will vary based on the pace you’re walking at. Generally, it will take about 4-5 hours to complete 15 miles of walking at a moderate speed of 3 miles per hour.
If you’re riding in a group, the time required to travel 15 miles will vary based on your riding speed and the terrain. For example, if the terrain is flat, an experienced cyclist may be able to cover 15 miles in about 30-45 minutes, assuming about 20-25mph on average. But if the terrain is steep or there’s a headwind, it may be longer.
If you’re using public transportation, the length of time required to travel 15 miles will vary based on the route taken and the frequency of service. For instance, a public bus may take approximately 45 minutes to cover 15 miles, whereas trains for commuters could take 20-25 minutes.
It is important to keep in mind that the time required for a 15-mile trip is contingent on many factors. These are only estimates. Ensure you allow enough time to account for any unexpected delays or changes to your travel plans.
### Things that could impact the duration of travel
Various factors affect the amount of time required fifteen miles to cover. A few of them are:
• Method of transportation: The amount of time required to cover 15 miles depends on the type of transport you’re employing.
• Conditions for traffic: Heavy traffic, accidents, road closures, and accidents could all affect your travel time and create major delays to your trip.
• Speed limitations: The speed restriction of the roadway you’re on may impact the time you travel. If the speed is less, it’ll take longer to cover the same distance.
• Surface: The terrain may affect the time required to travel 15 miles, especially if you’re cycling or walking. Headwinds or hills can make you slow down.
• Service frequency: If you’re traveling by public transport, the frequency of service may impact your travel time. Regular services have shorter waiting durations.
### Tips to estimate travel time
If you’re planning to travel and need to determine the time required to cover 15 miles of travel, here are some methods you can employ to get an accurate estimate. These include:
• Think about the transportation mode: Consider the type of transport you’ll use and the speed at which you can travel.
• Check the traffic conditions: Utilize an app or a website to search for road closures or delays on your route.
• Make allowances for the stops: If you must make stops during your travels like fuel or food, be sure to include that in your budget.
• Be real: Be aware that unexpected delays are possible, and it’s better to be in the direction of caution and allow yourself ample time to reach the destination you want to be at.
## How Long Are 15 Miles?
A mile is one of the length units widely employed throughout the United States and other countries using the Imperial system of measuring. It’s equivalent in length to 1.609 miles or 1760 yards. It is utilized to define distance in many situations, including road markers, race distances, and sporting events.
### What is the time in 15 miles?
The time required to travel for 15 miles will be contingent upon a number of aspects, including the mode of transport you’re using and the speed at which you travel. For example, if you’re driving a vehicle, the distance of 15 miles could be between 15 and 20 hours at a 60-mph average on a clean, not busy road. On the other hand, it may take longer if there’s a lot of construction, traffic, or road construction.
To determine how long it will take to cover 15 miles, you must determine the speed at which you’re traveling. For example, if you’re moving at 60 mph, it’ll take between 15 and 20 minutes to cover 15 miles. So, it is possible to say that it takes about 0.25 minutes or an hour to travel 15 miles at 60 mph.
If you’re traveling at a different speed and you want to calculate the distance, use an easy formula to determine the amount of time required for a journey of 15 miles. This formula works as follows:
Time = Distance / Speed
For instance, if moving at 30 mph, your calculation will be:
Time = 15 miles / 30 mph
Time is 0.5 hours or thirty minutes
If you’re walking, the amount of time required to walk 15 miles will vary based on your walking pace. Generally, it will take 4-5 hours to complete 15 miles of walking at a pace of 3 miles per hour.
If you’re riding for a long distance, the time needed to travel 15 miles will be contingent on your speed as well as the terrain. For example, in flat terrain, experienced cyclists may be able to complete 15 miles in about 30-45 minutes with an average speed of 20-25 mph. But if the terrain is steep or there’s a strong headwind, it might be longer.
If you’re taking public transportation, the length of time to travel 15 miles will be contingent on the route chosen and the frequency of service. For instance, buses can take 45 minutes to cover 15 miles, whereas trains for commuters could take 20-25 minutes.
It is important to remember that the time required for a 15-mile trip depends on various factors, and these are only estimates. Be prepared for unexpected delays or adjustments within your plan.
### The factors that influence the duration of travel
Many factors can impact the amount of time required for a 15-mile trip. A few of them are:
• The mode of transport: The amount of time required for a journey of 15 miles depends on the method of transportation you’re employing.
• Conditions for traffic: Traffic jams, accidents, and closures of roads could all affect your travel time, and cause substantial delays to your journey.
• Speed limitations: The speed limitation of the road you’re on may impact the time you travel. If the speed limit is higher, it’ll take more time to travel that same amount of distance.
• Surface: The terrain may affect the time required to travel 15 miles, especially if you’re cycling or walking. Headwinds or hills can make you slow down.
• Service frequency: If you’re taking public transport, The frequency of the service could impact your travel time. More frequent services will have shorter waiting durations.
## What Is The Average Time 15 Miles Of Walking Takes?
Walking is a fantastic method to exercise, discover your surroundings, and breathe fresh air. But, if you’re thinking about how long it takes to walk 15 miles, the answer is that it depends on many variables. Generally, 15 miles of walking could take between four and six hours, depending on your walking pace and the terrain.
### What are the factors that affect how long is required to complete 15 miles of walking?
A variety of factors can impact the amount of time required for fifteen miles of walking, for example:
• Your pace for walking: The speed you walk, the less time you’ll need to travel 15 miles. But, it’s important to determine a suitable pace that you can maintain for longer.
• The terrain: The flat terrain is more comfortable than walking downhill or uphill. If you’re walking over steep terrain, it’ll take longer to walk the same distance on flat terrain.
• It’s the weather: extreme weather, like excessive rain or high temperatures, could slow you to a crawl and cause it difficult to travel the distance.
• The level of fitness you have: If you’re in good health and used to walking for long distances, it’ll take you less time to walk 15 miles as opposed to if you’re a novice.
### How long is it to walk 15 miles at different speeds?
The time needed for walking 15 miles could depend on your walking pace. Here’s a breakdown on the duration at different speeds:
• 2 1 mph: It will take 7.5-8 hours to travel for 15 miles at a slow rate of 2mph.
• 3.0 mph: At a steady speed of 3 mph, it will take 5 to 6 hours to complete 15 miles.
• 4.75 5 mph: Walking at a fast rate of 4 mph is expected to take 3.75-4.5 hours to walk 15 miles.
• 5 miles per hour: Walking at a speed of 5 mph will take about 3-3.75 hours to travel 15 miles.
Be aware that these times are estimates and may differ according to the previously mentioned variables.
### Tips for walking 15 miles
15 miles is quite a distance, and preparing well to finish the walk safely and comfortably is crucial. Here are some suggestions to prepare:
1. Choose comfortable and appropriate clothes: Select breathable and comfortable clothing that lets you move around freely. Choose shoes that provide cushioning and support to minimize foot pain and blisters.
2. Drink plenty of water: Take fluids before, during, and following your walk to remain well-hydrated. Dehydration can cause headaches, fatigue as well as other health issues.
3. Get your energy levels up: Get a healthy and nutritious breakfast before your walk. Also, take snacks and water to fuel yourself on the journey.
4. Choose your route: Pick the route that best suits your level of fitness and knowledge. It’s recommended to start by taking shorter routes before gradually moving upwards to 15 miles.
5. Make breaks: Regularly take intervals to take a break, drink water and replenish your energy. You must pay attention to your body’s needs and take breaks whenever needed.
6. Take a walk with a companion: Walking with a partner or group can enhance enjoyment and provide support and motivation.
7. Be aware of the weather: Review the weather forecast before walking and dress accordingly. Avoid walking in extremely cold conditions.
8. Prepare for any emergency: Have a first-aid kit, a fully charged cell phone, and emergency contact numbers with you in an emergency.
## What Is The Distance Of 15 Miles In Miles?
If you’re familiar with measuring distances measured in km instead of miles, you’re probably thinking about how long 15 miles can be measured in kilometers. The answer is 15 miles corresponds to roughly 24.14 kilometers. This conversion is useful in a situation where you are traveling or communicating with others who utilize the metric system.
### Why is it crucial to know the conversions between kilometers and miles?
Understanding the conversion between miles and kilometers could be beneficial for a variety of reasons:
1. Traveling: A lot of countries utilize the metric system. Understanding the conversion system will help you plan your travels and comprehend the distances you need to travel.
2. Communications: If you interact with people using a metric measurement method, understanding the conversion will allow you to quickly comprehend distances and measurements.
3. Exercise: Certain fitness apps and gadgets use meters to gauge distance, and knowing how to convert it can aid in tracking your progress with greater precision.
4. Education: The study of various measurement systems is an important aspect of education. It will aid in understanding your surroundings more clearly.
### How do I convert miles into kilometers?
Converting kilometers to miles is an easy process. The following formula can be used: formula:
1 mile = 1.60934 kilometers
For the conversion of 15 miles into kilometers, multiply the number of miles multiplied by the rate:
15 miles x 1.60934 = 24.14 kilometers
So, 15 miles is equal to 24.14 kilometers.
### How do I convert miles into kilometers
To convert miles into kilometers, you can use the following formula:
1 kilometer = 0.621371 miles
To convert 24.14 miles to kilometers, multiply the number of km by the rate:
24.14 kilometers x 0.621371 = 15 miles
So, 24.14 kilometers is equivalent to 15 miles.
### How can you estimate distances between kilometers and miles
If you are looking to estimate distances between miles and kilometers, you can apply the following approximates:
• 1 mile is roughly equal to approximately 1.6 kilometers.
• One kilometer is roughly equal to the equivalent of 0.6 miles.
These estimates can be very useful when you have to do quick calculations or understand distances between various units.
# FAQ’s
### How long is 15 miles in hours driving?
26 minutes, or almost 0.43 hours.
### How far of a drive is 15 miles?
It would take 15 minutes to travel 15 miles. Another explanation is that 15 miles is 1/4 the distance a car travelling at 60 will cover in an hour, therefore it will take 15 minutes to travel 15 miles.
### How long is 15 miles in minutes in a car?
15 miles may be reached in around 15 minutes at a speed of 60 mph. It is the same as travelling a mile each minute!
### How many miles is 1 hour?
In light of this, you can anticipate travelling 40 miles in an hour of driving. If you spend the most of the trip travelling at speeds ranging from 55 to 70 miles per hour on lengthy stretches of roadway, you might cover more than 60 miles in an hour.
### How far is 8 miles in a car in minutes?
We now know that 8 miles at 30 mph requires 16 minutes to complete. | 3,807 | 17,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-23 | longest | en | 0.92939 |
https://math.stackexchange.com/questions/1687527/how-to-evaluate-the-integral-int-02-pi-mathrmd-theta-eia-cos-theta-th | 1,571,435,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684854.67/warc/CC-MAIN-20191018204336-20191018231836-00223.warc.gz | 574,991,365 | 32,638 | # How to evaluate the integral $\int_0^{2\pi}\mathrm{d}\theta e^{ia\cos(\theta-\theta_1)}\cos^2(\theta-\theta_2)$
I have an integral: $$\int_0^{2\pi}\mathrm{d}\theta e^{ia\cos(\theta-\theta_1)}\cos^2(\theta-\theta_2),$$ where $a, \theta_1$ and $\theta_2$ are reals. Any idea on how to evaluate this integral.
• This is surely interesting. For the moment, I tried to use Mathematica, and it says that the integral does exist and its result is $$\pi\left(J_0(a) - J_2(a)\cos(2\theta_1 - 2\theta_2)\right)$$ For $a\in\mathbb{R}$. $J_n(z)$ are the Bessel functions of the first kind --> mathworld.wolfram.com/BesselFunctionoftheFirstKind.html I will think something. Maybe you could start with some Taylor series for the cosine? – Von Neumann Mar 7 '16 at 21:51
Let $\theta \mapsto \theta + \theta_1$, so that the integral becomes
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} \, \left [ 1+\cos{(2 (\theta+\theta_1-\theta_2))} \right ]$$
The first term produces
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}}$$
Let $z=e^{i \theta}$; then the integral is
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z} e^{i (a/2) \left ( z+z^{-1} \right )}$$
To deal with the essential singularity in the exponential, we form the Laurent series of the integrand:
$$-\frac{i}{2} \sum_{n=0}^{\infty} \frac{i^n}{n!} \oint_{|z|=1} \frac{dz}{z} \left ( \frac{a}{2} \right )^{n} \left ( z+\frac1z \right )^n$$
The only nonzero terms will be the constant term of the expansion of the binomial terms, and those only appear in the even terms. Thus, the integral is
$$-\frac{i}{2} i 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!} \binom{2 n}{n} \left ( \frac{a}{2} \right )^{2 n} = \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left ( \frac{a}{2} \right )^{2 n} = \pi \, J_0(a)$$
The next term may be split up as $\cos{(2 (\theta+\theta_1-\theta_2))} = \cos{2 \theta} \cos{2 (\theta_1-\theta_2)} - \sin{2 \theta} \sin{2 (\theta_1-\theta_2)}$.
\begin{align} \frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} \cos{2 \theta} &= -\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} e^{i (a/2) \left ( z+z^{-1} \right )} \left (z^2+\frac1{z^2} \right )\\ &= -\frac{i}{4} \sum_{n=0}^{\infty} \frac{i^n}{n!} \oint_{|z|=1} \frac{dz}{z} \left ( \frac{a}{2} \right )^{n} \left ( z+\frac1z \right )^{n} \left (z^2+\frac1{z^2} \right ) \\ &= -\frac{i}{4} i 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!} \left [ \binom{2 n}{n-1} + \binom{2 n}{ n+1}\right ]\left ( \frac{a}{2} \right )^{2 n} \\ &= \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(n-1)! (n+1)!} \left ( \frac{a}{2} \right )^{2 n} \\ &= -\pi \left ( \frac{a}{2} \right )^{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{n! (n+2)!} \left ( \frac{a}{2} \right )^{2 n} \\ &= -\pi J_2(a)\end{align}
Note that the integral involving $\sin{2 \theta}$ has, upon expansion of the exponential, terms such as $\binom{2 n}{n-1} - \binom{2 n}{ n+1}$, which are all zero. Thus,
$$\int_0^{2 \pi} d\theta \, e^{i a \cos{(\theta-\theta_1)}} \, \cos^2{(\theta-\theta_2)} = \pi J_0(a) - \pi J_2(a) \cos{2 (\theta_1-\theta_2)}$$
• You know when I was just browsing around and saw this integral I thought to myself "I don't have time to do this one but, man this looks like a fun contour integral, I'll bet RG will have a jab at it". I'm impressed by this solution. – DaveNine Mar 8 '16 at 17:49
• @DaveNine: Thank you; yes, I find working with essential singularities kind of fun. – Ron Gordon Mar 8 '16 at 18:12 | 1,377 | 3,420 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.587068 |
http://www.dlxedu.com/detail/29/434669.html | 1,544,492,050,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823516.50/warc/CC-MAIN-20181210233803-20181211015303-00637.warc.gz | 381,590,174 | 11,710 | # C语言实现二叉树-02版
---恢复内容开始---
Problem
#include <stdio.h>#include <stdlib.h>typedef struct _node {int data;struct _node *link[2];}Node;typedef struct _tree{struct _node *root;}Tree;Tree * init_tree(){Tree *temp = (Tree*)malloc(sizeof(Tree));temp->root = NULL;return temp;}
int insert(Tree *tree, int data){ if(tree->root == NULL){ //TODO }else{ //TODO } return 1;}
int insert(Tree *tree, int data){ if(tree->root == NULL){ tree->root = make_node(data); }else{ //TODO } return 1;}
int insert(Tree *tree, int data){if(tree->root == NULL){tree->root = make_node(data);}else{Node * it = tree->root;for(;;){//TODO}} return 1;}
ok那么我们继续分析；
int insert(Tree *tree, int data){if(tree->root == NULL){tree->root = make_node(data);}else{Node * it = tree->root;int dir;for(;;){dir = it->data < data;if(it->data == data){return 0;}else if(it->link[dir] == NULL){break;}it = it->link[dir];}//TODO} return 1;}
6<8显然，成立；
dir = it->data < data;
if(it->data < data){ dir = 1;}else{ dir = 0;}
if(it->data == data){ return 0; }else if(it->link[dir] == NULL){ break; }
take break可能更加是人类需要的；
int insert(Tree *tree, int data){if(tree->root == NULL){tree->root = make_node(data);}else{Node * it = tree->root;int dir ;for(;;){dir = it->data < data;if(it->data == data){return 0;}else if(it->link[dir] == NULL){break;}it = it->link[dir];}//TODO}return 1;}
( ⊙ o ⊙ )！怎么还有一个TODO啊！你会惊讶到我们不是已经找到啦目标啦吗；
int insert(Tree *tree, int data){if(tree->root == NULL){tree->root = make_node(data);}else{Node * it = tree->root;int dir ;for(;;){dir = it->data < data;if(it->data == data){return 0;}else if(it->link[dir] == NULL){break;}it = it->link[dir];}it->link[dir] = make_node(data);}return 1;}
#include <stdio.h>#include <stdlib.h>typedef struct _node {int data;struct _node *link[2];}Node;typedef struct _tree{struct _node *root;}Tree;Tree * init_tree(){Tree *temp = (Tree*)malloc(sizeof(Tree));temp->root = NULL;return temp;}Node * make_node(int data){Node *temp = (Node*)malloc(sizeof(Node));temp->link[0] = temp->link[1] = NULL;temp->data = data;return temp;}int insert(Tree *tree, int data){if(tree->root == NULL){tree->root = make_node(data);}else{Node * it = tree->root;int dir ;for(;;){dir = it->data < data;if(it->data == data){return 0;}else if(it->link[dir] == NULL){break;} it = it->link[dir];} it->link[dir] = make_node(data);} return 1;}void print_inorder_recursive(Node *root){if(root){print_inorder_recursive(root->link[0]);printf("data:%d/n",root->data);print_inorder_recursive(root->link[1]);}return ;}void print_inorder(Tree *tree){print_inorder_recursive(tree->root);return ;}int main(void){Tree * tree = init_tree();insert(tree,6);insert(tree,7);insert(tree,5);insert(tree,8);print_inorder(tree);return 0;}
Problem
Solution
int remove(Tree *tree, int data){if(tree->root != NULL){//TODO}return 1;}
int remove(Tree *tree, int data){if(tree->root != NULL){Node *p = NULL;Node *succ;Node *it = tree->root;int dir;for(;;){//TODO}}return 1;}
p其实是parent（父亲）的缩写，succ是successor（继承者）的缩写；
int remove(Tree *tree, int data){if(tree->root != NULL){Node *p = NULL;Node *succ;Node *it = tree->root;int dir;for(;;){if( it == NULL){return 0;}else if(it->data == data){break;}dir = it->data < data;p = it;it = it->link[dir];}}return 1;}
p=it就是说p指向7；
int remove(Tree *tree, int data){if(tree->root != NULL){Node *p = NULL;Node *succ;Node *it = tree->root;int dir;for(;;){if( it == NULL){return 0;}else if(it->data == data){break;}dir = it->data < data;p = it;it = it->link[dir];}/***********************************************************************/if(it->link[0] != NULL && it->link[1] != NULL){//TODO}else{//TODO}}return 1;}
it->data = succ->data;
int remove(Tree *tree, int data){if(tree->root != NULL){Node *p = NULL;Node *succ;Node *it = tree->root;int dir;for(;;){if( it == NULL){return 0;}else if(it->data == data){break;}dir = it->data < data;p = it;it = it->link[dir];}/***********************************************************************/if(it->link[0] != NULL && it->link[1] != NULL){p = it;succ = it->link[1];while(succ->link[0] != NULL){p = succ;succ = succ->link[0];}it->data = succ->data;p->link[p->link[1] == succ] = succ->link[1];free(succ);/***********************************************************************/}else{//TODO}}return 1;}
Node *p = NULL;Node *succ;Node *it = tree->root;
it就是根的右子树； | 1,390 | 4,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-51 | longest | en | 0.207536 |
https://www.coursehero.com/file/6207484/sy13-oct17-07hc/ | 1,493,174,169,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121121.5/warc/CC-MAIN-20170423031201-00152-ip-10-145-167-34.ec2.internal.warc.gz | 878,014,536 | 294,167 | sy13_oct17_07hc
# sy13_oct17_07hc - Physics 207, Lecture 13, Oct. 15 Physics...
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Physics 207: Lecture 12, Pg 1 Physics 207, Physics 207, Lecture 13, Oct. 15 Lecture 13, Oct. 15 Agenda: Finish Chapter 10, start Chapter 11 Agenda: Finish Chapter 10, start Chapter 11 Assignment: Assignment: HW5 due tonight HW5 due tonight HW6 available today HW6 available today Monday, finish reading chapter 11 Monday, finish reading chapter 11 Chapter 10: Energy Chapter 10: Energy Potential Energy (gravity, springs) Kinetic energy Mechanical Energy Conservation of Energy Start Chapter 11, Work
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Physics 207: Lecture 12, Pg 2 Chapter 10: Energy Chapter 10: Energy Rearranging Newton’s Laws gives (Fd vs . ½ mv 2 relationship) -2mg (y f y i ) = m (v yf 2 - v yi 2 ) or ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f and adding ½ m v xi 2 + ½ m v zi 2 and ½ m v xf 2 + ½ m v zf 2 ½ m v i 2 + mgy i = ½ m v f 2 + mgy f where v i 2 = v xi 2 +v yi 2 + v zi 2
Physics 207: Lecture 12, Pg 3 Energy Energy If only “conservative” forces are present, the total energy If only “conservative” forces are present, the total energy ( ( sum of potential, U, and kinetic energies, K sum of potential, U, and kinetic energies, K ) of a system ) of a system is is conserved conserved . K ≡ ½ mv 2 U ≡ mgy K and U may change, but E = K + U mech remains constant. E mech = K + U = constant constant E mech is called “mechanical energy” K i + U i = K f + U f
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Physics 207: Lecture 12, Pg 4 Another example of a conservative system: Another example of a conservative system: The simple pendulum. The simple pendulum. Suppose we release a mass m from rest a distance h 1 above its lowest possible point. What is the maximum speed of the mass and where does this happen ? To what height h 2 does it rise on the other side ? v h 1 h 2 m
Physics 207: Lecture 12, Pg 5 Example: The simple pendulum. Example: The simple pendulum. y y=0 y= h 1 What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum.
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Physics 207: Lecture 12, Pg 6 Example: The simple pendulum. Example: The simple pendulum. v h 1 y y= h 1 y=0 What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing
Example: The simple pendulum. Example: The simple pendulum. y y= h 1 =h 2 y=0 To what height h2 does it rise on the other side? E = K + U = constant
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## This note was uploaded on 04/12/2011 for the course PHYS 204 taught by Professor Winoker during the Spring '11 term at University of Wisconsin - Fond du Lac.
### Page1 / 30
sy13_oct17_07hc - Physics 207, Lecture 13, Oct. 15 Physics...
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Ask a homework question - tutors are online | 967 | 3,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-17 | longest | en | 0.784287 |
https://gamedev.stackexchange.com/questions/176000/follow-3d-path-around-sphere | 1,631,980,343,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00710.warc.gz | 322,861,514 | 38,303 | I have currently have a path that wraps around a Sphere from one side to another - this is basically a 3D highway in the sky type object that is a textured road. I have an Aircraft model that I can get to follow this path, however I'm having trouble getting it to rotate correctly so that it is parallel to the surface of the globe
Using the following code:
this.transform.forward = Vector3.RotateTowards(this.transform.forward, this.transform.position - TargetWaypointPos, Airspeed * Time.deltaTime, 0.0f);
this.transform.position = Vector3.MoveTowards(this.transform.position, TargetWaypointPos, Airspeed * Time.deltaTime);
This gets the plane to correctly face the next waypoint and follow the path, however I cannot get it to rotate correctly in the Z-space to be parallel to the path/globe. I can compute the correct angle by:
Vector3 yAxis = this.transform.position;
yAxis.Normalize();
Vector3 zAxis = nextPoint - this.transform.position;
zAxis.Normalize();
xAxis = Vector3.Cross(yAxis, zAxis);
xAxis.Normalize();
float angleZ = Vector3.Angle(Vector3.right, xAxis);
However attempting to apply this to my model in any way appears to ruin the Vector3.RotateTowards logic. I must be missing something simple to combine these two.
Any ideas? I'm new to Unity, still figuring out all the available options.
This is easier than you think!
// Get the vector toward the next waypoint from here.
Vector3 forward = TargetWaypointPos - transform.position;
// Get the local "up" vector pointing out of the planet to here.
Vector3 up = transform.position - SphereCenterPos;
// Form an orientation that points the z+ axis along "forward",
// and the y+ axis as close as possible to "up".
Quaternion targetRotation = Quaternion.LookRotation(forward, up);
// Rotate toward this target at a controlled rate.
transform.rotation = Quaternion.RotateTowards(
transform.rotation,
targetRotation,
Time.deltaTime * DegreesPerSecond
); | 413 | 1,932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | longest | en | 0.843554 |
https://cboard.cprogramming.com/cplusplus-programming/49480-calendar-program-what-am-i-doing-wrong-printable-thread.html | 1,516,668,723,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891546.92/warc/CC-MAIN-20180122232843-20180123012843-00508.warc.gz | 649,044,071 | 4,011 | # Calendar Program. What am I doing wrong?
• 01-17-2004
Sektor
Calendar Program. What am I doing wrong?
Hey everyone. I'm typing up a program which prints out a calendar for any month and year (starting with 1 A D) which the user inputs. I found a general algorithm for calculating the first day of the month but when i cout it in my if statements, something's wrong. It prints out | | 1 no matter what B is equal to. The algorithm is
1) Subtract 1 from the year and store the result in a variable A.
2) Multiply A by 365
3) Find the quotient when you divide the (year - 1) by 4 (ignore remaindr) and add this to A
4) Find the quotient when you divide the (year - 1) by 100 (ignore the remainder) and subtract this from A.
5) Find the quotient when you divide the (year - 1) by 400 (ignore the remainder) and add this to A.
6) Add the days in the year prior to this month (dont worry about leap years here) to A.
7) Add 1 to A (for first day of month)
8) If the year is a leap year and the month is after February add 1 to A. Note: A year is a leap year if it is divisible by 4 AND if it is not a centennial OR it is a centennial divisible by 400.
9) Find the remainder when dividing A by 7
Here's my code so far:
Code:
```// Calendar Project // Ez Nadar #include <iostream> using std::cin; using std::cout; using std::endl; int day(int year, int month); int daysbefore(int month); int main() { int month = 0, year = 0, d = 0, b = 0; cout << "Enter the year of the calendar" << endl; cin >> year; cout << "Enter the month of the calendar" << endl; cin >> month; d = daysbefore(month); b = day(year, month); if ( month == 12) cout << " " << "Calendar of December " << year << endl; if ( month == 11) cout << " " << "Calendar of November " << year << endl; if ( month == 10) cout << " " << "Calendar of October " << year << endl; if ( month == 9) cout << " " << "Calendar of September " << year << endl; if ( month == 8) cout << " " << "Calendar of August " << year << endl; if ( month == 7) cout << " " << "Calendar of July " << year << endl; if ( month == 6) cout << " " << "Calendar of June " << year << endl; if ( month == 5) cout << " " << "Calendar of May " << year << endl; if ( month == 4) cout << " " << "Calendar of April " << year << endl; if ( month == 3) cout << " " << "Calendar of March " << year << endl; if ( month == 2) cout << " " << "Calendar of February " << year << endl; if ( month == 1) cout << " " << "Calendar of January " << year << endl; cout << " ----------------------------------------- " << endl; cout << " | Sun | Mon | Tue | Wed | Thu | Fri | Sat |" << endl; cout << " ----------------------------------------- " << endl; if ( b = 0) cout << " | 1|"; else if ( b = 1) cout << " | | 1|"; else if ( b = 2) cout << " | | | 1|"; else if ( b = 3) cout << " | | | | 1|"; else if ( b = 4) cout << " | | | | | 1|"; else if ( b = 5) cout << " | | | | | | 1|"; else if ( b = 6) cout << " | | | | | | | 1|"; return 0; } int daysbefore(int month) // calculates days in the year before this month { int d = 0; if ( month == 12) d = 334; if ( month == 11) d = 304; if ( month == 10) d = 273; if ( month == 9) d = 243; if ( month == 8) d = 212; if ( month == 7) d = 181; if ( month == 6) d = 151; if ( month == 5) d = 120; if ( month == 4) d = 90; if ( month == 3) d = 59; if ( month == 2) d = 31; if ( month == 1) d = 0; return d; } int day(int year, int month) // determines first day of month { int a = 0, b = 0; a = year - 1; a = a * 365; a = a + ((year - 1) / 4); a = a - ((year - 1) / 100); a = a + ((year - 1) / 400); a = a + daysbefore(month); a = a + 1; if ( ( year % 100 ) != 0 ) if ( (year % 4) != 0 ) { a = a + 1; } b = (a%7); return b; }```
I know the spacing is off in my cout statements
• 01-17-2004
joshdick
For those if statements involving b, you're using the assignment operator = instead of the comparison operator ==. | 1,428 | 5,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-05 | latest | en | 0.725492 |
https://www.esaral.com/q/in-the-given-figure-the-sides-cb-and-ba-of-abc-have-been-produced-to-d-and-e-95546/ | 1,659,963,614,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00748.warc.gz | 717,135,928 | 23,749 | In the given figure, the sides CB and BA of ∆ABC have been produced to D and E,
Question:
In the given figure, the sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 65°
(b) 45°
(c) 55°
(d) 35°
Solution:
(a) 65°
We have :
$\angle A B D+\angle A B C=180^{\circ} \quad[\because C B D$ is a straight line $]$
$\Rightarrow 100^{\circ}+\angle A B C=180^{\circ}$
$\Rightarrow \angle A B C=70^{\circ}$
Side AB of triangle ABC is produced to E.
$\therefore \angle C A E=\angle A B C+\angle A C B$
$\Rightarrow 135^{\circ}=70^{\circ}+\angle A C B$
$\Rightarrow \angle A C B=65^{\circ}$ | 237 | 662 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-33 | latest | en | 0.754427 |
https://www.teacherspayteachers.com/Product/Golden-Subtraction-Subtraction-Practice-4367125 | 1,558,350,963,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255943.0/warc/CC-MAIN-20190520101929-20190520123929-00285.warc.gz | 968,473,441 | 18,798 | # Golden Subtraction - Subtraction Practice
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2. This bundle contains 7 Literacy centers and 7 Math Centers. Each center focuses on multiple common core standards. Please download the preview to view the activities included. Each activity contains:· “I can” kid friendly statement posters for students and/or center· Both color and black & whit
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Product Description
This Golden Subtraction center focuses on solving a given subtraction problem and matching the correct answer to each problem. Please download the preview to see all that is included.
This activity contains:
· “I can” kid friendly statement poster for students and/or center
· Both color and black & white versions
· Recording sheets for student accountability
Using this Resource
· Print and laminate for long term use
· Small group work
· Centers for station time
· Good Morning Bins
This activity is aligned to these Common Core Standards:
· K.OA.A.1:Represent addition and subtraction with objects, fingers,mental images, drawings1, sounds, acting out situations, verbal explanations, expressions, or equations.
· K.OA.A.2:solve addition and subtraction word problems, and add and subtract within 10, by using objects or drawings to represent the problem.
· K.OA.A.3: Decompose numbers less than or equal to 10 into pairs in more than one way, e.g. by using objects or drawings, and record each decomposition by drawing or equation.
· K.OA.A.5:Fluently add and subtract within 5.
Questions before you purchase? Ask away in the Q&A! If you notice any errors, please email me before leaving negative feedback. My email address is located on the credits page. Thank you so much!
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\$2.00 | 644 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-22 | latest | en | 0.855635 |
https://goodriddlesnow.com/riddles/view/3932 | 1,579,557,492,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00455.warc.gz | 462,897,378 | 10,027 | # loss
Question: someone came into my shop and stole 100 dollars from the register without my knowledge. the person came back 5 minutes later with the same 100 dollars and bought 70 dollars in items and i gave him back 30 dollars change. How much did i lose?
Riddle Discussion
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Question: In a house of Kardak's family each brother have a sisters as double as their brothers and each sister have brothers as same as their sister. How many brothers and sisters in Kardak's family ? | 380 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-05 | longest | en | 0.969827 |
http://rabbitholereport.blogspot.com/2012/02/should-you-buy-powerball-ticket.html | 1,527,162,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866276.61/warc/CC-MAIN-20180524112244-20180524132244-00155.warc.gz | 248,415,359 | 12,697 | ## Selasa, 21 Februari 2012
### Should You Buy a Powerball Ticket?
Wouldn't it be nice to win the lottery? And since the multi-state Powerball lottery game has just changed to offer higher jackpots, including one that recently went over \$330 million, could it be a good idea to buy a Powerball ticket today?
Basic Lottery Game Data
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The Odds of Winning the Jackpot (1 in ...)
The Magic Jackpot (The Number That Makes Playing the Lottery Worth the Cost of a Ticket)
The answer is "it depends". Specifically, it depends upon the following factors:
• What are the odds of winning the jackpot?
• How much does a ticket cost?
• If you do win, how much of your prize will be taxed away from you?
Together, these three things, combined with the kind of math that an economist might do to calculate the environmental costs of a spill from an oil pipeline, will tell us how big the lottery jackpot needs to be to be worth the cost of the ticket to play!
And that's the math our tool today is here to do for you! Just enter the indicated data for the lottery game of your choice, and we'll calculate just how be the jackpot has to get to be worth the amount of money you might be willing to plunk down on a ticket.
Our default data is that for the new Powerball game, which was revised back on 15 January 2012 to double the price of a single ticket from \$1 to \$2, and which was also tweaked to increase the odds of winning.
Doing the math, we find that in order to fully justify the cost of a \$2 lottery ticket, and to also compensate for the negative effect of having the jackpot get taxed at just the current top federal income tax rate of 35%, the Powerball lottery would have to exceed \$539,149,262.
Will the Powerball jackpot ever exceed the more than half billion dollars it at least needs to be to be worth the price of its \$2 ticket to play? It's possible, but the odds are such that the average jackpot paid out will be around the \$175 million level, which means there's probably something else a lot better you can do with your \$2 than play the new Powerball game! | 590 | 2,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-22 | latest | en | 0.957807 |
https://www.physicsforums.com/threads/vlad-arnolds-book-on-mechanics.58199/ | 1,513,369,447,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00626.warc.gz | 822,547,969 | 14,410 | # Vlad Arnold's book on Mechanics
1. Dec 30, 2004
### homology
Hey Folks,
Anyone here ever deal with V.Arnold's Book "Mathematical Methods of Classical Mechanics?" A friend of mine and I have started working through it and we have a question. First I'll state it then comment:
Problem: A mechanical system consists to two points. At the initial moment their velocities (in some inertial coordinate system) are equal to zero. Show that the points will stay on the line which connected them at the initial moment.
Now sure this is pretty obvious. But all we have to work with are
1) Space is 3D and time is 1D
2) Galileo's prin. of relativity (inertial frames are good and any frame moving uniformly with respect to an inertial frame is inertial)
3) Newton's prin. of determinacy (initial positions and velocities uniquely determine all the motions of a system.
Then Arnold develops the Galilean structure, the Galilean transformations and draws conclusions about Newton's equations as a result. If you're interested in helping out, I'll post some more. Again, our intuition says "Duh" (at least in terms of classical mech) but we don't have much to work with.
Kevin
2. Dec 31, 2004
### Coelum
I'm not an expert but I believe we have something more than that. We have the hypothesys that the space is invariant under translation and rotation. Can the space be invariant under rotation if we break the initial rotational symmetry (e.g., by moving orthogonally to the joining line)? | 348 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-51 | longest | en | 0.944554 |
https://www.coursehero.com/file/245335/solution-pdf4/ | 1,495,807,965,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608665.33/warc/CC-MAIN-20170526124958-20170526144958-00147.warc.gz | 1,077,423,021 | 24,639 | solution_pdf4
# solution_pdf4 - chaney(glc568 – Laws of Motion – murthy...
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Unformatted text preview: chaney (glc568) – Laws of Motion – murthy – (21118) 1 This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Within a book on a table there are billions of forces pushing and pulling on all of the molecules. Why is it that these forces never by chance add up to a net force in one direction, causing the book to accelerate “spontaneously” across the table? 1. These forces between molecules are much smaller than the friction between the book and the table. 2. These forces are counteracted by grav- ity. 3. The billions of force pairs are internal to the book and exert no net force on the book. correct 4. The forces cause the book to move across the table spontaneously all the time, but the movement is too weak to observe. Explanation: The billions of force pairs are internal to the book, and exert no net force on the book. An external net force is necessary to accelerate the book. 002 10.0 points A heavily loaded freight train moves with constant velocity. What is the relationship between the net force on the first car ( F 1 ) and the net force on the last car ( F 2 )? 1. F 1 > F 2 2. F 1 < F 2 3. Unable to determine. 4. F 1 = F 2 correct Explanation: The net force on each car is zero because the train moves with constant velocity (no acceleration). 003 10.0 points An elevator is being lifted up an elevator shaft at a constant speed by a steel cable as shown in the figure below. All frictional effects are negligible. steel cable Elevator going up at constant speed In this situation, forces on the elevator are such that 1. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air. 2. the upward force by the cable is greater than the downward force of gravity. 3. None of these. (The elevator goes up because the cable is being shortened, not be- cause an upward force is exerted on the eleva- tor by the cable.) 4. the upward force by the cable is equal to the downward force of gravity. correct 5. the upward force by the cable is smaller than the downward force of gravity. Explanation: Since the elevator is being lifted at a con- stant speed, the net force on it is zero, there- fore, the upward force by the cable is equal to the downward force of gravity. chaney (glc568) – Laws of Motion – murthy – (21118) 2 004 (part 1 of 2) 10.0 points Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 41 . 6 N to the north. Jenny’s foot exerts a force of 88 . 6 N to the east. a) What is the magnitude of the resultant force on the ball? Correct answer: 97 . 8801 N....
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## This note was uploaded on 10/22/2008 for the course PHY 260 taught by Professor Murthy during the Spring '08 term at Kentucky.
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Ask a homework question - tutors are online | 840 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-22 | longest | en | 0.914728 |
http://spmmathematics.blog.onlinetuition.com.my/2016/10/spm-2016-mathematics-forecast-paper_71.html | 1,569,065,384,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574409.16/warc/CC-MAIN-20190921104758-20190921130758-00366.warc.gz | 177,909,672 | 13,142 | SPM Mathematics (Model Test Paper)
SPM Mathematics (Model Test Paper)
Section A
(52 marks)
Answer all questions in this section.
1. The Venn diagram in the answer space shows sets E, and G such that the universal set, $\xi =E\cup F\cup G.$
On the diagrams in the answer space, shade
$\begin{array}{l}\left(a\right)\text{}E\text{}\text{'}\cap G\text{'}\\ \left(b\right)\text{}\left(E\cap F\right)\text{'}\text{}\cup G\text{'}\end{array}$
[3 marks]
Answer
2. Calculate the value of x and of that satisfy the following simultaneous linear equations:
$\begin{array}{l}x+6y=12\\ \frac{2}{3}x+2y=6\end{array}$
[4 marks]
3.
Diagram 1
Diagram 1 shows a square ABCD and a right-angled triangle PQR with equal area.
Based on the information, find the value of x.
Hence, find the perimeter, in cm, of the square ABCD. [4 marks]
4 (a) For each of the following statements, determine whether the statement is true or false.
(i) 35 is multiple of 3 and 5
(ii) 7 is a factor of 42 or 16 is a multiple of 6.
(b) Write down two implications based on the following sentence:
p3 = –8 if and only if p = –2
(c) Make a general conclusion by induction for a list of numbers 2, 11, 26, 47, ……. which follows the following pattern.
2 = 3(1)2– 1
11 = 3(2)2– 1
26 = 3(3)2– 1
47 = 3(4)2– 1
……………
[6 marks]
5. Diagram 2 in the answer space shows a right prism. Trapezium ABGF is the uniform cross section of the prism.
(a) On Diagram 2, mark the angle between the plane ADE and the plane ADCB.
(b) Calculate the angle between the plane ADE and the plane ADCB.
[3 marks]
Answer:
(a)
Diagram 2
6. Diagram 3 shows a right cylinder with a diameter of ( y + 4 ) cm.
Given the volume of the cylinder is 269.5 cm3 and by using find the value of it’s radius.
[4 marks]
7. Diagram 4 shows a straight line ST and a straight line PQ drawn on a Cartesian plane. ST is parallel to PQ. Given that equation of the straight line ST is 2y = 8x + 3.
Diagram 4
Find,
(a) the equation of the straight line PQ,
(b) the x-intercept of the straight line PQ.
[ 5 marks ]
8. Diagram 5, shows a quadrant KLM with centre M and sector JMN with centre J.
Diagram 5
Using $\pi =\frac{22}{7}$ , calculate
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
[6 marks]
9. Diagram 6 shows the speed-time graph for the movement of two particles, J and K, for a period of t s. The graph ABCD represents the movement of J and the graph AE represents the movement of K. Both particles start at the same point and move along the same route.
Diagram 6
(a) State the uniform speed, in ms-1, of particle J.
(b) Calculate the rate of change of speed, in ms-2, of particle J for the first 13 s.
(c) At t s, the difference between the distance travelled by J and K is 169 m. Calculate the value of t.
[ 6 marks ]
10. Diagram 7.1 shows three cards labelled with letters in bag A and three numbered cards in bag B.
Diagram 7.1
A card is picked at random from bag A and then a card is picked at random from bag B.
(a) Diagram 7.2 in the answer space shows the incomplete possible outcomes of the event. Complete the possible outcomes in Diagram 7.2.
(b) Using the complete possible outcomes in 10(a), find the probability that
(i) a card labelled Q and the card with odd number are picked,
(ii) a card labelled P or the card with a number which is multiple of 3 are picked.
[5 marks]
Answer:
(a)
Diagram 7.2
11. It is given that matrix M is a 2 × 2 matrix such that
(a) Find matrix M.
(b) Write the following simultaneous linear equations as matrix equations:
–2x + y = 10
x + 3y= 9
Hence, using matrix method, calculate the value of x and of y. [6 marks] | 1,096 | 3,676 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-39 | latest | en | 0.754354 |
https://homework.cpm.org/category/CC/textbook/cca2/chapter/11/lesson/11.2.3/problem/11-64 | 1,718,552,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00257.warc.gz | 269,851,390 | 16,025 | ### Home > CCA2 > Chapter 11 > Lesson 11.2.3 > Problem11-64
11-64.
A copy machine company advertises that its copiers will make at least $25{,}000$ copies before requiring maintenance. A consumer research group tested the claim by collecting data from users of the particular copy machine in $30$ different regions of the country. The means for each of the $30$ regions are listed in the table below.
$24928$ $24574$ $24652$ $24758$ $24691$ $24893$ $25024$ $24767$ $24791$ $24609$ $25249$ $24914$ $24895$ $24656$ $24883$ $24551$ $24928$ $25025$ $24798$ $25041$ $24782$ $25020$ $24618$ $24904$ $24764$ $24705$ $24889$ $24656$ $24600$ $24735$ checksum $744300$
Find the upper and lower $5\%$ bounds of the sample-to-sample variability and predict the number of copies that can be made before a machine requires maintenance. Do you think the consumer research group will support the company’s claim?
Read the Math Notes box in Lesson 11.2.2. | 285 | 944 | {"found_math": true, "script_math_tex": 35, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.826985 |
https://www.coursehero.com/file/5669001/Notes-11-finance-5108-08/ | 1,521,540,985,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647327.52/warc/CC-MAIN-20180320091830-20180320111830-00573.warc.gz | 779,789,694 | 83,286 | {[ promptMessage ]}
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Notes_11_finance_5108_08
# Notes_11_finance_5108_08 - Notes 11 Interest Rate Swaps...
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Notes 11 Interest Rate Swaps Learning Objectives 1. Understand the basics of plain Vanilla interest rate swaps. You should be able to set up the swap. 2. Understand the ability of swaps to transform the liabilities of a firm 3. Be able to introduce a swap desk to the equation 4. Understand the comparative advantage argument for swaps and its criticisms 5. Understand the zero LIBOR/SWAP rate using bootstrapping 6. Obtain knowledge on the valuation of interest rate swaps and methodologies 7. Understand currency Swap and their structure 8. Understand the valuation of Currency swaps 9. Obtain knowledge of the credit risk in the swap market 10 Variation on the plain vanilla swaps Interest rate Swaps Company A Company B Plain Vanilla Interest Rate Swap Floating Fixed Company A LIBOR + 1 10% Company B LIBOR + 2 12% Pays 10% Pays 11.5% Pays LIBOR + 2 Assets Variable Rate Asset Fixed Rate Pays 10% Gets 11.5% Pays LIBOR + 2 Net LIBOR + .5 Pays LIBOR + 2 Gets LIBOR +2 Pays 11.5% Net 11..5% Amount Swap is based on \$100 million is the Notional Principal A and B are counter parties Term is 3 years Called the Tenor Assuming Settlement is 6 months and Libor is 10% A pays B (10 + 2)/(2)(100) X \$100,000,000 = \$6,000,000 B pays to A 11.5/(2)(100) X 100,000,000 = \$5,750,000 Netting A pays to be \$750,000
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The following table looks at what happens over the next 3 years as far as cash flows go In millions Date Libor A to B B to A Net Payment 1 10 6 5.75 0.25 2 12 7 5.75 1.25 3 8 5 5.75 -0.75 4 9 5.5 5.75 -0.25 5 13 7.5 5.75 1.75 6 14 8 5.75 2.25 The 100 million is called the notional principal and is the amount the swap is based on. As can be seen from the above figure the purpose of the swap is to transform the variable rate liability of B into a fixed rate liability and to transform the variable rate liability to a fixed rate liability. Plain vanilla swaps could be used to
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Ask a homework question - tutors are online | 704 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.764705 |
https://tssoj.ca/problem/tccc24feb4 | 1,713,462,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00360.warc.gz | 525,164,769 | 5,577 | ## TCCC '24 Feb P4 - Good Gifts
View as PDF
Points: 5 (partial)
Time limit: 0.5s
Memory limit: 128M
Author:
Problem type
Thornhill Computer Club 2024 - February Mock CCC - Problem 4
During a party, people exchange gifts. Every person has written their favourite number on the gifts that they will give. If the sum of the numbers of the gifts exchanged between two people is prime, their exchange is considered a good exchange. Every person can give infinitely many gifts, but any pair of two people may only exchange their gifts once.
Given the number of people at the party and their favourite numbers, determine how many good exchanges are possible.
#### Input Specification
The first line will contain an integer .
The next line will consist of space-separated integers, representing the favourite numbers of each person .
#### Output Specification
On a single line, output the maximum possible amount of good exchanges that can happen at this party.
#### Sample Input 1
4
1 2 3 4
#### Sample Output 1
4
#### Explanation for Sample Output 1
The possible good exchanges are between the pairs (1,2), (1,4), (2,3), and (3,4). The sums of these pairs are 3, 5, 5, and 7, respectively, which are all prime. Hence, there are 4 possible good exchanges.
#### Sample Input 2
6
1 1 4 5 7 9
#### Sample Output 2
5
#### Explanation for Sample Output 2
The possible good exchanges are between the pairs (1,1), (1,4), (1,4), (4,7), and (4,9). The sums of these pairs are 1, 5, 5, 11, and 13, respectively, which are all prime. Hence, there are 5 possible good exchanges.
Note that (1,4) occurs twice, as there are two people whose favourite number is 1. | 433 | 1,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.918221 |
http://www.go4expert.com/forums/reduced-randon-fraction-class-post64417/ | 1,469,549,269,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00326-ip-10-185-27-174.ec2.internal.warc.gz | 472,660,316 | 3,706 | Code:
```#include <iostream>
#include <ctime>
using namespace std;
class Fraction
{
private:
int num; // Numerator
int den; // Denominator
public:
void setFraction (int); //sets fraction
int Reduce(); //will reduce the fraction to lowest terms
void print(); //will print the fraction in the form: numerator / denominator
int rand(); //will create random fractions
void multiply(); //will multiply the fractions
Fraction(); // constructor
Fraction(int,int); // 2 Argument Constructor
};
Fraction::Fraction()
{
num=0;
den=1;
}
Fraction::Fraction(int n,int d)
{
num=n;
den=(d==0)? 1 :d;
}
int rand ()
{
int Num[3];
int Den[3];
//int Numerator;
//int Denominator;
srand (time(0));
{
Num[0] = rand() % 9 + 1; // load the first random number into Num[0]
Num[1] = rand() % 9 + 1; // load the second into Num[1]
Den[0] = rand() % 9 + 1; // load the first random number into Den[0]
Den[1] = rand() % 9 + 1; // load the second into Den[1]
cout << "\n\n""What is " << Num[0] << "/" << Den[0]<< "*" << Num[1] << "/" << Den[1] <<"?" << endl;
cout << "Enter Numerator / Denominator" << endl;
if ((answer == Num[0] * Num[1]) && (Den[0] * Den[1])); // check against the numbers array
{
cout << "\n""Correct!"<< endl; //Outcome for correct answer
}
//else //statement for while loop
// GIVE THE CORRECT ANSWER AN MOVE ON
//cout << "\nWrong! The correct answer was " << numbers[2] << endl; //Outcome for incorrect answer
}
return 0;
}
void multiply(int &num, int &den, int &num2, int &den2)
{
int calcnum;
int calcden;
calcnum = num * num2;
calcden = den * den2;
}
void Reduce(int &num, int &den, int &num2, int &den2)
{
int a, b, c, d, i, j = 0;
a = den;
b = num;
c = den2;
d = num2;
for (i = a * b; i > 1; i--)
{
if ((a % i == 0) && (b % i == 0))
{
a /= i;
b /= i;
}
}
for (j = 50; j > 1; j--)
{
if ((c % j == 0) && (d % j == 0))
{
c /= j;
d /= j;
}
}
den = a;
num = b;
den2 = c;
num2 = d;
}
int Reduce(int a, int b)
{
int calcden = a;
int calcnum = b;
if(b == 0)
{
return a;
}
else
{
return Reduce(b, a % b);
}
}
void print(int &num, int &den)
{
cout << "The reduced and added fraction is " << num << "/" << den << endl;
}
int main()
{
int num, den, num2, den2;
int calcnum = 0;
int calcden = 0;
int a = 0;
int b = 0;
int m = 0;
int n = 0;
//int i;
//int rand ();
cout << "What is 3/4 * 3/5?" << endl;
//cout << "What is" << num[i]<< "/" << den[i]<< "*" << num2[i]<<"/"<<den2[i]<<"?"<<endl;
cout << "enter numerator / denominator" << " " << "Your answer must be in reduced form" << endl;
cin >> num >> den;
multiply(num, den, num2, den2);
Reduce(a, b);
print(num, den);
cout << endl;
return(0);
}```
Last edited by shabbir; 18Feb2010 at 09:02.. Reason: Code blocks | 916 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-30 | latest | en | 0.419184 |
https://www.coursehero.com/file/6590906/21-The-Derivative-and-the-Tangent-Line-Problem/ | 1,519,368,389,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814493.90/warc/CC-MAIN-20180223055326-20180223075326-00410.warc.gz | 830,857,824 | 303,604 | {[ promptMessage ]}
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2.1 The Derivative and the Tangent Line Problem
# 2.1 The Derivative and the Tangent Line Problem - The...
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The Derivative The Derivative and the Tangent and the Tangent Line Problem Line Problem Section 2-1
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Slope of Secant Line Slope of Secant Line x x ) ( ) ( c f x c f - + (c, f(c)) (c + ,f(c + ) x x x c f x c f m - + = ) ( ) ( sec
Definition of Tangent Line with Definition of Tangent Line with Slope Slope m m If f is defined on an open interval containing c, and if the limit exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)). m x c f x c f x y x x = - + = ) ( ) ( lim lim 0 0
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Vertical Tangent Lines Vertical Tangent Lines The defn. of a tangent line to a curve does not cover the possibility of a vertical tangent line. If f is continuous at c and The vertical line x = c, passing through (c, f(c)) is a vertical tangent line to the graph of f. - = - + or x c f x c f x ) ( ) ( lim 0
Definition of the Derivative of Definition of the Derivative of a Function a Function
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Unformatted text preview: The derivative of f at x is given by provided the limit exists. x x f x x f x f x ∆-∆ + = → ∆ ) ( ) ( lim ) ( ' Notation for Derivatives Notation for Derivatives ) ( ' x f dx dy ' y [ ] ) ( x f dx d [ ] y D x Alternative form of the Alternative form of the Derivative Derivative c x c f x f c f c x--= → ) ( ) ( lim ) ( ' Thm. 2.1 Differentiability Thm. 2.1 Differentiability Implies Continuity Implies Continuity If f is differentiable at x = c, then f is continuous at x = c. It is possible for a function to be continuous at x = c and not be differentiable at x = c. Thus, continuity does not imply differentiability. Joke Time Joke Time What would America be called if we all drove pink cars? A pink carnation! Why should you never play cards in the jungle? It’s full of cheetahs!...
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Ask a homework question - tutors are online | 747 | 2,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-09 | latest | en | 0.853447 |
https://teamtreehouse.com/community/-im-not-sure-where-i-am-going-wrong | 1,718,675,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00420.warc.gz | 480,252,878 | 34,903 | ## Welcome to the Treehouse Community
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# Im not sure where I am going wrong.
I am following the tutorial on converting measurements but I am not sure exactly where I am going wrong. I have done the conversation by dividing inches by 12 but my areaInFeet is still not calculating the right amount
area.swift
```// the height and width of a wall
let height = 120.0 // in inches
let width = 144.0 // in inches
let area = height * width
// 1 sq. foot = 1 sq. inch / 12
let areaInFeet = area / 12
```
Hey!
I noticed you are dividing your final area by a Int type. This is probably converting your areaInFeet to an Int, which might make the system think that it's the wrong answer when it's expecting a Double. Be careful when you are using different number types!
Rio,
You are doing
```height * width / 12
```
What you need is
```(height / 12) * (width / 12)
```
Florin
There is no need to create a whole other constant (you named it "area") to store the square feet in inches.
```// the height and width of a wall
let height = 120.0 // in inches
let width = 144.0 // in inches
let areaInFeet = (height / 12) * (width / 12)
``` | 417 | 1,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.9178 |
http://www.ashishmathur.com/tag/rank/ | 1,606,688,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00151.warc.gz | 102,828,482 | 11,455 | Tags: RANK
Assume a five column dataset - ID, Age, Gender, Time and Class. For chosen ID's, the objective is to:
1. Assign a Rank (in ascending order of time i.e. lowest time will be rank 1 and so on) to each ID
2. Determine the overall place of each ID - Count of unique time entries lesser than equal to the chosen ID' time entry
These can be computed with the VLOOKUP(), RANK(), FREQUENCY(), INDIRECT() functions and array formulas. You may refer to range H3:K8 of the Sample worksheet. So far so good.
What adds to the problem is to meet the objectives outlined above after satisfying additional conditions. For e.g., one may want to give conditions such as Age between 20 and 35 and colours as Orange and Yellow. Carrying out computations for ranking and Overall place after satisfying these conditions will make the formulas fairly complex.
I have been able to solve this problem with the help of the PowerPivot. You may download my solution workbook from this link.
Assume a two column database with names in column A and numbers in column B. Names in column A may be repeated. If a user types a certain name in a cell, a formula should sum the highest three values from column B for that name.
Depending upon the version of MS Excel which you are using, there could be two ways to solve this problem
Solution for MS Excel 2010 and higher versions
If you are using the PowerPivot add-in, then a simple DAX formula can solve this problem.
Solution for all versions of MS Excel
While this solution works for all versions of MS Excel, it uses an array formula (Ctrl+Shift+Enter). Array formulas, if used extensively in the workbook, adversely effect the system's performance.
You may refer to my solution in this workbook. | 395 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-50 | latest | en | 0.878406 |
https://brainly.com/question/105369 | 1,484,580,913,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279189.36/warc/CC-MAIN-20170116095119-00332-ip-10-171-10-70.ec2.internal.warc.gz | 809,133,185 | 9,284 | 2014-09-01T05:23:53-04:00
### This Is a Certified Answer
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• Brainly User
2014-09-01T09:32:09-04:00
### This Is a Certified Answer
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(45/58)
=(45*(100/58))/(58*(100/58))
=(4,500/58)/100
=77.59/100 (to 2dp)
Answer: 77.59% (to 2dp) | 208 | 784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-04 | latest | en | 0.935042 |
http://happycoding.io/examples/processing/creating-functions/random-faces | 1,532,047,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00298.warc.gz | 168,269,319 | 3,984 | examples / processing / creating-functions / random-faces
Random Faces
example processing procedural-generation
Working from this simpler example that draws a single random face whenever the program runs, we can use functions to draw a new random face every second:
``````void setup() {
size(200, 200);
frameRate(1);
}
void draw() {
background(128);
drawFace();
}
void drawFace() {
//place face in middle of window
float faceX = width/2;
float faceY = height/2;
//smallest face is half the window, biggest face is whole window
float faceWidth = random(width/2, width);
float faceHeight = random(height/2, height);
//random face color
fill(random(255), random(255), random(255));
ellipse(faceX, faceY, faceWidth, faceHeight);
//random eye size
float eyeWidth = random(faceWidth*.1, faceWidth*.25);
float eyeHeight = random(faceHeight*.1, faceHeight*.25);
//random eye position
float spaceBetweenEyes = random(eyeWidth, eyeWidth*2);
float leftEyeX = faceX - spaceBetweenEyes/2;
float rightEyeX = faceX + spaceBetweenEyes/2;
float eyeY = faceY - random(faceHeight*.1, faceHeight*.25);
//white
fill(255);
//draw the eyes
ellipse(leftEyeX, eyeY, eyeWidth, eyeHeight);
ellipse(rightEyeX, eyeY, eyeWidth, eyeHeight);
//random pupil size
float pupilWidth = random(eyeWidth*.1, eyeWidth*.9);
float pupilHeight = random(eyeHeight*.1, eyeHeight*.9);
//black
fill(0);
//draw the pupils
ellipse(leftEyeX, eyeY, pupilWidth, pupilHeight);
ellipse(rightEyeX, eyeY, pupilWidth, pupilHeight);
//random mouth size and Y
float mouthWidth = random(faceWidth*.2, faceWidth*.8);
float mouthHeight = random(faceHeight*.1, faceHeight*.3);
float mouthY = faceY + random(faceHeight*.1, faceHeight*.25);
//random mouth color
fill(random(255), random(255), random(255));
//draw the mouth
arc(faceX, mouthY, mouthWidth, mouthHeight, 0, 3.14);
line(faceX - mouthWidth/2, mouthY, faceX + mouthWidth/2, mouthY);
}
``````
This code uses variables along with the `random()` function to procedurally generate random faces. Every time you run the program, you get a different face.
Code Editor
See the Pen by Happy Coding (@KevinWorkman) on CodePen.
Tweak Ideas
• Randomly generate an entire crowd.
• Randomly generate an entire scene by creating functions that generate different objects: one that generates trees, one that generates buildings, etc. | 614 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-30 | latest | en | 0.568099 |
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This resource is Volume 2 which focuses on subtraction within 10!
What are SOLVE ITS?
Solve Its: Number Sentence Puzzlers are engaging challenges your students will be determined to solve! In each volume, number sentences are provided on each of the 20 Solve Its but instead of numbers being displayed, there are pictures. Students are tasked with using the information provided to solve for the value of each of the pictures.
Each Solve It is provided in a full-page colored version, half-page colored version, half-page black/white version, and printable version with an area for students to show their work. An answer guide is included.
How to USE SOLVE ITS?
Solve Its can be used in many ways in your classroom. You know your students so make the choice that works best for them! These could be used as an independent activity, but would also be a great opportunity for students to collaborate!
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https://math.answers.com/Q/What_is_the_simplest_form_of_45_over_72 | 1,701,809,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00042.warc.gz | 432,548,566 | 43,609 | 0
What is the simplest form of 45 over 72?
Updated: 9/17/2023
Wiki User
8y ago
The simplest form of 45/72 is 5 over 8.
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8y ago
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Q: What is the simplest form of 45 over 72?
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Related questions
What is 45 over 72 in simplest form with its GCD?
GCD(45, 72) = 9 so the simplest form is (45/9) / (72/9) = 5/8
What is he simplest form of 45 and 72?
45/72 is 5/8 in simplest form.
45/72 = 5/8
8/5
What is 45 over 72 in simplest form?
45/72 = 9*5/9*8 = 5/8
What is the simplest form of 61 over 72?
61/72 is in its simplest form.
What is the simplest form of 67 over 72?
67/72 is in its simplest form.
What is 72 over 95 in simplest form?
72/95 is the simplest form.
What is 67 over 72 in simplest form?
67/72 is in its simplest form.
What is 35 over 72 in simplest form?
35/72 is in its simplest form.
What is 49 over 72 in simplest form?
49/72 is already in simplest form.
What is 72 over 75 in simplest form?
72/75 in simplest form is 24/25. | 340 | 1,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-50 | latest | en | 0.945207 |
http://forums.wolfram.com/mathgroup/archive/2000/Apr/msg00115.html | 1,529,507,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00616.warc.gz | 116,589,164 | 7,427 | Re: system of simultaneous equations
• To: mathgroup at smc.vnet.net
• Subject: [mg22948] Re: [mg22895] system of simultaneous equations
• From: "Tomas Garza" <tgarza at mail.internet.com.mx>
• Date: Fri, 7 Apr 2000 02:54:31 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```David Braunholtz wrote:
> WHile I've got the attention of a Mathematica expert, I am wanting to
> select a single root of a system of simultaneous equations (with
> positive values for all 3 variables) - and (being new to mathematica)
> cannot work out how to do this. I successfully get a list of roots, but
> cannot make the (you would think) easy step to reduce the list to
> only (positive) roots. Any help VERY gratefully received. I have
> searched in Mathgroup at Wolfram without finding anything.
Bear in mind that Solve gives solutions in terms of rules of the form x ->
sol. To give an example, let
In[1]:=
r := Random[Integer, {1, 9}];
In[2]:=
sols = Solve[
r x + r y + r z == r && r x + r y + r z == r &&
r x + r y + r z == r, {x, y, z}] // Flatten
Out[2]=
{x -> 4, y ->-49/3, z -> 25/3}
This is a list of rules, each giving the value of the corresponding
variable. Each member of this list is a list, too. For example,
In[3]:=
sols[[1]]
Out[3]=
x -> 4
and
In[4]:=
sols[[1, 2]]
Out[4]=
4
If you want a list of just the numerical values of the solutions, take
In[5]:=
vals = #[[2]] & /@ sols
Out[5]=
{4,-49/3, 25/3}
and if you want to pick only the positive ones, use Select:
In[3]:=
Select[vals, # > 0 &]
Out[3]=
{4, 25/3}
Tomas Garza
Mexico City
```
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• Next by thread: Edit Mathematica graphics in Adobe | 565 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-26 | latest | en | 0.8422 |
http://docplayer.net/21024471-Addition-and-subtraction-of-integers.html | 1,544,853,181,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00052.warc.gz | 85,349,708 | 23,771 | # Addition and Subtraction of Integers
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1 Addition and Subtraction of Integers Integers are the negative numbers, zero, and positive numbers Addition of integers An integer can be represented or graphed on a number line by an arrow. An arrow pointing to the right represents a positive number. An arro w pointing to the left represents a negative num ber. The absolute value of the number can be determined by counting the distance to zero, or by counting the distance between numbers in the arrow. The integers 5 and 4 are shown on the number line in the figure below Another example of integers or positive and negative numbers is used with a checking account or dealing with money. A deposit to your checking account or pocket is an example of an addition called a positive integer (number or amount); a deduction or expense is an example of a negative integer (number/amount). If there is a balance of \$25 in a checking account, and a check is written for \$30, the account will be overdrawn/overspent by \$5. The sum of two integers can be shown on a number line. To add two integers, find the point on the number line corresponding to the first addend (integer). The sum is the number directly below the tip of the arrow = (-2) = = -2 Student Learning Assistance Center - San Antonio College 1
4 Is 7 a solution of the equation x + 4 = - 3? Replace x by - 7 and then simplify = -3-3 = -3 The results are equal. 7 is a solution of the equation. Subtraction of integers Recall that the sign can indicate the sign of a number, as in 3 (negative 3), or can indicate the operation of subtraction, as in 9 3 (nine minus three). Subtraction of integers can be written as the addition of the opposite number. To subtract two integers, rewrite the subtraction expression as the first number plus the opposite of the second number. Some examples are shown below. First number - second number = First number + opposite of the second number 8-15 = 8 + (-15) = (-15) = = = -8 + (-15) = (-15) = = 7 Rule for Subtracting Two Integers To subtract two integers, add the opposite of the second integer to the first integer. This can be written symbolically as a - b = a + (-b). Subtract (-15) 75 Rewrite the subtraction operation as he sum of (-15) - 75 the first number and the opposite of the second = (-15) + (-75) number. The opposite of 75 is = - 90 Add. When subtraction o ccurs severa l times in an expression, rewrite each subtraction as add ition of the opposite and then add. Student Learning Assistance Center - San Antonio College 4
5 Subtract : (-8) (-8) Rewrite each subtraction as addition of the Opposite. = (-5) + 8 = Add. = - 10 Student Learning Assistance Center - San Antonio College 5
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Grade 7 Mathematics, Quarter 1, Unit 1.1 Integer Operations Overview Number of Instructional Days: 15 (1 day = 45 minutes) Content to Be Learned Describe situations in which opposites combine to make zero. | 6,894 | 28,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2018-51 | latest | en | 0.881971 |
http://www.cnblogs.com/aziint/p/9780437.html | 1,558,280,244,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00040.warc.gz | 249,166,136 | 5,361 | # 51nod 1355 斐波那契的最小公倍数
### Description
$f[n]= \begin{cases} 1 , & \text {if n is equal to 0 or 1} \\ f(n-1) + f(n-2), & \text{otherwise} \end{cases}$
$2\le N \le 50000,1 \le a_i\le 1000000$
### Solution
$lcm(S)=\prod_{T\subseteq S,T\ne\emptyset} \gcd(T) ^{(-1)^{|T|+1}}$
$f(\gcd(a,b)) = \gcd(f(a), f(b))$
\begin{align} lcm(f_S) &=\prod_{T\subseteq S,T\ne\emptyset} \gcd(f_T)^{(-1)^{|T|+1}} \\ &=\prod_{T\subseteq S,T\ne\emptyset} f_{\gcd(T)}^{(-1)^{|T|+1}} \end{align}
$f_n=\sum_{d|n}g_d$
\begin{align} lcm(f_S) &=\prod_{T\subseteq S,T\ne\emptyset} \gcd(f_T)^{(-1)^{|T|+1}} \\ &=\prod_{T\subseteq S,T\ne\emptyset} f_{\gcd(T)}^{(-1)^{|T|+1}} \\ &=\prod_{T\subseteq S,T\ne\emptyset} (\prod_{d|\gcd(T)}g_d)^{(-1)^{|T|+1}} \\ &=\prod_d g_d^{\sum_{T\subseteq S,T\ne \emptyset,d|\gcd(T)} (-1)^{|T|+1}} \end{align}
$\sum_{T\subseteq S,T\ne \emptyset,d|\gcd(T)} (-1)^{|T|+1} = \begin{cases} 1, & \exists x \in S,d|x\\ 0, & \text{otherwise} \end{cases}$
\begin{align} lcm(f_S) &=\prod_{T\subseteq S,T\ne\emptyset} \gcd(f_T)^{(-1)^{|T|+1}} \\ &=\prod_{T\subseteq S,T\ne\emptyset} f_{\gcd(T)}^{(-1)^{|T|+1}} \\ &=\prod_{T\subseteq S,T\ne\emptyset} (\prod_{d|\gcd(T)}g_d)^{(-1)^{|T|+1}} \\ &=\prod_d g_d^{\sum_{T\subseteq S,T\ne \emptyset,d|\gcd(T)} (-1)^{|T|+1}} \\ &=\prod_{\exists x \in S,d|x} g_d \end{align}
$g_n=f_n\times(\prod _{d|n,d\ne n} g_d)^{-1}$
#include<bits/stdc++.h>
using namespace std;
template <class T> void read(T &x) {
x = 0; bool flag = 0; char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar()) flag |= ch == '-';
for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48; flag ? x = 0 - x : 0;
}
#define N 10000010
#define rep(i, a, b) for (auto i = (a); i <= (b); ++i)
#define drp(i, a, b) for (auto i = (a); i >= (b); --i)
#define ll long long
#define P 1000000007
int n, a[N], f[N], g[N];
int qpow(int x, int k) {
int ret = 1;
for (; k; k >>= 1, x = 1ll * x * x % P) if (k & 1) ret = 1ll * ret * x % P;
return ret;
}
void init() {
f[1] = 1;
rep(i, 2, n) f[i] = (f[i - 2] + f[i - 1]) % P;
rep(i, 1, n) g[i] = f[i];
rep(i, 1, n) {
int inv = qpow(g[i], P - 2);
for (int j = i + i; j <= n; j += i) g[j] = 1ll * g[j] * inv % P;
}
}
bool vis[N];
int main() {
read(n); int _n = 0;
rep(i, 1, n) read(a[i]), _n = max(a[i], _n), vis[a[i]] = 1;
n = _n;
init();
int ans = 1;
rep(i, 1, n) {
bool tag = 0;
for (int j = i; j <= n; j += i) if (vis[j]) { tag = 1; break; }
if (tag) ans = 1ll * ans * g[i] % P;
}
printf("%d", ans);
return 0;
}
posted @ 2018-10-12 20:20 aziint 阅读(...) 评论(...) 编辑 收藏 | 1,150 | 2,537 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.178083 |
http://nrich.maths.org/public/leg.php?code=32&cl=1&cldcmpid=2353 | 1,493,102,883,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120187.95/warc/CC-MAIN-20170423031200-00047-ip-10-145-167-34.ec2.internal.warc.gz | 274,481,981 | 10,310 | # Search by Topic
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A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids. | 2,265 | 9,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-17 | longest | en | 0.905747 |
http://mathhelpforum.com/calculus/17602-calculating-angle-print.html | 1,527,241,401,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00411.warc.gz | 189,582,699 | 3,600 | # calculating angle
• Aug 8th 2007, 01:56 AM
SuperEletcric
calculating angle
hello
having position of 3 point i want to calculate the angle between them
• Aug 8th 2007, 03:28 AM
ThePerfectHacker
Quote:
Originally Posted by SuperEletcric
hello
having position of 3 point i want to calculate the angle between them
Given points A,B,C.
1)Find vector AB
2)Find vector AC
3)Find ||AB|| (the length)
4)Find ||AC|| (the length)
5)Find ||AB||*||AC|| (their product)
6)Find AB*AC (dot product)
7)Find #5 divided by #6
8)Take the arc cosine of #7
• Aug 8th 2007, 06:11 AM
ticbol
Quote:
Originally Posted by SuperEletcric
hello
having position of 3 point i want to calculate the angle between them
If in 2D only, plot the 3 points on the same rectangular axes, so that you'd know how to get the angle between the two rays. For simplicity, put the corner of the angle on the origin (0,0).
Use the slopes of the rays. The slope of a line is the tangent of the angle made by the line with the horizontal axis, so the angle of the line with the horizontal axis is the arctangent of the slope.
So the angle between the two rays is actan(m1) minus arctan(m2).
Be careful with the signs of the slopes m1 and m2.
• Aug 8th 2007, 06:15 AM
Soroban
Hello, SuperEletcric!
Another approach . . .
Quote:
Having the positions of 3 points,
calculate the angles of the triangle.
Suppose the points are: .$\displaystyle A(x_1,y_1),\:B(x_2,y_2),\:C(x_3,y_3)$
Let's find angle $\displaystyle A \:=\:\angle BAC$
Use the Slope Formula to find: .$\displaystyle m_{_{AB}} \text{ and }m_{_{AC}}$
Then: .$\displaystyle \tan A \:=\:\left|\frac{m_{_{AB}} - m_{_{AC}}}{1 + m_{_{AB}}m_{_{AC}}}\right|$
• Aug 8th 2007, 07:46 AM
Plato
Any three non-collinear points determine a triangle, even in $\displaystyle \Re^3$.
Hence we have three angles. If A, B, & C are the three points then $\displaystyle m\left( {\angle ABC} \right) = \arccos \left( {\frac{{{\overrightarrow {BA} \cdot \overrightarrow {BC} } }}{{\left\| {\overrightarrow {BA} } \right\|\left\| {\overrightarrow {BC} } \right\|}}} \right)$.
This is exactly what TPH posted above.
• Aug 15th 2007, 01:56 AM
SuperEletcric
thanx to all
but i have small problem.
(finding vectors and doing dot product).
would you mind solving algorithm for instance for this points : (0,0),(1,1),(1,0)
• Aug 15th 2007, 05:33 AM
ThePerfectHacker
Quote:
Originally Posted by SuperEletcric
thanx to all
but i have small problem.
(finding vectors and doing dot product).
would you mind solving algorithm for instance for this points : (0,0),(1,1),(1,0)
The vector from $\displaystyle (a,b,c)$ to $\displaystyle (x,y,z)$ is found by subtracting the endpoints from the starting points:
$\displaystyle (x-a)\bold{i}+(y-b)\bold{j}+(z-c)\bold{k}$
The dot product, $\displaystyle (a\bold{i}+b\bold{j}+c\bold{k})\cdot (x\bold{i}+y\bold{j}+z\bold{k})=ax+by+cz$ | 920 | 2,869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-22 | latest | en | 0.794688 |
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# Two-step equations with decimals and fractions
## Video transcript
let's do a few more examples solving equations and I think you're going to see that these equations require require a few more steps than the ones we did in the last video but the fun thing about these is that there's more than way one way to do it but as long as you do legitimate steps as long as anything you do to the left-hand side you also do to the right-hand side you should move in the correct direction or you shouldn't get the wrong answer so let's let's do a couple of these so the first one says one two I'll rewrite it one point three times X minus 0.7 times X is equal to 12 well here the the first thing that my instinct is to do is to to merge these two terms because I have one point three of something minus 0.7 of that same something this is the same variable if I have one point three apples minus 0.7 apples well why don't I subtract 0.7 from one point three and I will get so I will get one point three minus zero point seven if there's X's or apples or whatever you want to call them so is equal to twelve you can imagine that I did the reverse distributive property out here I factored out an X by the way my head thinks about it is I have one point three of something minus point seven of something that's going to be equal to one point three minus point seven of those some things that X and of course one point three minus 0.7 is 0.6 0.6 times X of my something's is equal to 12 and now this looks just like one of the problems we did in the last video we have a coefficient times X is equal to some other number well let's divide both sides of this equation by that coefficient divide both sides by 0.6 divided both sides by 0.6 and what is so the left-hand side will just become an X X is equal to and what is 12 divided by 0.6 0.6 goes into 12 let's add some decimal points here that's the same thing as 6 going into 120 6 goes into 12 2 times 2 times 6 is 12 subtract you get a Z row 6 goes into 0 zero times it's going to go 20 times 12 divided by 0.6 is 20 and we can verify what's let's verify this 1 point 3 let's substitute back 1.3 times 20 minus 0.7 times 20 let's verify that that is equal to 12 so I'll take the calculator out just so you don't have to trust my math so we have 1.3 times 20 is equal to 26 so this piece right here that piece right there is 26 and then 0.7 times 20 that's I don't need a calculator for that that is 14 26 minus 14 is 12 so it checks out we got the right answer for this equation X is equal to 20 let's do this one right here 5 X minus 3 X plus 2 is equal 1 this looks very complicated and whenever something looks daunting just do steps that look like they're simplifying the equation and over time as long as you do the legitimate steps you should be able to make some progress so the first thing I want to do is I want to distribute this negative one over here so this is the same thing as 5 X minus 3 X minus 3 X minus 2 alright I just did the distributive property on the 3 X which is a negative 1 times 3 X plus 2 so it's negative 1 times 3 X plus negative 1 times 2 or negative 3 X minus 2 and that is going to be equal to 1 now I have five of something minus 3 of that same something so that's going to be equal to 2 of that something 5x minus 3x is 2x 5 minus 3 is 2 and then I have the minus 2 is equal to 1 and now I like to get it to the form where I have 2 X or I have something times X is equal to something so I want to get rid of this negative 2 on the left hand side the best way I know how to do that is to add 2 to both sides so add it to on the left hand side if I do it to the left hand side I've got to do it to the right hand side plus 2 on the right hand side the these two guys will cancel out and you're going to get 2x you're going to get 2x is equal to one plus two is equal to three and now you can divide both sides by 2 and you get X is equal to X is equal to three halves and I'll leave it for you to verify that this is indeed the correct answer let me draw a little line here so that our work doesn't get messy although Matt might have made it even Messier so here we have to solve for s and let me have a fraction and 2's terms how do we do that well just do it the same way we have 1 times s - you could view this as 3/8 times s is equal to 5/6 you could view this as 1 times s minus 3/8 times s is equal to 5/6 you could factor out an S if you like maybe I'll do it this way I'll factor it onto the left hand side this is the same thing as s times 1 minus 3/8 is equal to 5/6 and 1 minus 3/8 what is that that's 1 I can rewrite as 8 over 8 right that's one so this is the same thing as 8/8 minus 3/8 is 5/8 5/8 times s you can switch the order of multiplication 5/8 times s is equal to 5/6 and you might be able to go straight from that if I have 1 of something minus 3/8 of that something I have 8/8 of that something minus 3/8 of that something I'm going to have 5/8 of that something and now to solve for s how I can multiply both sides by the inverse of this coefficient so I multiply 8 over 5 times 5 8 s if I do it to the left hand side I have to do it to the right hand side 8 over 5 I multiplied by 8 over 5 so that those cancel out and those cancel out and you are left with s is equal to right this was just a 1 is equal to well the fives we can divide divide the numerator denominator by 5 divide the numerator by 2 and the denominator by 2 you're left with sorry divide the denominator by 2 you get 6 divided by 2 is 3 you're left four-thirds s is equal to four-thirds let's do one more of these so here I have five times Q minus seven over twelve is equal to two thirds let me write this and like I could rewrite this it's just five over twelve times Q minus seven is equal to two thirds and what I want to do with this video is to show you that I can do it two different ways but as long as I do legitimate operations I should get the same answer so that what the first way I'm going to do it is I'm going to multiply both sides of this equation by the inverse of five twelve so I'm going to multiply both sides by 12 over 5 I'm going to multiply both sides by 12 over five because I want to get rid of this 512 s on the left-hand side it makes everything look a little bit messy and I multiply it by 12 over five because these are going to cancel out the 5 and the 5 cancel out the 12 and the twelve cancel out so the left-hand side of my equation becomes Q minus seven is equal to the right-hand side 2/3 times 512 so if you divide the 12 by 3 you get a 4 you divide the 3 by 3 you get a 1 so 2 times 4 is 8 over 5 and now we can add 7 to both sides of this equation so at let's add I want to do that in a different color add 7 to both sides of this equation these two 7s cancel out that was the whole point of adding the 7 and you are left with Q is equal to 8/5 plus 7 or we could write 8 fifths plus 7 can be written as 35 35 fifths and so this is going to be equal to 8 well the denominator is 5 8 plus 35 8 plus 35 is 43 so my answer going this way is Q is equal to 43 over 5 now I said I would do it two ways let's do it another way so let me write the same problem down so I have 512 512 actually let me do it a completely different way let me say let me write it the way they wrote it five times q minus 7 over 12 is equal to 2/3 let me just get rid of the 12 first let me multiply both sides of this equation by 12 I just don't like that 12 sitting there so I'm going to multiply both sides by 12 so these are going to cancel out and you're going to be left with 5 times Q minus 7 is equal to 2/3 times 12 that's the same thing as 24 over 3 so this is let me write this 2 over 3 times 12 over 1 is equal to what you get a if you divide that by 3 you get a 4 divide that by 3 you get a 1 it's equal to 8 so we get 5 times Q minus 7 is equal to 8 and then instead of dividing both sides by 5 which would get us something pretty close to what we were doing over here let me distribute this 5 I just want to show you you can do it multiple legitimate ways so 5 times Q is 5q 5 times negative 7 is minus or negative 35 is equal to 8/5 Q minus 35 is equal to 8 now if I want to get rid of that minus 35 or that negative 35 the best way to do it is to add 35 to both sides so 35 to both sides I did that so these cancel out and I'm left with 5 Q is equal to 8 plus 35 which is 43 now I can multiply both sides of this equation by 1/5 which is the same thing as dividing both sides by 5 and these cancel out you get Q is equal to 43 over 5 so there's a bunch of ways you can do these problems but as long as you do legitimate steps you will get the right answer and I'll leave it to you to verify that this truly is the right answer for Q this is the Q that will satisfy this equation let's do one word problem here Jade is stranded downtown with only $10 to get home taxis cost 75 cents per mile but there is an additional 235 higher charge write a formula and use it to calculate how many miles she can travel with her money all right so the total the total cost of a cab ride is going to be equal to just the initial higher charge the initial higher charge which is two dollars and thirty five plus the seventy five cents per mile plus the seventy five cents times the number of miles we're letting M is equal to the miles she travels miles traveled miles traveled so this is the equation we know that she can only she only has ten dollars to get home so her cost has to be ten dollars so we have to say the cost has to be$10 so 10 is equal to two point three five plus plus 0.75 M so how do we solve for M or the number of miles Jade can travel we can get rid of the two point three five on this right hand side by subtracting that amount from both sides of this equation so let's do that so let's subtract a minus two point three five let's subtract a minus two point three five from both sides these will cancel out that was the point the left-hand side what is 10 minus two point three and I would have three four this is ten minus ten minus two point three five now these will cancel out now what is 10 minus two point three five 10 minus two is 10 minus 2 is 8 10 minus two point three is seven point seven so it's going to be seven point six five if you want to believe me let's do it 10 minus two point three five seven point six five and that is going to be equal to 0.75 and let me write that in that same color it's nice to see where different things came from 0.75 and I have like five shades of this purple here so this is this is that that is that and then these two guys cancelled out now to solve for M I can just divide both sides by 0.75 so I divide that side by 0.75 I have to do it to the left hand side as well 0.75 that cancels out so on the right hand side I'm left with just an M and on the left hand side I'll have to get my calculator out for this one I have seven point six five divided by 0.75 which is equal to ten point two so Jade can travel ten point M is ten point two so Jade can travel ten point two miles | 2,927 | 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https://python.engineering/php-is_finite-is_infinite-is_nan-functions/ | 1,632,026,203,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00123.warc.gz | 522,039,400 | 6,342 | +
# PHP Is_finite (), is_infinite (), is_nan () functions
So this is a safe way to check if a given input is finite or not. is_finite() Syntax:
`bool is_finite (\$value)`
Parameters : the function takes one parameter, which is floating point to be checked.Return type : This function returns TRUE if the given value is finite, FALSE otherwise.Examples:
`Input: \$value = M_PI_4; Output: TRUE Input: \$value = log (0); Output: FALSE`
is_infinite() Syntax:
`bool is_infinite (\$value)`
Parameters : the function takes one parameter, which is a floating point that must be checked.Return type : This function returns TRUE if the given value is infinite, FALSE otherwise.Examples:
`Input: \$value = M_PI_4; Output: FALSE Input: \$value = log (0); Output: TRUE`
is_nan() function Syntax:
`bool is_nan (\$value)`
Parameters : the function takes one parameter, which is a floating point that must be checked. Return type : This function returns TRUE if the given value is not a number, FALSE otherwise.Examples:
`Input: \$value = M_PI_4; Output: FALSE Input: \$value = acos (1.1); // cos function can not be greater than 1 Output: TRUE`
The program below illustrates the operation of the is_finite(), is_infinite(), is_nan() functions in PHP: ``` // PHP -code to illustrate the work // Functions is_finite(), is_infinte() and is_nan() // Final value: PI \$val1 = M_PI; // Embedded INFINITY value \$val2 = INF; // Creates NaN because the COS value can be found // from -1 up to +1 both inclusive \$val3 = acos (- 1.01); echo var_dump ( is_finite ( \$val1 ), is_finite ( \$val2 ), is_finite ( \$val3 )). " " ; echo var_dump ( is_infinite ( \$val1 ), is_infinite ( \$val2 ), is_infinite ( \$val3 )). "" ; echo var_dump ( is_nan ( \$val1 ), is_nan ( \$val2 ), is_nan ( \$val3 )). "" ; ?> Output:bool (true ) bool (false) bool (false) bool (false) bool (true) bool (false) bool (false) bool (false) bool (true) to watch out for : This function can also check if an expression produces an end result, but in the case of expressions that result in NaN, PHP itself shows an error and returns FALSE by default , for example, when checking the Divided by Zero expression. The is_finite() function is used in many projects to make it safer and more efficient. These methods give very accurate results, but not very time efficient. ```
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``` Get Solution for free from DataCamp guru © 2021 Python.Engineering Best Python tutorials books for beginners and professionals Python.Engineering is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to amazon.com Computations Development Cryptography For dummies Machine Learning Big Data Loops Counters NumPy NLP PHP Regular Expressions File Handling Arrays String Variables Knowledge Database X Submit new EBook \$(document).ready(function () { \$(".modal_galery").owlCarousel({ items: 1, itemsCustom: false, itemsDesktop: [1300, 1], itemsDesktopSmall: [960, 1], itemsTablet: [768, 1], itemsTabletSmall: false, itemsMobile: [479, 1], singleItem: false, itemsScaleUp: false, pagination: false, navigation: true, rewindNav: true, autoPlay: true, stopOnHover: true, navigationText: [ "<img class='img_no_nav_mob' src='/wp-content/themes/nimani/image/prevCopy.png'>", "<img class='img_no_nav_mob' src='/wp-content/themes/nimani/image/nextCopy.png'>" ], }); \$(".tel_mask").mask("+9(999) 999-99-99"); }) ``` | 889 | 3,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | latest | en | 0.388849 |
https://www.physicsforums.com/threads/the-falling-rod-problem.760067/ | 1,508,235,457,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00612.warc.gz | 966,331,606 | 16,058 | The Falling Rod Problem
1. Jun 30, 2014
zemansky
1. The problem statement, all variables and given/known data
A uniform rod of mass M and length L is placed vertically on a flat surface. The coefficient of friction between the rod and the surface is μ. The rod is beginning to fall, initially without slipping. To describe the motion of the rod during this stage:
(a) Write down the Lagrangian and derive the equation of motion for the generalized coordinate
$\phi$.
b) From the Lagrangian derive the expression for the mechanical en- ergy of the system and write down the energy conservation law.
(c) Use Newton’s second law (or D’Alembert’s principle) to derive expressions for the normal (N) and friction (f) forces acting on the rod. Express the answers in terms of $\phi$ only. You can use the results of parts (a) and (b) for this step.
(d) Under what condition will the rod fall all the way to the ground without slipping?
2. Relevant equations
3. The attempt at a solution
No attempt to solve this problem has been done. I need help to set up the problem!
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited: Jun 30, 2014
2. Jul 9, 2014
ehild
Hi zemansky, welcome to PF!
You need to show what you know. What is "Lagrangian"?
ehild | 385 | 1,553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-43 | longest | en | 0.864916 |
https://quick-adviser.com/how-accurate-are-neural-network/ | 1,642,510,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00108.warc.gz | 552,685,221 | 12,777 | # How accurate are neural network?
## How accurate are neural network?
A survey of 96 studies comparing the performance between neural networks and statistical regression models in several fields, showed that neural networks outperformed the regression models in about 58% of the cases, whereas in 24% of the cases, the performance of the statistical models were equivalent to the neural …
## How is accuracy calculated in machine learning?
For Classification Model:
1. Precision = TP/(TP+FP)
2. Sensitivity(recall)=TP/(TP+FN)
3. Specificity=TN/(TN+FP)
4. Accuracy=(TP+TN)/(TP+TN+FP+FN)
## How is Top 5 accuracy calculated?
Top-5 accuracy means any of our model’s top 5 highest probability answers match with the expected answer. It considers a classification correct if any of the five predictions matches the target label. In our case, the top-5 accuracy = 3/5 = 0.6.
## What is good accuracy in machine learning?
If you are working on a classification problem, the best score is 100% accuracy. If you are working on a regression problem, the best score is 0.0 error. These scores are an impossible to achieve upper/lower bound. All predictive modeling problems have prediction error.
## Is 70% a good accuracy?
If your ‘X’ value is between 70% and 80%, you’ve got a good model. If your ‘X’ value is between 80% and 90%, you have an excellent model. If your ‘X’ value is between 90% and 100%, it’s a probably an overfitting case.
## What is a good accuracy for image classification?
While 91% accuracy may seem good at first glance, another tumor-classifier model that always predicts benign would achieve the exact same accuracy (91/100 correct predictions) on our examples.
## Which neural network is best for image classification?
convolutional neural networks
## What is the best model for image classification?
Pre-Trained Models for Image Classification
• Very Deep Convolutional Networks for Large-Scale Image Recognition(VGG-16) The VGG-16 is one of the most popular pre-trained models for image classification.
• Inception. While researching for this article – one thing was clear.
• ResNet50.
## How do you increase image classification accuracy?
Add More Layers: If you have a complex dataset, you should utilize the power of deep neural networks and smash on some more layers to your architecture. These additional layers will allow your network to learn a more complex classification function that may improve your classification performance. Add more layers!
## How do you increase the accuracy of a neural network?
Now we’ll check out the proven way to improve the performance(Speed and Accuracy both) of neural network models:
1. Increase hidden Layers.
2. Change Activation function.
3. Change Activation function in Output layer.
4. Increase number of neurons.
5. Weight initialization.
6. More data.
7. Normalizing/Scaling data.
## How can you increase the accuracy of convolutional neural network?
Train with more data: Train with more data helps to increase accuracy of mode. Large training data may avoid the overfitting problem. In CNN we can use data augmentation to increase the size of training set….
1. Tune Parameters.
2. Image Data Augmentation.
3. Deeper Network Topology.
4. Handel Overfitting and Underfitting problem.
## How do you increase the accuracy of CNN?
Class weights >> Used to train highly imbalanced (biased) database, class weights will give equal importance to all the classes during training. Fine tuning the model with train data >> Use the model to predict on training data, retrain the model for the wrongly predicted images.
## How do you improve validation accuracy?
We have the following options.
1. Use a single model, the one with the highest accuracy or loss.
2. Use all the models. Create a prediction with all the models and average the result.
3. Retrain an alternative model using the same settings as the one used for the cross-validation. But now use the entire dataset.
## What happens if validation accuracy is not increasing?
1. Use weight regularization. It tries to keep weights low which very often leads to better generalization.
2. Corrupt your input (e.g., randomly substitute some pixels with black or white).
4. Pre-train your layers with denoising critera.
5. Experiment with network architecture.
## Can validation accuracy be more than training accuracy?
Validation accuracy will be usually less than training accuracy because training data is something with which the model is already familiar with and validation data is a collection of new data points which is new to the model.
## What is training accuracy and validation accuracy?
In other words, the test (or testing) accuracy often refers to the validation accuracy, that is, the accuracy you calculate on the data set you do not use for training, but you use (during the training process) for validating (or “testing”) the generalisation ability of your model or for “early stopping”.
## Why is my validation accuracy higher than training accuracy?
Especially if the dataset split is not random (in case where temporal or spatial patterns exist) the validation set may be fundamentally different, i.e less noise or less variance, from the train and thus easier to to predict leading to higher accuracy on the validation set than on training.
## Why test accuracy is low?
A model that is selected for its accuracy on the training dataset rather than its accuracy on an unseen test dataset is very likely have lower accuracy on an unseen test dataset. The reason is that the model is not as generalized. For example, you may want to stop training your model once the accuracy stops improving.
## What is the difference between loss and accuracy?
Loss value implies how poorly or well a model behaves after each iteration of optimization. An accuracy metric is used to measure the algorithm’s performance in an interpretable way. It is the measure of how accurate your model’s prediction is compared to the true data.
## What is the relationship between loss and accuracy?
There is no relationship between these two metrics. Loss can be seen as a distance between the true values of the problem and the values predicted by the model. Greater the loss is, more huge is the errors you made on the data. Accuracy can be seen as the number of error you made on the data.
## What is Overfitting problem?
Overfitting is a modeling error in statistics that occurs when a function is too closely aligned to a limited set of data points. Thus, attempting to make the model conform too closely to slightly inaccurate data can infect the model with substantial errors and reduce its predictive power.
## What does loss represent in neural network?
The Loss Function is one of the important components of Neural Networks. Loss is nothing but a prediction error of Neural Net. And the method to calculate the loss is called Loss Function. In simple words, the Loss is used to calculate the gradients. And gradients are used to update the weights of the Neural Net.
## How is CNN loss calculated?
For example, using mean square error, the loss function is (output−expected)2. So if I had a binary classifier, say the class labels are (0,1) then the output of the neural network would need to be one dimension to compute the loss.
## How do neural networks reduce loss?
Solutions to this are to decrease your network size, or to increase dropout. For example you could try dropout of 0.5 and so on. If your training/validation loss are about equal then your model is underfitting. Increase the size of your model (either number of layers or the raw number of neurons per layer)
## What is the difference between loss and cost function?
The terms cost and loss functions almost refer to the same meaning. The cost function is calculated as an average of loss functions. The loss function is a value which is calculated at every instance. So, for a single training cycle loss is calculated numerous times, but the cost function is only calculated once.
## Is ReLU a cost function?
A ReLU is simply a function that converts any negative values to 0. Let’s rename that as the max(0,z) function, which returns z if z is positive and 0 if z is negative. This is the loss function that we have to find the slope to. In order to find the slope, we have to find the loss function’s derivative.
## What is cost function neural network?
Introduction. A cost function is a measure of “how good” a neural network did with respect to it’s given training sample and the expected output. It also may depend on variables such as weights and biases. A cost function is a single value, not a vector, because it rates how good the neural network did as a whole.
## What is the cost function?
Put simply, a cost function is a measure of how wrong the model is in terms of its ability to estimate the relationship between X and y. This is typically expressed as a difference or distance between the predicted value and the actual value. The cost function (you may also see this referred to as loss or error.)
Begin typing your search term above and press enter to search. Press ESC to cancel. | 1,906 | 9,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.91015 |
https://ifyoufeedme.com/question/1222/ | 1,660,515,314,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572077.62/warc/CC-MAIN-20220814204141-20220814234141-00569.warc.gz | 303,671,593 | 12,863 | # A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electric field. The potential difference between points A and B in the electric field is approximately
A recent school question and answer requested students to assert what they feel is the main important thing for a student to do if they wanted to get success. One that response stood out from the rest was practice. Successful persons commonly are not born successful; they become successful by just hard work and persistence. If you wish to complete your goals, keep this in mind! in this article, are one of the answer and question example that you could simply utilize to practice and enriches your information and also give you insights that could just assist you to preserve your study in school.
## Question:
A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electric field. The potential difference between points A and B in the electric field is approximately
The potential difference between points A and B in the electric field is approximately equal to
Given the following data:
• Kinetic energy =
Scientific data:
• Charge of proton =
To determine the potential difference between points A and B in the electric field:
Mathematically, kinetic energy is calculated by using this formula;
Where:
• K.E is the kinetic energy.
• q is the charge of a particle.
• is the potential difference of a particle.
Making the subject of formula, we have:
Substituting the given parameters into the formula, we have;
When is approximated, we have:
Potential difference = 522,000 Volts
Potential difference =
From the answer and question examples above, hopefully, they may help the student sort out the question they had been looking for and remember of all the things declared in the answer above. Then can certainly make some sharing in a group discussion and also study with the classmate about the topic, so another student also will have some enlightenment and still keeps up the school learning.
READ MORE Tasha assembled a picture frame that is advertised as rectangular. The completed frame is 14 inches long and 10 inches wide. She measured the diagonal length across the frame as 20 inches. Which best explains why the frame cannot actually be rectangular? | 479 | 2,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-33 | longest | en | 0.96325 |
https://www.instasolv.com/question/180-if-a-b-and-c-are-positive-numbers-such-that-a-b-c-18-also-if-a-b3c4-2-1rvfsl | 1,611,184,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00311.warc.gz | 811,425,275 | 11,293 | 180. If a, b and c are positive num...
Question
# 180. If a, b and c are positive numbers such that a + b + c = 18. Also if a-b3c4 = 219 x 33 then the a+ b2 is equal to (A) 50 (B) 52 (C) 100 (D) 48
JEE/Engineering Exams
Maths
Solution
117
4.0 (1 ratings)
( a+b+c=18, quad frac{a^{2} b^{3} c^{4}=2^{19} times 3^{3}}{1} ) From this eqn. we can cleculy assume that ( a, b, c ) are multiples of 2 and 3 only. so most switable values for these ( operatorname{eqn} operatorname{arcs} 8,6,4 ) And it also satisfy they ( operatorname{con} a+b+c=18 ) [ begin{array}{l} text { So let } a=4, b=6, quad c=8 qquad begin{array}{rl} frac{1}{2}^{2} & 2 times 3 & 1 left(2^{2}right)^{3} times(2 times 3)^{3} timesleft(2^{3}right)^{4} & 2^{3} =2^{6} times 2^{3} times 3^{3} times 2^{12} end{array} end{array} ] ( =2^{19} times 3^{3}left(begin{array}{l}2 t text { als of satisfy } text { this eqn } )end{array}right. ) Mence the values we assumed ar correct. so ( a^{2}+b^{2}=36+16= ) ine 52 well in these use options d 5 weur we can types of question and 5ee that option only option which are sum of two squares ret 16 , ( 100 rightarrow 64+36 ). ( 52 rightarrow 36+16=cos b e cos u y d theta ) so other options
Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free | 492 | 1,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-04 | latest | en | 0.588261 |
http://stackoverflow.com/questions/4924842/javascript-math-object-methods-negatives-to-zero/4924858 | 1,394,773,531,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678686979/warc/CC-MAIN-20140313024446-00097-ip-10-183-142-35.ec2.internal.warc.gz | 134,724,366 | 15,928 | # Javascript Math Object Methods - negatives to zero
in Javascript I can't seem to find a method to set negatives to zero?
-90 becomes 0
-45 becomes 0
0 becomes 0
90 becomes 90
Is there anything like that? I have just rounded numbers.
-
Just do something like
``````value = value < 0 ? 0 : value;
``````
or
``````if (value < 0) value = 0;
``````
or
``````value = Math.max(0, value);
``````
-
Geee that was fast!! accepted answer because of the 3 choises :-) – FFish Feb 7 '11 at 18:21
``````Math.positive = function(num) {
return Math.max(0, num);
}
// or
Math.positive = function(num) {
return num < 0 ? 0 : num;
}
``````
-
I suppose you could use `Math.max()`.
``````var num = 90;
num = Math.max(0,num); // 90
var num = -90;
num = Math.max(0,num); // 0
``````
-
`x < 0 ? 0 : x` does the job .
-
If you want to be clever:
``````num = (num + Math.abs(num)) / 2;
``````
However, `Math.max` or a conditional operator would be much more understandable.
Also, this has precision issues for large numbers.
- | 326 | 1,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-10 | latest | en | 0.857372 |
https://converter.ninja/volume/us-gallons-to-imperial-pints/311-usgallon-to-imperialpint/ | 1,600,780,556,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206133.46/warc/CC-MAIN-20200922125920-20200922155920-00276.warc.gz | 340,322,582 | 5,343 | # 311 US gallons in imperial pints
## Conversion
311 US gallons is equivalent to 2071.69337135692 imperial pints.[1]
## Conversion formula How to convert 311 US gallons to imperial pints?
We know (by definition) that: $1\mathrm{usgallon}\approx 6.66139347703191\mathrm{imperialpint}$
We can set up a proportion to solve for the number of imperial pints.
$1 usgallon 311 usgallon ≈ 6.66139347703191 imperialpint x imperialpint$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{imperialpint}\approx \frac{311\mathrm{usgallon}}{1\mathrm{usgallon}}*6.66139347703191\mathrm{imperialpint}\to x\mathrm{imperialpint}\approx 2071.693371356924\mathrm{imperialpint}$
Conclusion: $311 usgallon ≈ 2071.693371356924 imperialpint$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 imperial pint is equal to 0.000482696915395842 times 311 US gallons.
It can also be expressed as: 311 US gallons is equal to $\frac{1}{\mathrm{0.000482696915395842}}$ imperial pints.
## Approximation
An approximate numerical result would be: three hundred and eleven US gallons is about two thousand and seventy-one point six nine imperial pints, or alternatively, a imperial pint is about zero times three hundred and eleven US gallons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 408 | 1,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-40 | latest | en | 0.616457 |
http://roompbkr.com/essays/Lab-Report-For-The-Limiting-Factors-1676002.html | 1,548,190,880,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583874494.65/warc/CC-MAIN-20190122202547-20190122224547-00547.warc.gz | 186,451,828 | 25,593 | # Lab Report for the Limiting Factors of the Cedar Glade
Topics: Population, Hunting, Pressure sensitive tape Pages: 4 (1358 words) Published: May 8, 2013
Lab Report for the Limiting Factors of the Cedar Glade
Mrs. Hall
Introduction:
In the wild, populations are normally controlled by limiting factors that are either biotic or abiotic. These prevent the population from continually growing and reaching their carrying capacity. This experiment will teach one what it is like to study real populations and to help one use and learn the meanings of certain words. Hypothesis:
Investigation 1: If both the male and female owl hunt both 2 times a day in the spring with favorable mouse conditions then, the owls will live and eat well almost every day because there are many mice, 2 owls, and no predators Investigation 2: If both the male and female both hunt twice a day with 25% (20 mice) of the population dead then, the owls will eat good for the first few days, but will slowly decrease until death in the 7th through 9th day, because in order to live one must get a total of 4 mice within 3 days and after one hunts for a few days, the mice will get scarcer and scarcer, until there are none left. Investigation 3: If the male and female both hunt 2 times a day with snakes as competitors which eat 6 mice a day then the owls will survive for 6 to 8 days then die off because with the owls eating mice and then snakes taking out 6 mice every day, that is 36 mice the snakes have eaten in just 6 days. Eventually, the mice will completely die off.
Materials:
The materials one will need are: masking tape, yard stick or tape measure, pipe cleaners bent into 10 x 10 cm squares to represent the owls, colored pencils, graph paper, large bag of pinto beans (or other dried beans) to represent mice, and a student data sheet. Methods:
Procedures for investigation 1: First one would lay down a 4ft. x 4ft. square with the masking tape, on the cleared floor. Next, form a group of four: Bean Counter, who would scatter beans between rounds, removes and adds beans. Recorder, who would read directions and record the data collected.... | 497 | 2,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-04 | latest | en | 0.951389 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/209/2/j/a/ | 1,696,366,375,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00877.warc.gz | 923,988,288 | 61,651 | # Properties
Label 209.2.j.a Level $209$ Weight $2$ Character orbit 209.j Analytic conductor $1.669$ Analytic rank $0$ Dimension $6$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [209,2,Mod(23,209)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(209, base_ring=CyclotomicField(18))
chi = DirichletCharacter(H, H._module([0, 2]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("209.23");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$209 = 11 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 209.j (of order $$9$$, degree $$6$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$1.66887340224$$ Analytic rank: $$0$$ Dimension: $$6$$ Coefficient field: $$\Q(\zeta_{18})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{6} - x^{3} + 1$$ x^6 - x^3 + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{9}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{18}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \zeta_{18}^{5} + \zeta_{18}^{4} - \zeta_{18}^{3} + \zeta_{18}^{2} - \zeta_{18} + 1) q^{2} + 2 \zeta_{18} q^{3} + ( - \zeta_{18}^{5} + \zeta_{18}^{4} + \zeta_{18}^{3} + \zeta_{18} - 2) q^{4} + (\zeta_{18}^{5} - 2 \zeta_{18}^{4} - \zeta_{18}^{2} + 1) q^{5} + (2 \zeta_{18}^{5} - 2 \zeta_{18}^{4} - 2 \zeta_{18}^{2} + 2 \zeta_{18} + 2) q^{6} + (\zeta_{18}^{5} + \zeta_{18}^{4} - 2 \zeta_{18}^{3} - \zeta_{18} + 2) q^{7} + (3 \zeta_{18}^{5} - 2 \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 3 \zeta_{18}) q^{8} + \zeta_{18}^{2} q^{9}+O(q^{10})$$ q + (-z^5 + z^4 - z^3 + z^2 - z + 1) * q^2 + 2*z * q^3 + (-z^5 + z^4 + z^3 + z - 2) * q^4 + (z^5 - 2*z^4 - z^2 + 1) * q^5 + (2*z^5 - 2*z^4 - 2*z^2 + 2*z + 2) * q^6 + (z^5 + z^4 - 2*z^3 - z + 2) * q^7 + (3*z^5 - 2*z^4 + 2*z^3 - 2*z^2 + 3*z) * q^8 + z^2 * q^9 $$q + ( - \zeta_{18}^{5} + \zeta_{18}^{4} - \zeta_{18}^{3} + \zeta_{18}^{2} - \zeta_{18} + 1) q^{2} + 2 \zeta_{18} q^{3} + ( - \zeta_{18}^{5} + \zeta_{18}^{4} + \zeta_{18}^{3} + \zeta_{18} - 2) q^{4} + (\zeta_{18}^{5} - 2 \zeta_{18}^{4} - \zeta_{18}^{2} + 1) q^{5} + (2 \zeta_{18}^{5} - 2 \zeta_{18}^{4} - 2 \zeta_{18}^{2} + 2 \zeta_{18} + 2) q^{6} + (\zeta_{18}^{5} + \zeta_{18}^{4} - 2 \zeta_{18}^{3} - \zeta_{18} + 2) q^{7} + (3 \zeta_{18}^{5} - 2 \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 3 \zeta_{18}) q^{8} + \zeta_{18}^{2} q^{9} + (2 \zeta_{18}^{4} - 4 \zeta_{18}^{3} + 3 \zeta_{18}^{2} - 4 \zeta_{18} + 2) q^{10} - \zeta_{18}^{3} q^{11} + (2 \zeta_{18}^{5} + 2 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + 2 \zeta_{18}^{2} - 4 \zeta_{18} + 2) q^{12} + ( - 2 \zeta_{18}^{5} - 3 \zeta_{18}^{4} + 2 \zeta_{18}^{3} + 2 \zeta_{18}^{2} + \zeta_{18} + 1) q^{13} + ( - 3 \zeta_{18}^{5} + 4 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + \zeta_{18}^{2} - 1) q^{14} + ( - 4 \zeta_{18}^{5} + 2 \zeta_{18} - 2) q^{15} + (3 \zeta_{18}^{5} - 3 \zeta_{18}^{3} - \zeta_{18} + 3) q^{16} + (\zeta_{18}^{5} + 5 \zeta_{18}^{4} + \zeta_{18}^{3} - \zeta_{18}^{2} - 5 \zeta_{18} - 1) q^{17} + ( - \zeta_{18}^{5} + \zeta_{18}^{2} + \zeta_{18} - 1) q^{18} + (\zeta_{18}^{5} + 2 \zeta_{18}^{4} - 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} - 4 \zeta_{18} + 2) q^{19} + ( - 7 \zeta_{18}^{5} + 4 \zeta_{18}^{4} + 3 \zeta_{18}^{2} + 3 \zeta_{18} - 5) q^{20} + (2 \zeta_{18}^{5} - 4 \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 4 \zeta_{18} - 2) q^{21} + ( - \zeta_{18}^{2} + \zeta_{18} - 1) q^{22} + (3 \zeta_{18}^{5} + \zeta_{18}^{4} + \zeta_{18}^{3} - 2 \zeta_{18} + 1) q^{23} + ( - 4 \zeta_{18}^{5} + 4 \zeta_{18}^{4} + 2 \zeta_{18}^{3} + 6 \zeta_{18}^{2} - 6) q^{24} + (\zeta_{18}^{5} - 5 \zeta_{18}^{4} + 4 \zeta_{18}^{3} - \zeta_{18}^{2} + \zeta_{18} + 1) q^{25} + ( - 2 \zeta_{18}^{5} - 2 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + 5 \zeta_{18}^{2} - 3 \zeta_{18} + 2) q^{26} - 4 \zeta_{18}^{3} q^{27} + ( - 4 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 5 \zeta_{18}^{2} + 6 \zeta_{18} - 4) q^{28} + ( - 2 \zeta_{18}^{3} - 4 \zeta_{18}^{2} - 2 \zeta_{18}) q^{29} + (4 \zeta_{18}^{5} - 8 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 8 \zeta_{18}^{2} + 4 \zeta_{18}) q^{30} + ( - 4 \zeta_{18}^{5} - 4 \zeta_{18}^{4} + \zeta_{18}^{3} + \zeta_{18}^{2} + 3 \zeta_{18} - 1) q^{31} + (3 \zeta_{18}^{4} - 3 \zeta_{18} - 3) q^{32} - 2 \zeta_{18}^{4} q^{33} + ( - 3 \zeta_{18}^{5} + 5 \zeta_{18}^{4} + 5 \zeta_{18}^{3} - \zeta_{18} - 4) q^{34} + (3 \zeta_{18}^{5} - 3 \zeta_{18}^{3} + 2 \zeta_{18}^{2} - 5 \zeta_{18} + 5) q^{35} + (\zeta_{18}^{5} - \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + \zeta_{18} - 1) q^{36} + ( - \zeta_{18}^{4} + \zeta_{18}^{2} + \zeta_{18} - 3) q^{37} + ( - 4 \zeta_{18}^{5} + 7 \zeta_{18}^{4} - \zeta_{18}^{3} + \zeta_{18}^{2} - 4 \zeta_{18} - 2) q^{38} + ( - 6 \zeta_{18}^{5} + 4 \zeta_{18}^{4} + 2 \zeta_{18}^{2} + 2 \zeta_{18} + 4) q^{39} + (\zeta_{18}^{5} - 9 \zeta_{18}^{4} + 8 \zeta_{18}^{3} - 8 \zeta_{18}^{2} + 9 \zeta_{18} - 1) q^{40} + ( - \zeta_{18}^{5} + \zeta_{18}^{3} - \zeta_{18}^{2} - 2) q^{41} + (8 \zeta_{18}^{5} - 4 \zeta_{18}^{4} - 4 \zeta_{18}^{3} - 2 \zeta_{18} + 6) q^{42} + ( - 3 \zeta_{18}^{5} - 6 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + \zeta_{18}^{2} - 1) q^{43} + (\zeta_{18}^{5} - 2 \zeta_{18}^{4} + \zeta_{18}^{3} - \zeta_{18}^{2} + \zeta_{18} + 1) q^{44} + ( - 2 \zeta_{18}^{3} + \zeta_{18}^{2} - \zeta_{18} + 2) q^{45} + ( - 3 \zeta_{18}^{5} + 6 \zeta_{18}^{4} - 3 \zeta_{18}^{3} + 6 \zeta_{18}^{2} - 3 \zeta_{18}) q^{46} + (3 \zeta_{18}^{4} + 3 \zeta_{18}^{3} - 3 \zeta_{18}^{2} + 3 \zeta_{18} + 3) q^{47} + ( - 6 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 6 \zeta_{18} - 6) q^{48} + ( - \zeta_{18}^{5} + 4 \zeta_{18}^{4} + \zeta_{18}^{3} + 4 \zeta_{18}^{2} - \zeta_{18}) q^{49} + (\zeta_{18}^{5} + \zeta_{18}^{4} - 7 \zeta_{18}^{3} + 9 \zeta_{18}^{2} - 10 \zeta_{18} + 7) q^{50} + (10 \zeta_{18}^{5} + 2 \zeta_{18}^{4} - 10 \zeta_{18}^{2} - 2 \zeta_{18} - 2) q^{51} + ( - 4 \zeta_{18}^{5} + 7 \zeta_{18}^{4} + 4 \zeta_{18}^{2} - 4) q^{52} + (3 \zeta_{18}^{5} + 6 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 6 \zeta_{18}) q^{53} + ( - 4 \zeta_{18}^{2} + 4 \zeta_{18} - 4) q^{54} + (2 \zeta_{18}^{4} - \zeta_{18}^{3} + \zeta_{18}^{2} - 2 \zeta_{18}) q^{55} + (9 \zeta_{18}^{5} - 8 \zeta_{18}^{4} - \zeta_{18}^{2} - \zeta_{18} + 5) q^{56} + (4 \zeta_{18}^{5} - 4 \zeta_{18}^{4} - 2 \zeta_{18}^{3} - 8 \zeta_{18}^{2} + 4 \zeta_{18} - 2) q^{57} + (2 \zeta_{18}^{5} + 2 \zeta_{18}^{4} - 4 \zeta_{18}^{2} - 4 \zeta_{18}) q^{58} + ( - 2 \zeta_{18}^{5} + \zeta_{18}^{4} - 6 \zeta_{18}^{3} + 6 \zeta_{18}^{2} - \zeta_{18} + 2) q^{59} + (8 \zeta_{18}^{5} - 8 \zeta_{18}^{3} + 6 \zeta_{18}^{2} - 10 \zeta_{18} + 14) q^{60} + (6 \zeta_{18}^{5} + \zeta_{18}^{4} + \zeta_{18}^{3} + 3 \zeta_{18} - 4) q^{61} + (3 \zeta_{18}^{5} - 8 \zeta_{18}^{4} + \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 2) q^{62} + ( - 2 \zeta_{18}^{5} + \zeta_{18}^{4} + 2 \zeta_{18}^{2} - \zeta_{18} - 1) q^{63} + (4 \zeta_{18}^{3} + 3 \zeta_{18}^{2} - 3 \zeta_{18} - 4) q^{64} + (3 \zeta_{18}^{5} - 7 \zeta_{18}^{4} + \zeta_{18}^{3} - 7 \zeta_{18}^{2} + 3 \zeta_{18}) q^{65} + ( - 2 \zeta_{18}^{3} + 2 \zeta_{18}^{2} - 2 \zeta_{18}) q^{66} + ( - \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 5 \zeta_{18}^{2} + 6 \zeta_{18} - 1) q^{67} + (3 \zeta_{18}^{5} - 8 \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 8 \zeta_{18}^{2} + 3 \zeta_{18}) q^{68} + (2 \zeta_{18}^{5} + 2 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 4 \zeta_{18}^{2} + 2 \zeta_{18} - 6) q^{69} + ( - 12 \zeta_{18}^{5} + 13 \zeta_{18}^{4} - 8 \zeta_{18}^{3} + 12 \zeta_{18}^{2} - 5 \zeta_{18} - 5) q^{70} + ( - 7 \zeta_{18}^{5} + 2 \zeta_{18}^{4} + 7 \zeta_{18}^{2} - 7) q^{71} + (2 \zeta_{18}^{5} + \zeta_{18}^{4} + \zeta_{18}^{3} - 3 \zeta_{18} + 2) q^{72} + (\zeta_{18}^{5} - \zeta_{18}^{3} - 3 \zeta_{18}^{2} + 11 \zeta_{18} - 2) q^{73} + (3 \zeta_{18}^{5} - 4 \zeta_{18}^{4} + 2 \zeta_{18}^{3} - 2 \zeta_{18}^{2} + 4 \zeta_{18} - 3) q^{74} + ( - 10 \zeta_{18}^{5} + 8 \zeta_{18}^{4} + 2 \zeta_{18}^{2} + 2 \zeta_{18} - 2) q^{75} + ( - 3 \zeta_{18}^{5} - 6 \zeta_{18}^{4} + 5 \zeta_{18}^{3} - 5 \zeta_{18}^{2} + 9 \zeta_{18} - 4) q^{76} + ( - \zeta_{18}^{5} + \zeta_{18}^{2} + \zeta_{18} - 2) q^{77} + ( - 4 \zeta_{18}^{5} - 4 \zeta_{18}^{4} + 6 \zeta_{18}^{3} - 6 \zeta_{18}^{2} + 4 \zeta_{18} + 4) q^{78} + ( - 3 \zeta_{18}^{5} + 3 \zeta_{18}^{3} + 4 \zeta_{18} - 3) q^{79} + (8 \zeta_{18}^{5} - 3 \zeta_{18}^{4} - 3 \zeta_{18}^{3} - 7 \zeta_{18} + 10) q^{80} - 11 \zeta_{18}^{4} q^{81} + (3 \zeta_{18}^{5} - 3 \zeta_{18}^{4} + 3 \zeta_{18}^{3} - 3 \zeta_{18}^{2}) q^{82} + (4 \zeta_{18}^{5} + 4 \zeta_{18}^{4} - 2 \zeta_{18}^{3} - 7 \zeta_{18}^{2} + 3 \zeta_{18} + 2) q^{83} + ( - 8 \zeta_{18}^{5} + 12 \zeta_{18}^{4} - 10 \zeta_{18}^{3} + 12 \zeta_{18}^{2} - 8 \zeta_{18}) q^{84} + (4 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + 9 \zeta_{18}^{2} - 2 \zeta_{18} + 4) q^{85} + ( - 4 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + \zeta_{18}^{2} - 2 \zeta_{18} - 4) q^{86} + ( - 4 \zeta_{18}^{4} - 8 \zeta_{18}^{3} - 4 \zeta_{18}^{2}) q^{87} + ( - \zeta_{18}^{5} - \zeta_{18}^{4} - 2 \zeta_{18}^{3} + 3 \zeta_{18}^{2} - 2 \zeta_{18} + 2) q^{88} + ( - \zeta_{18}^{5} - \zeta_{18}^{4} - 5 \zeta_{18}^{3} + \zeta_{18}^{2} + 6 \zeta_{18} + 6) q^{89} + ( - 4 \zeta_{18}^{5} + 3 \zeta_{18}^{4} - 2 \zeta_{18}^{3} + 2 \zeta_{18}^{2} - 2) q^{90} + (\zeta_{18}^{4} + \zeta_{18}^{3} - 7 \zeta_{18} + 6) q^{91} + ( - 5 \zeta_{18}^{5} + 5 \zeta_{18}^{3} - 5 \zeta_{18}^{2} + 6 \zeta_{18} - 10) q^{92} + ( - 8 \zeta_{18}^{5} + 2 \zeta_{18}^{4} - 6 \zeta_{18}^{3} + 6 \zeta_{18}^{2} - 2 \zeta_{18} + 8) q^{93} + (3 \zeta_{18}^{5} - 3 \zeta_{18}^{2} - 3 \zeta_{18} + 12) q^{94} + (7 \zeta_{18}^{5} + \zeta_{18}^{4} + 2 \zeta_{18}^{2} - 6 \zeta_{18} + 4) q^{95} + (6 \zeta_{18}^{5} - 6 \zeta_{18}^{2} - 6 \zeta_{18}) q^{96} + ( - 3 \zeta_{18}^{5} - 2 \zeta_{18}^{4} + 3 \zeta_{18}^{3} - 3 \zeta_{18}^{2} + 2 \zeta_{18} + 3) q^{97} + ( - 5 \zeta_{18}^{5} + 5 \zeta_{18}^{3} + \zeta_{18}^{2} + 6 \zeta_{18} - 4) q^{98} - \zeta_{18}^{5} q^{99} +O(q^{100})$$ q + (-z^5 + z^4 - z^3 + z^2 - z + 1) * q^2 + 2*z * q^3 + (-z^5 + z^4 + z^3 + z - 2) * q^4 + (z^5 - 2*z^4 - z^2 + 1) * q^5 + (2*z^5 - 2*z^4 - 2*z^2 + 2*z + 2) * q^6 + (z^5 + z^4 - 2*z^3 - z + 2) * q^7 + (3*z^5 - 2*z^4 + 2*z^3 - 2*z^2 + 3*z) * q^8 + z^2 * q^9 + (2*z^4 - 4*z^3 + 3*z^2 - 4*z + 2) * q^10 - z^3 * q^11 + (2*z^5 + 2*z^4 - 2*z^3 + 2*z^2 - 4*z + 2) * q^12 + (-2*z^5 - 3*z^4 + 2*z^3 + 2*z^2 + z + 1) * q^13 + (-3*z^5 + 4*z^4 - 2*z^3 + z^2 - 1) * q^14 + (-4*z^5 + 2*z - 2) * q^15 + (3*z^5 - 3*z^3 - z + 3) * q^16 + (z^5 + 5*z^4 + z^3 - z^2 - 5*z - 1) * q^17 + (-z^5 + z^2 + z - 1) * q^18 + (z^5 + 2*z^4 - 2*z^3 - 2*z^2 - 4*z + 2) * q^19 + (-7*z^5 + 4*z^4 + 3*z^2 + 3*z - 5) * q^20 + (2*z^5 - 4*z^4 + 2*z^3 - 2*z^2 + 4*z - 2) * q^21 + (-z^2 + z - 1) * q^22 + (3*z^5 + z^4 + z^3 - 2*z + 1) * q^23 + (-4*z^5 + 4*z^4 + 2*z^3 + 6*z^2 - 6) * q^24 + (z^5 - 5*z^4 + 4*z^3 - z^2 + z + 1) * q^25 + (-2*z^5 - 2*z^4 - 2*z^3 + 5*z^2 - 3*z + 2) * q^26 - 4*z^3 * q^27 + (-4*z^4 + 6*z^3 - 5*z^2 + 6*z - 4) * q^28 + (-2*z^3 - 4*z^2 - 2*z) * q^29 + (4*z^5 - 8*z^4 + 6*z^3 - 8*z^2 + 4*z) * q^30 + (-4*z^5 - 4*z^4 + z^3 + z^2 + 3*z - 1) * q^31 + (3*z^4 - 3*z - 3) * q^32 - 2*z^4 * q^33 + (-3*z^5 + 5*z^4 + 5*z^3 - z - 4) * q^34 + (3*z^5 - 3*z^3 + 2*z^2 - 5*z + 5) * q^35 + (z^5 - z^4 + 2*z^3 - 2*z^2 + z - 1) * q^36 + (-z^4 + z^2 + z - 3) * q^37 + (-4*z^5 + 7*z^4 - z^3 + z^2 - 4*z - 2) * q^38 + (-6*z^5 + 4*z^4 + 2*z^2 + 2*z + 4) * q^39 + (z^5 - 9*z^4 + 8*z^3 - 8*z^2 + 9*z - 1) * q^40 + (-z^5 + z^3 - z^2 - 2) * q^41 + (8*z^5 - 4*z^4 - 4*z^3 - 2*z + 6) * q^42 + (-3*z^5 - 6*z^4 - 2*z^3 + z^2 - 1) * q^43 + (z^5 - 2*z^4 + z^3 - z^2 + z + 1) * q^44 + (-2*z^3 + z^2 - z + 2) * q^45 + (-3*z^5 + 6*z^4 - 3*z^3 + 6*z^2 - 3*z) * q^46 + (3*z^4 + 3*z^3 - 3*z^2 + 3*z + 3) * q^47 + (-6*z^4 + 6*z^3 - 2*z^2 + 6*z - 6) * q^48 + (-z^5 + 4*z^4 + z^3 + 4*z^2 - z) * q^49 + (z^5 + z^4 - 7*z^3 + 9*z^2 - 10*z + 7) * q^50 + (10*z^5 + 2*z^4 - 10*z^2 - 2*z - 2) * q^51 + (-4*z^5 + 7*z^4 + 4*z^2 - 4) * q^52 + (3*z^5 + 6*z^4 + 6*z^3 - 6*z) * q^53 + (-4*z^2 + 4*z - 4) * q^54 + (2*z^4 - z^3 + z^2 - 2*z) * q^55 + (9*z^5 - 8*z^4 - z^2 - z + 5) * q^56 + (4*z^5 - 4*z^4 - 2*z^3 - 8*z^2 + 4*z - 2) * q^57 + (2*z^5 + 2*z^4 - 4*z^2 - 4*z) * q^58 + (-2*z^5 + z^4 - 6*z^3 + 6*z^2 - z + 2) * q^59 + (8*z^5 - 8*z^3 + 6*z^2 - 10*z + 14) * q^60 + (6*z^5 + z^4 + z^3 + 3*z - 4) * q^61 + (3*z^5 - 8*z^4 + z^3 - 2*z^2 + 2) * q^62 + (-2*z^5 + z^4 + 2*z^2 - z - 1) * q^63 + (4*z^3 + 3*z^2 - 3*z - 4) * q^64 + (3*z^5 - 7*z^4 + z^3 - 7*z^2 + 3*z) * q^65 + (-2*z^3 + 2*z^2 - 2*z) * q^66 + (-z^4 + 6*z^3 - 5*z^2 + 6*z - 1) * q^67 + (3*z^5 - 8*z^4 + 2*z^3 - 8*z^2 + 3*z) * q^68 + (2*z^5 + 2*z^4 + 6*z^3 - 4*z^2 + 2*z - 6) * q^69 + (-12*z^5 + 13*z^4 - 8*z^3 + 12*z^2 - 5*z - 5) * q^70 + (-7*z^5 + 2*z^4 + 7*z^2 - 7) * q^71 + (2*z^5 + z^4 + z^3 - 3*z + 2) * q^72 + (z^5 - z^3 - 3*z^2 + 11*z - 2) * q^73 + (3*z^5 - 4*z^4 + 2*z^3 - 2*z^2 + 4*z - 3) * q^74 + (-10*z^5 + 8*z^4 + 2*z^2 + 2*z - 2) * q^75 + (-3*z^5 - 6*z^4 + 5*z^3 - 5*z^2 + 9*z - 4) * q^76 + (-z^5 + z^2 + z - 2) * q^77 + (-4*z^5 - 4*z^4 + 6*z^3 - 6*z^2 + 4*z + 4) * q^78 + (-3*z^5 + 3*z^3 + 4*z - 3) * q^79 + (8*z^5 - 3*z^4 - 3*z^3 - 7*z + 10) * q^80 - 11*z^4 * q^81 + (3*z^5 - 3*z^4 + 3*z^3 - 3*z^2) * q^82 + (4*z^5 + 4*z^4 - 2*z^3 - 7*z^2 + 3*z + 2) * q^83 + (-8*z^5 + 12*z^4 - 10*z^3 + 12*z^2 - 8*z) * q^84 + (4*z^4 - 2*z^3 + 9*z^2 - 2*z + 4) * q^85 + (-4*z^4 - 2*z^3 + z^2 - 2*z - 4) * q^86 + (-4*z^4 - 8*z^3 - 4*z^2) * q^87 + (-z^5 - z^4 - 2*z^3 + 3*z^2 - 2*z + 2) * q^88 + (-z^5 - z^4 - 5*z^3 + z^2 + 6*z + 6) * q^89 + (-4*z^5 + 3*z^4 - 2*z^3 + 2*z^2 - 2) * q^90 + (z^4 + z^3 - 7*z + 6) * q^91 + (-5*z^5 + 5*z^3 - 5*z^2 + 6*z - 10) * q^92 + (-8*z^5 + 2*z^4 - 6*z^3 + 6*z^2 - 2*z + 8) * q^93 + (3*z^5 - 3*z^2 - 3*z + 12) * q^94 + (7*z^5 + z^4 + 2*z^2 - 6*z + 4) * q^95 + (6*z^5 - 6*z^2 - 6*z) * q^96 + (-3*z^5 - 2*z^4 + 3*z^3 - 3*z^2 + 2*z + 3) * q^97 + (-5*z^5 + 5*z^3 + z^2 + 6*z - 4) * q^98 - z^5 * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6 q + 3 q^{2} - 9 q^{4} + 6 q^{5} + 12 q^{6} + 6 q^{7} + 6 q^{8}+O(q^{10})$$ 6 * q + 3 * q^2 - 9 * q^4 + 6 * q^5 + 12 * q^6 + 6 * q^7 + 6 * q^8 $$6 q + 3 q^{2} - 9 q^{4} + 6 q^{5} + 12 q^{6} + 6 q^{7} + 6 q^{8} - 3 q^{11} + 6 q^{12} + 12 q^{13} - 12 q^{14} - 12 q^{15} + 9 q^{16} - 3 q^{17} - 6 q^{18} + 6 q^{19} - 30 q^{20} - 6 q^{21} - 6 q^{22} + 9 q^{23} - 30 q^{24} + 18 q^{25} + 6 q^{26} - 12 q^{27} - 6 q^{28} - 6 q^{29} + 18 q^{30} - 3 q^{31} - 18 q^{32} - 9 q^{34} + 21 q^{35} - 18 q^{37} - 15 q^{38} + 24 q^{39} + 18 q^{40} - 9 q^{41} + 24 q^{42} - 12 q^{43} + 9 q^{44} + 6 q^{45} - 9 q^{46} + 27 q^{47} - 18 q^{48} + 3 q^{49} + 21 q^{50} - 12 q^{51} - 24 q^{52} + 18 q^{53} - 24 q^{54} - 3 q^{55} + 30 q^{56} - 18 q^{57} - 6 q^{59} + 60 q^{60} - 21 q^{61} + 15 q^{62} - 6 q^{63} - 12 q^{64} + 3 q^{65} - 6 q^{66} + 12 q^{67} + 6 q^{68} - 18 q^{69} - 54 q^{70} - 42 q^{71} + 15 q^{72} - 15 q^{73} - 12 q^{74} - 12 q^{75} - 9 q^{76} - 12 q^{77} + 42 q^{78} - 9 q^{79} + 51 q^{80} + 9 q^{82} + 6 q^{83} - 30 q^{84} + 18 q^{85} - 30 q^{86} - 24 q^{87} + 6 q^{88} + 21 q^{89} - 18 q^{90} + 39 q^{91} - 45 q^{92} + 30 q^{93} + 72 q^{94} + 24 q^{95} + 27 q^{97} - 9 q^{98}+O(q^{100})$$ 6 * q + 3 * q^2 - 9 * q^4 + 6 * q^5 + 12 * q^6 + 6 * q^7 + 6 * q^8 - 3 * q^11 + 6 * q^12 + 12 * q^13 - 12 * q^14 - 12 * q^15 + 9 * q^16 - 3 * q^17 - 6 * q^18 + 6 * q^19 - 30 * q^20 - 6 * q^21 - 6 * q^22 + 9 * q^23 - 30 * q^24 + 18 * q^25 + 6 * q^26 - 12 * q^27 - 6 * q^28 - 6 * q^29 + 18 * q^30 - 3 * q^31 - 18 * q^32 - 9 * q^34 + 21 * q^35 - 18 * q^37 - 15 * q^38 + 24 * q^39 + 18 * q^40 - 9 * q^41 + 24 * q^42 - 12 * q^43 + 9 * q^44 + 6 * q^45 - 9 * q^46 + 27 * q^47 - 18 * q^48 + 3 * q^49 + 21 * q^50 - 12 * q^51 - 24 * q^52 + 18 * q^53 - 24 * q^54 - 3 * q^55 + 30 * q^56 - 18 * q^57 - 6 * q^59 + 60 * q^60 - 21 * q^61 + 15 * q^62 - 6 * q^63 - 12 * q^64 + 3 * q^65 - 6 * q^66 + 12 * q^67 + 6 * q^68 - 18 * q^69 - 54 * q^70 - 42 * q^71 + 15 * q^72 - 15 * q^73 - 12 * q^74 - 12 * q^75 - 9 * q^76 - 12 * q^77 + 42 * q^78 - 9 * q^79 + 51 * q^80 + 9 * q^82 + 6 * q^83 - 30 * q^84 + 18 * q^85 - 30 * q^86 - 24 * q^87 + 6 * q^88 + 21 * q^89 - 18 * q^90 + 39 * q^91 - 45 * q^92 + 30 * q^93 + 72 * q^94 + 24 * q^95 + 27 * q^97 - 9 * q^98
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/209\mathbb{Z}\right)^\times$$.
$$n$$ $$78$$ $$134$$ $$\chi(n)$$ $$-\zeta_{18}^{5}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
23.1
−0.173648 − 0.984808i −0.173648 + 0.984808i 0.939693 − 0.342020i 0.939693 + 0.342020i −0.766044 − 0.642788i −0.766044 + 0.642788i
1.26604 + 0.460802i −0.347296 1.96962i −0.141559 0.118782i −0.358441 + 0.300767i 0.467911 2.65366i 1.17365 2.03282i −1.47178 2.54920i −0.939693 + 0.342020i −0.592396 + 0.215615i
100.1 1.26604 0.460802i −0.347296 + 1.96962i −0.141559 + 0.118782i −0.358441 0.300767i 0.467911 + 2.65366i 1.17365 + 2.03282i −1.47178 + 2.54920i −0.939693 0.342020i −0.592396 0.215615i
111.1 0.673648 + 0.565258i 1.87939 0.684040i −0.213011 1.20805i −0.286989 + 1.62760i 1.65270 + 0.601535i 0.0603074 + 0.104455i 1.41875 2.45734i 0.766044 0.642788i −1.11334 + 0.934204i
177.1 0.673648 0.565258i 1.87939 + 0.684040i −0.213011 + 1.20805i −0.286989 1.62760i 1.65270 0.601535i 0.0603074 0.104455i 1.41875 + 2.45734i 0.766044 + 0.642788i −1.11334 0.934204i
188.1 −0.439693 + 2.49362i −1.53209 1.28558i −4.14543 1.50881i 3.64543 1.32683i 3.87939 3.25519i 1.76604 + 3.05888i 3.05303 5.28801i 0.173648 + 0.984808i 1.70574 + 9.67372i
199.1 −0.439693 2.49362i −1.53209 + 1.28558i −4.14543 + 1.50881i 3.64543 + 1.32683i 3.87939 + 3.25519i 1.76604 3.05888i 3.05303 + 5.28801i 0.173648 0.984808i 1.70574 9.67372i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 23.1 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
19.e even 9 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 209.2.j.a 6
19.e even 9 1 inner 209.2.j.a 6
19.e even 9 1 3971.2.a.e 3
19.f odd 18 1 3971.2.a.f 3
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
209.2.j.a 6 1.a even 1 1 trivial
209.2.j.a 6 19.e even 9 1 inner
3971.2.a.e 3 19.e even 9 1
3971.2.a.f 3 19.f odd 18 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{6} - 3T_{2}^{5} + 9T_{2}^{4} - 24T_{2}^{3} + 36T_{2}^{2} - 27T_{2} + 9$$ acting on $$S_{2}^{\mathrm{new}}(209, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{6} - 3 T^{5} + 9 T^{4} - 24 T^{3} + \cdots + 9$$
$3$ $$T^{6} - 8T^{3} + 64$$
$5$ $$T^{6} - 6 T^{5} + 9 T^{4} - 3 T^{3} + \cdots + 9$$
$7$ $$T^{6} - 6 T^{5} + 27 T^{4} - 52 T^{3} + \cdots + 1$$
$11$ $$(T^{2} + T + 1)^{3}$$
$13$ $$T^{6} - 12 T^{5} + 75 T^{4} + \cdots + 5041$$
$17$ $$T^{6} + 3 T^{5} - 9 T^{4} + \cdots + 12321$$
$19$ $$T^{6} - 6 T^{5} - 12 T^{4} + \cdots + 6859$$
$23$ $$T^{6} - 9 T^{5} + 36 T^{4} - 153 T^{3} + \cdots + 81$$
$29$ $$T^{6} + 6 T^{5} - 240 T^{3} + \cdots + 576$$
$31$ $$T^{6} + 3 T^{5} + 45 T^{4} + \cdots + 16129$$
$37$ $$(T^{3} + 9 T^{2} + 24 T + 19)^{2}$$
$41$ $$T^{6} + 9 T^{5} + 36 T^{4} + 72 T^{3} + \cdots + 81$$
$43$ $$T^{6} + 12 T^{5} + 159 T^{4} + \cdots + 5329$$
$47$ $$T^{6} - 27 T^{5} + 324 T^{4} + \cdots + 210681$$
$53$ $$T^{6} - 18 T^{5} + 270 T^{4} + \cdots + 263169$$
$59$ $$T^{6} + 6 T^{5} + 126 T^{4} + \cdots + 45369$$
$61$ $$T^{6} + 21 T^{5} + 276 T^{4} + \cdots + 11449$$
$67$ $$T^{6} - 12 T^{5} + 246 T^{4} + \cdots + 128881$$
$71$ $$T^{6} + 42 T^{5} + 693 T^{4} + \cdots + 3249$$
$73$ $$T^{6} + 15 T^{5} + 228 T^{4} + \cdots + 1129969$$
$79$ $$T^{6} + 9 T^{5} + 18 T^{4} - 388 T^{3} + \cdots + 289$$
$83$ $$T^{6} - 6 T^{5} + 135 T^{4} + \cdots + 47961$$
$89$ $$T^{6} - 21 T^{5} + 207 T^{4} + \cdots + 12321$$
$97$ $$T^{6} - 27 T^{5} + 378 T^{4} + \cdots + 157609$$ | 12,151 | 20,786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.521652 |
rocketguy101.jimdofree.com | 1,695,950,838,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00554.warc.gz | 531,695,177 | 10,657 | Ogive Nose Cones
Since geocities closed down, the image files with all the equations are gone, so I created a pdf file showing the equations.
Ogive Nose Cones.pdf
Ogive Nose Cones
The ogive nose cone is probably one of the most common shapes used in model rocketry. It exhibits very good drag characteristics for general model rocketry use. If you are building a scale Tomahawk, Sandhawk, or Phoenix, then you will be dealing with an ogive. Ogives are classified by their "caliber”, or length to diameter ratio. Thus a nose that is one inch diameter and three inches long is a “3-to-1” (3:1) ogive and has a caliber of 3.
Once, while trying to build a Nike-Tomahawk, I tried to find the formula to draw a 3:1 ogive so that I could turn the cone on a lathe. I looked though Pete Alway’s The Art of Scale Model Rocketry, engineering and math textbooks, the CRC math handbook, old rocketry magazines, and the Internet. I could not find the formulas.
I decided to derive the equations myself. Step 1: What the heck is an ogive anyway? Webster’s Ninth Collegiate Dictionary defines “ogive” as “\‘oh-,jiv\ n 1. a: a diagonal arch or rib across a Gothic vault b: a pointed arch 2: a graph each of whose ordinates represents the sum of all the frequencies up to and including a corresponding frequency in a frequency distribution”.
That didn't help much! Searching math and statistics sites turned up an explanation of the term "ogive" which can be found here at MathWorld, or here. They weren't much help in figuring out what an ogive nose cone is.
Further searching turned up an old NAR Technical Report #8: "Ogive’ me a ring-tailed cylindrical bird says G. Harry Stine", and a description of ogive nose cones is given as follows:
"There are two basic types of ogival nose shapes: the tangent ogive and the secant ogive. A tangent ogive has an arc which meets the body contour smoothly, thereby creating no break in line where the ogive joins the cylindrical body. In other words, the center of rotation of the arc is in the plane of the base of the nose. If the center of rotation of the arc is aft of the plane of the base of the nose, you’ve got a secant ogive. Simple, no?"
This has been the only definition of a secant ogive I could find. The following diagrams illustrate what G. Harry was saying:
Figure 1. Figure 2.
Using Figure 1 and Figure 3, along with some mathematical maneuvering, you can derive equation (1a). As my math professors in college were so fond of saying, "the solution to that exercise will be left to the student to complete…" (or see Appendix A). So now we know what an ogive curve is. (as related to rocket nose cones)
The following methods describe ways to calculate the centers and radii of the tangent, and secant ogive curves.
Tangent Ogive (numerical method)
The formula to generate a tangent ogive cone is:
(1a)
where x, y are coordinates, x being along the length of the cone, and y being the "height" (or radius) of the cone taken from the centerline of the cone.
The caliber of the cone is:
(2)
Where:
L = cone length
d = cone base dia
A simpler approximation formula can also be used:
(1b) (This parabolic equation produces a slightly “flatter” curve than the arc equation.)
Note these equations will form 1/2 of the cone, you will need to mirror the plot to make the complete cone.
Equations 1a or 1b will provide x,y coordinates to machine a cone either manually, or by a numerical controlled (NC) lathe.
Tangent Ogive (graphical method, layout Figure 3)
This method was originally described in the Estes Technical Report TR-11: Aerodynamic Drag of Model Rockets, by Dr. Gerald Gregorek. In that report, the offset K and radius R were given as constants for 3:1 and 2:1 ogive contours. To obtain similar relations for any caliber of ogive curve, see Appendix A:
Figure 3. Vertical offset = (3) Radius = (4) Caliber = (5)
Secant Ogive (Derived from existing design, such as a scale rocket blueprint)
This is more involved to create than a tangent ogive. You have to find the center of rotation of the arc and determine how far aft it is from the plane of the base of the nose (xoff and yoff in the figure below). Then you will basically have a tangent ogive at an imaginary diameter, d2 (see layout, Figure 4).
1. Draw a line from the point of the nose to the intersection of the body tube (see close up Figure 5, below). This is called a chord, or a secant (imagine that!).
2. Find the center of the chordal length ( c ) and draw a perpendicular line (90 degrees) that crosses the curve of the nose cone.
3. Measure the perpendicular distance ( h ), and calculate the radius of the curve using the following formula:
(6)
4. Now calculate the angle between the secant and the centerline of the nose ( b ), and the angle between the perpendicular and the radial line running from the nose tip to the center of the arc ( a/2 ):
(7)
(8)
Figure 4.
Figure 5.
5. Now calculate xoff and yoff to determine where the center of the arc is in relation to the base of the nose, and the centerline of the body:
(9)
(10)
6. At this point, you can draw the secant ogive cone using a compass. If you want to determine the true tangent ogive diameter, use the following:
(11)
7. Now if you want to calculate x and y coordinates for a CAD program or NC lathe, use the tangent ogive equations from above, starting at x = xoff and use d2 for d.
Virginia Tech has a website on aerodynamics called "Configuration Aerodynamics Course" that has a pdf file called "Geometry for Aerodynamicists". This document contains formulas for other nose shapes, such as the Von Karmen ogive. It is 79 KB, and is available at http://www.dept.aoe.vt.edu/~mason/Mason_f/ConfigAeroAppA.pdf .
Another source of equations for other nose cone shape can be found at the VCP website. or from Wikipedia
David Stribling
NAR 18402 Sr.
# References
1. Ogive’ me a ring-tailed cylindrical bird says G. Harry Stine; Stine, G.H.; NAR Technical Report #8
2. Aerodynamic Drag of Model Rockets; Gregorek, Gerald; Estes Industries TR-11, 1970.
Derivation of the Ogive Equation
The equation of a circle is
or,
@ x=0, , or
@ x=L, Ybar = K,
Since R = R0 = RL
Square both sides and cancel terms:
becomes
Substitute caliber or L = C x d
and K becomes:
Therefore
Substitute K back into the equation for R0
Referring back to the figure, y = Ybar – K
substitute K and R from above,
[Home to Stribling's Main] [Strib's Rocketry Page] [More Rocket Stuff] [DropUsALine] | 1,670 | 6,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-40 | latest | en | 0.912368 |
https://www.bartleby.com/questions-and-answers/o-trigonometric-functions-trigonometric-functions-and-special-angles-problem-type-3-find-the-exact-v/91764136-c2bd-4fa4-b7c8-7fd8272a618e | 1,579,887,422,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00355.warc.gz | 787,882,067 | 21,056 | # O TRIGONOMETRIC FUNCTIONSTrigonometric functions and special angles: Problem type 3Find the exact values below. If applicable, click on "Undefined."JTcot 630°Undefined9п?sec4
Question
1 views
See attached
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Step 1
Part (a)
Every trigonometric function has a definite periodicity of 360 degrees.
Cot 630° = Cot (360° + 270°) = Cot 270° = 1 / tan 270...
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Tagged in | 170 | 648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | latest | en | 0.796233 |
https://www.scirra.com/forum/how-do-you-turn-seconds-into-minutes-and-seconds_t57054 | 1,484,719,754,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00028-ip-10-171-10-70.ec2.internal.warc.gz | 993,780,631 | 11,212 | how do you turn seconds into minutes and seconds
» Wed Nov 04, 2009 5:21 pm
Hello everyone,
I am making a timer for a game, but at the moment it can only be in seconds (i.e. 100 seconds). Does anyone know how I can turn 100 seconds into 1 min 40 secs. Is this possible using events or will I have to use a piece of python code?
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» Wed Nov 04, 2009 6:18 pm
It's easily possible in events! You just need a little math. If seconds is your number in seconds, floor(seconds / 60) is the minutes (divide by 60 and round down), and seconds % 60 is the seconds counter (modulo 60, ie. remainder after dividing by 60).
Scirra Founder
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» Wed Nov 04, 2009 7:03 pm
I made a example for you.
Press the A key to start the timer.
It is ascending up to infinite or in this case all the number reach 0:00 and count up again.
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» Thu Nov 05, 2009 1:09 pm
thanks toralord and Ashley,
Everybody states (quite rightly) how good construct is, and for me as a newbe to gamemaking it is the only platform I would attempt to write the game i am making at the moment. But the other great thing about construct is how supportive everyone is. I have asked three questions so far and I have never had to wait more than two hours for an answer. People like me are really grateful for the help!
thanks again.
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» Tue Nov 24, 2009 7:05 am
[quote="Ashley":1ds9taky]It's easily possible in events! You just need a little math. If seconds is your number in seconds, floor(seconds / 60) is the minutes (divide by 60 and round down), and seconds % 60 is the seconds counter (modulo 60, ie. remainder after dividing by 60).[/quote:1ds9taky]
what if you have above 59 minutes and you want hours too?
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» Tue Nov 24, 2009 9:41 am
[quote="alspal":2unqdj5p][quote="Ashley":2unqdj5p]It's easily possible in events! You just need a little math. If seconds is your number in seconds, floor(seconds / 60) is the minutes (divide by 60 and round down), and seconds % 60 is the seconds counter (modulo 60, ie. remainder after dividing by 60).[/quote:2unqdj5p]
what if you have above 59 minutes and you want hours too?[/quote:2unqdj5p]
hours: floor(seconds / 3600) (or: seconds / 60 / 60)
then use the result of seconds modulo 3600 for the rest of the calculations as seen above
totalseconds = 3980
hours = floor(totalseconds / 3600) = 1
totalseconds = totalseconds % 3600 = 380
minutes = floor(totalseconds / 60) = 6
seconds = totalseconds % 60 = 20
"01:06:20"
yeah, nice
EDIT: I forgot, in case someone asks for days too, just start with floor-dividing by 86400 (=3600 * 24) and use totalseconds % 86400 for the rest of the calculations.
Now try to find yourself the right value for weeks
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» Tue Nov 24, 2009 11:32 am
Cheers!
Also, I'd never noticed that zeropad system expression before - that's really useful!
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Reputation: 19,727 | 969 | 3,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-04 | longest | en | 0.901799 |
https://thescholtes.com/2019/02/25/fifty-four/ | 1,685,571,864,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00170.warc.gz | 623,459,598 | 24,780 | # Fifty-Four
A regular deck of cards, as we saw a few days ago, has fifty-two cards. However, most such decks also come with two extra cards: the Jokers. (This makes each deck hold 54 cards, obviously.) Jokers can be used in many games to keep things interesting. Some games use them as “wild cards,” cards you can use in any situation, in place of any other card. This adds a level of surprise and uncertainty. In poker, you may think that your straight is unbeatable, until you find out your opponent has five aces, thanks to wild cards! Maybe I should start playing Solitaire with jokers – maybe I’d finally win once in a while. It reminds me of the “green spaces” on a roulette wheel. If I understand the odds and the payouts of roulette properly, if there were no green spaces (0 and 00), then every bet in roulette would be perfectly fair; if you bet on red (50% of the spaces), you’d receive a payout of 1:2. If you bet on even, same thing. If you bet on a particular number, the payout would still match the odds. But the green spaces throw a wrinkle into that, making the odds no longer quite match the payout. Of course, in this case, the primary reason is to make sure the house comes out ahead. (And of course they should; there’s a lot of overhead in a casino!)
What’s interesting is that there are wild cards in the real world as well. Chaos theory is all about such wild cards; in fact, it states that there is nothing other than wild cards in the world. Chaos is the theory in physics from which the “butterfly effect” comes. Perhaps you’ve heard of that. The general idea is that a butterfly flapping its wings in India can create a hurricane in the Atlantic. The butterfly could flap its wings in a slightly different way than expected, which creates a tiny difference in the local weather, which bumps the jet stream just a fraction of a millimeter south, which changes the rainfall somewhere, which so on and so forth, eventually causes a hurricane. This is one of the reasons why weather forecasts might be good in the short run, say two or three days, but become nearly impossible past a week or so. There are just too many variables, too many possibilities to account for. Too many wild cards flapping their gossamer wings.
And you might think that this is just a matter of getting more precise measuring equipment, developing faster and more powerful computers to analyze more data. But — no. Now, perhaps there are some major improvements in the future in weather forecasting, but there will always, always be a limit, and that limit is predicted by quantum physics. Because quantum physics — the physics of the very, very small — shows that the universe as we know it is actually based on probabilities. Prior to the discovery of quantum physics, one could theoretically find out the location and velocity of every single particle in the universe, and with that knowledge, correctly predict every moment of both the past and the future. But that is now understood as impossible, because at the scale of the very small, it is impossible to know with certainty where a particle is, or what it’s doing. We can only measure things in terms of probabilities. There is a 95% chance that a particle will do such and such, or perhaps even a 99.9999999% chance of it. But it can never be 100%. And that means that wild cards are built into the very fabric of our universe. There is always a chance that something unexpected will happen. (Quantum physics even predicts that there is a chance that particles will spontaneously “tunnel” from one location to another, even thousands of miles away. So, it may be 99.9999999999999999999999999999999999999999999% likely that my entire body will suddenly “jump” to Tibet, but it’s not 100%, and it never can be.
Those jokers are everywhere. Steve Miller was right. | 852 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.963473 |
http://lumberjocks.com/StLouisWoodworker/blog | 1,545,153,363,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829542.89/warc/CC-MAIN-20181218164121-20181218190121-00029.warc.gz | 165,379,165 | 9,915 | # StLouisWoodworker's Blog
This blog is written by StLouisWoodworker | 7 entries so far
## laying out Steve Latta's lotus flower
04-19-2017 06:59 PM by | 2 comments »
I wrote a short note based on my experience in trying to make Steve Latta’s lotus flower as a string inlay project. In the note, I discuss one of the first steps in the project—locating 10 equally spaced points on a circle where the tips of the leaves are to be placed. The note can be read by clicking here.
## Older Entries
• ### an application of calculus to string inlay - 01-21-2017 09:14 PM | 2 comments
If you’re more than a little nerdy, you might like to learn how I used a bit of calculus in...
• ### converting between wire gauge and diameter/thickness - 06-30-2015 09:07 PM | 0 comments
I came across an article written by furniture maker Jane Swanson when reading about inlaying silv...
• ### The math of compound-angle joinery - 09-07-2014 06:09 PM | 3 comments
My friend, Bill Gottesman, and I recently finished writing a note about compound-angle joinery. ...
• ### using a table saw to make objects having polygonal shapes - 04-28-2012 10:09 PM | 2 comments
I posted a blog on my website that describes cutting polygons on a table saw and making objects t...
• ### How to cut hexagons on a table saw - 07-10-2011 02:59 AM | 3 comments
I wrote a note in Adobe Acrobat (.pdf) format explaining a way to cut hexagons on a table saw. I...
• ### how to cut compound miters on a table saw - 07-10-2011 02:51 AM | 1 comment
I wrote a note in Adobe Acrobat format about how to cut compound miters on a table saw. It is to...
DISCLAIMER: Any posts on LJ are posted by individuals acting in their own right and do not necessarily reflect the views of LJ. LJ will not be held liable for the actions of any user.
Contemporary Master Bath Project 09-26-2018 11:18 PM Elegant Modern Foyer 09-26-2018 10:27 PM Kitchen Makeover 08-27-2018 06:30 PM Custom Trapezoid Door in a built in bookshelf 06-09-2018 05:13 PM Fireplace Mantel 05-12-2018 09:20 PM Custom Staircase and Handrail 05-06-2018 06:30 PM 50 ft deck 04-27-2018 04:30 PM Bathroom renovation London 04-13-2018 01:29 AM Bathroom Vanity 02-01-2018 09:49 PM folding chair hangers 01-29-2018 01:02 AM | 645 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.871971 |
https://www.instructables.com/The-Never-ending-Maze/ | 1,712,945,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00879.warc.gz | 742,405,729 | 31,310 | ## Introduction: The Never-ending Maze
Welcome to the shifting maze!
I enjoy doing mazes, but they're only ever fun until you solve it. They're a one-time use puzzle. I wanted to make a maze that could be replayed for as long as you were willing to play it.
## Supplies
• Paper - any kind
• Pencil
• Marker/Pen - something that's not a pencil
• Scissors
• A box (optional)
• Ruler (optional)
## Step 1: Set-up
With your pencil, draw a box on the paper. This is going to be the area of your maze. (You could use any shape as your maze base. I went with a box to keep things simple.)
I used to ruler to make my maze precisely 3 x 3, however you can make the maze whatever size or shape you want.
After you've drawn your box, divide the area into sections. Later, these will be the pieces you cut. I made my pieces 1 x 1, but again, you can make the pieces whatever size or shape you want.
## Step 2: Start the Maze
When designing the maze, start with the end point, the center in this instance, and navigate back to the start. Despite cutting the maze up later, it's good to at least have a "winning road" in the base maze.
## Step 3: Finish the Maze
After you've made your "winning road", fill in the rest of the maze with dead-ends and misleading paths. Before you ink the walls, try testing the base maze on a friend or family member to see if anything needs to be changed such as, they didn't know where to end, or they solved it too quickly.
After you're satisfied with the layout, ink the walls. I used black here, but feel free to use whatever color you'd like.
## Step 4: Cut the Maze
Now that you've completed your maze, use your scissors to cut it up into the pieces you outlined before.
## Step 5: Box of Randomness
Put the pieces in your box and go find a test subject.
## Step 6: Have Fun!
Once you've located a challenger, shuffle your pieces and see how they do!
The rules can be up to you. My rules were:
• Reconstruct the maze using all 9 pieces. (Doesn't mean they have to cross all the pieces to get to the end.)
• The black borders don't have to match up perfectly.
• They can rearrange the maze into any shape. It does not have to be a square.
Participated in the
Puzzles Speed Challenge | 555 | 2,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.964374 |
https://rpg.stackexchange.com/questions/188354/can-i-cast-the-light-cantrip-on-an-11-foot-pole | 1,723,388,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00634.warc.gz | 405,421,339 | 44,713 | # Can I cast the Light cantrip on an 11-foot pole?
The description of the Light spell in D&D 5e says:
You touch one object that is no larger than 10 feet in any dimension. Until the spell ends, the object sheds bright light in a 20-foot radius and dim light for an additional 20 feet. (...)
Can I cast the Light cantrip on an 11-foot pole?
On one hand, 11 is greater than 10. This would suggest that the answer to my question is "no".
On the other hand, I could put an 11-foot pole in a 10×10×10-foot box. This would suggest that the answer is "yes".
I don't expect that the success of an adventure will hinge on the answer to this question any time soon, but I would like to understand how dimensions are supposed to work.
## No, you can't cast light on it.
If the spell intended to say "an object that fits within a 10-foot cube", it would say that, or something like it. The phrase "no larger than 10 feet in any dimension" is very clear: the object can't be more than 10 feet in its longest possible measure. You don't get to pick an arbitrary X-, Y-, and Z-axis and only measure against those three.
This does leave a hole in the rules where you cast light on an object that is less than 10 feet long, and then change it so that it is longer - for example, casting it on a ball of yarn that you then unroll, or a telescoping pole that you then extend. I would probably rule that when the object expands past 10 feet, it becomes an invalid target and the spell fizzles out, but that's just my ruling.
(That said, one of the classic uses of a light cantrip in literature was Gandalf lighting up the end of his staff in the mines of Moria, so if you wanted to rule that you can use light to put a point-source light on one spot on a larger object, it wouldn't be without precedent.)
• Comments are not for extended discussion; feel free to continue in chat. Commented Jun 21, 2021 at 20:09
• I guess one could rule that just the first 10 feet of the object are shining bright. Commented Jun 22, 2021 at 22:01
## RAW: No, you can't cast light on an 11-foot-long pole.
It appears to be pretty straightforward. The description of the light cantrip says:
You touch one object that is no larger than 10 feet in any dimension.
An 11-foot-long pole is greater than 10 feet in one of its dimensions, so it is not a valid target. The spell description specifically says "dimension" (length, width, depth), and not "volume" or whether it would fit in a 10-foot square.
That said, it is hardly balance-breaking if a DM were to allow it.
• Can you cast the spell on a 10x10x10 cube? Commented Jun 20, 2021 at 14:54
• Where does the spell description specifically list the dimensions as length, width, and depth? My phb doesn't mention that. Commented Jun 20, 2021 at 15:17
• @indigochild The PHB doesn't define "dimensions", but from Wictionary: "A measure of spatial extent in a particular direction, such as height, width or breadth, or depth." Commented Jun 20, 2021 at 21:46
• @ThomasMarkov no, because the 10x10x10 cube has at least one dimension that is longer than 10ft: its longest being the diagonal cutting across the opposite corners of the cube (10√3, ~17 ft). Commented Jun 21, 2021 at 20:54
• So the largest item you can cast it on is a sphere with a 10ft diameter. Commented Jun 21, 2021 at 22:18
## No, but you can cast it on the 10' cubic box the pole fits in
This is an interesting puzzle you present. It is clear, as you say, that an 11 foot pole is not a legitimate target for a light spell. And yet, you assume, a 10 x 10 x 10 box is a legitimate target for the spell. Since the pole will clearly fit inside the box, in some sense the box is larger. How can the larger box not be too big while the smaller pole is too big?
Like Light, a number of spells in the PHB target objects of limited size: Drawmij's Instant Summons ("an object weighing 10 pounds or less whose longest dimension is 6 feet or less"), Mending (a "break or tear [that] is no larger than 1 foot in any dimension"), Nondetection ("a willing creature or a place or an object no larger than 10 feet in any dimension"), Stone shape ("a stone object of Medium size or smaller or a section of stone no more than 5 feet in any dimension"), and Wish (an "object [of] no more than 300 feet in any dimension").
Referencing these descriptions themselves only, it is not clear what "any dimension" means. A number of commenters (on other answers and earlier versions of this answer) have pointed out that "any dimension" can include the 'space diagonal' of a three dimensional object, which is longer than any of its sides. For example, a 10' x 10' x 10' box has a face diagonal of over 14 feet on each surface and a space diagonal of over 17 feet running through its interior. Thus, they reason that a 10' cube is not a legitimate target for a light spell, since it has an interior dimension of nearly double ten feet.
However, two sections of the PHB make it clear "how dimensions are supposed to work": dimension does NOT include diagonals.
Move Earth says
Choose an area of terrain no larger than 40 feet on a side within range. You can reshape dirt, sand, or clay in the area in any manner you choose for the duration. You can raise or lower the area’s elevation, create or fill in a trench, erect or flatten a wall, or form a pillar. The extent of any such changes can’t exceed half the area’s largest dimension. So, if you affect a 40-foot square, you can create a pillar up to 20 feet high, raise or lower the square’s elevation by up to 20 feet, dig a trench up to 20 feet deep, and so on.
If the "area's largest dimension" were allowed to be the diagonal, you could lower the elevation by more than 28 feet. By stating that the maximum change is 20 feet, the spell is clearly indicating that only the "sides" (and not the diagonal) count as its largest dimension.
The Monk Disciple of the Elements Class Feature Shape the Flowing River says
As an action, you can spend 1 ki point to choose an area of ice or water no larger than 30 feet on a side within 120 feet of you. You can change water to ice within the area and vice versa, and you can reshape ice in the area in any manner you choose. You can raise or lower the ice’s elevation, create or fill in a trench, erect or flatten a wall, or form a pillar. The extent of any such changes can’t exceed half the area’s largest dimension. For example, if you affect a 30-foot square, you can create a pillar up to 15 feet high, raise or lower the square’s elevation by up to 15 feet, dig a trench up to 15 feet deep, and so on. You can’t shape the ice to trap or injure a creature in the area.
As with move earth, the description of this ability makes it clear that the "largest dimension" can be either length or width measured along the surface, but not an interior diagonal.
Thus, an 11 foot pole is not a valid target of the light spell because its largest dimension, its length, is longer than 10 feet. However, such a pole can easily fit within a 10' x 10' x 10' box. By choosing the dimensions of face diagonals and space diagonals, it is clear that some dimensions of the box are larger than 10 feet. However, according to the way the PHB indicates that dimensions work, these diagonal dimensions do not count for the application of the size limit of the spell, merely its exterior length, width, and height.
• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jun 21, 2021 at 22:34
• Wait, if I'm understanding correctly, this means that a 19-foot-tall icosohedron is a valid target? Commented Jun 22, 2021 at 22:15
• this doesn't feel right. even if a box is hollow it has dimensions larger than 10 feet along its three dimensional diagonal the wording "that is no larger than 10 feet in any dimension" should also be valid for the diagonal that has the 'hollow' part in between and that is larger than 10 feet. If you allow hollow parts, then you any wire frame structure would have holes and would be allowed to exceed the 10 feet Commented Jun 23, 2021 at 9:13
• @tuskiomi After finding context in another spell and ability in the PHB, I have substantially edited my answer - you might want to read the new version. To your point, I would now assume that the 19 foot height would render that target illegitimate.
– Kirt
Commented Jun 26, 2021 at 17:00
• Yeah the reasoning w/o hollowness is much more coherent. However I still interpret the phrase no larger than x feet in any dimension to mean, the worst case dimension. The object had to be rotated to reveal its worst case size. Thus with my understanding the biggest object that could be affected would be a sphere with 10 feet diameter. The biggest cube would be a cube with a diagonal (in 3 dimensions) having 10 feet which means a 5.77 x 5.77 * 5.77 foot cube. However I didn't check so far not whether rules with the above phrasing give examples violating my assumption Commented Jun 28, 2021 at 12:12 | 2,212 | 8,973 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.964221 |
https://wiki.haskell.org/index.php?title=99_questions/Solutions/8&oldid=60166 | 1,610,778,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00243.warc.gz | 638,453,541 | 6,176 | # 99 questions/Solutions/8
(**) Eliminate consecutive duplicates of list elements.
```compress :: Eq a => [a] -> [a]
compress = map head . group
```
We simply group equal values together (using Data.List.group), then take the head of each.
An alternative solution is
```compress (x:ys@(y:_))
| x == y = compress ys
| otherwise = x : compress ys
compress ys = ys
```
A variation of the above using `foldr` (note that GHC erases the `Maybe`s, producing efficient code):
```compress xs = foldr f (const []) xs Nothing
where
f x r a@(Just q) | x == q = r a
f x r _ = x : r (Just x)
```
Another possibility using foldr (this one is not so efficient, because it pushes the whole input onto the "stack" before doing anything else):
```compress :: (Eq a) => [a] -> [a]
compress = foldr skipDups []
where skipDups x [] = [x]
skipDups x acc
| x == head acc = acc
| otherwise = x : acc
```
A similar solution without using `foldr`.
```compress :: (Eq a) => [a] -> [a]
compress list = compress_acc list []
where compress_acc [] acc = acc
compress_acc [x] acc = (acc ++ [x])
compress_acc (x:xs) acc
| x == (head xs) = compress_acc xs acc
| otherwise = compress_acc xs (acc ++ [x])
```
A very simple approach:
```compress [] = []
compress (x:xs) = x : (compress \$ dropWhile (== x) xs)
```
Another approach, using foldr
```compress :: Eq a => [a] -> [a]
compress x = foldr (\a b -> if a == (head b) then b else a:b) [last x] x
```
Wrong solution using foldr
```compress :: Eq a => [a] -> [a]
compress xs = foldr (\x acc -> if x `elem` acc then acc else x:acc) [] xs
-- Main> compress [1, 1, 1, 2, 2, 1, 1]
-- [2,1] - must be [1,2,1]
```
and using foldl
```compress :: (Eq a) => [a] -> [a]
compress x = foldl (\a b -> if (last a) == b then a else a ++ [b]) [head x] x
compress' x = reverse \$ foldl (\a b -> if (head a) == b then a else b:a) [head x] x
```
A crazy variation that acts as a good transformer for fold/build fusion
```{-# INLINE compress #-}
compress :: Eq a => [a] -> [a]
compress xs = build (\c n ->
let
f x r a@(Just q) | x == q = r a
f x r _ = x `c` r (Just x)
in
foldr f (const n) xs Nothing)
``` | 682 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-04 | latest | en | 0.807535 |
http://eprints.umm.ac.id/64900/ | 1,610,718,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00185.warc.gz | 37,003,038 | 13,591 | ## PENGARUH MODEL PEMBELAJARAN PREDICT OBSERVE EXPLAIN DENGAN PENDEKATAN METAPHORICAL THINKING TERHADAP KEMAMPUAN KONEKSI MATEMATIS DAN BERPIKIR KRITIS SISWA SMP DALAM MEMECAHKAN MASALAH MATEMATIKA
Pradani, Ditma Ayu (2020) PENGARUH MODEL PEMBELAJARAN PREDICT OBSERVE EXPLAIN DENGAN PENDEKATAN METAPHORICAL THINKING TERHADAP KEMAMPUAN KONEKSI MATEMATIS DAN BERPIKIR KRITIS SISWA SMP DALAM MEMECAHKAN MASALAH MATEMATIKA. Undergraduate (S1) thesis, Universitas Muhammadiyah Malang.
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## Abstract
This study aims to analyze the effect of the Predict Observe Explain learning model with the Metaphorical Thinking approach to the ability of students' mathematical connections and critical thinking in solving mathematical problems. This study uses a quantitative approach. This research was conducted at Doko 1 Public Middle School with class VII E subjects as an experimental class using POE model learning with metaphorical thinking approach with online or online methods because it was carried out during the pandemic period. The instruments used in this study were tests and observation sheets. Tests in the form of pretest and posttest were validated based on indicators of achievement of mathematical connection abilities and critical thinking and observation sheets that were used to observe teacher and student activities while learning. Data analysis techniques used were normality test, homogeneity test and hpotesis test using independent sample t-test. The results of the study are 1) there is no effect of the use of the POE learning model with the metaphorical thinking approach to the mathematical connection ability of Grade VII students of SMP Negeri 1 Doko 2) There is an influence of using the POE learning model with the metaphorical thinking approach to critical thinking of Grade VII students of SMP Negeri 1 Doko. This can be seen from the average N-Gain of students who were given learning by the POE model using the metaphorical thinking approach that is 63.8752 higher than students who were given learning by the POE model which is 56.9113.
Item Type: Thesis (Undergraduate (S1)) 201610060311094 Mohammad Syaifuddin (0724086401), Mayang Dintarini (0728088902) POE Learning Model, Metaphorical Thinking Approach, Mathematical connection ability, critical thinking L Education > L Education (General)L Education > LB Theory and practice of education > LB2300 Higher EducationL Education > LC Special aspects of education Faculty of Teacher Training and Education > Department of Mathematics Education (84202) 201610060311094 24 Aug 2020 01:40 24 Aug 2020 01:40 http://eprints.umm.ac.id/id/eprint/64900 | 713 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.835243 |
https://in.mathworks.com/matlabcentral/cody/problems/1070-next-higher-power-of-b/solutions/556029 | 1,606,942,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141715252.96/warc/CC-MAIN-20201202175113-20201202205113-00375.warc.gz | 337,159,061 | 17,624 | Cody
# Problem 1070. Next Higher Power of B
Solution 556029
Submitted on 12 Jan 2015 by Gergely Patay
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
%% assert(nextpowb(2,126) == 7)
ans = 7
2 Pass
%% assert(nextpowb(3,6560) == 8)
ans = 8
3 Pass
%% assert(nextpowb(4,262141) == 9)
ans = 9
4 Pass
%% assert(nextpowb(5,21) == 2)
ans = 2
5 Pass
%% assert(nextpowb(6,1294) == 4)
ans = 4
6 Pass
%% assert(nextpowb(7,5) == 1)
ans = 1
7 Pass
%% assert(nextpowb(8,134217726) == 9)
ans = 9
8 Pass
%% assert(nextpowb(9,4782966) == 7)
ans = 7
9 Pass
%% assert(nextpowb(10,99993) == 5)
ans = 5
10 Pass
%% assert(nextpowb(11,1771559) == 6)
ans = 6
11 Pass
%% assert(nextpowb(12,429981693) == 8)
ans = 8
12 Pass
%% assert(nextpowb(13,2194) == 3)
ans = 3
13 Pass
%% assert(nextpowb(14,537814) == 5)
ans = 5
14 Pass
%% assert(nextpowb(15,2562890613) == 8)
ans = 8
15 Pass
%% assert(nextpowb(16,249) == 2)
ans = 2
16 Pass
%% assert(nextpowb(17,2015993900438) == 10)
ans = 10
17 Pass
%% assert(nextpowb(18,3570467226613) == 10)
ans = 10
18 Pass
%% assert(nextpowb(19,6131066257790) == 10)
ans = 10
19 Pass
%% assert(nextpowb(20,3199997) == 5)
ans = 5
20 Pass
%% assert(nextpowb(21,85766100) == 6)
ans = 6
21 Pass
%% assert(nextpowb(22,467) == 2)
ans = 2
22 Pass
%% assert(nextpowb(23,519) == 2)
ans = 2
23 Pass
%% assert(nextpowb(24,2641807540202) == 9)
ans = 9
24 Pass
%% assert(nextpowb(25,95367431640600) == 10)
ans = 10
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Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
(Help)
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```power = 1 watt (power) ``` Related Measurements: Try converting from "mechanical power" to horsepower, lusec, ton of refrigeration, ton refrigeration, tons of refrigeration, tons refrigeration, watt, or any combination of units which equate to "mass length squared / time cubed" and represent acoustical power, electrical power, heat flow rate, linear power, mechanical power, power, radiant flux, or radiant power. Sample Conversions: mechanical power = .00134102 horsepower, 7,500.62 lusec, .00028435 ton of refrigeration, .00028435 ton refrigeration, .00028435 tons of refrigeration, .00028435 tons refrigeration, 1 watt.
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https://www.reference.com/web?q=Lines+on+a+Ruler&qo=contentPageRelatedSearch&o=600605&l=dir&gc=1 | 1,508,497,900,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824068.35/warc/CC-MAIN-20171020101632-20171020121632-00206.warc.gz | 968,784,680 | 14,243 | articles
To use a ruler, place it alongside or on top of the object or line that is to be measured and count the number of marks from edge to edge of the object. For those who are working with a metric ruler, the longer lines are... More »
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www.reference.com Science Measurements | 392 | 1,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | longest | en | 0.864836 |
https://www.physicsforums.com/threads/energy-stored-in-a-spring.662013/ | 1,524,605,243,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947328.78/warc/CC-MAIN-20180424202213-20180424222213-00301.warc.gz | 842,293,312 | 14,319 | # Energy Stored in a spring
1. Jan 2, 2013
### GeneralOJB
Why is the energy stored in a spring 1/2 * distance * force? Isn't work just force * distance?
2. Jan 2, 2013
### jbriggs444
The full force is not applied for the full distance.
In the simple case of a constant force over a fixed distance you can just multiply force by distance. But in the case of a spring being compressed, the force is not constant.
You could approximate the answer by analyzing the situation in small steps... You compress the spring the first 1/10th of the distance using 1/10th of the full force, the next 1/10th of the instance using 2/10th of the total force and so on. This approach would give you an answer that is pretty close.
An exact answer can be obtained by using calculus and integrating instantaneous force over incremental distance. That answer turns out to be 1/2 * distance * maximum-force
Last edited: Jan 2, 2013
3. Jan 2, 2013
### GeneralOJB
Thanks, I understand now. | 250 | 978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-17 | longest | en | 0.904955 |
https://nerdytermpapers.com/there-are-two-identical-firms-eqt-corp-and-dbt-corp-except-for-their-capital-str/ | 1,623,594,685,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00090.warc.gz | 395,682,800 | 12,661 | There Are Two Identical Firms Eqt Corp And Dbt Corp Except For Their capital structure
There are two identical firms EQT Corp and DBT Corp except for their capital structure. For example none of the firms pay taxes, and earnings are \$22,000 per year. There is no growth for any of the firms, i.e., they pay out all earnings to shareholders as dividends. EQT is an all equity firm with 1,000 shares outstanding. In contrast, DBT has \$40,000 worth of debt outstanding with interest rate of 10%. The equity price of DBT Corp is \$120 with 500 shares outstanding.
(a) What is the equity value of EQT Corp? And, what is EQT’s stock price?
(b) What are the expected returns on equity of each firm? Which company is riskier and why?
(c) Suppose currently you own 300 shares of EQT stock and John owns 200 shares of DBT stock. What would you do if you want to have exactly the same total dividends as John’s? Assume you can borrow or lend at 10%.
2. Carlson Enterprise is financed entirely by common stock with a beta of 0.8. The stock is currently traded at \$100 per share. There are no taxes. The stock is expected to have constant earnings in perpetuity, has a price- earning multiple of 16. The market expected return is 12%. The company decides to repurchase one third of its common stock and substitute with an equal value of debt. If the debt yields the same as the risk-free rate of 4%, calculate:
(a) the beta of the common stock after the refinancing
(b) the required return and risk premium on the common stock before and after the refinancing
(c) the required return on the company (i.e., stock and debt combined) after the refinancingAssume that the EBIT of the firm is expected to remain constant in perpetuity.
Determine:
(d) the percentage increase in earnings per share after refinancing
(e) the new price-earnings multiple3. Backyard Utility is expected to have an EBIT of \$250,000 and is an all-equity firm. EBIT is expected to grow at 6% per year. In order to get this growth rate, Backyard needs to retain 50% of earnings for investment. The remaining will be paid out as dividends. The firm’s corporate tax rate is 20%. The discount rate for the firm is 10%.
(a) What is the market value of the firm?
(b) Now assume the firm issues \$1.5 million of debt paying interest of 5% per year, using the proceeds to retireequity. The debt is expected to be permanent. What is the equity value of the firm after refinancing?
(c) What is the required return on equity after refinancing?
(d) In order to get the same growth rate of 6%, Backyard only need to retain 29.41% of its earnings forinvestment. Compute the equity value by discounting all the future dividends as in (a).
(Hint: you shouldget the same result as in (b))(e) If Backyard’s manager only wants its total equity value to be \$1.7 million instead, how much debt shouldBackyard actually use?
4. APV Inc. is considering an investment project that will generate a level after-tax cash flow of \$235,000 a year for the next 5 years. The all-equity discount rate is 15%, and the project requires an investment of \$800,000.Should APV take on the project? Now, if the management desires to raise 50% of the initial investment in the form of debt at an interest rate of 8%. This debt is repaid in equal yearly installments; interest accrues on the unpaid balance. The corporation tax rate is 30%. What is the APV of the project?
5. Here is a simplified book balance sheet for Upjohn Company at the end of 2005 (dollars in millions):Current assetsNet property, plant,and equipment Other assets\$1,4181,240 472Current liabilities \$653 Long-term debt 457
(a) Calculate Upjohn’s WACC. Use the CAPM and the data above. Make additional assumptions and approximations as necessary. (hint: only use “Long-term debt” as debt)
(b) What would Upjohn’s WACC be if it moved to and maintained a debt to market value ratio of 20%? Assume the only taxes are corporate taxes.
(c) Assume that the bondholders and the equityholders have effective tax rates of 30% and 10%, respectively, and the current market price of \$25 reflects this capital structure. What would be the total equity value if Upjohn were 100% equity finance?
6. Managers of the Edge Corporation must choose between two mutually exclusive projects. Suppose that the project that Edge chooses will be the only business it does next year, and assume zero discount rate. Therefore, the payoff on the project will determine the value of the firm. Edge is obliged to make a \$500 payment to bondholders. The first project has low risk:State of the economy Slump Probability 0.3 Project Payoff 600Boom 0.7 700Low-risk ProjectDeferred taxes 199 Shareholders’ equity 1,821 Total 3,130 Total 3,130 Other information:There are 100 million shares outstanding with a per share price of \$25 at end of year. The company has a beta of 1.2. Historical average risk premium is 8.0% per year and the risk-free rate is 4%. Newly issued Upjohn long-term debt carries a spread of 1.0% to the risk-free rate. The company’s marginal tax rate is 30%.
If the firm does not undertake the low-risk project, it will choose the following high-risk project: High-risk Project
State of the economy ProbabilityProject PayoffSlump 0.3 100Boom 0.7 800
(a) Which project is the social optimum?
(b) Which project would be preferred by stockholders? Why? | 1,272 | 5,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-25 | latest | en | 0.954796 |
https://traveltweaks.com/travel-ideas-how-far-is-new-zealand-from-florida-what-to-visit-there-47102/ | 1,701,965,287,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00614.warc.gz | 663,362,057 | 34,106 | # Travel Ideas! How far is New Zealand from Florida & What to Visit There!
## How Far is New Zealand from Florida
When it comes to the distance between New Zealand and Florida, it’s safe to say that they are quite far apart. The two locations are situated on opposite sides of the world, separated by thousands of miles and numerous time zones. So, just how far is New Zealand from Florida? Let’s delve into the details.
In terms of direct flight distance, New Zealand and Florida are approximately 8,000 miles apart. This is a considerable journey that would require multiple layovers and long hours in the air. A non-stop flight would take around 17-18 hours, depending on factors such as wind patterns and aircraft speed.
## Distance between New Zealand and Florida
When it comes to the question of how far is New Zealand from Florida, the answer is quite astonishing. The two destinations are separated by an immense distance, spanning across several continents and vast bodies of water. Strap in and prepare for a journey that will take you thousands of miles away.
To truly grasp the magnitude of this distance, let’s dive into some numbers. The straight-line distance between Auckland, New Zealand, and Miami, Florida is approximately 8,500 miles (13,700 kilometers). That’s equivalent to flying around the Earth more than one-third times! It’s mind-boggling to think about covering such a tremendous expanse.
Now, bear in mind that this figure represents the shortest direct route between these two points. In reality, traveling from New Zealand to Florida involves multiple flights with layovers in various cities along the way. These layovers can extend travel time significantly and add extra mileage to your overall journey.
If you’re planning a trip from New Zealand to Florida or vice versa, be prepared for an extensive travel itinerary. Depending on flight schedules and connections, it could take anywhere from 20 to 30 hours or more to complete the journey. It’s advisable to plan your trip well in advance and consider factors such as jet lag and adjusting to different time zones along the way.
So there you have it – when pondering how far New Zealand is from Florida, remember that you’re looking at a staggering distance of approximately 8,500 miles. Embarking on this transcontinental adventure requires careful planning but promises incredible experiences awaiting at both ends of the globe.
## Flight Duration from New Zealand to Florida
When it comes to the distance between New Zealand and Florida, one might wonder how far these two places are from each other. While they may be on opposite sides of the world, modern air travel has made it possible to bridge this vast distance. So let’s delve into the flight duration from New Zealand to Florida.
The flight duration from New Zealand to Florida can vary depending on various factors such as the specific airports involved, layovers, and weather conditions. On average, a direct flight from Auckland, New Zealand to Miami or Orlando, Florida takes around 18-20 hours. This includes both the time spent in the air and any layovers along the way.
If you’re flying from other cities in New Zealand like Wellington or Christchurch, you’ll likely have a stopover in Auckland before continuing your journey towards Florida. In such cases, you can expect a slightly longer total travel time due to the additional domestic leg.
It’s worth noting that there are no direct flights available between all cities in New Zealand and all cities in Florida. Most commonly, travelers will need to make a connection at major international airports like Los Angeles International Airport (LAX) or San Francisco International Airport (SFO) before reaching their final destination in Florida.
Additionally, keep in mind that flight durations can also be influenced by factors like wind patterns and jet stream currents during different times of the year. These natural phenomena can either shorten or extend your overall travel time.
In summary, while there is no exact answer for how long it takes to fly from New Zealand to Florida as it depends on several variables including departure city and layovers, one should anticipate an average flight duration of around 18-20 hours for a direct journey. | 830 | 4,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.952261 |
https://www.hackmath.net/en/example/1418?tag_id=104 | 1,521,658,323,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647681.81/warc/CC-MAIN-20180321180325-20180321200325-00530.warc.gz | 842,742,230 | 21,885 | # Unknown integer
Find the smallest integer that: divided by 2, the remainder is 1 divided by 3 the remainder is 2, divided by 4 remainder is 3, ... divided by eight reminder is 7, by 9 reminder is 8.
Result
x = 2519
### x = LCM (2, 3, 4, 5, 6, 7, 8, 9) - 1 x = 5 . 7 . 8 . 9 - 1 = 2519
1 .. 1 1 1 1 1 1 1 1
2 .. 0 2 2 2 2 2 2 2
3 .. 1 0 3 3 3 3 3 3
4 .. 0 1 0 4 4 4 4 4
5 .. 1 2 1 0 5 5 5 5
6 .. 0 0 2 1 0 6 6 6
7 .. 1 1 3 2 1 0 7 7
8 .. 0 2 0 3 2 1 0 8
9 .. 1 0 1 4 3 2 1 0
10 .. 0 1 2 0 4 3 2 1
11 .. 1 2 3 1 5 4 3 2
12 .. 0 0 0 2 0 5 4 3
13 .. 1 1 1 3 1 6 5 4
14 .. 0 2 2 4 2 0 6 5
15 .. 1 0 3 0 3 1 7 6
16 .. 0 1 0 1 4 2 0 7
17 .. 1 2 1 2 5 3 1 8
18 .. 0 0 2 3 0 4 2 0
19 .. 1 1 3 4 1 5 3 1
20 .. 0 2 0 0 2 6 4 2
21 .. 1 0 1 1 3 0 5 3
22 .. 0 1 2 2 4 1 6 4
23 .. 1 2 3 3 5 2 7 5
24 .. 0 0 0 4 0 3 0 6
25 .. 1 1 1 0 1 4 1 7
26 .. 0 2 2 1 2 5 2 8
27 .. 1 0 3 2 3 6 3 0
28 .. 0 1 0 3 4 0 4 1
29 .. 1 2 1 4 5 1 5 2
30 .. 0 0 2 0 0 2 6 3
31 .. 1 1 3 1 1 3 7 4
32 .. 0 2 0 2 2 4 0 5
33 .. 1 0 1 3 3 5 1 6
34 .. 0 1 2 4 4 6 2 7
35 .. 1 2 3 0 5 0 3 8
36 .. 0 0 0 1 0 1 4 0
37 .. 1 1 1 2 1 2 5 1
38 .. 0 2 2 3 2 3 6 2
39 .. 1 0 3 4 3 4 7 3
40 .. 0 1 0 0 4 5 0 4
41 .. 1 2 1 1 5 6 1 5
42 .. 0 0 2 2 0 0 2 6
43 .. 1 1 3 3 1 1 3 7
44 .. 0 2 0 4 2 2 4 8
45 .. 1 0 1 0 3 3 5 0
46 .. 0 1 2 1 4 4 6 1
47 .. 1 2 3 2 5 5 7 2
48 .. 0 0 0 3 0 6 0 3
49 .. 1 1 1 4 1 0 1 4
50 .. 0 2 2 0 2 1 2 5
51 .. 1 0 3 1 3 2 3 6
52 .. 0 1 0 2 4 3 4 7
53 .. 1 2 1 3 5 4 5 8
54 .. 0 0 2 4 0 5 6 0
55 .. 1 1 3 0 1 6 7 1
56 .. 0 2 0 1 2 0 0 2
57 .. 1 0 1 2 3 1 1 3
58 .. 0 1 2 3 4 2 2 4
59 .. 1 2 3 4 5 3 3 5
60 .. 0 0 0 0 0 4 4 6
61 .. 1 1 1 1 1 5 5 7
62 .. 0 2 2 2 2 6 6 8
63 .. 1 0 3 3 3 0 7 0
64 .. 0 1 0 4 4 1 0 1
65 .. 1 2 1 0 5 2 1 2
66 .. 0 0 2 1 0 3 2 3
67 .. 1 1 3 2 1 4 3 4
68 .. 0 2 0 3 2 5 4 5
69 .. 1 0 1 4 3 6 5 6
70 .. 0 1 2 0 4 0 6 7
71 .. 1 2 3 1 5 1 7 8
72 .. 0 0 0 2 0 2 0 0
73 .. 1 1 1 3 1 3 1 1
74 .. 0 2 2 4 2 4 2 2
75 .. 1 0 3 0 3 5 3 3
76 .. 0 1 0 1 4 6 4 4
77 .. 1 2 1 2 5 0 5 5
78 .. 0 0 2 3 0 1 6 6
79 .. 1 1 3 4 1 2 7 7
80 .. 0 2 0 0 2 3 0 8
81 .. 1 0 1 1 3 4 1 0
82 .. 0 1 2 2 4 5 2 1
83 .. 1 2 3 3 5 6 3 2
84 .. 0 0 0 4 0 0 4 3
85 .. 1 1 1 0 1 1 5 4
86 .. 0 2 2 1 2 2 6 5
87 .. 1 0 3 2 3 3 7 6
88 .. 0 1 0 3 4 4 0 7
89 .. 1 2 1 4 5 5 1 8
90 .. 0 0 2 0 0 6 2 0
91 .. 1 1 3 1 1 0 3 1
92 .. 0 2 0 2 2 1 4 2
93 .. 1 0 1 3 3 2 5 3
94 .. 0 1 2 4 4 3 6 4
95 .. 1 2 3 0 5 4 7 5
96 .. 0 0 0 1 0 5 0 6
97 .. 1 1 1 2 1 6 1 7
98 .. 0 2 2 3 2 0 2 8
99 .. 1 0 3 4 3 1 3 0
100 .. 0 1 0 0 4 2 4 1
101 .. 1 2 1 1 5 3 5 2
102 .. 0 0 2 2 0 4 6 3
103 .. 1 1 3 3 1 5 7 4
104 .. 0 2 0 4 2 6 0 5
105 .. 1 0 1 0 3 0 1 6
106 .. 0 1 2 1 4 1 2 7
107 .. 1 2 3 2 5 2 3 8
108 .. 0 0 0 3 0 3 4 0
109 .. 1 1 1 4 1 4 5 1
110 .. 0 2 2 0 2 5 6 2
111 .. 1 0 3 1 3 6 7 3
112 .. 0 1 0 2 4 0 0 4
113 .. 1 2 1 3 5 1 1 5
114 .. 0 0 2 4 0 2 2 6
115 .. 1 1 3 0 1 3 3 7
116 .. 0 2 0 1 2 4 4 8
117 .. 1 0 1 2 3 5 5 0
118 .. 0 1 2 3 4 6 6 1
119 .. 1 2 3 4 5 0 7 2
120 .. 0 0 0 0 0 1 0 3
121 .. 1 1 1 1 1 2 1 4
122 .. 0 2 2 2 2 3 2 5
123 .. 1 0 3 3 3 4 3 6
124 .. 0 1 0 4 4 5 4 7
125 .. 1 2 1 0 5 6 5 8
126 .. 0 0 2 1 0 0 6 0
127 .. 1 1 3 2 1 1 7 1
128 .. 0 2 0 3 2 2 0 2
129 .. 1 0 1 4 3 3 1 3
130 .. 0 1 2 0 4 4 2 4
131 .. 1 2 3 1 5 5 3 5
132 .. 0 0 0 2 0 6 4 6
133 .. 1 1 1 3 1 0 5 7
134 .. 0 2 2 4 2 1 6 8
135 .. 1 0 3 0 3 2 7 0
136 .. 0 1 0 1 4 3 0 1
137 .. 1 2 1 2 5 4 1 2
138 .. 0 0 2 3 0 5 2 3
139 .. 1 1 3 4 1 6 3 4
140 .. 0 2 0 0 2 0 4 5
141 .. 1 0 1 1 3 1 5 6
142 .. 0 1 2 2 4 2 6 7
143 .. 1 2 3 3 5 3 7 8
144 .. 0 0 0 4 0 4 0 0
145 .. 1 1 1 0 1 5 1 1
146 .. 0 2 2 1 2 6 2 2
147 .. 1 0 3 2 3 0 3 3
148 .. 0 1 0 3 4 1 4 4
149 .. 1 2 1 4 5 2 5 5
150 .. 0 0 2 0 0 3 6 6
151 .. 1 1 3 1 1 4 7 7
152 .. 0 2 0 2 2 5 0 8
153 .. 1 0 1 3 3 6 1 0
154 .. 0 1 2 4 4 0 2 1
155 .. 1 2 3 0 5 1 3 2
156 .. 0 0 0 1 0 2 4 3
157 .. 1 1 1 2 1 3 5 4
158 .. 0 2 2 3 2 4 6 5
159 .. 1 0 3 4 3 5 7 6
160 .. 0 1 0 0 4 6 0 7
161 .. 1 2 1 1 5 0 1 8
162 .. 0 0 2 2 0 1 2 0
163 .. 1 1 3 3 1 2 3 1
164 .. 0 2 0 4 2 3 4 2
165 .. 1 0 1 0 3 4 5 3
166 .. 0 1 2 1 4 5 6 4
167 .. 1 2 3 2 5 6 7 5
168 .. 0 0 0 3 0 0 0 6
169 .. 1 1 1 4 1 1 1 7
170 .. 0 2 2 0 2 2 2 8
171 .. 1 0 3 1 3 3 3 0
172 .. 0 1 0 2 4 4 4 1
173 .. 1 2 1 3 5 5 5 2
174 .. 0 0 2 4 0 6 6 3
175 .. 1 1 3 0 1 0 7 4
176 .. 0 2 0 1 2 1 0 5
177 .. 1 0 1 2 3 2 1 6
178 .. 0 1 2 3 4 3 2 7
179 .. 1 2 3 4 5 4 3 8
180 .. 0 0 0 0 0 5 4 0
181 .. 1 1 1 1 1 6 5 1
182 .. 0 2 2 2 2 0 6 2
183 .. 1 0 3 3 3 1 7 3
184 .. 0 1 0 4 4 2 0 4
185 .. 1 2 1 0 5 3 1 5
186 .. 0 0 2 1 0 4 2 6
187 .. 1 1 3 2 1 5 3 7
188 .. 0 2 0 3 2 6 4 8
189 .. 1 0 1 4 3 0 5 0
190 .. 0 1 2 0 4 1 6 1
191 .. 1 2 3 1 5 2 7 2
192 .. 0 0 0 2 0 3 0 3
193 .. 1 1 1 3 1 4 1 4
194 .. 0 2 2 4 2 5 2 5
195 .. 1 0 3 0 3 6 3 6
196 .. 0 1 0 1 4 0 4 7
197 .. 1 2 1 2 5 1 5 8
198 .. 0 0 2 3 0 2 6 0
199 .. 1 1 3 4 1 3 7 1
200 .. 0 2 0 0 2 4 0 2
201 .. 1 0 1 1 3 5 1 3
202 .. 0 1 2 2 4 6 2 4
203 .. 1 2 3 3 5 0 3 5
204 .. 0 0 0 4 0 1 4 6
205 .. 1 1 1 0 1 2 5 7
206 .. 0 2 2 1 2 3 6 8
207 .. 1 0 3 2 3 4 7 0
208 .. 0 1 0 3 4 5 0 1
209 .. 1 2 1 4 5 6 1 2
210 .. 0 0 2 0 0 0 2 3
211 .. 1 1 3 1 1 1 3 4
212 .. 0 2 0 2 2 2 4 5
213 .. 1 0 1 3 3 3 5 6
214 .. 0 1 2 4 4 4 6 7
215 .. 1 2 3 0 5 5 7 8
216 .. 0 0 0 1 0 6 0 0
217 .. 1 1 1 2 1 0 1 1
218 .. 0 2 2 3 2 1 2 2
219 .. 1 0 3 4 3 2 3 3
220 .. 0 1 0 0 4 3 4 4
221 .. 1 2 1 1 5 4 5 5
222 .. 0 0 2 2 0 5 6 6
223 .. 1 1 3 3 1 6 7 7
224 .. 0 2 0 4 2 0 0 8
225 .. 1 0 1 0 3 1 1 0
226 .. 0 1 2 1 4 2 2 1
227 .. 1 2 3 2 5 3 3 2
228 .. 0 0 0 3 0 4 4 3
229 .. 1 1 1 4 1 5 5 4
230 .. 0 2 2 0 2 6 6 5
231 .. 1 0 3 1 3 0 7 6
232 .. 0 1 0 2 4 1 0 7
233 .. 1 2 1 3 5 2 1 8
234 .. 0 0 2 4 0 3 2 0
235 .. 1 1 3 0 1 4 3 1
236 .. 0 2 0 1 2 5 4 2
237 .. 1 0 1 2 3 6 5 3
238 .. 0 1 2 3 4 0 6 4
239 .. 1 2 3 4 5 1 7 5
240 .. 0 0 0 0 0 2 0 6
241 .. 1 1 1 1 1 3 1 7
242 .. 0 2 2 2 2 4 2 8
243 .. 1 0 3 3 3 5 3 0
244 .. 0 1 0 4 4 6 4 1
245 .. 1 2 1 0 5 0 5 2
246 .. 0 0 2 1 0 1 6 3
247 .. 1 1 3 2 1 2 7 4
248 .. 0 2 0 3 2 3 0 5
249 .. 1 0 1 4 3 4 1 6
250 .. 0 1 2 0 4 5 2 7
251 .. 1 2 3 1 5 6 3 8
252 .. 0 0 0 2 0 0 4 0
253 .. 1 1 1 3 1 1 5 1
254 .. 0 2 2 4 2 2 6 2
255 .. 1 0 3 0 3 3 7 3
256 .. 0 1 0 1 4 4 0 4
257 .. 1 2 1 2 5 5 1 5
258 .. 0 0 2 3 0 6 2 6
259 .. 1 1 3 4 1 0 3 7
260 .. 0 2 0 0 2 1 4 8
261 .. 1 0 1 1 3 2 5 0
262 .. 0 1 2 2 4 3 6 1
263 .. 1 2 3 3 5 4 7 2
264 .. 0 0 0 4 0 5 0 3
265 .. 1 1 1 0 1 6 1 4
266 .. 0 2 2 1 2 0 2 5
267 .. 1 0 3 2 3 1 3 6
268 .. 0 1 0 3 4 2 4 7
269 .. 1 2 1 4 5 3 5 8
270 .. 0 0 2 0 0 4 6 0
271 .. 1 1 3 1 1 5 7 1
272 .. 0 2 0 2 2 6 0 2
273 .. 1 0 1 3 3 0 1 3
274 .. 0 1 2 4 4 1 2 4
275 .. 1 2 3 0 5 2 3 5
276 .. 0 0 0 1 0 3 4 6
277 .. 1 1 1 2 1 4 5 7
278 .. 0 2 2 3 2 5 6 8
279 .. 1 0 3 4 3 6 7 0
280 .. 0 1 0 0 4 0 0 1
281 .. 1 2 1 1 5 1 1 2
282 .. 0 0 2 2 0 2 2 3
283 .. 1 1 3 3 1 3 3 4
284 .. 0 2 0 4 2 4 4 5
285 .. 1 0 1 0 3 5 5 6
286 .. 0 1 2 1 4 6 6 7
287 .. 1 2 3 2 5 0 7 8
288 .. 0 0 0 3 0 1 0 0
289 .. 1 1 1 4 1 2 1 1
290 .. 0 2 2 0 2 3 2 2
291 .. 1 0 3 1 3 4 3 3
292 .. 0 1 0 2 4 5 4 4
293 .. 1 2 1 3 5 6 5 5
294 .. 0 0 2 4 0 0 6 6
295 .. 1 1 3 0 1 1 7 7
296 .. 0 2 0 1 2 2 0 8
297 .. 1 0 1 2 3 3 1 0
298 .. 0 1 2 3 4 4 2 1
299 .. 1 2 3 4 5 5 3 2
300 .. 0 0 0 0 0 6 4 3
301 .. 1 1 1 1 1 0 5 4
302 .. 0 2 2 2 2 1 6 5
303 .. 1 0 3 3 3 2 7 6
304 .. 0 1 0 4 4 3 0 7
305 .. 1 2 1 0 5 4 1 8
306 .. 0 0 2 1 0 5 2 0
307 .. 1 1 3 2 1 6 3 1
308 .. 0 2 0 3 2 0 4 2
309 .. 1 0 1 4 3 1 5 3
310 .. 0 1 2 0 4 2 6 4
311 .. 1 2 3 1 5 3 7 5
312 .. 0 0 0 2 0 4 0 6
313 .. 1 1 1 3 1 5 1 7
314 .. 0 2 2 4 2 6 2 8
315 .. 1 0 3 0 3 0 3 0
316 .. 0 1 0 1 4 1 4 1
317 .. 1 2 1 2 5 2 5 2
318 .. 0 0 2 3 0 3 6 3
319 .. 1 1 3 4 1 4 7 4
320 .. 0 2 0 0 2 5 0 5
321 .. 1 0 1 1 3 6 1 6
322 .. 0 1 2 2 4 0 2 7
323 .. 1 2 3 3 5 1 3 8
324 .. 0 0 0 4 0 2 4 0
325 .. 1 1 1 0 1 3 5 1
326 .. 0 2 2 1 2 4 6 2
327 .. 1 0 3 2 3 5 7 3
328 .. 0 1 0 3 4 6 0 4
329 .. 1 2 1 4 5 0 1 5
330 .. 0 0 2 0 0 1 2 6
331 .. 1 1 3 1 1 2 3 7
332 .. 0 2 0 2 2 3 4 8
333 .. 1 0 1 3 3 4 5 0
334 .. 0 1 2 4 4 5 6 1
335 .. 1 2 3 0 5 6 7 2
336 .. 0 0 0 1 0 0 0 3
337 .. 1 1 1 2 1 1 1 4
338 .. 0 2 2 3 2 2 2 5
339 .. 1 0 3 4 3 3 3 6
340 .. 0 1 0 0 4 4 4 7
341 .. 1 2 1 1 5 5 5 8
342 .. 0 0 2 2 0 6 6 0
343 .. 1 1 3 3 1 0 7 1
344 .. 0 2 0 4 2 1 0 2
345 .. 1 0 1 0 3 2 1 3
346 .. 0 1 2 1 4 3 2 4
347 .. 1 2 3 2 5 4 3 5
348 .. 0 0 0 3 0 5 4 6
349 .. 1 1 1 4 1 6 5 7
350 .. 0 2 2 0 2 0 6 8
351 .. 1 0 3 1 3 1 7 0
352 .. 0 1 0 2 4 2 0 1
353 .. 1 2 1 3 5 3 1 2
354 .. 0 0 2 4 0 4 2 3
355 .. 1 1 3 0 1 5 3 4
356 .. 0 2 0 1 2 6 4 5
357 .. 1 0 1 2 3 0 5 6
358 .. 0 1 2 3 4 1 6 7
359 .. 1 2 3 4 5 2 7 8
360 .. 0 0 0 0 0 3 0 0
361 .. 1 1 1 1 1 4 1 1
362 .. 0 2 2 2 2 5 2 2
363 .. 1 0 3 3 3 6 3 3
364 .. 0 1 0 4 4 0 4 4
365 .. 1 2 1 0 5 1 5 5
366 .. 0 0 2 1 0 2 6 6
367 .. 1 1 3 2 1 3 7 7
368 .. 0 2 0 3 2 4 0 8
369 .. 1 0 1 4 3 5 1 0
370 .. 0 1 2 0 4 6 2 1
371 .. 1 2 3 1 5 0 3 2
372 .. 0 0 0 2 0 1 4 3
373 .. 1 1 1 3 1 2 5 4
374 .. 0 2 2 4 2 3 6 5
375 .. 1 0 3 0 3 4 7 6
376 .. 0 1 0 1 4 5 0 7
377 .. 1 2 1 2 5 6 1 8
378 .. 0 0 2 3 0 0 2 0
379 .. 1 1 3 4 1 1 3 1
380 .. 0 2 0 0 2 2 4 2
381 .. 1 0 1 1 3 3 5 3
382 .. 0 1 2 2 4 4 6 4
383 .. 1 2 3 3 5 5 7 5
384 .. 0 0 0 4 0 6 0 6
385 .. 1 1 1 0 1 0 1 7
386 .. 0 2 2 1 2 1 2 8
387 .. 1 0 3 2 3 2 3 0
388 .. 0 1 0 3 4 3 4 1
389 .. 1 2 1 4 5 4 5 2
390 .. 0 0 2 0 0 5 6 3
391 .. 1 1 3 1 1 6 7 4
392 .. 0 2 0 2 2 0 0 5
393 .. 1 0 1 3 3 1 1 6
394 .. 0 1 2 4 4 2 2 7
395 .. 1 2 3 0 5 3 3 8
396 .. 0 0 0 1 0 4 4 0
397 .. 1 1 1 2 1 5 5 1
398 .. 0 2 2 3 2 6 6 2
399 .. 1 0 3 4 3 0 7 3
400 .. 0 1 0 0 4 1 0 4
401 .. 1 2 1 1 5 2 1 5
402 .. 0 0 2 2 0 3 2 6
403 .. 1 1 3 3 1 4 3 7
404 .. 0 2 0 4 2 5 4 8
405 .. 1 0 1 0 3 6 5 0
406 .. 0 1 2 1 4 0 6 1
407 .. 1 2 3 2 5 1 7 2
408 .. 0 0 0 3 0 2 0 3
409 .. 1 1 1 4 1 3 1 4
410 .. 0 2 2 0 2 4 2 5
411 .. 1 0 3 1 3 5 3 6
412 .. 0 1 0 2 4 6 4 7
413 .. 1 2 1 3 5 0 5 8
414 .. 0 0 2 4 0 1 6 0
415 .. 1 1 3 0 1 2 7 1
416 .. 0 2 0 1 2 3 0 2
417 .. 1 0 1 2 3 4 1 3
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419 .. 1 2 3 4 5 6 3 5
420 .. 0 0 0 0 0 0 4 6
421 .. 1 1 1 1 1 1 5 7
422 .. 0 2 2 2 2 2 6 8
423 .. 1 0 3 3 3 3 7 0
424 .. 0 1 0 4 4 4 0 1
425 .. 1 2 1 0 5 5 1 2
426 .. 0 0 2 1 0 6 2 3
427 .. 1 1 3 2 1 0 3 4
428 .. 0 2 0 3 2 1 4 5
429 .. 1 0 1 4 3 2 5 6
430 .. 0 1 2 0 4 3 6 7
431 .. 1 2 3 1 5 4 7 8
432 .. 0 0 0 2 0 5 0 0
433 .. 1 1 1 3 1 6 1 1
434 .. 0 2 2 4 2 0 2 2
435 .. 1 0 3 0 3 1 3 3
436 .. 0 1 0 1 4 2 4 4
437 .. 1 2 1 2 5 3 5 5
438 .. 0 0 2 3 0 4 6 6
439 .. 1 1 3 4 1 5 7 7
440 .. 0 2 0 0 2 6 0 8
441 .. 1 0 1 1 3 0 1 0
442 .. 0 1 2 2 4 1 2 1
443 .. 1 2 3 3 5 2 3 2
444 .. 0 0 0 4 0 3 4 3
445 .. 1 1 1 0 1 4 5 4
446 .. 0 2 2 1 2 5 6 5
447 .. 1 0 3 2 3 6 7 6
448 .. 0 1 0 3 4 0 0 7
449 .. 1 2 1 4 5 1 1 8
450 .. 0 0 2 0 0 2 2 0
451 .. 1 1 3 1 1 3 3 1
452 .. 0 2 0 2 2 4 4 2
453 .. 1 0 1 3 3 5 5 3
454 .. 0 1 2 4 4 6 6 4
455 .. 1 2 3 0 5 0 7 5
456 .. 0 0 0 1 0 1 0 6
457 .. 1 1 1 2 1 2 1 7
458 .. 0 2 2 3 2 3 2 8
459 .. 1 0 3 4 3 4 3 0
460 .. 0 1 0 0 4 5 4 1
461 .. 1 2 1 1 5 6 5 2
462 .. 0 0 2 2 0 0 6 3
463 .. 1 1 3 3 1 1 7 4
464 .. 0 2 0 4 2 2 0 5
465 .. 1 0 1 0 3 3 1 6
466 .. 0 1 2 1 4 4 2 7
467 .. 1 2 3 2 5 5 3 8
468 .. 0 0 0 3 0 6 4 0
469 .. 1 1 1 4 1 0 5 1
470 .. 0 2 2 0 2 1 6 2
471 .. 1 0 3 1 3 2 7 3
472 .. 0 1 0 2 4 3 0 4
473 .. 1 2 1 3 5 4 1 5
474 .. 0 0 2 4 0 5 2 6
475 .. 1 1 3 0 1 6 3 7
476 .. 0 2 0 1 2 0 4 8
477 .. 1 0 1 2 3 1 5 0
478 .. 0 1 2 3 4 2 6 1
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480 .. 0 0 0 0 0 4 0 3
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483 .. 1 0 3 3 3 0 3 6
484 .. 0 1 0 4 4 1 4 7
485 .. 1 2 1 0 5 2 5 8
486 .. 0 0 2 1 0 3 6 0
487 .. 1 1 3 2 1 4 7 1
488 .. 0 2 0 3 2 5 0 2
489 .. 1 0 1 4 3 6 1 3
490 .. 0 1 2 0 4 0 2 4
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492 .. 0 0 0 2 0 2 4 6
493 .. 1 1 1 3 1 3 5 7
494 .. 0 2 2 4 2 4 6 8
495 .. 1 0 3 0 3 5 7 0
496 .. 0 1 0 1 4 6 0 1
497 .. 1 2 1 2 5 0 1 2
498 .. 0 0 2 3 0 1 2 3
499 .. 1 1 3 4 1 2 3 4
500 .. 0 2 0 0 2 3 4 5
501 .. 1 0 1 1 3 4 5 6
502 .. 0 1 2 2 4 5 6 7
503 .. 1 2 3 3 5 6 7 8
504 .. 0 0 0 4 0 0 0 0
505 .. 1 1 1 0 1 1 1 1
506 .. 0 2 2 1 2 2 2 2
507 .. 1 0 3 2 3 3 3 3
508 .. 0 1 0 3 4 4 4 4
509 .. 1 2 1 4 5 5 5 5
510 .. 0 0 2 0 0 6 6 6
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1911 .. 1 0 3 1 3 0 7 3
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1927 .. 1 1 3 2 1 2 7 1
1928 .. 0 2 0 3 2 3 0 2
1929 .. 1 0 1 4 3 4 1 3
1930 .. 0 1 2 0 4 5 2 4
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1933 .. 1 1 1 3 1 1 5 7
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1935 .. 1 0 3 0 3 3 7 0
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1937 .. 1 2 1 2 5 5 1 2
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1939 .. 1 1 3 4 1 0 3 4
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1944 .. 0 0 0 4 0 5 0 0
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1946 .. 0 2 2 1 2 0 2 2
1947 .. 1 0 3 2 3 1 3 3
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1949 .. 1 2 1 4 5 3 5 5
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1967 .. 1 2 3 2 5 0 7 5
1968 .. 0 0 0 3 0 1 0 6
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1975 .. 1 1 3 0 1 1 7 4
1976 .. 0 2 0 1 2 2 0 5
1977 .. 1 0 1 2 3 3 1 6
1978 .. 0 1 2 3 4 4 2 7
1979 .. 1 2 3 4 5 5 3 8
1980 .. 0 0 0 0 0 6 4 0
1981 .. 1 1 1 1 1 0 5 1
1982 .. 0 2 2 2 2 1 6 2
1983 .. 1 0 3 3 3 2 7 3
1984 .. 0 1 0 4 4 3 0 4
1985 .. 1 2 1 0 5 4 1 5
1986 .. 0 0 2 1 0 5 2 6
1987 .. 1 1 3 2 1 6 3 7
1988 .. 0 2 0 3 2 0 4 8
1989 .. 1 0 1 4 3 1 5 0
1990 .. 0 1 2 0 4 2 6 1
1991 .. 1 2 3 1 5 3 7 2
1992 .. 0 0 0 2 0 4 0 3
1993 .. 1 1 1 3 1 5 1 4
1994 .. 0 2 2 4 2 6 2 5
1995 .. 1 0 3 0 3 0 3 6
1996 .. 0 1 0 1 4 1 4 7
1997 .. 1 2 1 2 5 2 5 8
1998 .. 0 0 2 3 0 3 6 0
1999 .. 1 1 3 4 1 4 7 1
2000 .. 0 2 0 0 2 5 0 2
2001 .. 1 0 1 1 3 6 1 3
2002 .. 0 1 2 2 4 0 2 4
2003 .. 1 2 3 3 5 1 3 5
2004 .. 0 0 0 4 0 2 4 6
2005 .. 1 1 1 0 1 3 5 7
2006 .. 0 2 2 1 2 4 6 8
2007 .. 1 0 3 2 3 5 7 0
2008 .. 0 1 0 3 4 6 0 1
2009 .. 1 2 1 4 5 0 1 2
2010 .. 0 0 2 0 0 1 2 3
2011 .. 1 1 3 1 1 2 3 4
2012 .. 0 2 0 2 2 3 4 5
2013 .. 1 0 1 3 3 4 5 6
2014 .. 0 1 2 4 4 5 6 7
2015 .. 1 2 3 0 5 6 7 8
2016 .. 0 0 0 1 0 0 0 0
2017 .. 1 1 1 2 1 1 1 1
2018 .. 0 2 2 3 2 2 2 2
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2240 .. 0 2 0 0 2 0 0 8
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2243 .. 1 2 3 3 5 3 3 2
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2255 .. 1 2 3 0 5 1 7 5
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2303 .. 1 2 3 3 5 0 7 8
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2324 .. 0 2 0 4 2 0 4 2
2325 .. 1 0 1 0 3 1 5 3
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2327 .. 1 2 3 2 5 3 7 5
2328 .. 0 0 0 3 0 4 0 6
2329 .. 1 1 1 4 1 5 1 7
2330 .. 0 2 2 0 2 6 2 8
2331 .. 1 0 3 1 3 0 3 0
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2333 .. 1 2 1 3 5 2 5 2
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2340 .. 0 0 0 0 0 2 4 0
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2349 .. 1 0 1 4 3 4 5 0
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2351 .. 1 2 3 1 5 6 7 2
2352 .. 0 0 0 2 0 0 0 3
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2355 .. 1 0 3 0 3 3 3 6
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2357 .. 1 2 1 2 5 5 5 8
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2359 .. 1 1 3 4 1 0 7 1
2360 .. 0 2 0 0 2 1 0 2
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2431 .. 1 1 3 1 1 2 7 1
2432 .. 0 2 0 2 2 3 0 2
2433 .. 1 0 1 3 3 4 1 3
2434 .. 0 1 2 4 4 5 2 4
2435 .. 1 2 3 0 5 6 3 5
2436 .. 0 0 0 1 0 0 4 6
2437 .. 1 1 1 2 1 1 5 7
2438 .. 0 2 2 3 2 2 6 8
2439 .. 1 0 3 4 3 3 7 0
2440 .. 0 1 0 0 4 4 0 1
2441 .. 1 2 1 1 5 5 1 2
2442 .. 0 0 2 2 0 6 2 3
2443 .. 1 1 3 3 1 0 3 4
2444 .. 0 2 0 4 2 1 4 5
2445 .. 1 0 1 0 3 2 5 6
2446 .. 0 1 2 1 4 3 6 7
2447 .. 1 2 3 2 5 4 7 8
2448 .. 0 0 0 3 0 5 0 0
2449 .. 1 1 1 4 1 6 1 1
2450 .. 0 2 2 0 2 0 2 2
2451 .. 1 0 3 1 3 1 3 3
2452 .. 0 1 0 2 4 2 4 4
2453 .. 1 2 1 3 5 3 5 5
2454 .. 0 0 2 4 0 4 6 6
2455 .. 1 1 3 0 1 5 7 7
2456 .. 0 2 0 1 2 6 0 8
2457 .. 1 0 1 2 3 0 1 0
2458 .. 0 1 2 3 4 1 2 1
2459 .. 1 2 3 4 5 2 3 2
2460 .. 0 0 0 0 0 3 4 3
2461 .. 1 1 1 1 1 4 5 4
2462 .. 0 2 2 2 2 5 6 5
2463 .. 1 0 3 3 3 6 7 6
2464 .. 0 1 0 4 4 0 0 7
2465 .. 1 2 1 0 5 1 1 8
2466 .. 0 0 2 1 0 2 2 0
2467 .. 1 1 3 2 1 3 3 1
2468 .. 0 2 0 3 2 4 4 2
2469 .. 1 0 1 4 3 5 5 3
2470 .. 0 1 2 0 4 6 6 4
2471 .. 1 2 3 1 5 0 7 5
2472 .. 0 0 0 2 0 1 0 6
2473 .. 1 1 1 3 1 2 1 7
2474 .. 0 2 2 4 2 3 2 8
2475 .. 1 0 3 0 3 4 3 0
2476 .. 0 1 0 1 4 5 4 1
2477 .. 1 2 1 2 5 6 5 2
2478 .. 0 0 2 3 0 0 6 3
2479 .. 1 1 3 4 1 1 7 4
2480 .. 0 2 0 0 2 2 0 5
2481 .. 1 0 1 1 3 3 1 6
2482 .. 0 1 2 2 4 4 2 7
2483 .. 1 2 3 3 5 5 3 8
2484 .. 0 0 0 4 0 6 4 0
2485 .. 1 1 1 0 1 0 5 1
2486 .. 0 2 2 1 2 1 6 2
2487 .. 1 0 3 2 3 2 7 3
2488 .. 0 1 0 3 4 3 0 4
2489 .. 1 2 1 4 5 4 1 5
2490 .. 0 0 2 0 0 5 2 6
2491 .. 1 1 3 1 1 6 3 7
2492 .. 0 2 0 2 2 0 4 8
2493 .. 1 0 1 3 3 1 5 0
2494 .. 0 1 2 4 4 2 6 1
2495 .. 1 2 3 0 5 3 7 2
2496 .. 0 0 0 1 0 4 0 3
2497 .. 1 1 1 2 1 5 1 4
2498 .. 0 2 2 3 2 6 2 5
2499 .. 1 0 3 4 3 0 3 6
2500 .. 0 1 0 0 4 1 4 7
2501 .. 1 2 1 1 5 2 5 8
2502 .. 0 0 2 2 0 3 6 0
2503 .. 1 1 3 3 1 4 7 1
2504 .. 0 2 0 4 2 5 0 2
2505 .. 1 0 1 0 3 6 1 3
2506 .. 0 1 2 1 4 0 2 4
2507 .. 1 2 3 2 5 1 3 5
2508 .. 0 0 0 3 0 2 4 6
2509 .. 1 1 1 4 1 3 5 7
2510 .. 0 2 2 0 2 4 6 8
2511 .. 1 0 3 1 3 5 7 0
2512 .. 0 1 0 2 4 6 0 1
2513 .. 1 2 1 3 5 0 1 2
2514 .. 0 0 2 4 0 1 2 3
2515 .. 1 1 3 0 1 2 3 4
2516 .. 0 2 0 1 2 3 4 5
2517 .. 1 0 1 2 3 4 5 6
2518 .. 0 1 2 3 4 5 6 7
2519 .. 1 2 3 4 5 6 7 8
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#### To solve this example are needed these knowledge from mathematics:
Do you want to calculate least common multiple two or more numbers?
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On the final certificate have one quarter of the class 9A mark "C" of mathematics, seventh mark "C" from the Czech language and two students failed in chemistry. How many students attend class 9A? | 50,436 | 63,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-13 | longest | en | 0.234972 |
http://www.reddit.com/r/programming/comments/9brvx/dont_worry_most_smart_people_dont_get_this_easy/c0c5vxv | 1,409,143,523,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829210.91/warc/CC-MAIN-20140820021349-00406-ip-10-180-136-8.ec2.internal.warc.gz | 570,084,102 | 11,976 | you are viewing a single comment's thread.
[–] -2 points-1 points (2 children)
sorry, this has been archived and can no longer be voted on
These people are fucking idiots. A coin toss is random. You could get HTH every fucking time on Monday and get heads 1 million times in a row on Tuesday. It's RANDOM. FUCK YOU!
[–] 2 points3 points (0 children)
sorry, this has been archived and can no longer be voted on
Can someone explain to me why (aside from the colorful language) this is being downvoted? The probability of landing on a particular side of a coin (1/2) is per individual coin flip not across X number of coin flips. You can't just extrapolate that probability across two days worth of coin flips and say with absolute certainty that the result will be higher on day two.
[–] 3 points4 points (0 children)
sorry, this has been archived and can no longer be voted on
That bugged me a little too. They should have said "which number is most likely to be higher". Still, the probability of Tuesday's number coming out higher than Monday's is rather close to zero, assuming you aren't the world's slowest coin flipper. | 271 | 1,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-35 | latest | en | 0.958117 |
https://edustrings.com/physics/2271754.html | 1,642,633,123,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00213.warc.gz | 283,626,599 | 7,100 | 28 August, 23:36
# The following forces act on an object: 10 N north, 7 N south, and 4 N east. What is the magnitude of the net force?
+1
1. 29 August, 01:31
0
Given data;
Fn = 10 N
Fs = 7 N
Fe = 4 N
force in X direction (Fx) = 4 N
force in Y direction (Fy) = 10-7 = 3 N
Net force (Fnet) = Sq. root[ (Fx) ² + (Fy) ²]
= Sq root [ 4² + 3² ]
= 25 N
Net force acting = 25 N | 158 | 380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-05 | longest | en | 0.882684 |
https://numberworld.info/131113214 | 1,603,892,899,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00197.warc.gz | 450,380,976 | 3,943 | # Number 131113214
### Properties of number 131113214
Cross Sum:
Factorization:
2 * 17 * 607 * 6353
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
7d0a0fe
Base 32:
3t187u
sin(131113214)
0.65071561775
cos(131113214)
0.75932152927218
tan(131113214)
0.8569697982536
ln(131113214)
18.691571736942
lg(131113214)
8.1176464634528
sqrt(131113214)
11450.467850704
Square(131113214)
### Number Look Up
Look Up
131113214 (one hundred thirty-one million one hundred thirteen thousand two hundred fourteen) is a special figure. The cross sum of 131113214 is 17. If you factorisate the figure 131113214 you will get these result 2 * 17 * 607 * 6353. The figure 131113214 has 16 divisors ( 1, 2, 17, 34, 607, 1214, 6353, 10319, 12706, 20638, 108001, 216002, 3856271, 7712542, 65556607, 131113214 ) whith a sum of 208614528. The figure 131113214 is not a prime number. The figure 131113214 is not a fibonacci number. The figure 131113214 is not a Bell Number. The number 131113214 is not a Catalan Number. The convertion of 131113214 to base 2 (Binary) is 111110100001010000011111110. The convertion of 131113214 to base 3 (Ternary) is 100010201020111222. The convertion of 131113214 to base 4 (Quaternary) is 13310022003332. The convertion of 131113214 to base 5 (Quintal) is 232031110324. The convertion of 131113214 to base 8 (Octal) is 764120376. The convertion of 131113214 to base 16 (Hexadecimal) is 7d0a0fe. The convertion of 131113214 to base 32 is 3t187u. The sine of the figure 131113214 is 0.65071561775. The cosine of 131113214 is 0.75932152927218. The tangent of 131113214 is 0.8569697982536. The square root of 131113214 is 11450.467850704.
If you square 131113214 you will get the following result 17190674885409796. The natural logarithm of 131113214 is 18.691571736942 and the decimal logarithm is 8.1176464634528. that 131113214 is very unique number! | 716 | 2,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.698412 |
https://community.khronos.org/t/drawing-normals-as-lines/65717 | 1,656,850,791,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104240553.67/warc/CC-MAIN-20220703104037-20220703134037-00020.warc.gz | 222,391,881 | 4,150 | # Drawing Normals as Lines
Hi, I’ve been having real trouble figuring out how to draw normals as lines. Right now for my project I’m working on smoothing out a cylinder. I’ve created normals for the faces and so far it looks fine. I just can’t grasp how to draw lines to indicate where the normals are.
Here’s a snipped of code.
// draw bottom boundary:
glBegin ( GL_TRIANGLE_STRIP);
p1[0] = ctx; p1[1] = cty; p1[2] = ctz;
p2[0] = ntx; p2[1] = nty; p2[2] = ntz;
p3[0] = 0; p3[1] = -len; p3[2] = 0;
getNormals( p1, p2, p3, topTriangle);
glNormal3f(topTriangle[0],topTriangle[1],topTriangle[2]);
glVertex3d ( cbx, cby, cbz ); glVertex3d ( nbx, nby, nbz ); glVertex3d ( 0, -len, 0 );
glEnd();
This is for the bottom circle of my cylinder.
Any help would be greatly appreciated.
The starting point of the line will be the vertex.
The end point will be thevertex+normal * some_length_factor
This is not a end point. This is just a vector.
p2[0] = ntx; p2[1] = nty; p2[2] = ntz; | 328 | 977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.746209 |
https://www.coursehero.com/file/8951486/Table-Use-The-future-value-interest-factor-for-an-initial-principal/ | 1,484,640,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00033-ip-10-171-10-70.ec2.internal.warc.gz | 898,157,559 | 21,611 | # Table use the future value interest factor for an
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Unformatted text preview: initial principal of \$1 on deposit for 5 years at 6% interest compounded annually, FVIF6%, 5yrs, found in Table A–1, is 1.338. Using Equation 4.6, \$800 1.338 \$1,070.40. Therefore, the future value of Jane’s deposit at the end of year 5 will be \$1,070.40. Input 800 Function PV 5 N 6 I CPT FV Solution 1070.58 Calculator Use4 The financial calculator can be used to calculate the future value directly.5 First punch in \$800 and depress PV; next punch in 5 and depress N; then punch in 6 and depress I (which is equivalent to “i” in our notation)6; finally, to calculate the future value, depress CPT and then FV. The future value of \$1,070.58 should appear on the calculator display as shown at the left. On many calculators, this value will be preceded by a minus sign ( 1,070.58). If a minus sign appears on your calculator, ignore it here as well as in all other “Calculator Use” illustrations in this text.7 Because the calculator is more accurate than the future value factors, which have been rounded to the nearest 0.001, a slight difference—in this case,...
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Ask a homework question - tutors are online | 346 | 1,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-04 | longest | en | 0.860565 |
http://www.jiskha.com/display.cgi?id=1275116015 | 1,455,135,549,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701160582.7/warc/CC-MAIN-20160205193920-00093-ip-10-236-182-209.ec2.internal.warc.gz | 482,435,931 | 3,921 | Wednesday
February 10, 2016
# Homework Help: Physics
Posted by Richard on Saturday, May 29, 2010 at 2:53am.
Having trouble determ the formulas for the follow problems.
2. If an automobile travels at 30 km/hr. for 2 hours, how far does it go?
60 km
3. At time = 0 an automobile already traveling at 30 km/hr. accelerates at 3m/s2 for 2 seconds. During these two seconds, how far does it go?
4. An automobile is traveling at 30 km/hr. It accelerates at 3 m/s2 until it has covered 100 meters.. What is its new velocity?
• Physics - drwls, Saturday, May 29, 2010 at 5:32am
Let D = distance travelled and Vo be the initial velocity, and a be the acceleration
2. D = Vo * t (no acceleration; V remains Vo)
3. D = Vo*t + (1/2) a t^2
Vo = 8 1/3 m/s ; answer will be in meters
4. Vfinal^2 = Vo^2 + 2 a D
Vfinal = Vo + a t,
but you need to solve for t to use the second formula. The two formulas are consistent. | 282 | 911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2016-07 | longest | en | 0.902317 |
http://www.jiskha.com/display.cgi?id=1283894303 | 1,498,438,415,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320595.24/warc/CC-MAIN-20170625235624-20170626015624-00695.warc.gz | 557,329,778 | 3,755 | # physics
posted by .
If 30 N of force is exerted over an area of 20 m2, how much pressure is being applied?
• physics -
pressure= force/area
• physics -
oh yes, units. remember that pressure has a special name. You can use N/m^2, but in honor of Pascal, we call 1N/m^2 a Pascal, symbol P. This is not a very large unit, atmospheric pressure is 101.3kP | 102 | 358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-26 | latest | en | 0.922978 |
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410Hw04ansJeff-1
# 410Hw04ansJeff-1 - STAT 410 Fall 2011 Homework#4(due Friday...
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STAT 410 Fall 2011 Homework #4 (due Friday, September 23, by 3:00 p.m.) 1. Let X be a Uniform ( 0, 1 ) and Y be a Uniform ( 0, 3 ) independent random variables. Let W = X + Y. Find and sketch the p.d.f. of W. ( ) < < = otherwise 0 1 0 1 X x x f ( ) < < = otherwise 0 3 0 3 1 Y y y f ( ) ( ) ( ) - - + = dx x w f x f w f Y X Y X ( ) < < = otherwise 0 3 0 3 1 Y y y f ( ) < < - < - < - = = otherwise 0 3 3 1 otherwise 0 3 0 3 1 Y w x w x w x w f Case 1: 0 < w < 1. f W ( w ) = w dx 0 3 1 1 = w 3 1 . Case 2: 1 < w < 3. f W ( w ) = 1 0 3 1 1 dx = 3 1 . Case 3: 3 < w < 4. f W ( w ) = - 1 3 3 1 1 w dx = ( ) 4 3 1 w - .
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OR ( ) ( ) ( ) - - + = dy y f y w f w f Y X Y X ( ) < < = otherwise 0 1 0 1 X x x f ( ) < < - = < - < = - otherwise 0 1 1 otherwise 0 1 0 1 X w y w y w y w f Case 1: 0 < w < 1. f W ( w ) = w dy 0 3 1 1 = w 3 1 . Case 2: 1 < w < 3. f W ( w ) = - w w dy 1 3 1 1 = 3 1 . Case 3: 3 < w < 4. f W ( w ) = - 3 1 3 1 1 w dy = ( ) 4 3 1 w - .
OR F W ( w ) = P ( W w ) = P ( X + Y w ) = Case 1: 0 < w < 1. Case 2: 1 < w < 3. Case 3: 3 < w < 4. = - w x w dx dy 0 0 3 1 1 = - 1 0 0 3 1 1 dx dy x w = - - - 1 3 3 3 1 1 1 w x w dx dy = 2 6 1 w . = ( ) 1 2 6 1 - w . = ( ) 2 4 6 1 1 w - - . f W ( w ) = F W ' ( w ) = = w 3 1 , = 3 1 , = ( ) 4 3 1 w - , 0 < w < 1. 1 < w < 3. 3 < w < 4.
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2. Let random variables X and Y have an Exponential distribution with mean 1 and a Uniform distribution on ( 0 , 1 ) , respectively. Suppose X and Y are independent. Find the probability density function of W = X + Y. ( ) > = - otherwise 0 0 X x x f x e ( ) < < = otherwise 0 1 0 1 Y y y f ( ) < < - = < - < = - otherwise 0 1 1 otherwise 0 1 0 1 Y w x w x w x w f Case 1: w < 0. ( ) w f Y X + = 0. Case 2: 0 < w < 1. Then w – 1 < 0. ( ) ( ) ( ) ( ) - - + = - = w x dx dx x w f x f w f e 0 Y X Y X 1 = 1 – e w .
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{[ snackBarMessage ]} | 1,049 | 2,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-17 | latest | en | 0.687405 |
https://vladgkh.ru/ocr-physics-a-coursework-10082.html | 1,623,707,124,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00245.warc.gz | 545,333,463 | 9,678 | # Ocr Physics A Coursework
Tags: Amplifier Design ThesisOedipus Argument EssayTypes Of Plans In BusinessCritical Thinking Activities In Patterns Imagery LogicEssay Of Frederick Douglass LifeWork Life Balance Thesis ProposalReading Computer Science Research PaperNative American Thesis Statement
The basic concepts of Physical Chemistry will also be covered these will include Enthalpy changes, Reaction Rates and Chemical Equilibrium.
5 A* to C passes at GCSE which should include at least grade A’s in Double Science (or Chemistry) and a B in Mathematics.
I am therefore looking at the relationship between the distance the object fell, and its velocity as it passed through the light gate.
An average of the velocities measured in each experiment has been calculated and the height at which the weight was dropped has been multiplied by 1000 to convert it to metres. I added a curved line of best fit to the graph, showing positive correlation between the distance fallen and the velocity of the object as it passed through the light gate.
I have therefore added range bars to the velocity-time graph using the table below to show the maximum and minimum values for velocity recorded around the average.
The graph shows that both energy forms increase in direct proportion to the distance fallen.
I am therefore going to investigate whether there is any relationship between the distance fallen and the difference between energy forms, to see where the kinetic energy was lost – most likely as heat from friction.
There seems to be little correlation between the difference in potential and kinetic energy and the distance the object fell from.
Also the beam is being broken when the total mass has not travelled as far, which, in relation to the very first graph would also mean the velocity could be less.
Another source of error could potentially come from the way the metal strip is attached.
## Comments Ocr Physics A Coursework
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OCR AS and A Level Physics A qualification information including specification, exam materials, teaching resources, learning resources.… | 837 | 4,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.946117 |
https://www.teacherspayteachers.com/Product/School-Number-Sense-1-10-counting-matching-reading-writing-numbers-1-10-3557080 | 1,529,338,666,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00138.warc.gz | 922,549,137 | 18,207 | # School Number Sense 1-10 counting, matching, reading & writing numbers 1-10
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Product Description
Students mastering their numbers to 10 is an important part of developing their number sense. That's where this set can help. In this school themed pack your students reading, writing, 1 to 1 counting, matching and learning their numbers 1-10 as they work through the task cards.
Note this activity is part of a money saving Bundle which you can check out HERE
These activities are designed to give your students practice in counting and identifying numbers to 10 through;
◼️1 to 1 counting of pictures (this helps introduce and learn the patterns of each number – subitizing)
◼️Reading and understanding the standard number patterns in a 10’s frame
◼️Reading and writing numbers 1 to 10 (note the number squares for matching have an orientation dot to aid students in knowing which way up the number goes, which will help with number confusion).
The cards come in two versions;
►The first set (fill in) has your students counting the pictures and filling in the 10’s frame to show the correct number then writing the number in the number box.
►In the second set (matching) your students will count the pictures and finding the matching 10's frame and number card to complete the strip
You can choose to print out both sets and have your students working on the activity that suits them best or just the one style. Will work great as a math center or morning tub activity.
To prep the center; print (cardstock will make them sturdier) and laminate. Cut into strips and provide students with either dry erase marker or the cut out tens frame & numbers (depending on which set you are using) and have them complete the activity.
Click on any of these titles to check out the other sets in this series:
Number Sense 1-10 - BUNDLE - counting, matching, reading & writing numbers 1-10
Winter Number Sense 1-10 counting, matching, reading & writing numbers 1-10
New Year Number Sense 1-10 counting, matching, reading & writing numbers 1-10
Monster Number Sense 1-10 counting, matching, reading & writing numbers 1-10
Dinosaur Number Sense 1-10 counting, matching, reading & writing numbers 1-10
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 768 | 3,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-26 | latest | en | 0.916274 |
https://codereview.stackexchange.com/questions/46320/how-can-i-make-my-solution-for-chef-and-digits-run-faster | 1,721,368,455,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00393.warc.gz | 148,096,622 | 44,385 | # How can I make my solution for Chef and Digits run faster?
## Problem Statement
Given $n$ digits and $m$ indices $x$ from $1. \ldots,n$, calculate the difference $b_y = a_x - a_y$ for all indices $y \; (y < x)$. Then, calculate $B1$, the sum of all $b_y$ which are greater than $0$ and $B2$, the sum of all $b_y$ which are less than $0$. The answer for this step is $B1 - B2$.
Input
The first line contains two integers $n, m$ denoting the number of digits and number of steps. The second line contains $n$ digits (without spaces) $a_1, a_2, ..., a_n$. Each of next $m$ lines contains single integer $x$ denoting the index for current step. (ADIGIT: Chef and Digits from CodeChef)
## My Issue
I've tried quite hard to optimize the following solution, but online judge always shows Time Limit Exceeded.
## My Question
Is it possible to optimize this code further to bring time down less than 1 second, or should I change my approach or perhaps shift to C?
import sys
final=[]
for i in xrange(y):
b1=b2=0
curr=int(input())-1
for x in stra[:curr]:
mu=int(stra[curr])-int(x)
if(mu>0):
b1+=mu
else:
b2+=mu
final.append(str(b1-b2))
print("\n".join(final))
• Don't know if it will speed it up much, but put the int(stra[curr]) before the inner for loop. Commented Apr 4, 2014 at 20:40
• Did you master the problem? Commented Aug 27, 2014 at 10:15
You didn't put any effort into choosing good names. Given a code challenge, either stick to the variable names used in the problem statement (n, m, ...) or, preferably, choose explicit and descriptive names such as steps, digits and indices rather than z, y, etc.
Why should you bother with this? Because good names reveal intent, and knowing what the code is supposed to do makes it possible for you and others to understand and improve it easily.
Your algorithm uses a naive, brute-force approach. This is fine as a starting point!
But when you get a time limit exceeded message, you need to think about smarter solutions.
• Rene's answer gives you some good advice: He suggests that you employ caching, separate I/O from the calculation, and add up all the absolute values of by, thus skipping the calculation of B1 and B2. This is a good start but you can go much further.
• Janne's answer tells you the essence of what's wrong with your algorithm: you need to avoid looping over the previous digits for every step. Why? Because there will be up to 105 (one hundred thousand) digits and up to 105 steps. However, it is unnecessary to make a sorted copy of the indices — see below.
# How?
Start by separating your concerns. Python has functions, so use them to group relevant parts of your code together, while keeping different levels of abstraction separate. Start at the most general level:
def main():
print(*solve(sys.stdin), sep='\n')
We are saying: print each line of the result of solving the problem with the input from sys.stdin.
Our next concern is to parse the input:
def solve(lines):
_, digits, *indices = lines
We discard the first line _ because we can deduce the length of the digits and the number of steps from the rest of the input, so there is no need to waste cycles parsing them. We then store the digits which are given in the second line, as well as the indices from the following lines.
To get rid of the redundancies in your integer-conversion code, we extract a function:
def as_ints(iterable):
return [int(x) for x in iterable]
We can pass the digits (stripping the newline character at the end) and the indices to this function to obtain usable values which we can use to calculate_answer(digits, indices). Since we have already parsed the input, this function can be focused on our algorithm to solve the problem.
def calculate_answer(digits, indices):
1. Store the indices in a dictionary, which allows them to be looked up in constant time.
result = dict.fromkeys(indices)
2. Set up another dictionary to count the number of occurrences of each unique digit, which will help us avoid iterating over all previous digits on every step.
counts = defaultdict(int)
3. Iterate over the digits and keep track of each digit's index, starting with 1
for index, digit in enumerate(digits, start=1):
4. If the current index is in result and we have not calculated a value for it yet:
Assign to the current index the sum of all the absolute values of the differences between the current digit and each unique previous digit d multiplied by its occurrences n.
if index in result and result[index] is None:
result[index] = sum(n * abs(digit - d) for d, n in counts.items())
5. Increment the number of occurrences of the digit we just encountered:
counts[digit] += 1
6. In the order of the given indices, retrieve each calculated value from result.
return (result[i] for i in indices)
# Result
The end result is fast enough to be accepted by CodeChef. As you didn't specify a language version, this is in Python 3, which you should use as well. If you insist on sticking with Python 2.x, you will need to make the necessary adjustments.
import sys
from collections import defaultdict
def main():
print(*solve(sys.stdin), sep='\n')
def solve(lines):
_, digits, *indices = lines
result = dict.fromkeys(indices)
counts = defaultdict(int)
for index, digit in enumerate(digits, start=1):
if index in result and result[index] is None:
result[index] = sum(n * abs(digit - d) for d, n in counts.items())
counts[digit] += 1
return (result[i] for i in indices)
def as_ints(iterable):
return [int(x) for x in iterable]
if __name__ == '__main__':
main()
A better algorithm usually wins over micro-optimizations.
• Do you really need to do all these calculations every time? Maybe you could cache them and re-use the results.
• You could perform all the string to int conversions up front, afterwards only looking up numbers from array.
• This whole b1 & b2 thing could be replaced with simple abs() function call. Not sure if it would speed things up, but it sure would simplify the code.
This would avoid looping m times through the string:
1. Sort a copy of the list of indices x given in the input.
2. Sweep through the string, counting the occurrences of each digit as you go along. Each time you hit one of the x, compute the answer from the current counts. Put the answer in a dict indexed by x.
3. Output the answers in correct order by looping through the original list and looking up the value in the dict. | 1,523 | 6,429 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.874377 |
flooringinc.site | 1,582,139,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00175.warc.gz | 375,643,591 | 8,036 | /Which Are Constant in T?
What are constants in mathematics?
Indeed, it’s a little tricky. You need to be familiar with the concept of a consistent and know the reason why they are important for you to understand how to find an understanding of how to read and also write equations.
You need to realize that a consistent is just a concept which is going to be there and will never change. They are not individual created, paramountessays.com but there by character.
Certainly one of the concepts you ought to know is that there are two types of continuous. All these are what we predict negative and positive constants.
Constants are and you could notice that the gap between what’s being subtracted. Positive constants are such which you’ll be able to incorporate to one other.
In general would be that they exist and are never likely to change. Needless to say you need to be aware that there are many sorts of mathematics. You can find a few factors that are termed constants.
Some of these constants are typical and a few are not. college admissions essay help By way of example, a circle’s area would be a continuing. There are numerous far more of those matters that as a way to get to this degree of knowledge the way to to read and create equations you have to master.
Most math’s idea is there are things that are all many others yet constants aren’t. This can help you obtain to the idea where you are familiar with all the concepts of how to read and also create equations.
You can find a number of important concepts that you need to stay in mind Whenever you are starting to know how to browse and also write equations. You want to continue in mind that every equation can be a method of equations, if you wish to actually know what are constants in math. All specimens can be written in the kind of y = mx, in which y is that the existing variable and m is the factor that has been shifted.
Here is another concept that you want to keep in mind: when you’re reading the equation, you must try to remember the first section is your very first part as well http://csmbio.csm.jmu.edu/biology/danie2jc/urinalysis.htm as the part may be your work. That really is very important that you comprehend. In addition, you need to realize they each have a name and there are various types of equations.
At calculus, for example, the expression is sometimes called the derivative. In calculus, the definition of may be known as the derivative. It’s employed when we have been currently trying to learn the slope of the curve.
So, you can see that there are lots of constants you have to master. These are those which so you could comprehend the notions of the way to browse and also compose equations you will need to be aware. | 568 | 2,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-10 | latest | en | 0.97751 |
https://tireconverter.com/mm-to-inch | 1,696,011,396,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00607.warc.gz | 615,359,419 | 6,070 | # Convert Millimeters to Inches
Easy to use online tool to convert millimeters to inches with detailed formulas and definitions.
Millimeters
Inches
## How many inches in a millimeter?
There are 0.03937 inches in a millimeter.
## How to convert millimeters to inches ?
To convert millimeters to inches, you can use the conversion factor of 0.03937 inch per mm. That means you multiply the number of millimeters by 0.03937 to get the equivalent measurement in inches. Alternatively, you can divide the number of millimeters by 25.4 to get the same result.
Formula to convert millimeters to inches:
1 mm = (1 / 25.4)" = 0.03937".
Learn how to convert inches to millimeters.
## Inch to Millimeter Conversion Chart
• 10 mm to inches equals 0.39 inches
• 15 mm to inches equals 0.59 inches
• 20 mm to inches equals 0.79 inches
• 25 mm to inches equals 0.98 inches
• 30 mm to inches equals 1.18 inches
• 35 mm to inches equals 1.38 inches
• 40 mm to inches equals 1.57 inches
• 45 mm to inches equals 1.77 inches
• 50 mm to inches equals 1.97 inches
• 55 mm to inches equals 2.17 inches
• 60 mm to inches equals 2.36 inches
• 65 mm to inches equals 2.56 inches
• 70 mm to inches equals 2.76 inches
• 75 mm to inches equals 2.95 inches
• 80 mm to inches equals 3.15 inches
• 85 mm to inches equals 3.35 inches
• 90 mm to inches equals 3.54 inches
• 95 mm to inches equals 3.74 inches
• 100 mm to inches equals 3.94 inches
• 105 mm to inches equals 4.13 inches
• 110 mm to inches equals 4.33 inches
• 115 mm to inches equals 4.53 inches
• 120 mm to inches equals 4.72 inches
• 125 mm to inches equals 4.92 inches
• 130 mm to inches equals 5.12 inches
• 135 mm to inches equals 5.31 inches
• 140 mm to inches equals 5.51 inches
• 145 mm to inches equals 5.71 inches
• 150 mm to inches equals 5.91 inches
• 155 mm to inches equals 6.1 inches
• 160 mm to inches equals 6.3 inches
• 165 mm to inches equals 6.5 inches
• 170 mm to inches equals 6.69 inches
• 175 mm to inches equals 6.89 inches
• 180 mm to inches equals 7.09 inches
• 185 mm to inches equals 7.28 inches
• 190 mm to inches equals 7.48 inches
• 195 mm to inches equals 7.68 inches
• 200 mm to inches equals 7.87 inches
• 205 mm to inches equals 8.07 inches
• 210 mm to inches equals 8.27 inches
• 215 mm to inches equals 8.46 inches
• 220 mm to inches equals 8.66 inches
• 225 mm to inches equals 8.86 inches
• 230 mm to inches equals 9.06 inches
• 235 mm to inches equals 9.25 inches
• 240 mm to inches equals 9.45 inches
• 245 mm to inches equals 9.65 inches
• 250 mm to inches equals 9.84 inches
• 255 mm to inches equals 10.04 inches
• 260 mm to inches equals 10.24 inches
• 265 mm to inches equals 10.43 inches
• 270 mm to inches equals 10.63 inches
• 275 mm to inches equals 10.83 inches
• 280 mm to inches equals 11.02 inches
• 285 mm to inches equals 11.22 inches
• 290 mm to inches equals 11.42 inches
• 295 mm to inches equals 11.61 inches
• 300 mm to inches equals 11.81 inches
• 305 mm to inches equals 12.01 inches
• 310 mm to inches equals 12.2 inches
• 315 mm to inches equals 12.4 inches
• 320 mm to inches equals 12.6 inches
• 325 mm to inches equals 12.8 inches
• 330 mm to inches equals 12.99 inches
• 335 mm to inches equals 13.19 inches
• 340 mm to inches equals 13.39 inches
• 345 mm to inches equals 13.58 inches
• 350 mm to inches equals 13.78 inches
• 355 mm to inches equals 13.98 inches
• 360 mm to inches equals 14.17 inches
• 365 mm to inches equals 14.37 inches
• 370 mm to inches equals 14.57 inches
• 375 mm to inches equals 14.76 inches
• 380 mm to inches equals 14.96 inches
• 385 mm to inches equals 15.16 inches
• 390 mm to inches equals 15.35 inches
• 395 mm to inches equals 15.55 inches
• 400 mm to inches equals 15.75 inches
• 405 mm to inches equals 15.94 inches
• 410 mm to inches equals 16.14 inches
• 415 mm to inches equals 16.34 inches
• 420 mm to inches equals 16.54 inches
• 425 mm to inches equals 16.73 inches
• 430 mm to inches equals 16.93 inches
• 435 mm to inches equals 17.13 inches
• 440 mm to inches equals 17.32 inches
• 445 mm to inches equals 17.52 inches
• 450 mm to inches equals 17.72 inches
• 455 mm to inches equals 17.91 inches
• 460 mm to inches equals 18.11 inches
• 465 mm to inches equals 18.31 inches
• 470 mm to inches equals 18.5 inches
• 475 mm to inches equals 18.7 inches
• 480 mm to inches equals 18.9 inches
• 485 mm to inches equals 19.09 inches
• 490 mm to inches equals 19.29 inches
• 495 mm to inches equals 19.49 inches
• 500 mm to inches equals 19.69 inches
• 505 mm to inches equals 19.88 inches
• 510 mm to inches equals 20.08 inches
• 515 mm to inches equals 20.28 inches
• 520 mm to inches equals 20.47 inches
• 525 mm to inches equals 20.67 inches
• 530 mm to inches equals 20.87 inches
• 535 mm to inches equals 21.06 inches
• 540 mm to inches equals 21.26 inches
• 545 mm to inches equals 21.46 inches
• 550 mm to inches equals 21.65 inches
• 555 mm to inches equals 21.85 inches
• 560 mm to inches equals 22.05 inches
• 565 mm to inches equals 22.24 inches
• 570 mm to inches equals 22.44 inches
• 575 mm to inches equals 22.64 inches
• 580 mm to inches equals 22.83 inches
• 585 mm to inches equals 23.03 inches
• 590 mm to inches equals 23.23 inches
• 595 mm to inches equals 23.43 inches
• 600 mm to inches equals 23.62 inches
• 605 mm to inches equals 23.82 inches
• 610 mm to inches equals 24.02 inches
• 615 mm to inches equals 24.21 inches
• 620 mm to inches equals 24.41 inches
• 625 mm to inches equals 24.61 inches
• 630 mm to inches equals 24.8 inches
• 635 mm to inches equals 25 inches
• 640 mm to inches equals 25.2 inches
• 645 mm to inches equals 25.39 inches
• 650 mm to inches equals 25.59 inches
• 655 mm to inches equals 25.79 inches
• 660 mm to inches equals 25.98 inches
• 665 mm to inches equals 26.18 inches
• 670 mm to inches equals 26.38 inches
• 675 mm to inches equals 26.57 inches
• 680 mm to inches equals 26.77 inches
• 685 mm to inches equals 26.97 inches
• 690 mm to inches equals 27.17 inches
• 695 mm to inches equals 27.36 inches
• 700 mm to inches equals 27.56 inches
• 705 mm to inches equals 27.76 inches
• 710 mm to inches equals 27.95 inches
• 715 mm to inches equals 28.15 inches
• 720 mm to inches equals 28.35 inches
• 725 mm to inches equals 28.54 inches
• 730 mm to inches equals 28.74 inches
• 735 mm to inches equals 28.94 inches
• 740 mm to inches equals 29.13 inches
• 745 mm to inches equals 29.33 inches
• 750 mm to inches equals 29.53 inches
• 755 mm to inches equals 29.72 inches
• 760 mm to inches equals 29.92 inches
• 765 mm to inches equals 30.12 inches
• 770 mm to inches equals 30.31 inches
• 775 mm to inches equals 30.51 inches
• 780 mm to inches equals 30.71 inches
• 785 mm to inches equals 30.91 inches
• 790 mm to inches equals 31.1 inches
• 795 mm to inches equals 31.3 inches
• 800 mm to inches equals 31.5 inches
• 805 mm to inches equals 31.69 inches
• 810 mm to inches equals 31.89 inches
• 815 mm to inches equals 32.09 inches
• 820 mm to inches equals 32.28 inches
• 825 mm to inches equals 32.48 inches
• 830 mm to inches equals 32.68 inches
• 835 mm to inches equals 32.87 inches
• 840 mm to inches equals 33.07 inches
• 845 mm to inches equals 33.27 inches
• 850 mm to inches equals 33.46 inches
• 855 mm to inches equals 33.66 inches
• 860 mm to inches equals 33.86 inches
• 865 mm to inches equals 34.06 inches
• 870 mm to inches equals 34.25 inches
• 875 mm to inches equals 34.45 inches
• 880 mm to inches equals 34.65 inches
• 885 mm to inches equals 34.84 inches
• 890 mm to inches equals 35.04 inches
• 895 mm to inches equals 35.24 inches
• 900 mm to inches equals 35.43 inches
• 905 mm to inches equals 35.63 inches
• 910 mm to inches equals 35.83 inches
• 915 mm to inches equals 36.02 inches
• 920 mm to inches equals 36.22 inches
• 925 mm to inches equals 36.42 inches
• 930 mm to inches equals 36.61 inches
• 935 mm to inches equals 36.81 inches
• 940 mm to inches equals 37.01 inches
• 945 mm to inches equals 37.2 inches
• 950 mm to inches equals 37.4 inches
• 955 mm to inches equals 37.6 inches
• 960 mm to inches equals 37.8 inches
• 965 mm to inches equals 37.99 inches
• 970 mm to inches equals 38.19 inches
• 975 mm to inches equals 38.39 inches
• 980 mm to inches equals 38.58 inches
• 985 mm to inches equals 38.78 inches
• 990 mm to inches equals 38.98 inches
• 995 mm to inches equals 39.17 inches
• 1000 mm to inches equals 39.37 inches
• 1005 mm to inches equals 39.57 inches
All listed guides, data and/or calculations are for informational purposes only. TireConverter.com does not warrant or make any representations regarding the accuracy of or the results of the use of this information. | 2,810 | 8,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | latest | en | 0.644445 |
http://pissedoffteeacher.blogspot.com/2010/05/another-child-being-left-behind.html | 1,526,893,062,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00073.warc.gz | 236,398,436 | 20,493 | ## Monday, May 31, 2010
### Another Child Being Left Behind
Julia could do almost nothing on Thursday's exam. Even using notes, her mind went totally blank. I was sure she knew nothing because she spent most of every period laughing and carrying on with her two friends.
Most of the class did horribly, even the few that work so Friday I decided to just redo some of the same problems on the board. Julia's buddies decided to make the three day weekend a four day weekend, which left Julia bored enough to actually work.
Julia wrote down the questions and then called me over to help her. I asked her to read the question and then began questioning her about the words? "What is the relationship between the diameter of a circle and its radius?" Without flinching, and in a complete sentence she said, "The radius is equal to half the diameter." I then told her to look at the problem again and see what she could do. "The diameter is 4x + 12 and the radius is 48 so 2x + 6 = 48. She then wrote the equation and solved it.
I went over to her again a few times during the period and every question we discussed, she did correctly. I praised her and asked her if she would like to take her final exam orally as she seems to do much better that way. She smiled and said yes. (I don't know if she will actually show up to do it but I am hopeful she really wants to pass.)
Julia is in the lowest third of our school. She obviously has learning difficulties that were never addressed but I never doubted the very intelligent girl that lives within her. We have data up the kazoo, data showing that Julia has a hard time in school, both behaviorally and academically, yet nothing has ever been done to address her problems and help her overcome them. I feel awful. Julia will only be with me for another 9 days and I have done so little to help her. How long will it take her new teacher to pick up on her problem? Suppose I find the teacher, tell him about how best to work with her and the teacher hasn't the time or the ambition? Will Julia just become another statistic in the world of drop outs? Why isn't any money being used to help kids like her?
#### 1 comment:
Anonymous said...
This student never received any services?? I believe you have the right as one of her teachers to refer her. I was always amazed that kids I had in the upper elementary grades were never referred. And I would hear some teachers actually say they didn't want to do the paperwork. Hopefully someone will put in the paperwork for her next semester. | 558 | 2,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | latest | en | 0.99156 |
https://www.physicsforums.com/threads/does-gravity-affect-gravity.503480/ | 1,624,175,712,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487658814.62/warc/CC-MAIN-20210620054240-20210620084240-00518.warc.gz | 799,595,630 | 15,957 | # Does gravity affect gravity?
Perhaps this is a really dumb/simple/obvious question but it just came to mind...
Let's say there are two scenarios.
1) A satellite is in space far from Earth. There is nothing between it and the Earth.
1) A satellite in the same position as before except the moon is now directly between it and the Earth.
Is the gravitational effect/potential experienced by the satellite from Earth the same in both cases? I.e, does the moons gravity affect the Earths gravity (make it weaker for example..?)
bcrowell
Staff Emeritus
Gold Member
Your scenario is one in which Newtonian gravity is an excellent approximation. In Newtonian gravity, gravitational forces add like vectors, and gravitational potentials add like scalars. The force and potential experienced by the satellite are different in #2 than in #1. I wouldn't describe it as the moon's gravity affecting the earth's gravity. The moon's gravity and the earth's gravity just add.
Your scenario is one in which Newtonian gravity is an excellent approximation. In Newtonian gravity, gravitational forces add like vectors, and gravitational potentials add like scalars. The force and potential experienced by the satellite are different in #2 than in #1. I wouldn't describe it as the moon's gravity affecting the earth's gravity. The moon's gravity and the earth's gravity just add.
I realized that the force is different in both cases of course since the moon is there in one of them. I am asking specifically about the Earth contribution in both cases. I figured that in Newtonian gravity it would be the same but perhaps things would be different when you took a relativistic look at it...
Pengwuino
Gold Member
So you're basically asking if there is a "screening" effect if the moon gets in the way. In the Newtonian sense, no, gravity adds like normal vectors. Relativistically, they don't add up like that but I'm not sure if it would have a screening effect or if it would be greater than what you would expect from Newtonian gravity.
PAllen
Your scenario is one in which Newtonian gravity is an excellent approximation. In Newtonian gravity, gravitational forces add like vectors, and gravitational potentials add like scalars. The force and potential experienced by the satellite are different in #2 than in #1. I wouldn't describe it as the moon's gravity affecting the earth's gravity. The moon's gravity and the earth's gravity just add.
I think what is being asked is, suppose you measure moon's mass in isolation, and the earths (you can freely move planets fare away from everything else). Then you put them in position as described by the OP, and apply Newton's gravitation law. Is there a theoretical correction that should be applied due to the binding energy of the earth moon system? I think the answer is yes - the total attractions feld by the satellite will be ininitesimally smaller than predicted by applying Newton to the independently measured masses.
I think what is being asked is, suppose you measure moon's mass in isolation, and the earths (you can freely move planets fare away from everything else). Then you put them in position as described by the OP, and apply Newton's gravitation law. Is there a theoretical correction that should be applied due to the binding energy of the earth moon system? I think the answer is yes - the total attractions feld by the satellite will be ininitesimally smaller than predicted by applying Newton to the independently measured masses.
Yeah that's pretty much it. I used the Earth and Moon but it can be any two masses really. I figured if there was any difference it would be negligible.
bcrowell
Staff Emeritus | 765 | 3,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-25 | longest | en | 0.961093 |
http://forums.wolfram.com/mathgroup/archive/2007/May/msg01000.html | 1,591,147,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00519.warc.gz | 49,825,938 | 7,697 | Re: Weird result in Mathematica 6
• To: mathgroup at smc.vnet.net
• Subject: [mg76430] Re: [mg76393] Weird result in Mathematica 6
• From: Carl Woll <carlw at wolfram.com>
• Date: Tue, 22 May 2007 02:47:07 -0400 (EDT)
• References: <200705211001.GAA10071@smc.vnet.net>
```Sebastian Meznaric wrote:
>I was playing around with Mathematica 6 a bit and ran this command to
>solve for the inverse of the Moebius transformation
>
>FullSimplify[
> Reduce[(z - a)/(1 - a\[Conjugate] z) == w && a a\[Conjugate] < 1 &&
> w w\[Conjugate] < 1, z]]
>
>This is what I got as a result:
>-1 < w < 1 && -1 < a < 1 && z == (a + w)/(1 + w Conjugate[a])
>
>Why is Mathematica assuming a and w are real? The Moebius
>transformation is invertible in the unit disc regardless of whether a
>and w are real or not. Any thoughts?
>
>
From the help:
Reduce[expr,vars] assumes by default that quantities appearing
algebraically in inequalities are real, while all other quantities are
complex.
You can try adding a domain specification:
In[144]:= FullSimplify[
Reduce[(z - a)/(1 - Conjugate[a] z) == w && a Conjugate[a] < 1 &&
w Conjugate[w] < 1, z, Complexes]]
Out[144]= -1 < Re[w] < 1 && -Sqrt[1 - Re[w]^2] < Im[w] < Sqrt[
1 - Re[w]^2] && -1 < Re[a] < 1 && -Sqrt[1 - Re[a]^2] < Im[a] < Sqrt[
1 - Re[a]^2] && z == (a + w)/(w Conjugate[a] + 1)
Carl Woll
Wolfram Research
```
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• Next by thread: Re: Weird result in Mathematica 6 | 529 | 1,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-24 | longest | en | 0.786475 |
http://stackoverflow.com/questions/8500840/algorithm-to-compute-maximal-points-in-pointset?answertab=active | 1,394,894,054,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678697956/warc/CC-MAIN-20140313024457-00068-ip-10-183-142-35.ec2.internal.warc.gz | 135,324,559 | 15,715 | # Algorithm to Compute Maximal Points in Pointset
I had this as the final question on an algorithms final (now completed):
Given a set of (x,y) points P, let M(P) be the set of maximal points given the following partial ordering on P:
``````(x,y) < (x',y') if and only if x < x' and y < y'.
``````
Thus:
``````M({(0,0),(1,1)})={(1,1)}
M({(0,0),(0,1),(1,0)})={(0,1),(1,0)}
``````
Give algorithms that compute M(P) with time complexity O(nh), O(n log n), and O(n log h) (where n = |P| and h=|M(P)|)
My O(nh) algorithm:
``````Declare M as a set of points
foreach p in P:
foreach m in M:
if(p < m):
break
if(m < p):
M.remove(m)
return M
``````
My O(n log n) algorithm:
``````Declare M as a set of points
Sort P in reverse lexicographic order
maxY := -inf
foreach p in P:
if(p.y > maxY):
maxY = p.y
return M
``````
What is an O(n log h) algorithm?
My intuition is that it is a modification of the first algorithm, but using some clever data structure (perhaps a modification of a binary tree) that doesn't need to be scanned through for each point.
Is there such an algorithm for a general poset?
This would find the leaves of any directed tree given a list of vertices and constant time lookup of whether there is a directed path between two given vertices.
-
you first algorithm doesn't work - the inner loop never executes, because M starts as empty – Petar Ivanov Dec 14 '11 at 7:34
In addition to what @PetarIvanov wrote: the O(nh) solution is simply iterate over the entire set of points and add a point to the maximal set, until there is nothing more to add. – amit Dec 14 '11 at 7:40
@PeterSmith: Ah, I misremembered it. Fixed now. – QuicksilverJohny Dec 14 '11 at 7:42
@amit: I'm not sure what you mean by "add a point to the maximal set" – QuicksilverJohny Dec 14 '11 at 7:44
@QuicksilverJohny: simple O(nh) solution: iterate over the entire set of points, and choose the maximal point, which was not previously chosen [can be done by marking points that were chosen], let it be `p`. If there is no such point - end the algorithm. else: add `p` to the maximal set, and repeat. – amit Dec 14 '11 at 7:50 | 631 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2014-10 | latest | en | 0.914917 |
https://brighterly.com/questions/what-is-3-5-as-a-fraction/ | 1,726,420,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00124.warc.gz | 136,498,945 | 12,594 | # What is 3.5 as a Fraction?
## Decimals to Fractions
When converting decimals to fractions, it’s important to understand the place value of decimals. For 3.5, the digit 5 is in the tenths place, meaning it represents 5/10 or 1/2. Therefore, 3.5 as a fraction is 3 1/2. This conversion is useful in various scenarios, like in cooking, where precise measurements are needed. It also helps students grasp the concept of parts of a whole in a more tangible way.
2(1/4)
3/4
11/10 | 132 | 480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.938953 |
https://gmatclub.com/forum/canadians-now-increasingly-engage-in-33267.html | 1,511,029,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805008.39/warc/CC-MAIN-20171118171235-20171118191235-00740.warc.gz | 613,571,857 | 43,939 | It is currently 18 Nov 2017, 11:22
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10 Aug 2006, 19:33
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Canadians now increasingly engage in “out-shopping,” which is shopping across the national border, where prices are lower. Prices are lower outside of Canada in large part because the goods-and-services tax that pays for Canadian social services is not applied.
Which one of the following is best supported on the basis of the information above?
(A) If the upward trend in out-shopping continues at a significant level and the amounts paid by the government for Canadian social services are maintained, the Canadian goods-and-services tax will be assessed at a higher rate.
(B) If Canada imposes a substantial tariff on the goods bought across the border, a reciprocal tariff on cross-border shopping in the other direction will be imposed, thereby harming Canadian businesses.
(C) The amounts the Canadian government pays out to those who provide social services to Canadians are increasing.
(D) The same brands of goods are available to Canadian shoppers across the border as are available in Canada.
(E) Out-shopping purchases are subject to Canadian taxes when the purchaser crosses the border to bring them into Canada.
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10 Aug 2006, 21:04
[quote="rkatl"]Canadians now increasingly engage in “out-shopping,â€
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10 Aug 2006, 21:10
Will go with D.
Clearly the goods available in Canada are also available across the border. Hence the comparison between the prices.
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10 Aug 2006, 21:14
go with A.
D has nothing to do with Social service. we need a choice that includes something about social service. C also does but not in a manner that is supported by the passage..
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10 Aug 2006, 21:46
I agree with Professor and A for this one... D doesn't support the argument in any way and doesn't relate to Social Services. You must support both sentences, one of which deals with "out-shopping" and another with tax and Social Services...
A ties both together and I am 100% sure here.
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10 Aug 2006, 22:59
Should be A.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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10 Aug 2006, 23:22
[quote]Canadians now increasingly engage in “out-shopping,â€
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11 Aug 2006, 03:53
OA: A
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11 Aug 2006, 06:47
I originally chose (C) but after rereading, now understand why A is correct. "Will" seemed a little strong in A, while C was more conservatively stated. But C makes more of an assumpion than a prediction.
Good one!
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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®. | 1,277 | 4,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-47 | latest | en | 0.932935 |
http://educationdocbox.com/Homework_and_Study_Tips/78840579-Statistical-computing-interactive-education-a-framework-and-toolkit-statistical-graphics-a-word-from-our-chairs-special-feature-article.html | 1,547,672,889,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657867.24/warc/CC-MAIN-20190116195543-20190116221543-00152.warc.gz | 66,070,670 | 65,699 | # Statistical Computing. Interactive Education: A Framework and Toolkit. Statistical Graphics A WORD FROM OUR CHAIRS SPECIAL FEATURE ARTICLE
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## Transcription
7 Figure 1 (Topics Screen). Describes the practice and challenge experiments and allows the student to navigate between them. Figure 2 (Practice Round). Students add icons to the boxes in the design by clicking on phrases in the articles on the left. The phrases are higlighted in the color of the box to which they have been added. Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 7
8 Figure 3 (Animation). Subjects are rejected or enrolled, randomized, treated and measured, building the histograms incrementally. Figure 4 (Challenge Round). The student selects a design and specifies the details from available variables in the glossary on the right. 8 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
9 several designs for her experiment. Once the design is filled in, she runs the animation, which produces responses at random according to a simple model for biases and interactions that is determined by the design. Finally, after the virtual experiment is run, a news story that summarizes and critiques her design is dynamically generated. Toolkit We now turn our attention to implementing the example and other labs in general. Producing a lab consists of 4 steps: 1. designing the pedagogical ideas, 2. designing the interface for the student, 3. programming both of these, and 4. creating input files (e.g. experiments, images). The previous section discussed the first of these two steps for the Design of Experiments lab. Most designers will agree that these two steps are complicated and that using simple prototypes in designing labs would be beneficial. As we started to design the labs, the need for a prototyping language became apparent. Since resources were scarce, we wanted to be able to reuse as much of the prototypes as possible in the final version of the lab. From our earlier experience considering over a dozen different lab designs, we were able to distill components common to several applications. Accordingly, to aid the third development step, we decided to create a library of these common tools and components. To avoid having the lab designer and instructors return to the program developer for lab-specific changes, we decided that an important feature of the components was to allow the instructors to configure and script elements of the lab. A further constraint was to develop the library and labs in a platform-neutral manner since it was unclear then (and still is) what form computing environments might take in the near future. In short, the primary intent of the third step was to provide a flexible and extensible environment in which to develop serious portable teaching applications supporting several different levels of programming competence. This comes in the form of a toolkit which we call TILE a Toolkit for an Interactive Learning Environment. Much of the portability issue was solved by using Java as the programming language. Our initial experience with Java was frustrating as we started to use it within months of the first release, and we were coming from a C++ and X windows background. After its initial maturing, we have found it to be an extremely useful tool with which to develop these more involved graphical applications relatively quickly. The choice of Java also allows us to run the applications within browsers even though our focus is on stand-alone applications. The toolkit is a collection of 16 Java packages 1 or modules and consists of almost 300 reusable classes developed by us and depending on other freely distributable components. The modules provide a relatively comprehensive suite of tools with which one can create a complete lab. The work window has been described earlier and acts as a container into which the lab-specific components are placed. Its services include tooltips for any of the components, navigation between the different work pages, an assistant providing hints for these pages and the basic menu items (such as Print, Quit, etc.). Also, the internal framework takes care of processing command line arguments; providing hooks to connect to different course management tools; saving and restoring a student s session; finding input files; etc. The help window is a hypertext system that displays HTML files supplied by the instructor, etc. The other modules provide tools that can be used to build up the work areas of the lab. These include classes for different plot types, an SGML 2 rendering component and parser, a dynamic animation rendering facility, a scripting language for generating dynamic text, a quiz component that provides automated grading and feedback, statistical tools for sampling, random number generation, etc. Additionally, there are many lower level tools such as a generic mechanism for managing asynchronous object creation for improved responsiveness in a lab. Perhaps the most unique thing about the toolkit relates to the final clause in the italicized sentence above. An enormous effort has been made to allow the content of each lab to be specified at run time through simple configuration files. These typically involve pairing names that the application understands and values which specify the content. These are basically properties of the application. In the above example, they control application-level properties such as the list of experiments to select from, and experiment-level properties such as the files containing the newspaper article and the layout of the design area for a specific experiment. The application level properties are processed in a hierarchical manner allowing the toolkit properties to be overridden by the instructor at both the course and lab level. While the configuration allows the instructor to specify different fixed sets of inputs, often greater resolution is required. For these cases, we have developed Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 9
10 a general way of creating simple scripting languages which allow instructors to control the application without programming in Java. We have used this mechanism to provide several such languages to control dynamic feedback and visual animation, for example. The languages use SGML as the syntax since instructors are familiar with HTML. Each language introduces new tag names and attributes for those tags which are used to control the application in some prescribed manner. For example, a quiz uses the question tag, and elements of an animation scene might be introduced by the component tag, which has attributes such as image, bbox, etc. These tags and attributes in the input document are made available to the Java application using a single method. This approach has proven to be very effective, as it has allowed us to develop a lab concurrently with one of us doing the lower-level Java development to provide the framework and the other controlling the content. Given the relative ease of creating new input languages based on the SGML/XML syntax and the tree traversal mechanisms in the toolkit, we believe that such scripting languages significantly help in developing, maintaining, and customizing individual labs. In addition to the labs, we have developed small applications which are interactive interfaces to these languages and modules. These serve as debugging tools for instructors. They allow an instructor to visualize and interact with a succession of quiz pages or an animation scene, for example, and refine the inputs to obtain the desired effect. We now turn our attention to 3 of the modules in the toolkit. These are of interest because they illustrate how the different programming levels are integrated to create a lab. Also, the dynamic text module is quite unusual. Plotting Classes The toolkit provides different plot types such as time series, scatter plots, histograms, bar charts, etc. All plots can be optionally made interactive and support facilities for: identifying data elements by clicking and rubberbanding; stretching axes for zooming; etc. From an instructor s point of view, a useful feature of the plotting classes is the ability to specify attributes in input property files. The plots understand a common set of attributes such as background and plotting colors, background image, axes ranges, number of tick marks, legends, fonts for the different labels, etc. Also, each class understands attributes that have meaning just for it, such as number of bins or bin widths for a histogram. From a programming point of view, the plots are interesting because they can be used anywhere (e.g. in a button) since they are self describing objects. This approach differs from having a GUI component for one or more plots. The classes support the Java listener event model and can notify other objects when the plot has changed or the user has interacted with it in different ways. Also data can be added incrementally a feature used in animations to show histograms and scatter plots being constructed dynamically. Finally, the plot classes are designed to be extensible due to the rich class hierarchy making it easy to create new types that inherit all of the features mentioned in here. Dynamic Text Creation As mentioned earlier, a major goal in designing the labs is to provide personalized questions and feedback to the student based on what he has done up to that point. While the instructor can change which file to display before the student runs the lab, elements of the document cannot be changed to include information that is only known when it is displayed. An obvious way to do this is to program the dynamic construction of the text in Java since this is where the variables are stored. This however is inconvenient and makes the application relatively rigid as the instructor can t easily change the phrasing and formatting of the text without assistance from the developer. Our approach is to use an SGML tag through which the instructor identifies a location in the document at which text will be conditionally inserted at different times in the application. This replace tag supports different attributes that indicate the conditions under which the text should be replaced and the actual content of the additional text. Each of these attributes can refer to variables in the Java application that are easily made available by the developer as needed by the designer. The conditions and substitution are defined in a simple language that supports most of the usual comparison and logical operations. A few other attributes of the replace tag and features of the comparison language make this a very simple but powerful scripting language and allows the instructor to specify control flow of parts of the lab without using Java. The developer need only provide access to the variables by putting them in different tables and arranging to process the document at the appropriate times. This mechanism is used in several labs. Quiz grading can provide feedback to the student based on his responses. In the challenge round of the Design of Experiments lab, the newspaper article is created by determining the type of design selected and the icons present in the design and then constructing sentences accordingly. 10 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
11 A novel use of the mechanism in another lab allows us to transfer information from a plot to a table in a flexible and extensible manner with a minimal amount of programming. Animation Module We use the animation facility in a variety of places throughout the labs. For example, the sampler module illustrates the process of drawing marked tickets from a pool with or without replacement to generate a sample and a statistic. In the example here, the experiment is animated to generate the data for the subjects. Such interactive dynamic displays are complicated and time consuming to program. Accordingly, we provide a simple abstraction of the animation scene by categorizing the elements into 3 groups: background/global attributes, static components, and elements that move between the static components. This has been sufficient to characterize all of the animations we have considered so far. Each element in the scene is identified by a particular SGML tag defining to which of the 3 categories it belongs. Attributes in the tag define the characteristics of the element such as its position and size, or a sequence of images to be displayed when it is animated, etc. The coordinates of the components are interpreted relative to the instructors scale specified in the animation tag. When the scene is rendered, the coordinates are scaled relative to the available space on the screen. The instructor can specify whether a component s bounding box is also scaled or fixed. Most of the static components are simply images but there is also a mechanism to use an arbitrary Java object. We use this, for example, to allow the student to select a ticket from the sampler pool and to allow histograms to be placed as nodes in the scene (Figure 3). The attributes are passed to these objects which allows us to configure the histogram from the animation input file. This facility allows the animation to be used in a variety of situations as a primitive programming language. The SGML syntax can also be used to define relationships between components and is used to describe design trees in the Design of Experiments lab. The elements in the input file allow the scene to be rendered statically. The final part of the animation is the part that controls the dynamic action. This is programmed in Java where each of the moving components runs in its own thread. When each component arrives at an end point of its sub-animation it sends a message to the centralized manager which then decides globally what will happen next. This allows developers to extend the animation control relatively easily since the objects work independently with a single point of synchronization. Different variants of the basic manager class control along which paths the dynamic components move. In our example above, the subjects are dynamic components and the paths they travel relate to the treatment which in turn is specified in input files as a stream of data for the 2 histograms. Finally, the animation controller randomly selects values using the random generator module and associates them with different subjects. When the subject reaches the end of the design, its value is added to the histogram. The animation classes provide the student with controls to start, stop, pause, fast forward and restart the dynamic animation. Summary and Status There are a few lessons we have learned from our experiences with this project. The following are perhaps the most important. 1. Educational material to be used in a multimedia environment, and especially graphical interfaces, require a great deal of consideration during the design. 2. Statistical concepts are the important aspects to emphasize rather than the computational techniques associated with data analysis. 3. And finally, careful software design and development with an emphasis on external configuration and extensibility makes for a longer shelf-life and greater reuse. While the cost appears to be longer development time, developing labs individually without such a design takes almost as long, especially if maintenance and customization is considered. This project is still work in progress, and should be completed by the end of this year. Our efforts so far have concentrated almost exclusively on developing the labs and toolkit. Four labs have been designed and approximately two and a half have been implemented so far. We have made one of the labs available to a handful of students at Berkeley and received valuable suggestions. The responses have been favorable but emphasized the need for a polished and thorough visual interface. The pedagogy has been greeted enthusiastically. Strategies to evaluate the software more formally are also underway. All of the components of the toolkit are now in place with some of the minor details and enhanced features existing as stubs. The testing and debugging phases should be complete by this Fall. Some of the compo- Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 11
14 Command Frequency Uniqueness Index users 10 users 45 users Figure 2. Stepplot of Uniqueness Index (1, Fraction of Number of Users) versus total Command Frequency for 4, 10, and 45 users. emerges. Such a plot is shown in Figure 3 for each of 10 users. Plots of different users are clearly distinguishable from each other. The fourth, seventh and the tenth plot are more regular than the other ones. They belong to UNIX processes (rather than to human users). The same plot can be seen in Figure 4 for 45 users. (Figure 3 corresponds to the first 10 columns of Figure 4). While the time series aspect is now difficult to make out because each column is very narrow, the column patterns are clearly distinct. A classification of users based on which commands are used and how often they are used appears promising. 3. Intrusion Detection by Uniqueness We are about to define a test statistic that depends on two quantities, W ij and U j. These quantities are estimated from the training data set. The test data is then used to evaluate the test statistic. Let N i denote the length of the test data sequence for user i and N ic the number of times command c appears in user i s test data. Then N i = P c2t est(i) N ic. We now define Uniqueness Score User i = 1 N i X c2t est(i) W ic U c N ic (1) wherec indexes the set of (distinct) commands T est(i), the weights W ic are W ic = 8< : 1 if user i s training data contains command c,1 otherwise Figure 3. Plots of command ID vs command order in the audit stream for each user. Commands are grouped by the number of users that have used them. Commands used by only one user (unique commands) have lowest ID s, commands used by all users have the highest. Within each plot, horizontal lines are drawn to separate groups. and where the uniqueness index U c is defined as U c = 8>< >: 0 if c used by all users 1=U if c used by all but one users ::: (U, 2)=U if c used by only two users (U, 1)=U if c used by only one users 1 if never used by any users : The quantity U c is 1 minus the fraction of users that have used command c in the training data. It is the same quantity that has been plotted on the vertical axis in Figures 1 and 2. For each new command the score in (1) either increases or decreases, depending on whether the associated user has used that command before in the training data or not. The amount of increase/decrease U c is higher for rare commands than for common commands. Hence a user will tend to score high if he/she uses similar commands to the ones he/she used in the training data. The order in which the commands appear does not matter. Weighted Uniqueness Scores Discrimination among users based on (1) works well, but we improve on it for the following reason: Suppose user A uses a rare command only a few times, and user B uses that rare command often. In this case test data from user B with many incidences of that same rare 14 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
15 Figure 4. Plots of command ID vs command order in the audit stream for each user. Commands are grouped by the number of users that have used them. Commands used by only one user (unique commands) have the lowest ID s, commands used by all users have the highest. Within each plot, horizontal lines are drawn to separate groups. command will be assigned a high score when trained on data of user A. The use of that rare command is indicative of user A and should therefore lead to an increase in the score. However, since user A does not use that command very much relative to other users, the use of that rare command is only moderately indicative and the score increase should not be very large. We modify (1) by changing the definition of the weights: Uniqueness Score User i = 1 N i X c2t est(i) W ic U c N ic (2) where U c is defined as before, W ic = 8 <: p ic =p :c if user i s training data contains command c,1 otherwise and where p ic = N ic =N i: : Commands previously associated with a weight of 1 are now given a smaller weight how small depends on command usage relative to other users. The weight for a particular command c for user i is high when user i uses command c a lot relative to other users, i.e.when the use of command c is especially characteristic of user i. As a consequence, a user using a particular rare command much more often than the legitimate user will score more often for that command when tested as the legitimate user, but the impact on the overall score is lower due to a low weight. 4. Results To evaluate the method presented in the previous section, we compute the test data score (2) of user i (i = 1;:::;45) based on the training data for user j (j =1;:::;45) for all pairs of users. The resulting scores are graphically displayed in Figure 5 (only scores greater than zero are shown). Darker shading corresponds to higher scores, lighter shading to lower scores. For the diagonal entries both the test data and the training data correspond to the same user. Ideally, we would be able to perfectly discriminate between diagonal entries and off-diagonal entries, that is all diagonal entries should be darker than all offdiagonal elements. By varying the threshold at which an alarm is raised one can obtain different rates of false alarms (false positives) and missing alarms (false negatives). Figure 6 shows this tradeoff between false positives and false negatives for scores based on 100 commands and 1000 commands. Figure 6 shows two curves in both cases to give an idea of the variability of the curves for different data sets; the second curve in both cases is obtained by interchanging the training and the test data. (The fact that the curves based on 1000 commands do not go all the way to the vertical axis on the left is due to a discreteness effect. The midpoint between zero and one out of 45 possible false alarms corresponds to 1=90 = 1:1% on the horizontal axis.) The lower left corner on this plot represents the ideal situation: no false alarms, and no missing alarms. For our method no false alarms correspond to about 10% Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 15
16 Test False Negatives (%) commands 1000 commands Training False Positives (%) Figure 5 Visual display of the scores from (2) based on 1000 commands. Only scores greater than zero are shown. The darker the shading, the higher the score. missing alarms based on 100 commands or 5% based on 1000 commands. As expected the discrimination among users based on 1000 commands is easier than based on 100 commands. As the rate of false alarms increases, the rate of missing alarms initially drops only slowly. Incidentally, the threshold of score = 0 corresponds to the situation when no false alarms are generated. Decreasing the threshold below zero leads to an increase of missing alarms without changing the false alarm rate. This implies that based on this data the legitimate user always has a score above zero, whereas most but not all of the other users have a score smaller than zero. 5. Discussion Intrusion detection based on command uniqueness is conceptually very simple. It requires very little storage (W ij for all i; j and U j for all j) and, since (2) is a weighted average, it can be easily updated. Moreover, each update is computationally fast requiring only a few multiplications. We would also like to point out that while (2) constitutes a quantitative improvement over (1), the uniqueness index is qualitatively more important than the weights. When too many of the alarms generated are false, they tend to be ignored altogether after a while. Therefore, a low false alarm rate is particularly important in intrusion detection. Choosing a threshold of 0, based on 100 commands we are able to avoid false alarms altogether in our experiment. We have not seen another intrusion Figure 6 Tradeoff between false negatives and false positives (ROC curve) based on (2) for 100 and 1000 commands. Two curves are obtained by interchanging the training and the test data. Both axes are on a logarithmic scale. detection method that has virtually no false alarms at an acceptable rate of missing alarms. The uniqueness approach may work especially well in the diversity of a research environment. The UNIX operating system allows and even encourages this diversity. The uniqueness approach may not work as well in an environment where everybody s job description is very similar. The choice of using 100 commands for the scores was arbitrary. We are investigating whether we can discriminate users based on even fewer commands. It is difficult to compare various intrusion detection methods due to the lack of a common yardstick. Different methods are based on different data sources and are often too complicated to reimplement for comparison purposes. Based on Figure 6, the uniqueness approach compares favorably with DuMouchel and Schonlau (1998) who report false negative rates between 10% and 50% at 5% false positives for comparable data sources. Acknowledgements M. Schonlau s work is funded in part by NSF grants DMS and DMS We are grateful for feedback from our network intrusion group with members from AT&T Labs Research, The National Institute of Statistical Sciences and Rutgers University. Their comments have led to a number of improvements. 16 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
17 References Anderson, James P. (1980), Computer security threat monitoring and surveillance, Technical report, James P. Anderson Co., Fort Washington, PA, April Computer Immune Systems, (accessed June 23, 1998) COAST project, (accessed June 23, 1998) DuMouchel, W. and Schonlau, M. (1998), A comparison of test statistics for computer intrusion detection based on principal components regression of transition probabilities, In Proceedings of the 30th Symposium on the Interface: Computing Science and Statistics, (to appear). Emerald, phlox.csl.sri.com/emerald/ (accessed June 23, 1998) Intrusion Detection for Large Networks, seclab.cs.ucdavis.edu/arpa/ (accessed June 23, 1998) Netranger, (accessed June 23, 1998) Martin Theus AT&T Labs-Research Matthias Schonlau AT&T Labs-Research and National Institute of Statistical Sciences NEW SOFTWARE TOOLS Mosaic Displays in S-PLUS: A General Implementation and a Case Study. By John W. Emerson Introduction Hartigan and Kleiner (1981) introduced the mosaic as a graphical method for displaying counts in a contingency table. Later, they defined a mosaic as a graphical display of cross-classified data in which each count is represented by a rectangle of area proportional to the count (Hartigan and Kleiner 1984). Mosaics have been implemented in SAS (see Friendly 1992) as a graphical tool for fitting log-linear models. Interactive mosaic plots (see Theus 1997a, b) have been implemented in Java. A third implementation is available in MANET, a data-visualization software package specifically for the Macintosh. No general implementation has been available in S-PLUS, one of the most popular statistical packages. The implementation presented in this article, while lacking the modelling features of Friendly s SAS implementation, provides a simply specified function for mosaics displaying the joint distribution of any number of categorical variables. As an illustration, this article examines patterns in television viewer data. A four-way table of 825 (51153) cells represents Nielsen television ratings (number of viewers) broken down by day, time, network, and switching behavior (changing channels, turning the television off, or staying with the current channel) for the week starting November 6, Simple patterns in the data appearing in the mosaic support intuitive explanations of viewer behavior. The Data Nielsen Media Research maintains a sample of over 5,000 households nationwide, installing a Nielsen People Meter (NPM) for each television set in the household. The sample is designed to reflect the demographic composition of viewers nationwide, and uses 1990 Census data to achieve the desired result. Nielsen summarizes the stream of minute-by-minute measurements to provide quarter-hour viewing measurements (defined as the channel being watched at the midpoint of each Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 17
18 Monday Tuesday Wednesday Thursday Friday 8:00 8:15 8:30 8:45 9:00 9:15 9:30 9:45 10:00 10:15 10: % 20.9% 19.7% 20.1% 17.6% 9.1% 9.4% 9.5% 9.6% 9.6% 9.6% 9.4% 9.3% 8.6% 8.3% 7.6% Figure 1. (a) Mosaic of the week s aggregate audience by day (lefthand panel); and (b) mosaic of the week s aggregate audience by time (righthand panel). quarter-hour block) for each viewer in the sample. (Details are presented in Nielsen s National Reference Supplement 1995.) A TV guide of the prime-time programming of the four major networks (ABC, CBS, NBC, and FOX) for the weekdays starting Monday, November 6, 1995 appears in Figure 2. During any quarter-hour, the individual is observed watching a major network channel, a non-network channel, or not watching television. At 10:00 however, FOX ends its network programming, so Nielsen does not record individuals watching FOX after 10:00. I confine this study to a subset consisting of 6307 East coast viewers in 2328 households. Creating Mosaics in S-PLUS Friendly (1994) describes the complete algorithm used to construct a mosaic for a general four-way table, alternatively dividing horizontal and vertical strips of area into tiles of area proportional to the counts in the remaining sub-contingency table. Without repeating the description of a general mosaic display, I note the important features of my S-PLUS implementation, which help explore various aspects of any cross-classified data set: Any number of categorical variables may be included in the mosaic, though in practice even a five-way table may be sufficiently complicated to defy explanation. Empty cells of the contingency table are represented (where possible) by a dashed line segment. The order in which the variables are represented may be specified, allowing simple exploration of any marginal or conditional frequencies on any subset of variables without physically manipulating the raw contingency table itself. The direction (horizontal or vertical) used in dividing the mosaic by each variable may be specified, allowing more flexibility than the traditional alternating divisions. Shading of the tiles resulting from the inclusion of the final variable in the mosaic may be specified, if desired. The amount of space separating the tiles at each level of the mosaic may also be customized. The documentation and S-PLUS code are available details are provided at the end of the article. The basic algorithm, an efficient recursive procedure, proceeds as follows: 1. Initialize the parameters and the graphics device the lower left and upper right corners of the plot area are (x 1 ;y 1 ) and (x 2 ;y 2 ). The term parameters refers to a collection of counts from the contingency table, labels, and values associated with features discussed above. 2. Call the recursive function mosaic.cell((x 1 ;y 1 ); (x 2 ;y 2 ); all parameters). 3. Recursive function mosaic.cell((a 1 ;b 1 ); (a 2 ;b 2 ); selected parameters for the current tile): (a) Divide the current tile, given by (a 1 ;b 1 ) and (a 2 ;b 2 ), into sub-tiles, taking into account the spacing and split direction arguments of the parameters. (b) Add labels if the current variable is one of the first two divisions of the axis. (c) If this division corresponds to the last variable of the contingency table, draw the subtiles. Otherwise, call mosaic.cell() once for each of the current sub-tiles, with the appropriate sub-tile coordinates and subsets of the current parameters. Results: Television Viewer Behavior Simple mosaics dividing the week s aggregate audience by day and time are presented in Figures 1a and b, respectively. Though they serve the same purpose as histograms, their tile areas are more difficult to compare than the tile heights in histograms. The advantage of mosaics does not appear until at least two categorical 18 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
19 8:00 8:30 9:00 9:30 10:00 10:30 ABC The Marshal Pro Football: Philadelphia at Dallas M O N D A Y CBS NBC FOX The Nanny Can t Hurry Murphy Brown High Society Chicago Hope Fresh Prince In the House Movie: She Fought Alone Melrose Place Beverly Hills Affiliate Programming: News T U E S D A Y ABC CBS NBC FOX 8:00 8:30 9:00 9:30 10:00 10:30 Roseanne Hudson Street Home Imp Coach NYPD Blue The Client Movie: Nothing Lasts Forever Wings News Radio Frasier Pursuit Hap Dateline NBC Movie: Bram Stoker s Dracula Affiliate Programming: News W E D N E S D A Y ABC CBS NBC FOX 8:00 8:30 9:00 9:30 10:00 10:30 Ellen The Drew C.S. Grace Under The Naked T Prime Time Live Bless this H Dave s World Central Park West Courthouse Seaquest 2032 Dateline NBC Law & Order Beverly Hills Party of Five Affiliate Programming: News T H U R S D A Y ABC CBS NBC FOX 8:00 8:30 9:00 9:30 10:00 10:30 Movie: Columbo: It s All in the Game Murder One Murder, She Wrote New York News 48 Hours Friends The Single G Seinfeld Caroline E.R. Living Single TheCrew New York Undercover Affiliate Programming: News F R I D A Y ABC CBS NBC FOX 8:00 8:30 9:00 9:30 10:00 10:30 Family M Boy Meets Step by Step Hangin With 20/20 Here Comes the Bride Ice Wars: USA vs The World Unsolved Mysteries Dateline NBC Homicide: Life on the Street Strange Luck X-Files Affiliate Programming: News Figure 2. TV Guide, 11/6/95 11/10/95. Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 19
20 Monday Tuesday Wednesday Thursday Friday A C N F Other A C N F Other A C N F Other A C N F Other A C N F Other 1 10:30 10:15 10:00 9:45 9:30 9:15 9:00 8:45 8:30 8:15 8:00 Monday Tuesday Wednesday Thursday Friday 10:30 10:15 10:00 9:45 9:30 9:15 9:00 8:45 8:30 8:15 8:00 20 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
21 Monday Tuesday Wednesday Thursday Friday A C N F Other A C N F Other A C N F Other A C N F Other A C N F Other 8:45 P S O SO 8:15 P SO 8:00 P 8:30 P SO 4 9:00 SO P SO 5 5 9:15 P 3 9:30 SO P SO 5 5 9:45 P :15 10:00 P S O P SO SO 10:30 P Figure 5. Mosaic of network shares and audience transitions. P = persistent, S = switch, O = off. Numbered tiles are discussed in Section 4. Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 21
22 Table 1. Thursday 9:45 Contingency Table. Thursday Transition 9:45 Network Off Persist Switch 9:45 Network Total ABC CBS NBC FOX CABLE Transition Total variables are included. These one-way mosaics show that the aggregate audience is smaller later in the week (Figure 1a) and later in the evening (Figure 1b). The mosaic corresponding to the two-way table of the aggregate audience, divided first by day and then by time, appears in Figure 3 just for clarity of exposition in this example, interesting analysis begins with the addition of specific network counts by day and time. As we add the network variable (to simplify exposition, the term network will include the aggregate cable, or non-network, alternative) and the transition categories to the mosaic (Figures 4 and 5, respectively), several points illustrate the use of these mosaics in studying television viewer behavior. The following numbers are marked in the relevant tiles in the mosaics. 1. When the network variable is added to the twoway mosaic in Figure 3 to form a three-way contingency table, the resulting mosaic tiles at each day and time represent the network ratings, or share of the viewing audience (Figure 4). For example, on Thursday at 10:00, 685 of 1692 viewers watching television were tuned into NBC s hit E.R., so the NBC rectangle occupies 40.4% of the area in Thursday s 10:00 tile. 2. Figure 5 includes an additional variable with three categories: among the viewers watching a certain network (at time t on day d), some turn the TV off and do not watch anything at time t+1 (represented by the black tiles); others switch networks at time t +1(shaded tiles), while the remaining viewers watch the same network, or persist (unshaded tiles). For example, consider the NBC viewers in the Thursday 9:45 tile who watch the end of Caroline in the City: 523 of 1803 viewers watching television then tuned into the end of Caroline in the City the NBC tile is 30% of the area of the Thursday 9:45 tile. Of the 523 viewers, only 80 turned the television off at 10:00 (black tile), 94 switched to a different network at 10:00 (shaded tile), and the remaining 349 watched the beginning of E.R. on NBC (persisting in their viewing of NBC, the unshaded area). Table 1 presents the two-way contingency table for the viewers watching television at 9:45 classified by network choice and viewing behavior after the quarter-hour. Note that there can be no viewers persisting in watching FOX from the 9:45 quarter-hour these FOX viewers must either turn the TV off or switch channels. This empty cell corresponds to the empty transition tile in the FOX 9:45 tile. Similarly, all FOX tiles after 10:00 are empty. 3. A quick study of the TV schedule in Figure 2 and the mosaic in Figure 5 shows that viewer persistence is higher when there is show continuity. For example, on Tuesday night after the 9:15 quarter-hour, CBS and FOX have continuations of longer shows (both are movies) while ABC and NBC start new shows at 9:30 (competing halfhour comedies). This tile shows a striking example of high persistence with show continuity and lower persistence going into new programming: ABC and NBC have lower persistence rates of roughly 60% and 50%, while CBS and FOX enjoy high persistence rates of close to 90% each. Note the uniformly high degree of switching at 8:45 and 9:45 in Figure It is also evident from the mosaic that persistence during the odd quarter-hour transitions (that is, always during a show) is fairly uniform between the networks, and usually high compared to other transitions. The 8:30 frame on Monday, for example, shows uniformly high flow of viewers persisting into 8: These mosaics provide insight into different sources of viewer persistence. The primary trend appears to be higher persistence during shows (and lower persistence at end of shows), but more specific elements of persistence are also evident in the mosaics. First, consider Monday Night Football on ABC after 9:00. There is unusually 22 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
26 a light yellow spotlight is easily learned. Our purpose is to introduce a middle ground contour rather than two equal emphasis background contours. Figure 3 calls attention to just two contours, one above and one below the median. This reinforces the statistical concept of median and targets a broad portion of the public that has at least some interest in statistical summaries. Examination of the parallel dot plot panels shows many dots close to the median. The display of other contours would be more consistent with gaps in the unemployment rate distribution. We indicate three color options of many for showing more contours in Figure 3 to sophisticated audiences. The first uses a shifting light yellow spotlight to focus on the region that combines highlighted units in panels immediately above and below the current panel. The second uses a multicolor spotlight to cover units from more panels. The third approach uses a different color for each state. One such pattern starts with a spectral hue sequence for the panels. Warm colors are on top, bright yellow represents the median and cool colors are below. A lightness ramp within each panel attempts to distinguish the states. Distinguishing states is difficult with fifty-one color schemes, so at present we retain our five distinct hue approach. Readers might be surprised at the mention of spectral ordering. After citing extensive literature arguing against spectral order, Brewer (1997) cites perceptual studies indicating that some instances of spectral order work quite well. Lightness remains the primary basis for ordering. The spectral order works when used as divergent spectral scheme with bright yellow in the middle. Brewer also discusses color scales for the color blind so the paper is of considerable interest. In some cases the micromap design itself is less than ideal for showing spatial patterns. For example, Figure 1 illustrates a caricature developed to show OECD nations. The topological distortion and use of two insets is not conducive to properly observing spatial patterns. The Figure 1 micromap sequence might even be deleted since spatial location is not central to the story. Our notion of micromaps is meant to be general. The examples here involve area representations. Some environmental applications involve monitoring sites that are represented as points. A map may be something other than areas or points on the surface of the earth. For example locations might be position within a building, nodes or links in a formal graph, or even a position in a transition matrix. Students at George Mason University have developed some of their own micromap variations. A student in a Statistical Graphics and Data Exploration class won an external poster competition by showing sequences of Virginia maps overlaid with pie glyphs at county centroids. The pie glyphs represented crime rates for different classes of crimes summarized at the county level. (The class did not promote pie glyphs for making comparisons). A student in a Scientific and Statistical Visualization class provided a much better example. He redesigned a statistical summary from World War I concerning the effects of mustard gas. The micromaps were caricatures of the human body and clearly showed the susceptibility of exposed and moist locations. Micromaps can take many forms. 2.3 Statistical Summary Panels The statistical summary panels can take many forms such as dotplots, barplots, boxplots, times series plots, scatterplots, cdf plots, perspective views, stereo pairs plots and so on. While most of these plots are familiar, that does not mean there is a lack of graphical design issues to address. Statistical summary panels are typically small, so overplotting remains a problem even though the number of highlighted elements in a panel is small. In Figure 1, the time series overplot substantially. Since there are no missing values the reader can infer values for hidden points. When there is missing data, as in Figure 2, sometimes the overplotting is not too bad. When something must be done, less than elegant solutions include plotting dots of different size with large dots plotted first, plotting symbols that remain identifiable when overplotted, staggering plotting locations and so on. Space constraints continually come into play. It is advantageous to keep a LM plot to one page. We often forego Cleveland s (1993) guidance about banking to 45 degrees and are tempted to skimp on labeling. Uncomfortable compromise, of course, is not unique to LM plots. Scaling and resolution issues are recurrent in statistical summary panels. Figure 1 deals with the scaling and resolution issue in two ways. First, the selected unit of measure is tons per person. This reduces some of the disparity between small and large population countries. Second, the top panel has a different scale than the other panels. This compromise is problematic. It is difficult to compare time series between panels on different scales. Since we ordered the nations by the time series mean it is not necessarily the case that values for a specific year in the top panel are above those in the second panel. A helpful option is to show the times series from all panels 26 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
28 That the U.S. was not top on the list was also a surprise. However, those familiar with Luxembourg describe special circumstances consistent with high values. The unusual value for Iceland is a rounding artifact with the tabled numerator being represented by a single digit. An encouraging hint of declining values appears for Germany, France, and Sweden. (German unification leads to interesting accounting issues in regard to future improvement.) The figure suggests increasing per capita rates for nations such as Ireland, Japan, New Zealand, Greece and Portugal. In terms of total emissions, even flat per capita patterns are cause for concern when coupled with increasing population. For example the Mexican population is increasing rapidly. Of course the U.S population continues to increase due to immigration and U.S. per capita values are much higher than Mexican values. Our perhaps injudicious interpretation is that nations tend to have CO2 emission styles that are relatively stable for the reporting period, that total emissions are linked to population, and that as far as we know population growth is not under control. As indicated by Wood (1992) plots reflect some agenda. The agenda behind showing amount per capita was both to reduce the tremendous range of values and to present the information on a personal level. In a chance airplane conversation, a consultant for U.S. utilities looked at the plot and suggested reporting CO2 emissions per gross domestic product. His agenda was to make the U.S. appear as a waste minimizing (efficient) energy producer. The verbal battle over greenhouse gas emissions will continue. Developing a micromap for OECD nations was a design challenge. Figure 1, takes a variety of liberties, not only slicing a way most of the Atlantic and enlarging small countries, such as Luxembourg, at the expense of others, but also by using two insets, one for Australia and New Zealand and one for Japan. Is the distortion too much for those familiar with maps of Europe? Will the inset placing Japan on Russia arouse political sensitivities? We have not addressed such issues, but note the micromap will need to be revised to incorporate the three new OECD nations that appear in the 1997 compendium. Revising an already published OECD view (The State of the Environment, OECD 1991, page 134) may provide a solution, but showing small nations is still a challenge. 3.2 Figure 2 Figure 2 re-expresses a portion of a table published on page A-29 in Agriculture Prices (release date January 31, 1994) by The National Agriculture Statistics Service. The micromaps call attention to the higher wheat prices in the West Coast and Northern States. Are transportation costs involved? Is the total amount available at different times during the year a major factor determining price factor? The time series indicate missing data. Why is it missing? A big question that jumps out in the graphical representation concerns the mismatch between the marketing year average for each state (footnoted as being preliminary) and the monthly time series values. In some cases the marketing year average for a state is near extreme values for the state, and that implies heavy weighting of specific months. An explanation about the weighting seems appropriate. Likely the weighting explanation is available in related Department of Agriculture documents. 3.3Figure3 The primary source for Figure 3 is the Geographic Profile of Employment and Unemployment, 1995, U. S. Department of Labor Bureau of Labor Statistics, Bulletin Carr and Pierson (1996) propose a LM plot as a replacement for a choropleth map in that document. They discuss the relative merits of choropleth maps and LM plots and we encourage readers to read the article. Figure 3 takes a step further from the previous LM plot, showing contours, the U.S. average as a dashed line, and more labeling. The plot tells a reasonably complete story indicating estimates, estimate precision, estimate importance (the number unemployed), and estimate location. A deeper interpretation item that is not shown concerns the determination of who is excluded from the numerators and denominators in the determination of rates. 4. LM plot history, Connections to other Research and Challenges We claim that LM template is new, but there are, of course, many connections to previous graphics and conceptualizations. While we were intrigued by the thumbnail images of Eddie and Mockus (1996), a stronger connection is to the work of Edward Tufte. The LM plots belong to class of graphics that Tufte (1983, 1993, 1997) calls small multiples. In The Visual Display of Quantitative Data, his eloquent description of welldesigned small multiples include phrases such as inevitably comparative, deftly multivariate, efficient in interpretation, and often narrative in content. We designed LM plots with the hope that such phrases would apply. In Visual Explanations, Tufte calls particular attention to explanatory power of parallelism. While our use of parallelism precedes this book, Tufte s 28 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
29 Annual CO2 Emissions From Energy Use Units = Tons Per Person Wheat Prices Received By State: 1993 Dollars Per Bushel Luxembourg United States Canada Australia Germany Denmark Belgium Netherlands Finland United Kingdom Ireland Iceland Japan Austria Norway France Sweden New Zealand Italy Switzerland Greece Spain Mexico Portugal Turkey Year Highlighted States On Contours North Dakota Arizona Montana South Dakota Minnesota Washington Oregon California Yellow Above: High \$ Contour Colorado Kansas Michigan Nebraska Idaho Yellow Below: Low \$ Contour Ohio Oklahoma Arkansas Texas Indiana Illinois Georgia Missouri States Sorted By Marketing Year Average Marketing Year Average \$4.00 \$3.50 \$3.00 \$2.50 \$2.50 \$4.00 \$3.50 \$3.00 \$4.50 \$4.00 \$3.50 \$3.00 \$2.50 \$2.00 \$4.00 \$3.50 \$3.00 \$2.50 \$4.00 \$3.50 \$3.00 \$2.50 Monthly Time Series Jan Mar May July Sept Nov Feb Apr June Aug Oct Dec Black = U.S. Values
30 Labor Force Statistics By State, 1995 Average States Unemployment Number Above Median Contour Ratio Unemployed D.C. West Virginia California Alaska Rhode Island Louisiana Washington New Jersey New York New Mexico Alabama Mississippi Texas Pennsylvania Montana Hawaii Maine Florida Connecticut Nevada Massachusetts Kentucky Idaho Michigan Tennessee Median Illinois South Carolina Maryland Arizona Georgia Arkansas Wyoming Oregon Ohio Missouri Oklahoma Indiana Virginia Kansas North Carolina Delaware Vermont Colorado New Hampshire Wisconsin Minnesota Utah Iowa North Dakota South Dakota Nebraska Below Median Contour Percent 100,000 People U.S. Average 90% Confidence Interval Figure 3. Linked Micromap of 1995 unemployment figures taken from the Bureau of Labor Statistics. 30 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
31 earlier examples may well have guided us to make parallelism a fundamental part of the LM plot design. As indicated in the introduction, LM plots emerged as at way of linking row-labeled plots to maps. The rowlabeled plots in turn build upon of Cleveland (1985) and Cleveland and McGill (1984). In fact the development of row-labels plots was part of an effort to encourage EPA staff to use Cleveland s dotplots in EPA graphics. Dissatisfaction with the look of early S-plus dotplots (see Cleveland 1993b for an example) and the promise of multiple panel layouts for expressing complex tables as plots lead to development of new S-plus functions (Carr 1994a and Carr 1997). While the rowlabeled plot development was independent of the Trellis graphics development, there are similarities. This is not surprising since Cleveland s design ideas were important in both and S-plus was a common computing environment. The linking of maps and statistical graphics also builds upon the work of Monmonier (1988) who connected contemporary methodology from cartographic and statistical graphics communities. One of his many interesting examples has a map on the left, labels in the middle and bar plots on the right. This example is a precursor to our LM plots. We have followed Monmonier s work over the years. The statemap caricature in Carr and Pearson (1996) was specifically inspired by a more recent reading of Monmonier (1993) and adapted from coordinates that he graciously supplied. Despite connections to other research, we claim the LM template is new. Many people have said There is nothing new under the sun. The truth of this statement always depends on ignoring distinctions. Presumably the speakers of the statement are not identical but if so they collectively get just one vote. We note that the use of parallel sequences of small multiples is relatively uncommon and have called attention to defining features in Section 2. Ultimately others will have to judge the distinctiveness of specific examples and the template in general. What is most important, however, is not the newness of the template, but rather its utility, community awareness of its relative merits, plot production convenience, and statistical graphics literacy. Plot production convenience remains a big issue. If LM plots are to be used they need to be easily produced. The general S-plus tools we developed (anonymous ftp to galaxy.gmu.edu and change directory to pub/dcarr/newsletter/lmplots) are flexible building blocks but not easy push button tools. The software also includes a Visual Basic front end that Andrew Carr developed to simplify production of LM plots similar to Figure 3. This is a start toward simple production. Much work remains to design micromaps for new applications and to develop software that makes it easy to produce a wide range of LM plots. Much research is appropriate concerning compromises and variations that are motivated by plot purpose, audience, specific data and metadata. The other big recurrent issue is statistical literacy. Carr and Pierson (1996) suggest that if federal agencies distribute estimates with confidence boundaries, then the Web literate public will grow comfortable with the general idea. Similarly, medians and other statistics can become familiar. A big challenge is to start the ball rolling with federal statistical graphics distributed on the Web (see Carr, Valliant and Rope 1996), a topic to be revisited in future articles. Acknowledgements The majority of the work behind this paper was supported by the EPA under cooperative agreement No. CR Some facets of this work have been supported by BLS, NASS, and NCHS. The article has not been subject to review by EPA, BLS, NASS, and NCHS so does not necessarily reflect the view of the agencies, and no official endorsement should be inferred. We wish to thank Wing K. Chong who helped in developing the OECD micromap and the many people commenting at our past presentations. References Brewer, C. A. (1997), Spectral Schemes: Controversial Color Use on Maps, Cartography and Geographic Information Systems, Vol. 24, No. 4, pp Carr, D. B. (1994a), Converting Plots to Tables, Technical Report No. 101, Center for Computational Statistics, George Mason University, Fairfax, VA. Carr, D. B. (1994b), A Colorful Variation on Boxplots, Statistical Computing & Graphics Newsletter, Vol. 5, No. 3, pp Carr, D. B. (1997), Some Simple Splus Tools for Matrix Layouts, Bureau of Statistics Statistical Note Series, No. 42. Carr, D. B. and A. R. Olsen (1996), Simplifying Visual Appearance By Sorting: An Example Using 159 AVHRR Classes, Statistical Computing & Graphics Newsletter, Vol. 7 No. 1 pp Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 31
32 Carr, D. B., R. Somogyi and G. Michaels (1997), Templates for Looking at Gene Expression Clustering, Statistical Computing & Graphics Newsletter, Vol. 8, No. 1, pp Carr, D. B. and S. Pierson (1996), Emphasizing Statistical Summaries and Showing Spatial Context with Micromaps, Statistical Computing & Graphics Newsletter, Vol. 7, No. 3, pp Carr, D. B., R. Valliant, and D. Rope (1996), Plot Interpretation and Information Webs: A Time-Series Example From the Bureau of Labor Statistics, Statistical Computing & Graphics Newsletter, Vol. 7, No. 2, pp Cleveland, W. S. (1985), The Elements of Graphing Data, Hobart Press, Summit, NJ. Cleveland, W. S. (1993a), Visualizing Data, Hobart Press, Summit, NJ. Cleveland, W. S. (1993b), Display Methods of Statistical Graphics, Journal of Computational and Graphical Statistics, Vol. 2., No. 4, pp Cleveland, W. S. and R. McGill. (1984), Graphical Perception: Theory, Experimentation, and Application to the Development of Graphical Methods, Journal of the American Statistical Association, Vol. 79, pp Eddy, W. F. and A. Mockus (1996), An Interactive Icon Index: Images of the Outer Planets, Journal of Computational and Graphical Statistics, Vol. 5., No. 1, pp Kosslyn, S. M. (1994), Elements of Graph Design, W. H. Freeman and Company, New York, NY. Tufte, E. R. (1990), Envisioning Information, Graphics Press, Cheshire, CT. Tufte, E. R. (1997), Visual Explanations, Graphics Press, Cheshire, CT. Wood, D. (1992), The Power of Maps, The Guilford Press, NY. Dan Carr Institute for Computational Statistics and Informatics George Mason University Tony Olsen EPA National Health and Environmental Effects Research Laboratory Pip Courbois Oregon State University Suzanne M. Pierson OAO Corporation D. Andrew Carr Bureau of Labor Statistics carr Monmonier, M. (1988), Geographical Representations in Statistical Graphics: A Conceptual Framework, American Statistical Association 1988 Proceedings of the Section on Statistical Graphics, American Statistical Association, Alexandria VA. pp Monmonier, M. (1993), Mapping It Out, The University of Chicago Press, Chicago, IL. Olsen,A.R.,D.B.Carr,J.P.Courbois,andS.M.Pierson (1996), Presentation of Data in Linked Attribute and Geographic Space, Poster presentation, 1996 ASA Annual Meeting, Chicago, Il. Tufte, E. R. (1983), The Visual Display of Quantitative Information, Graphics Press, Cheshire, CT. 32 Statistical Computing & Statistical Graphics Newsletter Vol.9 No.1
33 NEWS CLIPPINGS AND SECTION NOTICES Statistical Computing Activities at the JSM By Russell D. Wolfinger The Statistical Computing Section is sponsoring an information-packed slate of invited and contributed sessions, posters, and roundtables for the Joint Statistical Meetings, August 9-13, 1998, in Dallas, Texas. The invited sessions are as follows: Data Mining, Sun 2:00-3:50, chaired by Don Sun, featuring Andreas Buja, Chidanand Apte, and Robert Chu Computing for Large Mixed Models, Mon 10:30-12:20, chaired by Russ Wolfinger, featuring David Harville, Yurii Bulavsky, and Stephen Smith Bootstrapping Time Series, Tues 8:30-10:20, chaired by Tim Hesterberg, featuring Joseph Romano, Hans-Ruedi Kuensch, and Lori Thombs Smoothing Methods and Data Analysis, Wed 10:30-12:20, chaired Michael O Connell, featuring Steve Marron, Doug Nychka, and Chong Gu Internet Developments, Wed 2:00-3:50, chaired by Balasubramanian Narasimhan, featuring P.B. Stark, R. Todd Ogden, Jan de Leeuw, Duncan Temple Lang, and Jim Rosenberger Computer Algebra Systems, Thur 8:30-10:20 chaired by John Kinney, featuring Eliot Tanis, John Kinney, and Barbara Heller The contributed sessions are as follows: Density Estimation and the Bootstrap, Sun 4:00-5:50, chaired by Ed Wegman, featuring David Scott Computing in Time Series, Mon 8:30-10:20, chaired by Michael Leonard Computing and Statistical Inference, Mon 2:00-3:50, chaired by Morteza Marzjarani Prediction and Simulation Algorithms, Tues 8:30-10:20, chaired by Anwar Hossain Computing in Psychometrics, Tues 10:30-12:20, chaired by Yiu-Fai Yung Regression Computation, Wed 8:30-10:20, chaired by Robert Cohen EM Algorithm and Imputation, Thurs 10:30-12:20, chaired by Dan McCaffrey The roundtables are as follows: The Future of Web-based Computing, R. Webster West Bayesian Computation, Merlise Clyde Information Theory and Statistical Reasoning, Bin Yu Directions in Computing for Statistics, John Chambers The Student Award Winners session will be Thurs 10:30-12:20 and chaired by Lionel Galway. The winners this year are Alessandra Brazzale, Matthew Calder, Yan Yu, and Steven Scott. Six posters will also be presented at the meetings. Finally, several other ASA sections are sponsoring invited and contributed sessions that may be of interest to you. Statistical Computing will be listed as a co-sponsor of these sessions in your JSM program. Please mark your schedules to attend these state-ofthe-art computing activities! Russell D. Wolfinger SAS Statistical Graphics at the JSM By Ed Wegman The Statistical Graphics Section is sponsoring three invited sessions, two special contributed sessions as well as co-sponsoring a regular invited session at JSM 98. Session 90, Real Success Stories of Statistical Graphics, was designed to highlight success stories for graphical methods. Often, graphics papers tend to show how graphical methods can uncover facts already known by other methodologies. Here we are intending to shed light on facts that have not or could not be known by other methods. With the ubiquitous presence of the World Wide Web and its potential for creating a revolution in the scientific investigation process, in Session 138, Statistical Graphics on the Web, we intended to explore the Web s potential for statistical graphics. Clearly animation, color and even three-dimensional, stereoscopic graphics are possible on the Web and our authors in this session explore Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 33
35 SECTION OFFICERS Statistical Graphics Section Michael M. Meyer, Chair Carnegie Mellon University Dianne H. Cook, Chair Elect Iowa State University Sally C. Morton, Past Chair ext 7360 The Rand Corporation Sally Edward J. Wegman, Program Chair George Mason University Deborah Swayne, Program Chair Elect AT&T Labs Research Antony Unwin, Newsletter Editor (98-00) Universität Augsburg Robert L. Newcomb, Secretary/Treasurer (97-98) University of California, Irvine Michael C. Minnotte, Publications Liaison Officer Utah State University Lorraine Denby, Rep.(96-98) to Council of Sections Bell Laboratories David W. Scott, Rep.(98-00) to Council of Sections Rice University Roy E. Welsch, Rep.(97-99) to Council of Sections MIT, Sloan School of Management Statistical Computing Section Karen Kafadar, Chair University of Colorado-Denver James L. Rosenberger, Chair Elect The Pennsylvania State University Daryl Pregibon, Past Chair AT&T Laboratories Russel D. Wolfinger, Program Chair SAS Mark Hansen, Program Chair Elect Bell Laboratories Mark Hansen, Newsletter Editor (96-98) Bell Laboratories Merlise Clyde, Secretary/Treasurer (97-98) Duke University James S. Marron, Publications Liaison Officer University of North Carolina, Chapel Hill Janis P. Hardwick, Rep.(96-98) Council of Sections University of Michigan Terry M. Therneau, Rep.(97-99) Council of Sections Mayo Clinic Naomi S. Altman, Rep.(97-99) to Council of Sections Cornell University naomi Vol.9 No.1 Statistical Computing & Statistical Graphics Newsletter 35
36 INSIDE A WORD FROM OUR CHAIRS StatisticalComputing... 1 StatisticalGraphics... 1 EDITORIAL... 2 SPECIAL FEATURE ARTICLE Interactive Education: A Framework and Toolkit.. 1 GETTING TO SLEEP AT NIGHT Intrusion Detection Based on Structural Zeroes.. 12 NEW SOFTWARE TOOLS Mosaic Displays in S-PLUS: A General ImplementationandaCaseStudy TOPICS IN INFORMATION VISUALIZATION Linked Micromap Plots: Named and Described.. 24 NEWS CLIPPINGS AND SECTION NOTICES Statistical Computing Activities at the JSM StatisticalGraphicsattheJSM Student Paper Competition SECTION OFFICERS Statistical Graphics Section Statistical Computing Section The Statistical Computing & Statistical Graphics Newsletter is a publication of the Statistical Computing and Statistical Graphics Sections of the ASA. All communications regarding this publication should be addressed to: Mark Hansen Editor, Statistical Computing Section Statistics Research Bell Laboratories Murray Hill, NJ (908) FAX: cm.bell-labs.com/who/cocteau Antony Unwin Editor, Statistical Graphics Section Mathematics Institute University of Augsburg Augsburg, Germany FAX: www1.math.uni-augsburg.de/unwin/ All communications regarding membership in the ASA and the Statistical Computing or Statistical Graphics Sections, including change of address, should be sent to: American Statistical Association 1429 Duke Street Alexandria, VA USA (703) FAX (703) Nonprofit Organization U. S. POSTAGE PAID Permit No. 50 Summit, NJ American Statistical Association 1429 Duke Street Alexandria, VA USA This publication is available in alternative media on request.
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### Effective Planning. Electrical Engineering and Computer Science
Effective Planning Witton Chou Computer Science University of California, Berkeley wittonchou@berkeley.edu Michael So Electrical Engineering and Computer Science University of California, Berkeley michaelso@berkeley.edu
### MLR Institute of Technology
MLR Institute of Technology Laxma Reddy Avenue, Dundigal, Quthbullapur (M), Hyderabad 500 043 Phone Nos: 08418 204066 / 204088, Fax : 08418 204088 UNIT WISE OBJECTIVE QUESTIONS Course Name : SOFT WARE | 19,502 | 91,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-04 | latest | en | 0.886221 |
https://livingstingy.blogspot.com/2011/10/nitpicking.html | 1,726,251,004,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00267.warc.gz | 339,103,291 | 16,914 | ## Friday, October 14, 2011
### Nitpicking
One way to attack an argument, when you have no ground to stand on, is to nitpick.
Oftentimes, in political and even legal arguments, you see a technique employed called nitpicking. And when you see this happen, you can be sure that the person making the argument hasn't got a leg to stand on, and in fact, is trying to lie to you.
How does it work? Well, suppose someone says something like, "The average return on investment in the stock market, over the last 30 years, was 10%".
The nitpicker immediately chimes in with, "That's not true at all! The actual rate of return of 10.36%! Your argument is flawed and your reasoning is sloppy!"
Now, to a rational human being (the two or three left on the planet at this point) this is an idiotic argument. It is an argument made by, and made to, someone who doesn't understand numbers or math. To such folks, numbers are absolute, and anything written out to four decimal places is, no doubt, God's word, and not to be approximated or rounded-off.
And increasingly, more and more Americans fall into this group - people who learned "fundamentals" of arithmetic in school, but have no real handle on what numbers mean or how they work. You plug in a number in a calculator, they think, and read off the answer to six decimal places, right?
And so, people fall for the nitpicker argument - not seeing the larger picture, but instead, picking apart arguments using imagined flaws.
And it is a popular sport, for example, in discussion groups and with conspiracy theories. And folks who participate in it, are weak-minded and weak thinkers.
The best way to deal with nit-pickers is to walk away from them. Trying to engage a nitpicker in a conversation is fruitless. And let's face it, these are not fun people to deal with, nor are they the life of the party. They are sad, lonely, and depressed people, who will go to any length to prove they are "right" and everyone else is wrong. And of course, they never, ever, make a mistake.
Once you try to engage a nitpicker, it is all over. They will get you to concede that yes, the number was 10.36%, not 10%. Then, they pounce. "AHA!" they scream, "Even YOU admit you are were wrong! That your logic and reasoning was specious! That everything you say is a deck of cards ready to COLLAPSE!"
And they will never let this go, of course. Days, weeks, even years later, they will come back to this. Again and again, they will return to your supposed "error" as "proof" that nothing you have to say is of consequence.
But of course, what would be the point in engaging in a conversation with a nit-picker, anyway? | 642 | 2,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.949562 |
https://uk.mathworks.com/matlabcentral/cody/problems/44486-vector-push/solutions/1560994 | 1,604,162,518,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107919459.92/warc/CC-MAIN-20201031151830-20201031181830-00669.warc.gz | 539,753,365 | 17,216 | Cody
Problem 44486. Vector push
Solution 1560994
Submitted on 15 Jun 2018
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
filetext = fileread('push.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
v = [1, 2, 7]; x = []; [v, n] = push(v, x); v_correct = [1, 2, 7]; n_correct = 3; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 3 x = [] n = 3 v = 1 2 7
3 Fail
v = []; x = [5, 6, 8]; [v, n] = push(v, x); v_correct = [5, 6, 8]; n_correct = 3; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 0
Output argument "n" (and maybe others) not assigned during call to "push". Error in Test3 (line 3) [v, n] = push(v, x);
4 Fail
v = []; x = [7; 3; 9; 4]; [v, n] = push(v, x); v_correct = [7; 3; 9; 4]; n_correct = 4; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 0
Output argument "n" (and maybe others) not assigned during call to "push". Error in Test4 (line 3) [v, n] = push(v, x);
5 Pass
v1 = randi(10, 1, 5); x = 5; [v, n] = push(v1, x); v_correct = [v1, 5]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 5 x = 5 n = 6 v = 1 5 1 2 4 5
6 Pass
v = [ones(4, 1)]; x = [3, 5]; [v, n] = push(v, x); v_correct = [1; 1; 1; 1; 3; 5]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 4 x = 3 5 n = 6 v = 1 1 1 1 3 5
7 Pass
v = [zeros(1, 4)]; x = [3; 0]; [v, n] = push(v, x); v_correct = [0, 0, 0, 0, 3, 0]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct)); v = [8, 2]; x = []; [v, n] = push(v, x); v_correct = [8, 2]; n_correct = 2; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
t = 4 x = 3 0 n = 6 v = 0 0 0 0 3 0 t = 2 x = [] n = 2 v = 8 2
Community Treasure Hunt
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Start Hunting! | 802 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-45 | latest | en | 0.569799 |
http://math.stackexchange.com/questions/25437/computing-the-break-even-point?answertab=active | 1,462,398,288,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860124045.24/warc/CC-MAIN-20160428161524-00025-ip-10-239-7-51.ec2.internal.warc.gz | 190,213,834 | 17,301 | # Computing the break-even point
### Problem
The Lake Shore Inn is trying to determine its break-even point. The inn has 50 rooms that are rented at \$60 per night. Operating costs are as follows, in dollars
Salaries 7,200 per month
Utilities 1,500 per month
Depreciation 1,200 per month
Maintenance 300 per month
Maid Service 8 per room
Other Costs 28 per room
Determine the inn's break-even point in (1) number of rented rooms per month and (2) dollars.
### Progress
I know the general formulas:
• Break-even point in Units is obtained by dividing Fixed Costs by Contribution Margin per Unit. The latter is (Unit selling price - Unit Variable Costs)
• Break-even point in Dollars is obtained by dividing Fixed Costs by Contribution Margin Ratio. The latter is (Contribution Margin per Unit / Unit selling price).
However, I cannot figure out the Unit Variable costs. This is is why this problem is so hard for me, even with the formulas.
-
I cannot figure out the Unit Variable costs that is why this problem has be so hard for me, even with the formulas. – Daniel Loisi Mar 7 '11 at 4:03
Please state your question in the body, not a comment. Since you are asking a question, it is only polite for you to ask, rather than post in the imperative mode. – Arturo Magidin Mar 7 '11 at 4:09
The unit variable cost would appear to be the extra cost you incur for each rented room. If you rent a room, you take in 60 dollars, but then you must spend, in addition to your fixed costs, 8 in maid service and 28 in "other costs", so you don't really get the entire 60 dollars free and clear, but rather only 60 - (8+28) = 24 dollars free and clear. – Arturo Magidin Mar 7 '11 at 4:28
@Arturo: still one extra zero early in the post. But I liked your answer. – Ross Millikan Mar 7 '11 at 4:30
## 1 Answer
Hint: Unit Variable costs are the costs that increase depending on the number of units that are rented, while the costs per month are fixed. The basic idea is that you spend a certain number of dollars to stay open (taxes, mortgage, desk staff) and there are costs that increase per room rented (cleaning rooms). So what would the Contribution Margin per Unit be? It is defined in the first formula.
- | 560 | 2,241 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-18 | latest | en | 0.920683 |
http://mathhelpforum.com/pre-calculus/113795-solving-logs.html | 1,481,364,173,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543030.95/warc/CC-MAIN-20161202170903-00434-ip-10-31-129-80.ec2.internal.warc.gz | 182,049,284 | 10,185 | 1. ## Solving with logs
How do you solve this problem leaving logs in there?
2. Originally Posted by Togechu64
How do you solve this problem leaving logs in there?
$\ln{6^{7x+1}}=\ln{7^{6x-1}}$
$(7x+1)\ln(6)=(6x-1)\ln(7)$
Now distribute the lns, then solve for x.
3. I tried that, but I'm not sure where to go after I distribute? :/
4. $(7x+1)\ln(6)=(6x-1)\ln(7)$
Put all logs on a side, and the $x$ terms on the other side (distribute the ln's, that is) :
$\frac{7x+1}{6x - 1} = \frac{\ln(7)}{\ln(6)}$
Say $a = \frac{\ln(7)}{\ln(6)}$ (for ease of manipulation and handling) :
$\frac{7x + 1}{6x - 1} = a$
Get rid of the fraction by multiplying each side by $(6x - 1)$.
$7x + 1 = a(6x - 1)$
Then expand and solve for $x$ (then substitute for $a$)
Does it help ?
5. yes it does, thank you very much!! | 301 | 814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-50 | longest | en | 0.823027 |
http://dev.silhouetteconsulting.com/2hlyipk/b32855-what-is-the-degree-of-a-zero-polynomial | 1,679,869,217,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00510.warc.gz | 13,038,860 | 8,513 | whose coefficients are all equal to 0. Using this theorem, it has been proved that: Every polynomial function of positive degree n has exactly n complex zeros (counting multiplicities). Monomials âAn algebraic expressions with one term is called monomial hence the name âMonomial. For example, f (x) = 8x3 + 2x2 - 3x + 15, g(y) = y3 - 4y + 11 are cubic polynomials. The constant polynomial whose coefficients are all equal to 0. Write the Degrees of Each of the Following Polynomials. Example: f(x) = 6 = 6x0 Notice that the degree of this polynomial is zero. A uni-variate polynomial is polynomial of one variable only. For example- 3x + 6x2 â 2x3 is a trinomial. At this point of view degree of zero polynomial is undefined. Let a â 0 and p(x) be a polynomial of degree greater than 2. The degree of a polynomial is nothing but the highest degree of its exponent(variable) with non-zero coefficient. it is constant and never zero. + cx + d, a â 0 is a quadratic polynomial. then, deg[p(x)+q(x)]=1 | max{$$1,{-\infty}=1$$} verified. Second Degree Polynomial Function. Mention its Different Types. So root is the same thing as a zero, and they're the x-values that make the polynomial equal to zero. For example, f(x) = x- 12, g(x) = 12 x , h(x) = -7x + 8 are linear polynomials. asked Feb 9, 2018 in Class X Maths by priya12 ( -12,629 points) polynomials Degree of a Constant Polynomial. Let P(x) = 5x 3 â 4x 2 + 7x â 8. Let us start with the general polynomial equation a x^n+b x^(n-1)+c x^(n-2)+â¦.+z The degree of this polynomial is n Consider the polynomial equations: 0 x^3 +0 x^2 +0 x^1 +0 x^0 For this polynomial, degree is 3 0 x^2+0 x^1 +0 x^0 Degree of ⦠Hence, the degree of this polynomial is 8. let P(x) be a polynomial of degree 3 where $$P(x)=x^{3}+2x^{2}-3x+1$$, and Q(x) be another polynomial of degree 2 where $$Q(x)=x^{2}+2x+1$$. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. The terms of polynomials are the parts of the equation which are generally separated by â+â or â-â signs. Every polynomial function with degree greater than 0 has at least one complex zero. And r(x) = p(x)+q(x), then degree of r(x)=maximum {m,n}. The degree of each term in a polynomial in two variables is the sum of the exponents in each term and the degree of the polynomial is the largest ⦠are equal to zero polynomial. Featured on Meta Opt-in alpha test for a new Stacks editor So technically, 5 could be written as 5x 0. A âzero of a polynomialâ is a value (a number) at which the polynomial evaluates to zero. This also satisfy the inequality of polynomial addition and multiplication. We ‘ll also look for the degree of polynomials under addition, subtraction, multiplication and division of two polynomials. A question is often arises how many terms can a polynomial have? They are as follows: Monomials âAn algebraic expressions with one term is called monomial hence the name âMonomial. What could be the degree of the polynomial? For example, 2x + 4x + 9x is a monomial because when we add the like terms it results in 15x. Let me explain what do I mean by individual terms. To find the degree of a term we ‘ll add the exponent of several variables, that are present in the particular term. A function with three identical roots is said to have a zero of multiplicity three, and so on. To recall an algebraic expression f(x) of the form f(x) = a. are real numbers and all the index of âxâ are non-negative integers is called a polynomial in x.Polynomial comes from âpolyâ meaning "many" and ânomialâ meaning "term" combinedly it means "many terms"A polynomial can have constants, variables and exponents. Definition: A polynomial is in standard form when its term of highest degree is first, its term of 2nd highest is 2nd etc.. All of the above are polynomials. ⇒ if m=n then degree of r(x) will m or n except for few cases. A polynomial of degree one is called Linear polynomial. Example #1: 4x 2 + 6x + 5 This polynomial has three terms. e is an irrational number which is a constant. For example, the polynomial $x^2â3x+2$ has $1$ and $2$ as its zeros. Polynomial degree can be explained as the highest degree of any term in the given polynomial. 1 answer. 2x 2, a 2, xyz 2). Cite. ⇒ same tricks will be applied for addition of more than two polynomials. Hence the degree of non zero constant polynomial is zero. Similar to any constant value, one can consider the value 0 as a (constant) polynomial, called the zero polynomial. If the polynomial is not identically zero, then among the terms with non-zero coefficients (it is assumed that similar terms have been reduced) there is at least one of highest degree: this highest degree is called the degree of the polynomial. The polynomial 0, which may be considered to have no terms at all, is called the zero polynomial. Furthermore, 21x. In other words deg[r(x)]= m if m>n or deg[r(x)]= n if m Gis Programming Certificate, Kpsc Sda Hall Ticket 2021, Fly High, My Angel Quotes, German Shorthaired Pointer Puppy, German Shorthaired Pointer Puppy, Factoring Quadratic Trinomials Examples, Kpsc Sda Hall Ticket 2021, Sba3 Brace Illegal, Resembling Silver Crossword Clue, Ford Engines Specs, | 1,430 | 5,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-14 | latest | en | 0.899301 |
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# If the polynomial x4−6x3+16x2−25x+10 is divided by another polynomial x2−2x+k, the remainder comes out to be x+a, find k and a.
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## To get the value of k and a, equate the obtained remainder with x+a. It can be observed that (−9+2k)x+(10−8k+k2)=x+a . By equating the coefficient of x and constant terms, we get Therefore, (−9+2k)=1 and (10−8k+k2)=a For (−9+2k)=1, 2k=10 k=5 For (10−8k+k2)=a 10−8×5+25=a −5=a Therefore, a=−5. Hence, k=5 and a=−5
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### Test of independence vs test of homogeneity
I am teaching a basic statistics course and today I will cover the chi-squared test of independence for two categories and the test for homogeneity. These two scenarios are conceptually different, but ... | 2,786 | 13,824 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-30 | latest | en | 0.930549 |
https://gis.stackexchange.com/questions/426616/converting-ogr-geometry-as-wkt-of-polygon-from-float-to-integer-in-the-wkt-strin | 1,656,691,149,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00553.warc.gz | 303,132,977 | 68,470 | # Converting OGR geometry as WKT of polygon from float to integer in the WKT string
(This is a repost into this community of my original question posted on SO, because there has been no answer so far. Hopefully in this community more luck. I'll ask for the other one to be closed.)
I am working with a hexagonal grid that needs to be exported to text and each feature needs to be represented by a WKT-string so it can be later imported in a next step.
I use OGR in Python (3.9.7).
Problem: The geometry data for the points that make up the polygon in the WKT string are all as floats. I want them to be as integers instead, because that is accurate enough for my purposes as the decimal places represent micrometers in the coordinate system used (SRID=28992). When all points in the polygons will be rounded the same way, my grid will remain a fully covering grid.
Example:
``````from osgeo import ogr
testpolywkt='POLYGON ((75694.8564184426 452182.812141684,75570.7757705624 451967.898155319,75322.6144748018 451967.898155319,75198.5338269215 452182.812141684,75322.6144748018 452397.726128049,75570.7757705624 452397.726128049,75694.8564184426 452182.812141684))'
geom = ogr.CreateGeometryFromWkt(testpolywkt)
``````
I want to get a WKT string like:
``````'POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 52398,75571 452398,75695 452183))'
``````
What could be the best way to do that?
--Update: Let me add what I have tried myself, which is to go point by point in the geometry and then round those numbers, like:
``````from shapely import wkt
for feat_hex in layer_hex:
strShape = str(feat_hex.GetField('receptor_i')) + '\tSRID=28992;POLYGON(('
geom_hex = feat_hex.GetGeometryRef()
strWkt = geom_hex.ExportToWkt()
# Get points from polygon
x, y = p1.exterior.coords.xy
# do this for each point
for i in range(0,len(x)):
if i > 0:
strShape += ' '
strShape += str(round(x[i])) + ' ' + str(round(y[i]))
if i < len(x)-1:
strShape += ',' # no comma at end
strShape += '))\n'
# write line to file
outputfile.write(strShape) # write line
``````
It works, though this feels like a workaround rather than a more elegant solution. The question is whether there are some settings in the `ogr` package that somehow instruct the underlying C-library to round it or set a lower precision.
Otherwise try this script:
``````from osgeo import ogr
testpolywkt = 'POLYGON ((75694.8564184426 452182.812141684,75570.7757705624 451967.898155319,75322.6144748018 451967.898155319,75198.5338269215 452182.812141684,75322.6144748018 452397.726128049,75570.7757705624 452397.726128049,75694.8564184426 452182.812141684))'
geom = ogr.CreateGeometryFromWkt(testpolywkt)
numRings = geom.GetGeometryCount()
# Create ring
newRing = ogr.Geometry(ogr.wkbLinearRing)
for i in range(numRings):
ring = geom.GetGeometryRef(i)
numPoints = ring.GetPointCount()
for j in range(numPoints):
coordinates = (round(ring.GetPoint(j)[0]), round(ring.GetPoint(j)[1]))
#also works as coordinates = (round(ring.GetX(j)), round(ring.GetY(j)))
# Create polygon
poly = ogr.Geometry(ogr.wkbPolygon)
print (poly.ExportToWkt())
``````
to get this:
``````POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 452398,75571 452398,75695 452183))
``````
• Thanks @Taras, this is somewhat similar as to what I have tried. However, could you explain what is the benefit of using these rings? One of the benefits is that your solution is all `ogr` and mine uses `shapely`. Do you have any opinion on what would be more efficient? I want to apply it to large datasets, so every percent could potentially matter. Mar 21 at 16:39
• I have now tried and compared this and it seems to be a tiny bit faster than my solution: 31.5s vs 32.5s on my test dataset. Every little helps! Maybe because now I can do without `shapely`? Mar 22 at 8:17
How about using the ST_SnapToGrid function that comes from SpatiaLite https://www.gaia-gis.it/gaia-sins/spatialite-sql-latest.html
``````return a new Geometry corresponding to the input Geometry; all points and vertices will be snapped to the grid defined by its origin and size(s).
Removes all consecutive points falling on the same cell.
All collapsed geometries will be stripped from the returned Geometry.
NULL is returned on failure.
``````
Usage example with ogrinfo follows. With Python you would use the ExecuteSQL function. Don't pay attention to `py.shp`, it is in the command only because ogrinfo requires some datasource or it gives an error.
``````ogrinfo -dialect sqlite -sql "SELECT ST_SnapToGrid(ST_GeomFromText('POLYGON ((75694.8564184426 452182.812141684,75570.7757705624 451967.898155319,75322.6144748018 451967.898155319,75198.5338269215 452182.812141684,75322.6144748018 452397.726128049,75570.7757705624 452397.726128049,75694.8564184426 452182.812141684))'),1) AS geom" py.shp
INFO: Open of 'py.shp'
using driver 'ESRI Shapefile' successful.
Layer name: SELECT
Geometry: Unknown (any)
Feature Count: 1
Extent: (75199.000000, 451968.000000) - (75695.000000, 452398.000000)
Layer SRS WKT:
(unknown)
Geometry Column = geom
OGRFeature(SELECT):0
POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 452398,75571 452398,75695 452183))
``````
• Are there ways of using this ST_SnapToGrid function in Python or would I need to call ogrinfo in a commandline? Mar 21 at 16:45
• I wrote `With Python you would use the ExecuteSQL function`. It is documented in gdal.org/python/osgeo.ogr.DataSource-class.html Mar 21 at 16:53
• I´m not sure how to use this one. When I use the same SELECT query (querystring) as in your example and use the ExecuteSQL method on the datasource (`driver_hex = ogr.GetDriverByName('ESRI Shapefile') datasource = driver_hex.Open('shapefile.shp', 0) datasource.ExecuteSQL(querystring)`), then I get: `ERROR 1: Undefined function 'ST_GeomFromText' used.` What am I doing wrong? Mar 22 at 9:09
• Did you set the SQL dialect SQLite? You can find usage examples from GDAL autotests github.com/OSGeo/gdal/blob/master/autotest/ogr/…. Mar 22 at 9:13
• Ok, changed to `result = ds_hex.ExecuteSQL(querystring, dialect = "SQLITE")` It returns a Layer object? `<osgeo.ogr.Layer; proxy of <Swig Object of type 'OGRLayerShadow *' at 0x7f50f1987870> >` Mar 22 at 9:21
Use the following script (includes the application of `json` package)
``````import json
from osgeo import ogr
testpolywkt = 'POLYGON ((75694.8564184426 452182.812141684,75570.7757705624 451967.898155319,75322.6144748018 451967.898155319,75198.5338269215 452182.812141684,75322.6144748018 452397.726128049,75570.7757705624 452397.726128049,75694.8564184426 452182.812141684))'
geom = ogr.CreateGeometryFromWkt(testpolywkt)
coordinates_rounded = [[round(pair[0]), round(pair[1])] for pair in coordinates]
# Create ring
ring = ogr.Geometry(ogr.wkbLinearRing)
for pair in coordinates_rounded:
# Create polygon
poly = ogr.Geometry(ogr.wkbPolygon)
print(poly.ExportToWkt())
``````
to achieve this:
``````POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 452398,75571 452398,75695 452183))
``````
Use simply the Regular expression module as in Find, round and replace numbers in a string with regexp for example
``````testpolywkt='POLYGON ((75694.8564184426 452182.812141684,75570.7757705624 451967.898155319,75322.6144748018 451967.898155319,75198.5338269215 452182.812141684,75322.6144748018 452397.726128049,75570.7757705624 452397.726128049,75694.8564184426 452182.812141684))'
import re
# 0 decimals
re.sub(r'\d*\.\d+', lambda m: format(float(m.group(0)), '.0f'), testpolywkt)
'POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 452398,75571 452398,75695 452183))'
# 2 decimals
re.sub(r'\d*\.\d+', lambda m: format(float(m.group(0)), '.2f'), testpolywkt)
'POLYGON ((75694.86 452182.81,75570.78 451967.90,75322.61 451967.90,75198.53 452182.81,75322.61 452397.73,75570.78 452397.73,75694.86 452182.81))'
from osgeo import ogr
wkt= re.sub(r'\d*\.\d+', lambda m: format(float(m.group(0)), '.0f'), testpolywkt)
poly = ogr.CreateGeometryFromWkt(wkt)
poly.ExportToWkt()
'POLYGON ((75695 452183,75571 451968,75323 451968,75199 452183,75323 452398,75571 452398,75695 452183))'
`````` | 2,573 | 8,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | longest | en | 0.799501 |
http://quizlet.com/30986099/accounting-201-exam-3-flash-cards/ | 1,427,569,650,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297689.58/warc/CC-MAIN-20150323172137-00237-ip-10-168-14-71.ec2.internal.warc.gz | 232,903,138 | 20,864 | # Accounting 201 Exam 3
### 35 terms by shelbyjeancrow
#### Study only
Flashcards Flashcards
Scatter Scatter
Scatter Scatter
## Create a new folder
### FIFO LIFO
which inventory costing method most closely approximates current cost for each of the following: ending inventory, cost of goods sold
a. FIFO FIFO
b. FIFO LIFO
c. LIFO FIFO
d. LIFO LIFO
### FIFO
Which inventory costing method most closely approximates current cost for ending inventory?
### LIFO
Which inventory costing method most closely approximates current cost for cost of goods sold?
### Aging of accounts receivable method
A method of estimating bad debts expense that involves a detailed examination of outstanding accounts and their length of time past due is the
### What is the depreciation base to use for the asset? What is the asset's useful life? What method of cost apportionment is best for this asset?
What are the basic questions that must be answered before the amount of depreciation charge can be computed?
### Straight line method or production or use method
Net income is understated if, in the first year, estimated salvage value is excluded from the depreciation computation using the:
### No effect on net income and no effect on total assets
On October 29 of the current year, a company concluded that a customer's \$4,400 account receivable was uncollectible and that the account should be written off. What effect will this write-off have on this company's net income and total assets assuming the allowance method is used to account for bad debts?
### Straight line method
depreciation is calculated by subtracting salvage value from the cost of the plant asset and then dividing the result by the useful life. When useful life is measured in years, the resulting amount is the annual depreciation expense for the asset.
### units of production method
depreciation is calculated by subtracting salvage value from the cost of the plant asset and then dividing the result by the useful life in units. The resulting amount is the depreciation expense per unit. That amount is multiplied by the number of units produced during each accounting period to determine the total amount of depreciation expense for that period.
### double declining
depreciation method uses twice the straight-line rate (100%/useful life in years) multiplied by the beginning of period book value of the asset. The resulting amount is the depreciation expense for that period.
### first step
Step in the accounting process for asset disposals: bring the amount of depreciation up to date
### second step
step in the accounting process for asset disposals: to then remove the cost of the asset and it's related accumulated depreciation from the accounting records
### third step
step in the accounting process for asset disposals: to recognize the amount of cash and/or other assets involved in the transaction
### fourth step
step in the accounting process for asset disposals: record any gain or loss from the asset disposal by comparing the asset book value to the market value of any asset received
### Credit
Allowance for doubtful accounts (D or C)
### FIFO
Which inventory method maximizes net income?
### FIFO, because the last bought one is the closest to current market price. Highest Ending inventory, because EI goes to balance sheet
Which inventory method produces the most meaningful inventory amount for the balance sheet?
### LIFO, because lowest taxes
inventory cost method that maximizes the company's cash flow, why?
### Ending Inventory
=Beginning Inventory + Purchases - Cost of goods sold
### = cost-salvage value / expected production value
Depreciation Rate used in production output depreciation method
### = depreciation rate x # of units
Deprecation expense used in production output depreciation method
### = (1/est. useful life) x 2
Depreciation ratio used in double declining balance depreciation method
### = book value x depreciation ratio
depreciation expense used in double declining balance method
### = (cost of asset @ aquisition - salvage value) / (est. useful life)
depreciation expense used in straight line depreciation method
### salvage value
cannot depreciate below
### balance sheet; assets
Accumulated depreciation goes to _____ ______ as a contra assets and decreases ____.
### net income
smallest depreciation expense leads to the highest ___ ___.
### allowance for doubtful accounts
is a contra asset
### accounts receivable
are amounts due from customers for credit sales
aka uncollectible accounts
### direct write-off method and allowance method
companies use two methods to account for uncollectible accounts:
1)
2)
### = (beginning net accounts receivable + ending net accounts receivable) / 2
Net Average Accounts Receivable formula
### = Net sales / Net average accounts receivable
Receivable turnover ratio formula
### Receivable turnover ratio
how many times you can create your sales in accounts receivable... do not want it to be low.
Example: | 1,023 | 5,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-14 | latest | en | 0.919999 |
https://astronomy.stackexchange.com/questions/53916/time-average-of-radial-velocities-of-multiple-galaxies | 1,702,079,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00363.warc.gz | 132,990,648 | 35,978 | # Time average of radial velocities of multiple galaxies
Equation (34) in this paper shows an average radial velocity. The author wrote that the value was obtained by measuring the radial velocities of many galaxies in the cluster of galaxy and by averaging the velocities over time and again over the different galaxies.
How do you obtain the "time average" of a radial velocity? When you measure the doppler shift of an object, does it give only the time average, or would you have to measure velocity over a long time? If the latter is the case, what would be the common time scale for averaging?
• @planetmaker, thank you for the reply! Then (1) are the galaxies (dots in the sky) tracked over the period? and (2) Do the time-averaged velocities over a few years sufficiently account for the virial of the cluster of galaxies? Jun 8 at 10:58
• It would actually take millions of years to see meaningful changes in the individual velocities of galaxies in a cluster (using modern-day technology, anyway). Eqn. 34 of that paper is an average over multiple galaxies within the cluster; it is not an "average over time". Jun 8 at 11:34
• I don't think Earth's annual motion is the issue, since that is well known and can be accounted for even in a single exposure. The average should in principle be taken over time, and are indeed earlier in the paper, as a "theoretical" answer. That's of course impossible given the timescales (~100 Myr), but using the galaxies' instantaneous velocities is a good approximation.
– pela
Jun 8 at 11:39
• @pela Thank you. Is the approximation justified by the assumption of dynamical equilibrium? I'm sorry for asking many questions... Jun 8 at 13:12
• No worries :) Yes, clusters are in general assumed to have virialized (at least the low-z ones that Zwicky was able to see).
– pela
Jun 8 at 13:34 | 424 | 1,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.952723 |
https://www.travellerpoint.in/news_center/2020_11_30/barite_density_in_ppg.html | 1,642,974,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00209.warc.gz | 979,607,371 | 9,320 | # Barite Density In Ppg
density of barite in ppg - newthingsnewtech.com,however, along with barite, hematite was added to the obf (11 ppg) or wbf (8.6 ppg) samples to get a density of 12.5, 14.5, and 16.5 ppg. a barite-weighted drilling fluid, with a density of 15 ppg, was used with the main drilling fluid additives..chapter three: drilling fluids | engineering360,example: determine the number of sacks of barite required to increase the density of 100bbl of 12.0ppg (w 1) mud to 14.0ppg (w 2): barite = 140 sk/100 bbl. metric calculation.
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• ### CONTROL OF MUD WEIGHT Drilling Fluid Management & Disposal
Ρ b = barite density = 35.82 ppg v o = original mud volume v b = barite volume wt b = barite weight nb: specific gravity of barite = 4.3. test procedure. calculate and list the amount of barite required to increase the density of each batch from 8.6 ppg to 9, 10, 11 and 12 ppg. obtain 400 cc of original base mud (density 8.6) add the calculated amount of barite to each batch, stir for about 2
• ### Barite Plugs - DrillingForGas.com
the effectiveness of a barite plug derives from the high density and fine particle size of the barite and its ability to form a tough impermeable barrier. a barite plug has the following advantages: the slurry has a high density (19.0 - 22.0 ppg) it can be pumped through the bit and offers a reasonable chance of recovering the drillstring
f vo + vb pb -p f where pf = final mud density po = original mud density pb = barite density = 35.82 ppg vo = original mud volume vb = barite volume
• ### Sample Calculation Drilling Fluid Management & Disposal
the volume of barite is the density (4.2 g/cc) divided by the weight of barite. 2. the volume of low-gravity solids is the total volume of dry solids minus the volume of barite.
• ### Investigation Of Mud Density And Weighting Materials
however, the quantity of barite that is needed as a weighting agent can be given using the following formula: (1) weighting agent = 1490 (w 2 − w 1) 35 − w 2 where, w 1 is the initial weight (ppg) and w 2 is the final required weight (ppg).
• ### Calculated Density Ppg Measured Density Ppg 600 Rpm
calculated density ppg measured density ppg 600 rpm reading deg 300 rpm reading from pge 430 at university of texas
• ### Drilling Fluid - Chapter 2 - SlideShare
assuming an average density for barite of 35. 4 ppg (4.25 sp.gr.) then a barrel of barite weighs 35.4 x 42 or 1490 lbs. barite for oil field applications is packaged in 100 lb sacks. therefore, if v1 is 100 bbls and wb is 35.4 ppg, then sacks of barite / 100 bbl of mud = [1490 (w2 – w1)] / (35.4 – w2) example-1 how much barite is required to increase the density of 300 bbls of mud from 4 ppg to 15 ppg.
• ### The Mud Density Test Report - Free Essay Example | EduZaurus
take readings in ppg, sg, lb/ft3, psi/1000ft ; procedures (sample 2) measure 10g of barite on the weighing balance ; add the 10g of barite to the prepared sample 1 ; put it into the mixer to mix the homogeneous mixture ; pour the mixture of sample 2 into the 4 scale mud balance cup until it is full ; cover the cup with the lid
• ### Mud Weight - Wikipedia
in the oil industry, mud weight is the density of the drilling fluid and is normally measured in pounds per gallon (lb/gal) (ppg) or pound cubic feet (pcf) . in the field it is measured using a mud scale or mud balance. mud can weigh up to 22 or 23 ppg. a gallon of water typically weighs 8.33 pounds. in conventional drilling fluids, barite is used
• ### Barite Sag – Drilling Fluids Explored
mud density : the full range of mud weight tests (11.6-16.0 ppg) showed sagging. the mud weight did not significantly affect the degree of sag in the mud weight range investigated. most problematic wells to date have occurred with a mud weight of over 13 ppg. hole angle : sag has been shown to occur at any angle between 30 and 90 degrees.
• ### The Measurement Of Mud Density - Free Essay Example
mud density ppg lb/ft3 s.g psi/1000ft sample 1 (water and bentonite) 8.7651.04450sample 2 (with barite, water and bentonite) 8.966.551.11460.05 conclusion at the end of this experiment, i have come to the conclusion that this method is very accurate in the measurement of mud density and also barite can be used to increase mud density.
• ### Evaluation Of Centrifuge - Formulas And Calculations - Rig
f) mass rate for api barite: where : mw = mud density into centrifuge, ppg pu qm = mud volume into centrifuge, gal/m pw = qw = dilution water volume, gal/mm po =
• ### DENSITY UNIT CONVERSION CALCULATOR
density is the mass of material per unit volume. specific gravity is a ratio of the mass of a material to the mass of an equal volume of water at 4 o c. because specific gravity is a ratio, it is a unitless quantity. for example, the specific gravity of water at 4 o c is 1.0 while its density is 1.0 gcm -3.
• ### Drilling Fluid - AAPG Wiki
the density can be adjusted with soluble salts or by addition of solids, termed weight material (for example, barite is added to the mud to increase the density). density values can be expressed as one of the following: ppg = pounds per gallon (united states) s.g. = specific gravity (dimensionless) (international)
• ### Compute The Volume In Bbl And Density In Ppg Of Mu
compute the volume in bbl and density in ppg of mud composed of 20 ibm of quikgel , 45 ibm of barite and 1 bbl of fresh water.
• ### Rig_Systems_Component_part 2 .pptx - Functions Of Drilling
barite sk/100 bbls = 1470 x (mw2 - mw1) 35 – mw2 barite sk = 100 lbs barite sg = 4.2 gm/cc water sg = 1.0 gm/cc barite sg x 8.33 ppg = 35 ppg volume increase per 100 bbs = 100 x (mw2 - mw1) 35 – mw2 158 sks 10.8 bbls
• ### Introduction To Completion Brines And Its Properties – Oil
the ph of 19.2 ppg stock zinc-calcium bromide brine is about 1 – 1.5. by ‘rule of thumb’, the ph of a lower density blend of zinc bromide is approximately proportional to the volume fraction of 19.2 ppg fluid in the blend, i.e., a 17.15 ppg blend of 19.2 ppg zn-cabr 2 and 15.1 ppg cacl 2 -br 2 should have a ph equal to about 3.5.
• ### THE MAJOR WATER BASED DRILLING FLUID ADDITIVES - Drilling
b) density control additives. barite (barium sulphate, baso4) is the primary weighting material used in muds. densities of 9 ppg to 19 ppg can be achieved by mixing water, clay and barite. the api specification for barite is shown in table 5. other weighting materials are calcium carbonate and galena (lead sulphide).
• ### MUD ENGINEERING This Chapter Covers The Following Items
¾ high gravity are added to increase mud weight or density ¾ referred to as weighting materials ¾ mud named as weighted mud, they are: barite (barium sulphate, baso4) sp.gr. 4.2 ¾ used to prepare mud in excess of 10 ppg ¾ referred to as weighting agent for low cost and
• ### All Mud Formulations | MAWAD DRILLING FLUIDS
brine density, ppg; vol % brine; vol % corrected solids; a s g of solids % low gravity solids; low gravity solids, ppb % high gravity solids; high gravity solids, ppb; oil ratio*** water (h20) ratio*** oil ratio (oil to brine) water/brine ratio (oil to brine)
• ### Barite Mineral | Uses And Properties
this gives the paper a very high density that allows the cards to be 'dealt' easily to players around a card table. barite is used as a weighting filler in rubber to make 'anti-sail' mudflaps for trucks. barite is the primary ore of barium, which is used to make a wide variety of barium compounds. some of these are used for x-ray shielding.
• ### Micronized Weight Material Optimizes ERD Drilling | Offshore
occurrences of barite sag have been observed over a relatively wide range of fluid densities from 11.7 to 20.0 ppg with weight variations as much as ±4.0 ppg exhibited in the flowline. further, barite settling can occur under static or circulating conditions, or under both, and is affected by mud density, rheological properties, temperature, hole angle, size of suspended solids, and other
• ### The Effect Of Testing Conditions On The Performance Of
in the second set, obf at the original density of 11 ppg and the wbf raised up to 11 ppg using barite were used. both sets of 8.6 ppg and 11 ppg drilling fluids were treated with three different formulations of lcm before being tested in the hpa.
• ### Barite Sag | 5 Ways To Improve Sag Problem - Drilling Manual
a symptom of barite sag would be when density checks at the shakers, see a drop in density, followed by an increase, to be then followed eventually, by mud weight of the correct density. this idealized cycle of mud weights seen at the shale shakers may not bee seen as it could be masked by new barite | 2,364 | 8,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-05 | latest | en | 0.877523 |
https://www.gamedev.net/profile/27287-ouranos/ | 1,542,663,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00173.warc.gz | 885,065,932 | 22,085 | # Ouranos
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1. ## How to do this collision response?
You pretty much described sliding planes. After calculating this plane (its normal is the vector between the point of contact and object center) and projecting your original destination onto it, you can calculate a new velocity vector to pass through your collision routines again. This is much more robust and realistic than messing with your velocity based on the colliding surfaces normal. Here's a link to an excellent paper by Paul Nettle that goes into the subject in detail: http://www.informatik.uni-oldenburg.de/~trigger/papers/good/Fluid_Studios_Generic_Collision_Detection_for_Games_Using_Ellipsoids.pdf
2. ## Drawing 2D tiled map of different textures
Choice #1 could work for you, but it's more difficult to maintain and add on to. It's also not very cache friendly depending on the size of the final texture and how many other textures your sampling from. I would do a pre-process as part of level loading where you render the map using your current technique or choice #2 onto a texture. Rather than doing this at a high frequency, you incur the cost once up front and use the result every frame.
3. ## Problem trying to set basic camera angle
You can still use the D3DX function to build your view matrix, just rotate the view direction vector before you pass it in rather than rotate the entire matrix.
4. ## Rotating camera
The link provided is just a blank page in your blog... ...A camera is essentially a view matrix and it sounds like you want to rotate the view direction (a vector in this matrix) around the Y-axis when a user hits a direction. You would also have to update the "right" vector which is an axis in your camera space by computing the cross product of the new direction vector and the up vector. This new view matrix should provide the desired effect.
5. ## NVidia Drivers don't like DX???
Make sure the nVidia driver is signed, since sometimes they like to release non-logo drivers with better performance but are not necessarily D3D-conformant. They usually will say so on the download screen, Windows will also complained if it's unsigned. Also make sure you are using the latest D3D runtime. Another cause might be that the game might be using some DX features not completely supported on your card, OpenGL might render it fine because the technique may vary in the game code depending on the graphics API it's using.
6. ## Engine doesn't work on some pc's
Crashed on mine when I clicked new game... You need to post the source and required files so that we can help you out, otherwise all we can do is say that it crashed. Since it only crashes on certain machines, it seems that you are assuming support for some hardware capability without checking for it. For example, I know youre not using it but in DirectX if you assemble a pixel shader and it fails because the hardware doesnt support that version, the pointer to that shader will be invalid and your program will crash when you try to use it, so you should always check the device caps for that kind of stuff. Again, there's no way to know unless you post the code.
7. ## longhorn
Most if not all the .NET classes you know and love will still be there. Stuff like Avalon will be available in a different namespace ( i.e. Windows.Avalon or whatever ). One big change to the way WinForms are developed is XaML which is a markup language that will be used to control the layout and some functionality of your forms, think along the lines of using XML to place your controls. There are a few videos on msdn.microsoft.com that highlight all these things (MSDNtv, The .NET Show), namely the .NET Show with Don Box is the most technical.
8. ## decoupled input
It goes beyond simply wrapping up an input API. You want your objects to receive notification of some action that is mapped to a certain key somewhere else. This makes things such as the user changing key assignments and AI a lot easier. For example, an AI control class telling an entity to move forward would look the same to that entity as a player pressing a key to move it forward, even if the player uses a different key to move forward.
9. ## Cubic Environment Mapping Problem
Post the code that actually updates the cube map, looks like something may be wrong there. | 927 | 4,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-47 | longest | en | 0.933792 |
https://ell.stackexchange.com/questions/75913/any-differences-between-i-didnt-have-many-apples-and-i-had-few-apples | 1,722,723,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00274.warc.gz | 183,715,168 | 40,464 | # Any differences between "I didn't have many apples" and "I had few apples"?
We all know that "I had a few apples" and "I had few apples" are different.
I wondered what the possible differences would be in the meaning of these two sentences:
I didn't have many apples.
Well, we know that "few" is a negative quantifier and it's the opposite of 'many'. But do the two sentences convey precisely the same meaning?
Yes, they convey different meanings.
## Down-to-Earth
1. I didn't have many apples.
Simply put, sentence one conveys the meaning that the speaker didn't have as many apples as it would take to call the apples "many". Sentence two tells us that the speaker had a small number of apples.
They could mean the same, but they do not. "I didn't have many" means "I had less than what could be called many". That has a broad meaning: I could be having a relatively large amount of apples, but not as many that would fit the label "many". However, "I had few" means "I only had a small number". It couldn't imply the meaning "I didn't have many" represents. In other words, sentence two implies sentence 1, but sentence 1 has a more general meaning. Take a look at the diagram I draw to get a better picture. A more elaborate description is given below.
## Technically . . .
More info on this can be found on Cambridge Grammar of English Language §5.2 — scalar entailments and implicatures; page 366–368.
CGEL points out that negating "many" results in a paucal implicature. It's useful to compare this pair to the more restrictive negative quantifier "no". Indeed, "few" is the opposite of many; but "not many" is not as restrictive as them. Compare these four instances:
1. None of the sailors tried to defend their captain.
2. Few of the sailors tried to defend their captain.
3. Not many of the sailors tried to defend their captain.
4. Not all of the sailors tried to defend their captain.
We can apply what's been said here. "None" is the most restrictive,; "few" is less restrictive than "none" but still more than the other pairs, "not many" is only more restrictive than "not all".
Hence, we can say that if sentence one is true, so are others. If sentence two is true, then so are 3 and 4, but we can't judge whether one is correct or not; and the same story goes for three and four. If we know that not all of the sailors helped their captain, we can't necessarily judge whether none of them helped the captain, or there were only a few exceptional sailors that didn't help the captain.
However, it should be noted that this difference in meaning is very subtle and may go unnoticed, since this was all about implicatures: context-dependent. We can conclude that the sentences do not convey 'precisely' the same meaning, although the difference is passable in most cases.
• Well my answer wasn't far then. Commented Dec 14, 2015 at 17:24
• Yes @Sub, and I upvoted it. However, I felt the need to share what I learned and hence I posted a self-answer. Commented Dec 14, 2015 at 17:25
• I love the graph. Moreover, you hit on a key point: terms like "many" can vary according to the situation. If my son brought home three bushels of apples from the supermarket, for example, that's many apples. (At least, it would be in my house!) However, if I owned an orchard, and my son only managed to harvest three bushels of apples, that same quantity would now be very few.
– J.R.
Commented Dec 14, 2015 at 21:09
They convey the meaning of quantity but in a different way.
By using didn't have many we specify that we had a certain amount of apples, not much as expected, but we did have apples to accomplish our goal.
If we use I had few apples, it means we didn't even have a certain quantity, just a short quantity that didn't allow us to make what we wanted. Compare:
I didn't have many apples but I managed to do a cake.
I had few apples, so I couldn't make the cake.
• This is closer to the correct answer than the other answers. Commented Dec 14, 2015 at 17:04
• I disagree. In the two example sentences, the phrases can be interchanged without changing the meaning. "I had few apples, but I managed to make a cake." "I didn't have many apples, so I couldn't make the cake."
– Era
Commented Dec 14, 2015 at 17:11
• @Era I think you can interchange them if you use I had a few apples, pretty different from I had few apples. Commented Dec 14, 2015 at 17:15
• @Subjunctive I don't know what you mean. What's wrong with the sentence I had few apples, but I managed to make a cake? I don't think it would be at all correct to say I had a few apples, but I managed to make a cake. It doesn't make any sense because you are using two positive statements and connecting them with but.
– Era
Commented Dec 14, 2015 at 17:21
• I agree with @Era here. The only way we can say for sure that few means "not enough" is to say "too few": We had too few apples, so we couldn't make the cake.
– J.R.
Commented Dec 14, 2015 at 21:06
In my opinion,
"few" means "I don't have any apple."; another explanation is "I almost don't have any apples."
and
"many" means "I have several apples."
Since I'm not a native speaker, I strongly advise you to get the answer from your teacher. The definitions of "few, a few, little, a little" will be a nuisance in future. | 1,322 | 5,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.992042 |
http://www.mablee.com/tag/1-2-3-in-english/ | 1,544,616,325,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00011.warc.gz | 436,659,213 | 13,248 | # 1 2 3 In English Tag
Counting Method / June 6, 2018 / Penelope Chow
Wetting and soiling are not likely to become entrenched habits. Limited use of diapers results in. Cost effective. Prevents diaper rash. Better for the environment. More hygienic.
Counting Method / June 6, 2018 / Lilyana Geary
Pros of the Infant Potty Training Method. Promotes parent child bonding. Start early, finish early
Counting Method / June 6, 2018 / Ximena Lucas
There are many different ways to potty train your toddler. At last count, I had found 35 different books for parents on potty training and I own most of those books. Based on my research and understanding, I break HOW
Counting Method / January 27, 2018 / Kenzie Bowser
The integers on the left side of zero (0) are called negative numbers, and are represented by a negative sign (−) placed before them, as in −5, −10, and −15.* The integers on the right side of 0 are called
Counting Method / February 24, 2018 / Abby Lemon
The researchers tested two counting methods in a study of 40 children who played a 100-space board game designed by the researchers to mimic products like Chutes & Ladders. In the first method, referred to as "count-from-1," children started counting
Counting Method / February 26, 2018 / Jaelynn Adler
Pros of the Infant Potty Training Method. Promotes parent child bonding. Start early, finish early
Counting Method / January 24, 2018 / Noor Burnette
Snacks are other properties that attract the child's attention. Count aloud the pieces of snacks before giving them. Ask them to count each piece they put into their mouth and continuously, encourage them for their good work. The child loves
Counting Method / February 15, 2018 / Faye Laws
he symbol Z can be annotated to denote various sets, with varying usage amongst different authors: Z+, Z+ or Z> for the positive integers, Z≥ for non-negative integers, Z≠ for non-zero integers. Some authors use Z* for non-zero integers, others
Counting Method / February 13, 2018 / Abby Lemon
Allow The Kid To Shop Around. Inside the home, build up a vegetable market with several potatoes, tomatoes and onions. Give them money and ask them to bring two potatoes, three tomatoes and five onions. Correct them, if they bring
Counting Method / March 2, 2018 / Lilyana Geary
Cons of the Infant Potty Training Method. Time consuming; nearly impossible for working mothers. Have to stay on top of it and stick with it until child can potty independently. Accidents and accident clean up. It's potty time!! -Practice until | 614 | 2,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-51 | latest | en | 0.948592 |
https://math.stackexchange.com/questions/3379012/algorithm-to-check-if-an-equivalent-node-exists-on-a-different-graph-with-differ | 1,702,310,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00729.warc.gz | 409,698,786 | 35,942 | # Algorithm to check if an equivalent node exists on a different graph with different labelings
Let's say I have an undirected graph defined by a set of edges:
$$\{(1, 2), (0, 2, 3), (0, 1, 3), (1, 2)\}$$
This could be drawn as follows:
Now, assume that I drew the same graph, but changed the numbers:
$$\{(1, 2, 3), (0, 3), (0, 3), (0, 1)\}$$
to form this:
It's clear that these two graphs are identical, at least visually. This question goes into more detail about this problem, but I'm looking for a solution to a more specific focus.
Now, we'll focus in on the node that is called $$3$$ in graph 1, and $$1$$ in graph 2.
This node is represented as $$(1, 2)$$ in the first graph, and $$(0, 1)$$ in the second. Now, here's my problem:
Suppose I am given a list of edges $$b$$, and one specific node $$b_n$$ from that list of edges. Also suppose I am given another list of edges $$z$$, which may or may not be equivalent in size. I now need to find out whether $$z$$ contains some node $$z_n$$ which, ignoring labelling, is equivalent to $$b_n$$.
By equivalent, I mean that if we only focus on the node $$b_n$$, its direct neighbours, and the edges connecting to the neighbours, it would be identical to that of $$z_n$$ in the context of the given graph, ignoring the numbering/labelling.
E.g:
Given $$b=\{(1, 2), (0, 2, 3), (0, 1, 3), (1, 2)\}$$, $$n=3$$ (0-indexed), and $$z=\{(1, 2, 3), (0, 3), (0, 3), (0, 1)\}$$. We can find that the result is true in this case, since there exists $$n=1\Rightarrow z_n$$ such that, removing all labels, $$z_n$$ is essentially equal to $$b_n$$.
In a more complex example, take this graph:
and this one:
Here, $$b=\{(1, 2, 3), (3, 4), (0, 3), (0, 1, 2), (1)\}$$, $$n=4$$ and $$z=\{(4), (2, 3, 4), (1, 3), (1, 2), (1, 0)\}$$
We'd once again return true.
$$b=\{(1, 2, 3), (0), (0, 3), (0, 2, 4, 5), (3, 5), (3, 4, 6), (5)\}$$, $$n=3$$
$$z=\{(1, 2, 3), (0, 7, 8), (0, 3), (0, 2, 4, 5), (3, 5), (3, 4, 6), (5, 9), (1, 8, 9), (1, 7, 9), (6, 7, 8)\}$$
Once again, this is true, because if we only do a shallow check then the node's direct neighbours are the same nodes despite the different numbering, even if the secondary neighbours are different.
My question is: Is there a generalised algorithm to check this, given the two graphs and an integer $$n$$? If so, what is the most computationally efficient method?
• It sounds like you consider two vertices equivalent if their neighbourhoods are isomorphic. So you still have to solve a graph isomorphism problem, but if the degree of the vertex is small then it might not be too expensive.
– user856
Oct 3, 2019 at 6:48
• @Rahul thanks! I know virtually nothing about the subject, so I'm sorry if the question seems quite obvious/easy. I'll look into what you said! Oct 3, 2019 at 6:56 | 937 | 2,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.887042 |
https://au.mathworks.com/matlabcentral/cody/solutions/1982758 | 1,601,356,589,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00616.warc.gz | 273,231,862 | 16,222 | Cody
# Problem 1946. Fibonacci-Sum of Squares
Solution 1982758
Submitted on 19 Oct 2019 by kranthi kumar
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
n = 5; S = 40; assert(isequal(FibSumSquares(n),S))
2 Pass
n = 8; S = 714; assert(isequal(FibSumSquares(n),S))
3 Pass
n = 11; S = 12816; assert(isequal(FibSumSquares(n),S))
4 Pass
n = 15; S = 602070; assert(isequal(FibSumSquares(n),S))
5 Pass
n = 21; S = 193864606; assert(isequal(FibSumSquares(n),S))
6 Pass
n = 26; S = 23843770274; assert(isequal(FibSumSquares(n),S)) | 242 | 659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.632477 |
http://forum.allaboutcircuits.com/threads/find-led-current-in-switching-circuit.30332/ | 1,481,278,240,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00426-ip-10-31-129-80.ec2.internal.warc.gz | 100,042,423 | 14,345 | # Find LED current in switching circuit
Discussion in 'Homework Help' started by ShowerOf, Nov 17, 2009.
1. ### ShowerOf Thread Starter Member
May 4, 2009
13
0
For the switching circuit shown:
Vcc = 6V
Base supply voltage (when ON) is 3.6V
The LED has a forward voltage drop of 2V
The transistor has a Vbe = 0.7V
Question: Calculate the current through the LED when the switch is on.
My calculations:
Ve = Vbb - Vbe
Ve = 3.6 - 0.7
Ve = 2.9V
Ie = Ve / Re
Ie = 2.9 / 180
Ie = 16.11mA
So can I assume that current through the LED is approximately equal to Ie ?
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2. ### Jony130 AAC Fanatic!
Feb 17, 2009
3,957
1,097
Yes, you can assume that Ic=β/(β+1)*Ie
So Ie≈Ic in active region.
3. ### ShowerOf Thread Starter Member
May 4, 2009
13
0
To find the power dissipated by the transistor.
using calculation based on another thread
Vc = Vcc - VLED
Vc = 6 - 2
Vc = 4V
Vce = Vc - Ve
Vce = 4 - 2.9
Vce = 1.1V
P = Vce * Ie
P = 1.1 * 16.11mA
P = 17.72 mW
Is this correct ?
4. ### Jony130 AAC Fanatic!
Feb 17, 2009
3,957
1,097
Yes, look ok for me.
5. ### ShowerOf Thread Starter Member
May 4, 2009
13
0
Thank you Jony130, I am grateful to have somewhere to get help and
Cheers
6. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
If you have 3.6V on the base of this transistor you will have 16.7ma through the collector. This configuration is a constant current source (BJT excel at those).
If the power supply were 2V greater you could use any color LED, as it will still be a constant current of 16.7ma.
Figure 3.6V (base) - 0.6 (BE drop) = 3.0V at the emitter
3.0V/180Ω = 16.7ma. | 590 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-50 | longest | en | 0.886279 |
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# Handbook of Monte Carlo Methods. Wiley Series in Probability and Statistics
## John Wiley and Sons Ltd, April 2011, Pages: 772
A comprehensive overview of Monte Carlo simulation that explores the latest topics, techniques, and real-world applications
More and more of today’s numerical problems found in engineering and finance are solved through Monte Carlo methods. The heightened popularity of these methods and their continuing development makes it important for researchers to have a comprehensive understanding of the Monte Carlo approach. Handbook of Monte Carlo Methods provides the theory, algorithms, and applications that helps provide a thorough understanding of the emerging dynamics of this rapidly-growing field.
The authors begin with a discussion of fundamentals such as how to generate random numbers on a computer. Subsequent chapters discuss key Monte Carlo topics and methods, including:
- Random variable and stochastic process generation
- Markov chain Monte Carlo, featuring key algorithms such as the Metropolis-Hastings method, the Gibbs sampler, and hit-and-run
- Discrete-event simulation
- Techniques for the statistical analysis of simulation data including the delta method, steady-state estimation, and kernel density estimation
- Variance reduction, including importance sampling, latin hypercube sampling, and conditional Monte Carlo
- Estimation of derivatives and sensitivity analysis
- Advanced topics including cross-entropy, rare events, kernel density estimation, quasi Monte Carlo, particle systems, and randomized optimization
The presented theoretical concepts are illustrated with worked examples that use MATLAB®, a related Web site houses the MATLAB® code, allowing readers to work hands-on with the material and also features the author's own lecture notes on Monte Carlo methods. Detailed appendices provide background material on probability theory, stochastic processes, and mathematical statistics as well as the key optimization concepts and techniques that are relevant to Monte Carlo simulation.
Handbook of Monte Carlo Methods is an excellent reference for applied statisticians and practitioners working in the fields of engineering and finance who use or would like to learn how to use Monte Carlo in their research. It is also a suitable supplement for courses on Monte Carlo methods and computational statistics at the upper-undergraduate and graduate levels.
Preface.
Acknowledgments.
1 Uniform Random Number Generation.
1.1 Random Numbers.
1.2 Generators Based on Linear Recurrences.
1.3 Combined Generators.
1.4 Other Gnerators.
1.5 Tests for Random Number Generators.
References.
2 Quasirandom Number Generation.
2.1 Multidimensional Integration.
2.2 Van der Corput and Digital Sequences.
2.3 Halton Sequences.
2.4 Faure Sequences.
2.5 Sobol’ Sequences.
2.6 Lattice Methods.
2.7 Randomization and Scrambling.
References.
3 Random Variable Generation.
3.1 Generic Algorithms Based on Common Transformations.
3.2 Copulas.
3.3 Generation Methods for Various Random Objects.
References.
4 Probability Distributions.
4.1 Discrete Distributions.
4.2 Continuous Distributions.
4.3 Multivariate Distributions.
References.
5 Random Process Generation.
5.1 Gaussian Processes.
5.2 Markov Chains.
5.3 Markov Jump Processes.
5.4 Poisson Processes.
5.5 Wiener Process and Brownian Motion.
5.6 Stochastic Differential Equations and Diffusion Processes.
5.7 Brownian Bridge.
5.8 Geometric Brownian Motion.
5.9 Ornstein-Uhlenbeck Process.
5.10 Reflected Brownian Motion.
5.11 Fractional Brownian Motion.
5.12 Random Fields.
5.13 Lévy Processes.
5.14 Time Series.
References.
6 Markov Chain Monte Carlo.
6.1 Metropolis-Hastings Algorithm.
6.2 Gibbs Sampler.
6.3 Specialized Samplers.
6.4 Implementation Issues.
6.5 Perfect Sampling.
References.
7 Discrete Event Simulation.
7.1 Simulation Models.
7.2 Discrete Event Systems.
7.3 Event-Oriented Approach.
7.4 More Examples of Discrete Event Simulation.
References.
8 Statistical Analysis of Simulation Data.
8.1 Simulation Data.
8.2 Estimation of Performance Measures for Independent Data.
8.3 Estimation of Steady-State Performance Measures.
8.4 Emprical Cdf.
8.5 Kernal Density Estimation.
8.6 Resampling and the Bootstrap Method.
8.7 Goodness of Fit.
References.
9 Variance Reduction.
9.1 Variance Reduction Example.
9.2 Antithetic Random Variables.
9.3 Control Variables.
9.4 Conditional Monte Carlo.
9.5 Stratified Sampling.
9.6 Latin Hypercube Sampling.
9.7 Importance Scaling.
9.8 Quasi Monte Carlo
References.
10 Rare-Event Simulation.
10.1 Efficiency of Estimators.
10.2 Importance Sampling Methods for Light Tails.
10.3 Conditioning Methods for Heavy Tails.
10.4 State-Dependent Importance Sampling.
10.5 Cross-Entropy Method for Rare-Event Simulation.
10.6 Splitting Method.
References.
11 Estimation of Derivatives.
11.2 Finite Difference Method.
11.3 Infinitesimal Perturbation Analysis.
11.4 Score Function Method.
11.5 Weak Deriatives.
11.6 Sensitivity Analysis for Regenerative Processes.
References.
12 Randomized Optimization.
12.1 Stochastic Approximation.
12.2 Stochastic Counterpart Method.
12.3 Simulated Annealing.
12.4 Evolutionary Algorithms.
12.5 Cross-Entropy Method for Optimization.
12
6 Other Randomized Optimization Techniques.
References.
13 Cross-Entropy Method.
13.1 Cross-Entropy Method.
13.2 Cross-Entropy Method for Estimation.
13.3 Cross-Entropy Method for Optimization.
References.
14 Particle Methods.
14.1 Sequential Monte Carlo.
14.2 Particle Splitting.
14.3 Splitting for Static Rare-Event Probability Estimaton.
14.5 Estimation of Multidimensional Integrals.
14.6 Combinatorial Optimization via Splitting.
14.7 Markov Chain Monte Carlo With Splitting.
References.
15 Applications to Finance.
15.1 Standard Model.
15.2 Pricing via Monte Carlo Simulation.
15.3 Sensitivities.
References.
16 Applications to Network Reliability.
16.1 Network Reliability.
16.2 Evolution Model for a Static Network.
16.3 Conditional Monte Carlo.
16.4 Importance Sampling for Network Reliability.
16.5 Splitting Method.
References.
17 Applications to Differential Equations.
17
1 Connections Between Stochastic and Partial Di_erential Equations.
17.2 Transport Processes and Equations.
17.3 Connections to ODEs Through Scaling.
References.
Appendix A: Probability and Stochastic Processes.
Appendix B: Elements of Mathematical Statistics.
Appendix C: Optimization.
Appendix D: Miscellany.
References.
Acronyms and Abbreviations.
List of Symbols.
List of Distributions.
Index.
“Statisticians Kroese, Thomas Taimre (both U. of Queensland), and Zdravko I. Botev (U. of Montreal)
methods, which are statistical methods that involve random experiments on a computer. There are a
great many such methods being used for so many kinds of problems in so many fields that such an
overall view is hard to find. Combining theory, algorithms, and applications, they consider such topics
as uniform random number generation, probability distributions, discrete event simulation, variance
reduction, estimating derivatives, and applications to network reliability.” (Annotation ©2011 Book News
Inc. Portland, OR) | 1,623 | 7,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-20 | latest | en | 0.858079 |
https://kr.mathworks.com/matlabcentral/answers/609521-finding-the-mean-based-on-a-specific-value-in-other-column | 1,652,993,459,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00154.warc.gz | 406,175,213 | 30,375 | # finding the mean based on a specific value in other column
조회 수: 19(최근 30일)
Najam us Saqib Fraz 2020년 10월 9일
댓글: Jon 2020년 10월 9일
Guys I have following data as an example.The data contain 4 coloumns.want to average 4th coloumn when 1st couloumn is equal to 527.1235 and third coloumn is 927.5
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### 채택된 답변
Asad (Mehrzad) Khoddam 2020년 10월 9일
편집: Asad (Mehrzad) Khoddam 2020년 10월 9일
You can find the rows with the first condition and the other rows for the second condition. The intersection of the two rows, are the row numbers that satisfy both conditions:
rows = intersect(find(a(:,1)==527.1235), find(a(:,3)==927.5));
% average of the above rows
avg = mean(a(rows,4));
disp(avg)
or simpler :
rows = find(a(:,1)==527.1235 & a(:,3)==927.5);
% average of the above rows
avg = mean(a(rows,4));
disp(avg)
##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
Asad (Mehrzad) Khoddam 2020년 10월 9일
When data is missing in the data file, it shows as NaN. You can remove the lines that are NaN
use this option:
avg = mean(a(rows,4),'omitnan');
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### 추가 답변(2개)
madhan ravi 2020년 10월 9일
편집: madhan ravi 2020년 10월 9일
ix = (abs(column_1 - 527.1235) < 1e-4) &...
(abs(column_1 - 927.5) < 1e-1);
M = mean(column_4(ix))
##### 댓글 수: 0표시숨기기 이전 댓글 수: -1
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Jon 2020년 10월 9일
편집: Jon 2020년 10월 9일
Lets say you have put your data into an array X.
Find a logical index where the rows match your criteria using:
criteria = [527.1235 927.5]
idl = ismember(X(:,[1,3]),criteria,'rows')
then do the averaging on the 4th colulmn for the rows where the criteria matches
xMean = mean(X(idl,4))
##### 댓글 수: 5표시숨기기 이전 댓글 수: 4
Jon 2020년 10월 9일
I think you are trying to show that floating point comparisons could be a problem if they are not exact. I assumed they were exact, in any case given your example I get idl = 1 not 0
>> X = [527.1235 1.0000 927.5000],criteria =[527.1235 927.5000]
X =
527.1235 1.0000 927.5000
criteria =
527.1235 927.5000
>> idl = ismember(X(:,[1,3]),criteria,'rows')
idl =
logical
1
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