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# Tips For Outsmarting Your Fat Cells Part Two!
There are no quick fixes when it comes to fat-loss because our bodies love it. Find out how to outsmart your fat cells by knowing how they work...
An active cell is a fat-burning cell. That's why cardiovascular exercise has been touted as a fast and efficient way of losing excess body fat.
The question of the superiority of low intensity/extended duration over high intensity/short duration exercise has been Vigorously debated. Current exercise literature uses the questionable term "intensity." "High density/low density" or high/low work rate would be more suitable terms. To keep things simple, I'll be using the term "high/low density" throughout this article.
On one side of the debate you have aerobic instructors and the so-called exercise 'gurus' of our times spewing forth propaganda which claims that jumping up-and-down for 30-60 minutes to the sound of Richard Simmons is more efficient than leaving that small reservoir of perspiration on the floor and equipment following a high density work-out.
What Is The Difference?
To understand the difference between the two methods we must first distinguish between the way they work. Low Density/Long Duration (LD/LD) workouts are typically performed in the range of 50-65% of one's max heart rate ( 220-age = Max Heart Rate).
There are two different ways to calculate your maximum heart rate and your target heart rates. The method I just explained is the simple method. Read the full article here.
Target Heart Rate Calculator
Using the 220 - Age formula.
Enter Your Age - Then press Calculate. MaximumHeart Rate TargetHeart Rate per minute(75% - 85% of Max) TargetHeart Rate15 sec count
The LD/LD method usually derives its source of energy from the oxidative, otherwise known as the Aerobic, system. The Aerobic (i.e., in the presence of oxygen) system uses fat (tryglycerides found in adiposities) as its primary source of fuel during exercise via the Krebs Cycle.
Part of the reason fat is used as a primary source of fuel during LD/LD is because fat molecules contain large quantities of energy per unit of weight. Slow twitch fibers, commonly known as Type I fibers are stressed during this type of activity.
These contain a large amount of oxidative enzymes, and are also surrounded by many capillaries which supply the muscle with myoglobin (oxygen-binding protein). During LD/LDs both carbohydrates and fats are being used; hence, the saying, "fats burn in the flames of carbohydrates." Low muscle and liver stores of glycogen will limit the effect LD/LD exercise has on the "fuel mix" of fat/carbohydrates.
[Note: the body rarely uses blood glucose directly as a means of fuel as its purpose is to maintain the glucose level of the blood for other body tissues, especially the brain].
High Density/Short Duration (HD/SD) workouts are usually performed in the range of 70-90% of one's max heart rate, so they can also be labeled as interval and anaerobic training methods. The HD/SD method derives its source of energy primarily from the phosphagen and glycolitic systems.
[Note: all systems are used simultaneously, and at no time during exercise or rest does a single energy system provide the complete supply of energy].
There appears to be a shift from fat metabolism to carbohydrate metabolism as the density of the exercise increases. The movement is towards muscle glycogen being used as the primary source of fuel while muscle tryglicerides, Plasma FFA (free fatty acids) and blood glucose take a back seat. The sudden shift, as the density of exercise increases, is due, in part, to a increase in fast twitch fibers as well as a release into the bloodstream of certain hormones like "epinephrine".
As the density of exercise rises there is a progressive rise in this hormone. "Epinephrine" increases muscle glycogen breakdown as well as an increase in lactate levels (lactic acid). Lactate increase will inhibit fat metabolism as a substrate.
Another factor is that fast twitch fibres have an abundance of glycolitic enzymes but few mitochondria and lypolitic enzymes (enzymes responsible for fat breakdown). In other words, fast twitch fibers are more efficient at metabolizing carbohydrates than fats. So an increase in fast twitch recruitment will bring about a greater metabolism of carbohydrates and less fat metabolism.
It has been recommended that LD/LD training sessions be the optimal method for reducing one's body fat stores. Though studies have shown that LD/LD may be the choice of many, the body is very adaptive and will become more efficient in conserving energy in relation to energy expenditure.
Have you ever wondered why aerobic instructors, in general, don't sport a lean physique? It's a bit of an exercise mystery since they usually teach several, often many, classes a week which often last for upwards of an hour. Charles Poliquin, a Canadian strength coach, identifies this as the "Chunky Aerobics Instructor Syndrome."
Despite the frequency of aerobic exercise most aerobic instructors carry a high percentage of body fat. The fitness magazine, IDEA, released a study several years ago indicating that aerobic instructors had, on an average, a body-fat percentage of 20%... yes, that's right... 20%! This is surprisingly high for people who train for extended periods of time at a low density level. An individual who trains using primarily LD/LD is like a Honda Civic which can stretch its fuel consumption for extended periods [making fuel consumption sufficient.]
Even though there is a large emphasis placed upon carbohydrate metabolism and low levels of fat metabolism in HD/SD exercise there are studies showing that it is the density of exercise which should be taken into consideration:
"Impact of high-intensity exercise on energy expenditure, lipid oxidation and body fatness".
Author: Yoshioka, M., Doucet, E., St-Pierre, S., Almeras, N., Richard, D., Labrie, A., Despres, J.P., Bouchard, C., Tremblay, A.,
Volume: 25 Issue:3, Page: 332-9 Y ear: 2001 Mar.
"Impact of exercise intensity on body fatness and skeletal muscle metabolism".
Author: Tremblay, A., Simoneau, J.A., Bouchard, C.,
Volume: 43 Issue:7, Page: 814-8 Year:1994 Jul.
It has also been found that, possibly, HD/SD workouts do a better job of promoting the release of human growth hormone (HGH) from the pituitary gland. HGH is a potent stimulator of fat metabolism.
"A number of studies indicate that LD/LD exercise increases energy expenditure in the period of time immediately after activity."
Research has not supplied firm proof for this, so there may yet be a threshold which one may have to work over before benefiting from this effect. LD/LD exercise (around 50% VO2 max) seems less likely to make a difference, while moderate or intense are more likely to.
For example, one review concluded that light exercise could be expected to lead to burning an extra 5-10 calories afterwards; moderate exercise to an extra 12-35 calories. In contrast, strenuous exercise was shown to increase post-exercise energy burning by a huge 180 calories ("Excess post-exercise oxygen consumption - magnitude, mechanisms and practical implications", Bahr, et al, Acta Physiol. Scand., suppl. 605, pp 1-70).
The reduction in body fat, observed in results from high intensity exercise could be explained, perhaps, by an increase in the post-exercise Resting Metabolic Rate. Most studies that have found an elevated resting metabolic rate have performed their measurements within 24-39 hours of the exercise session.
[Note: studies were preformed on men and woman and it was found that women did not benefit from a post-exercise rise in their metabolic rate. This indicates that women have to work twice as hard as men when it comes to fat-loss.It appears that women conserve energy more efficiently (ie, they burn fewer calories) at rest and in response to exercise.]
So, the verdict is in. It has been proven that LD/LD exercise metabolizes fat more efficiently than HD/SD exercise. On the other hand, HD/SD exercise has, over an extended period of time, shown a greater decreases in fat stores. Each method is useful depending upon individual circumstances, and factors such as time, current fitness level and the presence of potential cardiovascular problems.
Individuals who want to lose weight (who are fit incardiovascular terms) should profit from a step-up in their workout intensities. ("lmpact of Exercise Intensity on Body Fatness and Skeletal Muscle Metabolism", "Metabolism", vol. 43(7), pp 814-818, 1994).
The time at which to perform each method is another topic. Some studies show that higher density exercise will elicit a better effect if performed in the afternoon (Bernard et al. 1998, "European Journal of Applied Physiology", 77, pp. 133-138).
These findings suggest that the best time of day for high density training is in the afternoon or early evening. This is consistent with other research into circadian rhythms showing that heart rate, body temperature and muscular strength are all higher in the afternoon than in the morning.
How Does Exercise Work Its Effects?
At the website of Peak Performance Online (http://www.pponline.co.uk), Janet Stansfeld explains how the mechanisms of the body relate to energy expenditure.
1. Increased hormonal activity Some investigations have found that increased resting metabolic rate following endurance exercise is associated with higher blood levels of adrenaline and noradrenaline. These hormones are controlled by the nerves embedded within muscles, known as sympathetic nerves.
Some scientists have looked directly at sympathetic nerves in muscles and found that exercise stimulates the nerve activity. These hormones tend in turn to stimulate various metabolic processes which have the net effect of raising RMR. Other hormones may also play a role. There's some evidence that exercise leads to an increased production of thyroid hormone, which in turn steps up general metabolic activity.
2. Protein resynthesis several lines of evidence suggest that increased metabolic rate following exercise is associated with altered protein metabolism. Some data show that exercise increases protein breakdown. To keep protein status constant, protein synthesis would need to be stepped up following exercise.
There's some indication that exercise causes an increase in the levels of enzymes involved with making proteins. However, so far there's no direct experimental evidence confirming a direct link between increased protein turnover and metabolic rate.
Conclusion
So now you can see how a combined effort using either HD/SD (high density /short duration), or LD/LD (low density long duration) exercises combined with resistance training and a proper eating regimen can impact a individual and help him/her towards a leaner, healthier body.
References:
1. Westcott, W. Be Strong, 1993.
2. King, J. Brad, Fat Wars, 2000. | 2,304 | 10,995 | {"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-2017-17 | latest | en | 0.925775 |
https://documen.tv/question/the-same-eplosive-charge-is-used-so-the-total-energy-of-the-cannon-plus-cannonball-system-remain-15016824-68/ | 1,632,137,799,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00029.warc.gz | 278,424,222 | 17,077 | ## The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how muc
Question
The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s.
in progress 0
3 weeks 2021-09-02T08:42:46+00:00 2 Answers 0 views 0
a. Speed of cannon = 0.902 m/s b. speed of cannonball = 29.44 m/s
Explanation:
Here is the complete question
A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?
The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same?
Solution
From the law of conservation of momentum,
momentum of cannon = momentum of cannonball.
Let m₁,v₁ and m₂,v₂ represent the masses and velocities of the cannon and cannonball respectively.
So, m₁v₁ = m₂v₂.
The speed of the cannon is thus v₁ = m₂v₂/m₁
m₁ = 2090 kg, m₂ = 16.7 kg and v₂ = 113 m/s
v₁ = m₂v₂/m₁ = 16.7 × 113/2090 m/s = 1887.1/2090 m/s = 0.902 m/s
Since the same charge is used, and the cannon mounted rigidly, the total kinetic energy of cannon + cannon ball = kinetic energy of cannonball.
Since all other parameters remain the same,
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₂v₃²
m₁v₁² + m₂v₂² = m₂v₃²
2090 × 0.902² + 16.7 × 113² = 16.7v₃²
1700.43 + 12769 = 16.7v₃²
14469.43 = 16.7v₃²
v₃² = 14469.43/16.7 = 866.43
v₃ = √866.43 = 29.44 m/s
So the speed of the cannon ball is now 29.44 m/s
Completed question
A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immedi-ately after it was fired?
The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters re-mained the same?
Explanation:
1. Mass of cannon, =2090kg
Mass of ball fires=16.7kg
The speed of the ball after being fired, = 113m/s.
Using the recoil of a gun.
Mass of cannon × recoil velocity = mass of ball fires × velocity of ball
2090×v=16.7×113
Then, v=16.7×113/2090
v=0.903m/s
2. The second problem deals with the conservation of energy and all the energy are Kinetic energy
If the explosive where mounted rigidly then, the recoil velocity will be zero, I.e the cannon won’t move
K.E of cannon=1/2mv²
K.E=0.5×2090×0.903²
k.E=851.94J
K.E of ball=1/2mv²
K.E=0.5×16.7×113²
K.E=106,621.15J
Total energy =106,621.15+851.94
Total energy =107,473.09J
Now all this energy is transfer to the ball alone because the canon is mounted
Total energy =1/2mv²
107,473.09=0.5×16.7×v²
v²=107,473.09/(0.5×16.7)
v²=12,871.029
v=√12871.029
v=113.45m/s
The velocity if the ball increases because the canon was mounted | 1,055 | 3,448 | {"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} | 4.21875 | 4 | CC-MAIN-2021-39 | longest | en | 0.848744 |
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Abstract (English):
The receiving antenna is an important part of a radio channel that requires electrodynamic approach in mathematical simulation of its characteristics. Since the invention of radio, and due to further theoretical studies of radio signal transmission, the following situation has arisen: researchers’ attention to receiving antennas is inversely proportional to the factor by which their number exceeds the number of transmitting antennas. We address the problem of building a receiving antenna electrodynamic model in terms of a waveguide representation of HF field. Structurally, the antenna is considered as metal wires of a finite length and arbitrary configuration. Current distribution in antenna is calculated using the long-line theory and normal-mode approach. The mathematical representation of the receiving antenna electrodynamic model is calculation expressions for receiving coefficients of normal modes. They reflect the effects of receiving antenna characteristics, including its directional pattern, on effectiveness of the incident HF field energy conversion into the energy of the driven current waves and final distribution of net current in antenna. These expressions are used to derive the expression to calculate the effective length of the receiving antenna. The obtained mathematical expressions of the receiving antenna electrodynamic model do not contradict the principle of antenna reciprocity. We present calculation formulas for the receiving coefficients and excitation of the isotropic antenna electromagnetic model.
Keywords:
HF field, receiving antenna, Earth — ionosphere waveguide, normal-mode approach
1. Aizenberg G.Z., Belousov S.P., Zhurbenko E.M., Kliger G.A., Kurashov A.G. Korotkovolnovye antenny [Shortwave antennas]. M .: Radio i Svyaz Publ. 1985, 536 p. (In Russian).
2. Altyntseva V.I., Ilyin N.V., Kurkin V.I., Orlov A.I, Orlov I.I., Polekh N.M., Ponomarchuk S.N., Khakhinov V.V. Modeling a decameter radiochannel based on normal-mode approach. Tekhnika sredstv svyazi [Technology of communications assets]. Ser. SS. Moscow, Ekos Publ., 1987, iss. 5, pp. 28–34. (In Russian).
3. Bremmer H. Terrestrial Radio Waves. Theory of Propagation. Amsterdam, 1949, 343 p.
4. Feinberg E.L. Rasprostranenie radiovoln vdol zemnoi poverkhnosti [Propagation of radiowaves along the terrestrial surface]. Moscow, USSR Academy of Sciences Publ., 1961, 548 p. (In Russian).
5. Khakhinov V.V. Calculation of current in receiving antenna in HF field generated by a series of normal modes. Issledovaniya po geomagnetizmu, aeronomii I fizike Solntsa [Res. on Geomagnetism, Aeronomy and Solar Physics]. 2000a, iss. 111, pp. 74–83. (In Russian).
6. Khakhinov V.V. Analyzing the HF field in the wave zone of the antenna using the normal-mode approach. Proc. VIII International Conference on Mathematical Methods in Electromagnetic Theory. IEEE: Kharkov, Ukraine, 2000b, pp. 298–300.
7. Khakhinov V.V. Electromagnetic model of the receiving antenna in terms of a waveguide representation of the HF field. Proc. IX International Conference on Mathematical Methods in Electromagnetic Theory. IEEE: Kiev, Ukraine, 2002, vol. 2, pp. 617–619.
8. Khakhinov V.V. The electrodynamical model of decameter radiochannel with isotropic receiving-transmitting antennas. Proc. X International Conference on Mathematical Methods in Electromagnetic Theory. IEEE: 04EX840. Dnepropetrovsk, Ukraine. 2004, pp. 372–374.
9. Khakhinov V.V., Kurkin V.I. Waveguide approach to modeling of the ionosphere radio channel. Proc. of the XI International Conference on Mathematical Methods in Electromagnetic Theory. IEEE: 06EX1428, Kharkov, Ukraine. 2006, pp. 284–286.
10. Kurkin V.I., Khakhinov V.V. On excitation of a spherical Earth — ionosphere waveguide using arbitrary current distribution. Issledovaniya po geomagnetizmu, aeronomii I fizike Solntsa [Res. on Geomagnetism, Aeronomy and Solar Physics]. 1984, iss. 69, pp. 16–22. (In Russian).
11. Kurkin V.I., Orlov I.I., Popov V.N. Metod normalnykh voln v problem korotkovolnovoi radiosvyazi [Normal-Mode Approach in the Problem of HF Radio Communication]. Moscow, Nauka Publ., 1981, 121 p. (In Russian).
12. Kurkin V.I., Ilyin N.V., Penzin M.S., Ponomarchuk S.N., Potekhin A.P., Khakhinov V.V. Calculation of characteristics of normal modes in a decameter Earth — ionosphere waveguide. Certificate of Computer Program State Registration No. 2017613880 of 03.04.2017. (In Russian).
13. Lavrov G.A., Knyazev A.S. Prizemnye i podzemnye antenny [Near-Surface and Subsurface Antennas. Moscow, Nauka Publ., 1965, 472 p. (In Russian).
14. Leontovich M.A. Izbrannye trudy. Teoreticheskaya Fizika [Selected works. Theoretical Physics]. Moscow, Nauka Publ., 1985, 432 p. (In Russian).
15. Ponomarchuk S.N., Ilyin N.V., Penzin M.S. The model of radio wave propagation in 1–10 MHz frequency range on the base of normal wave technique. Solnechno-zemnaya fizika [Solar-Terrestrial Physics]. 2014, iss. 25, pp. 33–39. (In Russian).
16. Popov V.N., Potekhin A.P. Field structure of a pulse decameter signal in the Earth-ionosphere waveguide. Issledovaniya po geomagnetizmu, aeronomii I fizike Solntsa [Res. on Geomagnetism, Aeronomy and Solar Physics]. 1982, iss. 59, pp. 68–76. (In Russian).
17. Vainshtein L.A. Elektromagnitnye volny [Electromagnetic waves]. Moscow, Radio i Svyaz Publ. 1988, 440 p. (In Russian). | 1,455 | 5,473 | {"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-2023-14 | latest | en | 0.86072 |
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# PS Questions with solutions_Forums
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Thanks to all the guys who posted these question and especially to those who solved them with explanation.
I wanted to share it later but shared it right now because it may be useful to those who are appearing for GMAT tomorrow or in this month. So that they can get all of these questions in 1 file and practice it whether they are online or not.
Hope it helps!
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18 Mar 2010, 07:56
AtifS wrote:
Thanks to all the guys who posted these question and especially to those who solve them with explanation.
I wanted to share it later but shared it right now because it may be useful to those who are appearing for GMAT tomorrow or in this month. So that they can get all of these questions in 1 file and practice it whether they are online or not.
Hope it helps!
+1. Nice post.
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Thanks for the collection
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Thank you. This is just amazing!
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# 3.5 Scalar Types
1
Scalar types comprise enumeration types, integer types, and real types. Enumeration types and integer types are called discrete types; each value of a discrete type has a position number which is an integer value. Integer types and real types are called numeric types. All scalar types are ordered, that is, all relational operators are predefined for their values.
#### Syntax
2
range_constraint ::= range range
3
4
A range has a lower bound and an upper bound and specifies a subset of the values of some scalar type (the type of the range). A range with lower bound L and upper bound R is described by “L .. R”. If R is less than L, then the range is a null range, and specifies an empty set of values. Otherwise, the range specifies the values of the type from the lower bound to the upper bound, inclusive. A value belongs to a range if it is of the type of the range, and is in the subset of values specified by the range. A value satisfies a range constraint if it belongs to the associated range. One range is included in another if all values that belong to the first range also belong to the second.
#### Name Resolution Rules
5
For a subtype_indication containing a range_constraint, either directly or as part of some other scalar_constraint, the type of the range shall resolve to that of the type determined by the subtype_mark of the subtype_indication. For a range of a given type, the simple_expressions of the range (likewise, the simple_expressions of the equivalent range for a range_attribute_reference) are expected to be of the type of the range.
#### Static Semantics
6
The base range of a scalar type is the range of finite values of the type that can be represented in every unconstrained object of the type; it is also the range supported at a minimum for intermediate values during the evaluation of expressions involving predefined operators of the type.
7
A constrained scalar subtype is one to which a range constraint applies. The range of a constrained scalar subtype is the range associated with the range constraint of the subtype. The range of an unconstrained scalar subtype is the base range of its type.
#### Dynamic Semantics
8
A range is compatible with a scalar subtype if and only if it is either a null range or each bound of the range belongs to the range of the subtype. A range_constraint is compatible with a scalar subtype if and only if its range is compatible with the subtype.
9
The elaboration of a range_constraint consists of the evaluation of the range. The evaluation of a range determines a lower bound and an upper bound. If simple_expressions are given to specify bounds, the evaluation of the range evaluates these simple_expressions in an arbitrary order, and converts them to the type of the range. If a range_attribute_reference is given, the evaluation of the range consists of the evaluation of the range_attribute_reference.
10
Attributes
11
For every scalar subtype S, the following attributes are defined:
12
S'First
S'First denotes the lower bound of the range of S. The value of this attribute is of the type of S.
13
S'Last
S'Last denotes the upper bound of the range of S. The value of this attribute is of the type of S.
14
S'Range
S'Range is equivalent to the range S'First .. S'Last.
15
S'Base
S'Base denotes an unconstrained subtype of the type of S. This unconstrained subtype is called the base subtype of the type.
16
S'Min
S'Min denotes a function with the following specification:
17
function S'Min(LeftRight : S'Base)
return S'Base
18
The function returns the lesser of the values of the two parameters.
19
S'Max
S'Max denotes a function with the following specification:
20
function S'Max(LeftRight : S'Base)
return S'Base
21
The function returns the greater of the values of the two parameters.
22
S'Succ
S'Succ denotes a function with the following specification:
23
function S'Succ(Arg : S'Base)
return S'Base
24
For an enumeration type, the function returns the value whose position number is one more than that of the value of Arg; Constraint_Error is raised if there is no such value of the type. For an integer type, the function returns the result of adding one to the value of Arg. For a fixed point type, the function returns the result of adding small to the value of Arg. For a floating point type, the function returns the machine number (as defined in 3.5.7) immediately above the value of Arg; Constraint_Error is raised if there is no such machine number.
25
S'Pred
S'Pred denotes a function with the following specification:
26
function S'Pred(Arg : S'Base)
return S'Base
27
For an enumeration type, the function returns the value whose position number is one less than that of the value of Arg; Constraint_Error is raised if there is no such value of the type. For an integer type, the function returns the result of subtracting one from the value of Arg. For a fixed point type, the function returns the result of subtracting small from the value of Arg. For a floating point type, the function returns the machine number (as defined in 3.5.7) immediately below the value of Arg; Constraint_Error is raised if there is no such machine number.
27.1/2
S'Wide_Wide_Image
S'Wide_Wide_Image denotes a function with the following specification:
27.2/2
function S'Wide_Wide_Image(Arg : S'Base)
return Wide_Wide_String
27.3/2
The function returns an image of the value of Arg, that is, a sequence of characters representing the value in display form. The lower bound of the result is one.
27.4/2
The image of an integer value is the corresponding decimal literal, without underlines, leading zeros, exponent, or trailing spaces, but with a single leading character that is either a minus sign or a space.
27.5/2
The image of an enumeration value is either the corresponding identifier in upper case or the corresponding character literal (including the two apostrophes); neither leading nor trailing spaces are included. For a nongraphic character (a value of a character type that has no enumeration literal associated with it), the result is a corresponding language-defined name in upper case (for example, the image of the nongraphic character identified as nul is “NUL” — the quotes are not part of the image).
27.6/2
The image of a floating point value is a decimal real literal best approximating the value (rounded away from zero if halfway between) with a single leading character that is either a minus sign or a space, a single digit (that is nonzero unless the value is zero), a decimal point, S'Digits–1 (see 3.5.8) digits after the decimal point (but one if S'Digits is one), an upper case E, the sign of the exponent (either + or –), and two or more digits (with leading zeros if necessary) representing the exponent. If S'Signed_Zeros is True, then the leading character is a minus sign for a negatively signed zero.
27.7/2
The image of a fixed point value is a decimal real literal best approximating the value (rounded away from zero if halfway between) with a single leading character that is either a minus sign or a space, one or more digits before the decimal point (with no redundant leading zeros), a decimal point, and S'Aft (see 3.5.10) digits after the decimal point.
28
S'Wide_Image
S'Wide_Image denotes a function with the following specification:
29
function S'Wide_Image(Arg : S'Base)
return Wide_String
30/3
The function returns an image of the value of Arg as a Wide_String. The lower bound of the result is one. The image has the same sequence of graphic characters as defined for S'Wide_Wide_Image if all the graphic characters are defined in Wide_Character; otherwise, the sequence of characters is implementation defined (but no shorter than that of S'Wide_Wide_Image for the same value of Arg).
Paragraphs 31 through 34 were moved to Wide_Wide_Image.
35
S'Image
S'Image denotes a function with the following specification:
36
function S'Image(Arg : S'Base)
return String
37/3
The function returns an image of the value of Arg as a String. The lower bound of the result is one. The image has the same sequence of graphic characters as that defined for S'Wide_Wide_Image if all the graphic characters are defined in Character; otherwise, the sequence of characters is implementation defined (but no shorter than that of S'Wide_Wide_Image for the same value of Arg).
37.1/2
S'Wide_Wide_Width
S'Wide_Wide_Width denotes the maximum length of a Wide_Wide_String returned by S'Wide_Wide_Image over all values of the subtype S. It denotes zero for a subtype that has a null range. Its type is universal_integer.
38
S'Wide_Width
S'Wide_Width denotes the maximum length of a Wide_String returned by S'Wide_Image over all values of the subtype S. It denotes zero for a subtype that has a null range. Its type is universal_integer.
39
S'Width
S'Width denotes the maximum length of a String returned by S'Image over all values of the subtype S. It denotes zero for a subtype that has a null range. Its type is universal_integer.
39.1/2
S'Wide_Wide_Value
S'Wide_Wide_Value denotes a function with the following specification:
39.2/2
function S'Wide_Wide_Value(Arg : Wide_Wide_String)
return S'Base
39.3/2
This function returns a value given an image of the value as a Wide_Wide_String, ignoring any leading or trailing spaces.
39.4/3
For the evaluation of a call on S'Wide_Wide_Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Wide_Wide_Image for a nongraphic character of the type), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised.
39.5/3
For the evaluation of a call on S'Wide_Wide_Value for an integer subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an integer literal, with an optional leading sign character (plus or minus for a signed type; only plus for a modular type), and the corresponding numeric value belongs to the base range of the type of S, then that value is the result; otherwise, Constraint_Error is raised.
39.6/2
For the evaluation of a call on S'Wide_Wide_Value for a real subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of one of the following:
39.7/2
numeric_literal
39.8/2
39.9/2
39.10/2
39.11/2
39.12/3
with an optional leading sign character (plus or minus), and if the corresponding numeric value belongs to the base range of the type of S, then that value is the result; otherwise, Constraint_Error is raised. The sign of a zero value is preserved (positive if none has been specified) if S'Signed_Zeros is True.
40
S'Wide_Value
S'Wide_Value denotes a function with the following specification:
41
function S'Wide_Value(Arg : Wide_String)
return S'Base
42
This function returns a value given an image of the value as a Wide_String, ignoring any leading or trailing spaces.
43/3
For the evaluation of a call on S'Wide_Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Wide_Image for a value of the type), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised. For a numeric subtype S, the evaluation of a call on S'Wide_Value with Arg of type Wide_String is equivalent to a call on S'Wide_Wide_Value for a corresponding Arg of type Wide_Wide_String.
Paragraphs 44 through 51 were moved to Wide_Wide_Value.
52
S'Value
S'Value denotes a function with the following specification:
53
function S'Value(Arg : String)
return S'Base
54
This function returns a value given an image of the value as a String, ignoring any leading or trailing spaces.
55/3
For the evaluation of a call on S'Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Image for a value of the type), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised. For a numeric subtype S, the evaluation of a call on S'Value with Arg of type String is equivalent to a call on S'Wide_Wide_Value for a corresponding Arg of type Wide_Wide_String.
#### Implementation Permissions
56/2
An implementation may extend the Wide_Wide_Value, Wide_Value, Value, Wide_Wide_Image, Wide_Image, and Image attributes of a floating point type to support special values such as infinities and NaNs.
56.1/3
An implementation may extend the Wide_Wide_Value, Wide_Value, and Value attributes of a character type to accept strings of the form “Hex_hhhhhhhh” (ignoring case) for any character (not just the ones for which Wide_Wide_Image would produce that form — see 3.5.2), as well as three-character strings of the form “'X'”, where X is any character, including nongraphic characters.
#### Static Semantics
56.2/3
For a scalar type, the following language-defined representation aspect may be specified with an aspect_specification (see 13.1.1):
56.3/3
Default_Value
This aspect shall be specified by a static expression, and that expression shall be explicit, even if the aspect has a boolean type. Default_Value shall be specified only on a full_type_declaration.
56.4/3
If a derived type with no primitive subprograms inherits a boolean Default_Value aspect, the aspect may be specified to have any value for the derived type.
#### Name Resolution Rules
56.5/3
The expected type for the expression specified for the Default_Value aspect is the type defined by the full_type_declaration on which it appears.
NOTES
57
24 The evaluation of S'First or S'Last never raises an exception. If a scalar subtype S has a nonnull range, S'First and S'Last belong to this range. These values can, for example, always be assigned to a variable of subtype S.
58
25 For a subtype of a scalar type, the result delivered by the attributes Succ, Pred, and Value might not belong to the subtype; similarly, the actual parameters of the attributes Succ, Pred, and Image need not belong to the subtype.
59
26 For any value V (including any nongraphic character) of an enumeration subtype S, S'Value(S'Image(V)) equals V, as do S'Wide_Value(S'Wide_Image(V)) and S'Wide_Wide_Value(S'Wide_Wide_Image(V)). None of these expressions ever raise Constraint_Error.
#### Examples
60
Examples of ranges:
61
-10 .. 10
X .. X + 1
0.0 .. 2.0*Pi
Red .. Green -- see 3.5.1
1 .. 0 -- a null range
Table'Range -- a range attribute reference (see 3.6)
62
Examples of range constraints:
63
range -999.0 .. +999.0
range S'First+1 .. S'Last-1 | 3,530 | 14,944 | {"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-2024-10 | latest | en | 0.874114 |
https://mafiadoc.com/cbse-class-x-real-number-solved-questions-jsunil-tutorial-cbse-_59f24cf81723dd677d01c590.html | 1,591,505,891,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00388.warc.gz | 424,770,928 | 10,668 | ## CBSE Class X Real Number Solved questions - jsunil tutorial cbse ...
10TH Chapter: Real Number http://jsuniltutorial.weebly.com/. CBSE Class X Real Number Solved questions. JSUNIL. P age. 1. Q.1. Based on Euclid's algorithm: ...
JSUNIL
CBSE Class X Real Number Solved questions
Q.1. Based on Euclid’s algorithm: a = b q + r; Using Euclid’s algorithm: Find the HCF of 825 and 175. Ans: Since 825>175, we use division lemma to 825 and 175 to get 825 = 175 x 4 + 125. Since r ≠ 0, we apply division lemma to 175 and 125 to get Again applying division lemma to 125 and 50 we get,
175 = 125 x 1 + 50
125 = 50 x 2 + 25.
Once again applying division lemma to 50 and 25 we get.
50 = 25 x 2 + 0.
Since remainder has now become 0, this implies that HCF of 825 and 125 is 25. Q.2. Based on Showing that every positive integer is either of the given forms: Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q. Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r 657, we apply the division lemma to 963 and 657 to obtain HCF. 963 = 657 x 1 + 306 ; 657 = 306 x 2 + 45
; 306 = 45 x 6 + 36 ; 45 = 36 x 1 + 9 ; 36 = 9 x 4 + 0
In this step the remainder is zero. Thus, the divisor i.e. 9 So, The H.C.F. of 657 and 963 is 9. Now, We can write 9 = 45 – 36 x 1 [from last steps of HCF];
36 = 306 – 45 x 6 ;
9 = 45 – (306 – 45 x 6) x 1 = 45 x 7 – 306 x 1 ; 9 = (657 – 306 x 2) x 7 – 306 x 1 = 657 x 7 – 306 x 15 9 = 657 x 7 – (963 – 657 x 1) x 15 = 657 x 22 – 963 x 15 ⇒ HCF of 657 and 963 = 657 x 22 – 963 x 15 ∴ HCF, 9 can be expressed as linear combination of 657 and 963 as 9 = 657x + 963y, where x and y are not unique. Hence In linear combination, x = 22 and y = –15 Q. A rectangular courtyard measures 18m 72cm long and 13m 20 cm broad. It is to be paved with square tiles of same size. Find the least possible no. of such tiles. Ans: Now, HCF of 1872 and 1320 = 24
Therefore, the side of the required square tile = 24 cm.
Thus, no. of such square tile required to pave the courtyard = [Area if courtyard/area of 1 tiles]= [1872x 1320]/[24x24] = 4290 Hence, least possible no. of such tiles = 4290 Q. show that any positive odd integer is of the form 6q+1 ,or 6q+3 ,or 6q+5 where q is some integer. Ans: Let 'a' be any positive odd integer. and b=6. Let 'q' be quotient and 'r' be remainder. a=6q+r where 0≤r r=0,1,2,3,4,5 But Now, here odd integer are 6q+1, 6q+3, and 6q+5 Hence proved that any odd integer is of the form 6q+1, 6q+3 and 6q+5 Page=14
Page
3
Q. Show that 5n can not end with the digit 2 for any natural number n. Ans: Here Given Number=5n Now Prime factors of this number=(5)n
10TH Chapter: Real Number
http://jsuniltutorial.weebly.com/
JSUNIL
CBSE Class X Real Number Solved questions Now, For the number to end with 2, it should be a factor of 2 which is not so in this case. So it cannot end with the digit as 2.
Q. Show that the cube of any positive integer is either of the form 2m OR 2m+1 for some integer Ans: Let a be any positive integer. So, a is either an even positive integer or an odd positive integer. ∴ a = 2n or a = 2n + 1 a 3 = (2n)3 = 8n 3 = 2(4n 3) = 2m, where m = 4n 3 Or a 3 = (2n + 1)3 = 8n 3 + 12n 2 + 6n + 1 = 2(4n 3 + 6n 2 + 3n) + 1 = 2m + 1, where m = 4n 3 + 6n 2 + 3n So, the cube of any positive integer is either of the form 2m or 2m+ 1, where m is integer. Q. Show that any positive odd integer is of the form 6q+1 ,or 6q+3 ,or 6q+5 where q is some integer. Ans: Let 'a' be any positive odd integer. and b=6. Let 'q' be quotient and 'r' be remainder. a=6q+r where 0≤r | 1,341 | 3,671 | {"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} | 4.75 | 5 | CC-MAIN-2020-24 | latest | en | 0.795505 |
https://www.transtutors.com/questions/1-calculate-x-for-the-shaded-region-knowing-that-68-54-mm-2-compute-y-for-the-region-1999828.htm | 1,618,836,145,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00586.warc.gz | 1,134,913,567 | 15,855 | 1. Calculate x for the shaded region, knowing that = 68.54 mm. 2. Compute y for the region shown,... 1 answer below »
1. Calculate x for the shaded region, knowing that = 68.54 mm.
2. Compute y for the region shown, given that = 25.86 mm.
Sathya
Given: th Gomm- C111 In=? y = 68.54 mm Az- 815 Iy - ? ä = 25.86 mm. 25 Az G (25.%, gomm 68.54) Centroid G = (25-86,68.54). k 80mm y Ai (1) 80 x 20 + 20% 80%...
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https://www.learn-c.com/experiment5.htm | 1,642,503,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00370.warc.gz | 945,947,243 | 7,322 | Controlling The Real World With Computers
::. Control And Embedded Systems .::
Experiment 5 - Controlling Motors
Home Order Let me know what you think
Previous: Experiment 4 - The Multiple Closure Problem And Basic Outputs With The PPI
Next: Experiment 6 More Precise Control Of Motors
More current is needed for things such as motors, relays and solenoids. Even a small motor draws more current than the PPI or a small transistor can provide. A power device is called for.
One such device is the TIP120, a darlington transistor (seeTIP120.PDF):
A darlington is actually two transistors. Notice that the emitter of the first transistor is connected to the base of the second. That causes their current gains to be multiplied, giving the TIP120 an hfe of about 1000. The base to emitter resistors are built into the package.
The TO-220 package has a metal tab with a hole in it which can be used to bolt the TIP120 to a piece of metal called a heat sink, used to pull heat away from the package. The tab is connected to the collectors of the transistors though, so if the heat sink is grounded the tab must be insulated from the heat sink but still conduct heat to it. That's accomplished with special materials made for the job. A heat sink is not needed for this experiment. Notice that the positions of the Base, Collector and Emitter are not the same as they were on the PN2222. The diode connected from collector to emitter will be discussed shortly.
Let's say we want to power an electric motor that needs about 1 Amp, such as the Radio Shack 273-223. With a current gain of 1000, we need only drive the base with 1ma (1ma * 1000 = 1A). The full 2.5ma capability of the 82C55 will be used however, in order to make certain plenty of current is delivered to the motor. The voltage between the base and emitter is about two diode drops, or about 1.4V. If we say we can provide 3V, our resistor has 3 - 1.4 = 1.6V across it. The value of the resistor would then be:
R = V/I = 1.6/2.5ma = 640
The closest standard value is 620 ohms, but 560 ohms is easier to find.
The power is: P = 1.62/560 = 4.57mw so any normal rating is OK.
Re-calculate the current:
I = V/R = 1.6/560 = 2.857ma plus a little
This circuit should deliver the full 2.5ma to the base of the the TIP120. That times 1000 is 2.5 amps. The TIP120 is being asked to provide only half of its maximum 5 amps current. A 4.7K pulldown resistor from the output of the PPI to ground is added to make certain the PPI line goes low when it's supposed to:
NOTE: Please be sure to read the Warranty And Disclaimer before working with the hardware!
A short discussion of magnetism and motors will be helpful to explain the need for the diodes in the circuit. Incidentally, diodes such as these handle high powers and are usually called rectifiers. A 1N4001 for example, can handle 1 amp at 50V or 50W (although I wouldn't advise pushing one that hard), compared to the 1N4148, which is rated at only about 500mw maximum.
Just about everybody played with magnets in grade school. From that experience it was discovered that opposite poles of magnets attract and like poles repel. We were told later that an electromagnet could be formed by running electrical current through a wire wrapped around a piece of iron. By using a direct current, such as that from a battery or the 5V power supply of a computer, the piece of iron will have fixed North and South poles just like a permanent magnet. It will be attracted to and repelled from a permanent magnet in the same manner as another permanent magnet. A motor can be constructed using that knowledge.
I will cover only the basic ideas of motor operation here. A more detailed explanation can be found in How Stuff Works. Notice what will happen when an electromagnet is placed inside a permanent magnet (wrap the fingers of your left hand around the electromagnet in the direction of the current and your thumb will point to North). The electromagnet is mounted on an imaginary shaft to allow it to rotate:
The opposite poles will attract, and the like poles will repel, so the electromagnet will rotate:
Now turn off the power long enough to let the electromagnet coast a little, then turn it back on, but with the polarity reversed. The whole process starts over:
The rotating electromagnet is called the armature. The switch used to reverse polarity is formed by brushes which connect to the coil by sliding over curved pieces of metal. Together the curved pieces of metal are called the commutator. They are bonded to an insulator on the shaft going through the armature.
In addition to producing mechanical movement, magnetism can also produce voltage. If a wire moves at right angles through a magnetic field or a magnetic field moves across a wire, a voltage will be produced in the wire. The magnitude of the voltage depends on the velocity of the wire or field, among other things. A magnet is formed when voltage is applied to the motor. A magnetic field exists around the armature just as it would around a permanent magnet. When the voltage is turned off, the magnetic field rapidly collapses. Since the field collapses in the opposite direction from which it was initiated, the voltage is reversed. It can be quite high compared to the powering voltage; enough to damage components connected to it.
We are providing one side of the motor with 5 volts. The TIP120 will take the other side near ground (actually about 1 volt above ground). Note that the 1N4001 will be reverse biased when the motor is turned on since its cathode is more positive than its anode. (Forgot about cathodes and anodes? See How To Read A Schematic.)
A relatively large voltage spike of short duration will occur when the motor is turned off that is opposite in polarity to the applied voltage. The TIP120 side will be significantly more positive than the 5V side. This will forward bias the 1N4001 which will short out the spike, protecting the TIP120 and other circuitry. I measured peaks as high as 100 volts across the Radio Shack motor. Please do not leave out spike protection rectifiers! The diode that is built into the TIP120 also helps by shorting negative spikes to ground. Another term for voltage is Electromotive Force or EMF. The large reverse voltage spike seen when power is removed from a coil is often called Back EMF.
Now add the following to the bottom of digital.c:
``` void motor(long on, long off) { int x; long y,z; printf("ON "); outp(ppi_porta, 0xff); for(y=0L; y>=6; // shift the mode value to the right 6 places command = (mode & 0x0c) << 1; // shift bits 2 and 3 into positions 4 and 5 command += (mode & 3); // add in bits 0 and 2 command |= 0x80; // OR in bit 7 for PPI set up outp(control, command); // set according to mode command } // end set_up_ppi() // end digi5a.c ```
The motor(..) function turns the motor on then off for the counts you send it. The functions portaon() and portaoff() turn Port A on and off. The set_up_ppi(..) function works the same as before except that it now accepts the new enumeration values as inputs, so it shifts the mode value right 6 places as required.
Now try the following to test it:
``` // experi5a.c #include #include #include // include header with constants #include "const5a.h" // include header with external declarations #include "extern5a.h" void main(void) { int x,y,r,c; get_port(); // get the port number and establish register locations // make A an output and B and C inputs set_up_ppi(Aout_CUin_Bin_CLin); // uses the new enumeration while(!kbhit()) // stay in loop until key is hit { motor(1000000L, 1000000L); // motor on, off times } // end while(!kbhit()) portaoff(); } // end experi5a.c ```
Notice the change in the while() loop. It works just like the previous loops that have been used, except that the test for a pressed key is now part of the while() function.
The extern5a.h file contains the external declarations:
```// external prototypes extern void set_up_ppi(int mode); extern void get_port(void); extern int is_closure(int closurenumber); extern void motor(long on, long off); extern void portaoff(void); extern void portaon(void); ```
The const5a.h file contains the new PPI setup enumeration from Experiment 4 .
Back to childhood (nice place). Did you ever turn a bicycle upside-down and spin its wheels? If not, go outside and do it; we'll wait. It's right up there with mud puddles -- well close, anyway. Most kids spin the wheel by slapping it. The more slaps per unit of time, the faster it turns. Also, the longer the hand is in contact with the tire, the faster it turns. This is how something called Pulse Width Modulation (PWM) works. Pulsing the motor with the full voltage and current produces the same magnetic force in the armature, but for a shorter time. Thus, the force provided by the motor is about the same. Hit the tire or pulse the motor only half the time however, and the average voltage is cut in half. The motor will turn at half its full-voltage speed but still supply the same force. The signal to the TIP120 from the PPI would look something like this for a 50% duty cycle, which is to say it's on half of the time:
The motor will run at half its normal speed, providing the pulses are fast enough to take advantage of the motor's moving inertia. If you try to slow the shaft down with your fingers however, you will find that the motor has about the same force (torque) as it does at full speed.
Here's another one. If the transistor is on for 30% of a cycle, the duty cycle is 30% and the motor will run at about 30% of its full speed:
Now add this at the bottom of the digital C module:
```void motor2(long on, long off) { int x; long y,z; outp(ppi_porta, 0xff); for(y=0L; y
Notice it's the same as motor(..), but without the print statements. Now put this in something called experi5b.c:
``` // experi5b.c #include #include #include // include header with constants #include "const5a.h" // include header with external declarations #include "extern5a.h" void main(void) { long on,off=5000L,r,c; get_port(); // get the port number and establish register locations // make A an output and B and C inputs set_up_ppi(Aout_CUin_Bin_CLin); // uses the new enumeration while(1) // stay in loop forever { for(on=1000L; on<=10000L; on+=100L) { // "keyboard hit" is a function that checks // to see if a key has been pressed. if(kbhit()) break;// A key was pressed -- break out of the loop printf("on=%5ld off=%5ld total = %12.6f -- on = %9.6f percent of total\n" ,on,off,(double)on+(double)off,100.0*((double)on/((double)on+(double)off))); motor(on, off); } if(on<10000L) break;// A key was pressed -- break out of the loop for(on=10000L; on>1000L; on-=100L) { // "keyboard hit" is a function that checks // to see if a key has been pressed. if(kbhit()) break;// A key was pressed -- break out of the loop printf("on=%5ld off=%5ld total = %12.6f -- on = %9.6f percent of total\n" ,on,off,(double)on+(double)off,100.0*((double)on/((double)on+(double)off))); motor(on, off); } if(on>1000L) break;// A key was pressed -- break out of the loop } // end while(1) portaoff(); } // end experi5b.c ``` | 2,758 | 11,152 | {"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-2022-05 | latest | en | 0.934581 |
https://uniapaclisbon2018.com/blog/are-rugs-measured-length-by-width/ | 1,675,887,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00028.warc.gz | 624,232,407 | 13,898 | # Are Rugs Measured Length By Width
Are Rugs Measured Length By Width. Outline area rug area with masking tape: Click to see full answer.
Carpet Seaming Diagram Floor Matttroy from floor.matttroy.net
Click to see full answer. If you are measuring for a rug for a room in your home that has furniture in it. for example your living room or. When i was shopping for a new living room area rug. i knew i wanted all of my furniture to sit on the rug so i measured the length and width of my existing furniture to get a minimum size.
Source: floor.matttroy.net
But. there is no harm even if you indicate the size in the form of width x length also. as under: In many instances. they work interchangeably and give the designer a ton of unique options.
mohawkhome.com
Use a measuring tape that can be hooked onto the edge of the rug. Round rugs are sold by the diameter.
pinterest.com
In many instances. they work interchangeably and give the designer a ton of unique options. Both terms refer to the longest side of the shape.
pinterest.com
In a standard setup. the furniture starts to separate itself from the rug. especially in 5×8. This equals 95.5cm by 190.50cm and 3 feet 2 inches by 6 feet 3 inches respectively.
pinterest.com
We usually express these dimensions by writing them out separated by a multiplication sign. Round rugs are sold by the diameter.
pinterest.com
Calculate volume capacity from length. width and height; If you want the rug centered within a single room. measure the length of the base of each wall.
#### If Youre Considering A Round Rug. Run A Piece Of Tape From The Center Of The Space You Want To Cover To The Outside Edge And Measure It.
Make a list of your measurements and multiply the length by the width of each room. A step by step guide This extra size provides a comfortable amount of space around the table for pulling out chairs.
#### Outline Area Rug Area With Masking Tape:
Two twin bed sizes can be combined to form a king size mattress. Start with the longest measurement of the rug. Start with the longest measurement of the rug.
#### T He Length (20 Cm) And The Width (10 Cm) Correspond To The Horizontal Dimension.
When you measure your rug. you always need to remember to use the exact measurements. Both terms refer to the longest side of the shape. Use a measuring tape that can be hooked onto the edge of the rug.
#### Before Going Shopping. Understand What Size You’ll Need. Length And Width.
This is just a suggestion. and you may find you like the look of a different size better. As people are downsizing. 5×8 and 6×9 rugs are becoming extremely popular. Using a measuring tape. measure the length and width of the space for which you want a rug.
#### This Equals 95.5Cm By 190.50Cm And 3 Feet 2 Inches By 6 Feet 3 Inches Respectively.
These standard rug sizes. which are measured in feet. are rectangular in shape. Heres how to pick the right rug sizes for your bedroom. living room. kitchen. and dining room. If you are measuring for a rug for a room in your home that has furniture in it. for example your living room or. | 710 | 3,102 | {"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-2023-06 | latest | en | 0.91435 |
https://stat.ethz.ch/pipermail/r-help/2010-October/257468.html | 1,725,870,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00076.warc.gz | 531,688,862 | 2,983 | # [R] question in using nlme and lme4 for unbalanced data
Chi Yuan cyuan at email.arizona.edu
Mon Oct 25 20:44:49 CEST 2010
```Hello:
I have an two factorial random block design. It's a ecology
experiment. My two factors are, guild removal and enfa removal. Both
are two levels, 0 (no removal), 1 (removal). I have 5 blocks. But
within each block, it's unbalanced at plot level because I have 5
plots instead of 4 in each block. Within each block, I have 1 plot
with only guild removal, 1 plot with only enfa removal, 1 plot for
control with no removal, 2 plots for both guild and enfa removal. I am
looking at how these treatment affect the enfa mortality rate. I
decide to use mixed model to treat block as random effect. So I try
both nlme and lme4. But I don't know whether they take the unbalanced
data properly. So my question is, does lme in nlme and lmer in lme4
take unbalanced data? How do I know it's analysis in a proper way?
Here is my code and the result for each method.
I first try nlme
library(nlme)
m=lme(enfa_mortality~guild_removal*enfa_removal,random=~1|block,data=com_summer)
It gave me the result as following
Linear mixed-effects model fit by REML
Data: com_summer
AIC BIC logLik
8.552254 14.81939 1.723873
Random effects:
Formula: ~1 | block
(Intercept) Residual
StdDev: 9.722548e-07 0.1880945
Fixed effects: enfa_mortality ~ guild_removal * enfa_removal
Value Std.Error DF t-value p-value
(Intercept) 0.450 0.0841184 17 5.349603 0.0001
guild_removal -0.100 0.1189614 17 -0.840609 0.4122
enfa_removal -0.368 0.1189614 17 -3.093441 0.0066
guild_removal:enfa_removal 0.197 0.1573711 17 1.251818 0.2276
Correlation:
(Intr) gld_rm enf_rm
guild_removal -0.707
enfa_removal -0.707 0.500
guild_removal:enfa_removal 0.535 -0.756 -0.756
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.7650706 -0.7017751 0.1594943 0.7974717 1.9139320
Number of Observations: 25
Number of Groups: 5
I kind of heard the P value does not matter that much in the mixed
model. Is there any other way I can tell whether the treatment has a
significant effect or not?
I then try lme4, it give similar result, but won't tell me the p value.
library(lme4)
m<-lmer(enfa_mortality ~ guild_removal*enfa_removal +(1|block), data=com_summer)
here is the result
Linear mixed model fit by REML
Formula: enfa_mortality ~ guild_removal * enfa_removal + (1 | block)
Data: com_summer
AIC BIC logLik deviance REMLdev
8.552 15.87 1.724 -16.95 -3.448
Random effects:
Groups Name Variance Std.Dev.
block (Intercept) 0.000000 0.00000
Residual 0.035380 0.18809
Number of obs: 25, groups: block, 5
Fixed effects:
Estimate Std. Error t value
(Intercept) 0.45000 0.08412 5.350
guild_removal -0.10000 0.11896 -0.841
enfa_removal -0.36800 0.11896 -3.093
guild_removal:enfa_removal 0.19700 0.15737 1.252
Correlation of Fixed Effects:
(Intr) gld_rm enf_rm
guild_remvl -0.707
enfa_removl -0.707 0.500
gld_rmvl:n_ 0.535 -0.756 -0.756
I really appreciate any suggestion!
Thank you!
--
Chi Yuan | 1,089 | 3,193 | {"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-2024-38 | latest | en | 0.801792 |
https://www.teachoo.com/7000/882/Exponent-Law/category/Laws-of-exponents/ | 1,726,352,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00296.warc.gz | 932,059,370 | 22,547 | Laws of exponents
Chapter 1 Class 9 Number Systems
Concept wise
Let's first see what exponents are
In 2 5 ,
2 is the base
5 is the exponent (or index)
Let's study the law of indices
• √a = a 1/2
• ∛a = a 1/3
• n √a = a 1/n
• a p. a q = a p + q
• a p / a q = a p - q
• a p. b p = (ab) p
• (a p ) q. = a pq
• a 0 . = 1
• a –n. = 1/a n
## Exponent law with examples
Click next for more questions on exponents. You can also check Algebra Formulas
### Transcript
Exponents Exponents are also called powers What you need to know 3 things Exponents means power Negative exponents mean dividing A fractional exponent means nth root 𝟐^(−𝟓 )=𝟏/𝟐^𝟓 𝒙^(𝟏/𝟐)= √𝒙 Laws of Exponents √(𝑛&𝑎)=𝑎^(1/𝑛) 𝑎^𝑝.𝑎^𝑞=𝑎^(𝑝 + 𝑞) 𝑎^𝑝/𝑎^𝑞 =𝑎^(𝑝 − 𝑞) (𝑎^𝑝 )^𝑞=𝑎^(𝑝 𝑞) 𝑎^𝑝 𝑏^𝑝=(𝑎𝑏)^𝑝 𝑎^0=1 𝑎^(−𝑛)=1/𝑎^𝑛 Laws of Exponents What about x = 0 in xn ? Positive power – Negative power - 0^𝑛 0 power -
Made by
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 505 | 1,140 | {"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} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.837331 |
https://cm2feet.com/converter/common/area/are-to-square-yard/ | 1,708,825,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00059.warc.gz | 186,166,893 | 6,151 | # =
Precision:
#### Are to Square yard Conversion Formula:
Square yard (yd²) = Are × 119.5989989085
#### How to Convert Are to Square yard (yd²)?
To get Square yard area, simply multiply Are by 119.5989989085. With the help of this area converter, we can easily convert Are to Square yard. Here you are provided with the converter, proper definitions,relations in detail along with the online tool to convert Are to Square yard (yd²).
#### How many Square yard in one Are?
1 Are is 119.59899890854 Square yard (yd²).
Are to Square yard (yd²) converter is the area converter from one unit to another. It is required to convert the unit of area from Are to Square yard, in area. This is the very basic unit conversion, which you will learn in primary classes. It is one of the most widely used operations in a variety of mathematical applications. In this article, let us discuss how to convert Are to Square yard (yd²), and the usage of a tool that will help to convert one unit from another unit, and the relation between Are and Square yard with detailed explanation.
#### Are Definition
Are, basic unit of area in the metric system, equal to 100 square metres and the equivalent of 0.0247 acre. Its multiple, the hectare (equal to 100 ares), is the principal unit of land measurement for most of the world.
#### Square yard Definition
A unit of area (abbreviation sq yd or sq. yd.) equal to the area of a square the sides of which are one yard long. ( i.e. 3 feet by 3 feet or 9 square feet).
#### Are to Square yard (yd²) Conversion table:
Common Convertersions | 374 | 1,578 | {"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-2024-10 | latest | en | 0.918879 |
https://www.physicsforums.com/threads/energy-problem-chain-on-low-friction-surface.147893/ | 1,553,513,733,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00005.warc.gz | 867,495,855 | 16,756 | # Energy problem - Chain on low friction surface.
#### sixstringartist
1. The problem statement, all variables and given/known data
Consider a chain of total mass M = 5kg that is coiled into a tight ball on a low friction floor(like an ideal ice rink). You pull on a link at one end with a constant horizontal force F=20 N, just until it reaches its full length L = 3 m. What is the final velocity of the chain? (Hint: consider the point particle system. Make reasonable assumptions, like treating the coiled ball as if it has negligible size.)
2. Relevant equations
Im pretty sure this involved the Energy Principle so delta_E = Wext + Q
the enlongated chain is 3 meters, with the center of mass having moved 1.5 m
I wish I had more to go on, but I just need a bump in the right direction.
3. The attempt at a solution
ah, well thats kindof in section 2. although its only a feable attempt to solve it, its as far as I can get right now.
#### OlderDan
Homework Helper
If the chain were stretched out, energy principles would work fine, but how are you going to find Q in this problem? (Although the equation you wrote is not quite what you would need here.) I think you will do better looking at the change in momentum of the system in some small amount of time Δt. There are two contributions to this change. One contribution is the gradual change in velocity of the mass that is already moving. The other is the abrupt change in velocity of the next little piece of the chain of mass Δm from zero to the current velocity of the moving part of the chain.
You could look at the motion of the center of mass, but although it is relatively easy to find the position of the center of mass as a function of the postion of the end of the chain, it is not quite so easy to find it as a function of time. There is the derivative chain rule that could help if you choose this approach.
#### sixstringartist
Thank you for your reply. Here we are to consider the ball of chain as a point mass.
W = F *d (here) so W = 20(1.5) = 30 J
Im guessing that the Energy when enlongated is nearly all kinetic so
30 = 1/2 m v^2 = (.5)(5)(v^2)
v^2 = 12
v ~ 3.46 m/s
Which is the correct answer. Thanks for the help!
#### OlderDan
Homework Helper
Thank you for your reply. Here we are to consider the ball of chain as a point mass.
W = F *d (here) so W = 20(1.5) = 30 J
Im guessing that the Energy when enlongated is nearly all kinetic so
30 = 1/2 m v^2 = (.5)(5)(v^2)
v^2 = 12
v ~ 3.46 m/s
Which is the correct answer. Thanks for the help!
It is not a point mass if part of the chain starts to move before all of it moves. If you were treating it as a point mass, all parts of the chain would have the same speed at all times.
How do you justify W = 20N(1.5m)? The point where you apply the force moves 3m, not 1.5m. The work you would have to do while applying this force is twice what you calculated. Where did the energy go?
#### sixstringartist
This problem was discussed in lecture and I think was a poor problem. The way I did it was how it was intended. But you are correct in the apparent loss of energy which was due to friction?!? Thats what the prof said.... then he discussed how the microscopic model of friction where "peaks" slide across each other, resulting in a shorter distance. Thats what I got from the explanation but it doesnt all make sense to me.
#### OlderDan
Homework Helper
This problem was discussed in lecture and I think was a poor problem. The way I did it was how it was intended. But you are correct in the apparent loss of energy which was due to friction?!? Thats what the prof said.... then he discussed how the microscopic model of friction where "peaks" slide across each other, resulting in a shorter distance. Thats what I got from the explanation but it doesnt all make sense to me.
There was no friction included in the problem. It was done as if the floor were perfectly frictionless. What you have in this problem can be considered to be a series of inelastic collisions between mass in motion and mass at rest. Each time a bit of mass dm starts to move, the momentum change of the system is exactly equivalent to an inelastic collision between that bit of mass (dm) initially at rest and the part of the chain already in motion while accelerataing under the action of the applied force. Energy is not conserved in this process. A lot of the work done gets converted into thermal energy of the chain.
A justification for your calculation follows:
x is the position of the end of the chain where F is applied, with x = 0 at the coiled end. When an external force is applied to a system of particles, the CM of the system accelerates obeying Newton's second law.
$$x_{CM} = \frac{1}{M}\frac{x}{2}\frac{{Mx}}{L} = \frac{{x^2 }}{{2L}}$$
but that is not explicitly needed to do the problem
$$F_{ext} = Ma_{CM} = M\frac{{dv_{CM} }}{{dt}} = M\frac{{dv_{CM} }}{{dx_{CM} }}\frac{{dx_{CM} }}{{dt}} = Mv_{CM} \frac{{dv_{CM} }}{{dx_{CM} }}$$
where the chain rule has been used so the variables can be separated as follows
$$F_{ext} dx_{CM} = Mv_{CM} dv_{CM}$$
$$F_{ext} \int_0^{L/2} {dx_{CM} } = M\int_0^V {v_{CM} dv_{CM} }$$
$$F_{ext} \frac{L}{2} = M\frac{{V^2 }}{2}$$
$$F_{ext} L = MV^2$$
The work done is twice the final kinetic energy of the chain.
Last edited:
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• Solo and co-op problem solving | 1,462 | 5,641 | {"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": 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} | 3.796875 | 4 | CC-MAIN-2019-13 | latest | en | 0.960057 |
https://wiki.alquds.edu/?query=Portal:Arithmetic | 1,660,394,300,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00191.warc.gz | 544,645,932 | 54,088 | # Portal:Arithmetic
## The Arithmetic Portal
Arithmetic tables for children, Lausanne, 1835
Arithmetic (from Ancient Greek ἀριθμός (arithmós) 'number', and τική [τέχνη] (tikḗ [tékhnē]) 'art, craft') is an elementary part of mathematics that consists of the study of the properties of the traditional operations on numbers—addition, subtraction, multiplication, division, exponentiation, and extraction of roots. In the 19th century, Italian mathematician Giuseppe Peano formalized arithmetic with his Peano axioms, which are highly important to the field of mathematical logic today. (Full article...)
The Egyptians and Babylonians used all the elementary arithmetic operations as early as 2000 BC. Later Roman numerals, descended from tally marks used for counting. The continuous development of modern arithmetic starts with ancient Greece, although it originated much later than the Babylonian and Egyptian examples. Euclid is often credited as the first mathematician to separate study of arithmetic from philosophical and mystical beliefs. Greek numerals were used by Archimedes, Diophantus and others in a positional notation not very different from ours. The ancient Chinese had advanced arithmetic studies dating from the Shang Dynasty and continuing through the Tang Dynasty, from basic numbers to advanced algebra. The ancient Chinese used a positional notation similar to that of the Greeks. The gradual development of the Hindu–Arabic numeral system independently devised the place-value concept and positional notation, which combined the simpler methods for computations with a decimal base and the use of a digit representing zero (0). This allowed the system to consistently represent both large and small integers. This approach eventually replaced all other systems. In the Middle Ages, arithmetic was one of the seven liberal arts taught in universities. The flourishing of algebra in the medieval Islamic world and in Renaissance Europe was an outgrowth of the enormous simplification of computation through decimal notation.
## General images
The following are images from various arithmetic-related articles on Wikipedia.
## Need help?
Do you have a question about Arithmetic that you can't find the answer to?
Consider asking it at the Wikipedia reference desk.
## Subcategories
Select [►] to view subcategories
## Associated Wikimedia
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Discover Wikipedia using portals | 480 | 2,486 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 19, "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} | 3.390625 | 3 | CC-MAIN-2022-33 | latest | en | 0.933435 |
http://www.traditionaloven.com/tutorials/distance/convert-china-fen-unit-to-chain-unit-of-length.html | 1,529,789,959,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00144.warc.gz | 513,407,655 | 11,720 | Convert 市分 to chain unit | Chinese fēn to chains
# length conversion
## Amount: 1 Chinese fēn (市分) of length Equals: 0.00017 chains (chain unit) in length
Converting Chinese fēn to chains value in the length units scale.
TOGGLE : from chains into Chinese fēn in the other way around.
## length from Chinese fēn to chain Conversion Results:
### Enter a New Chinese fēn Amount of length to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
Conversion calculator for webmasters.
## Length, Distance, Height & Depth units
Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
Convert length measuring units between Chinese fēn (市分) and chains (chain unit) but in the other reverse direction from chains into Chinese fēn.
conversion result for length: From Symbol Equals Result To Symbol 1 Chinese fēn 市分 = 0.00017 chains chain unit
# Converter type: length units
This online length from 市分 into chain unit converter is a handy tool not just for certified or experienced professionals.
First unit: Chinese fēn (市分) is used for measuring length.
Second: chain (chain unit) is unit of length.
## 0.00017 chain unit is converted to 1 of what?
The chains unit number 0.00017 chain unit converts to 1 市分, one Chinese fēn. It is the EQUAL length value of 1 Chinese fēn but in the chains length unit alternative.
How to convert 2 Chinese fēn (市分) into chains (chain unit)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.000165698984597 * 2 (or divide it by / 0.5)
QUESTION:
1 市分 = ? chain unit
1 市分 = 0.00017 chain unit
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Chinese fēn and chains ( 市分 vs. chain unit ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with length's values and properties.
International unit symbols for these two length measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Chinese fēn is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for chain is:
chain unit
### One Chinese fēn of length converted to chain equals to 0.00017 chain unit
How many chains of length are in 1 Chinese fēn? The answer is: The change of 1 市分 ( Chinese fēn ) unit of length measure equals = to 0.00017 chain unit ( chain ) as the equivalent measure for the same length type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 市分 - Chinese fēn for length amount, the rule is that the Chinese fēn number gets converted into chain unit - chains or any other length unit absolutely exactly.
Conversion for how many chains ( chain unit ) of length are contained in a Chinese fēn ( 1 市分 ). Or, how much in chains of length is in 1 Chinese fēn? To link to this length Chinese fēn to chains online converter simply cut and paste the following.
The link to this tool will appear as: length from Chinese fēn (市分) to chains (chain unit) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 926 | 3,852 | {"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-2018-26 | latest | en | 0.756516 |
https://fr.mathworks.com/matlabcentral/cody/problems/43028-convert-decimal-to-hex-as-shown-in-test-cases/solutions/2274625 | 1,601,334,277,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00641.warc.gz | 388,939,057 | 16,648 | Cody
# Problem 43028. Convert decimal to hex as shown in test cases
Solution 2274625
Submitted on 12 May 2020 by Karl Ezra Pilario
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1023; y_correct = '3FF'; assert(isequal(dectohex(x),y_correct))
2 Pass
x = 10; y_correct = '00A'; assert(isequal(dectohex(x),y_correct))
3 Pass
x = 5563; y_correct = '15BB'; assert(isequal(dectohex(x),y_correct))
4 Pass
x = 0; y_correct = '000'; assert(isequal(dectohex(x),y_correct)) | 191 | 599 | {"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-2020-40 | latest | en | 0.612235 |
http://mathforum.org/wagon/fall95/p796.html | 1,524,402,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945596.11/warc/CC-MAIN-20180422115536-20180422135536-00417.warc.gz | 198,744,087 | 2,546 | Hosted by The Math Forum
# Gold in ratios
Fall 95 Archive || MacPOW Home || Math Forum POWs || Search MacPOW
Can you find positive real numbers a <= b such that a/b is closer to the golden ratio than b/(a+b)?
Recall that the golden ratio phi is (Sqrt[5] - 1)/2 (approx. 0.618) and it has the property that if x/y = phi then also y/(x+y) = phi.
Source: Putz (and Mozart?), very pretty article in October issue of Mathematics Magazine.
(The title of course refers to the \$1.00 my students get for a correct solution.)
See Jeff Erickson's solution. | 150 | 553 | {"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-2018-17 | latest | en | 0.920449 |
https://physics.stackexchange.com/questions/61745/in-terms-of-the-doppler-effect-what-happens-when-the-source-is-moving-faster-th | 1,709,327,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00277.warc.gz | 454,319,471 | 41,153 | # In terms of the Doppler effect, what happens when the source is moving faster than the wave?
I'm just trying to understand this problem from a qualitative perspective. The Doppler effect is commonly explained in terms of how a siren sounds higher in pitch as it is approaching a particular observer. I understand this is because the velocity of the wave is constant and so the frequency of the waves increase as they are bunched together. What would happen if a siren was mounted on say a plane traveling at a supersonic speed? To clarify what would the observer observe/hear? Apologies if my question is not phrase very well my knowledge of physics is very rudimentary.
• Apr 21, 2013 at 1:20
• The thread linked to by Nicolau is about a moving observer. This question is about a moving source. The equation given in Eudoxos's comment on the other question is the one for the moving source case, so it would actually be more relevant here.
– user4552
Apr 21, 2013 at 2:29
The first image shows an object traveling at Mach 1 ($v=c$). The second one shows the object traveling at some supersonic velocity ($v>c$). For both the cases, the longitudinal pressure waves pile up. Say the observer is standing in the ground and the object is traveling at $c$. The observer can't hear the pitch of sound because, the waves reach him all at once and hence, he'd hear a loud "bash". The most necessary thing is that he had to wait until the source arrives. When the source is directly overhead, he hears the shock waves. | 343 | 1,515 | {"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} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.961485 |
https://cdn.codeproject.com/Forums/326859/Algorithms?df=90&mpp=25&sort=Position&view=Normal&spc=Relaxed&prof=True&select=5731916&fr=426&fid=326859 | 1,675,410,082,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00320.warc.gz | 174,498,292 | 25,714 | 15,564,684 members
Home / Discussions / Algorithms
# Algorithms
Re: Figuring out when one algorithm will be slower than another algorithm. Gerry Schmitz10-Jul-19 7:39 Gerry Schmitz 10-Jul-19 7:39
what does this algorithm do ? Member 145190962-Jul-19 14:34 Member 14519096 2-Jul-19 14:34
Re: what does this algorithm do ? Maciej Los3-Jul-19 9:33 Maciej Los 3-Jul-19 9:33
Re: what does this algorithm do ? Patrice T6-Jul-19 13:33 Patrice T 6-Jul-19 13:33
Better operating system than 64 bit? Member 145095551-Jul-19 20:39 Member 14509555 1-Jul-19 20:39
Re: Better operating system than 64 bit? OriginalGriff1-Jul-19 20:51 OriginalGriff 1-Jul-19 20:51
Re: Better operating system than 64 bit? Richard Deeming2-Jul-19 2:28 Richard Deeming 2-Jul-19 2:28
Puchi and Luggage. Wrong Answer for some test cases. Member 144776851-Jun-19 10:24 Member 14477685 1-Jun-19 10:24
Re: Puchi and Luggage. Wrong Answer for some test cases. Richard MacCutchan1-Jun-19 22:32 Richard MacCutchan 1-Jun-19 22:32
What kind of structure can it be? Member 1447661631-May-19 1:24 Member 14476616 31-May-19 1:24
Re: What kind of structure can it be? Richard MacCutchan31-May-19 2:09 Richard MacCutchan 31-May-19 2:09
Algorithm to find maximum sum in an array given that the elements have unique digits akshit bhatia28-May-19 10:02 akshit bhatia 28-May-19 10:02
Re: Algorithm to find maximum sum in an array given that the elements have unique digits Patrice T1-Jun-19 11:12 Patrice T 1-Jun-19 11:12
Xamarin Platform Member 1447108926-May-19 6:50 Member 14471089 26-May-19 6:50
Re: Xamarin Platform OriginalGriff26-May-19 6:51 OriginalGriff 26-May-19 6:51
[solved] Find the sublist of objects with the highest value without exceeding weight limit Member 1433642827-Apr-19 3:23 Member 14336428 27-Apr-19 3:23
Re: Find the sublist of objects with the highest value without exceeding weight limit OriginalGriff27-Apr-19 3:25 OriginalGriff 27-Apr-19 3:25
Re: Find the sublist of objects with the highest value without exceeding weight limit Member 1433642827-Apr-19 6:29 Member 14336428 27-Apr-19 6:29
Re: Find the sublist of objects with the highest value without exceeding weight limit ChrisFromWales1-May-19 1:09 ChrisFromWales 1-May-19 1:09
Re: Find the sublist of objects with the highest value without exceeding weight limit Member 143364281-May-19 7:49 Member 14336428 1-May-19 7:49
Re: Find the sublist of objects with the highest value without exceeding weight limit Richard Deeming1-May-19 9:09 Richard Deeming 1-May-19 9:09
Re: Find the sublist of objects with the highest value without exceeding weight limit Member 143364282-May-19 5:19 Member 14336428 2-May-19 5:19
Re: Find the sublist of objects with the highest value without exceeding weight limit Member 143364281-May-19 12:09 Member 14336428 1-May-19 12:09
Algorithm to compress Image to RGB565 for embedded screen XxKeldecknightxX24-Apr-19 9:10 XxKeldecknightxX 24-Apr-19 9:10
Re: Algorithm to compress Image to RGB565 for embedded screen Gerry Schmitz24-Apr-19 9:56 Gerry Schmitz 24-Apr-19 9:56
Last Visit: 31-Dec-99 19:00 Last Update: 2-Feb-23 21:41 Refresh ᐊ Prev1...14151617181920212223 Next ᐅ | 1,009 | 3,165 | {"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-2023-06 | longest | en | 0.69596 |
http://www.coursehero.com/file/6756338/Project-2-Report-Results/ | 1,368,962,844,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697420704/warc/CC-MAIN-20130516094340-00058-ip-10-60-113-184.ec2.internal.warc.gz | 399,209,051 | 18,505 | # Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.
12 Pages
### Project 2 Report - Results
Course: ME 352, Fall 2011
School: Purdue
Rating:
Word Count: 1941
#### Document Preview
and Results Discussion This section presents the results obtained through the analysis described above for the geared five-bar linkage. Three different designs of the geared five-bar linkage have been considered. For case one, the radius of gear 2 is 2 inches and the radius of gear 3 is 3 inches. For case two, the radius of gear 2 is 3 inches and the radius of gear 3 is 3 inches. For case three, the radius of gear...
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Contemporary Moral Issues Fall 2010 Final Exam Section One: Theory 1. What should be the proper relationship between morality and law? Should we defer to morality to resolve questions of law? Should we defer to law to resolve questions of morality? Commen | 4,473 | 18,936 | {"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-2013-20 | latest | en | 0.865824 |
http://www.thefreedictionary.com/Roman+numbers | 1,548,331,887,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584520525.90/warc/CC-MAIN-20190124100934-20190124122934-00427.warc.gz | 397,670,690 | 17,294 | # Roman numerals
(redirected from Roman numbers)
Also found in: Encyclopedia.
## Roman numerals
pl n
(Mathematics) the letters used by the Romans for the representation of cardinal numbers, still used occasionally today. The integers are represented by the following letters: I (= 1), V (= 5), X (= 10), L (= 50), C (= 100), D (= 500), and M (= 1000). If a numeral is followed by another numeral of lower denomination, the two are added together; if it is preceded by one of lower denomination, the smaller numeral is subtracted from the greater. Thus VI = 6 (V + I), but IV = 4 (V – I). Other examples are XC (= 90), CL (= 150), XXV (= 25), XLIV (= 44). Multiples of a thousand are indicated by a superior bar: thus, V̅ = 5000, X̅ = 10 000, X̅D̅ = 490 000, etc
## Ro′man nu′merals
n.pl.
the numerals in the ancient Roman system of notation, still used occasionally, as in pagination and dates on buildings. The basic symbols are I(=1), V(=5), X(=10), L(=50), C(=100), D(=500), and M(=1000). If a letter is immediately followed by one of equal or lesser value, the two values are added; if followed by one of greater value, the first is subtracted from the second; thus, XX equals 20 and IV equals 4. The year 1914 would appear as MCMXIV.
Translations
الأرْقام الرّومانيَّه
římské číslovky
romertal
római számok
rómverskar tölur
rímske číslice
Romen rakamları
## Roman numerals
nnumeri mpl romani
## Roman
1. connected with Rome, especially ancient Rome. Roman coins.
2. (no capital) (of printing) in ordinary upright letters like these.
noun
a person belonging to Rome, especially to ancient Rome.
Roman alphabet
the alphabet in which Western European languages such as English are written.
Roman Catholic (also Catholic)
(a member) of the Christian church which recognizes the Pope as its head.
Roman Catholicism (also Catholicism)
the beliefs, government etc of the Roman Catholic Church.
Roman numerals
I,II,III etc, as opposed to the Arabic numerals 1,2,3 etc.
References in periodicals archive ?
The Roman numbers stand for the date it was made, 1964.
where we also find Annuit Coeptis, Novus Ordo Seclorum, and the date MDCCLXXVI (1776) in Roman numbers.
The Guide for overcoming obsessions and compulsions using mindfulness and cognitive behavioral therapy, by Jon Hershfield and Tom Corboy, ISBN 978-1608828784, numbering 220 pages, with eight pages numbered with Roman numbers and two nienumerowanymi parties at the beginning of the book, and a translation of the text on cover of the book, along with the transfer of the whole of property copyrights to the Employer, commissioned by the Jagiellonian University.
Rola's wrist tattoo consisted of Roman numbers, which according to Al Jaras Magazine represent the date of birth of her late brother Georges, whom she lost over 18 years ago.
Now it supports Roman numbers style for heading in PDF documents and allows adding document level JavaScript.
22 [tweeentwintig] the number-name is the sum of its Roman numbers (setting W = VV)
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Open / Close | 787 | 3,035 | {"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-2019-04 | latest | en | 0.831933 |
http://www.evi.com/q/1_foot_in_cm | 1,419,116,497,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770415.113/warc/CC-MAIN-20141217075250-00011-ip-10-231-17-201.ec2.internal.warc.gz | 514,671,992 | 8,378 | # 1 foot in cm
• 1 foot is 30.48 centimeters.
• tk10npubl tk10ncanl
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• 1 feet to centimeter | 4,836 | 14,449 | {"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} | 2.875 | 3 | CC-MAIN-2014-52 | latest | en | 0.918797 |
https://www.indiabix.com/aptitude/clock/discussion-645-1 | 1,716,569,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00193.warc.gz | 710,004,936 | 8,937 | Aptitude - Clock - Discussion
Discussion Forum : Clock - General Questions (Q.No. 9)
9.
At what angle the hands of a clock are inclined at 15 minutes past 5?
58 1 ° 2
64°
67 1 ° 2
72 1 ° 2
Answer: Option
Explanation:
Angle traced by hour hand in 21 hrs = 360 x 21 ° = 157 1 ° 4 12 4 2
Angle traced by min. hand in 15 min. = 360 x 15 ° = 90°. 60
Required angle = 157 1 ° - 90° = 67 1 ° 2 2
Discussion:
41 comments Page 1 of 5.
Srinitha said: 1 month ago
Use formula( d × 30 + min/2).
D = distance b/w 5 and 15 minutes = 2,
= 2 × 30 + 15/2.
= 60+7 1/2.
= 67 1/2.
Udai said: 2 months ago
It is 5:15.
we can assume 15 as "3".
So 5-3 = 2*30+15/2= 60+7.5.
Answer= 67.5.
(1)
Srikanth said: 1 year ago
11/2 *m -30*h.
= 11/2*15 -30 * 5.
= 165/2 -150,
= 82.5-150,
= 67.5.
(2)
Shraddha said: 2 years ago
Use the formula: 30*(H)-11/2*(M).
(2)
Laxmi Hosamani said: 3 years ago
30 * 5 - 11/2 = 67.5.
(2)
Alekhya said: 3 years ago
Just put the formula :
θ = abs | 30*(H) - 11/2* (M)|.
= 30*5 - 5.5 * 15,
= 67.5°.
(2)
Deki Zangmo said: 4 years ago
We can use the below formula to find angle when the hour and minute values are given:
A=30H-5.5M.
H=5 M=15 as per the above questions;
So now we just have to substitute values in the formula
A=(30*5) -5.5*15.
150 - 82.5.
=67.5°.
(1)
Ibrahim Payak said: 4 years ago
Simply put this eq:
∠ = |60H-11M/2|.
= |60*5-11*15/2|,
= |300-165/2|,
= |135/2|,
= 67.5.
Deepak menaria said: 5 years ago
5:15 time.
5 hr angle = 5x30 = 150.
15 min in hour =(15/60)x30 =7.5.
150+7.5=157.5.
In minute hand for 15 min.
Minute hand angle is 6 degree so,
For 15 min = 15x6 =90.
So,
157.5 - 90 = 67.5 degree, Answer.
(5)
Richin said: 5 years ago
It's a simple logic.
Angle between two adjasent numbers is 30.
So for 5: 15, the hands of the clock will be in 3 and 5 but the hour hand will be moved a little from 5.
So we have to find that only. ie, angle b/w 3&5 is 60 (30+30).
60+ half of the given minute will be the answer.
ie, 60+ 15/2= 67.5.
1 minute is 1/2°
(3)
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• ## Help to check if element in [array A] is made up by sum of some elements in [array B]
Thanks Mr/Ms Carim
Because my English is not well, so I don't know how to choose right key words to search
Thank you so much many many times
• ## Help to check if element in [array A] is made up by sum of some elements in [array B]
Good evening! Members in forum
Can you help me to solve this problem by VBA
I have two arrays like picture below (example), I want to check if element in array A is made up of sum of two or more elements in array B. Sum of elements in array A equal to elements in array B.
This example can be described:
Option Base 1 [The first ordinary element in array is 1]
Loop through elements in array A, we find that
A[1] = B[1] + B[3] + B[4], ==> A[1] & B[1]; B[3]; B[4] will be eliminated.
A[2] = 30 can't not be result of sum of two or more remains. [5,10,10,24,13] ==> They are ignored
A[3] = B[2] + B[5] ==> A[3] & B[2]; B[5] will be eliminated.
A[4] = 17 can't not be result of sum of two or more remains. [10,24,13] ==> They are ignored
So we have A[1] = 0, A[2] = 30 , A[3] = 0, A[4] = 17
B[1] = B[2] = B[3] = B[4] = B[5] = 0, B[6] = 10, B[7] = 24, B[8] = 13
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M0IITU19 - Probability qns
# M0IITU19 - Probability qns - Probability 1 6 coins are...
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1 QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439 1. 6 coins are tossed simultaneously . The propability of getting atleast 6 heads is : (A) 57 64 (B) 229 256 (C) 7 64 (D) 37 256 2. The probabilities of three mutually exclusive events are 2 3 , 1 4 & 1 6 . The statement is : True Wrong Could be either Do not know 3. A & B toss a coin alternatively, the first to show a head being the winner. If A starts the game, the chance of his winning is : 5 8 1 2 1 3 2 3 4. A & B are two events such that and P (A) = 0.4, P (A + B) = 0.7 and P (AB) = 0.2, then P (B) = 0.1 0.3 0.5 None of these 5. Suppose that A, B, C are events such that P (A) = P (B) = P (C) = 1 4 , P (AB) = P (CB) = 0, P (AC) = 1 8 , then P (A + B) = 0.125 0.25 0.375 6. A single letter is selected at random from the word “PROBABILITY” . The probability that the selected letter is a vowel is : 2 11 3 11 4 11 0 7. If A & B are two events such that, P (A B) = P (A B), then the true relation is : P (A) = P (B) = 0 P (A) + P (B) = P (A) P B A P (A) + P (B) = 2 P (A) P B A 8. A coin is tossed and a dice is rolled . The probability that the coin shows the head and the dice shows 6 is : 1 8 1 12 1 2 1 9. A card is drawn at random from a pack of cards . The probability of this card being a red or a queen is : 1 13 1 26 1 2 7 13 10. The probability of happening an even A is 0.5 and that of B is 0.3 . If A & B are mutually exclusive events, then Probability
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2 QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439 the probability of happening neither A nor B is : (A) 0.6 (B) 0.2 (C) 0.21 (D) None of these 11. If P (A) = 0.4, P (B) = x, P (A B) = 0.7 and the events A & B are independent, then x = 1 3 1 2 2 3 12. A box contains 6 nails and 10 nuts . Half of the nails and half of the nuts are rusted . If one item is chosen at random, what is the probability that it is rusted or is a nail . 3 16 5 16 11 16 14 16 13. A man draws a card from a pack of 52 playing cards, replaces it and shuffles the pack . He continues this process until he gets a card of spade . The probability that he will fail the first two times is : 9 16 1 16 9 64 14. In a box of 10 electric bulbs, two are defective . Two bulbs are selected at random one after the other from the box . The first bulb after selection being put back in the box before making the second selection . The probability that both the bulbs are without defect is : 9 25 16 25 4 5 8 25 15. If A & B are any two events, then the true relation is : P (A B) > P (A) + P (B) - 1 B) < P (A) + P (B) (C) P (A B) = P (A) + P (B) - P (A B) 16. A box contains 15 tickets numbered 1, 2, . .... , 15 . Seven tickets are drawn at random one after the other with replacement . The probability that the greatest number on a drawn ticket is 9, is : 9 10 6 8 15 7 3 5 7 17. A purse contains 4 copper coins & 3 silver coins, the second purse contains 6 copper coins & 2 silver coins . If a coin is drawn out of any purse, then the probability that it is a copper coin, is : 4 7 3 4 37 56
3 QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439 18. If A & B are any two events, then the probability that exactly one of them occur is : (A) P (A) + P (B) - P (A B) (B) P (A) + P (B) - 2 P (A B) (C) (D)
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Ask a homework question - tutors are online | 1,365 | 4,021 | {"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} | 4.25 | 4 | CC-MAIN-2018-09 | latest | en | 0.863799 |
https://numbermatics.com/n/953247/ | 1,591,051,350,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00126.warc.gz | 481,401,541 | 5,998 | # 953247
## 953,247 is an odd composite number composed of three prime numbers multiplied together.
What does the number 953247 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.
953247 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.
## Prime factorization of 953247:
### 3 × 61 × 5209
See below for interesting mathematical facts about the number 953247 from the Numbermatics database.
### Names of 953247
• Cardinal: 953247 can be written as Nine hundred fifty-three thousand, two hundred forty-seven.
### Scientific notation
• Scientific notation: 9.53247 × 105
### Factors of 953247
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 5273
### Divisors of 953247
• Number of divisors d(n): 8
• Complete list of divisors:
• Sum of all divisors σ(n): 1292080
• Sum of proper divisors (its aliquot sum) s(n): 338833
• 953247 is a deficient number, because the sum of its proper divisors (338833) is less than itself. Its deficiency is 614414
### Bases of 953247
• Binary: 111010001011100111112
• Base-36: KFJ3
### Squares and roots of 953247
• 953247 squared (9532472) is 908679843009
• 953247 cubed (9532473) is 866196334308800223
• The square root of 953247 is 976.3436894863
• The cube root of 953247 is 98.4166283287
### Scales and comparisons
How big is 953247?
• 953,247 seconds is equal to 1 week, 4 days, 47 minutes, 27 seconds.
• To count from 1 to 953,247 would take you about four days!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 953247 cubic inches would be around 8.2 feet tall.
### Recreational maths with 953247
• 953247 backwards is 742359
• The number of decimal digits it has is: 6
• The sum of 953247's digits is 30
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https://biz.libretexts.org/Under_Construction/Book%3A_Introduction_to_Financial_Accounting_(Dauderis_and_Annand)/05%3A_Accounting_for_the_Sale_of_Goods/5.13%3A_Problems | 1,680,162,916,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00768.warc.gz | 159,630,450 | 32,092 | # 5.13: Problems
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## PROBLEM 5–1 (LO1,2,3,4)
Salem Corp. was incorporated on July 2, 2015 to operate a merchandising business. It uses the perpetual inventory system. All its sales are on account with terms: 2/10, n30. Its transactions during July 2015 are as follows:
July 2 Issued share capital for $5,000 cash. 2 Purchased$3,500 merchandise on account from Blic Pens Ltd. for terms 2/10, n30. 2 Sold $2,000 of merchandise on account to Spellman Chair Rentals Inc. (Cost to Salem:$1,200). 3 Paid Sayer Holdings Corp. $500 for July rent. 5 Paid Easton Furniture Ltd.$1,000 for equipment. 8 Collected $200 for a cash sale made today to Ethan Matthews Furniture Ltd. (Cost:$120). 8 Purchased $2,000 merchandise on account from Shaw Distributors Inc. for terms 2/15, n30. 9 Received the amount due from Spellman Chair Rentals Inc. for the July 2 sale. 10 Paid Blic Pens Ltd. for the July 2 purchase. 10 Purchased$200 of merchandise on account from Peel Products Inc. for terms n30. 15 Sold $2,000 of merchandise on account to Eagle Products Corp. (Cost:$1,300). 15 Purchased $1,500 of merchandise on account from Bevan Door Inc. for terms 2/10, n30. 15 Received a memo from Shaw Distributors Inc. to reduce accounts payable by$100 for defective merchandise included in the July 8 purchase. 16 Eagle Products Corp. returned $200 of defective merchandise which was scrapped (Cost to Salem:$150). 20 Sold $3,500 of merchandise on account to Aspen Promotions Ltd. (Cost:$2,700). 20 Paid Shaw Distributors Inc. for half the purchase made July 8. 24 Received half the amount due from Eagle Products Corp. in partial payment for the July 15 sale. 24 Paid Bevan Doors Ltd. for the purchase made July 15. 26 Sold $600 merchandise on account to Longbeach Sales Ltd. (Cost:$400). 26 Purchased $800 of merchandise on account from Silverman Co. for terms 2/10, n30. 31 Paid Speedy Transport Co.$350 for transportation to Salem's warehouse during the month (all purchases are fob shipping point).
Required:
1. Prepare journal entries to record the July transactions. Include general ledger account numbers and a brief description.
2. Calculate the unadjusted ending balance in merchandise inventory.
3. Assume the merchandise inventory is counted at July 31 and assigned a total cost of $2,400. Prepare the July 31 adjusting entry. ## PROBLEM 5–2 (LO1,5,6) The following closing entries were prepared for Whirlybird Products Inc. at December 31, 2015, the end of its fiscal year. General Journal Date Account/Explanation F Debit Credit Dec. 31 Sales 37,800 Income Summary 37,800 31 Income Summary 32,800 Cost of Goods Sold 26,800 Sales Returns and Allowances 690 Sales Discounts 310 Salaries Expenses 5,000 31 Income Summary 5,000 Retained Earnings 5,000 Required: Calculate gross profit. ## PROBLEM 5–3 (LO1,5,6) The following alphabetized adjusted trial balance has been extracted from the records of Acme Automotive Inc. at December 31, 2015, its third fiscal year-end. All accounts have a normal balance. Accounts Payable 9,000 Accounts Receivable 15,000 Accumulated Depreciation – Equipment 36,000 Advertising Expense 14,000 Bank Loan 14,000 Cash 2,000 Commissions Expense 29,000 Cost of Goods Sold 126,000 Delivery Expense 14,800 Depreciation Expense 12,000 Dividends 11,000 Equipment 120,000 Income Taxes Expense 4,200 Income Taxes Payable 4,200 Insurance Expense 10,400 Interest Expense 840 Merchandise Inventory 26,000 Office Supplies Expense 3,100 Rent Expense 32,400 Rent Revenue 19,200 Retained Earnings 12,440 Sales 310,000 Sales Discounts 1,300 Sales Returns and Allowances 2,900 Sales Salaries Expense 26,400 Share Capital 70,000 Supplies 3,200 Telephone Expense 1,800 Utilities Expense 4,200 Wages Expense – Office 14,300 Required: 1. Prepare a classified multi-step income statement and statement of changes in equity for the year ended December 31, 2015. Assume 40% of the Rent Expense is allocated to general and administrative expenses with the remainder allocated to selling expenses. Additionally, assume that$20,000 of shares were issued during the year ended December 31, 2015.
2. Prepare closing entries.
## PROBLEM 5–4 (LO1,2,3,4) Challenge Question – Pulling It All Together
Calculating Purchases, Inventory Shrinkage, Net Sales, Cost Goods Sold, Gross Profit, and Net Income/(Loss)
The information below is a summary of the merchandise inventory and sales transactions for 2016.
Total cost of purchases $250,000 Total sales 580,000 Purchases shipping costs 500 Merchandise inventory, opening balance 55,000 Purchase discounts 3,500 Sales discounts 200 Total sales returns to inventory 100 Merchandise inventory, closing GL balance 90,000 Merchandise inventory, physical inventory count 88,500 Sales allowances 600 Operating expenses 250,000 Sales returns 200 Purchase returns and allowances 200 Net purchases ? Inventory shrinkage adjustment amount ? Cost of goods sold ? Net sales ? Gross profit ? Net income/(loss) ? Gross profit ratio ? Required: Calculate and fill in the blanks. (Hint: Refer to the merchandising company illustration in Section 5.3 and the T-account summary illustrations for inventory and cost of goods sold at the end of Section 5.6.) ## PROBLEM 5–5 (LO1,2,3,5,6) Preparing a Classified Multiple-step Income Statement and Closing Entries Below is the adjusted trial balance presented in alphabetical order for Turret Retail Ltd., for 2016. Their year-end is December 31. Turret Retail Ltd. Trial Balance At December 31, 2016 Accounts payable$31,250 Accounts receivable $140,000 Accrued salaries and benefits payable 12,000 Accumulated depreciation, furniture 4,300 Cash 21,000 Cash dividends 10,000 Cost of goods sold 240,000 Bank loan payable (long-term) 40,320 Depreciation expense 3,200 Copyright 20,000 Furniture 20,000 Income tax expense 2,028 Income taxes payable 8,000 Insurance expense 5,000 Interest expense 200 Interest payable 550 Land 140,000 Merchandise inventory 120,000 Prepaid insurance expense 6,000 Rent expense 30,240 Rental income 6,000 Retained earnings 307,748 Salaries expense 57,000 Sales 360,000 Sales discounts 3,600 Sales returns and allowances 9,600 Share capital 20,000 Shop supplies expense 2,400 Shop supplies expense 1,000 Travel expense 2,100 Unearned revenue 50,500 Utilities expense 7,300$840,668 $840,668 Required: 1. Prepare a classified multiple-step income statement in good form, reporting operating expenses by nature, for the year ended December 31, 2016. 2. Prepare the closing entries for the year-ended December 31, 2016. 3. Calculate the gross profit ratio to two decimal places and comment on what this ratio means. ## PROBLEM 5–6 (LO1,2,3,4,5) Challenge Question – Preparing Adjusting Entries and a Classified Multiple-step Income Statement Below are the unadjusted accounts balances for Yuba Yabi Enterprises Ltd., for the year ended March 31, 2017. All account balances are normal. Yuba Yabi's business involves selling frozen food to restaurants as well as providing consulting services to assist restaurant businesses with their daily operations. Yuba Yabi Enterprises Ltd. Unadjusted Trial Balance March 31, 2017 Accounts payable 68,750 Accounts receivable 308,000 Accrued salaries and benefits payable 26,400 Accumulated depreciation, furniture 9,460 Cash 46,200 Cash dividends 22,000 Cost of goods sold 528,000 Advertising expense 9,900 Bank loan payable (long-term) 88,704 Depreciation expense 7,040 Copyright 44,000 Franchise 66,000 Furniture 44,000 Income tax expense - Income taxes payable 17,600 Insurance expense 11,000 Interest expense 440 Interest payable 1,210 Land 308,000 Merchandise inventory 264,000 Prepaid insurance expense 13,200 Prepaid advertising expense 8,800 Rent expense 66,528 Rental income 13,200 Retained earnings 265,364 Salaries expense 125,400 Sales 792,000 Sales discounts 7,920 Sales returns and allowances 21,120 Service revenue 495,000 Share capital 44,000 Shop supplies 8,360 Shop supplies expense 2,200 Travel expense 4,620 Unearned service revenue 111,100 Utilities expense 16,060 Additional information: The following are adjusting entries that have not yet been recorded: Accrued salaries$12,000 Accrued interest on the bank loan 5,600 Inventory shrinkage 7,800 Prepaid insurance expense 5,000 has expired Prepaid advertising expense no change Unearned revenue 30,000 has been earned Income tax rate 30%
Required:
1. Update the affected accounts by the adjustments, if any. Round all adjustments to the nearest whole dollar.
2. Prepare a classified multiple-step income statement in good form for the year ended March 31, 2017.
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posted by .
An object is thrown upward at a certain angle above the ground, eventually returning to erth.1.Is there any place along the trajectory where the velocity and acceleration are perpendicular? If so, where? 2. Is there any place where the velocity and acceleration are parallel? if so, where? in each case, explain.
• physical science -
See my other response to the same question
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Sorry if it is alot of questions 1. A crate falls from an airplane flying horizontally at an altitude of 2000 m. Neglecting air resistance, about how long will the crate take to strike the ground?
More Similar Questions | 596 | 2,631 | {"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} | 2.78125 | 3 | CC-MAIN-2018-05 | latest | en | 0.909596 |
https://www.quizzes.cc/calculator/time/days/400/years | 1,597,148,760,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00269.warc.gz | 822,237,361 | 4,588 | ### How long is 400 days in years?
Convert 400 days to years. How many years are there in 400 days? How long is 400 days in years? To calculate, enter the information into the calculator. Then, press Calculate. Finally, change the inputted data to calculate a different amount. Convert between days and years.
### Summary
Convert 400 days to years. Browse other values below.
#### More Values
400.01 days equals 1.095190182 years 400.02 days equals 1.095217561 years 400.03 days equals 1.09524494 years 400.04 days equals 1.095272319 years 400.05 days equals 1.095299698 years 400.06 days equals 1.095327077 years 400.07 days equals 1.095354456 years 400.08 days equals 1.095381835 years 400.09 days equals 1.095409214 years 400.1 days equals 1.095436594 years 400.11 days equals 1.095463973 years 400.12 days equals 1.095491352 years 400.13 days equals 1.095518731 years 400.14 days equals 1.09554611 years 400.15 days equals 1.095573489 years 400.16 days equals 1.095600868 years 400.17 days equals 1.095628247 years 400.18 days equals 1.095655626 years 400.19 days equals 1.095683005 years 400.2 days equals 1.095710384 years 400.21 days equals 1.095737763 years 400.22 days equals 1.095765142 years 400.23 days equals 1.095792521 years 400.24 days equals 1.095819901 years 400.25 days equals 1.09584728 years 400.26 days equals 1.095874659 years 400.27 days equals 1.095902038 years 400.28 days equals 1.095929417 years 400.29 days equals 1.095956796 years 400.3 days equals 1.095984175 years 400.31 days equals 1.096011554 years 400.32 days equals 1.096038933 years 400.33 days equals 1.096066312 years 400.34 days equals 1.096093691 years 400.35 days equals 1.09612107 years 400.36 days equals 1.096148449 years 400.37 days equals 1.096175828 years 400.38 days equals 1.096203208 years 400.39 days equals 1.096230587 years 400.4 days equals 1.096257966 years 400.41 days equals 1.096285345 years 400.42 days equals 1.096312724 years 400.43 days equals 1.096340103 years 400.44 days equals 1.096367482 years 400.45 days equals 1.096394861 years 400.46 days equals 1.09642224 years 400.47 days equals 1.096449619 years 400.48 days equals 1.096476998 years 400.49 days equals 1.096504377 years 400.5 days equals 1.096531756 years 400.51 days equals 1.096559135 years 400.52 days equals 1.096586514 years 400.53 days equals 1.096613894 years 400.54 days equals 1.096641273 years 400.55 days equals 1.096668652 years 400.56 days equals 1.096696031 years 400.57 days equals 1.09672341 years 400.58 days equals 1.096750789 years 400.59 days equals 1.096778168 years 400.6 days equals 1.096805547 years 400.61 days equals 1.096832926 years 400.62 days equals 1.096860305 years 400.63 days equals 1.096887684 years 400.64 days equals 1.096915063 years 400.65 days equals 1.096942442 years 400.66 days equals 1.096969821 years 400.67 days equals 1.096997201 years 400.68 days equals 1.09702458 years 400.69 days equals 1.097051959 years 400.7 days equals 1.097079338 years 400.71 days equals 1.097106717 years 400.72 days equals 1.097134096 years 400.73 days equals 1.097161475 years 400.74 days equals 1.097188854 years 400.75 days equals 1.097216233 years 400.76 days equals 1.097243612 years 400.77 days equals 1.097270991 years 400.78 days equals 1.09729837 years 400.79 days equals 1.097325749 years 400.8 days equals 1.097353128 years 400.81 days equals 1.097380508 years 400.82 days equals 1.097407887 years 400.83 days equals 1.097435266 years 400.84 days equals 1.097462645 years 400.85 days equals 1.097490024 years 400.86 days equals 1.097517403 years 400.87 days equals 1.097544782 years 400.88 days equals 1.097572161 years 400.89 days equals 1.09759954 years 400.9 days equals 1.097626919 years 400.91 days equals 1.097654298 years 400.92 days equals 1.097681677 years 400.93 days equals 1.097709056 years 400.94 days equals 1.097736435 years 400.95 days equals 1.097763815 years 400.96 days equals 1.097791194 years 400.97 days equals 1.097818573 years 400.98 days equals 1.097845952 years 400.99 days equals 1.097873331 years
401 days equals 1.09790071 years 402 days equals 1.100638617 years 403 days equals 1.103376524 years 404 days equals 1.106114431 years 405 days equals 1.108852338 years 406 days equals 1.111590245 years 407 days equals 1.114328152 years 408 days equals 1.117066059 years 409 days equals 1.119803966 years 410 days equals 1.122541873 years 411 days equals 1.12527978 years 412 days equals 1.128017687 years 413 days equals 1.130755594 years 414 days equals 1.133493501 years 415 days equals 1.136231408 years 416 days equals 1.138969315 years 417 days equals 1.141707222 years 418 days equals 1.144445129 years 419 days equals 1.147183036 years 420 days equals 1.149920943 years 421 days equals 1.15265885 years 422 days equals 1.155396757 years 423 days equals 1.158134664 years 424 days equals 1.160872571 years 425 days equals 1.163610478 years 426 days equals 1.166348385 years 427 days equals 1.169086292 years 428 days equals 1.171824199 years 429 days equals 1.174562106 years 430 days equals 1.177300013 years 431 days equals 1.18003792 years 432 days equals 1.182775827 years 433 days equals 1.185513734 years 434 days equals 1.188251641 years 435 days equals 1.190989548 years 436 days equals 1.193727455 years 437 days equals 1.196465362 years 438 days equals 1.199203269 years 439 days equals 1.201941176 years 440 days equals 1.204679083 years 441 days equals 1.20741699 years 442 days equals 1.210154897 years 443 days equals 1.212892804 years 444 days equals 1.215630711 years 445 days equals 1.218368618 years 446 days equals 1.221106525 years 447 days equals 1.223844432 years 448 days equals 1.226582339 years 449 days equals 1.229320246 years 450 days equals 1.232058153 years 451 days equals 1.23479606 years 452 days equals 1.237533967 years 453 days equals 1.240271874 years 454 days equals 1.243009781 years 455 days equals 1.245747688 years 456 days equals 1.248485595 years 457 days equals 1.251223502 years 458 days equals 1.253961409 years 459 days equals 1.256699316 years 460 days equals 1.259437223 years 461 days equals 1.26217513 years 462 days equals 1.264913037 years 463 days equals 1.267650944 years 464 days equals 1.270388851 years 465 days equals 1.273126758 years 466 days equals 1.275864665 years 467 days equals 1.278602572 years 468 days equals 1.281340479 years 469 days equals 1.284078386 years 470 days equals 1.286816293 years 471 days equals 1.2895542 years 472 days equals 1.292292107 years 473 days equals 1.295030014 years 474 days equals 1.297767921 years 475 days equals 1.300505828 years 476 days equals 1.303243735 years 477 days equals 1.305981642 years 478 days equals 1.308719549 years 479 days equals 1.311457456 years 480 days equals 1.314195363 years 481 days equals 1.31693327 years 482 days equals 1.319671177 years 483 days equals 1.322409084 years 484 days equals 1.325146991 years 485 days equals 1.327884898 years 486 days equals 1.330622805 years 487 days equals 1.333360712 years 488 days equals 1.336098619 years 489 days equals 1.338836526 years 490 days equals 1.341574433 years 491 days equals 1.34431234 years 492 days equals 1.347050247 years 493 days equals 1.349788155 years 494 days equals 1.352526062 years 495 days equals 1.355263969 years 496 days equals 1.358001876 years 497 days equals 1.360739783 years 498 days equals 1.36347769 years 499 days equals 1.366215597 years 500 days equals 1.368953504 years
500 days equals 1.368953504 years 600 days equals 1.642744204 years 700 days equals 1.916534905 years 800 days equals 2.190325606 years 900 days equals 2.464116306 years 1000 days equals 2.737907007 years 1100 days equals 3.011697708 years 1200 days equals 3.285488408 years 1300 days equals 3.559279109 years 1400 days equals 3.83306981 years
1400 days equals 3.83306981 years 2400 days equals 6.570976817 years 3400 days equals 9.308883824 years 4400 days equals 12.04679083 years 5400 days equals 14.78469784 years 6400 days equals 17.52260485 years 7400 days equals 20.26051185 years 8400 days equals 22.99841886 years 9400 days equals 25.73632587 years 10400 days equals 28.47423287 years | 2,715 | 8,128 | {"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} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.471089 |
https://www.stata.com/statalist/archive/2012-01/msg00192.html | 1,669,803,425,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00569.warc.gz | 1,079,835,281 | 3,572 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# RE: st: How to implement Discrete Principal Component Analysis by using POLYCHORICPCA
From Cameron McIntosh To STATA LIST Subject RE: st: How to implement Discrete Principal Component Analysis by using POLYCHORICPCA Date Thu, 5 Jan 2012 15:38:35 -0500
```What's the purpose here? You want to compare two ways of doing PCA with discrete variables? Doesn't the polychoric pca module give you all you need to know?Cam
> Date: Fri, 6 Jan 2012 00:26:50 +0400
> Subject: st: How to implement Discrete Principal Component Analysis by using POLYCHORICPCA
> From: chejen.wang@gmail.com
> To: statalist@hsphsun2.harvard.edu
>
> Dear all,
>
> Happy New year.
>
> Is there anyone who knows how to implement Discrete Principal
> Component Analysis after running POLYCHORICPCA.ado. My question is
> that, after obtaining the eigenvalues, how can I know the
> corresponding eigenvector which is similar to those eigenvector
> obtained after running PCA procedure, shown below.
>
>
> Principal components (eigenvectors)
>
> ------------------------------------------------
> Variable | Comp1 Comp2 | Unexplained
> -------------+--------------------+-------------
> foreign | 0.7071 0.7071 | 0
> gear_ratio | 0.7071 -0.7071 | 0
> ------------------------------------------------
>
> with kind regards,
>
>
> Charles Wang
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
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``` | 491 | 1,816 | {"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-2022-49 | latest | en | 0.726745 |
http://gluedideas.com/content-collection/Masonry-Structures-1921/T-Beams-with-Tension-Reinforcement_P1.html | 1,643,413,452,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00492.warc.gz | 26,697,691 | 3,289 | Home >> Masonry-structures-1921 >> Properties Of Clay And to Watertight Concrete 87 >> T Beams with Tension Reinforcement_P1
T-Beams with Tension Reinforcement 111
beam, flange, steel, width, compression and inches
Page: 1 2
T-BEAMS WITH TENSION REINFORCEMENT 111. Flexure Formulas.—In a rectangular reinforced concrete beam, in which the steel carries all the tension, the area of concrete below the neutral axis does not affect the resisting moment of the beam. The office of this concrete is to hold the steel in place and carry the shear, thus connecting the steel with the compression area of concrete.
In a '1'-beam, the flange carrying the compression is connected with a narrow web which holds the steel, as shown in Fig. 52. When the neutral axis is in the flange, such a beam may be computed by the formula_ and tables used for a rectangular beam, using the width of the flange, b, as the width of the beam.
When the neutral axis is below the bottom of the flange of the T-beam, the compression area is less than that of the rectangular beam, and special formulas are necessary. Fig. .52 represents a beam of this kind. The amount of compression on the web is usually very small and may be neglected without material thus greatly simplifying the formulas.
The same notation will be employed as in the rectangular beam, letting b =width of flange; b'= width of web; t =thichmess of flange.
The position of the neutral axis in terms of the unit stresses may be found as in the rectangular beam, giving (22) f k ' and (23) The average unit compression on the flange is the half sum of the compressions at the top and bottom of the flange, or 1kd—t kd .
The total compression on the concrete is C= f ` 2kd (24)This is the equal to the total tension on the steel, T =f,A=f,pbd (25) From the equality of (24) and (25) we find (2k—t d) t 26p_ 2n(1-k) d' (26) and (t!d)2 (27) pn+t!cl • The distance of the centroid of compression from the upper face of the beam is t 2k—t: d therefore jd=d-3k.-2t'd t (28) 3 The resisting moment of the beam is 11I = Tjcl = = (29) or 31= Cjd = f, 3k • btjd. (30) Examples.—The use of these formulas in the solution of problems arising in the design or investigation of T-beams are illustrated in the following examples: 9. A T-beam has the following dimensions, b=48 inches, t=4
inches, d=22 inches, inches. The steel reinforcement con sists of six .-,-inch round rods. If the safe unit stresses of steel and concrete are 15,000 and 600 lb., respectively, and n=15, what is the safe resisting moment of the beam? Solution.—From Table X, A =2.65 and formula (27) gives _ . 0025X 15-} Using (28) we find 2X.247— ' From (22), 15(1—.247) =-l5.7. If f,= 600 X45.7 =27,420 lb. This is greater than the safe unit stress on steel, and the safe moment will be that which causes a stress of 15,000 lbs., on the steel, or from (29), M = 2.65 X 15000X 20.39 = 810000 in.-11).
10. The flange of the T-beam is 26 inches wide and 4 inches thick. The beam is to carry a bending moment of 520,000 in.-lb. The safe unit stresses for concrete and steel are 600 and 16,000 lb. in." respectively. What area of steel and depth of beans are needed.
(23) l:'XGO0 —.360. We must now X 600 find d by assuming values and testing their suitability. Try d= 18; from (28) we have ) 2X.360— 3 (9) gives C = /,' jd = 520000/16.3 = 31900.
From (24) fc= C 2kd — t lb. in.'- This is a safe value, but 2kd a less depth will answer. Trying 15 inches, we find C=38,000 pounds, and f =580 lb., in."; 15 inches is, therefore, approximately the minimum value for d. For this value of d, Formula (25) gives, Width of without lateral reinforcement in the flanges should have a width of flange not more than three times the width of web, b = 3b'. When the flange is reinforced at right angles to the length of beam, as in a slab floor with T-beams support, experi ence indicates that the flange may overhang the web on each side to a distance equal to five or six times the thickness of flange, and still act satisfactorily as compression area for the beam. If the width of flange be greater than this, the extra width is of little value and should not be considered in estimating the strength of the beam.
Page: 1 2 | 1,173 | 4,221 | {"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} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.880475 |
https://help.libreoffice.org/latest/gl/text/scalc/guide/multioperation.html | 1,669,905,316,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00179.warc.gz | 325,238,603 | 4,779 | # Applying Multiple Operations
## Multiple Operations in Columns or Rows
The Data - Multiple Operations command provides a planning tool for "what if" questions. In your spreadsheet, you enter a formula to calculate a result from values that are stored in other cells. Then, you set up a cell range where you enter some fixed values, and the Multiple Operations command will calculate the results depending on the formula.
In the Formulas field, enter the cell reference to the formula that applies to the data range. In the Column input cell/Row input cell field, enter the cell reference to the corresponding cell that is part of the formula. This can be explained best by examples:
### Exemplos
You produce toys which you sell for \$10 each. Each toy costs \$2 to make, in addition to which you have fixed costs of \$10,000 per year. How much profit will you make in a year if you sell a particular number of toys?
### Calculating With One Formula and One Variable
1. To calculate the profit, first enter any number as the quantity (items sold) - in this example 2000. The profit is found from the formula Profit=Quantity * (Selling price - Direct costs) - Fixed costs. Enter this formula in B5.
2. In column D enter given annual sales, one below the other; for example, 500 to 5000, in steps of 500.
3. Select the range D2:E11, and thus the values in column D and the empty cells alongside in column E.
4. Choose Data - Multiple operations.
5. With the cursor in the Formulas field, click cell B5.
6. Set the cursor in the Column input cell field and click cell B4. This means that B4, the quantity, is the variable in the formula, which is replaced by the selected column values.
7. Close the dialog with OK. You see the profits for the different quantities in column E.
### Calculating with Several Formulas Simultaneously
1. Delete column E.
2. Enter the following formula in C5: = B5 / B4. You are now calculating the annual profit per item sold.
3. Select the range D2:F11, thus three columns.
4. Choose Data - Multiple Operations.
5. With the cursor in the Formulas field, select cells B5 thru C5.
6. Set the cursor in the Column input cell field and click cell B4.
7. Close the dialog with OK. You will now see the profits in column E and the annual profit per item in column F.
## Multiple Operations Across Rows and Columns
LibreOffice allows you to carry out joint multiple operations for columns and rows in so-called cross-tables. The formula cell has to refer to both the data range arranged in rows and the one arranged in columns. Select the range defined by both data ranges and call the multiple operation dialog. Enter the reference to the formula in the Formulas field. The Row input cell and the Column input cell fields are used to enter the reference to the corresponding cells of the formula.
### Calculating with Two Variables
Consider columns A and B of the sample table above. You now want to vary not just the quantity produced annually, but also the selling price, and you are interested in the profit in each case.
Expand the table shown above. D2 thru D11 contain the numbers 500, 1000 and so on, up to 5000. In E1 through H1 enter the numbers 8, 10, 15 and 20.
1. Select the range D1:H11.
2. Choose Data - Multiple Operations.
3. With the cursor in the Formulas field, click cell B5.
4. Set the cursor in the Row input cell field and click cell B1. This means that B1, the selling price, is the horizontally entered variable (with the values 8, 10, 15 and 20).
5. Set the cursor in the Column input cell field and click in B4. This means that B4, the quantity, is the vertically entered variable.
6. Close the dialog with OK. You see the profits for the different selling prices in the range E2:H11.
Precisamos da súa axuda! | 880 | 3,792 | {"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-2022-49 | longest | en | 0.852614 |
https://www.teacherspayteachers.com/Product/Number-Sense-Math-Centers-for-the-ENTIRE-YEAR-2110472 | 1,485,006,282,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00207-ip-10-171-10-70.ec2.internal.warc.gz | 1,004,905,520 | 51,518 | # Number Sense Math Centers for the ENTIRE YEAR
Subjects
Resource Types
Product Rating
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Be sure that you have an application to open this file type before downloading and/or purchasing.
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5.51 MB | 49 pages
### PRODUCT DESCRIPTION
These 36 print and go centers for practicing number sense are created to help build number sense skills in 3rd and 4th grade students. Students use number tiles to “build” numbers as they practice place value, arithmetic, and problem solving.
Skills Included:
1. Writing numbers in standard, expanded, and word form
2. Rounding numbers
3. Addition number bonds (which can be checked with the inverse operation so students are practicing both addition and subtraction)
4. Multiplication
5. Number Riddle problem solving
This resource contains 36 centers, making it ideal to use as a weekly number sense center.
Here's what's included:
1. Instructions Page
2. 36 construction themed number sense centers
3. Two pages of number tiles in different styles (Copying one page per student is recommended. You may have students place the tiles in a Ziploc bag to use all year, along with the mini sheets of work paper. Dice are also required.)
4. Construction themed work paper for scratch work
This resource is great for differentiation! Students may build a number containing between 1 and 6 digits, depending upon their ability level for many of the activities. Learners who struggle with number sense may build smaller numbers and students who need a challenge may build larger numbers to practice the skills. Also, extra digits are included for students who are able to figure out the missing number in a number bond by arranging tiles by place value to add.
This resource may also be used for early finishers and for homework!
*See the Preview for a closer look!
I hope you choose to download this product. If you do and find it helpful, please leave a great rating, follow me, and check out the other products in my store. Thank you!
~For personal and single classroom use only.~
Total Pages
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Total:
11 ratings | 505 | 2,249 | {"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-2017-04 | longest | en | 0.920872 |
http://mathhelpforum.com/number-theory/107543-crt-system-congruences.html | 1,480,794,980,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541134.5/warc/CC-MAIN-20161202170901-00057-ip-10-31-129-80.ec2.internal.warc.gz | 176,454,490 | 11,010 | # Thread: CRT and System of Congruences
1. ## CRT and System of Congruences
$x\equiv1\ (mod\ 3)$
$x\equiv3\ (mod\ 5)$
$x\equiv5\ (mod\ 7)$
I did this exercise based on this post: http://www.mathhelpforum.com/math-he...ongruence.html
And I just want to know if I did this properly and if this is the right approach.
$x\equiv1\ (mod\ 3)\ \Rightarrow\ 5x\ \equiv5\ (mod\ 15)\ \Rightarrow\ 10x\ \equiv10\ (mod\ 15)$
$x\equiv3\ (mod\ 5)\ \Rightarrow\ 3x\ \equiv9\ (mod\ 15)\ \Rightarrow\ -9x\ \equiv-27\ (mod\ 15)$
Adding gives $x\equiv-17 \ (mod\ 15)$
Comparing with the third equation:
$x\equiv-17\ (mod\ 15)\ \Rightarrow\ 7x\ \equiv-119\ (mod\ 105)\ \Rightarrow\ -14x\ \equiv238\ (mod\ 105)$
$x\equiv5\ (mod\ 7)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)$
Adding gives $x\equiv313\ (mod\ 15)$
So the solution is $\{313+15k:k\in\mathbb{Z}\}$
While I'm here, can anyone tell me what the easiest way to find the inverse of a congruence is? I seem to struggle with that.
2. Originally Posted by kyldn6
$x\equiv1\ (mod\ 3)$
$x\equiv3\ (mod\ 5)$
$x\equiv5\ (mod\ 7)$
I did this exercise based on this post: http://www.mathhelpforum.com/math-he...ongruence.html
And I just want to know if I did this properly and if this is the right approach.
$x\equiv1\ (mod\ 3)\ \Rightarrow\ 5x\ \equiv5\ (mod\ 15)\ \Rightarrow\ 10x\ \equiv10\ (mod\ 15)$
$x\equiv3\ (mod\ 5)\ \Rightarrow\ 3x\ \equiv9\ (mod\ 15)\ \Rightarrow\ -9x\ \equiv-27\ (mod\ 15)$
Adding gives $x\equiv-17 \ (mod\ 15)$
Comparing with the third equation:
$x\equiv-17\ (mod\ 15)\ \Rightarrow\ 7x\ \equiv-119\ (mod\ 105)\ \Rightarrow\ -14x\ \equiv238\ (mod\ 105)$
$x\equiv5\ (mod\ 7)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)$
Adding gives $x\equiv313\ (mod\ 15)$
So the solution is $\{313+15k:k\in\mathbb{Z}\}$
While I'm here, can anyone tell me what the easiest way to find the inverse of a congruence is? I seem to struggle with that.
Let me suggest you a nice observation here.
x+2 = 0(mod 3)
x+2 = 0(mod 5)
x+2 = 0(mod 7)
Now can you take it fwd?
Also haven't seen your eqs in details but answer appear wrong. Put k=1 and try the last congruence?
3. Originally Posted by kyldn6
$x\equiv1\ (mod\ 3)$
$x\equiv3\ (mod\ 5)$
$x\equiv5\ (mod\ 7)$
I did this exercise based on this post: http://www.mathhelpforum.com/math-he...ongruence.html
And I just want to know if I did this properly and if this is the right approach.
$x\equiv1\ (mod\ 3)\ \Rightarrow\ 5x\ \equiv5\ (mod\ 15)\ \Rightarrow\ 10x\ \equiv10\ (mod\ 15)$
$x\equiv3\ (mod\ 5)\ \Rightarrow\ 3x\ \equiv9\ (mod\ 15)\ \Rightarrow\ -9x\ \equiv-27\ (mod\ 15)$
Adding gives $x\equiv-17 \ (mod\ 15)$
Comparing with the third equation:
$x\equiv-17\ (mod\ 15)\ \Rightarrow\ 7x\ \equiv-119\ (mod\ 105)\ \Rightarrow\ -14x\ \equiv238\ (mod\ 105)$
$x\equiv5\ (mod\ 7)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)$
Adding gives $x\equiv313\ (mod\ 15)$
I think you mean $x\equiv 313 (mod 105)$ here.
So the solution is $\{313+15k:k\in\mathbb{Z}\}$
And $\{313+105k:k\in\mathbb{Z}\}$ here.
While I'm here, can anyone tell me what the easiest way to find the inverse of a congruence is? I seem to struggle with that.
What do you mean by "inverse of a congruence"?
4. Originally Posted by HallsofIvy
I think you mean $x\equiv 313 (mod 105)$ here.
And $\{313+105k:k\in\mathbb{Z}\}$ here.
What do you mean by "inverse of a congruence"?
Yes this looks correct. Thanks.
@kyldn6 But you don't have to perform any of these modular calculations if you look at the observations I said earlier. It comes directly from there, atleast in this question.
5. Yes, those were typos. And thanks everyone! | 1,351 | 3,717 | {"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": 34, "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} | 4.0625 | 4 | CC-MAIN-2016-50 | longest | en | 0.6736 |
https://www.jiskha.com/display.cgi?id=1241757831 | 1,516,326,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887692.13/warc/CC-MAIN-20180119010338-20180119030338-00594.warc.gz | 929,112,509 | 4,020 | # math
posted by .
how would you determine the eccentricity of the conic section when represented by the equation
ax^2+cy^2+dx+ey+f=0
• math -
Put
x_1 = x
x_2 = y
Then the quadratic form is of the form
Sum over i and j of
M_{i,j}x_i x_j = constant.
Then, since this equation is invariant under orhtogonal transformations, we can write it in diagonal form. If you define new coordinates x' and y' that are in the direction of the two egienvectors, the equation becomes:
lambda_1 x'^2 + lambda_2 y'2 = constant
where lambda_1 and lambda_2 are the two eigenvalues of M.
• math -
I now see that there is no xy term in your expression, so the problem is trivial. You can simply complete the square, no need to diagonalize a matrix.
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More Similar Questions | 785 | 2,910 | {"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-2018-05 | latest | en | 0.856072 |
http://cerco.cs.unibo.it/export/248/Deliverables/D4.1/Matita/Arithmetic.ma | 1,611,478,007,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00508.warc.gz | 19,329,334 | 1,587 | include "Universes.ma". include "Equality.ma". include "Connectives.ma". include "Nat.ma". include "Exponential.ma". include "Bool.ma". include "BitVector.ma". include "List.ma". ndefinition eight ≝ exponential (S(S(Z))) (S(S(S(Z)))). ndefinition sixteen ≝ eight + eight. ndefinition one_hundred_and_twenty_eight ≝ sixteen * eight. ndefinition two_hundred_and_fifty_six ≝ one_hundred_and_twenty_eight + one_hundred_and_twenty_eight. ndefinition nat_of_bool ≝ λb: Bool. match b with [ False ⇒ Z | True ⇒ S Z ]. ndefinition add_n_with_carry: ∀n: Nat. ∀b, c: BitVector n. ∀carry: Bool. Cartesian (BitVector n) (List Bool) ≝ λn: Nat. λb: BitVector n. λc: BitVector n. λcarry: Bool. let b_as_nat ≝ nat_of_bitvector n b in let c_as_nat ≝ nat_of_bitvector n c in let carry_as_nat ≝ nat_of_bool carry in let result_old ≝ b_as_nat + c_as_nat + carry_as_nat in let ac_flag ≝ ((modulus b_as_nat ((S (S Z)) * n)) + (modulus c_as_nat ((S (S Z)) * n)) + c_as_nat) ≥ ((S (S Z)) * n) in let bit_xxx ≝ (((modulus b_as_nat ((S (S Z))^(n - (S Z)))) + (modulus c_as_nat ((S (S Z))^(n - (S Z)))) + c_as_nat) ≥ ((S (S Z))^(n - (S Z)))) in let result ≝ modulus result_old ((S (S Z))^n) in let cy_flag ≝ (result_old ≥ ((S (S Z))^n)) in let ov_flag ≝ exclusive_disjunction cy_flag bit_xxx in ? (mk_Cartesian (BitVector n) ? (? (bitvector_of_nat n result)) (cy_flag :: ac_flag :: ov_flag :: Empty Bool)). //. nqed. (* ndefinition sub_8_with_carry ≝ λb: BitVector eight. λc: BitVector eight. λcarry: Bool. let b_as_nat ≝ nat_of_bitvector eight b in let c_as_nat ≝ nat_of_bitvector eight c in let carry_as_nat ≝ nat_of_bool carry in let result_old_1 ≝ subtraction_underflow b_as_nat c_as_nat in match result_old_1 with [ Nothing ⇒ let ac_flag ≝ True in | Just result_old_1' ⇒ ] *) ndefinition add_8_with_carry ≝ add_n_with_carry eight. ndefinition add_16_with_carry ≝ add_n_with_carry sixteen. (* ndefinition increment ≝ λn: Nat. λb: BitVector n. let b_as_nat ≝ (nat_of_bitvector n b) + (S Z) in let overflow ≝ b_as_nat ≥ (S (S Z))^n in match overflow with [ False ⇒ bitvector_of_nat n b_as_nat | True ⇒ bitvector_of_nat n Z ]. ndefinition decrement ≝ λn: Nat. λb: BitVector n. let b_as_nat ≝ nat_of_bitvector n b in match b_as_nat with [ Z ⇒ max n | S o ⇒ bitvector_of_nat n o ]. alias symbol "greater_than_or_equal" (instance 1) = "Nat greater than or equal prop". ndefinition bitvector_of_bool: ∀n: Nat. ∀b: Bool. BitVector n ≝ λn: Nat. λb: Bool. ? (pad (n - (S Z)) (S Z) (Cons Bool ? b (Empty Bool))). //. nqed. *) | 830 | 2,490 | {"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} | 2.875 | 3 | CC-MAIN-2021-04 | latest | en | 0.47357 |
https://myassignments-help.com/2022/10/08/machine-learning-dai-xie-comp3670/ | 1,679,626,274,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00241.warc.gz | 462,885,135 | 36,321 | # 计算机代写|机器学习代写machine learning代考|COMP3670
## 计算机代写|机器学习代写machine learning代考|Model Complexity and Regularization
So far we have talked vaguely about what it means for a model to be ‘too complex’ (or too simple) and suggested that we should choose a model that is complex enough to fit the data but simple enough not to overfit. This idea is often referred to as Occam’s Razor, a philosophical principle which states that among several alternate hypotheses that explain some phenomenon, one should favor the simplest.
However, for these notions to be useful, we must be precise about what it means for a model to be ‘complex.’ We would like to define complexity in terms of the parameters $\theta$, such that given a fixed set of features and labels, we could select the ‘simplest’ $\theta$ that adequately explains (or models) the data.
We will discuss two candidate notions of ‘simplicity’ as follows:
(i) A simple model is one that includes only a few terms, that is, in which only a few values $\theta_k$ are nonzero.
(ii) A simple model is one in which all terms are about equally important, that is, one in which particularly large values of $\theta_k$ are rare.
These two potential notions of ‘complexity’ are captured by the following expressions:
\begin{aligned} &\Omega_1(\theta)=|\theta|_1=\sum_k\left|\theta_k\right|, \ &\Omega_2(\theta)=|\theta|_2^2=\sum_k \theta_k^2, \end{aligned}
that is, the sum of absolute values and the sum of squares, also called the $\ell_1$ and (squared) $\ell_2$ norms of $\theta$. We state without proof that these expressions penalize models that have many nonzero parameters (eq. (3.29)) or large parameters (eq. (3.30)), though we further characterize their behavior later.
## 计算机代写|机器学习代写machine learning代考|Regularization
In order to fit a model which simultaneously explains the data but is not overly complex (corresponding to our goal above), we write down a new objective that combines our original accuracy objective with one of the complexity expressions above (in this case the squared $\ell_2$ norm). For a regression model we add the regularizer to the expression from Equation (2.16):
$$\underbrace{\frac{1}{|y|} \sum_{i=1}^{|y|}\left(x_i \cdot \theta-y_i\right)^2}{\text {accuracy }}+\underbrace{\sum_k \theta_k^2}{\text {model complexity } |\left.\theta\right|2 ^2} .$$ For a classification model, we subtract the regularizer, since we seek to maximize accuracy rather than minimizing error (so we maximize $-\lambda|\theta|_2^2$ rather than minimizing $\lambda|\theta|_2^2$ ): $$\sum_i-\log \left(1+e^{-x_i \cdot \theta}\right)+\sum{y_i=0}-x_i \cdot \theta-\lambda|\theta|_2^2 .$$
This procedure-where we add a penalty term to control model complexityis known as regularization; the parameter $\lambda$, which controls the extent to which complexity is penalized, is termed a regularization parameter.
Note that we can straightforwardly adapt the derivatives (from eqs. (2.54) and (3.9)) to include the regularization term, $\lambda|\theta|_2^2$ by noting that $\frac{\partial}{\partial \theta_k}$ $\lambda|\theta|_2^2=2 \lambda \theta_k$.
# 机器学习代考
## 计算机代写|机器学习代写machine learning代考|Model Complexity and Regularization
. Model Complexity and Regularization
(i)一个简单模型是一个只包含少数项的模型，也就是说，其中只有少数值$\theta_k$是非零。
(ii)一个简单模型是一个所有项都同等重要的模型，也就是说，其中$\theta_k$特别大的值是罕见的
\begin{aligned} &\Omega_1(\theta)=|\theta|_1=\sum_k\left|\theta_k\right|, \ &\Omega_2(\theta)=|\theta|_2^2=\sum_k \theta_k^2, \end{aligned}
## 计算机代写|机器学习代写machine learning代考|Regularization
.
$$\underbrace{\frac{1}{|y|} \sum_{i=1}^{|y|}\left(x_i \cdot \theta-y_i\right)^2}{\text {accuracy }}+\underbrace{\sum_k \theta_k^2}{\text {model complexity } |\left.\theta\right|2 ^2} .$$对于一个分类模型，我们减去正则化器，因为我们寻求最大化的精度而不是最小化的错误(所以我们最大化$-\lambda|\theta|_2^2$而不是最小化$\lambda|\theta|_2^2$): $$\sum_i-\log \left(1+e^{-x_i \cdot \theta}\right)+\sum{y_i=0}-x_i \cdot \theta-\lambda|\theta|_2^2 .$$
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Math作业代写、数学代写常见问题
myassignments-help擅长领域包含但不是全部: | 1,442 | 4,321 | {"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} | 3.5 | 4 | CC-MAIN-2023-14 | longest | en | 0.920534 |
https://www.jiskha.com/questions/45058/Hello-I-was-wondering-how-to-explain-the-potential-energy-in-a-Ferris-wheel-and | 1,534,638,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213903.82/warc/CC-MAIN-20180818232623-20180819012623-00057.warc.gz | 903,816,875 | 4,824 | # PHYSCIS FO SHIZZLE
Hello, I was wondering how to explain the potential energy in a Ferris wheel, and if there were any reputable (aka not something such as Wikipedia, etc) that had an explanation for this? The only thing I have gathered is that mgh=Eg and that Eg is the energy stored in the wheel. The energy is stored as the result of the gravitational attraction of the Earth for the object...I need help, and are there any equations I should know about?
posted by drwls
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7. ### trig question
A Ferris wheel has a deameter of 50m. The platform at the bottom, where you load the ferris wheel, is 3 m above the ground. The Ferris wheel rotates three times every two minutes. A stopwatch is started and you notice you are even | 542 | 2,268 | {"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-2018-34 | latest | en | 0.971056 |
https://socratic.org/questions/a-projectile-is-shot-at-a-velocity-of-1-5-m-s-and-an-angle-of-pi-8-what-is-the-p | 1,611,505,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00328.warc.gz | 560,801,098 | 6,031 | # A projectile is shot at a velocity of 1 5 m/s and an angle of pi/8 . What is the projectile's peak height?
Jul 4, 2017
The peak height is $= 1.68 m$
#### Explanation:
Resolving in the vertical direction ${\uparrow}^{+}$
The initial velocity is ${u}_{y} = v \sin \theta = 15 \cdot \sin \left(\frac{1}{8} \pi\right)$
The acceleration is $a = - g$
At the maximum height, $v = 0$
We apply the equation of motion
${v}^{2} = {u}^{2} + 2 a s$
to calculate the greatest height
$0 = \left(15 \sin {\left(\frac{1}{8} \pi\right)}^{2}\right) - 2 g h$
The greatest height is
$h = {\left(15 \sin \left(\frac{1}{8} \pi\right)\right)}^{2} / \left(2 g\right) = 1.68 m$ | 241 | 666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.34375 | 4 | CC-MAIN-2021-04 | longest | en | 0.737053 |
http://stackoverflow.com/questions/14336827/how-to-return-unique-and-sorted-array-values-useing-jquery?answertab=active | 1,454,954,331,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00227-ip-10-236-182-209.ec2.internal.warc.gz | 199,682,165 | 19,292 | # How to return unique and sorted array values useing jQuery? [duplicate]
Possible Duplicate:
Javascript array sort and unique
I was surprised to see that there is no built in jQuery function for that.
I've seen many solutions on stackoverflow, but the questions were polluted with not-working answers (to find a working one I had to test them all).
So, for future reference and to spare other users the trouble I decided to post this Q&A style.
# How to return unique and sorted array values useing jQuery?
### Numbers:
``````// input array
var inputArray = [10, 5, 15, 10, 5, 15];
// expected result array
var resultArray = [5, 10, 15];
``````
### Strings:
``````// input array
var inputArray = ['b', 'a', 'c', 'b', 'a', 'c'];
// expected result array
var resultArray = ['a', 'b', 'c'];
``````
-
## marked as duplicate by Felix Kling, Kanishka Panamaldeniya, Denys Séguret, Yoshi, ithcyJan 15 '13 at 14:09
wooh you answered your own question at the same time ? – Sibu Jan 15 '13 at 11:37
@Sibu That's totally legit on SO : if you solve a problem and have a solution that may help others, you can do what OP did. But in this specific case I don't think it's interesting enough as it's yet frequently answered. – Denys Séguret Jan 15 '13 at 11:39
@dystroy agreed but why ask an question when you know the answer, if you see OP question time and answer time, both are nearly same..amusing isn't it ? – Sibu Jan 15 '13 at 11:43
The reason is to help other people : if the question is new and interesting and you think you have special knowledge, then you may bring it to the masses. See this. – Denys Séguret Jan 15 '13 at 11:50
Now, to give you some more intuition into the solution that I propose: Sorting is O(n log n), removing duplicates is O(n), so we conclude that the entire operation should be no more complex then n log n. However, if you think of it, n will never increase, but will most likely decrease if you first remove duplicates, and then sort. So, while on the surface of it, it is still O(n log n), it will be generally faster. You could probably improve it (in other languages) by collecting values into a tree instead of a hash-table, but given the tremendous difference in performance between the "native" data structures and custom ones in JavaScript - the solution below should be optimal:
``````function sortUnique(array) {
"use strict";
var table = {}, key, i;
for (i = 0; i < array.length; i++) {
table[[array[i]]] = '';
}
i = 0;
for (key in table) {
array[i++] = key;
}
array.length = i;
return array.sort();
}
sortUnique(['b', 'a', 'c', 'b', 'a', 'c']);
// [ 'a', 'b', 'c' ]
``````
-
Are you sure it's faster ? Especially for a not gigantic array ? – Denys Séguret Jan 15 '13 at 12:04
I will test your solution, if it works and is faster I will obviously mark it as correct, no need for comments like "but I don't hope for the author to alter the decision". – loostro Jan 19 '13 at 12:27
I'm marking this as accepted becouse it works and also i've learned a lesson about performance. tyvm wvxvw – loostro Jan 27 '13 at 13:52
If the array contains numbers or Dates, this alg converts them to strings in IE or node.js. The reason is (if I understand well) that `table[[...]]` is a nonstandard feature, present in ECMA6 draft only, not implemented in IE "yet" (2 years later than the above answer!). In ECMA5, object keys must always be converted to strings. – robert4 Mar 2 '15 at 23:17
hope this will work for you
``````var dummy = [10, 5, 15, 10, 5, 15];
var arr = [];
\$.map(arr, function(n, i){
if(\$.inArray(n, arr) == -1)
arr.push(n);
});
``````
now in arr values are unique. now you can apply any sorting algo on this.
-
As a function:
``````function sort_unique(arr) {
return arr.sort(function(a,b){
return (a > b) ? 1 : -1;
}).filter(function(el,i,a) {
return (i==a.indexOf(el));
});
}
``````
-
I never said my solution was the fastest one. I posted a simple and working solution. The question was not "what is the fastest way to..."? – loostro Jan 19 '13 at 12:25 | 1,118 | 4,019 | {"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} | 2.796875 | 3 | CC-MAIN-2016-07 | latest | en | 0.919188 |
https://praadisedu.com/selina-solutions-for-class-6-maths-chapter-19/Fundamental-Operations/1018/85 | 1,611,278,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703528672.38/warc/CC-MAIN-20210121225305-20210122015305-00644.warc.gz | 505,621,525 | 41,898 | # Concise Selina Solutions for Class 6 maths Chapter 19 Fundamental Operations
The fundamental operation in simplifying mathematical expressions consisting of the same type of operation, we perform one operation at a time generally starting from the left towards the right. If an expression has more than one fundamental operation, you cannot perform operations in the order they appear in the given question. We need to follow the rules to perform the operations. Some operations have to be performed before the others. That is, each operation has its precedence.
Generally, the order in which we perform operations sequentially from left to right is division, multiplication, addition, subtraction. This order is expressed in short as ‘DMAS’ where ‘D’ stands for division, ‘M’ stands for multiplication, ‘A’ stands for addition and, ‘S’ for subtraction. Here are one more fundamental operations in mathematics: and it is modular forms. | 182 | 939 | {"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-2021-04 | longest | en | 0.951711 |
https://busynessgirl.com/giving-up-calculation-by-hand/ | 1,723,205,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00181.warc.gz | 109,759,453 | 69,686 | This is scary stuff for math professors, but with the arrival of amazing programs like Wolfram Alpha, we’re going to have to start paying attention to the signs of change. I talked to Conrad Wolfram (at Wolfram Alpha Homework Day) when he was still formulating what he wanted to say at this TED Talk. I think it’s worth 18 minutes of your time to watch Teaching kids real math with computers.
Here’s an outline of the Conrad Wolfram’s argument (which I am paraphrasing/quoting here):
What’s the point of teaching people math?
1. Technical jobs (critical to the development of our economies)
2. Everyday living (e.g. figuring out mortgage, being skeptical of government statistics)
3. Logical mind training / logical thinking (math is a great way to learn logic)
What IS math?
1. Posing the right questions.
2. Convert from real world to mathematical formulation
3. Computation
4. Convert from mathematical formulation BACK to real world
The problem? In math education, we’re spending about 80% of the time teaching students to do step 3 by hand.
Math is not equal to calculating, math is a much broader subject than calculating. In fact, math has been liberated from calculating.
Should we have to “Get the basics first”? Are the “basics” of driving a car learning how to service or design the car? Are the “basics” of writing learning how to sharpen a quill?
People confuse the order of the invention of the tools with the order in which they should use them in teaching. Just because paper was invented before computers, it doesn’t necessarily mean you get more to the basics of the subject by using paper instead of a computer to teach mathematics.
What about this idea that “Computer dumb math down” … that somehow, if you use a computer, it’s all mindless button-pushing. But if you do it by hand it’s all intellectual. This one kind of annoys me, I must say. Do we really believe that the math that most people are actually doing in school practically today is more than applying procedures to problems they don’t really understand for reasons they don’t get? … What’s worse … what they’re learning there isn’t even practically useful anymore. It might have been 50 years ago, but it isn’t anymore. When they’re out of education, they do it on a computer.
Understanding procedures and processes IS important. But there’s a fantastic way to do that in the modern world … it’s called programming.
We have a unique opportunity to make math both more practical and more conceptual simultaneously.
Personally, I’m all for it. But how? That’s the question. How to shift and incredibly complex and interconnected system of education? How to train tens of thousands of teachers and faculty to teach a new curriculum that they themselves never learned? Hmmm … it seems that we might need some help, maybe a new paradigm for education itself. It’s coming. | 636 | 2,879 | {"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-2024-33 | latest | en | 0.921439 |
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Home » Topics » Nutrition and Anthropometry » Stunting rate with accurate observation number
Stunting rate with accurate observation number Thu, 31 May 2018 11:46
anikhpg42@gmail.com Messages: 38Registered: December 2017 Location: Bangladesh Member
Hi,
In order to find the stunting rate among children under age 5 using BDHS 14 data, I have used the following Stata command.
I used this command both in KR & PR dataset. But the total observation number is not exactly matching with the BDHS-14 report.
Prevalence of Stunting in 2014 (Children under age 5)
My estimation,
Using KR file, 36.24 % (N = 6965)
Using PR file, 36.09 % (N = 7256)
From BDHS'14 Report
36.1 % (N = 7318)
Here, my question is, how can I obtain the same prevalence rate and same observation for calculating child stunting rate?
Where is my mistakes, that I've made in my Stata command?
I have to get the exact observations (N = 7318) and the exact stunting rate (36.10%).
Stata command for stunting (just for the KR file)
*SVY command
gen strata=v023
gen psu=v021
gen sampwt=v005/1000000
svyset psu [pw=sampwt], strata (strata)
//child stunting calculation
codebook hw70
tab hw70 if hw70>9990,m
tab hw70 if hw70>9990, m nolabel
gen HAZ=hw70
replace HAZ=. if HAZ>=9996
*****
gen stunted=.
replace stunted=0 if HAZ~=.
replace stunted=1 if HAZ<-200
svy: tab stunted
Best regards,
Anik
ASIBUL ISLAM ANIK
Stunting rate with accurate observation number By: anikhpg42@gmail.com on Thu, 31 May 2018 11:46 Re: Stunting rate with accurate observation number By: Trevor-DHS on Thu, 07 June 2018 10:45 Re: Stunting rate with accurate observation number By: hamidine on Mon, 25 June 2018 14:25 Re: Stunting rate with accurate observation number By: Trevor-DHS on Mon, 25 June 2018 15:27 Re: Stunting rate with accurate observation number By: dhsforum on Thu, 06 September 2018 20:20 Re: Stunting rate with accurate observation number By: Trevor-DHS on Fri, 07 September 2018 00:00 Re: Stunting rate with accurate observation number By: dhsforum on Fri, 07 September 2018 11:14 Re: Stunting rate with accurate observation number By: Trevor-DHS on Fri, 07 September 2018 11:19 Re: Stunting rate with accurate observation number By: dhsforum on Fri, 07 September 2018 11:29
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source: Kaplan Premier Although the square root of a
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source: Kaplan Premier
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications
C) equations involving these inapplicable imaginary numbers are practical
D) equations involving imaginary numbers such as these are inapplicable practically
E) there is no practical applications for equations involving such imaginary numbers as these
So, you thought I put this question in the wrong forum?
If you have any questions
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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10 Jun 2012, 23:34
1
KUDOS
diogoguitarrista wrote:
source: Kaplan Premier
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications
C) equations involving these inapplicable imaginary numbers are practical
D) equations involving imaginary numbers such as these are inapplicable practically
E) there is no practical applications for equations involving such imaginary numbers as these
So, you thought I put this question in the wrong forum?
these, this those must be followed by noun so A and D are out. C is just reversed the intended meaning. E is just wordy. so B wins
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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08 Mar 2013, 02:12
1
KUDOS
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications
-I eliminated options C, D and E for the following reasons. After eliminating them, I was left with A and B. B is more precise, and clear in meaning than B is.
C) equations involving these inapplicable imaginary numbers are practical
-This option changes the meaning by making the numbers inapplicable, Eliminated
D) equations involving imaginary numbers such as these are inapplicable practically
-Practically inapplicable is much better than inapplicable practically
E) there is no practical applications for equations involving such imaginary numbers as these
- Wordy. "As these" is an unnecessary add-on
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04 Nov 2009, 15:37
B
Practically inapplicable...
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18 Dec 2009, 18:49
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications - CORRECT
C) equations involving these inapplicable imaginary numbers are practical - misplaced modifier
D) equations involving imaginary numbers such as these are inapplicable practically - awkward
E) there is no practical applications for equations involving such imaginary numbers as these - lengthy and awkward
B is the best option.
It contains 'such' not like and it is short and concise and is in active voice also.
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01 Sep 2011, 12:15
Quote:
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications
C) equations involving these inapplicable imaginary numbers are practical
D) equations involving imaginary numbers such as these are inapplicable practically
E) there is no practical applications for equations involving such imaginary numbers as these
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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22 Feb 2012, 20:44
B is the winner here.
We need the adjective "practical" to qualify applications.
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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06 Mar 2012, 15:48
Can "such as these" be correct in a sentence?
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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29 Aug 2012, 06:28
I went with "B". Could anyone explain if option "E" is grammatically incorrect?
Thanks
Vishwa
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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29 Aug 2012, 06:56
E) There is no practical applications for equations involving such imaginary numbers as these ----- there is no practical applications– SV error: it should be - there are no practical applications, since the real subject applications is plural
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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29 Aug 2012, 08:07
daagh wrote:
E) There is no practical applications for equations involving such imaginary numbers as these ----- there is no– SV error: it should be - there are no practical applications, since the real subject applications is plural
Oops, I should have noticed that!
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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29 Aug 2012, 10:18
The word "these" must be followed by a noun, am I right? In that case the only valid answer would be B.
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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20 Apr 2013, 10:31
diogoguitarrista wrote:
source: Kaplan Premier
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
B) equations involving such imaginary numbers have no practical applications
C) equations involving these inapplicable imaginary numbers are practical
D) equations involving imaginary numbers such as these are inapplicable practically
E) there is no practical applications for equations involving such imaginary numbers as these
So, you thought I put this question in the wrong forum?
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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05 Aug 2014, 04:12
I could come down to a and b but chose a.Isn't these referring to the noun 'Square root...' and is therefore correct?Or like should only be followed by a Noun,never a pronoun?
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Re: source: Kaplan Premier Although the square root of a [#permalink]
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source: Kaplan Premier Although the square root of a [#permalink]
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14 Dec 2016, 14:51
Although the square root of a negative number has no real value, it is not necessarily true that equations involving imaginary numbers like these are practically inapplicable
A) equations involving imaginary numbers like these are practically inapplicable
to state example 'such as' should be used not like
B) equations involving such imaginary numbers have no practical applications
correct
C) equations involving these inapplicable imaginary numbers are practical
changes the meaning. here the point is about equations' , which use imaginary numbers, practical application
D) equations involving imaginary numbers such as these are inapplicable practically
these should be followed by imaginary numbers (noun)
E) there is no practical applications for equations involving such imaginary numbers as these
'is' wrong 'are' should have been used. Wrong usage of these. Changes the original meaning
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source: Kaplan Premier Although the square root of a [#permalink] 14 Dec 2016, 14:51
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Display posts from previous: Sort by | 3,319 | 13,159 | {"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} | 3.3125 | 3 | CC-MAIN-2017-04 | latest | en | 0.895549 |
https://www.nagwa.com/en/explainers/678151837434/ | 1,709,342,481,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00201.warc.gz | 901,189,388 | 25,599 | Lesson Explainer: Two-Step Equations | Nagwa Lesson Explainer: Two-Step Equations | Nagwa
# Lesson Explainer: Two-Step Equations Mathematics • 7th Grade
In this explainer, we will learn how to solve two-step equations.
Equations are like an enigma—they give us information about an unknown number, denoted with a letter (often ) that we can use to find the value of this number. A simple equation tells us that when the unknown number has undergone some operations, it led to a certain result. The equation can be solved by reversing all the operations and in the contrary order, to finally find the value of the starting number. More complex equations cannot be solved in that way; however, we are going to derive with this first approach a general method that can be applied to solve any equation.
Let us first imagine that when a given number is doubled and the result is added to 5, we get 19. Let be this unknown number. The corresponding equation is . How can we find the value of this unknown number? We can use diverse models to visualize what is happening. First, we can use a bar diagram.
The bar diagram suggests that if we can split the 19 in and 5, we find that .
This shows us that the first step to solve the equation is to subtract 5 from both its sides. We find then that . The second step is to split both sides in two, that is, dividing by two.
We find that .
We can also write the operations that have been performed on as given in the question.
And, from there, we simply go the other way around and perform the reverse operations on 19 to get to the value of .
Both models show us that the first step to solve the equation is to subtract 5 from each side, which can be formally written as
that is,
The second step is to divide both sides by two:
which gives the solution of
We have illustrated here a general method that will be applied to solve all equations, not only the type that we have just taken as example. This method is called the balance method and is based on the additive and multiplicative properties of equalities.
### Balance Method: Additive and Multiplicative Properties of Equalities
The balance method is a general method for solving equations based on the additive and multiplicative properties of equalities. These properties are very intuitive. The first states that if the same amount is added to (or subtracted from) both sides of the equality, then the equality is still true. The second states that if both sides of the equation are multiplied by the same number (or both sides are divided by the same number), then the equality is still true.
The additive and multiplicative properties of equalities can be written in symbols as follows:
This balance method can be illustrated with a balance scale.
We have ; we can see numbers as weights put on a balance scale. We place each side of the equation on either side of the scale. Since , the scale is balanced. It is clear that if we add or remove a weight from only one side of the scale, it will not be balanced any more. It is exactly the same with the equation. For it to remain balanced, we need to do exactly the same operation on each side.
Let us now look at some examples.
### Example 1: Solving an Equation from a Description of the Operations on an Unknown Number
A number is tripled and 7 is subtracted from the result. If the answer is 17, what is the number?
Let us call the unknown number . When this number is tripled, we get three times , that is, . Then, 7 is subtracted from , which is written as
And this number is 17, so we have .
To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the term on one side of the equation, so we need to add 7 to both sides to get rid of the “” on the left-hand side. We have
which gives
Then, we need to divide both sides by three (which is the reverse operation of tripling) to get the value of :
which is
Hence, we find that the number is 8.
We can check our result by performing the operations as given in the question. First, the number is tripled, so we get 24. Then, seven is subtracted from the result, so we get 17. It is indeed the result given in the question, so our answer is correct.
### Example 2: Solving a Two-Step Equation
Find the solution set of in .
Here, we are asked to find the solution set of an equation in the set of natural numbers. It means that we need to find the value(s) of so that is true and so that is a natural number.
Let us first find the value of that is the solution of . For this, we want to get rid of the “” on the left-hand side to have the term on its own. So, we need to add 9 to each side of the equation. We have
that is,
It is quite obvious that if 3 times a number equals 3, then this number is 1. We can, however, write this step formally as dividing both sides by three:
that is,
We can check our answer by plugging 1 into the equation . We find
This equality is true, so our answer is correct.
Next, we need to check that our solution is a natural number. The number 1 is a natural number; therefore, the solution set of is .
### Example 3: Solving a Two-Step Equation
Find the solution set of the equation .
To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the term on one side of the equation, so we start by subtracting 3 from each side. We have
which gives
The second step is to divide both sides of the equation by :
which gives
We can check our answer by plugging it into , evaluating it, and comparing with 4. We find
Therefore, the value of is the solution to the equation ; our answer is correct.
The solution set of is .
### Example 4: Solving a Two-Step Equation with a Fractional Leading Coefficient
Solve .
To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the term on one side of the equation, so our first step is to add 13 to both sides. We have
which gives
We now need to multiply by 3 both sides of the equation to get the value of . We find
hence,
We can check our answer by plugging it into and evaluating it. We find
Therefore, the value of 72 is the solution to the equation ; our answer is correct.
So, the solution set of the equation is .
### Example 5: Solving a Two-Step Equation Involving a Fraction
Find the solution set of in .
To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We want to isolate the term on one side of the equation. For this, our first step is to multiply both sides by 4. We have
which gives
We now need to add 5 to both sides of the equation.
We find
hence,
We can check our answer by plugging it into and evaluating this expression. We find
Therefore, the value of 25 is the solution to the equation . We now need to check that 25 is a natural number: it is. Therefore, the solution set to is .
### Key Points
• A simple equation such as can be solved by reversing all the operations and in the contrary order, to finally find the value of the starting number.
• More complex equations cannot be solved in that way.
• The balance method is a general method for solving equations based on the additive and multiplicative properties of equalities.
• These properties state the following:
• If the same amount is added to (or subtracted from) both sides of the equality, then the equality is still true.
• If both sides of the equation are multiplied by the same number (or if both sides are divided by the same number), then the equality is still true.
• It is written in symbols as follows: | 1,839 | 8,055 | {"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} | 4.9375 | 5 | CC-MAIN-2024-10 | latest | en | 0.962299 |
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Utility maximization choice with current income & prices of two goods - wine & cheese
Utility maximization choice with current income & prices of two goods - wine & cheese
Published by: ClassOf1.com on Sep 10, 2009
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ClassOf1provides exert guidance for College, Graduate and High schoolhomework and liveonline tutoring on subjects likeFinance, Marketing,Statistics,Economicsand others. Check out more solved problems in our Solution Library.
Sub: Economics Topic: Micro Economics
Question:
Utility maximization choice with current income & pricesof two goods - wine & cheese
Bridget has limited income and consumers only wine and cheese; her currentconsumption choice is four bottles of wine and 10 pounds of cheese. Theprice of wine is \$10 per bottle, and the price of cheese is \$4 per pound. Thelast bottle of wine added 50 units to Bridget’s utility, while the last pound of cheese added 40 units.a. Is Bridget making the utility maximizing choice? Why or why not.b. If not, what should she do instead.
Solution:
Part aWe assume well behaved preference ordering for Bridget, i.e. a concaveutility function and hence convex to the origin Indifference curves.Let Marginal Utility of wine be MUwLet Marginal Utility of cheese be MUcApplying the Ordinal approach to consumer utility, the MRS = (MUw/MUc) The budget constraint of Bridget isPwQw + PcQc=M
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Sub: Economics Topic: Micro Economics
WherePw=price of WinePc=price of cheeseQw=quantity of Wine consumedQc=quantity of cheese consumedGiven constant income and prices, we writePwdQw + PcdQc=0Or PwdQw=-PcdQcOr dQc/ dQw = - (Pw/ Pc) The above is the slope of the budget line.At optimality we must have: MRS=|-(Pw/ Pc)|At the point of Bridget’s consumption MUw=50,MUc=40.Hence MRS=5/4. The slope of his budget line (Pw/ Pc) =10/4=5/2.Hence MRS is not equal to Price ratio.
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## Activity (6)
You've already reviewed this. Edit your review. | 715 | 2,766 | {"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-2015-22 | latest | en | 0.843049 |
http://slideplayer.com/slide/5176038/ | 1,600,811,219,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206763.24/warc/CC-MAIN-20200922192512-20200922222512-00078.warc.gz | 129,889,778 | 21,815 | # What happens to solar energy ? 1.Absorption (absorptivity= ) Results in conduction, convection and long-wave emission 2.Transmission (transmissivity=
## Presentation on theme: "What happens to solar energy ? 1.Absorption (absorptivity= ) Results in conduction, convection and long-wave emission 2.Transmission (transmissivity="— Presentation transcript:
What happens to solar energy ? 1.Absorption (absorptivity= ) Results in conduction, convection and long-wave emission 2.Transmission (transmissivity= ) 3.Reflection (reflectivity= ) + + = 1
Response varies with the surface type Snow reflects 40 to 95% of solar energy and requires a phase change to increase above 0°C Forests and oceans absorb more than dry lands Then why do dry lands still “heat up” more? Oceans transmit solar energy and have a high heat capacity
Characteristics of Radiation Energy due to rapid oscillations of electromagnetic fields, transferred by photons The energy of a photon is equal to Planck’s constant, multiplied by the speed of light, divided by the wavelength All bodies above 0 K emit radiation Black body emits maximum possible radiation per unit area. Emissivity, = 1.0 All bodies have an emissivity between 0 and 1 E = hv
Stefan-Boltzmann Law As the temperature of an object increases, more radiation is emitted each second
Temperature determines E, emitted Higher frequencies (shorter wavelengths) are emitted from bodies at a higher temperature Max Planck determined a characteristic emission curve whose shape is retained for radiation at 6000 K (Sun) and 300 K (Earth) Energy emitted = (T 0 ) 4 Radiant flux or flux density refers to the rate of flow of radiation per unit area (eg., W m -2 ) Irradiance=incident radiant flux density Emittance =emitted radiant flux density
Wien’s Displacement Law As the temperature of a body increases, so does the total energy and the proportion of shorter wavelengths max = (2.88 x 10 -3 )/(T 0 ) *wavelength in metres Sun’s max = 0.48 m Ultraviolet to infrared - 99% short-wave (0.15 to 3.0 m) Earth’s max = 10 m Infrared - 99% longwave (3.0 to 100 m)
8-11 m window
ALBEDO: April, 2002 White and red are high albedo, green and yellow are low albedo
white snow0.80-0.95 old snow0.40-0.60 vegetation0.15-0.30 light colour soil0.25-0.40 dark colour soil0.10 clouds0.50-0.90 calm water 0.10 (noon) March 3, 2009
DAYTIME: Q* = K - K + L - L Q* = K* + L* NIGHT: Q* = L* K = solar (shortwave) radiation ↓ = incoming L = longwave (terrestrial radiation)↑ = outgoing Q* = net all-wave radiation* = net Radiation Balance
Source: NOAA L
Conduction The transfer of heat from molecule to molecule within a substance
Convection and Thermals
Convection The transfer of heat by the mass movement of a substance (eg. air) Rising air expands and cools Sinking air is compressed and warms
The Hydrological Cycle
Heat capacity The amount of heat energy absorbed (or released) by unit volume of a substance for a corresponding temperature rise (or fall) of 1 °C Specific heat The amount of heat energy absorbed (or released) by unit mass of a substance for a corresponding temperature rise (or fall) of 1 °C
Latent heat The heat energy required to change a substance from one state to another Sensible heat Heat energy that we can feel and sense with a thermometer
Thermometer and radiation shield SENSIBLE HEAT Radiation Sensors (PAR and K ) Raingauge Datalogger Photo: Weather station, Tausa, Cundinamarca, Colombia (3,243 m asl)
http://www.jgiesen.de/sunshine/index.htm Check this out:
N
Dec 15, 2004 Jan 19, 2005 Temperature ( C)
Dec 15, 2004 Jan 19, 2005
Dec 15, 2004Jan 19, 2005 Temperature ( C)
10 – 100 m ●
0.0001 – 0.001 m ●
Mie scattering 0.01 to 1.0 m ●
LONG PATH LENGTH OF LIGHT THROUGH THE EARTH’S ATMOSPHERE MOST OF THE THE VIOLET, BLUE AND GREEN LIGHT IS SCATTERED
(from Pacific) (prairie cold)
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Similar presentations | 1,144 | 4,098 | {"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-2020-40 | latest | en | 0.858156 |
https://www.physicsforums.com/threads/combining-poisson-error.792535/ | 1,656,906,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00733.warc.gz | 982,422,881 | 17,962 | # Combining Poisson error
I have low counting stats and need to subtract background, account for efficiency, and divide by volume. How do I combine the asymmetrical (Poisson) errors?
Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
Exactly what kind of analysis are you looking at performing? Are you planning to test a hypothesis? Are you going to make a confidence interval for some parameter?
I want to eventually test the hypothesis that one sample is is greater than the controls. And if two samples are different from each other. This I am ok with- but I have to show all of my calculations for how I can mathematically prove the values are different.
I have small counts in 60 fields of view on a scope and I was propagating error following Gaussian error propagation- which I now know is wrong. But what do I do with these asymmetrical error bars when I want to know sample (+/- error) minus control (+/- error)?
Stephen Tashi
I suggest you have another try at stating your question, unless you are writing to someone on the forum who already knows what kind of experiment you are doing.
How do I subtract a Poisson background from a Poisson sample and propagate the error associated with each?
Stephen Tashi
How do I subtract a Poisson background from a Poisson sample and propagate the error associated with each?
That isn't a description of an experiment. As far as I know, it isn't a description of a specific problem in statistics.
I am counting the number of particles in 60 fields of view on a scope. I count three pieces of a filter for a sample and three pieces of a filter for a control. All of my counts in 60 fields of view are <50 and Poisson distributed.
Stephen Tashi
I am counting the number of particles in 60 fields of view on a scope. I count three pieces of a filter for a sample and three pieces of a filter for a control. All of my counts in 60 fields of view are <50 and Poisson distributed.
Estimation and "proving a difference" are technically two different statistical tasks. Statistics doesn't actually "prove" a difference. There are statistical procedures that make a decision about whether a difference in two situations exists, but these procedures are not proofs. These are regarded as "evidence". They are not a mathematical proof.
With respect to the task of estimation, are you trying to estimate the parameters of a Poission distribution that would account for the difference between the counts on the control filters and the counts on the non-control samples?
With respect to the task of giving evidence for a difference (a task called Hypothesis Testing), how many different situations are there? Are all the non-controls from the same general situation (e.g. from the livers of rats treated with drug X) or are they from different situations (e.g. some from the livers of rats treated with drug X and some from the livers of rats treated with drug Y).
chiro
As others have hinted you need to specify what you are trying to test in terms of parameters (this is what estimators do - they model the parameters with random variables and you use this to make inferences) and also supply assumptions and the kind of data you have.
If you are doing a difference of means then you will be basically doing a hypothesis of something like H0: lambda1 = lambda2 => lambda1 - lambda2 = 0 vs H1: lambda1 > lambda2 or lambda1 != lambda2 or something else.
To use a normal distribution on the mean you need a large sample size. If you are not confident about that then you need to derive the joint distribution for your random variable of lambda1 - lambda2 and then get an interval (using say the likelihood ratio test) and use that to test the hypothesis.
You can do this kind of thing in SAS or R if you have it (R is free and open source and if you've done any statistical or mathematical programming then it will be fairly straightforward) and you can find the site by typing in R project in google.
jim mcnamara
Mentor
What everyone is asking - from a wholly different perspective: You seem to have an XY problem here. You did X and you think Y will solve it. The problem is that you are looking at Y assuming it will fix things. We think that we, that being all of us, need to get to X and start there. Please tell us precisely what you did, and what hypothesis you want to test. And importantly: why? There are lots of smart folks here, it is a virtually given that one of them can help.
http://mywiki.wooledge.org/XyProblem
Svein
I want to eventually test the hypothesis that one sample is is greater than the controls. And if two samples are different from each other. This I am ok with- but I have to show all of my calculations for how I can mathematically prove the values are different.
I have small counts in 60 fields of view on a scope and I was propagating error following Gaussian error propagation- which I now know is wrong. But what do I do with these asymmetrical error bars when I want to know sample (+/- error) minus control (+/- error)?
The standard way of testing for significant difference is:
1. Calculate the mean and standard deviation of both your samples. Call them m1, m2, s1 and s2. Assume that m1 is the mean of the sample you are interested in.
2. Then calculate the mean and standard deviation of the total data set (both samples merged). Call them M and S
3. State the null hypothesis: There is no significant difference
4. Then calculate $\frac{(M-m1)}{S}$. This tells you how many standard deviations your sample is from the merged mean
5. From that number, you can calculate the probability of the null hypothesis being true.
Stephen Tashi
From that number, you can calculate the probability of the null hypothesis being true.
You can't calculate the probability that the null hypothesis is true.
You can only assume the null hypothesis is true and calculate the probability that a number computed from the data is in some subset of the real numbers.
Svein | 1,296 | 5,967 | {"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} | 3.171875 | 3 | CC-MAIN-2022-27 | latest | en | 0.943227 |
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tigger,2000. 1400 if you're running.There's an old Roman Empire army trick (I didn't learn it myself in the ancient Roman army. I learned it in history studies--it's amazing what you can learn when... Read More »
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https://encyclopediaofmath.org/wiki/Nephroid | 1,721,702,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00212.warc.gz | 194,640,553 | 5,848 | # Nephroid
2020 Mathematics Subject Classification: Primary: 52A10 [MSN][ZBL]
An epicycloid with parameter $m=2$; an algebraic plane curve with equation $$x= 3r \cos\theta-r\cos\left[3\theta\right] \,,$$ $$y= 3r \sin\theta-r\sin\left[3\theta\right] \ .$$
The nephroid is the catacaustic of the cardioid with respect to a cusp, and of a circle with respect to a point at infinity; the evolute of a nephroid is another nephroid.
The nephroid of Freeth is the strophoid of a circle with respect to its centre and a point on the circumference. It has equation $$r = a(1 + 2\sin(\theta/2)) \ .$$
#### References
• J.D. Lawrence, "A catalog of special plane curves" , Dover (1972) ISBN 0-486-60288-5 Zbl 0257.50002
How to Cite This Entry:
Nephroid. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Nephroid&oldid=54470 | 269 | 849 | {"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} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.738502 |
https://dokumen.tips/documents/thursday-feb-12-th-p-170-171-thursday-feb-12-th-170-21215-thurs-lt.html | 1,563,217,321,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00206.warc.gz | 381,507,726 | 18,680 | # Thursday, Feb. 12 th p. 170, 171. Thursday, Feb. 12 th 170 2/12/15 Thurs. L.T.: I can collaboratively use claim and evidence to interpret graphs of speed.
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Thursday, Feb. 12th
Thursday, Feb. 12th p. 170, 171Thursday, Feb. 12th 1702/12/15 Thurs.L.T.: I can collaboratively use claim and evidence to interpret graphs of speed and acceleration.DO NOW: What is this graph trying to tell you? How do you know?
..Reflection:Did I work collaboratively with my group? Did I contribute to making claim and evidence statements? Did I write on the notecard? 171Title: Interpreting GraphsA. B.C.D.
Claim and Evidence Card RotationGoal: to collaboratively use claim and evidence to interpret graphs of speed and acceleration.Success Criteria: 1. Everyone discusses the talking points to interpret the graph together.2. Everyone writes on the index card to turn in to Ms. Kim.3. Everyone writes the complete claim and evidence in their SNB.
Claim and EvidenceClaim: Explaining what you think something is telling you
Ex: Chocolate is the best ice cream flavor.
Evidence: Proof or back up of why you think this... the data that shows what youre claiming is accurate.
Ex: A poll that asked 1,000 people showed that 856 people chose chocolate as the best ice cream flavor.Not an example: Because its way better than vanilla. JobsClaim Person: leads the discussion on the claim and writes it on the index card.Evidence 1: leads the discussion on 1st piece of evidence and writes it on the card.Evidence 2: leads the discussion on the 2nd piece of evidence and writes it on the card.Person 4 (Synthesizer): leads the discussion on whether or not you have a full claim and evidence. You make sure that the final product is what you want to turn in.Sketch this graph into the top half of Box A.At Ms. Kims signal, we will start our discussion circles.
Sketch this graph into the top half of Box B.At Ms. Kims signal, we will start our discussion circles.
Sketch this graph into the top half of Box C.At Ms. Kims signal, we will start our discussion circles.
Sketch this graph into the top half of Box D.At Ms. Kims signal, we will start our discussion circles.
Ignore this; look at the red line at A, B, and C! | 550 | 2,257 | {"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-2019-30 | longest | en | 0.899839 |
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Parking Cars (Posted on 2004-07-01)
There is an east-west street of length L units. And we park cars of unit length along the north side until we can't place any more cars. Each car is placed randomly (uniformly).
What is the expected number of cars that can be parked (as a function of L)?
__________________________
I'll start you off...
For 0 <= L < 1, F(L) = 0
For 1 <= L < 2, F(L) = 1
Okay... now the easy ones are out of the way, can you describe the function for L>=2?
No Solution Yet Submitted by SilverKnight Rating: 4.0000 (2 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re: Different? | Comment 8 of 14 |
(In reply to Different? by Gamer)
For L = 2.9, why would it be hard to park only 2 cars? Certainly no more than 2 will fit. Simulation shows that 95% of the time two would fit, and the other 5% of the time only one. Here's a finer breakdown between L=1 and L=3:
`1.00 1.001.10 1.001.20 1.001.30 1.001.40 1.001.50 1.001.60 1.001.70 1.001.80 1.001.90 1.002.00 1.002.10 1.192.20 1.342.30 1.462.40 1.572.50 1.662.60 1.762.70 1.822.80 1.892.90 1.953.00 2.00`
Note that a length of 3 is assured of fitting two cars, and there's zero probability of fitting 3. The latter would require that the first car to arrive would place its front exactly at the beginning of the curb, exacty 1/3 of the way back or exactly 2/3 of the way back, which, being a finite number of points on a line, add to zero probability.
Edited on July 2, 2004, 12:45 am
Posted by Charlie on 2004-07-02 00:34:05
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https://www.geeksforgeeks.org/order-statistic-tree-using-fenwick-tree-bit/ | 1,709,602,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00108.warc.gz | 768,051,544 | 61,529 | # Order statistic tree using fenwick tree (BIT)
Given an array of integers with limited range (0 to 1000000). We need to implement an Order statistic tree using fenwick tree.
It should support four operations: Insert, Delete, Select and Rank. Here n denotes the size of Fenwick tree and q denotes number of queries.
Each query should be one of the following 4 operations.
• insertElement(x) – Insert element x into Fenwick tree, with O(log n) worst case time complexity
• deleteElement(x) – Delete element x from fenwick tree, with O(log n) worse case time complexity
• findKthSmallest(k) – Find the k-th smallest element stored in the tree, with O(log n * log n) worst case time complexity
• findRank(x) – Find the rank of element x in the tree, i.e. its index in the sorted list of elements of the tree, with O(log n) time complexity
Prerequisite: Binary Indexed Tree or Fenwick Tree
The idea is to create a BIT of size with maximum limit. We insert an element in BIT using it as an index. When we insert an element x, we increment values of all ancestors of x by 1. To delete an element, we decrement values of ancestors by 1. We basically call standard function update() of BIT for both insert and delete. To find rank, we simply call standard function sum() of BIT. To find k-th smallest element, we do binary search in BIT.
## C++
`// C++ program to find rank of an element``// and k-th smallest element.``#include ``using` `namespace` `std;` `const` `int` `MAX_VAL = 1000001;` `/* Updates element at index 'i' of BIT. */``void` `update(``int` `i, ``int` `add, vector<``int``>& BIT)``{`` ``while` `(i > 0 && i < BIT.size())`` ``{`` ``BIT[i] += add;`` ``i = i + (i & (-i));`` ``}``}` `/* Returns cumulative sum of all elements of`` ``fenwick tree/BIT from start upto and`` ``including element at index 'i'. */``int` `sum(``int` `i, vector<``int``>& BIT)``{`` ``int` `ans = 0;`` ``while` `(i > 0)`` ``{`` ``ans += BIT[i];`` ``i = i - (i & (-i));`` ``}` ` ``return` `ans;``}` `// Returns lower bound for k in BIT.``int` `findKthSmallest(``int` `k, vector<``int``> &BIT)``{`` ``// Do binary search in BIT[] for given`` ``// value k.`` ``int` `l = 0;`` ``int` `h = BIT.size();`` ``while` `(l < h)`` ``{`` ``int` `mid = (l + h) / 2;`` ``if` `(k <= sum(mid, BIT))`` ``h = mid;`` ``else`` ``l = mid+1;`` ``}` ` ``return` `l;``}` `// Insert x into BIT. We basically increment``// rank of all elements greater than x.``void` `insertElement(``int` `x, vector<``int``> &BIT)``{`` ``update(x, 1, BIT);``}` `// Delete x from BIT. We basically decreases``// rank of all elements greater than x.``void` `deleteElement(``int` `x, vector<``int``> &BIT)``{`` ``update(x, -1, BIT);``}` `// Returns rank of element. We basically``// return sum of elements from start to``// index x.``int` `findRank(``int` `x, vector<``int``> &BIT)``{`` ``return` `sum(x, BIT);``}` `// Driver code``int` `main()``{`` ``vector<``int``> BIT(MAX_VAL);`` ``insertElement(20, BIT);`` ``insertElement(50, BIT);`` ``insertElement(30, BIT);`` ``insertElement(40, BIT);` ` ``cout << ``"2nd Smallest element is "`` ``<< findKthSmallest(2, BIT) << endl;` ` ``cout << ``"Rank of 40 is "`` ``<< findRank(40, BIT) << endl;` ` ``deleteElement(40, BIT);` ` ``cout << ``"Rank of 50 is "`` ``<< findRank(50, BIT) << endl;` ` ``return` `0;``}`
## Java
`// Java program to find rank of an element``// and k-th smallest element.``import` `java.util.Arrays;` `class` `GFG{` `static` `int` `MAX_VAL = ``1000001``;` `// Updates element at index 'i' of BIT``static` `void` `update(``int` `i, ``int` `add, `` ``Integer[] BIT) ``{`` ``while` `(i > ``0` `&& i < BIT.length)`` ``{`` ``BIT[i] += add;`` ``i = i + (i & (-i));`` ``}``}` `// Returns cumulative sum of all elements``// of fenwick tree/BIT from start upto``// and including element at index 'i'.``static` `int` `sum(``int` `i, Integer[] BIT)``{`` ``int` `ans = ``0``;`` ``while` `(i > ``0``) `` ``{`` ``ans += BIT[i];`` ``i = i - (i & (-i));`` ``}`` ``return` `ans;``}` `// Returns lower bound for k in BIT.``static` `int` `findKthSmallest(``int` `k,`` ``Integer[] BIT) ``{`` ` ` ``// Do binary search in BIT[] `` ``// for given value k.`` ``int` `l = ``0``;`` ``int` `h = BIT.length;`` ` ` ``while` `(l < h) `` ``{`` ``int` `mid = (l + h) / ``2``;`` ``if` `(k <= sum(mid, BIT))`` ``h = mid;`` ``else`` ``l = mid + ``1``;`` ``}`` ``return` `l;``}` `// Insert x into BIT. We basically``// increment rank of all elements ``// greater than x.``static` `void` `insertElement(``int` `x, Integer[] BIT)``{`` ``update(x, ``1``, BIT);``}` `// Delete x from BIT. We basically``// decreases rank of all elements ``// greater than x.``static` `void` `deleteElement(``int` `x, Integer[] BIT) ``{`` ``update(x, -``1``, BIT);``}` `// Returns rank of element. We basically``// return sum of elements from start to``// index x.``static` `int` `findRank(``int` `x, Integer[] BIT) ``{`` ``return` `sum(x, BIT);``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ``Integer[] BIT = ``new` `Integer[MAX_VAL];`` ``Arrays.fill(BIT, ``0``);`` ` ` ``insertElement(``20``, BIT);`` ``insertElement(``50``, BIT);`` ``insertElement(``30``, BIT);`` ``insertElement(``40``, BIT);` ` ``System.out.println(``"2nd Smallest element is "` `+ `` ``findKthSmallest(``2``, BIT));`` ``System.out.println(``"Rank of 40 is "` `+ `` ``findRank(``40``, BIT));` ` ``deleteElement(``40``, BIT);`` ` ` ``System.out.println(``"Rank of 50 is "` `+ `` ``findRank(``50``, BIT));``}``}` `// This code is contributed by sanjeev2552`
## Python3
`from` `typing ``import` `List` `# Maximum value of elements``MAX_VAL ``=` `1000001` `def` `update(i: ``int``, add: ``int``, BIT: ``List``[``int``]) ``-``> ``None``:`` ``"""Updates element at index 'i' of BIT."""`` ``while` `i > ``0` `and` `i < ``len``(BIT):`` ``BIT[i] ``+``=` `add`` ``i ``=` `i ``+` `(i & (``-``i))` `def` `sum``(i: ``int``, BIT: ``List``[``int``]) ``-``> ``int``:`` ``"""Returns cumulative sum of all elements of fenwick tree/BIT from start upto`` ``and including element at index 'i'."""`` ``ans ``=` `0`` ``while` `i > ``0``:`` ``ans ``+``=` `BIT[i]`` ``i ``=` `i ``-` `(i & (``-``i))` ` ``return` `ans` `def` `find_kth_smallest(k: ``int``, BIT: ``List``[``int``]) ``-``> ``int``:`` ``"""Returns lower bound for k in BIT."""`` ``# Do binary search in BIT[] for given value k.`` ``l ``=` `0`` ``h ``=` `len``(BIT)`` ``while` `l < h:`` ``mid ``=` `(l ``+` `h) ``/``/` `2`` ``if` `k <``=` `sum``(mid, BIT):`` ``h ``=` `mid`` ``else``:`` ``l ``=` `mid``+``1` ` ``return` `l` `def` `insert_element(x: ``int``, BIT: ``List``[``int``]) ``-``> ``None``:`` ``"""Insert x into BIT. We basically increment rank of all elements greater than x."""`` ``update(x, ``1``, BIT)` `def` `delete_element(x: ``int``, BIT: ``List``[``int``]) ``-``> ``None``:`` ``"""Delete x from BIT. We basically decreases rank of all elements greater than x."""`` ``update(x, ``-``1``, BIT)` `def` `find_rank(x: ``int``, BIT: ``List``[``int``]) ``-``> ``int``:`` ``"""Returns rank of element. We basically return sum of elements from start to index x."""`` ``return` `sum``(x, BIT)` `# Driver code``BIT ``=` `[``0``] ``*` `MAX_VAL``insert_element(``20``, BIT)``insert_element(``50``, BIT)``insert_element(``30``, BIT)``insert_element(``40``, BIT)` `print``(f``"2nd Smallest element is {find_kth_smallest(2, BIT)}"``)` `print``(f``"Rank of 40 is {find_rank(40, BIT)}"``)` `delete_element(``40``, BIT)` `print``(f``"Rank of 50 is {find_rank(50, BIT)}"``)` `#This code is contributed by Shivam Tiwari`
## Javascript
`const MAX_VAL = 1000001;` `function` `update(i, add, BIT) {`` ``while` `(i > 0 && i < BIT.length) {`` ``BIT[i] += add;`` ``i = i + (i & (-i));`` ``}``}` `function` `sum(i, BIT) {`` ``let ans = 0;`` ``while` `(i > 0) {`` ``ans += BIT[i];`` ``i = i - (i & (-i));`` ``}`` ``return` `ans;``}` `function` `findKthSmallest(k, BIT) {`` ``let l = 0;`` ``let h = BIT.length;`` ``while` `(l < h) {`` ``let mid = Math.floor((l + h) / 2);`` ``if` `(k <= sum(mid, BIT)) {`` ``h = mid;`` ``} ``else` `{`` ``l = mid + 1;`` ``}`` ``}`` ``return` `l;``}` `function` `insertElement(x, BIT) {`` ``update(x, 1, BIT);``}` `function` `deleteElement(x, BIT) {`` ``update(x, -1, BIT);``}` `function` `findRank(x, BIT) {`` ``return` `sum(x, BIT);``}` `let BIT = ``new` `Array(MAX_VAL).fill(0);` `insertElement(20, BIT);``insertElement(50, BIT);``insertElement(30, BIT);``insertElement(40, BIT);` `console.log(`2nd Smallest element is \${findKthSmallest(2, BIT)}`);``console.log(`Rank of 40 is \${findRank(40, BIT)}`);` `deleteElement(40, BIT);` `console.log(`Rank of 50 is \${findRank(50, BIT)}`);``// This code is contributed by Shivam Tiwari`
## C#
`using` `System;``using` `System.Collections.Generic;``public` `class` `GFG{`` ``const` `int` `MAX_VAL = 1000001;` ` ``// Updates element at index 'i' of BIT.`` ``static` `void` `Update(``int` `i, ``int` `add, ``int``[] BIT)`` ``{`` ``while` `(i > 0 && i < BIT.Length)`` ``{`` ``BIT[i] += add;`` ``i = i + (i & (-i));`` ``}`` ``}` ` ``// Returns cumulative sum of all elements of BIT from start upto and`` ``// including element at index 'i'.`` ``static` `int` `Sum(``int` `i, ``int``[] BIT)`` ``{`` ``int` `ans = 0;`` ``while` `(i > 0)`` ``{`` ``ans += BIT[i];`` ``i = i - (i & (-i));`` ``}` ` ``return` `ans;`` ``}` ` ``// Returns lower bound for k in BIT.`` ``static` `int` `FindKthSmallest(``int` `k, ``int``[] BIT)`` ``{`` ``// Do binary search in BIT[] for given value k.`` ``int` `l = 0;`` ``int` `h = BIT.Length;`` ``while` `(l < h)`` ``{`` ``int` `mid = (l + h) / 2;`` ``if` `(k <= Sum(mid, BIT))`` ``h = mid;`` ``else`` ``l = mid + 1;`` ``}` ` ``return` `l;`` ``}` ` ``// Insert x into BIT. We basically increment rank of all elements greater than x.`` ``static` `void` `InsertElement(``int` `x, ``int``[] BIT)`` ``{`` ``Update(x, 1, BIT);`` ``}` ` ``// Delete x from BIT. We basically decreases rank of all elements greater than x.`` ``static` `void` `DeleteElement(``int` `x, ``int``[] BIT)`` ``{`` ``Update(x, -1, BIT);`` ``}` ` ``// Returns rank of element. We basically return sum of elements from start to index x.`` ``static` `int` `FindRank(``int` `x, ``int``[] BIT)`` ``{`` ``return` `Sum(x, BIT);`` ``}` ` ``static` `void` `Main(``string``[] args)`` ``{`` ``int``[] BIT = ``new` `int``[MAX_VAL];`` ``InsertElement(20, BIT);`` ``InsertElement(50, BIT);`` ``InsertElement(30, BIT);`` ``InsertElement(40, BIT);` ` ``Console.WriteLine(``"2nd Smallest element is "` `+ FindKthSmallest(2, BIT));` ` ``Console.WriteLine(``"Rank of 40 is "` `+ FindRank(40, BIT));` ` ``DeleteElement(40, BIT);` ` ``Console.WriteLine(``"Rank of 50 is "` `+ FindRank(50, BIT));`` ``}``}``// This code is contributed by Shivam Tiwari`
Output:
```2nd Smallest element is 30
Rank of 40 is 3
Rank of 50 is 3```
```Time Complexity:- O(log N)
Space Complexity:- O(N) ```
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https://quorvita.com/fear/climate-change/q-7/q-7-suppb/ | 1,603,906,700,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900200.97/warc/CC-MAIN-20201028162226-20201028192226-00338.warc.gz | 490,627,251 | 230,307 | ## B. Does Mars Support a Greenhouse Effect?
The IPCC Table relating to Mar indicates that the temperature above a black-body temperature was caused by the Greenhouse Effect. Again no calculation, analysis, proof, or discussion was presented to support the statement.
Mars has a reported average temperature of minus -60 degrees C and an atmosphere containing 95% carbon dioxide. Although the above Table indicates that Mars had an 80% CO2 concentration, most studies place it at 95%. NASA has it at 80% CO2. The difference is sizable suggesting that there was either a math error in the reporting or subsequent Mar’s investigations have improved the data.
The IPCC Table indicates that Mar’s black body temperature is minus -57°C. The average actual temperature on Mars is significantly different. Using the -60°C value reported by multiple sources, then it shows Mars to be colder (4°C) than it would be if it had no atmosphere. That does not make sense. According to Kirchhoff’s Radiation Law, the existence of an atmosphere should slow the cooling/heating process. Using the IPCC value of -47°C from the above Table above then it shows that Mars is about 10°C warmer than the black-body temperature. This would be consistent with what the expected should be, i.e. higher than the black body temperature.
Wikipedia reports an average high of minus -5.7°C and an average low of minus -78.5°C with an arithmetic average being minus -42.5. [ https://en.wikipedia.org/wiki/Climate_of_Mars%5D A measurement at one crater does not translate to the entire Mars surface. However, it does provide some support for the IPCC minus -47°C value.
Wikipedia reports that the Mars atmosphere is much thinner than on Earth, i.e. less than 1% (0.006atm) of Earth’s atmosphere. Because the IPCC contends the entire 523°C temperature effect on Venus was caused by the CO2 atmosphere, and Mars has a similar concentration of CO2. One of the reasons given why Mars is not warmer via the Greenhouse Effect is because the atmosphere is too thin. The Earth’s Stratosphere is similar to Mars, i.e. about 0.006 atm on Mars and .01atm at mid-height of the Stratosphere. Both have similar temperatures. The probability of an infrared photons colliding with CO2 molecules in the Stratosphere is determining by multiplying the density (pressure) and the mole ratio of CO2 in each atmosphere. This shows that the probability of infrared absorption on Mars is 140 times higher than in the Earth’s Stratosphere. Therefore, if the thin atmosphere on Mar’s is the justification why Mars is not warmer, then the arguments that the Stratosphere on Earth is where CO2 does its Greenhouse Effect work are inconsistent by over 100 times. The physical gas and radiation laws on Mars should be the same as they are on Earth.
NASA also supports the argument that a thin atmosphere allows the radiation to escape unimpeded in the Stratosphere. Actually NASA suggests that a thin atmosphere where absorption is unimpeded occurs at an elevation below the Stratosphere.
“Heat radiated upward continues to encounter greenhouse gas molecules; those molecules absorb the heat, their temperature rises, and the amount of heat they radiate increases. At an altitude of roughly 5-6 kilometers, the concentration of greenhouse gases in the overlying atmosphere is so small that heat can radiate freely to space.” Emphasis added https://earthobservatory.nasa.gov/features/EnergyBalance/page6.php
The Stratosphere starts at about 12 kilometers. At 5 to 6 kilometers, the pressure is about .5 atmospheres and the water vapor drops to about 1600 ppmv (mixing ratio of 1 g/kg). This concentration remains dominates over second place CO2.
Often common sense explains observations. Mars is cold because it is too far away from the sun, has a mass of one-tenth that of Earth and cannot hold on to any significant atmosphere, and has a wide elliptical orbit. It also has no ocean to modulate the temperature. The Greenhouse Effect plays no measurable role. | 887 | 4,033 | {"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} | 2.875 | 3 | CC-MAIN-2020-45 | latest | en | 0.900673 |
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Thoroughness:
3.9
Creativity:
3.9
Clarity:
Hard Goods
N/A
Shipping Efficiency:
N/A
Overall Quality:
N/A
Accuracy:
N/A
Practicality:
N/A
Thoroughness:
N/A
Creativity:
N/A
Clarity:
Used Goods
N/A
Shipping Efficiency:
N/A
3.9
Total:
382 total vote(s)
TEACHING EXPERIENCE
I have had the privilege of working in education for over 20 years, spending time in both small and large primary school settings.
MY TEACHING STYLE
I promote hands on and engaging tasks for point of need. I want to support all students to be challenged at their own level.It is an honor to work with students and teachers everyday to be the best we can be.
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
I have taught in small and large schools, in both city and rural areas. I have been a classroom teacher, Assistant Principal and a Curriculum/Mathematics Coach.
Welcome to iSURF! Kerry is an Assistant Principal, she spend her days at a regional primary school in Victoria, Australia. Three ways to sum Kerry up, professionally and in my spare time… Professional Kerry: curriculum juggler, Professional Learning Team (PLT) support crew and brainteaser. Kerry’s spare time: urban escapologist, passport stamper and Rogainer (Google that!). Kerry has a passion for great curriculum that engages, captivates and inspires. iSURF promotes Mathematical thinking and supports individual achievement of the proficiencies... Solving problems, Understanding, Reasoning and Fluency (iSURF!). | 1,364 | 5,549 | {"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-2018-05 | latest | en | 0.894358 |
http://www.resistors-and-diodes-and-picchips-oh-my.co.uk/?cat=17 | 1,726,645,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00593.warc.gz | 48,779,996 | 5,953 | My Own Personal Yellow Brick Road
I just thought I’d put together a collection of some of the cool things I’ve come across on looking into mechanisms and structures, it might be a bit disjointed but hopefully it’ll provide some inspiration The self balancing bookshelf: What a conversation starter, I think when I move next I might just have to make […]
So here he is in all his glory: And finally… in flight! (sorry Dean didn’t have access to my video editing software!) You can see he’s still got a bit of clank and clunk in the movement but I think that it’s caused buy the weight rather than the mechanism. The Pegasus was made out of […]
Categories: Mechanisms | Add a Comment
So, first, what is going to make my dragon fly? As a said before I think that the best mechanism to replicate the movement of the Pegasus will be a crank. A crank is basically a lever attached to a rotating shaft, the diagrams below are from a great book ‘Cabaret Mechanical Movement’ by Aidan […]
I’ve always loved dragons so I’m going to have a go at making an automata of a dragon flying. I got most of my inspiration from an automata made by a guy called Keith Newstead: I love the smooth motion! I think I’ve worked out that it’s made from a crank to describe the elliptical […]
Right, like linkages drive mechanisms (the collective term for pulley and gear systems) connect other mechanisms together but this time using rotary motion. This may also involve gearing or changing the angle or direction of the motion. There are two types of drives: positive and friction. Gears fall into the positive category as they are […]
Now we’re getting into the details… So, ready for more mathsy goodness? There are three important formulas to remember when dealing with mechanisms… Mechanical Advantage This is basically a comparison of the effort put in to the load moved, this leverage is the ration of the distances of the effort and load to the fulcrum. […]
Turning motions are really easy to understand, take this set of scales: The right hand side is turnign clockwise and the left hand side is turning anti clockwise. To balance the scales or bring them into equilibrium the clockwise turning motion has to balance the anticlockwise motion. To work this out you need to use […] | 480 | 2,270 | {"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-2024-38 | latest | en | 0.946095 |
https://quant.stackexchange.com/questions/72165/volatility-modelling-negative-gjr-garch-x-coefficient?noredirect=1 | 1,718,980,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00668.warc.gz | 425,686,634 | 41,297 | # Volatility Modelling negative GJR-GARCH-X coefficient
I have estimated GARCH and GJR-GARCH with several exogenous variables. Some of the exogenous variables have negative coefficients that are statistically significant. For instance, I can write my GJR-GARCH estimate as:
$$h_t = 0.213 + 0.011 u_{t-1}^2 + 0.847 h_{t-1} + 0.196 u^2_{t-1}I_{t-1<0} - 0.026 X_{t-1}$$
Where $$u_{t-1}^2$$ is the lagged ARCH term, $$I_t$$ is the dummy variable that models the asymmetric leverage effect and $$X_{t-1}$$ is the exogenous variable.
My main concern is that the coefficient of $$X_{t-1}$$ is negative. How can I verify that my volatility is not negative? Is just a plot of the conditional variance over the in-sample period is enough?
I have tried to estimate the GJR-GARCH-X model, however, I obtained some negative coefficients. I am not sure how to verify that these negative coefficients will cause the volatility to be negative.
### Positivity of GARCH-X models:
When it comes to GARCH models with exogenous regressors, it is more "tricky" to ensure positivity of the model, due to the unspecified model-dynamics of your exogenous regressor (moreover exogenous regressors can have drastically different characteristics). Without any model-specification on the exogenous regressor $$X_{t-1}$$, it is common to restrict the parameter-space of the model in order to ensure positivity.
For simplicity, let us vaguely define the GJR-GARCH(1,1)-X model with demeaned returns, $$r_t$$: \begin{align*} r_t \vert \mathcal{F}_{t-1} &= \varepsilon_t\\ \varepsilon_t &= \sigma_t \cdot z_t\\ \sigma^2_t &= \omega + \alpha \varepsilon_{t-1}^2 + \beta \sigma_{t-1}^2 + \gamma I_{t-1} \varepsilon_{t-1}^2 + \nu X_{t-1}, \end{align*} where $$z_t \overset{iid}{\sim} D(0,1)$$ is a standardized distribution and
$$I_{t-1} =\begin{cases} 1 & \text{if } \varepsilon_{t-1} < 0 \\ 0 & \text{if } \varepsilon_{t-1} \geq 0 \end{cases}.$$
I have detailed some of my observations:
• When working under the GJR-GARCH(1,1) positivity is satisfied when we impose $$\omega, \beta,\alpha > 0$$ and $$\alpha + \gamma > 0$$. The latter condition is a broader statement than imposing $$\alpha, \gamma >0$$, since we can allow one of the parameters to become negative (in this case, $$\gamma$$). I have made an in-depth answer detailing the GJR-GARCH(1,1) model, positivity, covariance stationarity and economical interpretations of the parameter estimates.
• When working under the GJR-GARCH(1,1)-X we can further ensure positivity by additionally restricting the exogenous regressor such that $$\omega + \nu X_{t-1} \geq 0$$. Here, we allow $$\nu$$ to vary freely (since $$\omega > 0$$) as long as the above condition is satisfied. The primary motivation for the extra restriction, comes from observing the unconditional variance (calculated under assumed covariance stationarity) of the GJR-GARCH-X model:
$$$$\mathbb{V}ar(r_t) = \frac{\omega + \nu \mathbb{E}\left[X_{t-1}\right]}{1 - \alpha - \beta - \kappa \gamma}.$$$$ In order to ensure non-negative unconditional variance of the return process, we specifically need $$\omega + \nu \mathbb{E}\left[X_{t-1}\right] \geq 0$$, which is satisfied when imposing $$\omega + \nu X_{t-1} \geq 0$$ for all $$t$$.
In conclusion, imposing $$\omega, \beta, \alpha > 0$$, $$\alpha + \gamma > 0$$ and $$\omega + \nu X_{t-1}>0$$, ensures that you obtain non-negative volatility estimates.
When $$X_{t-1}$$ is strictly positive, it is common to let $$\nu \geq 0$$. This is also emphasized in the article of Han, H. (2015) that investigates asymptotic results of the GARCH-X model when $$X_{t-1}$$ follows a fractionally integrated process. In general, it is common in academia to assume a functional form on the exogenous regressor, (see for instance (C1 - C4) in this paper p.699). This is also done in the Realized GARCH model that incorporates intraday data to procure better forecasts (I have detailed this model here, if you're interested). I hope this helps.
• This is very helpful response. However, I have couple of question to ask and will be grateful if you can answer for me: 1- My exogenous variable $X_{t}$ is a news sentiment variable that takes negative values with negative news and positive with positive news i.e. it is not strictly positive. My second question comes in a different comment due to the number of characters restrictions. Commented Sep 9, 2022 at 18:40
• 2- I am using rugarch package on R for my model, this is my code: ug_spec_x <- ugarchspec(mean.model = list(armaOrder=c(0,0)), variance.model = list(model="sGARCH", garchOrder = c(1,1), external.regressors = matrix(C\$all.news)) ,distribution.model = "std" ) setbounds(ug_spec_x) <- list(vxreg1=c(-1,1)) ugfit_x <- ugarchfit(spec = ug_spec_x, data = C\$GSPC.Adjusted, solver= "solnp") How can I impose the restriction you advised me in my code ? i.e. $\omega + \nu X_{t-1} \geq 0$ Commented Sep 9, 2022 at 18:41
• Thank you Pleb! I will wait to hear back from you Commented Sep 10, 2022 at 0:08
• @Moataz Most of the papers I have been skim-reading works with a positive exogenous covariate. In that regard, I would transform my exogenous regressor into a GJR-type equation, $\nu I_{\{{X_{t-1}<0}\}} \varepsilon_{t-1}^2$. This will intuitively, give you an understanding on how negative shocks in your sentimental news covariate, affects future volatility of returns. This is very similar to the original GJR-term. Therefore, be aware that bad news are already "incorporated" into past squared returns and in such case, the original GJR-term will likely catch most of the asymmetrical response.
– Pleb
Commented Sep 10, 2022 at 10:11
• Hi Pleb, thank you for your comemnt and your suggestion. If I followed your suggestion, then my unconditional variance will be $\frac{\omega}{1- \alpha - \beta - \frac{\gamma}{2} - \nu }$ ? Commented Sep 10, 2022 at 12:30 | 1,679 | 5,864 | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.774185 |
https://essayusa.net/math810-mini-project-3-statistics-homework-help/ | 1,721,061,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00263.warc.gz | 219,940,378 | 12,998 | # Math810 mini project 3 | Statistics homework help
* Word Document and Excel Sheet Attached
Synopsis
Marketing and Advertising Analysis
You are the regional marketing vice president overseeing all US marketing for an international pharmaceutical distributor. Your team has recently submitted a proposed budget for advertising and marketing spending for the upcoming year to support 10% annual revenue growth for your company’s best-selling product, Dilomatox. A summary of that budget along with this year’s forecasted data (forecasted since your fiscal year isn’t quite complete yet) is below:
DILOMATOX – Proposed Marketing Budget (DATA TABLE IN WORD DOCUMENT)
Proposed Budget (Next Year) Current Forecast (This Year) % Change
Advertising and Marketing Spending (total) \$64,250,000 \$56,860,000 +13%
Product Revenue \$1,164,471,000 \$1,058,610,000 +10%
Marketing Spending as % of Revenue 5.52% 5.37%
CURRENT YEAR FINANCIAL FORECASTS (TABLE IN ATTACHED WORD DOC);
Dilomatox Zoraffil % Change
Advertising and Marketing Spending (total) \$56,860,000 \$52,040,000 -8.5%
Product Revenue \$1,058,610,000 \$1,138,510,000 +7.5%
Marketing Spending as % of Revenue 5.37% 4.57%
To support their findings, the committee has supplied your team with the attached data file, providing weekly marketing spending and revenue (in millions of dollars) for the last 52 weeks for both brands.
Your task is to analyze this data, ‘uncover the story’ behind how advertising spend and revenue for these brands are related (or not!), and to write a managerial summary that you can use to justify your proposed advertising and marketing budget. You should organize your summary in a way that provides a strong and coherent argument, but in that argument your analysis should answer all of the following questions:
1. Answer parts a-b below:
1. Describe the relationship between advertising and revenue for Dilomatox. Would you characterize these relationships as strong or weak? Support your response with relevant graphs and statistics.
2. Describe the relationship between advertising and revenue for Zoraffil. Would you characterize these relationships as strong or weak? Support your response with relevant graphs and statistics.
2. Analyze the multivariate relationship between Dilomatox’s revenue and the other variables provided (Dilomatox’s marketing spend, Zoraffil’s revenue, and Zorafill’s marketing spend). Is there a significant relationship between Dilomatox’s sales and any (or all) of these variables? Support your response with relevant charts or statistics.
3. What percent of the variation in revenue does advertising and marketing spend explain for both brands? Explain.
4. Based on your analysis, if both brands ceased all advertising and marketing spend, how much revenue would be lost? Explain.
5. What impact will the CFO’s proposed \$11 million dollar cut to your budget have on Dilomatox revenue next year?
Your managerial summary should include a description of the statistical tests or processes used to answer each question, an explanation of the necessary results (appropriate descriptive or graphical summaries, statistics like r-values and least-squares regression equations, predicted values – and if appropriate estimates of error for any parameters or predictions made). It should also show that any required assumptions for any statistical procedures used are valid. Use a 95% level of significance for any statistical tests.
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At Essay USA, we prioritize on all aspects that bring about a good grade such as impeccable grammar, proper structure, zero-plagiarism and conformance to guidelines. Our experienced team of writers will help you completed your essays and other assignments. | 1,304 | 5,985 | {"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-2024-30 | latest | en | 0.833808 |
https://oeis.org/A141723 | 1,632,167,299,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00210.warc.gz | 486,606,851 | 4,010 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A141723 Triangle T(n, k) = Sum_{j=0..n} (2*n)!/((2*n-k-j)!*j!*k!), read by rows. 1
1, 3, 4, 11, 28, 24, 42, 156, 225, 160, 163, 792, 1596, 1736, 1120, 638, 3820, 9855, 14400, 13230, 8064, 2510, 17832, 55968, 102520, 122265, 100584, 59136, 9908, 81368, 300482, 661024, 968968, 1005004, 765765, 439296, 39203, 365104, 1549320, 3975440, 6910540, 8653008, 8112104, 5845840, 3294720 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Rows n = 0..50 of the triangle, flattened FORMULA T(n, k) = Sum_{j=0..n} (2*n)!/((2*n-k-j)!*j!*k!). EXAMPLE Triangle begins as: 1; 3, 4; 11, 28, 24; 42, 156, 225, 160; 163, 792, 1596, 1736, 1120; 638, 3820, 9855, 14400, 13230, 8064; 2510, 17832, 55968, 102520, 122265, 100584, 59136; 9908, 81368, 300482, 661024, 968968, 1005004, 765765, 439296; MATHEMATICA Table[Sum[Multinomial[2*n-k-j, k, j], {j, 0, n}], {n, 0, 12}, {k, 0, n}]//Flatten PROG (Magma) F:= Factorial; [(&+[F(2*n)/(F(k)*F(j)*F(2*n-k-j)): j in [0..n]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 28 2021 (Sage) f=factorial; flatten([[sum(f(2*n)/(f(k)*f(j)*f(2*n-k-j)) for j in (0..n)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 28 2021 CROSSREFS Sequence in context: A198443 A041231 A042129 * A268478 A180363 A100845 Adjacent sequences: A141720 A141721 A141722 * A141724 A141725 A141726 KEYWORD nonn,tabl AUTHOR Roger L. Bagula, Sep 12 2008 EXTENSIONS Edited by G. C. Greubel, Mar 28 2021 STATUS approved
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Last modified September 20 14:44 EDT 2021. Contains 347586 sequences. (Running on oeis4.) | 860 | 2,056 | {"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-2021-39 | latest | en | 0.468409 |
https://tutorbin.com/questions-and-answers/q4-a-test-rocket-starting-from-rest-at-point-a-is-launched-by-accelera | 1,657,214,597,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00577.warc.gz | 627,772,167 | 17,213 | Question
Modern Physics
Q4) A test rocket starting from rest at point A is launched by accelerating it along a 530.0-mincline at 6.85 m/s/s. The incline rises at 25.0 degrees above the horizontal, and at the instant the rocket leaves it, the engines turn off and the rocket is subject to gravity only (ignore air resistance). Find (a) the maximum height above the ground that the rocket reaches, and (b) the rocket's greatest horizontal range beyond point A.
Verified
### Question 52422
Modern Physics
1A cruiser moves away from the planet Hoth at a speed of 0.5c. A frigate follows, catching up at a speed of 0.25c as viewed from the cruiser. With what speed does the frigate appear to be moving according to observers on the planet? You should use the relativistic velocity transformation, clearly defining all the parameters used.
u^{\prime}=\frac{u-v}{1-\frac{u v}{c^{2}}}
### Question 44658
Modern Physics
Question 5: A spaceman goes out on a trip. First, he travels at 1/3 the speed of light in one direction. Then he comes back home at 2/3 of the speed of light. On Earth 9 years have passed,by how much has the spaceman aged? (Make sure you draw a space-time diagram).
### Question 44657
Modern Physics
Question 4: The typical pole for pole vaulting is 17 feet long. How fast would a pole vaulter have to run (in the direction of the pole) in order for a spectator sitting in the stands to see it 15feet long?
### Question 44656
Modern Physics
Question 3: If you weigh 150 lb in Chicago, how much do you weigh on top of mount Everest?(mount Everest is 8,848 m high).
### Question 44655
Modern Physics
Question 2: As a result of an explosion a stationary object breaks into three pieces. One piece flies east, one west and one north.
Explain why this is impossible.
Now, there are only two pieces; one flies east the other west. One piece weighs three times as much as the other. What is the ratio of the kinetic energies of the two pieces?
### Question 44654
Modern Physics
Question 1: The international Space Station (ISS) orbits the Earth once every 90 mins at an-altitude of 409 km. How high would it have to be in order to be to be in geosynchronous orbit?(make sure you show how you worked it out).
### Question 44179
Modern Physics
Question 6: A "star" made of ideal gas radiates away 10% of its energy while remaining in hydrostatic balance. Assuming that there are no nuclear reactions
Does the radius of the "star" increases or decreases? By how much?
Does the (average) temperature of the "star" increases or decreases? By how much?
### Question 44178
Modern Physics
5: There are two stars, one has twice the luminosity of the other and half the radius.
What is the ratio of their temperatures?
What is the ratio of their (apparent) brightness if the more luminous one is 100 times more distant than the other.
### Question 44177
Modern Physics
Question 4: According to general relativity people who live in hi-rise buildings age at differentrates depending on their floor.
Is that true?
Which floor ages fastest?
### Question 44176
Modern Physics
Question 3: Consider two time-like events P and Q. P is at the space-time origin P=(0,0); Q is at the space origin but one second later Q=(0,1).
Show that the order of events is preserved by all observers travelling at less than the speed of light.
Now events P and Q are space-like. P is still at the origin, but Q is simultaneous with P but one meter to the right 0=(1.0)
.Show that the order of events is not preserved by observers travelling at less than the speed of light. | 871 | 3,577 | {"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-2022-27 | longest | en | 0.900634 |
https://caramenggugurkankandungann.com/qa/question-how-many-joules-does-an-aed-deliver.html | 1,632,522,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00020.warc.gz | 204,017,436 | 8,075 | Question: How Many Joules Does An AED Deliver?
How many joules does a defibrillator deliver?
Typically, biphasic defibrillation begins at 120 joules, with the level increased as needed.
When used for synchronized electrical cardioversion, however, the device delivers a much lower energy level; for example, 30 joules..
Why are defibrillators measured in joules?
Joules are the amount of energy an automated external defibrillator, or AED, delivers in a life-saving shock to the heart of a victim of sudden cardiac arrest (SCA). … Joules are important in AEDs because they determine how much of an electric shock is delivered from the AED through the pads and into the victim.
How many joules does it take to defibrillate a child?
Pediatric Defibrillation — Current AHA Guidelines With a manual defibrillator (monophasic or biphasic), use a dose of 2 J/kg for the first attempt and 4 J/kg for subsequent attempts.”
How many volts is 200 joules?
Joules To Volts ConversionEnergy in Joules (J)Charge in coulombs (C )Voltage in volts (V)200 joules to volts6 coulombs33.33 volts250 joules to volts7 coulombs35.71 volts300 joules to volts8 coulombs37.5 volts350 joules to volts9 coulombs38.88 volts7 more rows
Can a defibrillator kill you?
No, you can do no harm with a defibrillator (AED). They will only allow an electrical shock to be delivered to the heart of someone who needs it. A shock cannot be delivered in error. When someone has a cardiac arrest, life cannot be sustained.
How many joules do you shock with?
Apply defibrillator pads (or paddles) and shock the patient with 120-200 Joules on a biphasic defibrillator or 360 Joules using a monophasic. Continue High Quality CPR for 2 minutes (while others are attempting to establish IV or IO access).
How many joules is in a Volt?
Where: voltage is in Volts, J is the work or energy in Joules and C is the charge in Coulombs. Thus if J = 1 joule, C = 1 coulomb, then V will equal 1 volt.
How much shock does an AED deliver?
An AED delivers a 3000-volt charge in less than 0.001 of a second. That’s enough electricity to light a 100-watt bulb for 23 seconds. The unit then instructs the user to immediately begin CPR. After two minutes, the unit will perform another analysis to see if defibrillation is needed again.
How many volts is 360 joules?
Defibrillators need clear, reliable power that rapidly accumulates in capacitors. This can be between 200 volts and 100 volts, at 360 joules, and 45 amps. The shock lasts for nearly about 45 seconds. After that, the capacitors needed to be freshly charged.
Can an AED machine kill you?
A manual defibrillator can cause Cardiac Arrest and then death if it is not reversed. An AED will not discharge or deliver a shock to anyone awake (or not) with a non-shockable rhythm.
What are the 3 shockable rhythms?
Shockable Rhythms: Ventricular Tachycardia, Ventricular Fibrillation, Supraventricular Tachycardia.
How much voltage is in a defibrillator?
Simply speaking, a defibrillator works by using a moderately high voltage (something like 200–1000 volts) to pass an electric current through the heart so it’s shocked into working normally again. | 797 | 3,169 | {"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-2021-39 | latest | en | 0.878856 |
https://scuteee.com/courses/sophomore/semiconductor/%E8%B4%B9%E7%B1%B3%E8%83%BD%E7%BA%A7%E4%B8%8E%E8%BD%BD%E6%B5%81%E5%AD%90%E7%BB%9F%E8%AE%A1%E5%88%86%E5%B8%83 | 1,632,509,450,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00206.warc.gz | 536,716,978 | 17,874 | $\newcommand{\dif}{\mathop{}\!\mathrm{d}} \newcommand{\p}{\partial}$
# 费米能级
## 定义与性质
$f(E)=\frac{1}{1+\exp\left( \frac{E-E_f}{kT} \right)}$
1. T=0K 时，$E<E_f, f(E)=1$；$E>E_f, f(E)=0$
2. T>0K 时，
$f(E)= \begin{cases} 1/2<f<1 & E<E_F\\ 1/2 & E=E_F\\ 0<f<1/2 &E>E_F \end{cases}$
## 费米能级的物理意义
$由近似关系：\\ \begin{cases} \frac{\sqrt{\chi^2+4}-\chi}{4\chi}=\frac{\sqrt{1+4/\chi^2}-1}{4}\approx\frac{1+\frac{1}{2}\frac{4}{\chi^2}-1}{4}=\frac{1}{2\chi^2}（泰勒展开）\\ \frac{2\chi}{\sqrt{\chi^2+4}+\chi}=\frac{2}{\sqrt{1+4/\chi^2}+1}\approx1 \end{cases} \\ 从而，\\ \begin{cases} E_f=E_C+kT\ln\left( \frac{N_D}{N_C} \right)\\ n_0=N_D \end{cases}$ | 340 | 630 | {"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": 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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-39 | latest | en | 0.133049 |
https://scienceoxygen.com/how-many-kilocalories-are-in-a-calorie-chemistry/ | 1,685,574,933,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00041.warc.gz | 577,475,364 | 24,286 | How many kilocalories are in a calorie chemistry?
Therefore, you don’t need to convert them, as 1 kilocalorie equals 1 calorie in nutrition. Calories may also be expressed as kilojoules (kJ). One calorie (kcal) equals 4.18 kJ or 4,184 joules (J) ( 1 ).
How do you calculate food value in chemistry?
1. You product is made of 25% ingredient A and 75% ingredient B. Ingredient A contains 10% fat and 50% carbohydrates. Ingredient B contains 7% protein.
2. The overall composition will be: Fat: 10 / 100 * 25 + 0 =2,5% Carbohydrates: 50 / 100 * 25 = 12,5% Protein: 7 / 100 * 75 = 5,25%
How do you calculate kcal in physics?
Here’s your equation: MET value multiplied by weight in kilograms tells you calories burned per hour (MET*weight in kg=calories/hour). If you only want to know how many calories you burned in a half hour, divide that number by two. If you want to know about 15 minutes, divide that number by four.
What is kilocalorie in chemistry?
A kilocalorie (kcal) is the total amount of energy that is required in order to raise the temperature of one kilogram of water by 1 degree Celsius at one atmosphere of pressure. The kilocalorie is still used in chemistry.
Is kcal the same as cal?
The “calorie” we refer to in food is actually kilocalorie. One (1) kilocalorie is the same as one (1) Calorie (uppercase C). A kilocalorie is the amount of heat required to raise the temperature of one kilogram of water one degree Celsius.
How do you calculate heat in kcal?
Calculate specific heat as c = Q / (mΔT) . In our example, it will be equal to c = -63,000 J / (5 kg * -3 K) = 4,200 J/(kg·K) .
How do you convert kcal to grams?
To convert a gram measurement to a calorie measurement, multiply the weight by the conversion ratio. The weight in calories is equal to the grams multiplied by 7.716179.
How do you find energy in joules in chemistry?
Multiply the mass of the object by its specific heat capacity and by the amount of temperature change. This formula is written H = mcΔT, where ΔT means “change in temperature.” For this example, this would be 500g x 4.19 x 20, or 41,900 joules.
What is kJ mol in chemistry?
kj/mol is the abbreviation of kilojoules per mole which is a term used to describe the amount of energy in kilojoules that is put into a number of moles of atoms.
How do you convert kJ mol to kJ?
Re: How to convert kJ/mol to J Answer: Cancel out the 1/mol unit by dividing by the Avogadro constant. Then convert kJ to J by multiplying the kJ value by 1000 (because of the conversion factor 1 kJ = 1000 J).
How do you calculate calories from energy?
1. Multiply grams of carbohydrate in the food by 4 calories per gram. A calorie is a unit of how much energy is in a given amount of food, also called a kcal.
2. Multiply grams of protein in the food by 4 calories per gram.
3. Multiply grams of fat in the food by 9 calories per gram.
What is the kinetic energy formula?
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.
How do you calculate ke?
Kinetic energy formula KE = 0.5 * m * v² , where: m – mass, v – velocity.
Which unit is kcal?
Heat Energy One kilocalorie (1 kcal or 1000 calories) is the amount of heat (energy) needed to raise the temperature of one kg of water by one degree Celsius (°C). The SI standard unit for energy is Joule (J).
How do you convert kcal to KG?
To convert a calorie measurement to a kilogram measurement, multiply the weight by the conversion ratio. The weight in kilograms is equal to the calories multiplied by 0.00013.
What is the difference between 1cal and 1cal?
To ease calculations, energy is expressed in 1000-calorie units known as kilocalories. That is, 1 Calorie is equivalent to 1 kilocalorie; the capital C in Calories denotes kcal on food labels, calories and kilocalories are used interchangeably to mean the same thing.
Is kcal an SI unit?
In physics and chemistry the word calorie and its symbol usually refer to the small unit; the large one being called kilocalorie. However, this unit is not officially part of the metric system (SI), and is regarded as obsolete, having been replaced in many uses by the SI unit of energy, the joule (J).
What is Q in Q MC ∆ T?
Q = mc∆T. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K)
How many kcal are in a kg?
There are 7,700kcals (kcal=calorie) worth of energy in 1kg of fat. That means in order to burn 1kg of fat, you must have a calorie deficit of 7,700.
How many kcal are in a gram of protein?
Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram.
Is kJ mol the same as kJ?
Usually it seems that delta H for an entire reaction is in kJ, since there are varying numbers of moles depending on which product or reactant it is. If the question asks for kJ per mole of product, then you would have to divide the enthalpy by the specific number of moles and give the answer in kJ/mole.
How do you convert kJ to moles?
The prefix “kilo” means 1,000, so one kJ = 1,000 J. As the energies associated with a single molecule or atom are quite small, we often find it easier to discuss the energy found in one mole of the substance, hence “per mole”. To get the energy for one molecule, divide kJ/mol by Avogadro’s number, 6.022 x 1023. | 1,403 | 5,367 | {"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} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.875836 |
http://math.stackexchange.com/questions/444358/the-birthday-problem | 1,469,795,147,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00314-ip-10-185-27-174.ec2.internal.warc.gz | 159,386,778 | 20,708 | # The Birthday Problem
I've been reading about the birthday problem which, as I'm sure many of you will know, is a statistical problem which aims at finding out the how many people you would need in a random group to be certain that two of them shared a birthday. I've read the wikipedia article and am happy with the concept and the answers to this problem. What I'm interested in doing is expanding the principle. I've been trying to work out the answer to a similar problem, but where you simply wanted to know the probability that two people were born in the same week and in the same month. I'm not really sure how to go about this, though, so my first question is: is there a general equation I can use to extend the problem to these cases? But, I know there are many articles and stackexchange questions on this, so I wouldn't ask unless I had a specific problem, which is this:
Suppose a person has met 500 people in their lifetime. What is the probability that seven of those 500 share a birthday in the same two month period?
I think the answer to my last question is that it's certain. But could I ask what is the smallest number of people you would need for the probability that - in a group of 500 people - the probability of them sharing a birthday in a two month period is less that 50%? If that makes sense?
Okay, thank you everybody, edited to tidy up:
Question 1: What is the smallest group of randomly selected people required such that the probability that two of them share a birthday within one week of each other is at least 75%?
Question 2: What is the smallest group of randomly selected people required such that the probability that two of them share a birthday within thirty days of each other is at least 75%?
Question 3: What is the smallest group of randomly selected people required such that the probability that seven of them share a birthday within sixty days of each other is at least 75%?
Question 4: In a group of 30 randomly selected people, what is the probability that seven of them will share a birthday with fifty days?
I hope that's a lot clearer. I had no idea how to word these questions until I posted this and am grateful to everyone who's contributed for helping me do so :)
-
The typical Birthday problem is something along the lines "What is the probability of two people share a birthday in a group of 23 people?", "Determine the size of the group, such that the probability exceeds 50% (or 90%)". You will not get certainity unless the group has more people than the year has days. Please specify, what you are looking for exactly. – Tomas Jul 15 '13 at 17:55
Please ask your question first, then get into the motivation. Readers are trying to figure out if they can help you, and having to read a long paragraph about you doesn't help them help you in the least. – Thomas Andrews Jul 15 '13 at 18:00
As suggested by Henry, questions 1 and 2 are answered by one generalization Wikipedia gives for the birthday problem, for near-misses. Take $m=365$, $k=7$ or $k=30$ respectively, and determine the smallest $n$ where $$p(n,k,m)=1-\frac{(m-nk-1)!}{m^{n-1}(m-n(k+1))!}$$ is greater than $0.75$.
If you wanted to count 7 people sharing the same birthday, there is another generalization that is relevant, for multiple collisions.
However questions 3 and 4 are combining both generalizations, asking for a "multiple near-miss". This seems rather tricky, and I don't know if it's been solved exactly in general.
-
Okay, for question 1: Wolfram|Alpha spits out four values of $n$. One is negative and two are in the nineties, but $n \approx 8.35709466412574\dots$ is good. And if I try it with $n = 9$, I get $P = 0.81$ (3sf) which seems to work. For question 2: I get a very low value of $n$ with $k = 30$, specifically $n \approx 0.00783908832757675\dots$ which I'm not really sure how to interpret. (I should clarify I merely set the equation above equal to 0.75 and filled in the values of $m$ and $k$, thus giving the exact value of $n$ that would give a probability of 75%) – Au101 Jul 15 '13 at 20:53
But you want integer $n$; just try $n=8, 9$ to see what happens in q1. Same for q2, just try $n=1,2,3,4,\ldots$. – vadim123 Jul 15 '13 at 21:01
Question 1 I can completely understand. If I try $n = 9$ I get $P(9,7,365) = 0.81$ (3sf). So, from that, we can see that if we randomly select 9 people, the probability that two of them will share birthdays within 7 days of each other is 0.81. Which is really cool. – Au101 Jul 15 '13 at 21:11
For question 2: If I set $n = 1$ $P = 1$ (well, not quite, but W|A gives the answer as so close to 1 it makes no difference. Well, this is odd, I don't really understand what this result means, but $n = 2$ gives an answer fantastically close to 1 again, so I can only assume that even with just 2 people, their birthdays are almost certainly within a month of each other. This is plainly false? – Au101 Jul 15 '13 at 21:12
I think you mistyped the formula. For $n=5$ I get $0.885$, while for $n=4$ I get $0.705$. – vadim123 Jul 15 '13 at 21:15
You may need to be more careful with your wording.
"To be certain that two of them shared a birthday" you need $367$ people (or $366$ if you ignore 29 February) by the pigeon-hole principle
Since $\frac{500}{6} \gt 83$, you can be certain that at least $84$ of the $500$ have birthdays within some two month period, which is rather more than seven.
I am not sure what your final question is asking.
-
That's a very good answer. Okay, let me put it to you this way. I was having a discussion with a friend of mine, who noted with surprise that she could name seven of her friends whose birthdays fell in the period August-September and five of them had birthdays in February. And I started to think about this and I wondered to myself: well, if the probability of two people sharing a birthday in a group of 23 is 50%, how many people would you need for a probability of 50% of them sharing a birthday in the same week, then in the same month. – Au101 Jul 15 '13 at 17:56
With regard to my last question. I thought: well, imagine she has met 500 people whom she has had a chance to get to know well. I then wondered to myself, just how remarkable is it that seven of them have a birthday between August and September. What I was struggling to do is to frame a mathematical question which would allow me to answer the question: how big would your group of randomly selected people have to be until you could be sure that they would share a birthday in the same two months and, also, suppose you have a group of 500, what are the odds of seven have birthdays in this range – Au101 Jul 15 '13 at 17:58
Wikipedia gives $7$ people for the probability that a pair of them have birthdays separated by six days or fewer to exceed $0.5$. – Henry Jul 15 '13 at 17:59
@Au101: if you read the Wikipedia article you should be able to figure out weeks and months, if you make some simplifying assumptions. Weeks can overlap, unlike days, and months have different numbers of days. If you assume people are equally likely to be born any given month, you need 5 to have better than $50\%$ chance (actually you have $63\%$ chance) that two share a birth month. – Ross Millikan Jul 15 '13 at 18:00
Okay thanks everyone, I'm really sorry, I don't have a very statistical brain and I was struggling to word the question clearly. Let me try again (sorry for the vagueness): Suppose you have a group of - let's say 30 people - what is the probability that seven of those 30 have a birthday within 50 days (i.e. less than two months)? – Au101 Jul 15 '13 at 18:02 | 1,945 | 7,594 | {"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} | 3.921875 | 4 | CC-MAIN-2016-30 | latest | en | 0.981539 |
http://sci.tech-archive.net/Archive/sci.physics.relativity/2007-12/msg00105.html | 1,493,360,725,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122865.36/warc/CC-MAIN-20170423031202-00604-ip-10-145-167-34.ec2.internal.warc.gz | 323,377,215 | 3,936 | # relativistic thermodynamics (temperature in inertial frames)
Hi, I'm a layman with an interest in physics. I just read an
interesting report on Physics News Update (focus.aps.org) concerning
unresolved questions in reconciling thermodynamics with special
relativity. In particular it seems that there is no agreement on how
temperature would transform between inertial observers. Apparently,
Planck and Einstein were of the opinion that moving observers would
find temperature reduced, but other physicists thought the reverse
would be true. And the question is still causing confusion.
If I recall correctly from my intro physics course, the temperature of
a gas is just (proportional to) the statistical variance of the speeds
of the atoms. Do I have this right? I also recall that the
distribution of speeds (actually, velocities) is normal (Gaussian).
(Actually, contrary to what our teachers told us, the distribution can
not be strictly normal, since the speeds of the atoms must be less
than the speed of light but the normal distribution is unbounded).
So, what would be the distribution of velocities perceived by a moving
observer? I would think we could get an answer by using the formula
for the relativistic addition of velocities. Consider first those
atoms moving most rapidly in the opposite direction to the moving
observer; the formula would compress the range of those velocities to
keep them all less than the speed of light. Now consider those atoms
moving more rapidly than the observer, and in his direction; in this
case the moving observer would measure a wider range of speeds than
would a stationary observer, since the atoms would not be pushing up
against the universal speed limit in his frame. So, we end up with a
skewed distribution as measured by the moving observer that is tighter
on one end and looser on the other, compared to the symmetric
distribution measured by the stationary observer. Is the variance
(temperature) increased or decreased? Hasn't anyone actually done
this calculation? Why is it so hard? Or, do I have it all wrong.
Thanks a bunch,
SRWenner
. | 442 | 2,107 | {"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-2017-17 | latest | en | 0.969939 |
https://conwaylife.com/forums/viewtopic.php?f=2&t=3962&start=100&sid=8b386ba81bb352af4be546fd2f81f68e&view=print | 1,621,090,528,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991370.50/warc/CC-MAIN-20210515131024-20210515161024-00403.warc.gz | 211,023,980 | 15,074 | Page 5 of 13
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 18th, 2019, 8:57 am
Goldtiger997 wrote:Here's a 3 glider reduction using my cheaper method of making the B-heptomino. The cleanup may still be reducible:
Thanks! That reduces the hardest xs17 from 37 gliders to 35. Shaving off the extra two gliders might be possible by removing the three cleanup gliders and gencolsing it with another glider.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 18th, 2019, 11:50 am
I need a two-sided method of making the dove at generation 9 here so as to synthesise the resulting SL efficiently.
Code: Select all
``````x = 21, y = 35, rule = LifeHistory
A\$.2A\$2A12\$18.A\$18.A.A\$18.2A\$13.A\$11.A.A\$12.2A3\$9.D\$8.D.D5.A\$7.D2.D3.
A.A\$7.D.D5.2A\$7.2D10.2A\$12.A5.2A\$12.2A6.A\$11.A.A3\$16.A\$15.2A\$15.A.A!``````
And this might be a way to another one in question – the one with the most soups (35):
Code: Select all
``````x = 20, y = 15, rule = B3/S23
5b2o\$4bo2bo\$4bo2bo\$5b2o\$10b2o\$9bo2bo\$10bobo\$11bo3\$18b2o\$2o9bobo3bobo\$
2o9b3o3b2o\$12bo\$15b2o!``````
Edit:: Actually, never mind for the second one...
Code: Select all
``````x = 91, y = 30, rule = B3/S23
12bo\$10b2o\$11b2o3\$8bobo\$8b2o78bo\$9bo78bobo\$88b2o2\$74b2o9b3o\$74bobo\$13b
2o60bo7bo5bo\$12b2o57b3o9bo5bo\$14bo58bo9bo5bo\$72bo\$85b3o\$78b2o\$78bobo\$b
3o76bo\$79bob2ob2o\$79bo2bob2o\$78b2obo5b2o\$77bo2bo6bobo\$10b2o8b2o4b2o50b
2o7bo\$9bobo8bobo2b2o\$11bo8bo6bo\$3o\$2bo\$bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 18th, 2019, 3:59 pm
Freywa wrote:Edit:: Actually, never mind for the second one...
Code: Select all
``rle``
Code: Select all
``````x = 90, y = 39, rule = B3/S23
27bo\$25b2o\$26b2o7\$12bo\$10b2o\$11b2o3\$8bobo\$8b2o\$9bo5\$13b2o\$12b2o\$14bo3\$
78b2o\$78bobo\$b3o76bo\$79bob2ob2o\$79bo2bob2o\$78b2obo5b2o\$77bo2bo6bobo\$
10b2o8b2o4b2o50b2o7bo\$9bobo8bobo2b2o\$11bo8bo6bo\$3o\$2bo\$bo!
``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 18th, 2019, 5:59 pm
Freywa wrote:I need a two-sided method of making the dove at generation 9 here so as to synthesise the resulting SL efficiently.
Code: Select all
``````x = 21, y = 35, rule = LifeHistory
A\$.2A\$2A12\$18.A\$18.A.A\$18.2A\$13.A\$11.A.A\$12.2A3\$9.D\$8.D.D5.A\$7.D2.D3.
A.A\$7.D.D5.2A\$7.2D10.2A\$12.A5.2A\$12.2A6.A\$11.A.A3\$16.A\$15.2A\$15.A.A!``````
Here's a method, though a cheaper one almost certainly exists:
Code: Select all
``````x = 21, y = 35, rule = B3/S23
o\$b2o\$2o12\$18bo\$18bobo\$18b2o\$13bo\$11bobo\$12b2o3\$9bo\$8bobo5bo\$7bo2bo3bo
bo\$7bobo5b2o\$3bo4bo10b2o\$bobo8bo5b2o\$2b2o8b2o6bo\$11bobo2\$3b3o\$5bo10bo\$
4bo10b2o\$15bobo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 18th, 2019, 11:49 pm
Kazyan wrote:Here's a method, though a cheaper one almost certainly exists:
Code: Select all
``````x = 21, y = 35, rule = B3/S23
o\$b2o\$2o12\$18bo\$18bobo\$18b2o\$13bo\$11bobo\$12b2o3\$9bo\$8bobo5bo\$7bo2bo3bo
bo\$7bobo5b2o\$3bo4bo10b2o\$bobo8bo5b2o\$2b2o8b2o6bo\$11bobo2\$3b3o\$5bo10bo\$
4bo10b2o\$15bobo!``````
Doesn't work. One of the gliders would interact with the mango beforehand.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 12:40 am
calcyman wrote:Thanks! That reduces the hardest xs17 from 37 gliders to 35. Shaving off the extra two gliders might be possible by removing the three cleanup gliders and gencolsing it with another glider.
Here's a 2G reduction to the creation of the carrier on paperclip, bringing the cost down to 33 gliders:
Code: Select all
``````x = 117, y = 50, rule = B3/S23
2bo\$obo\$b2o\$40bo\$38b2o\$39b2o10\$114bo\$114bobo\$40bobo71b2o\$40b2o\$41bo69b
o\$110bobo\$110bobo\$37bo63b2o8bo\$35b2o63bo2bob2o\$36b2o62bob2obo2bo\$99b2o
bo4b2o\$99bo2bo\$100b2o7\$39bo\$38b2o\$38bobo8\$2b2o\$3b2o\$2bo\$40b2o\$40bobo\$
40bo!``````
Now all 17-bit still-lifes can be synthesized in less than 2 gliders per bit!
Freywa wrote:Doesn't work. One of the gliders would interact with the mango beforehand.
Among other things, this can be solved with the addition of a single kickback:
Code: Select all
``````x = 29, y = 42, rule = B3/S23
obo\$b2o\$bo11\$26bo\$25bo\$25b3o\$13bo\$14bo\$12b3o\$18bo\$19b2o\$18b2o4\$3bo9bo\$
4bo7bobo\$2b3o6bo2bo7bo\$11bobo7b2o\$12bo8bobo3\$27bo\$26b2o\$12b2o12bobo\$
13b2o\$12bo\$4b2o\$3bobo\$5bo17b2o\$22b2o\$24bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 4:18 am
Goldtiger997 wrote:
calcyman wrote:Thanks! That reduces the hardest xs17 from 37 gliders to 35. Shaving off the extra two gliders might be possible by removing the three cleanup gliders and gencolsing it with another glider.
Here's a 2G reduction to the creation of the carrier on paperclip, bringing the cost down to 33 gliders:
Code: Select all
``````x = 117, y = 50, rule = B3/S23
2bo\$obo\$b2o\$40bo\$38b2o\$39b2o10\$114bo\$114bobo\$40bobo71b2o\$40b2o\$41bo69b
o\$110bobo\$110bobo\$37bo63b2o8bo\$35b2o63bo2bob2o\$36b2o62bob2obo2bo\$99b2o
bo4b2o\$99bo2bo\$100b2o7\$39bo\$38b2o\$38bobo8\$2b2o\$3b2o\$2bo\$40b2o\$40bobo\$
40bo!``````
Now all 17-bit still-lifes can be synthesized in less than 2 gliders per bit!
Well done!
Indeed, that's something of an understatement; we actually have the stronger global result:
Now all synthesisable still-lifes can be synthesized in less than 2 gliders per bit!
by use of the 35-glider RCT mechanism for >= 18-bit still lifes.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 7:33 am
Carrier on paperclip is two still-lifes, not one.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 7:43 am
Gamedziner wrote:Carrier on paperclip is two still-lifes, not one.
It's on the critical path to synthesise this 17-bit still-life: https://catagolue.appspot.com/object/xs ... z39c/b3s23
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 11:19 am
calcyman wrote:Now all synthesisable still-lifes can be synthesized in less than 2 gliders per bit!
by use of the 35-glider RCT mechanism for >= 18-bit still lifes.
You know, at some point we really ought to put together a working sample of the glider-producing crystallization and decay mechanisms in the 35-glider RCT design. The wiki article isn't specific enough yet to reassure a casual reader that the construction mechanism will really work.
EDIT: Looks like there's some out-of-date wording still in the wiki article, like
... the receding block-laying switch engine means that the construction arm must drag each temporary elbow increasingly far, such that the time taken to effect an n-slow-glider synthesis is 2^((6 + o(1))n^2). By self-modifying circuitry, it is possible to improve this to 2^(O(n)) without increasing the number of gliders in the initial synthesis.
The first sentence isn't true for the latest 35-glider crystallization/decay design, but my big-O allergies are acting up again so I'm not sure exactly how to correct the wording.
Meanwhile, congratulations to everyone on getting past a really improbable milestone! As recently as a year ago this result would have been pretty far beyond the bounds of believability.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 1:06 pm
xs17_gbdz122c871 looks like one of the bashiest and least efficient syntheses I've seen. Is there no way to directly convert from the snake to the eater?
Code: Select all
``````#CSYNTH xs17_gbdz122c871 costs 18 gliders (true).
#CLL state-numbering golly
x = 207, y = 29, rule = B3/S23
142bo\$140bobo\$141b2o3\$190bobo\$98bo92b2o\$99b2o4bo85bo\$obo11bo78bo4b
2o5bobo\$b2o9b2o80bo10b2o\$bo11b2o77b3o\$46b2o54b2o37bo\$47bo55bo35bob
o11bob2o45b2o\$46bo42b3o10bo37b2o11b2obo42bobobo\$12bo33b2o43bo3b3o
4b2o53b2o40b2o3b2o\$12bobo33bo41bo6bo6bo54bo46bo\$12b2o34bo47bo7bo
33b2o19bo46bo\$46b2o54b2o35b2o16b2o36bo8b2o\$46bo9bo45bo35bo18bo38b
2o6bo\$14b2o31bo8bobo44bo54bo36b2o8bo\$13b2o31b2o8b2o44b2o53b2o45b2o
\$15bo36b2o\$51bo2bo141b2o\$52b2o141bobo\$197bo\$140b2o\$141b2o50b2o\$
140bo51bobo\$194bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 1:30 pm
Yes, the standard method would interfere with the other snake.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 4:13 pm
This might help for a different expensive object. Apply a common phi spark predecessor and a good dot spark from the right (a bad one is pictured).
Code: Select all
``````x = 15, y = 17, rule = B3/S23
14bo\$12b2o\$13b2o\$3bo3b2o\$4b3o\$10bobo\$2bo7b2o\$bobo7bo\$bobo\$2ob3o4b2o\$6b
o2bo2bo\$5bo4b2o\$5b2o2\$10b3o\$12bo\$11bo!``````
EDIT: Yep. This step can probably be done in 7 gliders, but here's the proof of concept in 10:
Code: Select all
``````x = 32, y = 32, rule = B3/S23
obo15bo\$b2o5bo8bo\$bo7bo2bo4b3o\$7b3o3b2o\$12b2o2\$19bobo\$19b2o\$8bobo9bo\$
9b2o\$9bo\$23bobo\$23b2o\$24bo3\$10bo\$9bobo\$9bobo\$8b2ob3o13b3o\$14bo12bo\$13b
o14bo\$13b2o3\$23b3o\$23bo\$24bo2\$30bo\$29b2o\$29bobo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 19th, 2019, 8:22 pm
Kazyan wrote:EDIT: Yep. This step can probably be done in 7 gliders, but here's the proof of concept in 10:
Code: Select all
``````x = 32, y = 32, rule = B3/S23
obo15bo\$b2o5bo8bo\$bo7bo2bo4b3o\$7b3o3b2o\$12b2o2\$19bobo\$19b2o\$8bobo9bo\$
9b2o\$9bo\$23bobo\$23b2o\$24bo3\$10bo\$9bobo\$9bobo\$8b2ob3o13b3o\$14bo12bo\$13b
o14bo\$13b2o3\$23b3o\$23bo\$24bo2\$30bo\$29b2o\$29bobo!``````
I'll raise you 9:
Code: Select all
``````x = 31, y = 32, rule = B3/S23
30bo\$8bobo17b2o\$9b2o18b2o\$9bo2\$11bobo\$o10b2o\$b2o9bo\$2o2\$15bo\$13b2o\$2bo
11b2o\$3b2o\$2b2o14bo\$18bobo\$18b2o4\$22bo\$4bo17bobo\$3bobo16b2o\$3bobo\$2b2o
b3o\$8bo\$7bo\$7b2o2\$18b2o\$17b2o\$19bo!
``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 12:02 am
This looks too simple but I can't find a synthesis from it:
Code: Select all
``````x = 9, y = 16, rule = B3/S23
4b2o\$4b2o3\$3b2o\$3b2o4\$7bo\$bo3bo2bo\$bo3bo\$bo3bo\$bo2\$3o!``````
Code: Select all
``````x = 11, y = 19, rule = B3/S23
5b2o\$5b2o4\$5b3o\$8b2o\$5b2o\$5bo\$5bobo2bo\$4b2obobo\$5bo3bo\$5b2ob2o\$6bobo\$
7bo\$bo\$obo\$o2bo\$b2o!``````
On the other hand, a synthesis that does work:
Code: Select all
``````x = 207, y = 57, rule = B3/S23
87bo\$88b2o\$87b2o5\$97bo\$98b2o\$97b2o\$125bo\$124bo\$124b3o3\$147bo\$145b2o\$
146b2o8\$91bobo\$b2o89b2o\$o2bo88bo\$bobo\$2bo108b2o86bo\$110bo2bo85bobo\$97b
o13bobo85b2o\$98b2o12bo84bo6bo\$97b2o97bobo5bobo\$195bo2bo5b2o\$196b2o3b2o
\$191b2o8b2o\$4bo185bobo\$2b2o185bo3b2o2b2o\$3b2o185b3o2bo2bo\$192bo2b2o\$4b
o\$3b2o\$3bobo3\$112b3o2\$103bo\$103b2o\$102bobo\$99bo\$99b2o\$98bobo\$89bo\$89b
2o\$88bobo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 1:39 am
calcyman wrote:Now all synthesisable still-lifes can be synthesized in less than 2 gliders per bit!
by use of the 35-glider RCT mechanism for >= 18-bit still lifes.
All synthesisable strict still-lifes. We don't know that we can place every 8-bitter next to every 9-bitter in any orientation with only 34 gliders.
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 2:28 am
Another one that does work:
Code: Select all
``````x = 255, y = 91, rule = B3/S23
bo97bo\$2bo95bo\$3o95b3o9\$87bo\$85b2o\$86b2o9\$31bo\$32bo\$30b3o11\$56b2o\$55bo
2bo\$56bobo185b3o\$57bo\$242bo5bo\$242bo5bo\$242bo5bo\$235b2obo\$234bob2obo4b
3o\$234bo4bo12b3o\$235b4o13bo\$236bo16bo\$234bo\$43bobo188b2o\$44b2o\$44bo5\$
86bo\$85b2o\$85bobo\$31b2o19bo\$32b2o17b2o\$31bo15b2o2bobo\$46bobo\$28bo19bo\$
28b2o\$27bobo24\$15b3o\$17bo\$16bo!``````
The remaining SLs as of now have at most 23 C1 soups.
Edit: Natural component:
Code: Select all
``````x = 26, y = 31, rule = B3/S23
7bo5bo\$5bobo4bo\$6b2o4b3o2\$9bo\$9bobo\$9b2o6\$20b3o4\$22b2o\$22bobo\$23bobo\$
24bo5\$6b2o\$5bobo\$7bo\$b2o\$obo19b2o\$2bo19bobo\$22bo!``````
Edit 2: This looks tantalisingly simple. There are a few similar soups for the same SL that have a mirrored dock intermediate.
Code: Select all
``````x = 23, y = 25, rule = B3/S23
2o2b2o\$o4bo\$b4o5bo\$10bo\$b4o5bo\$o4bo\$2o2b2o4\$8bo\$7b4o\$5bo5bo\$4bo2b2o2bo
\$8b4o\$4b2ob2o\$7b2o2\$14bo\$13bobo\$13bob2o\$13bo2bo\$14bob2o2b2o\$16bo2bo2bo
\$15bo4b2o!``````
Edit 3: The current table of expensives. 271 remain.
Code: Select all
``````x = 604, y = 217, rule = B3/S23
218bo198b3o\$217b2o197bo3bo\$218bo201bo\$218bo200bo\$218bo199bo\$218bo198bo
\$217b3o196b5o4\$17b3o17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b3o17b3o
17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b3o17b3o17bo19b3o17b3o19bo16b
5o16b3o16b5o16b3o17b3o\$16bo3bo15b2o18bo3bo15bo3bo17b2o16bo19bo3bo19bo
15bo3bo15bo3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo19bo3bo19bo15bo3bo15b
o3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo19bo3bo19bo15bo3bo15bo3bo\$16bo
3bo16bo22bo19bo16bobo16bo19bo22bo16bo3bo15bo3bo15bo3bo16bo22bo19bo16bo
bo16bo19bo22bo16bo3bo15bo3bo15bo3bo16bo22bo19bo16bobo16bo19bo22bo16bo
3bo15bo3bo\$16bobobo16bo21bo18b2o16bo2bo17b3o16b4o19bo17b3o17b4o15bobob
o16bo21bo18b2o16bo2bo17b3o16b4o19bo17b3o17b4o15bobobo16bo21bo18b2o16bo
2bo17b3o16b4o19bo17b3o17b4o\$16bo3bo16bo20bo21bo15b5o19bo15bo3bo17bo17b
o3bo19bo15bo3bo16bo20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo15bo3bo16bo
20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo\$16bo3bo16bo19bo18bo3bo18bo16bo
3bo15bo3bo17bo17bo3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo
17bo17bo3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo17bo17bo3bo
15bo3bo\$17b3o16b3o17b5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b
5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b5o16b3o19bo17b3o17b3o
18bo18b3o17b3o9\$16b2o18b2o19bo18b2o20bo19bobo15b2o18b2obo15b2o19bo19bo
18b2o2bo3bo11bo3b2o34b2o2b2o14b2obob2o16bo18b2o19b2o19bo18bo17b2o2b2o
15b2o18b2o19b2o35b2o20bo18bo22b2o\$b3o13bo18bobob2o14bobo17bo21b3o16bob
2o16bo18bob2o16bo18bobo2b2o13bobo17bo2bobobobo10b3o2bo34bo3bobo13bob2o
bo2bo12b3o17bo2bo15bo2bo16b2obobo16bobo15bo2bo2bo14bo2bobo14bobo18bo2b
o34bobo19b3o15bobo2b2o16bobo\$o3bo12bob2o17b2obo14bo2bobo15bo18b2o3bo
15bo19bob2o19b2o14bobo17bo2bo2bo13bobo2b2o14b2obobo2bo12b2o37bobobo17b
o2b2o11bo19bobo2bo14b3o2bo14bob2o2bo14bobobo15bobobo14bo2bob2o13bobo
19bobobo36bo16b2o3bo14bo2bo2bo16bobobo\$o3bo13bo2bo15bo19b2ob2o14b2o2bo
15bo2b2o17b2obo16bobo16b2o2bo15bobo17b3obo16bo3bo18bo2b2o11bo38b2o2bo
15b3o17b5o15bob3o17b3o18b2o14bo2bobo14b2o2bo15b2obo16bo2b3o15b2o2bo35b
2o18bo2bobo14bob2o17b2o2bobo\$obobo15bo16bo20bo19b3o16b2o17bobob2o14bob
o17bob2o19bo20bo16bob3o35bo39bo18bo23bo16bo19bo18b2o18b2obo16bo21bo17b
2o2bo14bo2b2o35bo19bo4bo14b2o21bo3bo\$o3bo11b4o15b2o19bobo17b2o20bo16bo
bo17bobobo16bo22b2o18bo18bo39b2o36bo41b2o18bo19bo17bo20bo17bo19b3o19bo
17bobo37bob3o15bobo18bo18b3o\$o3bo12bo17bo19bobo17bo2bo16b3o17b2o18bo2b
o16b2o20b2o2bo16bo20bo39bo36b2o40bo16b3o17b3o19bo17bobo17b2o18bo19bo
20b2o38bobo17bobo16bo19bo\$b3o11bo20bo18b2o19bobo16bo40b2o39bo2bo17b2o
18b2o38bo80bo15bo19bo20b2o17b2o58b2o63bo16bo17b2o\$15b2o18b2o40bo100b2o
78b2o78b2o199b2o12\$17bo18b2ob2o19b2o16b2o38b2o16b2o23b2o12b2o19bo19bo
18b2o2b2o14bo19b2o4b2o12b2o2b2o14b2obo19bo2bo15b2o18b2o21b2o16b2o15b2o
20b2o18bo41bo15b2o21bo17bo24b2o\$bo14bobo17bob2o16b2o3bo17bo39bo2b2o13b
o19b2o2bo14bo2b2o14bobo4b2o11bobo17bo2bobo14b3o17bobo3bobo11bo3bobo13b
ob2o17b6o14bo2bo15bo2bo16b2o2bobo15bo2bo14bo2b2o16bo2bo16bobo39bobo14b
obo18b3o16bobo20bobobo\$2o15bo2bo20bo14bo2bo16b3o40bobobo13bobo16bo2bob
o14bobo2bo14bo2bo2bo13bobo18b2o19bo17bo2b2o2bo13bobobo17b2o14bo19bobob
o15b2obob2o13bobo2bo15bo2bobo14bobobo14bo2bobo14bo2bo38bobobo16bo16bo
19bo2bob2o15bob2o\$bo16b2obo15b2ob2o15bob2o15bo2b3o35bobo2bo15bobo16bob
2o16bob2o16b2obobo15bo2bo17bo17bo19b2obobo13b2o2bo15b2obo2bo13b2o2b2o
14bo2b2o16bobobo16b2o15bo4bo14b2obobo14b2obo2bo13bobob2o36bobobo14b2o
18bo19bob2obo15bo3bo\$bo18bo16bo20bo19bo2bo34bob2o16bobobo15b2o21bo19bo
bo15bob3o17bo17b2o22bo15bo18bobo2b2o15bo2bo15b2o18bo18b2o18b4o17bobo
17bob2o15b2o2bo34b2o3bo14bo19b2o18b2o21bob2o\$bo14b3o16bobo18bobo17bobo
37bo18bobobo17bo19bobo19bo18bo18b2o20bo35bo21bo20b2o18bo19bo18bo39bo
19bo20bo36bo19bo19bo4bo15bo18b3o\$bo13bo2b2o15b2o18bobo17bobo37b2o18bob
o17bo20bobo19b2o19bo17bo21bo35b2o58b3o21bo16bo19b2o18b2o18b2o17bobo38b
3o17b2o2b2o14bo2bobo13bo19bo\$3o12b2o38b2o19bo59b2o17b2o20bo39bobo18bo
21b2o93bo22b2o16b2o18b2o57b2o41bo19bo2bo15bo2bo14b2o\$217b2o18b2o22bo
276b2o18b2o\$258b3o\$258bo10\$17bo20b2o21b2o17b2o35b2o22b2o14bo17b2o19bo
19b2o17b2o3b2o13b2o2b2o14b2o18b2o2b2o14b2o21bo18b2o18b2ob2o18b2o16b2o
15b2o20b2o18bobo18b2o19bo15b2o23b2o16bo21bo\$b3o12bobo18bobo16b2o3bo19b
o35bo2b2o14b2o2bobo13bobo2b2o13bo2bo15bobo17bo2bobo14bo2bo2bo13bo2bo2b
o13bo2bobo14bo3bo15bo20b3o2bo14bo2bo15bo2bobo14b2o2bobo15bo2bo14bo2b2o
16bo2bo17b2obo16bo2bo17bobo14bobo18b2o2bo16bobo19bobo\$o3bo12bo2bo15bo
19bo2bobo14bo2bo38bobobo14bo2bo15bo2bo2bo13bobobob2o12bo2b2o15b2ob2o
16b2obo15bobob2o14b2ob2o16bobobo15bo4bo12bo3b3o13bobo2bo14b2obo2bo13bo
3bo16bo2bobo14bobo16bo2bobo14b2o3bo17bo2bo15bobobo16bo16bo2bobo16bob3o
15bo2bo\$4bo13b2obo15b2o18b2obo15b5o36b2o2bo15bob2o16bob2o16b2obobo2bo
11b2o2bo15bo20bob2o15bobo17bo18b2o2b2o14b2o3bobo12b2o18bo4bo15bob2o16b
ob2o14bo4bo14b2obobo14bob2o2bo13bo2b2o16bob2obo15bobobo14b2o18bob2o15b
2o5bo14b2obobo\$3bo16bo18bo18bo57bo2b2o15b2o18b2o23bo2b2o13b2o16bo19bo
21bo17bo19bo18bo3bobo15b2o17b4o16bo18b2o18b4o17bob2o15bo2bobo14b2o17bo
bo2bo14b2o3bo14bo19b2o18bo5b2o15bo2b2o\$2bo13b4o15b4o17bobo19bo36bobo
19bo19bo39b2o19b2o16bo21bo19b2o15bo21bo2bo17bo19bo17b2o19bo19bo19bo20b
obo16bo17bo2bo16bo19bo20bo19bo19bobo\$bo15bo17bo19bobo18b3o37bo18bo19bo
42bo20bo16b2o20b2o19bo15b2o21bobo19bo15bo38bo19bo20b2o21bo15bo20b2o18b
o19b2o17bo21bo17bobo\$5o10bo20bo19bo18bo59b2o18b2o39bo21bo59bo40bo19b2o
15b2o37b2o18b2o58b2o40bo20bob2o13b2o19b2o18bo\$15b2o20bo37b2o119b2o20b
2o58b2o236b2o20b2obo\$36b2o11\$17bo21b2o38bo37b2o17b2o19bo17b2ob2o16bo
38b2o18b2o3bo14b2o2bo15b2o60b2o17bo23bo18b2o14bo2b2o19b2o14b2o21bo20bo
15b2o21bo19bo2b2o17bo\$b3o12bobo2b2o14bo2bo37bobo36bo18bo19bobo17bobo
16bobo37bo2bob2o13bo3bobo13bo2bobo14bo60bo2bo15bobo3b2o12b2o2bobo16bob
o14b4obobo13b2o2bo15bo20bobo18bobo14bo3b2o14bobobo17bobo2bo16bobo\$o3bo
12bobo2bo13bob2o36bo2bobo36bo18bo18bo2bob2o13bo2b3o14bo2b2o36b2obo15bo
2bo2bo13b2obo17bo4bo52bobo2bo14bobo3bo13bo3bobo14b3o22b2o12bo2b2o15bo
22bobo16bobo17bo2bo14b2obobo16bobobo17bo2bo\$4bo14b2o16bo2b3o33b2obo2bo
34b2o19bo18bob2obo14b2o2bo15b2o2bo36bo2bo15bob3o15bo18b2o2b3o53bo4bo
14bob2obo15bob2o14bo3bo15b2o18bobo2b3o12b2o19bobo2bo15bobobo14b2obo18b
obo14b2o3bo16b2obobo\$2b2o14bo19bo3bo35bo2b2o33bo19bobo17b2o20bo20b2o
36bo2b2o16bo18bo19bobo57b4o17bobo15b2o18b2obo16bo19bobo3bo14bo17bobob
2o14b2o3b2o13bo3b2o14b2o2bo15bo21bo2b2o\$4bo11b3o16b3o37bobo38bo18bobob
2o16bo18bo20b2o37bo22bo15b2o20bo2bo75bo20bo20bo16bo21bo17b2o18bo2bo16b
o20b3o16bo20bo21bo\$o3bo10bo19bo39b2o38b2ob2o15bobo2bo14bo20b2o17bo2bo
37b2o20b2o15bo22bobo56b2o17b2o17bo19bobo17b2o38bo3b2o15bobo17b3o19bo
18bo19bo17b3o\$b3o11b2o100bo2bo15bobo16b2o38b2o79bo22bo57b2o36b2o18b2o
59bo2bo17bo20bo37b2o18b2o17bo\$117bobo17bo137b2o200bobo\$118bo359bo11\$
16b2o20b2o20bo16b2o17b2o19b2o17b2o22b2o14bo2b2o15bo24bo13b2o2b2o14b2o
3b2o33b2obob2o13b2o21b2o18bobo15bobo39b2o15b2o2b2o34b2o23b2o17bo38bo
19b2o20bo\$3bo12bobo18bo2bo17b3o17bo18bo19bo3b2o13bo19b2o3bo13bobo2bo
14bobo17b2o3bobo12bo2bo2bo13bo2bo2bo33bob3obo13bobo18b3obo15b4obo13bob
2o39bo16bo3bo36bo22bobo16bobo34bobobo17bo2bo18bobo\$2b2o15bo16bo2bobo
15bo18bo2b2o15bo22bo2bo15bo17bo2bo16bob2o16bo2b2o15bo3bo2bo13b2obo15bo
b3o56b3o15bo5bo13bo5bo13bo3b2o35b2o2bo15bo3bo34bo21b3o17bobo35b2obobo
17b2obo18bobo\$bobo13b2o2bo15b2obo15bo2b2o15b2obo2bo13b4o17bob2o16b2o
18bob2o17bo18b2o2bo14bob2obobo14bob2o15bo41b2o15bo3bo15bo5bo13b2o3b2o
13b2o2bo34bo2b3o14b2o2b2o34b2o19bo3bo16bobobo36bobo14b3o2bo16b3o2bo\$o
2bo14bob2o16bo18b2obo17bo2b2o17bo15bobo17bo19b2o2bo17bo20b2o16bobob2o
14bo20bo37bobobo15b3obo16bo3b2o14bo20b2o36b2o18bobo38bo17bob3o15b2o3b
2o33b2o2bo15bo2b2o17bo2b2o\$5o13bo17bobo19bo16bobo18b4o15bo2bo16bo2b3o
14bo2bo17b2o19b2o37bo18b3o38b2o21bo18bobo15bo21bo39bo17bobo38bo18bo18b
o39bo21bo20bo\$3bo11b2obo16bobo18bobo16b2o20bo17bobo18b2obo16b2o18bo21b
o37b2o17bo62bo20b2o15b2o21bo35bobo19bo40b2o17bo18bo39bo19b2o17b3o\$3bo
11bobo17b2o18bobo37bo20bo24bo35bo19bo120b2o58b2o35b2o63bo15b2o19bo37b
2o38bo\$56bo38b2o43b2o34b2o19b2o279b2o36b2o\$478bo\$479bo\$478b2o9\$16b2o
23b2o15bo20bo16b2o22bo15b2o21bo16b2o18bo21b2o15b2o3b2o33b2o18bo3b2o16b
2o3bo15bob2o13b2o3bo15bo20bo43bo15bo19b2obo19bo18b2o14b2o21bo18b2o22bo
\$5o11bo3b2o15b2o2bo15bobobo15b3o17bo21bobo14bo21bobo14bo2bo16bobo2b2o
13b2o2bo15bo2bo2bo33bo19b3o2bo15bo2bobobo12b3obo15bo2bobo13bobo3b2o13b
obo41bobo13bobo18bob2o18bobo16bobo14bo19bobobo17bo21b3o\$o17bo2bo14bo2b
obo15bob2obo13bo19bo21bobobo15bo19bo2bo14b2obobo14bobo2bo14bobo18b2obo
36bob2o17b2o15bobo2b2o2bo10bo5bo13bo3bobo13bo2bo2bo14bo2bo35b2o3bobo
12bobobo21b2o14b3obo15bo2bobo14bo17b2obo19bo19bo\$o16b2obo16b2obo15b2o
4bo13b2o18b2o20bobobo14b2o17b2o2b2o16bob2o16b2o16bob4o17bo36b3o2bo15bo
2b3o13bo5bobo11bo5bo13b3obo15b2obobo13b2ob2o35bo2bob2o13bobo2bo17b2o2b
o13bo3bo16b2o2b2o13b2o20bobo14b3obo19bo\$b3o14bob2o16bo18bo4b2o15bo18bo
17b2o3bo14bo19bo21bo18bo19bo3bo15b2o41bo15bobo3bo19b2o13bob3o17bo18bob
o16bo39b2o17bob2o15bobob2o14bob3o15b2o18bo19b2o2b2o14bo2bo19b2o\$4bo10b
3o18bobo16bo19bob2obo17bo16bo19bo2b3o16bo18bobo19b2o18bo18bo40b2o17bo
40b2o18bo19bo18bo40bo18bo17b2o19bo19bo18bo4b2o13bo22bobo16bo\$o3bo10bo
19bobo17b2o18b2obobo14b2ob2o16bobo17b2o2bo17bo16bobo21bo17b2o19bo39bo
79b2o17b2o16b2o40bo17bobo39bo17bo20b2o2bo16bo21bobo15bo\$b3o32bo42bo15b
o2bo18b2o19bo17bobo17bo20bo39b2o40bo115bo41b2o16b2o39b2o17b2o21bobo15b
2o23bo13b2o\$96bobo39b2o15bobo39b2o79b2o116bo141b2o41b2o12bo\$97bo58bo
238b2o199bo\$597bo\$596b2o9\$16b2o22b2o16b2o18b2o16b2o22bo15b2o23bo14b2o
18bo38bo2bo16b2o2bo15b2o3b2o13b2o20bo20bo16b2ob2o16bo21b2o19b2o18bo17b
o18b2ob2o15bo38b2o3b2o16bo18b2o20b2o\$b3o12bobo17b2o3bo15bobo17bobo17bo
21bobo14bo23bobo12bo2bob2o13bobo2b2o33b6o14bo2bobo14bo4bo14bo2b2o16bob
ob2o14b3o2b2o13bobo2bo13bobo20bo19bobo17bobo15bobo17bob2obo14b3o36bo2b
o2bo13bobobo17bo21bobo\$o3bo14bo16bo2bo17bobobo14bo19bo21bobo17bo20bobo
14bob2obo14bobo2bo39bo14bobo2bo15bo3bo14b2o2bo14bo2b2obo13bo5bo13bo4b
2o13bobo18b2o2bo15b3o16b2o2bo15bobobo21bo17bo37b2obo14b2obo19bo23bo\$o
19bo17b3o15b2o2bobo13bob4o14b2o20bobobo14b2o17bo2bo17bo20b2o37b4o16bob
2o15b5o17bobo15b2o18bo5bo13b2o18b2ob2o14bo2b3o14bo3bo15bob2o16bobo2bo
16b2obo15b2o2bo37bob2o16bobo14b3o2bo19b2o\$4o13b4o36bo3bo15bo3bo15bo18b
2o3b2o13bo19b3obo18b2o16bo40bo19bo37b2obo19b2o16bo3b2o15bo19bobo15bo
19b3obo16bo18bob3o14bo2b2o15bo2b2o35b3o17b2o2b2o14bo2b3o18bo\$o3bo12bo
18b3o17bo21bo17bo3b2o13bo20bo22bo20bo17b2o36bo19bobo19bo16bo2bo21bo17b
obo14bobo18b2o19b2o20bo17bo19bo17b2o19b2o37bo19bo21bo21bo\$o3bo13bo16bo
2bo16bo19b3o18bobo2bo14bobo16b2obo19bo19bo20bo36b2o18b2o19bobo16b2o20b
o20b2o14b2o19bo21bo19bo16b2o21bo38bo58bo21bo18b2o\$b3o12bobo17b2o17b2o
18bo19b2ob2o17b2o18bobo15bobo20b2o18bo79bo39b2o57bo19bo20b2o15bo21b2o
36bo59b2o22bo17bo\$15bobo119bo2bo14b2o41b2o176b2o19b2o37bo58b2o81b2o18b
o\$16bo121b2o295b2o158b3o\$595bo10\$16b2o2b2o18b2o14b2o20bo17b2o22b2o14b
2o21bo16b2o2b2o14bo38bo39b2o2b2o14b2o20b2o22bo16b2ob2o13bo42b2o18bo20b
2o17bo16b2o37b2o21b2o17b2o\$5o11bobo2bo14b2o3bo15bo19bobo17bo19b2o2bo
14bo21bobo14bo2bo2bo13bobo37b3o2b2o33bo3bo15bo20bobo17b2o2bobo12b2o2bo
bo13bobo39bo2bo17bobo15bo2bobo16bobo15bo38bo2bob2o13bo2bobo16bo2bo\$4bo
13b2o16bo2bo17bob2o15bo2bo16bo21bobo17bo18bo2bo15bob3o15bobo2bo36bo2bo
35bobo16bo18bo2bob2o13bo2bo2bo13bob2o3bo12bo2bo2b2o34b2o16bo2bobo14bob
o2bo17bo2bo15bo39b2obo14b3o2bo17b2obo\$3bo13bo20b3o15b2o2bo15bob2obo14b
2o20bob2o15b2o17bo3bob2o13bo20b4o35b2obo35b2o2b2o15bo18b3obo15bo2b2o
18b2obo13b2obo2bo32b2o18b3obo15bo2b2o16b2o2b2o14b2o40bo2bo16b2o15b3o2b
o\$3bo13bo40bo18bo2b2o15bo18b2o18bo19b2o3bobo14bo18bo40bob2o36bobo15b2o
23bo15b2o20bobo15bob2o35bo21bo17b2o17bo19bo39b3o2b2o13b2o18bo2b2o\$2bo
12b2o19b3o17b2o20bo17bo18bo20bo20bo17b3o20b2o36bo41bobo14bo2b2o18b3o
17bo36bo39bo19b2o20bo17bo19bob3o35bo19bo21bo\$2bo12bo19bo2bo16bo19b3o
18bobob2o14bobo18b2obo14bo19bo23bo36b2o41bo15bobo2bo17bo18bo37b2o38bob
o17bo19bo20b2o18bobobo56bo18b2o\$2bo14bo17b2o19bobo16bo19b2ob2obo15b2o
16bobob2o14b2o40bo98bo2bo37b2o77bobo17bo18b2o20bo22bo55b2o\$16b2o39b2o
76b2o60b2o98b2o119bo16b2o39bo23b2o\$418b2o56b2o11\$16b2o2b2o15bo19b2obo
17b2o20bo35b2o17b2o4b2obo11b2o2b2o14bo19b2o17bo3b2o14b2o18b2o18b2o20b
2o22b2o16b2o38bo17bo20b2o17bo21bo16b2o18b2o17b2o2b2o37bobo\$b3o12bobo2b
o14bobo17bo2b2o16bo2bo17b3o35bo19bo4bob2o10bo2bo2bo13bobo17bo2bo16b3o
2bo14bo2b2o15bo3b2o14bobob2o15bobo17bo3bobo12b2o3bo36b3o16bobobo17bo
17bobo19bobo15bo18bo2bobo14bo2bo2bo35bob2o\$o3bo13b2o16bo2bo17bo18bobo
2bo15bo3b2o35bo17bob2obo14bob2o16bobo2bo14b2o2b2o16b2o16bobo2bo15bo2bo
16b2obo14bo19bobo2bo14bo2bo37bo3b2o14bob2obo13bo3b3o13bobo2b2o16bo2bo
15bo18bob2obo15b2obo36bo\$o3bo12bo19b2obo15b2ob2o15bo2b2o17b2o2bo34b2o
18bobobo15bo20b4o16b2obo15bo19bo2b2o14b2obo16bo19b2o18bo2b2o16b3o35bob
2o2bo13b2o4bo13b4o2bo13bo2b2obo14b2o2b2o14b2o19bo3bo16bob2o33bobob2o\$b
3o13bo21bo18bobo16b2o17bobob2o34bo23bo17b2o17bo18b2o19bo20b2o18bo17bo
21bo18b2o57bo2b2o15bo4b2o15bo17b2o17bo19bo19bobo3b2o12b3o37b2obobo\$o3b
o10b2o18b4o16bobo20bo16bobo38bo42bo18b2o17bo17b2o22bo16b2o19b2o19bo19b
o16b3o40bo16bo21bo19bo17bo19bo4b2o13b2o18bo42bo\$o3bo10bo19bo19b2o18b3o
18bo40b2o39bo20bo16bo18bo22bo17bo21bo18b2obo16bo17bo2bo38b2o16b2o19bo
18bo20b2o18b2o2bo77b2o\$b3o12bo19bo38bo59bobo2bo37b2o18bo17b2o19bo20b2o
17bo19bo21bobo15b2o18b2o77b2o17b2o20bo20bobo\$15b2o18b2o98b2o2b2o57b2o
36b2o38b2o19b2o20bobo153bo22b2o\$320bo154b2o11\$16b2o19bo18b2o20b2o17b2o
b2o17bo16b2obo15b2o19bo4b2o13bo18b2obo16bo3b2o14b2o18b2o3b2o35bo21b2o
18b2o18b2o18bo16b2o18b2o19bo22b2o14b2o17b2o20bob2o16b2o20b2o\$b3o12bobo
b2o14bobo17bo2b2o16bo2bo17bobo17bobo15bob2o16bo18bobo4bo12bobo2b2o13bo
b4o14b3o2bo14bo2b2o15bobobobo34bobo17bo2bo15b2o3bo14b2obo2bo15b3o15bo
2bo18bo18bobo21bo15bo18bo2bobo16b2obo14bo2bo19bo2bo\$o3bo13b2obo14bo2bo
17bobo2bo13bob2obo15bo3bo15bobo36bob2o16bo3bo15bobo2bo19bo16b2o16bobob
o16bobo35bo2b3o14bobobo15bo2bo16bob2obo15bo19bob3o14bo19bobo2b2o16bobo
16bo18bob2obo13b2o18b3o20bobo\$o3bo12bo19b2obo15b2o3b2o13bo3bo17b2obo
15bobobo15b5o15bobo17b5o16b3o18bo2bo14bo19bo3bo14bobobo35b2o3bo14bo2b
2o16b3o19bo15bo2b2o15b2o4bo13b2o18bo2b2obo14b4o16b2o19bo3bo14bo21b3o
16b2o2b3o\$b4o12bo21bo18bo18b3o16bobobo15b2o2bobo13bobo2bo18b2o35bo20bo
b2o15bo20b3o16bo2bo37bob2o15b2o37b2o18b2o2bo15bo3b2o15bo18b2o2bo14bo
19bo19bobo3b2o12bo19b2o3bo17bo4bo\$4bo10b2o20b2o16bobo17bobo17bobo19bo
3bo14bo25bo16bo18bo19bo16b2o19bobo19b2o35bobo20bo16b3o18bo21b2o15bo21b
o20bo16b3o18bobo16b2o18bobo17bo19b3o\$o3bo10bo21bo17b2o18b2o19bo18bo19b
2o25b3o13bobo14b3o19b2o16bo20b2o57b2o18b3o17bo2bo18bo20bo16b2o21b2o18b
2o18bo16b2ob3o35bobo17bo18bo\$b3o13bo20bo76b2o48bo13bo15bo40bo98bo20b2o
18b2o21bo40bo36bo23bo36bo16b2o\$16b2o17b3o126b2o69b2o161b2o38b2o37b2o
21b2o36b2o\$35bo402bo\$439bo\$438b2o!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 3:28 pm
A component that solves a family of related expensive objects:
Code: Select all
``````x = 138, y = 16, rule = B3/S23
8b2o38b2o37b2o38b2o\$8bo39bo38bo39bo\$9bo2b2obo33bo2b2o35bo2b2obo33bo2b
2o\$8b2o2bob2o32b2o2bo2bo32b2o2bob2o32b2o2bo2bo\$2bo4bo34bo4bo6b2o26bo4b
o34bo4bo6b2o\$3b2o3bobo32b2o3bobo32b2o3bobo32b2o3bobo\$2b2o5b2o31b2o5b2o
31b2o5b2o31b2o5b2o2\$13b2o38b2o38b2o38b2o\$5b3o5bobo29b3o5bobo29b3o5bobo
29b3o5bobo\$bo5bo5bo27bo5bo5bo27bo5bo5bo27bo5bo5bo\$b2o3bo34b2o3bo34b2o
3bo34b2o3bo\$obo37bobo37bobo37bobo\$15b2o38b2o38b2o38b2o\$15bobo37bobo37b
obo37bobo\$15bo39bo39bo39bo!``````
EDIT: Actually, it takes 4 gliders, not 5.
Code: Select all
``````x = 15, y = 13, rule = B3/S23
7b2o\$7bo\$8bo2b2obo\$7b2o2bob2o\$6bo\$bobo3bobo\$2b2o4b2o\$2bo9b2o\$5b2o5bobo
\$4bobo5bo\$2o4bo\$b2o\$o!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 10:05 pm
#185 in fourteen gliders:
Code: Select all
``````x = 80, y = 28, rule = B3/S23
35bo\$36bo\$34b3o2\$37bo\$b2o34bo\$obo34bo\$2bo9bo\$10b2o\$11b2o59bo5bo\$73bo3b
2o\$35b2o29b2o3b3o3bobo\$34bo2bo27bo2bo\$9b3o23bobo28bobo\$11bo2b2o20bob2o
27bob2o\$10bo2b2o22bo2bo27bo2bo\$15bo22b2o29b2o\$60bobo\$5b2o54b2o\$6b2o29b
2ob2o19bo6b2ob2o\$5bo3b2o25bobobobo20bo3bobobobo\$9bobo25bo3bo21b2o3bo3b
o2b3o\$9bo52bobo10bo\$76bo\$3b2o\$4b2o67bo\$3bo68b2o\$72bobo!
``````
#108 in sixteen:
Code: Select all
``````x = 225, y = 101, rule = B3/S23
2bo\$obo\$b2o39\$145bo36bo35bo\$144bobobob2o29bobobob2o28bobobob2o\$144bobo
b2obo29bobob2obo28bobob2obo\$145bo36bo35bobo\$220bo\$53b2o2b2o131b2o27bo\$
52bobob2o95b2o34bo2bo25bo\$54bo3bo95b2ob3o29bo2bo25b2o\$153bo3bo32b2o\$
158bo64bo\$214b2o6b2o\$213bobo6bobo\$183bobo29bo\$184b2o33b3o\$184bo36bo\$
175b3o42bo\$177bo6b2o10b2o\$176bo6bobo9b2o\$185bo4b2o5bo\$190bobo\$190bo4\$
43bo\$43b2o\$42bobo15\$4b2o\$3bobo\$5bo14\$109b2o\$109bobo\$109bo!
``````
#213 in fifteen:
Code: Select all
``````x = 181, y = 32, rule = B3/S23
18bobo\$18b2o\$19bo5\$76bo\$74bobo\$75b2o2\$73b2o35bo59bobo\$46bo22bo3bobo32b
2o27bo32b2o\$45bobo20bobo2bo29bo2bo2b2o20bo2bobobo2bo16bo2bob2o6bo\$45b
2o21b2o33b4o24b4ob2o2bo17b4ob2o\$obo137b3o37bo\$b2o42b2o21b2o33b2o26b2o
25b2o18b2o\$bo44bo22bo34bo27bo26bo19b2o\$45bo22bo34bo27bo26bo15b3o\$45b2o
21b2o33b2o26b2o25b2o3bobo8bo\$163b2o10bo\$164bo\$8bo\$9bo154b2o5b2o\$7b3o
155b2o3b2o\$164bo7bo2\$8b3o35b2o\$10bo35b2o\$9bo39b2o\$49bobo\$49bo!
``````
EDIT: #186 in fifteen:
Code: Select all
``````x = 109, y = 46, rule = B3/S23
62bobo\$63b2o\$63bo20bo\$82b2o7bo\$80bo2b2o6bobo\$78bobo10b2o\$79b2o2\$obo\$b
2o67bo\$bo69bo\$69b3o2\$35b2o52b2o\$36bo53bo\$7bo28bob2o50bob2o\$6bo30bobo
51bobo\$6b3o2\$8bo24bobo52b3o\$7b2o25b2o\$7bobo24bo\$b2o\$2b2o30b2o\$bo33b2o\$
34bo7\$72b2o\$71bobo\$73bo5\$70bo36b2o\$70b2o34b2o\$69bobo36bo2\$104b2o\$103b
2o\$105bo!
``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 20th, 2019, 11:28 pm
Kazyan wrote:A component that solves a family of related expensive objects:
EDIT: Actually, it takes 4 gliders, not 5.
The truth is a little more complicated than that. The four-glider method does not work when the smaller SL is a carrier, and then there are no 4-glider methods for inserting a snake that close to the feathered carrier/snake, so each SL costs 13 gliders:
Code: Select all
``````x = 178, y = 114, rule = B3/S23
43bo85bo\$41b2o84b2o\$42b2o84b2o2\$36b2o5bo34b2o13bo27b2o6bo29b2o14bo\$36b
o5b2o34bo14bobo25bo6b2o29bo15bobo\$37bo4bobo34bo13b2o28bo4bobo30bo13b2o
\$36b2o40b2o42b2o36b2o\$35bo41bo43bo37bo\$36bobo39bobo8bo32bobo35bobo8bo\$
37b2o40b2o7bo34b2o36b2o7bo\$88b3o79b3o2\$91b2o80b2o\$91bobo79bobo\$91bo81b
o30\$35bo\$33bobo88bo\$34b2o86bobo\$123b2o\$36bo\$36bobo86bo\$36b2o87bobo\$79b
o45b2o37bo\$78bobo82bobo\$25b2o42b2o7bobo72b2o8bobo\$25bo5bo37bo9bo33b2o
38bo10bo\$26bo4bo38bo2b2o6b2o30bo6bo34bo2b2o6b2o\$25b2o4bo37b2o2bo2bo4bo
bo31bo4bo33b2o2bo2bo4bobo\$24bo17b2o24bo6b2o4bo32b2o4bo32bo6b2o4bo\$25bo
bo13b2o26bobo41bo17b2o21bobo\$26b2o15bo26b2o42bobo13b2o23b2o\$115b2o15bo
37\$7b2o40b2o41b2o36b2o\$7bo41bo42bo37bo\$8bo2b2obo35bo2b2o39bo2b2obo31bo
2b2o\$7b2o2bob2o34b2o2bo2bo36b2o2bob2o30b2o2bo2bo\$6bo36bo4bo6b2o35bo32b
o4bo6b2o\$bobo3bobo34b2o3bobo35bobo3bobo30b2o3bobo\$2b2o4b2o33b2o5b2o36b
2o4b2o29b2o5b2o\$2bo9b2o74bo9b2o\$5b2o5bobo39b2o35b2o5bobo35b2o\$4bobo5bo
33b3o5bobo33bobo5bo29b3o5bobo\$2o4bo35bo5bo5bo31b2o4bo31bo5bo5bo\$b2o39b
2o3bo39b2o35b2o3bo\$o40bobo42bo36bobo\$56b2o80b2o\$56bobo79bobo\$56bo81bo!``````
Also, #102 in 10:
Code: Select all
``````x = 56, y = 19, rule = B3/S23
bo52bo\$2b2o49bo\$b2o14bo35b3o\$15b2o\$6bo9b2o31b2o\$4bobo42bobo\$5b2o10bo
32bo\$3o8bobo2b2o\$2bo8b2o3bobo\$bo10bo\$51b2o\$9bo40bo2bobo\$10b2o39b2ob2o\$
9b2o41bo\$52bo\$53b2o\$54bo\$13b2o38bo\$13b2o38b2o!``````
#18 in 10:
Code: Select all
``````x = 145, y = 24, rule = B3/S23
141bo\$142bo\$140b3o2\$143b2o\$75b2o66b2o\$74bo2bo\$bo73b2o9bo\$2bo81b2o\$3o5b
2o46bo7bo20b2o\$8bobo43bobo7bobo\$8bo46b2o7b2o67bo\$3bobo54b3o69bobo\$3b2o
57bo68bobobo\$4bo47b2o7bo9b2o57bo2bobo\$51bobo17b2o58b2o2bob2o\$53bo81b2o
bo5\$83b3o\$83bo\$84bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 21st, 2019, 12:55 am
#98 can be done in 16 by using this known converter instead of the two-step converter in the current synthesis:
Code: Select all
``````x = 28, y = 18, rule = B3/S23
13bobo\$13b2o\$obo11bo\$b2o\$bo7bobo\$9b2o6bobo\$10bo6b2o\$18bo\$25bo\$25bobo\$
21b2o2b2o\$21bobo\$9b2o10bo\$8bo2bo\$4b2obob2o\$4bob2obo\$9bo\$9b2o!``````
I'm surprised this isn't already in Shinjuku, as I recall it being a fairly common trick. Are synthesis components submitted through Catagolue automatically detected and applied to other still lifes?
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 21st, 2019, 2:15 am
Finally finished that SL I was asking for (#203) in 10:
Code: Select all
``````x = 133, y = 56, rule = B3/S23
50bo\$48b2o\$49b2o2\$6bobo\$7b2o\$7bo6\$7bo\$8b2o\$7b2o4\$9bo115bo\$10bo113bobo\$
8b3o112bo2bo\$124b2o2\$119b3o\$121bo\$120bo3\$131b2o\$37b2o91bobo\$36b2o91bo\$
38bo89bob2o\$128bo3bo\$34bo94b3o\$34b2o91bobo\$33bobo4bo86b2o\$39b2o\$39bobo
15\$b2o\$obo5bo\$2bo5b2o\$7bobo!``````
Surely this paves the way for other SLs? (#232)
Code: Select all
``````x = 103, y = 28, rule = B3/S23
21bo\$21bobo\$21b2o2\$2bo\$obo\$b2o\$26bo\$24b2o\$25b2o2\$81bo\$79bobo\$80b2o\$26b
o58b2o\$3bo21bo59b2o\$4b2o19b3o69b2o2b2o\$3b2o80b2o10bobo2bo\$85b2o8b2o2b
2o\$96bobo\$b3o91bo2bo\$3bo92b2o\$2bo3\$3b2o\$2b2o\$4bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 21st, 2019, 4:44 am
xs17_4aaj2arzx11 (#161) in 10G:
Code: Select all
``````x = 96, y = 24, rule = B3/S23
o\$b2o24bobo\$2o26b2o\$28bo65bo\$17bo75bo\$18b2o73b3o\$17b2o\$89b2o\$88bo2bo\$
89b2o2\$83bo2bo\$83b6o\$28bo60bo\$28bobo52b2o2b2o\$14bobo6bo4b2o53bo2bo\$15b
2o4bobo61b2o\$11bo3bo6b2o14b2o\$11b2o24b2o50b2o\$10bobo26bo49b2o2\$92b3o\$
92bo\$93bo!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 21st, 2019, 5:31 am
#139 (the other that SL I was inquiring above) in 10:
Code: Select all
``````x = 112, y = 46, rule = B3/S23
10bobo\$11b2o\$11bo9\$110b2o\$108bo2bo\$107bob2o\$106bobo\$97bo8bo2b2o\$96bobo
8b2o2bo\$97b2o11b2o\$33bo\$19bobo9b2o4b2o53b3o\$19b2o11b2o2b2o56bo\$20bo17b
o54bo2\$19b2o12bo\$20b2o10b2o\$19bo12bobo4\$28b3o\$28bo\$29bo2\$21b3o\$23bo\$
22bo8\$3o\$2bo\$bo!``````
And thus, with all other improvements known so far, 240 remain:
Code: Select all
``````#CLL state-numbering golly
x = 604, y = 217, rule = B3/S23
218bo198b3o\$217b2o197bo3bo\$218bo201bo\$218bo200bo\$218bo199bo\$218bo
198bo\$217b3o196b5o4\$17b3o17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b
3o17b3o17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b3o17b3o17bo19b3o
17b3o19bo16b5o16b3o16b5o16b3o17b3o\$16bo3bo15b2o18bo3bo15bo3bo17b2o
16bo19bo3bo19bo15bo3bo15bo3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo
19bo3bo19bo15bo3bo15bo3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo19bo3b
o19bo15bo3bo15bo3bo\$16bo3bo16bo22bo19bo16bobo16bo19bo22bo16bo3bo
15bo3bo15bo3bo16bo22bo19bo16bobo16bo19bo22bo16bo3bo15bo3bo15bo3bo
16bo22bo19bo16bobo16bo19bo22bo16bo3bo15bo3bo\$16bobobo16bo21bo18b2o
16bo2bo17b3o16b4o19bo17b3o17b4o15bobobo16bo21bo18b2o16bo2bo17b3o
16b4o19bo17b3o17b4o15bobobo16bo21bo18b2o16bo2bo17b3o16b4o19bo17b3o
17b4o\$16bo3bo16bo20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo15bo3bo16b
o20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo15bo3bo16bo20bo21bo15b5o
19bo15bo3bo17bo17bo3bo19bo\$16bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3b
o17bo17bo3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo17bo17b
o3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo17bo17bo3bo15bo
3bo\$17b3o16b3o17b5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b
5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b5o16b3o19bo17b3o
17b3o18bo18b3o17b3o9\$36b2o19bo18b2o40bobo15b2o18b2obo15b2o19bo19bo
38bo3b2o34b2o2b2o14b2obob2o35b2o19b2o19bo18bo17b2o2b2o15b2o18b2o
19b2o35b2o20bo18bo22b2o\$b3o32bobob2o14bobo17bo40bob2o16bo18bob2o
16bo18bobo2b2o13bobo37b3o2bo34bo3bobo13bob2obo2bo32bo2bo15bo2bo16b
2obobo16bobo15bo2bo2bo14bo2bobo14bobo18bo2bo34bobo19b3o15bobo2b2o
16bobo\$o3bo33b2obo14bo2bobo15bo39bo19bob2o19b2o14bobo17bo2bo2bo13b
obo2b2o35b2o37bobobo17bo2b2o31bobo2bo14b3o2bo14bob2o2bo14bobobo15b
obobo14bo2bob2o13bobo19bobobo36bo16b2o3bo14bo2bo2bo16bobobo\$o3bo
32bo19b2ob2o14b2o2bo37b2obo16bobo16b2o2bo15bobo17b3obo16bo3bo34bo
38b2o2bo15b3o37bob3o17b3o18b2o14bo2bobo14b2o2bo15b2obo16bo2b3o15b
2o2bo35b2o18bo2bobo14bob2o17b2o2bobo\$obobo32bo20bo19b3o35bobob2o
14bobo17bob2o19bo20bo16bob3o35bo39bo18bo40bo19bo18b2o18b2obo16bo
21bo17b2o2bo14bo2b2o35bo19bo4bo14b2o21bo3bo\$o3bo30b2o19bobo17b2o
37bobo17bobobo16bo22b2o18bo18bo39b2o36bo61bo19bo17bo20bo17bo19b3o
19bo17bobo37bob3o15bobo18bo18b3o\$o3bo30bo19bobo17bo2bo36b2o18bo2bo
16b2o20b2o2bo16bo20bo39bo36b2o57b3o17b3o19bo17bobo17b2o18bo19bo20b
2o38bobo17bobo16bo19bo\$b3o32bo18b2o19bobo57b2o39bo2bo17b2o18b2o38b
o96bo19bo20b2o17b2o58b2o63bo16bo17b2o\$35b2o40bo100b2o78b2o279b2o
12\$17bo18b2ob2o19b2o16b2o38b2o16b2o23b2o12b2o19bo19bo38bo19b2o4b2o
12b2o2b2o14b2obo38b2o18b2o21b2o16b2o15b2o20b2o18bo41bo15b2o21bo17b
o24b2o\$bo14bobo17bob2o16b2o3bo17bo39bo2b2o13bo19b2o2bo14bo2b2o14bo
bo4b2o11bobo37b3o17bobo3bobo11bo3bobo13bob2o37bo2bo15bo2bo16b2o2bo
bo15bo2bo14bo2b2o16bo2bo16bobo39bobo14bobo18b3o16bobo20bobobo\$2o
15bo2bo20bo14bo2bo16b3o40bobobo13bobo16bo2bobo14bobo2bo14bo2bo2bo
13bobo39bo17bo2b2o2bo13bobobo17b2o34bobobo15b2obob2o13bobo2bo15bo
2bobo14bobobo14bo2bobo14bo2bo38bobobo16bo16bo19bo2bob2o15bob2o\$bo
16b2obo15b2ob2o15bob2o15bo2b3o35bobo2bo15bobo16bob2o16bob2o16b2obo
bo15bo2bo35bo19b2obobo13b2o2bo15b2obo2bo33bo2b2o16bobobo16b2o15bo
4bo14b2obobo14b2obo2bo13bobob2o36bobobo14b2o18bo19bob2obo15bo3bo\$b
o18bo16bo20bo19bo2bo34bob2o16bobobo15b2o21bo19bobo15bob3o35b2o22bo
15bo18bobo2b2o34b2o18bo18b2o18b4o17bobo17bob2o15b2o2bo34b2o3bo14bo
19b2o18b2o21bob2o\$bo14b3o16bobo18bobo17bobo37bo18bobobo17bo19bobo
19bo18bo40bo35bo21bo40bo19bo18bo39bo19bo20bo36bo19bo19bo4bo15bo18b
3o\$bo13bo2b2o15b2o18bobo17bobo37b2o18bobo17bo20bobo19b2o19bo39bo
35b2o58b3o21bo16bo19b2o18b2o18b2o17bobo38b3o17b2o2b2o14bo2bobo13bo
19bo\$3o12b2o38b2o19bo59b2o17b2o20bo39bobo40b2o93bo22b2o16b2o18b2o
57b2o41bo19bo2bo15bo2bo14b2o\$217b2o42bo276b2o18b2o\$258b3o\$258bo10\$
17bo20b2o21b2o17b2o35b2o22b2o14bo17b2o19bo38b2o3b2o13b2o2b2o14b2o
18b2o2b2o14b2o21bo18b2o18b2ob2o18b2o16b2o15b2o20b2o39b2o19bo15b2o
23b2o16bo21bo\$b3o12bobo18bobo16b2o3bo19bo35bo2b2o14b2o2bobo13bobo
2b2o13bo2bo15bobo37bo2bo2bo13bo2bo2bo13bo2bobo14bo3bo15bo20b3o2bo
14bo2bo15bo2bobo14b2o2bobo15bo2bo14bo2b2o16bo2bo37bo2bo17bobo14bob
o18b2o2bo16bobo19bobo\$o3bo12bo2bo15bo19bo2bobo14bo2bo38bobobo14bo
2bo15bo2bo2bo13bobobob2o12bo2b2o36b2obo15bobob2o14b2ob2o16bobobo
15bo4bo12bo3b3o13bobo2bo14b2obo2bo13bo3bo16bo2bobo14bobo16bo2bobo
37bo2bo15bobobo16bo16bo2bobo16bob3o15bo2bo\$4bo13b2obo15b2o18b2obo
15b5o36b2o2bo15bob2o16bob2o16b2obobo2bo11b2o2bo36bob2o15bobo17bo
18b2o2b2o14b2o3bobo12b2o18bo4bo15bob2o16bob2o14bo4bo14b2obobo14bob
2o2bo34bob2obo15bobobo14b2o18bob2o15b2o5bo14b2obobo\$3bo16bo18bo18b
o57bo2b2o15b2o18b2o23bo2b2o13b2o36bo21bo17bo19bo18bo3bobo15b2o17b
4o16bo18b2o18b4o17bob2o15bo2bobo33bobo2bo14b2o3bo14bo19b2o18bo5b2o
15bo2b2o\$2bo13b4o15b4o17bobo19bo36bobo19bo19bo39b2o37bo21bo19b2o
15bo21bo2bo17bo19bo17b2o19bo19bo19bo20bobo34bo2bo16bo19bo20bo19bo
19bobo\$bo15bo17bo19bobo18b3o37bo18bo19bo42bo37b2o20b2o19bo15b2o21b
obo19bo15bo38bo19bo20b2o21bo36b2o18bo19b2o17bo21bo17bobo\$5o10bo20b
o19bo18bo59b2o18b2o39bo81bo40bo19b2o15b2o37b2o18b2o100bo20bob2o13b
2o19b2o18bo\$15b2o20bo37b2o119b2o80b2o236b2o20b2obo\$36b2o11\$17bo21b
2o38bo37b2o17b2o19bo17b2ob2o16bo38b2o18b2o3bo14b2o2bo77b2o17bo23bo
58b2o14b2o42bo15b2o21bo19bo2b2o17bo\$b3o12bobo2b2o14bo2bo37bobo36bo
18bo19bobo17bobo16bobo37bo2bob2o13bo3bobo13bo2bobo75bo2bo15bobo3b
2o12b2o2bobo54b2o2bo15bo41bobo14bo3b2o14bobobo17bobo2bo16bobo\$o3bo
12bobo2bo13bob2o36bo2bobo36bo18bo18bo2bob2o13bo2b3o14bo2b2o36b2obo
15bo2bo2bo13b2obo75bobo2bo14bobo3bo13bo3bobo53bo2b2o15bo41bobo17bo
2bo14b2obobo16bobobo17bo2bo\$4bo14b2o16bo2b3o33b2obo2bo34b2o19bo18b
ob2obo14b2o2bo15b2o2bo36bo2bo15bob3o15bo78bo4bo14bob2obo15bob2o54b
obo2b3o12b2o40bobobo14b2obo18bobo14b2o3bo16b2obobo\$2b2o14bo19bo3bo
35bo2b2o33bo19bobo17b2o20bo20b2o36bo2b2o16bo18bo79b4o17bobo15b2o
58bobo3bo14bo37b2o3b2o13bo3b2o14b2o2bo15bo21bo2b2o\$4bo11b3o16b3o
37bobo38bo18bobob2o16bo18bo20b2o37bo22bo15b2o99bo20bo59bo17b2o38bo
20b3o16bo20bo21bo\$o3bo10bo19bo39b2o38b2ob2o15bobo2bo14bo20b2o17bo
2bo37b2o20b2o15bo81b2o17b2o17bo79bo3b2o35b3o19bo18bo19bo17b3o\$b3o
11b2o100bo2bo15bobo16b2o38b2o79bo80b2o36b2o79bo2bo38bo37b2o18b2o
17bo\$117bobo17bo137b2o200bobo\$118bo359bo11\$16b2o20b2o20bo16b2o17b
2o38b2o22b2o34bo24bo13b2o2b2o14b2o3b2o33b2obob2o13b2o21b2o18bobo
57b2o15b2o2b2o34b2o23b2o17bo38bo19b2o20bo\$3bo12bobo18bo2bo17b3o17b
o18bo38bo19b2o3bo33bobo17b2o3bobo12bo2bo2bo13bo2bo2bo33bob3obo13bo
bo18b3obo15b4obo56bo16bo3bo36bo22bobo16bobo34bobobo17bo2bo18bobo\$
2b2o15bo16bo2bobo15bo18bo2b2o15bo41bo17bo2bo36bo2b2o15bo3bo2bo13b
2obo15bob3o56b3o15bo5bo13bo5bo54b2o2bo15bo3bo34bo21b3o17bobo35b2ob
obo17b2obo18bobo\$bobo13b2o2bo15b2obo15bo2b2o15b2obo2bo13b4o37b2o
18bob2o36b2o2bo14bob2obobo14bob2o15bo41b2o15bo3bo15bo5bo13b2o3b2o
52bo2b3o14b2o2b2o34b2o19bo3bo16bobobo36bobo14b3o2bo16b3o2bo\$o2bo
14bob2o16bo18b2obo17bo2b2o17bo35bo19b2o2bo38b2o16bobob2o14bo20bo
37bobobo15b3obo16bo3b2o14bo58b2o18bobo38bo17bob3o15b2o3b2o33b2o2bo
15bo2b2o17bo2b2o\$5o13bo17bobo19bo16bobo18b4o35bo2b3o14bo2bo38b2o
37bo18b3o38b2o21bo18bobo15bo61bo17bobo38bo18bo18bo39bo21bo20bo\$3bo
11b2obo16bobo18bobo16b2o20bo38b2obo16b2o40bo37b2o17bo62bo20b2o15b
2o57bobo19bo40b2o17bo18bo39bo19b2o17b3o\$3bo11bobo17b2o18bobo37bo
45bo55bo120b2o95b2o63bo15b2o19bo37b2o38bo\$56bo38b2o43b2o55b2o279b
2o36b2o\$478bo\$479bo\$478b2o9\$16b2o23b2o15bo20bo16b2o22bo15b2o21bo
16b2o18bo21b2o15b2o3b2o33b2o18bo3b2o16b2o3bo15bob2o13b2o3bo36bo43b
o35b2obo19bo18b2o14b2o21bo18b2o22bo\$5o11bo3b2o15b2o2bo15bobobo15b
3o17bo21bobo14bo21bobo14bo2bo16bobo2b2o13b2o2bo15bo2bo2bo33bo19b3o
2bo15bo2bobobo12b3obo15bo2bobo34bobo41bobo34bob2o18bobo16bobo14bo
19bobobo17bo21b3o\$o17bo2bo14bo2bobo15bob2obo13bo19bo21bobobo15bo
19bo2bo14b2obobo14bobo2bo14bobo18b2obo36bob2o17b2o15bobo2b2o2bo10b
o5bo13bo3bobo34bo2bo35b2o3bobo38b2o14b3obo15bo2bobo14bo17b2obo19bo
19bo\$o16b2obo16b2obo15b2o4bo13b2o18b2o20bobobo14b2o17b2o2b2o16bob
2o16b2o16bob4o17bo36b3o2bo15bo2b3o13bo5bobo11bo5bo13b3obo34b2ob2o
35bo2bob2o36b2o2bo13bo3bo16b2o2b2o13b2o20bobo14b3obo19bo\$b3o14bob
2o16bo18bo4b2o15bo18bo17b2o3bo14bo19bo21bo18bo19bo3bo15b2o41bo15bo
bo3bo19b2o13bob3o17bo37bo39b2o36bobob2o14bob3o15b2o18bo19b2o2b2o
14bo2bo19b2o\$4bo10b3o18bobo16bo19bob2obo17bo16bo19bo2b3o16bo18bobo
19b2o18bo18bo40b2o17bo40b2o18bo38bo40bo36b2o19bo19bo18bo4b2o13bo
22bobo16bo\$o3bo10bo19bobo17b2o18b2obobo14b2ob2o16bobo17b2o2bo17bo
16bobo21bo17b2o19bo39bo79b2o35b2o40bo59bo17bo20b2o2bo16bo21bobo15b
o\$b3o32bo42bo15bo2bo18b2o19bo17bobo17bo20bo39b2o40bo115bo41b2o57b
2o17b2o21bobo15b2o23bo13b2o\$96bobo39b2o15bobo39b2o79b2o116bo141b2o
41b2o12bo\$97bo58bo238b2o199bo\$597bo\$596b2o9\$16b2o22b2o16b2o18b2o
16b2o22bo15b2o23bo34bo38bo2bo16b2o2bo15b2o3b2o35bo20bo16b2ob2o38b
2o19b2o18bo17bo18b2ob2o15bo38b2o3b2o16bo18b2o20b2o\$b3o12bobo17b2o
3bo15bobo17bobo17bo21bobo14bo23bobo32bobo2b2o33b6o14bo2bobo14bo4bo
35bobob2o14b3o2b2o13bobo2bo36bo19bobo17bobo15bobo17bob2obo14b3o36b
o2bo2bo13bobobo17bo21bobo\$o3bo14bo16bo2bo17bobobo14bo19bo21bobo17b
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2bo15b3o16b2o2bo15bobobo21bo17bo37b2obo14b2obo19bo23bo\$o19bo17b3o
15b2o2bobo13bob4o14b2o20bobobo14b2o17bo2bo38b2o37b4o16bob2o15b5o
35b2o18bo5bo13b2o37bo2b3o14bo3bo15bob2o16bobo2bo16b2obo15b2o2bo37b
ob2o16bobo14b3o2bo19b2o\$4o13b4o36bo3bo15bo3bo15bo18b2o3b2o13bo19b
3obo36bo40bo19bo60b2o16bo3b2o15bo37bo19b3obo16bo18bob3o14bo2b2o15b
o2b2o35b3o17b2o2b2o14bo2b3o18bo\$o3bo12bo18b3o17bo21bo17bo3b2o13bo
20bo22bo38b2o36bo19bobo19bo41bo17bobo14bobo39b2o20bo17bo19bo17b2o
19b2o37bo19bo21bo21bo\$o3bo13bo16bo2bo16bo19b3o18bobo2bo14bobo16b2o
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21bo18b2o\$b3o12bobo17b2o17b2o18bo19b2ob2o17b2o18bobo15bobo40bo79bo
39b2o77bo20b2o15bo21b2o36bo59b2o22bo17bo\$15bobo119bo2bo14b2o41b2o
197b2o37bo58b2o81b2o18bo\$16bo121b2o295b2o158b3o\$595bo10\$16b2o2b2o
18b2o14b2o20bo17b2o22b2o37bo16b2o2b2o53bo39b2o2b2o14b2o20b2o22bo
16b2ob2o13bo42b2o18bo20b2o17bo16b2o37b2o21b2o17b2o\$5o11bobo2bo14b
2o3bo15bo19bobo17bo19b2o2bo36bobo14bo2bo2bo53b3o2b2o33bo3bo15bo20b
obo17b2o2bobo12b2o2bobo13bobo39bo2bo17bobo15bo2bobo16bobo15bo38bo
2bob2o13bo2bobo16bo2bo\$4bo13b2o16bo2bo17bob2o15bo2bo16bo21bobo36bo
2bo15bob3o57bo2bo35bobo16bo18bo2bob2o13bo2bo2bo13bob2o3bo12bo2bo2b
2o34b2o16bo2bobo14bobo2bo17bo2bo15bo39b2obo14b3o2bo17b2obo\$3bo13bo
20b3o15b2o2bo15bob2obo14b2o20bob2o34bo3bob2o13bo59b2obo35b2o2b2o
15bo18b3obo15bo2b2o18b2obo13b2obo2bo32b2o18b3obo15bo2b2o16b2o2b2o
14b2o40bo2bo16b2o15b3o2bo\$3bo13bo40bo18bo2b2o15bo18b2o38b2o3bobo
14bo59bob2o36bobo15b2o23bo15b2o20bobo15bob2o35bo21bo17b2o17bo19bo
39b3o2b2o13b2o18bo2b2o\$2bo12b2o19b3o17b2o20bo17bo18bo41bo17b3o58bo
41bobo14bo2b2o18b3o17bo36bo39bo19b2o20bo17bo19bob3o35bo19bo21bo\$2b
o12bo19bo2bo16bo19b3o18bobob2o14bobo36bo19bo60b2o41bo15bobo2bo17bo
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35bo402bo\$439bo\$438b2o!``````
### Re: 17 in 17: Efficient 17-bit synthesis project
Posted: June 21st, 2019, 9:33 am
Freywa wrote:#139 (the other that SL I was inquiring above) in 10:
Code: Select all
``````x = 112, y = 46, rule = B3/S23
10bobo\$11b2o\$11bo9\$110b2o\$108bo2bo\$107bob2o\$106bobo\$97bo8bo2b2o\$96bobo
8b2o2bo\$97b2o11b2o\$33bo\$19bobo9b2o4b2o53b3o\$19b2o11b2o2b2o56bo\$20bo17b
o54bo2\$19b2o12bo\$20b2o10b2o\$19bo12bobo4\$28b3o\$28bo\$29bo2\$21b3o\$23bo\$
22bo8\$3o\$2bo\$bo!``````
9:
Code: Select all
``````x = 40, y = 39, rule = B3/S23
11bobo\$12b2o\$12bo16\$34bo\$20bobo9b2o4b2o\$20b2o11b2o2b2o\$21bo17bo2\$20b2o
12bo\$21b2o10b2o\$20bo12bobo4\$29b3o\$29bo\$30bo2\$22b3o\$24bo\$23bo\$3o\$2bo\$bo
!
`````` | 27,132 | 48,152 | {"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-2021-21 | latest | en | 0.801422 |
http://mathhelpforum.com/differential-geometry/180858-using-generating-function-legendre-polynomials.html | 1,524,806,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00361.warc.gz | 191,306,111 | 10,435 | # Thread: Using the generating function for the Legendre polynomials
1. ## Using the generating function for the Legendre polynomials
Using the identity:
$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:
$\displaystyle x \in [-1,1], r \in (-1,1)$
Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.
Any help here will be amazing as I absolutely haven't got a clue!
Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$
Thank you
2. Originally Posted by garunas
Using the identity:
$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:
$\displaystyle x \in [-1,1], r \in (-1,1)$
Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.
Any help here will be amazing as I absolutely haven't got a clue!
Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$
Thank you
you need to expand the right hand side in a Taylor series around zero
so let
$\displaystyle f(r)=(1-2xr+r^2)^{-\frac{1}{2}}$
Now you get
$\displaystyle 1+f'(0)r+\frac{f''(0)}{2!}r^2+\frac{f'''(0)}{3!}r^ 3+...$
Now just equate coefficients this gives
$\displaystyle P_2(x)r^2=\frac{f''(0)}{2!}r^2 \implies P_2(x)=\frac{f''(0)}{2!}$
Can you finish from here?
3. ...i can sure finish from here. Thanks a lot!
4. Originally Posted by garunas
Using the identity:
$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:
$\displaystyle x \in [-1,1], r \in (-1,1)$
Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.
Any help here will be amazing as I absolutely haven't got a clue!
Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$
$\displaystyle \frac{1}{\sqrt {1-2xr+r^2} } = \bigl(1-(2xr-r^2)\bigr)^{-1/2}$. Use the first few terms of the binomial expansion of $\displaystyle (1-t)^{-1/2}$ (with $\displaystyle t=2xr-r^2$) to find the coefficient of $\displaystyle r^2$.
You could get $\displaystyle P_3'(0)$ by the same method. Alternatively, it might be slightly easier to start by differentiating $\displaystyle \frac{1}{\sqrt {1-2xr+r^2} }$ with respect to x, then put x=0, and finally use the binomial series to find the coefficient of $\displaystyle r^3$. | 892 | 2,758 | {"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} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.75831 |
https://eanshub.com/community/circle/a-triangle-abc-is-drawn-to-circumscribe-a-circle-of-radius-4-cm-such-that-the-segments-bd-and-dc-into-which-bc-is-divided-by-the-point-of-contact-d-are-of-lengths-8-cm-and-6-cm-respectively-see-figur/ | 1,720,946,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00111.warc.gz | 196,076,461 | 36,953 | # Forum
A triangle ABC is d...
Clear all
# A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Figure).
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A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Figure). Find the sides AB and AC.
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Given
Consider the triangle ABC,
We know that the length of any two tangents which are drawn from the same point to the circle is equal.
(i) CF = CD = 6 cm
(ii) BE = BD = 8 cm
(iii) AE = AF = x
Now, it can be observed that,
(i) AB = EB + AE = 8 + x
(ii) CA = CF + FA = 6 + x
(iii) BC = DC + BD = 6 + 8 = 14
Now the semi perimeter 's' will be calculated as follows
2s = AB + CA + BC
By putting the respective values we get,
2s = 28 + 2x
s = 14+x
Area of ΔABC = $$\sqrt{S(s-a)(s-b)(s-c)}$$
By solving this we get,
= √(14 + x)48x ……… (i)
Again, the area of ΔABC = 2 × area of (ΔAOF + ΔCOD + ΔDOB)
= 2 × [(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2 × 1/2(4x + 24 + 32) = 56 + 4x …………..(ii)
Now from (i) and (ii) we get,
√(14 + x)48x = 56 + 4x
Now, square both the sides,
48 x (14 + x) = (56 + 4x)2
48x = [4(14 + x)]2/(14 + x)
48x = 16(14 + x)
48x = 224 + 16x
32x = 224
x = 7 cm
So, AB = 8 + x
i.e. AB = 15 cm
CA = x + 6 = 13 cm
This post was modified 2 years ago by admin
Share: | 570 | 1,555 | {"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": 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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.89622 |
https://www.kylesconverter.com/area-density/short-tons-per-thousand-square-feet-to-kilograms-per-square-meter | 1,670,110,447,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00389.warc.gz | 884,467,400 | 5,797 | Convert Short Tons Per Thousand Square Feet to Kilograms Per Square Meter
Kyle's Converter > Area Density > Short Tons Per Thousand Square Feet > Short Tons Per Thousand Square Feet to Kilograms Per Square Meter
Short Tons Per Thousand Square Feet (t (US)/MSF) Kilograms Per Square Meter (kg/m2) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Kilograms Per Square Meter to Short Tons Per Thousand Square Feet
(or just enter a value in the "to" field)
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Unit Descriptions
1 Short Ton per Thousand Square Feet:
Mass of short tons per area of a thousand square feet. A short ton (US) having 2 000 international pounds of 0.45359237 kilograms. An international square foot having 0.3048 meters per side. 1 t (US)/MSF ≈ 9.764 855 272 7661 kg/m2.
1 Kilogram per Square Meter:
Mass of kilograms per an area of a square meter. 1 kg/m2 = 1 kg/m2.
Conversions Table
1 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 9.764970 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 683.5399
2 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 19.529780 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 781.1884
3 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 29.294690 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 878.837
4 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 39.0594100 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 976.4855
5 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 48.8243200 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 1952.9711
6 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 58.5891300 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 2929.4566
7 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 68.354400 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 3905.9421
8 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 78.1188500 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 4882.4276
9 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 87.8837600 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 5858.9132
10 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 97.6486800 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 7811.8842
20 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 195.2971900 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 8788.3697
30 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 292.94571,000 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 9764.8553
40 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 390.594210,000 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 97648.5527
50 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 488.2428100,000 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 976485.5273
60 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 585.89131,000,000 Short Tons Per Thousand Square Feet to Kilograms Per Square Meter = 9764855.2728 | 884 | 3,323 | {"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-2022-49 | latest | en | 0.570177 |
http://www.jiskha.com/display.cgi?id=1321567751 | 1,496,032,307,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612013.52/warc/CC-MAIN-20170529034049-20170529054049-00246.warc.gz | 665,527,191 | 3,778 | # Physics
posted by on .
A parachutist descending at a speed of 10.0 m/s loses a shoe at an altitude of 40.0 m. (Assume the positive direction is upward.) when does the shoe reach the ground??
• Physics - ,
hf=hi+vi*t-4.9t^2
vi=-10 hf=0 hi=40 solve for time t
• Physics - , | 89 | 278 | {"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-2017-22 | latest | en | 0.803865 |
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A320648 a(n) is the number of connected Veblen hypergraphs (i.e., k-uniform hypergraphs where the degree of each vertex is divisible by k) with n edges up to isomorphism. 0
0, 0, 1, 1, 2, 11, 26, 122, 781 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,5 LINKS Gregory Clark, Joshua Cooper, A Harary-Sachs Theorem for Hypergraphs, arXiv:1812.00468 [math.CO], 2018. J. Cooper, A. Dutle, Spectra of uniform hypergraphs, Linear Algebra Appl. 436 (2012) 3268-3292. EXAMPLE The only 3-uniform Veblen hypergraph with 3 edges is the single edge with multiplicity 3, {(1,2,3)^3}. The only 3-uniform Veblen hypergraph with 4 edges is the 4-uniform simplex (i.e., the tetrahedron) as shown in Cooper and Dutle. There are two 3-uniform Veblen hypergraphs with 5 edges: the crown, {(1,2,3),(1,2,4),(1,2,5),(3,4,5)^2}, and the tight 5-cycle, {(1,2,3),(2,3,4),(3,4,5),(4,5,1),(5,1,2)}. PROG (Sage) e = n #Given a 3-uniform hypergraph H, returns true if H is 3-valent. def is_3_valent(H = IncidenceStructure([[]])): return (Set([H.degree(i) % 3 for i in range(len(H.ground_set()))]) == Set([0])) #Returns a list of all connected 3-uniform Veblen hypergraphs with exactly e edges, up to isomorphism. def Veblen_3graphs(e=1): V = [] for n in range(3, e+2): #might be able to give a better bound for H in hypergraphs.nauty(e, n, uniform =3, multiple_sets = True, vertex_min_degree = 3, set_min_size = 3, connected = True): if is_3_valent(IncidenceStructure(H)): V.append(IncidenceStructure(H)) return V len(Veblen_3graphs(e)) CROSSREFS Sequence in context: A104085 A080663 A248118 * A141464 A218471 A139211 Adjacent sequences: A320645 A320646 A320647 * A320649 A320650 A320651 KEYWORD nonn,more AUTHOR Gregory J. Clark, Oct 18 2018 STATUS approved
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https://origin.geeksforgeeks.org/construct-a-tree-whose-sum-of-nodes-of-all-the-root-to-leaf-path-is-not-divisible-by-the-count-of-nodes-in-that-path/?ref=lbp | 1,686,005,407,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00634.warc.gz | 485,337,199 | 45,089 | GFG App
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# Construct a Tree whose sum of nodes of all the root to leaf path is not divisible by the count of nodes in that path
Given an N-ary tree consisting of N nodes numbered from 1 to N rooted at node 1, the task is to assign values to each node of the tree such that the sum of values from any root to the leaf path which contains at least two nodes is not divisible by the number of nodes along that path.
Examples:
Input: N = 11, edges[][] = {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 6}, {2, 10}, {10, 11}, {3, 7}, {4, 8}, {5, 9}}
Output: 1 2 1 2 2 1 1
Explanation:
According to the above assignment of values, below are all the possible paths from the root to leaf:
• Path 1 → 2 → 6, sum = 1 + 2 + 1 = 4, length = 3.
• Path 1 → 2 → 10 → 11, sum = 1 + 2 + 1 + 2 = 6, length = 4
• Path 1 → 3 → 7, sum = 1 + 2 + 1 = 4, length = 3.
• Path 1 → 4 → 8, sum = 1 + 2 + 1 = 4, length = 3.
• Path 1 → 5 → 9, sum = 1 + 2 + 1 = 4, length = 3.
From all the above paths, none of the paths exists having the sum of values divisible by their length.
Input: N = 3, edges = {{1, 2}, {2, 3}}
Output: 1 2 1
Approach: The given problem can be solved based on the observation that for any root to leaf path with a number of nodes at least 2, say K if the sum of values along this path lies between K and 2*K exclusive, then that sum can never be divisible by K as any number over the range (K, 2*K) is never divisible by K. Therefore, for K = 1, assign node values of odd level nodes as 1, and rest as 2. Follow the steps below to solve the problem:
• Initialize an array, say answer[] of size N + 1 to store the values assigned to the nodes.
• Initialize a variable, say K as 1 to assign values to each node.
• Initialize a queue that is used to perform BFS Traversal on the given tree and push node with value 1 in the queue and initialize the value to the nodes as 1.
• Iterate until then the queue is non-empty and perform the following steps:
• Pop the front node of the queue and if the value assigned to the popped node is 1 then update the value of K to 2. Otherwise, update K as 1.
• Traverse all the child nodes of the current popped node and push the child node in the queue and assigned the value K to the child node.
• After completing the above steps, print the values stored in the array answer[] as the result.
Below is the implementation of the above approach:
## C++
// C++ program for the above approach #include "bits/stdc++.h" using namespace std; // Function to assign values to nodes // of the tree s.t. sum of values of // nodes of path between any 2 nodes // is not divisible by length of path void assignValues(int Edges[][2], int n) { // Stores the adjacency list vector tree[n + 1]; // Create a adjacency list for(int i = 0; i < n - 1; i++) { int u = Edges[i][0]; int v = Edges[i][1]; tree[u].push_back(v); tree[v].push_back(u); } // Stores whether node is // visited or not vector visited(n + 1, false); // Stores the node values vector answer(n + 1); // Variable used to assign values to // the nodes alternatively to the // parent child int K = 1; // Declare a queue queue q; // Push the 1st node q.push(1); // Assign K value to this node answer[1] = K; while (!q.empty()) { // Dequeue the node int node = q.front(); q.pop(); // Mark it as visited visited[node] = true; // Upgrade the value of K K = ((answer[node] == 1) ? 2 : 1); // Assign K to the child nodes for (auto child : tree[node]) { // If the child is unvisited if (!visited[child]) { // Enqueue the child q.push(child); // Assign K to the child answer[child] = K; } } } // Print the value assigned to // the nodes for (int i = 1; i <= n; i++) { cout << answer[i] << " "; } } // Driver Code int main() { int N = 11; int Edges[][2] = {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 6}, {2, 10}, {10, 11}, {3, 7}, {4, 8}, {5, 9}}; // Function Call assignValues(Edges, N); return 0; }
## Python3
# Python3 program for the above approach from collections import deque # Function to assign values to nodes # of the tree s.t. sum of values of # nodes of path between any 2 nodes # is not divisible by length of path def assignValues(Edges, n): # Stores the adjacency list tree = [[] for i in range(n + 1)] # Create a adjacency list for i in range(n - 1): u = Edges[i][0] v = Edges[i][1] tree[u].append(v) tree[v].append(u) # Stores whether any node is # visited or not visited = [False]*(n+1) # Stores the node values answer = [0]*(n + 1) # Variable used to assign values to # the nodes alternatively to the # parent child K = 1 # Declare a queue q = deque() # Push the 1st node q.append(1) # Assign K value to this node answer[1] = K while (len(q) > 0): # Dequeue the node node = q.popleft() # q.pop() # Mark it as visited visited[node] = True # Upgrade the value of K K = 2 if (answer[node] == 1) else 1 # Assign K to the child nodes for child in tree[node]: # If the child is unvisited if (not visited[child]): # Enqueue the child q.append(child) # Assign K to the child answer[child] = K # Print the value assigned to # the nodes for i in range(1, n + 1): print(answer[i],end=" ") # Driver Code if __name__ == '__main__': N = 7 Edges = [ [ 1, 2 ], [ 4, 6 ], [ 3, 5 ], [ 1, 4 ], [ 7, 5 ], [ 5, 1 ] ] # Function Call assignValues(Edges, N) # This code is contributed by mohit kumar 29.
## Javascript
Output:
1 2 2 2 2 1 1 1 1 1 2
Time Complexity: O(N), where N is the total number of nodes in the tree.
Auxiliary Space: O(N)
My Personal Notes arrow_drop_up | 1,989 | 6,343 | {"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} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.900466 |
https://www.teachwire.net/news/the-best-factorising-quadratics-worksheets-and-resources-for-ks4-maths/ | 1,716,577,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00063.warc.gz | 877,493,494 | 37,390 | SecondaryMaths
# Factoring quadratics worksheet – Best KS4 resources and lessons
Improve your secondary students’ algebra skills by factorising quadratic equations with these lesson plans, activities, ideas and printouts…
by Teachwire
Looking for a factoring quadratics worksheet, lesson plan or other resource? We’ve got you covered…
This free eight-page, exam-style factoring quadratics worksheet is from Maths Genie.
### Corbettmaths worksheets
The website corbettmaths.com features lots of factoring quadratics worksheets that take the form of exam-style questions and textbook exercises. Check out these links:
### Let students construct their own quadratic equations
When learners are bored in maths lessons they sometimes ask, “When will we ever actually use this?”. But even topics that are unlikely to feature in everyday life can still be fun for learners to work on if they experience them as a puzzle to solve.
This lesson plan from Colin Foster offers a scenario where students need to construct their own quadratic equations in order to solve a bigger problem. Lots of practice happens along the way…
Quadratic expressions are considerably more complicated to work with than linear expressions, and students often find them hard to handle.
In this lesson plan (with worksheet), students approach factorising non-monic quadratics by trying to find factorisations which will expand and simplify to produce a quadratic expression of a specified form.
This entails lots of useful practice at expanding pairs of brackets and collecting like terms. It also gives an opportunity for students to unpick what is going on, to gain insight into how the inverse process of factorising works.
### PowerPoint lesson
This is a two-part factorising quadratics lesson from PixiMaths. The first activity focuses on factorising quadratics where the coefficient of x^2 is 1, followed by a second where the coefficient is greater than 1.
Both lesson parts are fully differentiated with examples and answers. The lesson starts with a product and sum activity to get students thinking about possible pairs of numbers.
There’s a PowerPoint lesson, and then printable Word doc worksheets for activity one and activity two.
In this unit of work from Mathcentre, students will learn how many quadratic expressions can be factorised. It’s all available in one ten-page PDF, with examples and exercises at each stage.
This video from the ever-reliable Corbettmaths is a good guide for students on factorising quadratic. It runs through some examples in a straightforward and easy-to-follow manner. You can also watch the more-advanced follow-up in part two.
### The three types of quadratics
This Maths Made Easy video covers factorising quadratics (a = 1) plus an example of factorising simple quadratics. | 550 | 2,812 | {"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} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.933284 |
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# Curator: Without question this sculpture that Mr. Rodianov
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Curator: Without question this sculpture that Mr. Rodianov [#permalink]
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25 Jan 2013, 16:23
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Curator: Without question this sculpture that Mr. Rodianov has proposed to donate to the museum is an accurate replica of the famous Egyptian Boat of the Dead, but despite its remarkable fidelity and its high artistic merit, our museum must decline the donation.
Which of the following, if true, would best explain the curator’s decision to decline the donation of the replica sculpture?
A. Rodianov has already donated a number of replicas to the museum.
B. The curator does not have sole authority to accept or decline public donations.
C. The museum currently faces a severe shortage of storage space.
D. The museum is not located in Egypt.
E. The curator and Rodianov are famously on bad terms.
Thought this was a good question, tricky upon further review. Take a stab and I will reveal OA.
Edit: by carcass
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25 Jan 2013, 17:13
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Weird question. Never seen an official question similar to this one.
Anyway, I pick C
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Re: Curator: Without question this sculpture that Mr. Rodianov [#permalink]
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25 Jan 2013, 20:49
agreed with carcass , weird . I'll pick C and definitely doesn't look like a good practice source IMO.
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Re: Curator: Without question this sculpture that Mr. Rodianov [#permalink]
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Re: Curator: Without question this sculpture that Mr. Rodianov [#permalink]
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05 Aug 2014, 06:02
Even I would go with C.
Do these types of questions come in main GMAT... The answer choices are quite strange
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Re: Curator: Without question this sculpture that Mr. Rodianov [#permalink] 05 Aug 2014, 06:02
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Display posts from previous: Sort by | 1,212 | 4,465 | {"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} | 3.59375 | 4 | CC-MAIN-2017-22 | longest | en | 0.921181 |
https://wiki.treasurers.org/w/index.php?title=Compound_Annual_Growth_Rate&oldid=10651 | 1,719,106,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00671.warc.gz | 546,293,637 | 7,192 | # Compound Annual Growth Rate
(CAGR).
The compound annual growth rate (CAGR) is calculated from total growth over a longer period as:
CAGR = (End amount / Starting amount)(1/n) - 1
Where:
n = number of years between the two points sampled
### Example 1: Sales growth over two years
Sales have grown from \$100m to \$150m over the most recent 2-year period.
The CAGR is:
= (150 / 100)(1/2) - 1
= 1.5(1/2) - 1
= 22.5%.
During this particular 2-year historical period, sales were growing at an average rate of 22.5% per annum.
However, this is not evidence about any other periods, particularly not future periods.
### Example 2: Sales growth over three months
The same formula can be used to calculate a compound annual growth rate, based on a shorter sampling period.
Sales grew from \$100m to \$115m over a historical period of 3 months (= 0.25 years).
The CAGR caclulated from this data is:
= (115 / 100)(1/0.25) - 1
= 1.154 - 1
= 74.9%.
During this particular 3-month period, sales grew at a rate of 74.9% per annum.
On its own, this is NOT evidence that sales will continue to grow at this rate during the remaining 9 months of the year, nor indeed in any other period.
Proper use of this kind of analysis will investigate the reasons for the figures, and then respond appropriately. | 349 | 1,309 | {"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} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.939477 |
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# Example 2
The principal at Park High School records the total number of students each year. The following table shows the number of students for each of the last 8 years. Create a scatter plot showing the relationship between the year and the total number of students. Show that the function is a good estimate for the relationship between the year and the population. Approximately how many students will attend the high school in year 9? | 92 | 462 | {"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-2024-22 | latest | en | 0.951276 |
https://edulissy.com/shop/projects/assignment-2-solar-system-in-a-box-solution/ | 1,611,259,918,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00057.warc.gz | 330,068,800 | 25,072 | # Assignment 2: Solar System in a Box SOlution
\$30.00
Category:
## Description
Objectives
The first objective of this assignment is to practice using transformations to set up a 3D solar system scene and animate it. The second objective of this assignment is to implement the common “trackball interface” through an understanding of transformations.
Starting early is highly recommended, as this assignment make take significantly more time than the first assignment.
Part 1: Solar System (60 points)
In this part you will create a “mini solar system” consisting of four planets, two of which have moons. A static screen shot is shown above. You will create an animation of all the planets revolving around the sun, with the moons revolving around their respective planets.
All planets and moons are spherical. The orbit of every planet is a circle. The plane of this circle is given by two angles. Let the “default” orbit be centered at the origin and lie in the X-Z plane, with the default position of the planet at the point where this default orbit intersects the +X axis. Then ф is the angle (in degrees) that a given orbit makes with the X-Z plane, and Ѳ is the angle (in degrees) made by the X-Z projection of the line joining the planet to the origin with the X-axis. That is, the orbit is the circle obtained by rotating the default orbit by ф about the Z axis, followed by Ѳ about the Y axis.
You should use the following parameters for every planet:
Planet Radius Color Orbit radius Ф Ѳ Speed Mercury 10 Light pink 200 30 30 3 Venus 20 Light yellow 300 10 335 5 Earth 40 Light blue 550 -60 50 7 Jupiter 60 Orange 900 60 20 10
In addition, the following are the parameters for the moons (ф and Ѳ for the moon’s orbit are relative to the orbit of the planet they revolve around):
Belongs to Radius Color Orbit radius Ф Ѳ Earth 20 White 100 20 10 Jupiter 20 White 120 -80 50 Jupiter 20 White 120 30 100
The radius of the sun is 70. All the orbits themselves should be drawn in white.
The whole solar system is centered at the origin of your world coordinate system. The entire solar system should be clearly visible in your screen window.
To produce the animation, you must use the orbital speeds mentioned above. Note that these are relative speeds, so you can use any scale factor to calculate the actual movement of every planet in your solar system, provided you use the same scale factor for all planets and moons. For example, you may increment the angle of rotation of Mercury by 3⁰ and Jupiter by 10⁰ respectively, and so on. Make sure that the speed is fast enough to observe multiple cycles of revolution, but slow enough to actually verify that each planet and moon is moving correctly.
Hint: Proceed in small steps and mentally work out how you should apply various transformations and the correct order. You may find it easy to start with the orbits, then the planets and finally their moons. Finally, animate the planets before animating the moons. You can use “sphere.obj” from the ObjViewer example to draw a sphere.
Part 2: Solar System in a Box (30 points)
In this part, you will pile on another set of transformations that will rotate the whole system according to mouse movements. Proceed as follows:
1. Draw a cube around the solar system using only lines. You can use the bounds of the solar system calculated in part 1 to determine the size of this cube. The solar system should be fully enclosed by the box at all times!
1. Scale the whole solar system and the cube so that you can see both clearly in the window.
1. Now implement a track-ball interface using the mouse. The trackball is an imaginary sphere surrounding your model. Imagine your model “pinned” to this sphere. As you rotate the sphere, you rotate the model accordingly.
The trackball behaves as follows: if you press the left mouse and drag the mouse vertically downwards, the cube with the solar system rotates counter-clockwise around +X axis. If you drag it to the right, it will rotate counter-clockwise around the +Y axis, and so on. Dragging it diagonally will cause similar rotations about a “middle” axis between X and Y.
A user should be able to press the mouse button, drag the mouse nonlinearly before releasing with the model rotating correctly. That is, you may not assume that a mouse drag follows a straight line.
Hint: Every mouse movement will cause a rotation. Try out various mouse movements and try to determine how specifically to use rotation.
Part 3: Documentation (10 points)
Prepare a document detailing the features and details of this program. This document should include:
1. How to use your program once it is running.
1. An explanation of how you got the trackball to work. This description should be point-wise and algorithmic in nature, rather than paragraphs of text. Show Math and transformations wherever possible (matrix notation, not code snippets).
Program details: Your program should be robust to screen refreshing and resizing (the solar system should not be stretched or squeezed). Also the cube in Part 2 should always look like a cube irrespective of how you resize the window (i.e. the cube should not stretch/squeeze either). Under no circumstances should your program crash.
What to submit: Submit the entire JOGL/QT project as a single zipped file.
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The Central Limit Theorem The applets in this section of Statistical Java allow you to see how the Central Limit Theorem works. The main page gives the characteristics of five non-normal distributions (Bernoulli, Poisson, Exponential, U-shaped, and Uniform). Users then select one of the distributions and change the sample size to see how the distribution of the sample mean approaches normality. Users can also change the... https://www.causeweb.org/repository/statjava/CLT.html
Statistical Java This is a collection of applets regarding various topics in statistics. Topics include central limit theorem, probability distributions, hypothesis testing, power, confidence intervals, correlation, control charts, experimental design, data analysis, and regression. Each topic has a description page and links to one or more applets. https://www.causeweb.org/repository/statjava/
Introductory Statistics: Concepts, Models, and Applications This goal of this resource is to aid in the understanding of the relationship between statistics and the scientific method and how it applies to psychology and the behavioral sciences. The learner will learn how to read and understand the statistics presented in the professional literature and will learn how to calculate and communicate statistical information to others. http://www.psychstat.missouristate.edu/sbk00.htm
HyperStat Online Textbook This resource contains a broad range of information concerning statistics. It is divided up into eighteen chapters and also includes links to other resources pertaining to statistics. Some of these chapters include: introduction to statistics, describing univariate data, describing bivariate data, introduction to probability, normal distribution, sampling distribution, point estimation,... http://davidmlane.com/hyperstat/
Tools for Teaching and Assessing Statistical Inference: Sampling SIM This website helps students learn concepts underlying statistical inference, through the simulation software, sampling SIM. This software lets students explore sampling distributions by building population distributions, taking random samples, and exploring the behavior of sampling distributions and confidence intervals. The site includes instructional modules and assessment instruments. http://www.tc.umn.edu/~delma001/stat_tools/
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Subscribe to bulletins | 442 | 2,436 | {"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} | 3.046875 | 3 | CC-MAIN-2015-48 | latest | en | 0.858857 |
https://icsecbsemath.com/2017/04/30/sample-questions-compound-interest-exercise-1b/ | 1,556,181,602,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578711882.85/warc/CC-MAIN-20190425074144-20190425100144-00034.warc.gz | 443,134,039 | 34,415 | Question 1: A sum is invested at compound interest compounded yearly. If the interest for two successive years be $Rs. \ 5700$ and $Rs. \ 7410$. calculate the rate of interest.
$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$
$= 7410-5700 = Rs. \ 1710$
$\Rightarrow Rs. \ 1710$ $\ is \ the \ interest \ on \ Rs. \ 5700$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 1710}{5700 \times 1} \% = 30\%$
$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(7410-5700) \times 100}{5700 \times 1} \% = 30\%$
$\\$
Question 2: A certain sum of money is put at compound interest, compounded half-yearly. If the interests for two successive half-years are $Rs. \ 650$ and $Rs. \ 760.50$; find the rate of interest.
$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$
$= 760.50-650 = Rs. \ 110.50$
$\Rightarrow Rs. \ 110.50$ is the interest on $Rs. \ 650$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times \frac{1}{2}} \% = \frac{100 \times 110.50}{650 \times \frac{1}{2}} \% = 34\%$
$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(760.50-650) \times 100}{650 \times \frac{1}{2}} \% = 34\%$
$\\$
Question 3: A certain sum amounts to $Rs. \ 5292$ in two years and $Rs. \ 5556.60$ in three years, interest being compounded annually. Find; The rate of interest. The original sum.
$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$
$= 5556.60-5292 = Rs. \ 264.60$
$\Rightarrow Rs. \ 264.60$ is the interest on $Rs. \ 5292$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 264.60}{5292 \times 1} \% = 5\%$
$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(5556.60-5292) \times 100}{5292 \times 1} \% = 5\%$
Let the sum of money $= Rs. \ 100$
Therefore Interest on it for 1st Year $= 5\% \ of \ Rs. \ 100 = Rs. \ 5$
$\Rightarrow Amount \ in \ one \ year = 100 + 5 = Rs. \ 105$
$\therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 5\% \ of \ 105 = Rs. \ 5.25$
$\Rightarrow Amount \ in \ 2nd \ year = 105 + 5.25 = Rs. \ 110.25$
When amount in two year $= Rs. \ 110.25, sum \ is \ Rs. \ 100$
$\Rightarrow \ When \ amount \ in \ two \ years= Rs. \ 5292$, then
$sum= \frac{100}{110.25} \times 5292 = Rs. \ 4800$
$\\$
Question 4: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 1089$ and for the third year it is $Rs. \ 1197.90$. Calculate the rate of interest and the sum of money.
$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$
$= 1197.90-1089 = Rs. \ 108.90$
$\Rightarrow Rs. \ 108.90$ is the interest on $Rs. \ 1089$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 108.90}{1089 \times 1} \% = 10\%$
$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(1197.90-1089) \times 100}{1089 \times 1} \% = 10\%$
Let the sum of money $= Rs. \ 100$
Therefore Interest on it for 1st Year $= 10\% \ of \ Rs. \ 100 = Rs. \ 10$
$\Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110$
$\therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 10\% \ of \ 110 = Rs. \ 11$
When interest of 2nd year $= Rs. \ 11, sum \ is \ Rs. \ 100$
$\Rightarrow \ When \ interest \ in \ two \ years= Rs. \ 1089$, then
$sum= \frac{100}{11} \times 1089 = Rs. \ 9900$
$\\$
Question 5: A person invests $Rs. \ 8000$ for $3$ years at a certain rate of interest, compounded annually. At the end of one year it amounts to $Rs. \ 9440$. Calculate;
1. The rate of interest per annum.
2. The amount at the end of the second year.
3. The interest accrued in the third year.
For 1st year: $P = Rs. \ 8000; \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = \frac{8000 \times x \times 1}{100} = Rs. \ 80x$
and, $Amount = 8000+ 80x = Rs. \ 9440 \Rightarrow x=\frac{9440-8000}{80}=18\%$
For 2nd year: $P = Rs. \ 9440; \ R=18\% \ and \ T=1 \ year$
Therefore $Interest = \frac{9440 \times 18 \times 1}{100} = Rs. \ 1699.20$
and, $Amount = 9440+1699.20 = Rs. \ 11139.2$
For 3rd year: $P = Rs. \ 11139.2; \ R=18\% \ and \ T=1 \ year$
Therefore $Interest = \frac{11139.2 \times 18 \times 1}{100} = Rs. \ 2005.05$
$\\$
Question 6: A person borrowed $Rs. \ 15,000$ for $18$ months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to $Rs. \ 15600$; calculate:
1. The rate of interest per annum;
2. The total amount of money that person must pay at the end of $18$ months in order to clear the account.
For 1st year: $P = Rs. \ 15000; \ R=x\% \ and \ T=\frac{1}{2} \ year$
Therefore $Interest =600 = \frac{15000 \times x \times \frac{1}{2}}{100} \Rightarrow x=8\%$
For 2nd year: $P = Rs. \ 15600; \ R=8\% \ and \ T=\frac{1}{2} \ year$
Therefore $Interest = \frac{15600 \times 8 \times \frac{1}{2}}{100} = Rs. \ 624$
and, $Amount = 15600+624 = Rs. \ 16224$
For 3rd year: $P = Rs. \ 16224; \ R=8\% \ and \ T=\frac{1}{2} \ year$
Therefore $Interest = \frac{16224 \times 8 \times \frac{1}{2}}{100} = Rs. \ 648.96$
and, $Amount = 16224+648.96 = Rs. \ 16872.96$
$\\$
Question 7: Ramesh invests $Rs. \ 12800$ for three years at the rate of $10\%$ per annum compound interest. Find;
1. The sum due to that person at the end of the first year.
2. The interest he earns for the second year.
3. The total amount due to him at the end of the third year. [2007]
For 1st year: $P = Rs. \ 12800; \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{12800 \times 10 \times 1}{100} = Rs. \ 1280$
and, $Amount = 12800+1280 = Rs. \ 14080$
For 2nd year: $P = Rs. \ 14080; \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{14080 \times 10 \times 1}{100} = Rs. \ 1408$
and, $Amount = 14080+1408 = Rs. \ 15488$
For 3rd year: $P = Rs. \ 15488; \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{15488 \times 10 \times 1}{100} = Rs. \ 1548.80$
and, $Amount = 15488+1548.80 = Rs. \ 17036.80$
$\\$
Question 8: The simple interest on a certain sum computes to $Rs. \ 256 in 2$ years, whereas the compound interest on the same sum at the same rate and for the same time computes to $Rs. \ 276.48$. Find the rate per cent and the sum.
Let the amount be $P$ and the rate be $x\%$
Simple Interest
$P \times \frac{x}{100} \times 2 = 256$
$\Rightarrow \frac{Px}{100}=128$
Compound Interest
For 1st year: $P = Rs. \ P; \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = \frac{P \times x \times 1}{100} = Rs. \ \frac{Px}{100}$
and, $Amount = P+\frac{Px}{100} = Rs. \ P(1+\frac{x}{100})$
For 2nd year: $P = Rs. \ P(1+\frac{x}{100}); \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = P(1+\frac{x}{100})\times \frac{x}{100} \times 1 = 276.48$
$\Rightarrow x=\frac{276.48}{128}\times 100 = 16\%$
Therefore $P \times \frac{16}{100}= 128 \Rightarrow P=Rs. \ 900$
$\\$
Question 9: On the certain sum and at a certain rate percent, the simple interest for the first year is $Rs. \ 270$ and the compound interest for the first two years is $Rs. \ 580.50$. Find the sum and the rate per cent.
Let the amount be $P$ and the rate be $x\%$
Simple Interest
$P \times \frac{x}{100} \times 1 = 270$
$\Rightarrow \frac{Px}{100}=270$
Compound Interest
For 1st year: $P = Rs. \ P; \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = \frac{P \times x \times 1}{100} = Rs. \ \frac{Px}{100}$
and, $Amount = P+\frac{Px}{100} = Rs. \ P(1+\frac{x}{100})$
For 2nd year: $P = Rs. \ P(1+\frac{x}{100}); \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = P(1+\frac{x}{100})\times \frac{x}{100} \times 1$
$\frac{Px}{100} + P(1+\frac{x}{100})\times \frac{x}{100} \times 1=580.50$
$\frac{Px}{100}(1+1+\frac{x}{100}) = 580.50$
$\Rightarrow x=15\%$
$\frac{P \times 15}{100}=270$
$\Rightarrow P=Rs. \ 1800$
$\\$
Question 10: The interest charged on a certain sum is $Rs. \ 720$ for one year and $Rs. \ 1497.60$ for two years. Find, whether the interest is simple or compound. Also, calculate the rate per cent and the sum.
Interest charged for 1st Year $Rs. \ 720$
Interest Charged for 2 years $Rs. \ 1497.60$
Interest charged for the 2nd year $1497.60-720 = \ Rs. 777.60$
Since the interest for the 2nd year is more than the 1st year, it is not simple interest. It is compound interest.
Difference between the Compound interest of two successive years
$= 777.60-720 = Rs. \ 57.60$
$\Rightarrow Rs. \ 57.60$ is the interest on $Rs. \ 720$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 57.60}{720 \times 1} \% = 8\%$
or you could also do it directly using the following approach:
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(777.60-720) \times 100}{720 \times 1} \% = 8\%$
Sum borrowed is $P = \frac{720 \times 100}{8} = 9000$
$\\$
Question 11: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 864$ and for the third year is $Rs. \ 933.12$. Calculate the rate of interest and the compound interest on the same sum and at the same rate, for the fourth year.
Difference between the Compound interest of two successive years
$= 933.12-864 = Rs. \ 69.12$
$\Rightarrow Rs. \ 69.12$ is the interest on $Rs. \ 864$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 69.12}{864 \times 1} \% = 8\%$
or you could also do it directly using the following approach:
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(933.12-864) \times 100}{864 \times 1} \% = 8\%$
For 1st year: $P = Rs. \ P; \ R=8\% \ and \ T=1 \ year$
Therefore $Interest = \frac{P \times 8 \times 1}{100} = Rs. \ \frac{8P}{100}$
and, $Amount = P+\frac{8P}{100} = Rs. \ P(1+\frac{8}{100})$
For 2nd year: $P = Rs. \ P(1+\frac{8}{100}); \ R=8\% \ and \ T=1 \ year$
Therefore $Interest = P(1+\frac{8}{100})\times \frac{8}{100} \times 1$
Given $P(1+\frac{8}{100}) \times \frac{8}{100} \times 1=864$
$\Rightarrow P (sum for 1st year)=Rs. \ 10000$
For 3rd year: $P = Rs. \ 11664; \ R=8\% \ and \ T=1 \ year$
Therefore $Interest = \frac{11664 \times 8 \times 1}{100} = Rs. \ 933.12$
and, $Amount = 11664+933.12 = Rs. \ 12597.12$
For 4th year: $P = Rs. \ 12597.12; \ R=8\% \ and \ T=1 \ year$
Therefore $Interest = \frac{12597.12 \times 8 \times 1}{100} = Rs. \ 1007.77$
$\\$
Question 12: A sum of money to $Rs. \ 20160$ in $3$ years and to $Rs. \ 24192$ in $4$ years. Calculate:
1. The rate of interest.
2. Amount in $2$ years and
3. Amount in $5$ years.
Difference between the Compound interest of two successive years
$= 24192-20160 = Rs. \ 4032$
$\Rightarrow Rs. \ 4032$ is the interest on $Rs. \ 20160$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 4032}{20160 \times 1} \% = 20\%$
For 1st year: $P = Rs. \ P; \ R=20\% \ and \ T=1 \ year$
Therefore $Interest = \frac{P \times 20 \times 1}{100} = Rs. \ \frac{20P}{100}$
and, $Amount = P+\frac{20P}{100} = Rs. \ 1.2P$
For 2nd year: $P = Rs. \ 1.2P; \ R=20\% \ and \ T=1 \ year$
Therefore $Interest = 1.2P \times \frac{20}{100} \times 1 =0.24P$
and, $Amount = 1.2P+0.24P = Rs. \ 1.44P$
For 3rd year: $P = Rs. \ 1.44P; \ R=20\% \ and \ T=1 \ year$
Therefore $Interest =1.44P \times \frac{20}{100} \times 1 =0.288P$
and, $Amount = 1.44P+0.288P = Rs. \ 1.728P$
$1.728P=20160 \Rightarrow P= Rs. \ 11666.70$
$Amount \ in \ 2 \ years = 1.44 \ 11666.70 = Rs. 16800.00$
For 4th year: $P = Rs. \ 24192; \ R=20\% \ and \ T=1 \ year$
Therefore $Interest = 24192 \times \frac{20}{100} \times 1 = Rs. \ 4838.40$
and, $Amount = 24192+4838.40 = Rs. \ 29030.4$
For 5th year: $P = Rs. \ 24192; \ R=20\% \ and \ T=1 \ year$
Therefore $Interest = 24192 \times \frac{20}{100} \times 1 = Rs. \ 4838.40$
and, $Amount = 24192+4838.40 = Rs. \ 29030.4$
$\\$
Question 13: $Rs. \ 8000$ is lent out at $7\%$ compound interest for $2$ years. At the end of the first year $Rs. \ 3560$ are returned. Calculate:
1. The interest paid for the second year.
2. The total interest paid in two years.
3. The total amount of money paid in two years to clear the debt.
For 1st year: $P = Rs. \ 8000; \ R=7\% \ and \ T=1 \ year$
Therefore $Interest = \frac{8000 \times 7 \times 1}{100} = Rs. \ 560$
and, $Amount = 8000 + 560 = Rs. \ 8560$
For 2nd year: $P = Rs. \ (8560-3560)=5000; \ R=7\% \ and \ T=1 \ year$
Therefore $Interest = \frac{5000 \times 7 \times 1}{100} = Rs. \ 350$
and, $Amount = 5000 + 350 = Rs. \ 5350$
1. i) Interest paid for 2nd year $= Rs. \ 350$
2. ii) Total interest paid in 2 years $= 560+350 = Rs. \ 910$
iii) Total money paid to clear the dues $= 8000 + 910 = Rs. \ 8910$
$\\$
Question 14: A sum of $Rs. \ 24000$ is lent out for $2$ years at compound interest, the rate of interest being $10\%$ per year. The borrower returns some money at the end of the first year and on paying $Rs. \ 12540$ at the end of the second year the total debt is cleared. Calculate the amount of money returned at the end of the first year.
For 1st year: $P = Rs. \ 24000; \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{24000 \times 10 \times 1}{100} = Rs. \ 2400$
and, $Amount = 24000+ 2400 = Rs. \ 26400$
For 2nd year: $P = Rs. \ (26400-x); \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{(26400-x) \times 10 \times 1}{100} = Rs. \ (2640-\frac{x}{10})$
and, $Amount = (26400-x)+(2640-\frac{x}{10})=12540 \Rightarrow x=Rs. \ 15000$
$\\$
Question 15: A man invests $Rs. \ 1200$ for two years at compound interest. After one year his money amounts to $Rs. \ 1275$. Find the interest for the second year correct to the nearest rupee.
For 1st year: $P = Rs. \ 1200; \ R=x\% \ and \ T=1 \ year$
Therefore $Interest = \frac{1200 \times x \times 1}{100} = Rs. \ 12x$
and, $Amount = 1200+ 12x = Rs. \ 1275 \Rightarrow x=\frac{75}{12}$
For 2nd year: $P = Rs. \ 1275; \ R=\frac{75}{12} \% \ and \ T=1 \ year$
Therefore $Interest = \frac{1275 \times\frac{75}{12} \times 1}{100} = Rs. \ 79.6875 \ or\ 80$
$\\$
Question 16: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 880$ and for the third year is $Rs. \ 968$. Calculate the rate of interest and the sum of money. [1995]
Difference between the Compound interest of two successive years
$= 968-880 = Rs. \ 88$
$\Rightarrow Rs. \ 88$ is the interest on $Rs. \ 880$
$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 88}{880 \times 1} \% = 10\%$
or you could also do it directly using the following approach:
$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$
$= \frac{(968-880) \times 100}{880 \times 1} \% = 10\%$
For 1st year: $P = Rs. \ x; \ R=10\% \ and \ T=1 \ year$
Therefore $Interest = \frac{x \times 10 \times 1}{100} = Rs. \ 0.1x$
and, $Amount = x+ 0.1x = Rs. \ 1.1x$
For 2nd year: $P = Rs. \ 1.1x; \ R=10 \% \ and \ T=1 \ year$
Therefore $Interest = \frac{1.1x \times 10 \times 1}{100} = Rs. \ 0.11x$
Given $0.11x = 880 \Rightarrow x=Rs. \ 8000$
$\\$
Question 17: The cost of a machine depreciated by $Rs. \ 4000$ during the first year and by $Rs. \ 3600$ during the second year. Calculate:
1. The rate of depreciation.
2. The original cost of the machine.
3. Its cost at the end of the third year.
Let the value of the machine is Rs. x and the rate of depreciation is r%.
For 1st year: $P = Rs. \ x; \ R=r\% \ and \ T=1 \ year$
Therefore $Depreciation = \frac{x \times r \times 1}{100} = Rs. \ \frac{xr}{100}$
and, $Value = x-\frac{xr}{100} = Rs. x(1-\frac{r}{100})$
For 2nd year: $P = Rs. x(1-\frac{r}{100}); \ R=r \% \ and \ T=1 \ year$
Therefore $Interest = x(1-\frac{r}{100}) \times \frac{r}{100} \times 1$
Given
$\frac{xr}{100} = 4000 ... ... ... ... ... ... (i)$
and
$x(1-\frac{r}{100}) \times \frac{r}{100} = 3600... ... ... ... ... ... (i)$
$Solving \ i) \ and \ ii) \ we \ get \ r=10 \% \ and \ x= Rs. \ 40000$
$Given \ 0.11x = 880 \Rightarrow x=Rs. \ 8000$
$\\$ | 6,248 | 17,021 | {"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": 249, "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} | 4.625 | 5 | CC-MAIN-2019-18 | longest | en | 0.808542 |
https://www.speedsolving.com/threads/theepiccubers-progression-thread.85043/#post-1449189 | 1,638,382,672,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00312.warc.gz | 1,059,629,050 | 20,443 | #### TheEpicCuber
##### Member
Day 1: I start by going onto the cubeskills website (developed by 2 time world champ Feliks himself). I already have an account, so I just go to the beginners area and start there. I begin by "learning" how to start a Rubik's cube, and days 2-5 will be me practicing what I have learned.
#### TheEpicCuber
##### Member
Day two: I have decided just standard beginners method is too slow. So I optimized it a bit with fingertrick, color neutrality and better cross techniques, plus four look last layer.
#### TheEpicCuber
##### Member
So it's been a while, but I have been practicing quite a bit with beginner's lookahead, 4LLL, and some F2L. I thought my brain was messed up this morning, but then I remembered I cleaned my cube and used MAX Command on the pieces. I will finish my lubing job, and crank some more solves.
#### TheEpicCuber
##### Member
So I learned something today: the D2 flick is kinda hard to learn. You might mistaken it with Ring followed by middle, but the proper way to do it is Pinkie followed by ring, or ring followed by pinkie. I learned it from JPerm's fingertrick video.
Also, I am "learning" Pll. Got Edges, corners T, F, Ja and Jb. Need to finish up adjacent corner swap and move on to opposite corner swap.
That's nice.
#### TheEpicCuber
##### Member
Been a while since an update, but I finished pll and started on some oll.
Also, I have been practicing look ahead a but more, and it is turning up nice.
Also, did a 5x5 average of 5.
Generated By csTimer on 2021-08-17
solves/total: 5/5
single
best: 1:37.27
worst: 1:58.33
mean of 3
current: 1:54.51 (σ = 5.77)
best: 1:48.86 (σ = 10.39)
avg of 5
current: 1:52.40 (σ = 4.74)
best: 1:52.40 (σ = 4.74)
Average: 1:52.40 (σ = 4.74)
Mean: 1:50.56
Time List:
1. 1:37.27 R Uw R2 B' F' Lw2 Rw2 B2 Fw2 Lw Fw' B2 U2 Rw2 B' Lw' D F2 D' Lw' L' B Lw F' D Fw2 U2 Bw2 Fw2 Uw2 B F2 U Rw' Fw2 D' Bw B2 Rw R2 Fw F2 Uw2 B' Lw' D' Dw2 L B Rw U' R' Uw L R2 Dw' L2 Bw' Dw' L
2. 1:51.99 F2 Uw D2 U B' U D' Bw' D2 Fw F Rw Bw' U2 F L2 Lw2 Dw' B' Fw Dw' D U Uw' R' D2 Rw B' U' Uw' Lw' Fw Bw2 L' B' Lw2 B' Fw2 Uw' B' Bw Dw L2 R' Dw R D Lw2 Bw2 U D2 Dw L2 D R2 U' B Lw R' U2
3. 1:57.33 D' R' F Lw' B2 U' B Dw' U Rw2 Uw F L2 Lw Uw U L2 D2 Rw' Fw U Lw' U' Uw2 F' U2 D2 Fw' U2 Bw' U B2 Dw F' Uw' Fw2 D2 Lw' Fw' U' Lw2 B' U Uw2 B Dw' Rw F' Dw2 Fw L U2 F Lw2 D2 B' D R Uw Lw
4. 1:47.87 Fw F2 B U2 F' Uw2 Dw2 D' Rw' Uw' Bw Uw F Dw2 Rw F L' Uw' F Lw' B' Lw U2 Lw Bw' Uw2 B Dw' Fw2 Bw2 D' L2 Lw2 Bw2 D2 L2 Fw' R' Bw2 D' Dw Lw B' Fw' D' Bw' B2 Fw L2 Rw' Fw' F' Dw D F' D Uw' R2 Rw' U'
5. 1:58.33 Uw' D2 Dw2 B2 F R B' Dw2 Bw' Uw' R L2 Bw U2 Lw' F' Bw' L' Lw2 Fw Rw Bw Fw U2 Dw Uw2 Rw Fw2 F' Rw2 B2 Lw' L' B L2 Dw2 U2 Bw2 Rw' R2 L U2 D Dw' Bw' B' D2 Rw Bw2 Fw Lw' Dw' Rw R' Bw R' Rw2 Lw' B2 D
#### Attachments
• Screenshot 2021-08-17 at 8.32.09 PM.png
114.1 KB · Views: 3
#### TheEpicCuber
##### Member
That last 5x5 solve had 2 look-ups. I was suprised with sub-2 average.
Yay for me.
#### TheEpicCuber
##### Member
Okay, started on megaminx solving again. I want to be average sub-1 minute by the end of the year, and following that is 6x6 goal of sub-2 minutes and 7x7 goal of sub-4 minutes.
#### TheEpicCuber
##### Member
Basically full oll on 3x3, just learning better finger tricks for them algorithms
#### TheEpicCuber
##### Member
Bruh, you have a problem for me phrasing it that way?
#### TheEpicCuber
##### Member
Did the competition for 5x5.
Wait, hol' up, is that a sub-1:30 solve?
Yeah, I smashed my PB during that solve by 8 seconds.
#### TheEpicCuber
##### Member
Alright, many people say that the Yuxin Little Magic is horrible, and trash, and very slow… but today, I am creating this progression thread to break the world record… with a puzzle that may be a bit slow, but with what I believe has a big chance
(Maybe I am over-exaggerating, but whatever)
Last edited:
#### TheCubingCuber347
##### Member
Alright, many people say that the Yuxin Little Magic is horrible, and trash, and very slow… but today, I am creating this progression thread to break the world record… with a puzzle that may be a bit slow, but with what I believe has a big chance
By the time you're that fast everybody will be using the CH XMD Riptide [magnetic core/internal pieces] primary internals UV coated size changing 4x4 V2m (WB edition) and the YLM will be a thing of the past.
#### TheEpicCuber
##### Member
By the time you're that fast everybody will be using the CH XMD Riptide [magnetic core/internal pieces] primary internals UV coated size changing 4x4 V2m (WB edition) and the YLM will be a thing of the past.
Dang, that’s harsh.
Look, if Yuxin ever makes a 2nd edition of the cube by the time I break a world record, then be it.
#### cuberswoop
##### Member
Yuxin little magic isn't trash though.
#### CFOP2020
##### Member
Alright, many people say that the Yuxin Little Magic is horrible, and trash, and very slow… but today, I am creating this progression thread to break the world record… with a puzzle that may be a bit slow, but with what I believe has a big chance
What do you average right now? And are you going for WR Single or Avg?
#### TheEpicCuber
##### Member
What do you average right now? And are you going for WR Single or Avg?
50-60 seconds. And I am going for average.
Which hasn't been broken for a little less than 2 years.
#### GodCubing
##### Member
Try learning the OBLBL method it is more efficient than Yau, or use Triforce which is also very efficient and ergonomic. | 1,912 | 5,514 | {"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-2021-49 | latest | en | 0.884102 |
http://www.lmfdb.org/Genus2Curve/Q/971/a/971/1 | 1,563,591,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526408.59/warc/CC-MAIN-20190720024812-20190720050812-00491.warc.gz | 236,981,723 | 28,102 | # Properties
Label 971.a.971.1 Conductor 971 Discriminant -971 Sato-Tate group $\mathrm{USp}(4)$ $$\End(J_{\overline{\Q}}) \otimes \R$$ $$\R$$ $$\End(J_{\overline{\Q}}) \otimes \Q$$ $$\Q$$ $$\overline{\Q}$$-simple yes $$\mathrm{GL}_2$$-type no
# Related objects
Show commands for: Magma / SageMath
## Minimal equation
magma: R<x> := PolynomialRing(Rationals()); C := HyperellipticCurve(R![0, 1, 0, -2, 0, 1], R![1]);
sage: R.<x> = PolynomialRing(QQ); C = HyperellipticCurve(R([0, 1, 0, -2, 0, 1]), R([1]))
$y^2 + y = x^5 - 2x^3 + x$
## Invariants
magma: Conductor(LSeries(C)); Factorization($1); $$N$$ = $$971$$ = $$971$$ magma: Discriminant(C); Factorization(Integers()!$1); $$\Delta$$ = $$-971$$ = $$-1 \cdot 971$$
### G2 invariants
magma: G2Invariants(C);
$$I_2$$ = $$1024$$ = $$2^{10}$$ $$I_4$$ = $$16384$$ = $$2^{14}$$ $$I_6$$ = $$5139456$$ = $$2^{10} \cdot 3 \cdot 7 \cdot 239$$ $$I_{10}$$ = $$-3977216$$ = $$-1 \cdot 2^{12} \cdot 971$$ $$J_2$$ = $$128$$ = $$2^{7}$$ $$J_4$$ = $$512$$ = $$2^{9}$$ $$J_6$$ = $$2000$$ = $$2^{4} \cdot 5^{3}$$ $$J_8$$ = $$-1536$$ = $$-1 \cdot 2^{9} \cdot 3$$ $$J_{10}$$ = $$-971$$ = $$-1 \cdot 971$$ $$g_1$$ = $$-34359738368/971$$ $$g_2$$ = $$-1073741824/971$$ $$g_3$$ = $$-32768000/971$$
Alternative geometric invariants: G2
## Automorphism group
magma: AutomorphismGroup(C); IdentifyGroup($1); $$\mathrm{Aut}(X)$$ $$\simeq$$ $$C_2$$ (GAP id : [2,1]) magma: AutomorphismGroup(ChangeRing(C,AlgebraicClosure(Rationals()))); IdentifyGroup($1); $$\mathrm{Aut}(X_{\overline{\Q}})$$ $$\simeq$$ $$C_2$$ (GAP id : [2,1])
## Rational points
magma: f,h:=HyperellipticPolynomials(C); g:=4*f+h^2; HasPointsEverywhereLocally(g,2) and (#Roots(ChangeRing(g,RealField())) gt 0 or LeadingCoefficient(g) gt 0);
This curve is locally solvable everywhere.
magma: [C![-1,-1,1],C![-1,0,1],C![0,-1,1],C![0,0,1],C![1,-1,1],C![1,0,0],C![1,0,1]];
All rational points: (-1 : -1 : 1), (-1 : 0 : 1), (0 : -1 : 1), (0 : 0 : 1), (1 : -1 : 1), (1 : 0 : 0), (1 : 0 : 1)
magma: #Roots(HyperellipticPolynomials(SimplifiedModel(C)));
Number of rational Weierstrass points: $$1$$
## Invariants of the Jacobian:
Analytic rank: $$1$$
magma: TwoSelmerGroup(Jacobian(C)); NumberOfGenerators($1); 2-Selmer rank: $$1$$ magma: HasSquareSha(Jacobian(C)); Order of Ш*: square Tamagawa numbers: 1 (p = 971) magma: TorsionSubgroup(Jacobian(SimplifiedModel(C))); AbelianInvariants($1);
Torsion: $$\mathrm{trivial}$$
### Sato-Tate group
$$\mathrm{ST}$$ $$\simeq$$ $\mathrm{USp}(4)$ $$\mathrm{ST}^0$$ $$\simeq$$ $$\mathrm{USp}(4)$$
### Decomposition
Simple over $$\overline{\Q}$$
### Endomorphisms
Not of $$\GL_2$$-type over $$\Q$$
Endomorphism ring over $$\Q$$:
$$\End (J_{})$$ $$\simeq$$ $$\Z$$ $$\End (J_{}) \otimes \Q$$ $$\simeq$$ $$\Q$$ $$\End (J_{}) \otimes \R$$ $$\simeq$$ $$\R$$
All $$\overline{\Q}$$-endomorphisms of the Jacobian are defined over $$\Q$$. | 1,185 | 2,887 | {"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} | 3.234375 | 3 | CC-MAIN-2019-30 | latest | en | 0.371634 |
https://openwetware.org/wiki/BME100_f2015:Group1_8amL2 | 1,606,178,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00408.warc.gz | 430,429,742 | 7,444 | # BME100 f2015:Group1 8amL2
BME 100 Fall 2015 Home
People
Lab Write-Up 1 | Lab Write-Up 2 | Lab Write-Up 3
Lab Write-Up 4 | Lab Write-Up 5 | Lab Write-Up 6
Course Logistics For Instructors
Photos
Wiki Editing Help
# OUR TEAM
Name: Maria Dooling Name: Tiffany Nguyen Name: Nathaniel Olsen Name: Nolveto Gutierrez Name: Brenden Smith Name: Dalia Sarbolandi
# LAB 2 WRITE-UP
## Descriptive Statistics
Experiment 1
The Mean and Variability Values in Humans
Dosage 0mg ' 5mg ' 10mg ' 15mg Average 3.834 8.932 61.622 657.941 Standard Deviation 1.523010177 1.593931547 30.11069386 212.9429762 Endpoint # 10 10 10 10 Standard Error 0.481618106 0.504045412 9.521837451 67.33848166
Experiment 2
The Mean and Variability Values in Rats
Dosage 0mg 10mg Average 10.516 11.112 Standard Deviation 2.225551617 7.402885924 Endpoint # 5 5 Standard Error 0.995296941 3.310671231
Experiment 1
Experiment 2
## Analysis
Experiment 1
Human Study Inferential Statistics: P-Value: 1.4E-16
Post-hoc Tests T-test value Corrected Significant? 0mg vs 5mg 8.60E-07 0.0167 Yes 5mg vs 10mg 3.02E-05 0.0167 Yes 10mg vs 15msg 6.48E-08 0.0167 Yes
There is an observable difference between levels of Inflammotin levels for each dosage in comparison to each other, since the p value for the comparisons is less than .05. The use of ANOVA is needed because we are investigating variance between several groups. Experiment 2
Rat Study Inferential Statistics:
Dosage 0mg 10mg Average 10.516 11.112 Standard Deviation 2.225551617 7.402885924 Endpoint # 5 5 Standard Error 0.995296941 3.310671231 T-test 0.861259
There is no significant difference between the 2 groups of rats, 0mg and 10mg, because the p value is .86, which is higher than .05. The T-test was needed because it was comparing 2 groups.
## Summary/Discussion
The data of the experiment has a purpose of showing the effects of lipopolysaccharide in 2 different species, rats and humans. The data showed that there was no statistically significant effects on rats, however there was a significant effect on humans. Because there were 2 sample groups of rats, a t-test was needed for the experiment. This showed a p value of .867 for rats, showing that the effect on inflammation concentration in rats was minimal. An ANOVA test was used to compare the several groups of human samples. The p-value for the ANOVA test was 1.40083E-16. Because that number is less than .05, there is a statistically significant difference in the data for humans. All of the data sets were significantly different from each other. Some reasons why the effectiveness results varied between species may be because: the number of samples in each group, the dosage difference in each test, and many other possibilities. | 784 | 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} | 3.46875 | 3 | CC-MAIN-2020-50 | latest | en | 0.794505 |
http://eclecticlvng.blogspot.com/2011/02/hands-on-equations.html | 1,660,919,687,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00315.warc.gz | 13,678,183 | 14,058 | ## Tuesday, February 1, 2011
### Hands-On Equations
While I have Algebra on my mind from the recent post... The olders use along with Montessori materials. However, dd took a break between Pre-Algebra and Algebra to do It has a video (or DVD) for instruction and short leveled workbooks. It uses a balance, pawns, and number cubes to do the problems with manipulatives. This curriculum could be set out as shelf work or done in a structured way. We did it both ways. Honestly, she didn't choose it for shelf work ahead of time (she also already knew we'd be tackling it so I guess she didn't feel compelled to start early on it). Then she did that as her Math time as mentioned above.
Since dd was older and had already done Pre-Algebra she zipped through it pretty quickly. If we had done it before Pre-Alg she would've gone slower.
They say that it can be used from Grade 3 on up. Youngest may or may not be ready but we have so much else going on in math right now that I'd rather not put it out at the moment. She is still using the Dot Game almost every day :). [I posted a picture of the Dot Game in progress below.]
You may be able to find some used products here:
Search Amazon.com for Borenson algebra
Here is their site to see it better:
http://www.borenson.com/
Dot Game: We use a grease pencil to write on the glass. It's simply a printout in a frame from the dollar store. For the grid paper I printed some 1/2 in square grid paper (instead of the 4 squares per inch that the quadrille notebooks have) and put it in a portfolio folder for her to use with Math shelf work:
What's Cookin'?
Beef Stew: I adjusted a recipe with what I had on hand. I ended up tossing the following in the crockpot: 5 small coarsely chopped (organic) potatoes; 2 coarsely chopped (organic) carrots; 1 fresh garlic clove ("smooshed" on the counter with the side of a knife); 2 small (organic) beets; 1 pound stew meat (grass-fed); 1 bay leaf. The stew was good but a little dry. I should've added a bit of broth to it. Unfortunately it wasn't enough for my hungry dc who wanted seconds and we had to supplement. Fortunately middle dd had just shot a kitchen video of her making strawberry cream cheese (she made the cream cheese from yogurt in the video) and I had tossed some yogurt dough in the toaster oven for crackers, so we ate that up too!
[The reason I use organic potatoes and carrots is because they are some of the biggies for absorbing pesticides and herbicides. In fact, carrots are actually used as a throw-away crop to clear fields of (natural) arsenic in the soil. The carrots are just a few more pennies, literally, than non-organic at the local Kroger. Potatoes are more but I was finding them for almost the same price as non-organic for a while at Wal-Mart. At Kroger they are just over \$4 for 5 pounds (I think)] [And yet, I buy non-organic potato chips for me 'cuz they don't use corn oil and I can eat them - rolling eyes at my own incongruencies.]
Mini Pesto Pizzas: I used yogurt dough to make mini individual pesto pizzas. I rolled out small circles of yogurt dough and baked them first. I put a dab of pesto, some cooked chicken, and some mozzarella on each. On dh's and mine I put some tomato also. Then I broiled them for just a bit until the cheese melted.
Mayonnaise: Yeah, it worked this time! I talked about my flop here but this time I used Katie's technique with the following ingredients: 2 egg yolks (room temperature); 2 tablespoons of lemon juice; salt; pinch of sugar; 1 Cup oil (I think I used sunflower - I still switch these around because I'm never sure which to use). I would've put mustard but since I had just had a flop I didn't want to add any variables. It's such a pretty yellow color but I'm not sure if you'll see it well in the picture:
Vanilla Beans: A friend is ordering these vanilla beans and offered to split the order with several people. I can't wait. I completely trust her judgement on this choice. My batch of homemade vanilla should be ready to use soon. | 1,012 | 4,061 | {"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} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.952005 |
https://stats.libretexts.org/Courses/Concord_University/Elementary_Statistics/09%3A_Hypothesis_Testing_with_One_Sample/9.03%3A_A_Single_Population_Mean_using_the_Normal_Distribution | 1,660,012,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00558.warc.gz | 525,377,861 | 32,859 | # 9.3: A Single Population Mean using the Normal Distribution
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All hypotheses tests have the same basic steps:
1. Determine the hypothesis: What are we trying to figure out? This is formally written as the null and alternative hypotheses.
1. The alternative hypothesis, $$H_{a}$$, never has a symbol that contains an equal sign.
2. The alternative hypothesis, $$H_{a}$$, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
2. Calculate the evidence: This will be a test statistics and either a critical value or a p-value.
1. In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset $$\alpha$$. The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. If no level of significance is given, a common standard to use is $$\alpha = 0.05$$.
2. When you calculate the $$p$$-value and draw the picture, the $$p$$-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
3. Make a decision: The options will be Reject the Null Hypothesis or Do not Reject the Null Hypothesis.
1. Never, ever, Accept the Null Hypothesis.
2. Thinking about the meaning of the $$p$$-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller $$p$$-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a $$p$$-value of 0.056 ($$\alpha = 0.05$$ is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
4. Determine the conclusion: What does the decision mean in terms of the problem given?
## Direction of Tail
##### Example $$\PageIndex{1}$$
$$H_{0}: \mu \geq 5, H_{a}: \mu < 5$$
Test of a single population mean. $$H_{a}$$ tells you the test is left-tailed. The picture of the $$p$$-value is as follows:
##### Exercise $$\PageIndex{1}$$
$$H_{0}: \mu \geq 10, H_{a}: \mu < 10$$
Assume the $$p$$-value is 0.0935. What type of test is this? Draw the picture of the $$p$$-value.
left-tailed test
##### Example $$\PageIndex{2}$$
$$H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2$$
This is a test of a single population proportion. $$H_{a}$$ tells you the test is right-tailed. The picture of the p-value is as follows:
##### Exercise $$\PageIndex{2}$$
$$H_{0}: \mu \leq 1, H_{a}: \mu > 1$$
Assume the $$p$$-value is 0.1243. What type of test is this? Draw the picture of the $$p$$-value.
right-tailed test
##### Example $$\PageIndex{3}$$
$$H_{0}: \mu = 50, H_{a}: \mu \neq 50$$
This is a test of a single population mean. $$H_{a}$$ tells you the test is two-tailed. The picture of the $$p$$-value is as follows.
##### Exercise $$\PageIndex{3}$$
$$H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5$$
Assume the p-value is 0.2564. What type of test is this? Draw the picture of the $$p$$-value.
two-tailed test
### Full Hypothesis Test Examples
##### Example $$\PageIndex{4}$$
Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.
$$P$$-value Solution
Determine the hypothesis:
Since the problem is about a mean, this is a test of a single population mean.
For Jeffrey to swim faster, his time will be less than 16.43 seconds. So the claim will be that he can swim it in less time than 16.43 seconds.
$$H_{0}: \mu \geq 16.43$$
$$H_{a}: \mu < 16.43$$ (claim)
The "$$<$$" in the alternative hypothesis tells you this is left-tailed.
Calculate the evidence:
Use the Standard Normal Distribution since the population standard deviation is given.
Calculate the test statistic using the same formula as a $$z$$-score using the Central Limit Theorem.
$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\nonumber$
$$\mu = 16.43$$ comes from $$H_{0}$$ and not the data. $$\sigma = 0.8$$ and $$n = 15$$. Which gives
$z=\frac{16-16.43}{\frac{0.8}{\sqrt{15}}}=\frac{-0.43}{\frac{0.8}{3.87298}}=\frac{-0.43}{0.20656}=-2.0817\nonumber$
Now calculate the p-value based on the test statistic found.
This is a left-tailed test, so use the Excel formula $$=\text{NORM.S.DIST}(z,\text{true})$$.
In this case, we found $$z$$, which is the test statistic, to be $$z=-2.0817$$.
Use the Excel formula $$=\text{NORM.S.DIST}(-2.0817,\text{true})=0.0187$$.
So the $$p\text{-value} = 0.0187$$. This is the area to the left of the sample mean, which is given as 16.
Make a decision:
Interpretation of the $$p-\text{value}$$: If $$H_{0}$$ is true, there is a 0.0187 probability (1.87%) that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.
$$\mu = 16.43$$ comes from $$H_{0}$$. Our assumption gives $$\mu = 16.43$$.
$$\alpha$$ is the minimum area that could be considered to make our result significant.
Compare $$\alpha$$ and the $$p\text{-value}$$
• If $$p\text{-value}$$ is less than the $$\alpha$$ then we will Reject $$H_0$$.
• If $$\alpha$$ is less than the $$p\text{-value}$$ then we will Fail to Reject $$H_0$$.
$$\alpha = 0.05$$ and $$p\text{-value} = 0.0187$$, so $$p\text{-value}<\alpha$$
Since $$p\text{-value}<\alpha$$, reject $$H_{0}$$.
Conclusion:
This means that you reject $$\mu \geq 16.43$$.
There is sufficient evidence to support the claim that Jeffrey's mean swim time for the 25-yard freestyle is less than 16.43 seconds.
Critical Value Solution
Determine the hypothesis (Same as the $$P$$-value solution):
Since the problem is about a mean, this is a test of a single population mean.
For Jeffrey to swim faster, his time will be less than 16.43 seconds. So the claim will be that he can swim it in less time than 16.43 seconds.
$$H_{0}: \mu \geq 16.43$$
$$H_{a}: \mu < 16.43$$ (claim)
The "$$<$$" in the alternative hypothesis tells you this is left-tailed.
Calculate the evidence:
Use the Standard Normal Distribution since the population standard deviation is given.
Calculate the critical value. Use the Standard Normal Distribution, Critical Value, Right-tail Excel formula: $$=\text{NORM.S.INV}(\alpha)$$.
In this problem, the $$\alpha=0.05$$, so use $$=\text{NORM.S.INV}(0.05)=-1.64485$$
Calculate the test statistic using the same formula as a $$z$$-score using the Central Limit Theorem.
$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\nonumber$
$$\mu = 16.43$$ comes from $$H_{0}$$ and not the data. $$\sigma = 0.8$$ and $$n = 15$$. Which gives
$z=\frac{16-16.43}{\frac{0.8}{\sqrt{15}}}=\frac{-0.43}{\frac{0.8}{3.87298}}=\frac{-0.43}{0.20656}=-2.0817\nonumber$
Make a decision:
Graph the critical value and the test statistic along the number line of the Standard Normal Distribution graph.
Since this is left-tailed, everything less than the critical value, $$\text{CV}=-1.64485$$ will be the rejection region.
Since the test statistic, $$z=-2.0817$$ is less than the critical value, $$\text{CV}=-1.64485, the decision will be to Reject the Null Hypothesis. Conclusion (Same as the \(P$$-value solution):
This means that you reject $$\mu \geq 16.43$$.
There is sufficient evidence to support the claim that Jeffrey's mean swim time for the 25-yard freestyle is less than 16.43 seconds.
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
##### Exercise $$\PageIndex{4}$$
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset $$\alpha = 0.01$$. Assume the throw distances for footballs are normal. Use the critical value method.
Determine the hypothesis:
Since the problem is about a mean, this is a test of a single population mean.
For Marco to throw farther, his distance will be greater than 40 yards. So the claim will be that he can throw farther than 40 yards.
$$H_{0}: \mu \leq 40$$
$$H_{a}: \mu > 40$$ (claim)
The "$$>$$" in the alternative hypothesis tells you this is right-tailed.
Calculate the evidence:
Use the Standard Normal Distribution since the population standard deviation is given.
Calculate the critical value. Use the Standard Normal Distribution, Critical Value, Right-tail Excel formula: $$=\text{NORM.S.INV}(1-\alpha)$$.
In this problem, the $$\alpha=0.01$$, so use $$=\text{NORM.S.INV}(1-0.01)=2.3263$$
Calculate the test statistic using the same formula as a $$z$$-score using the Central Limit Theorem.
$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\nonumber$
$$\mu = 40$$ comes from $$H_{0}$$ and not the data. $$\sigma = 2$$ and $$n = 20$$. Which gives
$z=\frac{45-40}{\frac{2}{\sqrt{20}}}=\frac{5}{\frac{2}{4.4721}}=\frac{5}{0.4472}=11.1803\nonumber$
Make a decision:
Graph the critical value and the test statistic along the number line of the Standard Normal Distribution graph.
Since this is right-tailed, everything greater than the critical value, $$\text{CV}=2.3263$$ will be the rejection region.
Since the test statistic, $$z=11.1803$$ is greater than the critical value, $$\text{CV}=2.3263$$, the decision will be to Reject the Null Hypothesis.
Conclusion:
This means that you reject $$\mu \leq 40$$.
There is sufficient evidence to support the claim that the change in Marco's grip improved his throwing distance to give a mean throw distance is greater than 40 yards.
##### Example $$\PageIndex{5}$$
A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was great than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights are given below
205 215 225 252 265 313 316 316 341 368 205 215 241 252 275 313 316 338 345 368 205 215 241 265 275 316 316 338 345 385
Conduct a $$p$$-value hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Determine the hypothesis:
Since the problem is about a mean weight, this is a test of a single population mean.
$$H_{0}: \mu \leq 275$$
$$H_{a}: \mu > 275$$ (claim)
The "$$>$$" in the alternative hypothesis tells you this is a right-tailed test.
Calculate the evidence:
Use the Standard Normal Distribution since the population standard deviation is given.
Calculate the test statistic using the same formula as the $$z$$-score using the Central Limit Theorem.
$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\nonumber$
$$\mu = 275$$ comes from $$H_{0}$$ and not the data. $$\sigma=55$$ and $$n=30$$. The problem does not give the sample mean, so that will need to be calculated using the data.
Enter the data into Excel, and use the Excel formula $$=\text{AVERAGE}()$$ to find $$\bar{x}=286.2$$.
$z=\frac{286.2-275}{\frac{55}{\sqrt{30}}}=\frac{11.2}{\frac{2}{5.4772}}=\frac{11.2}{10.04}=1.11536\nonumber$
Now calculate the $$p$$-value based on the test statistic found.
This is a right-tailed test, so use the Excel formula $$=1-\text{NORM.S.DIST}(z,\text{true})$$.
In this case, we found $$z$$, which is the test statistic, to be $$z=1.11536$$.
Use the Excel formula $$=1-\text{NORM.S.DIST}(1.11536,\text{true})=0.132348$$.
So the $$p\text{-value} = 0.132348$$.
Interpretation of the p-value: If $$H_{0}$$ is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
Make a decision:
$$\alpha$$ is the minimum area that could be considered to make our result significant.
Compare $$\alpha$$ and the $$p\text{-value}$$
• If $$p\text{-value}$$ is less than the $$\alpha$$ then we will Reject $$H_0$$.
• If $$\alpha$$ is less than the $$p\text{-value}$$ then we will Fail to Reject $$H_0$$.
$$\alpha = 0.025$$ and $$p$$-value $$= 0.1323$$
Since $$\alpha < p\text{-value}$$, do not reject $$H_{0}$$.
Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
### Review
The hypothesis test itself has an established process. This can be summarized as follows:
1. Determine $$H_{0}$$ and $$H_{a}$$. Remember, they are contradictory.
2. Find the evidence: Draw a graph, calculate the test statistic, and use the test statistic to calculate the $$p\text{-value}$$. (A z-score and a t-score are examples of test statistics.)
3. Compare the preconceived α with the p-value, make a decision (reject or do not reject H0).
4. Write a clear conclusion using English sentences.
Notice that in performing the hypothesis test, you use $$\alpha$$ and not $$\beta$$. $$\beta$$ is needed to help determine the sample size of the data that is used in calculating the $$p\text{-value}$$. Remember that the quantity $$1 – \beta$$ is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.
##### Exercise $$\PageIndex{5}$$
Assume $$H_{0}: \mu = 9$$ and $$H_{a}: \mu < 9$$. Is this a left-tailed, right-tailed, or two-tailed test?
This is a left-tailed test.
##### Exercise $$\PageIndex{6}$$
Assume $$H_{0}: \mu \leq 6$$ and $$H_{a}: \mu > 6$$. Is this a left-tailed, right-tailed, or two-tailed test?
##### Exercise $$\PageIndex{7}$$
Assume $$H_{0}: p = 0.25$$ and $$H_{a}: p \neq 0.25$$. Is this a left-tailed, right-tailed, or two-tailed test?
This is a two-tailed test.
##### Exercise $$\PageIndex{8}$$
Draw the general graph of a left-tailed test.
##### Exercise $$\PageIndex{9}$$
Draw the graph of a two-tailed test.
##### Exercise $$\PageIndex{10}$$
A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?
##### Exercise $$\PageIndex{11}$$
Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?
a right-tailed test
##### Exercise $$\PageIndex{12}$$
A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?
##### Exercise $$\PageIndex{13}$$
You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?
a left-tailed test
##### Exercise $$\PageIndex{14}$$
If the alternative hypothesis has a not equals ( $$\neq$$ ) symbol, you know to use which type of test?
##### Exercise $$\PageIndex{15}$$
Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?
This is a left-tailed test.
##### Exercise $$\PageIndex{16}$$
Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?
##### Exercise $$\PageIndex{17}$$
Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?
This is a two-tailed test.
### References
1. Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
3. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
4. Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
5. Data from Growing by Degrees by Allen and Seaman.
6. Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
7. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
8. Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
9. Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm.
10. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
11. Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
12. Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
13. Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1.
14. Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
15. Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
16. “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
17. Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
18. Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).
## Glossary
Central Limit Theorem
Given a random variable (RV) with known mean $$\mu$$ and known standard deviation $$\sigma$$. We are sampling with size $$n$$ and we are interested in two new RVs - the sample mean, $$\bar{X}$$, and the sample sum, $$\sum X$$. If the size $$n$$ of the sample is sufficiently large, then $$\bar{X} - N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$$ and $$\sum X - N \left(n\mu, \sqrt{n}\sigma\right)$$. If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal $$n$$ times the population mean. The standard deviation of the distribution of the sample means, $$\frac{\sigma}{\sqrt{n}}$$, is called the standard error of the mean. | 5,990 | 21,139 | {"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} | 4.0625 | 4 | CC-MAIN-2022-33 | latest | en | 0.612428 |
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### Resistors & capacitors
1. 1. ResistorsExample: Circuit symbol:FunctionResistors restrict the flow of electric current, for example a resistor is placed in series with alight-emitting diode (LED) to limit the current passing through the LED.Connecting and solderingResistors may be connected either way round. They are not damaged by heat when soldering. The Resistor Colour Code Colour Number Black 0 Brown 1 Red 2 Orange 3 Yellow 4 Green 5 Blue 6 Violet 7 Grey 8
2. 2. Resistor values - the resistor colour code White 9Resistance is measured in ohms, the symbol for ohm is an omega .1 is quite small so resistor values are often given in k and M .1k = 1000 1M = 1000000 .Resistor values are normally shown using coloured bands.Each colour represents a number as shown in the table.Most resistors have 4 bands: • The first band gives the first digit. • The second band gives the second digit. • The third band indicates the number of zeros. • The fourth band is used to shows the tolerance (precision) of the resistor, this may be ignored for almost all circuits but further details are given below.This resistor has red (2), violet (7), yellow (4 zeros) and gold bands.So its value is 270000 = 270 k .On circuit diagrams the is usually omitted and the value is written 270K.Find out how to make your own Resistor Colour Code CalculatorSmall value resistors (less than 10 ohm)The standard colour code cannot show values of less than 10 . To show these small valuestwo special colours are used for the third band:gold which means × 0.1 and silver whichmeans × 0.01. The first and second bands represent the digits as normal.For example:red, violet, gold bands represent 27 × 0.1 = 2.7green, blue, silver bands represent 56 × 0.01 = 0.56Tolerance of resistors (fourth band of colour code)The tolerance of a resistor is shown by the fourth band of the colour code. Tolerance isthe precision of the resistor and it is given as a percentage. For example a 390 resistor with
3. 3. a tolerance of ±10% will have a value within 10% of 390 , between 390 - 39 = 351 and 390 +39 = 429 (39 is 10% of 390).A special colour code is used for the fourth band tolerance:silver ±10%, gold ±5%, red ±2%, brown ±1%.If no fourth band is shown the tolerance is ±20%.Tolerance may be ignored for almost all circuits because precise resistorvalues are rarely required.Resistor shorthandResistor values are often written on circuit diagrams using a code system which avoids using adecimal point because it is easy to miss the small dot. Instead the letters R, K and M are used inplace of the decimal point. To read the code: replace the letter with a decimal point, thenmultiply the value by 1000 if the letter was K, or 1000000 if the letter was M. The letter R meansmultiply by 1.For example: 560R means 560 2K7 means 2.7 k = 2700 39K means 39 k 1M0 means 1.0 M = 1000 kReal resistor values (the E6 and E12 series)You may have noticed that resistors are not available with every possible value, for example22k and 47k are readily available, but 25k and 50k are not!Why is this? Imagine that you decided to make resistors every 10 giving 10,20, 30, 40, 50 and so on. That seems fine, but what happens when you reach1000? It would be pointless to make 1000, 1010, 1020, 1030 and so onbecause for these values 10 is a very small difference, too small to benoticeable in most circuits. In fact it would be difficult to make resistorssufficiently accurate.
4. 4. To produce a sensible range of resistor values you need to increase the sizeof the step as the value increases. The standard resistor values are based onthis idea and they form a series which follows the same pattern for everymultiple of ten.The E6 series (6 values for each multiple of ten, for resistors with 20%tolerance)10, 15, 22, 33, 47, 68, ... then it continues 100, 150, 220, 330, 470, 680, 1000etc.Notice how the step size increases as the value increases. For this series thestep (to the next value) is roughly half the value.The E12 series (12 values for each multiple of ten, for resistors with 10%tolerance)10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82, ... then it continues 100, 120,150 etc.Notice how this is the E6 series with an extra value in the gaps.The E12 series is the one most frequently used for resistors. It allows you tochoose a value within 10% of the precise value you need. This is sufficientlyaccurate for almost all projects and it is sensible because most resistors areonly accurate to ±10% (called their tolerance). For example a resistor marked390 could vary by ±10% × 390 = ±39 , so it could be any value between351 and 429 .Resistors in Series and ParallelFor information on resistors connected in series and parallel please seethe Resistance page,Power Ratings of Resistors
5. 5. Electrical energy is converted to heat when current flowsthrough a resistor. Usually the effect is negligible, but if theresistance is low (or the voltage across the resistor high) a largecurrent may pass making the resistor become noticeably warm.The resistor must be able to withstand the heating effect andresistors have power ratings to show this.Power ratings of resistors are rarely quoted in parts High power resistorslists because for most circuits the standard power (5W top, 25W bottom)ratings of 0.25W or 0.5W are suitable. For the rarecases where a higher power is required it should be Photographs © Rapid Electronicsclearly specified in the parts list, these will becircuits using low value resistors (less than about 300 ) or highvoltages (more than 15V).The power, P, developed in a resistor is given by:P = I² × R where: P = power developed in the resistor in watts (W)or I = current through the resistor in amps (A) R = resistance of the resistor in ohms ( )P = V² / R V = voltage across the resistor in volts (V)Examples: • A 470 resistor with 10V across it, needs a power rating P = V²/R = 10²/470 = 0.21W. In this case a standard 0.25W resistor would be suitable. • A 27 resistor with 10V across it, needs a power rating P = V²/R = 10²/27 = 3.7W. A high power resistor with a rating of 5W would be suitable.
6. 6. Variable ResistorConstructionVariable resistors consist of a resistance track withconnections at both ends and a wiper which moves alongthe track as you turn the spindle. The track may be madefrom carbon, cermet (ceramic and metal mixture) or a coilof wire (for low resistances). The track is usually rotary butstraight track versions, usually called sliders, are alsoavailable.Variable resistors may be used asa rheostat with two connections (the wiper and justone end of the track) or as a potentiometer withall three connections in use. Miniature versionscalled presets are made for setting up circuits whichwill not require further adjustment. Standard Variable ResistorVariable resistors are often called potentiometers in Photograph © Rapid Electronicsbooks and catalogues. They are specified by their maximum resistance, linearor logarithmic track, and their physical size. The standard spindle diameter is6mm.The resistance and type of track are marked on the body: 4K7 LIN means 4.7 k linear track. 1M LOG means 1 M logarithmic track.Some variable resistors are designed to be mounted directly on the circuitboard, but most are for mounting through a hole drilled in the case containingthe circuit with stranded wire connecting their terminals to the circuit board.Linear (LIN) and Logarithmic (LOG) tracks
7. 7. Linear (LIN) track means that the resistance changes at a constant rate as you movethe wiper. This is the standard arrangement and you should assume this type isrequired if a project does not specify the type of track. Presets always have lineartracks.Logarithmic (LOG) track means that the resistance changes slowly at oneend of the track and rapidly at the other end, so halfway along the trackis not half the total resistance! This arrangement is used for volume(loudness) controls because the human ear has a logarithmic response toloudness so fine control (slow change) is required at low volumes and coarsercontrol (rapid change) at high volumes. It is important to connect the ends ofthe track the correct way round, if you find that turning the spindle increasesthe volume rapidly followed by little further change you should swap theconnections to the ends of the track.RheostatThis is the simplest way of using a variable resistor. Twoterminals are used: one connected to an end of the track,the other to the moveable wiper. Turning the spindlechanges the resistance between the two terminals from zero Rheostat Symbolup to the maximum resistance.Rheostats are often used to vary current, for example to control thebrightness of a lamp or the rate at which a capacitor charges.If the rheostat is mounted on a printed circuit board you may find that all three terminalsare connected! However, one of them will be linked to the wiper terminal. This improvesthe mechanical strength of the mounting but it serves no function electrically.PresetsThese are miniature versions of the standard variableresistor. They are designed to be mounted directly onto thecircuit board and adjusted only when the circuit is built. Forexample to set the frequency of an alarm tone or the Preset Symbolsensitivity of a light-sensitive circuit. A small screwdriver orsimilar tool is required to adjust presets.
8. 8. Presets are much cheaper than standard variable resistors so they aresometimes used in projects where a standard variable resistor would normallybe used.Multiturn presets are used where very precise adjustments must be made.The screw must be turned many times (10+) to move the slider from one endof the track to the other, giving very fine control. Preset Presets Multiturn preset (open style) (closed style) Photographs © Rapid Electronics
9. 9. CapacitorsFunctionCapacitors store electric charge. They are used with resistors in timing circuits because ittakes time for a capacitor to fill with charge. They are used to smooth varying DC supplies byacting as a reservoir of charge. They are also used in filter circuits because capacitors easilypass AC (changing) signals but they block DC (constant) signals.CapacitanceThis is a measure of a capacitors ability to store charge. A large capacitance means that morecharge can be stored. Capacitance is measured in farads, symbol F. However 1F is very large,so prefixes are used to show the smaller values.Three prefixes (multipliers) are used, µ (micro), n (nano) and p (pico): • µ means 10-6 (millionth), so 1000000µF = 1F • n means 10-9 (thousand-millionth), so 1000nF = 1µF • p means 10-12 (million-millionth), so 1000pF = 1nFCapacitor values can be very difficult to find because there are many types ofcapacitor with different labelling systems!There are many types of capacitor but they can be splitinto two groups, polarised and unpolarised. Each grouphas its own circuit symbol.Polarised capacitors (large values, 1µF +)
10. 10. Examples: Circuit symbol:Electrolytic CapacitorsElectrolytic capacitors are polarised and they must be connected the correct way round, atleast one of their leads will be marked + or -. They are not damaged by heat when soldering.There are two designs of electrolytic capacitors; axial where the leads areattached to each end (220µF in picture) and radial where both leads are atthe same end (10µF in picture). Radial capacitors tend to be a little smallerand they stand upright on the circuit board.It is easy to find the value of electrolytic capacitors because they are clearlyprinted with their capacitance and voltage rating. The voltage rating can bequite low (6V for example) and it should always be checked when selecting anelectrolytic capacitor. If the project parts list does not specify a voltage,choose a capacitor with a rating which is greater than the projects powersupply voltage. 25V is a sensible minimum for most battery circuits.Tantalum Bead CapacitorsTantalum bead capacitors are polarised and have low voltage ratings like electrolytic capacitors.They are expensive but very small, so they are used where a large capacitance is needed in asmall size.Modern tantalum bead capacitors are printed with their capacitance, voltageand polarity in full. However older ones use a colour-code system which hastwo stripes (for the two digits) and a spot of colour for the number of zeros togive the value in µF. The standard colour code is used, but for thespot, grey is used to mean × 0.01 and white means × 0.1 so that values ofless than 10µF can be shown. A third colour stripe near the leads shows thevoltage (yellow 6.3V, black 10V, green 16V, blue 20V, grey 25V, white 30V,pink 35V). The positive (+) lead is to the right when the spot isfacing you: when the spot is in sight, the positive is to theright.For example: blue, grey, black spot means 68µFFor example: blue, grey, white spot means 6.8µFFor example: blue, grey, grey spot means 0.68µF
11. 11. Unpolarised capacitors (small values, up to 1µF)Examples: Circuit symbol:Small value capacitors are unpolarised and may be connected either wayround. They are not damaged by heat when soldering, except for one unusualtype (polystyrene). They have high voltage ratings of at least 50V, usually250V or so. It can be difficult to find the values of these small capacitorsbecause there are many types of them and several differentlabelling systems!Many small value capacitors have their value printed but without amultiplier, so you need to use experience to work out what themultiplier should be!For example 0.1 means 0.1µF = 100nF.Sometimes the multiplier is used in place of the decimal point:For example: 4n7 means 4.7nF.Capacitor Number CodeA number code is often used on small capacitors where printing is difficult: • the 1st number is the 1st digit, • the 2nd number is the 2nd digit, • the 3rd number is the number of zeros to give the capacitance in pF. • Ignore any letters - they just indicate tolerance and voltage rating.For example: 102 means 1000pF = 1nF (not 102pF!)For example: 472J means 4700pF = 4.7nF (J means 5%tolerance). Colour Code
12. 12. Capacitor Colour Code Colour NumberA colour code was used on polyester capacitors for many years. It is nowobsolete, but of course there are many still around. The colours should be read Black 0like the resistor code, the top three colour bands giving the value in pF. Ignorethe 4th band (tolerance) and 5th band (voltage rating). Brown 1For example: Red 2 brown, black, orange means 10000pF = 10nF Orange 3= 0.01µF. Yellow 4Note that there are no gaps between the colourbands, so 2 identical bands actually appear as a Green 5wide band. Blue 6For example: Violet 7 wide red, yellow means 220nF = 0.22µF. Grey 8 White 9Polystyrene CapacitorsThis type is rarely used now. Their value (in pF) is normally printedwithout units. Polystyrene capacitors can be damaged by heat whensoldering (it melts the polystyrene!) so you should use a heat sink(such as a crocodile clip). Clip the heat sink to the lead between the capacitor and the joint.Real capacitor values (the E3 and E6 series)You may have noticed that capacitors are not available with every possible value, for example22µF and 47µF are readily available, but 25µF and 50µF are not!Why is this? Imagine that you decided to make capacitors every 10µF giving10, 20, 30, 40, 50 and so on. That seems fine, but what happens when youreach 1000? It would be pointless to make 1000, 1010, 1020, 1030 and so onbecause for these values 10 is a very small difference, too small to benoticeable in most circuits and capacitors cannot be made with that accuracy.
13. 13. To produce a sensible range of capacitor values you need to increase the sizeof the step as the value increases. The standard capacitor values are basedon this idea and they form a series which follows the same pattern for everymultiple of ten.The E3 series (3 values for each multiple of ten)10, 22, 47, ... then it continues 100, 220, 470, 1000, 2200, 4700, 10000 etc.Notice how the step size increases as the value increases (values roughlydouble each time).The E6 series (6 values for each multiple of ten)10, 15, 22, 33, 47, 68, ... then it continues 100, 150, 220, 330, 470, 680, 1000etc.Notice how this is the E3 series with an extra value in the gaps.The E3 series is the one most frequently used for capacitors because manytypes cannot be made with very accurate values.Variable capacitorsVariable capacitors are mostly used in radio tuning circuits andthey are sometimes called tuning capacitors. They have verysmall capacitance values, typically between 100pF and 500pF(100pF = 0.0001µF). The type illustrated usually has trimmers builtin (for making small adjustments - see below) as well as the main Variable Capacitor Symbolvariable capacitor.Many variable capacitors have very short spindleswhich are not suitable for the standard knobs used forvariable resistors and rotary switches. It would bewise to check that a suitable knob is available beforeordering a variable capacitor. Variable CapacitorVariable capacitors are not normally used in timing Photograph © Rapid Electronicscircuits because their capacitance is too small to bepractical and the range of values available is very limited. Instead timingcircuits use a fixed capacitor and a variable resistor if it is necessary to varythe time period.
14. 14. Trimmer capacitorsTrimmer capacitors (trimmers) are miniature variable capacitors.They are designed to be mounted directly onto the circuit boardand adjusted only when the circuit is built.A small screwdriver or similar tool is required to adjust Trimmer Capacitor Symboltrimmers. The process of adjusting them requirespatience because the presence of your hand and thetool will slightly change the capacitance of the circuitin the region of the trimmer!Trimmer capacitors are only available with very smallcapacitances, normally less than 100pF. It isimpossible to reduce their capacitance to zero, sothey are usually specified by their minimum and Trimmer Capacitor Photograph © Rapid Electronicsmaximum values, for example 2-10pF.Trimmers are the capacitor equivalent of presets which are miniature variableresistors.Capacitance and uses of capacitorsCapacitanceCapacitance (symbol C) is a measure of a capacitors abilityto store charge. A large capacitance means that more unpolarised capacitor symbolcharge can be stored. Capacitance is measured in farads,symbol F. However 1F is very large, so prefixes (multipliers)are used to show the smaller values: • µ (micro) means 10-6 (millionth), so 1000000µF polarised capacitor symbol = 1F • n (nano) means 10-9 (thousand-millionth), so 1000nF = 1µF • p (pico) means 10-12 (million-millionth), so 1000pF = 1nF
15. 15. Charge and Energy StoredThe amount of charge (symbol Q) stored by a capacitor is given by: Q = charge in coulombs (C)Charge, Q = C × V where: C = capacitance in farads (F) V = voltage in volts (V)When they store charge, capacitors are also storing energy:Energy, E = ½QV = ½CV² where E = energy in joules (J).Note that capacitors return their stored energy to the circuit. They do not useup electrical energy by converting it to heat as a resistor does. The energystored by a capacitor is much smaller than the energy stored by a battery sothey cannot be used as a practical source of energy for most purposes.Capacitive Reactance XcCapacitive reactance (symbol Xc) is a measure of a capacitors opposition to AC(alternating current). Like resistance it is measured in ohms, , but reactance is morecomplex than resistance because its value depends on the frequency (f) of the electricalsignal passing through the capacitor as well as on the capacitance, C. 1 Xc = reactance in ohms ( )Capacitive reactance, Xc = 2 fC where: f = frequency in hertz (Hz) C = capacitance in farads (F)The reactance Xc is large at low frequencies and small at high frequencies.For steady DC which is zero frequency, Xc is infinite (total opposition), hencethe rule that capacitors pass AC but block DC.For example a 1µF capacitor has a reactance of 3.2k for a 50Hz signal, butwhen the frequency is higher at 10kHz its reactance is only 16 .Note: the symbol Xc is used to distinguish capacitative reactance frominductive reactance XL which is a property of inductors. The distinction isimportant because XL increases with frequency (the opposite of Xc) and if
16. 16. both XL and Xc are present in a circuit the combined reactance (X) isthe difference between them. For further information please see the pageon Impedance.Capacitors in Series and ParallelCombined capacitance 1 1 1 1(C) of C = C1 C2 C3 +capacitors connected + + ...in series:Combined capacitance(C) ofcapacitors connected C = C1 + C2 + C3 + ...in parallel:Two or more capacitors are rarelydeliberately connected in series in realcircuits, but it can be useful to connectcapacitors in parallel to obtain a verylarge capacitance, for example to smooth a power supply.Note that these equations are the opposite way round for resistors in seriesand parallel.Charging a capacitorThe capacitor (C) in the circuit diagram is being chargedfrom a supply voltage (Vs) with the current passingthrough a resistor (R). The voltage across the capacitor(Vc) is initially zero but it increases as the capacitorcharges. The capacitor is fully charged when Vc = Vs.The charging current (I) is determined by the voltageacross the resistor (Vs - Vc):Charging current, I = (Vs - Vc) / R (note that Vcis increasing)At first Vc = 0V so the initial current, Io = Vs / R
17. 17. Vc increases as soon as charge (Q) starts to build up (Vc = Q/C), this reducesthe voltage across the resistor and therefore reduces the charging current.This means that the rate of charging becomes progressively slower. time constant is in seconds (s)time constant = R × C where: R = resistance in ohms ( ) C = capacitance in farads (F)For example:If R = 47k and C = 22µF, then the time constant, RC = 47k × 22µF = 1.0s.If R = 33k and C = 1µF, then the time constant, RC = 33k × 1µF = 33ms.A large time constant means the capacitor charges slowly. Note that the timeconstant is a property of the circuit containing the capacitance andresistance, it is not a property of acapacitor alone. Graphs showing the current and voltage for a capacitor charging time constant = RC
18. 18. The time constant is the time taken for thecharging (or discharging) current (I) to fallto 1/e of its initial value (Io). e is the baseof natural logarithms, an important numberin mathematics (like ). e = 2.71828 (to 6significant figures) so we can roughly saythat the time constant is the time taken forthe current to fall to 1/3 of its initial value.After each time constant the current fallsby 1/e (about 1/3). After 5 timeconstants (5RC) the current has fallen toless than 1% of its initial value and we canreasonably say that the capacitor is fullycharged, but in fact the capacitor takes forever to charge fully!The bottom graph shows how the voltage (V) Time Voltage Chargeincreases as the capacitor charges. At first the voltagechanges rapidly because the current is large; but as 0RC 0.0V 0%the current decreases, the charge builds up moreslowly and the voltage increases more slowly. 1RC 5.7V 63%After 5 time constants (5RC) the capacitor is almost 2RC 7.8V 86%fully charged with its voltage almost equal to the supplyvoltage. We can reasonably say that the capacitor is 3RC 8.6V 95%fully charged after 5RC, although really chargingcontinues for ever (or until the circuit is changed). 4RC 8.8V 98% 5RC 8.9V 99%Discharging a capacitor Graphs showing the current and voltage for a capacitor discharging time constant = RC
19. 19. The top graph shows how the current (I)decreases as the capacitor discharges.The initial current (Io) is determined by theinitial voltage across the capacitor (Vo)and resistance (R):Initial current, Io = Vo / R.Note that the current graphs are the sameshape for both charging and discharging acapacitor. This type of graph is anexample of exponential decay.The bottom graph shows how Time Voltage Chargethe voltage (V) decreases as the capacitordischarges. 0RC 9.0V 100%At first the current is large because the 1RC 3.3V 37%voltage is large, so charge is lost quicklyand the voltage decreases rapidly. As 2RC 1.2V 14%charge is lost the voltage is reducedmaking the current smaller so the rate of 3RC 0.4V 5%discharging becomes progressivelyslower. 4RC 0.2V 2%After 5 time constants (5RC) the voltage across the 5RC 0.1V 1%capacitor is almost zero and we can reasonably saythat the capacitor is fully discharged, although reallydischarging continues for ever (or until the circuit is changed).Uses of CapacitorsCapacitors are used for several purposes: • Timing - for example with a 555 timer IC controlling the charging and discharging. • Smoothing - for example in a power supply.
20. 20. • Coupling - for example between stages of an audio system and to connect a loudspeaker. • Filtering - for example in the tone control of an audio system. • Tuning - for example in a radio system. • Storing energy - for example in a camera flash circuit.Capacitor Coupling(CR-coupling)Sections of electroniccircuits may be linked witha capacitor becausecapacitors passAC (changing) signalsbut block DC (steady)signals. This iscalled capacitorcoupling or CR-coupling.It is used between thestages of an audio systemto pass on the audio signal(AC) without any steadyvoltage (DC) which maybe present, for example toconnect a loudspeaker. Itis also used for the ACswitch setting onan oscilloscope.The precise behaviourof a capacitor couplingis determined by itstime constant (RC).Note that the resistance (R) may be inside the next circuit section rather thana separate resistor.For successful capacitor coupling in an audio system the signals must passthrough with little or no distortion. This is achieved if the time constant (RC) islarger than the time period (T) of the lowest frequency audio signals required(typically 20Hz, T = 50ms).
21. 21. Output when RC >> TWhen the time constant is much larger than the time period of the input signalthe capacitor does not have sufficient time to significantly charge or discharge,so the signal passes through with negligible distortion.Output when RC = TWhen the time constant is equal to the time period you can see that thecapacitor has time to partly charge and discharge before the signal changes.As a result there is significant distortion of the signal as it passes through theCR-coupling. Notice how the sudden changes of the input signal pass straightthrough the capacitor to the output.Output when RC << TWhen the time constant is much smaller than the time period the capacitorhas time to fully charge or discharge after each sudden change in the inputsignal. Effectively only the sudden changes pass through to the output andthey appear as spikes, alternately positive and negative. This can be usefulin a system which must detect when a signal changes suddenly, but mustignore slow changes. | 6,746 | 27,292 | {"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} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.677354 |
https://www.coursehero.com/tutors-problems/Economics/11903748-Three-firms-compete-in-Cournot-competition-in-a-market-where-the-i/ | 1,544,705,121,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824822.41/warc/CC-MAIN-20181213123823-20181213145323-00031.warc.gz | 865,280,800 | 22,019 | View the step-by-step solution to:
# Three firms compete in Cournot competition in a market where the inverse demand function is P(q1, q2, q3) = 50 q1 q2 q3. Each has per-unit cost 10...
Three firms compete in Cournot competition in a market where the inverse demand function is P(q1, q2, q3) = 50 − q1− q2− q3. Each has per-unit cost 10 and zero fixed cost. They simultaneously choose quantities. What is the Nash equilibrium quantity for firm 3? Round your answer to three decimal places.
Let A = firm 3's profit in the Cournot equilibrium.
Now suppose that the firms form a cartel, i.e., they act as a monopoly and split the profit evenly. If the total quantity produced by the cartel is Q, then the inverse demand is P(Q) = 50 - Q.
Let B = firm 3's profit in the cartel.
Calculate C = B - A. Round your answer to 3 decimal places.
In this case, C = B-A = 133.333-100 = 33.33 Therefore... View the full answer
Attached is a detailed explanation... View the full answer
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Browse Documents | 337 | 1,361 | {"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} | 3.140625 | 3 | CC-MAIN-2018-51 | latest | en | 0.889471 |
http://us.metamath.org/mpeuni/dvdsflf1o.html | 1,638,004,995,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00080.warc.gz | 81,137,651 | 9,814 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dvdsflf1o Structured version Visualization version GIF version
Theorem dvdsflf1o 24808
Description: A bijection from the numbers less than 𝑁 / 𝐴 to the multiples of 𝐴 less than 𝑁. Useful for some sum manipulations. (Contributed by Mario Carneiro, 3-May-2016.)
Hypotheses
Ref Expression
dvdsflf1o.1 (𝜑𝐴 ∈ ℝ)
dvdsflf1o.2 (𝜑𝑁 ∈ ℕ)
dvdsflf1o.f 𝐹 = (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ↦ (𝑁 · 𝑛))
Assertion
Ref Expression
dvdsflf1o (𝜑𝐹:(1...(⌊‘(𝐴 / 𝑁)))–1-1-onto→{𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})
Distinct variable groups: 𝑥,𝑛,𝐴 𝑛,𝑁,𝑥 𝜑,𝑛
Allowed substitution hints: 𝜑(𝑥) 𝐹(𝑥,𝑛)
Proof of Theorem dvdsflf1o
Dummy variable 𝑚 is distinct from all other variables.
StepHypRef Expression
1 dvdsflf1o.f . 2 𝐹 = (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ↦ (𝑁 · 𝑛))
2 dvdsflf1o.2 . . . . 5 (𝜑𝑁 ∈ ℕ)
3 elfznn 12309 . . . . 5 (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) → 𝑛 ∈ ℕ)
4 nnmulcl 10988 . . . . 5 ((𝑁 ∈ ℕ ∧ 𝑛 ∈ ℕ) → (𝑁 · 𝑛) ∈ ℕ)
52, 3, 4syl2an 494 . . . 4 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ∈ ℕ)
6 dvdsflf1o.1 . . . . . . . . 9 (𝜑𝐴 ∈ ℝ)
76, 2nndivred 11014 . . . . . . . 8 (𝜑 → (𝐴 / 𝑁) ∈ ℝ)
8 fznnfl 12598 . . . . . . . 8 ((𝐴 / 𝑁) ∈ ℝ → (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ↔ (𝑛 ∈ ℕ ∧ 𝑛 ≤ (𝐴 / 𝑁))))
97, 8syl 17 . . . . . . 7 (𝜑 → (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ↔ (𝑛 ∈ ℕ ∧ 𝑛 ≤ (𝐴 / 𝑁))))
109simplbda 653 . . . . . 6 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑛 ≤ (𝐴 / 𝑁))
113adantl 482 . . . . . . . 8 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑛 ∈ ℕ)
1211nnred 10980 . . . . . . 7 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑛 ∈ ℝ)
136adantr 481 . . . . . . 7 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝐴 ∈ ℝ)
142nnred 10980 . . . . . . . 8 (𝜑𝑁 ∈ ℝ)
1514adantr 481 . . . . . . 7 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑁 ∈ ℝ)
162nngt0d 11009 . . . . . . . 8 (𝜑 → 0 < 𝑁)
1716adantr 481 . . . . . . 7 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 0 < 𝑁)
18 lemuldiv2 10849 . . . . . . 7 ((𝑛 ∈ ℝ ∧ 𝐴 ∈ ℝ ∧ (𝑁 ∈ ℝ ∧ 0 < 𝑁)) → ((𝑁 · 𝑛) ≤ 𝐴𝑛 ≤ (𝐴 / 𝑁)))
1912, 13, 15, 17, 18syl112anc 1327 . . . . . 6 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → ((𝑁 · 𝑛) ≤ 𝐴𝑛 ≤ (𝐴 / 𝑁)))
2010, 19mpbird 247 . . . . 5 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ≤ 𝐴)
212nnzd 11425 . . . . . . 7 (𝜑𝑁 ∈ ℤ)
22 elfzelz 12281 . . . . . . 7 (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) → 𝑛 ∈ ℤ)
23 zmulcl 11371 . . . . . . 7 ((𝑁 ∈ ℤ ∧ 𝑛 ∈ ℤ) → (𝑁 · 𝑛) ∈ ℤ)
2421, 22, 23syl2an 494 . . . . . 6 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ∈ ℤ)
25 flge 12543 . . . . . 6 ((𝐴 ∈ ℝ ∧ (𝑁 · 𝑛) ∈ ℤ) → ((𝑁 · 𝑛) ≤ 𝐴 ↔ (𝑁 · 𝑛) ≤ (⌊‘𝐴)))
2613, 24, 25syl2anc 692 . . . . 5 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → ((𝑁 · 𝑛) ≤ 𝐴 ↔ (𝑁 · 𝑛) ≤ (⌊‘𝐴)))
2720, 26mpbid 222 . . . 4 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ≤ (⌊‘𝐴))
286flcld 12536 . . . . . 6 (𝜑 → (⌊‘𝐴) ∈ ℤ)
2928adantr 481 . . . . 5 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (⌊‘𝐴) ∈ ℤ)
30 fznn 12347 . . . . 5 ((⌊‘𝐴) ∈ ℤ → ((𝑁 · 𝑛) ∈ (1...(⌊‘𝐴)) ↔ ((𝑁 · 𝑛) ∈ ℕ ∧ (𝑁 · 𝑛) ≤ (⌊‘𝐴))))
3129, 30syl 17 . . . 4 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → ((𝑁 · 𝑛) ∈ (1...(⌊‘𝐴)) ↔ ((𝑁 · 𝑛) ∈ ℕ ∧ (𝑁 · 𝑛) ≤ (⌊‘𝐴))))
325, 27, 31mpbir2and 956 . . 3 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ∈ (1...(⌊‘𝐴)))
33 dvdsmul1 14922 . . . 4 ((𝑁 ∈ ℤ ∧ 𝑛 ∈ ℤ) → 𝑁 ∥ (𝑁 · 𝑛))
3421, 22, 33syl2an 494 . . 3 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑁 ∥ (𝑁 · 𝑛))
35 breq2 4622 . . . 4 (𝑥 = (𝑁 · 𝑛) → (𝑁𝑥𝑁 ∥ (𝑁 · 𝑛)))
3635elrab 3351 . . 3 ((𝑁 · 𝑛) ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥} ↔ ((𝑁 · 𝑛) ∈ (1...(⌊‘𝐴)) ∧ 𝑁 ∥ (𝑁 · 𝑛)))
3732, 34, 36sylanbrc 697 . 2 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → (𝑁 · 𝑛) ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})
38 breq2 4622 . . . . . . 7 (𝑥 = 𝑚 → (𝑁𝑥𝑁𝑚))
3938elrab 3351 . . . . . 6 (𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥} ↔ (𝑚 ∈ (1...(⌊‘𝐴)) ∧ 𝑁𝑚))
4039simprbi 480 . . . . 5 (𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥} → 𝑁𝑚)
4140adantl 482 . . . 4 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑁𝑚)
42 elrabi 3347 . . . . . . 7 (𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥} → 𝑚 ∈ (1...(⌊‘𝐴)))
4342adantl 482 . . . . . 6 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑚 ∈ (1...(⌊‘𝐴)))
44 elfznn 12309 . . . . . 6 (𝑚 ∈ (1...(⌊‘𝐴)) → 𝑚 ∈ ℕ)
4543, 44syl 17 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑚 ∈ ℕ)
462adantr 481 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑁 ∈ ℕ)
47 nndivdvds 14908 . . . . 5 ((𝑚 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑁𝑚 ↔ (𝑚 / 𝑁) ∈ ℕ))
4845, 46, 47syl2anc 692 . . . 4 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝑁𝑚 ↔ (𝑚 / 𝑁) ∈ ℕ))
4941, 48mpbid 222 . . 3 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝑚 / 𝑁) ∈ ℕ)
50 fznnfl 12598 . . . . . . 7 (𝐴 ∈ ℝ → (𝑚 ∈ (1...(⌊‘𝐴)) ↔ (𝑚 ∈ ℕ ∧ 𝑚𝐴)))
516, 50syl 17 . . . . . 6 (𝜑 → (𝑚 ∈ (1...(⌊‘𝐴)) ↔ (𝑚 ∈ ℕ ∧ 𝑚𝐴)))
5251simplbda 653 . . . . 5 ((𝜑𝑚 ∈ (1...(⌊‘𝐴))) → 𝑚𝐴)
5342, 52sylan2 491 . . . 4 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑚𝐴)
5445nnred 10980 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑚 ∈ ℝ)
556adantr 481 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝐴 ∈ ℝ)
5614adantr 481 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑁 ∈ ℝ)
5716adantr 481 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 0 < 𝑁)
58 lediv1 10833 . . . . 5 ((𝑚 ∈ ℝ ∧ 𝐴 ∈ ℝ ∧ (𝑁 ∈ ℝ ∧ 0 < 𝑁)) → (𝑚𝐴 ↔ (𝑚 / 𝑁) ≤ (𝐴 / 𝑁)))
5954, 55, 56, 57, 58syl112anc 1327 . . . 4 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝑚𝐴 ↔ (𝑚 / 𝑁) ≤ (𝐴 / 𝑁)))
6053, 59mpbid 222 . . 3 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝑚 / 𝑁) ≤ (𝐴 / 𝑁))
617adantr 481 . . . 4 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝐴 / 𝑁) ∈ ℝ)
62 fznnfl 12598 . . . 4 ((𝐴 / 𝑁) ∈ ℝ → ((𝑚 / 𝑁) ∈ (1...(⌊‘(𝐴 / 𝑁))) ↔ ((𝑚 / 𝑁) ∈ ℕ ∧ (𝑚 / 𝑁) ≤ (𝐴 / 𝑁))))
6361, 62syl 17 . . 3 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → ((𝑚 / 𝑁) ∈ (1...(⌊‘(𝐴 / 𝑁))) ↔ ((𝑚 / 𝑁) ∈ ℕ ∧ (𝑚 / 𝑁) ≤ (𝐴 / 𝑁))))
6449, 60, 63mpbir2and 956 . 2 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → (𝑚 / 𝑁) ∈ (1...(⌊‘(𝐴 / 𝑁))))
6545nncnd 10981 . . . . 5 ((𝜑𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥}) → 𝑚 ∈ ℂ)
6665adantrl 751 . . . 4 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → 𝑚 ∈ ℂ)
672nncnd 10981 . . . . 5 (𝜑𝑁 ∈ ℂ)
6867adantr 481 . . . 4 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → 𝑁 ∈ ℂ)
6911nncnd 10981 . . . . 5 ((𝜑𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁)))) → 𝑛 ∈ ℂ)
7069adantrr 752 . . . 4 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → 𝑛 ∈ ℂ)
712nnne0d 11010 . . . . 5 (𝜑𝑁 ≠ 0)
7271adantr 481 . . . 4 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → 𝑁 ≠ 0)
7366, 68, 70, 72divmuld 10768 . . 3 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → ((𝑚 / 𝑁) = 𝑛 ↔ (𝑁 · 𝑛) = 𝑚))
74 eqcom 2633 . . 3 (𝑛 = (𝑚 / 𝑁) ↔ (𝑚 / 𝑁) = 𝑛)
75 eqcom 2633 . . 3 (𝑚 = (𝑁 · 𝑛) ↔ (𝑁 · 𝑛) = 𝑚)
7673, 74, 753bitr4g 303 . 2 ((𝜑 ∧ (𝑛 ∈ (1...(⌊‘(𝐴 / 𝑁))) ∧ 𝑚 ∈ {𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})) → (𝑛 = (𝑚 / 𝑁) ↔ 𝑚 = (𝑁 · 𝑛)))
771, 37, 64, 76f1o2d 6841 1 (𝜑𝐹:(1...(⌊‘(𝐴 / 𝑁)))–1-1-onto→{𝑥 ∈ (1...(⌊‘𝐴)) ∣ 𝑁𝑥})
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 384 = wceq 1480 ∈ wcel 1992 ≠ wne 2796 {crab 2916 class class class wbr 4618 ↦ cmpt 4678 –1-1-onto→wf1o 5849 ‘cfv 5850 (class class class)co 6605 ℂcc 9879 ℝcr 9880 0cc0 9881 1c1 9882 · cmul 9886 < clt 10019 ≤ cle 10020 / cdiv 10629 ℕcn 10965 ℤcz 11322 ...cfz 12265 ⌊cfl 12528 ∥ cdvds 14902 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1841 ax-6 1890 ax-7 1937 ax-8 1994 ax-9 2001 ax-10 2021 ax-11 2036 ax-12 2049 ax-13 2250 ax-ext 2606 ax-sep 4746 ax-nul 4754 ax-pow 4808 ax-pr 4872 ax-un 6903 ax-cnex 9937 ax-resscn 9938 ax-1cn 9939 ax-icn 9940 ax-addcl 9941 ax-addrcl 9942 ax-mulcl 9943 ax-mulrcl 9944 ax-mulcom 9945 ax-addass 9946 ax-mulass 9947 ax-distr 9948 ax-i2m1 9949 ax-1ne0 9950 ax-1rid 9951 ax-rnegex 9952 ax-rrecex 9953 ax-cnre 9954 ax-pre-lttri 9955 ax-pre-lttrn 9956 ax-pre-ltadd 9957 ax-pre-mulgt0 9958 ax-pre-sup 9959 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1883 df-eu 2478 df-mo 2479 df-clab 2613 df-cleq 2619 df-clel 2622 df-nfc 2756 df-ne 2797 df-nel 2900 df-ral 2917 df-rex 2918 df-reu 2919 df-rmo 2920 df-rab 2921 df-v 3193 df-sbc 3423 df-csb 3520 df-dif 3563 df-un 3565 df-in 3567 df-ss 3574 df-pss 3576 df-nul 3897 df-if 4064 df-pw 4137 df-sn 4154 df-pr 4156 df-tp 4158 df-op 4160 df-uni 4408 df-iun 4492 df-br 4619 df-opab 4679 df-mpt 4680 df-tr 4718 df-eprel 4990 df-id 4994 df-po 5000 df-so 5001 df-fr 5038 df-we 5040 df-xp 5085 df-rel 5086 df-cnv 5087 df-co 5088 df-dm 5089 df-rn 5090 df-res 5091 df-ima 5092 df-pred 5642 df-ord 5688 df-on 5689 df-lim 5690 df-suc 5691 df-iota 5813 df-fun 5852 df-fn 5853 df-f 5854 df-f1 5855 df-fo 5856 df-f1o 5857 df-fv 5858 df-riota 6566 df-ov 6608 df-oprab 6609 df-mpt2 6610 df-om 7014 df-1st 7116 df-2nd 7117 df-wrecs 7353 df-recs 7414 df-rdg 7452 df-er 7688 df-en 7901 df-dom 7902 df-sdom 7903 df-sup 8293 df-inf 8294 df-pnf 10021 df-mnf 10022 df-xr 10023 df-ltxr 10024 df-le 10025 df-sub 10213 df-neg 10214 df-div 10630 df-nn 10966 df-n0 11238 df-z 11323 df-uz 11632 df-fz 12266 df-fl 12530 df-dvds 14903 This theorem is referenced by: dvdsflsumcom 24809 logfac2 24837
Copyright terms: Public domain W3C validator | 6,164 | 8,759 | {"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} | 3.140625 | 3 | CC-MAIN-2021-49 | latest | en | 0.288273 |
http://www.wyzant.com/answers/1231/what_property_would_x22_be | 1,371,635,959,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708664942/warc/CC-MAIN-20130516125104-00064-ip-10-60-113-184.ec2.internal.warc.gz | 792,232,874 | 7,577 | Search 73,804 tutors
what property would x/2=2 be?
I am going to spell out each property along the way.
x/2 = 2
Multiply both sides by 2. The Multiplication Property of Equality allows us to multiply both sides by the same number. The Multiplication Property of Equality says if a = b, then a·c = b·c. In this case, a is x/2 and b is the 2 on the right hand side of the equal sign. Let c also be 2.
a·c = b·c
(x/2)·2 = 2·2
(x/2)·2 = 4
Now, we want to cancel out the 2's on the left. First, we need to reorder the left side.
2·(x/2) = 4
This reordering can be done by the commutative property of multiplication (a·b = b·a).
The associative property of multiplication allows us to do the multiplication between 2 and x first. The associative property states a(b·c) = (a·b)c
2·x/2 = 4
The commutative property of multiplication then allows:
x·2/2 = 4
The inverse property of multiplication states a·1/a = 1. The inverse property of multiplication then allows:
x·1 = 4
The identity property of multiplication states a·1 = a. The identity property of multiplication then allows:
x = 4.
Now, personally, I find this kind of work very tedious. But being able to work stepwise through a problem using one property at a time is a useful skill for higher math courses such as geometry and linear algebra. Most of the time though, we just use these properties without remembering why we can use them or what they are called.
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Algebra 1 | 496 | 1,844 | {"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} | 3.890625 | 4 | CC-MAIN-2013-20 | latest | en | 0.866968 |
https://explog.in/aoc/2019/AoC21.html | 1,585,658,129,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00353.warc.gz | 475,652,559 | 48,803 | # Advent of Code Day 21¶
## Intcode¶
In [1]:
def readfile(path):
with open(path) as input_file:
def params(program, i, count):
modes = str(program[i] // 100)[::-1]
result = []
for x in range(count - 1):
arg = program[i + 1 + x]
if x >= len(modes) or modes[x] == "0":
result.append(program[arg])
elif modes[x] == "2":
result.append(program[program.offset + arg])
else:
result.append(arg)
output = program[i + count]
if len(modes) >= count and modes[count - 1] == "2":
output += program.offset
result.append(output) # output
return result
def add(program, i, **kwargs):
assert program[i] % 100 == 1
a, b, out = params(program, i, 3)
program[out] = a + b
return i + 4, None
def mul(program, i, **kwargs):
assert program[i] % 100 == 2
a, b, out = params(program, i, 3)
program[out] = a * b
return i + 4, None
def inp(program, i, input_value, **kwargs):
assert program[i] % 100 == 3
out, = params(program, i, 1)
program[out] = input_value
# print(f"> {program[i]} program[{out}] = {program[out]}")
return i + 2, None
def out(program, i, **kwargs):
assert program[i] % 100 == 4
val, _ = params(program, i, 2)
# print(f"> {val}")
return i + 2, val
def jit(program, i, **kwargs):
assert program[i] % 100 == 5
test, ip, _ = params(program, i, 3)
if test != 0:
return ip, None
return i + 3, None
def jif(program, i, **kwargs):
assert program[i] % 100 == 6
test, ip, _ = params(program, i, 3)
if test == 0:
return ip, None
return i + 3, None
def lt(program, i, **kwargs):
assert program[i] % 100 == 7
a, b, out = params(program, i, 3)
program[out] = int(a < b)
return i + 4, None
def eq(program, i, **kwargs):
assert program[i] % 100 == 8
a, b, out = params(program, i, 3)
program[out] = int(a == b)
return i + 4, None
def rbo(program, i, **kwargs):
assert program[i] % 100 == 9
delta, _ = params(program, i, 2)
program.offset += delta
# print (f"> rbo = {program.offset}" )
return i + 2, None
operations = { 1: add, 2: mul, 3: inp, 4: out, 5: jit, 6: jif, 7: lt, 8: eq, 9: rbo }
class Program(list):
def __init__(self, *args, **kwargs):
super(Program, self).__init__(args[0])
self.offset = 0
def compute(program, debug=False):
program = Program(program)
l = len(program)
i = 0
output_value = None
last_output = None
while i < l:
op = program[i]
if op == 99:
if debug:
print(f">> exit {output_value}")
return
if debug:
print(f">> {i}, {list(enumerate(program))[i:i+5]}")
if op % 100 == 3:
input_value = yield
else:
input_value = None
i, output_value = operations[op % 100](program, i, input_value=input_value)
if output_value is not None:
yield output_value
raise Exception("Didn't terminate properly")
def process(listing):
lst = list(map(int, listing.split(",")))
lst.extend([0] * 100000)
return lst
In [3]:
input21 = readfile("AoC/input21.txt")
In [40]:
def out(program):
while (output := next(program)):
if output < 128:
print(chr(output), end="")
else:
print(output, end="")
def inp(program, text):
result = None
for ch in text:
result = program.send(ord(ch))
if result:
print(chr(result), end="")
return result
def jumper(input21):
program = compute(process(input21))
out(program)
inp(program, "NOT A J\n") # J = A is a hole
inp(program, "NOT B T\n") # T = B is a hole
inp(program, "OR T J\n") # J = A OR B is a hole
inp(program, "NOT C T\n") # T = C is a hole
inp(program, "OR J T\n") # T = A or B or C is a hole
inp(program, "AND D T\n") # T = (A or B or C is a hole) AND D is not.
inp(program, "NOT T J\n") # J = !T
inp(program, "NOT J J\n") # J = !J = !!T
inp(program, "WALK\n")
out(program)
def jumper2(input21):
program = compute(process(input21))
out(program)
# Jump if there is ground at 4, repeated twice
inp(program, "NOT A J\n") # J = A is a hole
inp(program, "NOT B T\n") # T = B is a hole
inp(program, "OR T J\n") # J = A OR B is a hole
inp(program, "NOT C T\n") # T = C is a hole
inp(program, "OR J T\n") # T = A or B or C is a hole
inp(program, "AND D T\n") # T = (A or B or C is a hole) AND D is not.
inp(program, "NOT I J\n") # J = !I
inp(program, "NOT J J\n") # J = I # Can jump from E
inp(program, "OR F J\n") # J = I OR F # Can move forward twice
inp(program, "AND E J\n") # J = E && (F || I) # Can move to E
inp(program, "OR H J\n") # J = H || (E && (F || I))
inp(program, "AND T J\n") # J = !T
inp(program, "RUN\n")
out(program)
In [41]:
jumper(input21)
Input instructions:
Walking...
19359316
StopIterationTraceback (most recent call last)
<ipython-input-41-a06bdcb033c8> in <module>
----> 1 jumper(input21)
<ipython-input-40-f55dbcfc39c5> in jumper(input21)
31 inp(program, "WALK\n")
32
---> 33 out(program)
34
35
<ipython-input-40-f55dbcfc39c5> in out(program)
1 def out(program):
----> 2 while (output := next(program)):
3 if output < 128:
4 print(chr(output), end="")
5 else:
StopIteration:
In [42]:
jumper2(input21)
Input instructions:
Running...
1141281622
StopIterationTraceback (most recent call last)
<ipython-input-42-20a5bbd1ec50> in <module>
----> 1 jumper2(input21)
<ipython-input-40-f55dbcfc39c5> in jumper2(input21)
61 inp(program, "RUN\n")
62
---> 63 out(program)
64
<ipython-input-40-f55dbcfc39c5> in out(program)
1 def out(program):
----> 2 while (output := next(program)):
3 if output < 128:
4 print(chr(output), end="")
5 else:
StopIteration:
In [ ]: | 1,760 | 5,334 | {"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} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.459215 |
https://webapps.stackexchange.com/questions/111813/sum-all-numbers-converting-all-negative-numbers-into-positive-numbers | 1,569,225,547,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576122.89/warc/CC-MAIN-20190923064347-20190923090347-00038.warc.gz | 725,260,274 | 29,793 | # Sum all numbers converting all negative numbers into positive numbers
I have a column of numbers and would like to know the total after everybody has paid me back, but I have negative numbers currently in there:
So the total of this should be `95`, currently if I do a sum it will be `75`.
This is what I currently have:
``````=SUM(IF(B2:B27<0,-1*B2:B27,B2:B27))
``````
but this doesn't work of course as it does it on the total of all the rows within the range in the column.
``````=ArrayFormula(sum(abs(B2:B27)))
(For an answer of `105` if all the numbers displayed fall into the range (and no others).) | 159 | 613 | {"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-2019-39 | latest | en | 0.934721 |
https://www.onlinemathlearning.com/statistics-center-spread-hss-id2.html | 1,586,445,207,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00535.warc.gz | 1,055,549,873 | 12,005 | # Center and Spread of Data
Examples, solutions, videos, and lessons to help High School students learn how to use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
Related Topics:
Common Core Statistics
Common Core Mathematics
Common Core: HSS-ID.A.2
The following figures show the measures of center (mean, median, mode) and outliers. Scroll down the page for more examples and solutions.
Measures of Central Tendency and Spread for One Variable Data
CC.9-12.S.ID.2 -- Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
Example:
LaTanja went to Park Plaza Mall to go shopping. She spent \$12.49, \$13.75, \$12.49, \$14.50, %16.65, \$17.75, \$12.49 and \$32.35 on different items she wanted for school. Find the following information: Minimum, Maximum, Range, Mean, Mode, Outliers and Median. Using Measures of Central Tendency - 9th-12th Grades Common Core Mathematics
Determine the effects of changes in a data set on the measures of central tendency.
Examples:
1. The ages of ten seventh-grade students are given. Find the mean, median and mode.
11,12,11,11,12,11,11,12,11,13
2. Find the mean, median, and mode of the ages if the teacher's age, 50, is added to the data set.
11,12,11,11,12,11,11,12,11,13
Now, compare the results of examples 1 and 2. Which measure of central tendency was most affected by the change in the data set? By comparing the results of examples 1 and 2, we see that the mean was most affected by a change in the data set. For this reason, the mean is not the best measure of central tendency when outliers are involved.
3. Companies A and B each have ten employees. Their hourly pay, in dollars, is shown below.
A: 5,5,5,5,5,5,5,5,5,40
B: 5,5,8,8,8,8,8,10,10,15
a) Find the mean hourly pay for each company.
b) If you are starting to work, would you rather work for Company A or Company B? Why?
Use statistics to compare center and spread of two different data sets
Examples:
1. This data set shows the number of people who attended a movie theater over a period of 16 days.
{14,23,10,21,7,80,32,30,92,14,26,21,38,20,35,21}
a. Find the measures of center.
b. The theater's management wants to compare its attendance to that of other theaters in the area. Which measure of center best represents the data?
2. Find the five-number summary for the number of points scored by the Chicago Bulls during the 1997-98 season The following table compares median and mean. Scroll down the page for more examples and solutions.
Comparing Measures of Center and Spread
Standard Deviation
The standard deviation measures dispersion by determining how much, on average, the data values vary (or deviate) about the mean.
When the data values are clustered near the mean, the standard deviation is small.
When the data values are scattered far from the mean, the standard deviation is large.
Median Absolute Deviation
The mean absolute deviation is obtained by finding the median of the absolute values of the deviations of the data values from the median.
Mean Absolute Deviation
Common Core State Standard S.ID.2.
Calculating the Mean Absolute Deviation to understand the variance of a set of data.
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[?] Subscribe To This Site | 948 | 3,956 | {"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} | 4.65625 | 5 | CC-MAIN-2020-16 | longest | en | 0.894829 |
https://www.bookmyessay.com/definition-of-algorithm-and-few-important-steps/ | 1,653,109,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00339.warc.gz | 769,989,151 | 58,371 | Programming and computer science are reliant on algorithms. When writing long, efficient and high-quality code, algorithms become an essential part of the process. In software, an algorithm consists of step-by-step instructions for performing a specific task. You can also get Algorithm Development Assignment Help through this write-up company. An algorithm is a representation of how a computer program will perform important operations step by step. An algorithm should be the starting point of any program, followed by flowcharts and pseudocode.
## What Are The Steps To Writing A Good Algorithm?
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### Types of Algorithms
• Dividing and Conquering Algorithm: An algorithm design paradigm is Divide and Conquer. Dividing a problem into subproblems and solving each of them is a Divide and conquer algorithm. In most cases, the sub-problems are identical or related to the original issue. Using the sub-problem solutions, the original problem is solved.
• Algorithm For Dynamic Programming: Dynamic Programming is a paradigm for designing algorithms. Optimization/overlapping problems are usually addressed using Dynamic Programming.
• Algorithm Greedy: An effective algorithm, the Greedy Algorithm is a simple one. Usually, it is used to solve optimization problems. The Greedy Algorithm selects whatever method appears to be the best for solving the problem.
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Various types of algorithms exist based on their functionality and application. We go over a few of the most important algorithms below along with Assignment Writing Tips.
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BookMyEssay is the best platform for Algorithm Development Assignment Help. The team is constantly working on new outcomes and assignment solutions that students usually face during their academic careers. All the students are well-guided from the BookMyEssay platform and are highly motivated for attaining further achievements. | 766 | 4,165 | {"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-2022-21 | latest | en | 0.926331 |
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xiaomubiao 2004-04-30 12:15:06
...全文
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#include<stdlib.h>
#include<time.h>
#include<stdio.h>
#define Swap(a,b) {int t=b; b=a; a=t;}
int partition(int a[],int L, int r)
{
int i = L-1,j=r;
int v = a[r];
while(1)
{
while(a[++i] < v);
while(v < a[--j]) if(j==L) break;
if(i>=j) break;
Swap(a[i],a[j]);
}
Swap(a[i],a[r]);
return i;
}
int RandomPartition( int a[], int l, int r)
{
srand(time(0));
int i = rand()%(r-l+1)+l;//随机划分:
Swap(a[i],a[r]);
return partition(a,l,r);
}
int RandomSelect(int a[], int l, int r, int k)
{
if(r<=l) return a[r];
int i= RandomPartition(a,l,r);
int j = i-l+1;
if(j == k)
return a[i];
if(j>k)
return RandomSelect(a,l,i-1,k);
else
return RandomSelect(a,i+1,r,k-j);
}
void disp(int a[],int n)
{
for(int i=0; i<n; i++)
printf("%d ", a[i]);
}
int main()
{
int a[]={1,12,3,-4,81,16,100,17};
disp(a,8);
int k;
scanf("%d",&k);
printf("\n the %d's Big Number is : %d ",k,RandomSelect(a,0,7,k));
system("pause");
}
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http://expert.csdn.net/Expert/topic/2642/2642994.xml?temp=.6752436
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2004-04-30 12:15 | 453 | 1,112 | {"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-2022-21 | latest | en | 0.140108 |
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# Paint on a new airliner is usually applied in two stages:
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25 Feb 2013, 08:21
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Paint on a new airliner is usually applied in two stages: first, a coat of primer, and then a top coat. A new process requires no primer, but instead uses two layers of the same newly developed coating, with each layer of the new coating having the same thickness and weight as a traditional top coat. Using the new process instead of the old process increases the price of a new aircraft considerably.
Which of the following, if true, most strongly indicates that it is in an airline's long-term economic interest to purchase new airliners painted using the new process rather than the old process?
(A) Although most new airliners are still painted using the old process, aircraft manufacturers now offer a purchaser of any new airliner the option of having it painted using the new process instead.
(B) A layer of primer on an airliner weighs more than a layer of the new coating would by an amount large enough to make a difference to that airliner's load-bearing capacity.
(C) A single layer of the new coating provides the aluminum skin of the airliner with less protection against corrosion than does a layer of primer of the usual thickness.
(D) Unlike the old process, the new process was originally invented for use on spacecraft, which are subject to extremes of temperature to which airliners are never exposed.
(E) Because the new coating has a viscosity similar to that of a traditional top coat, aircraft manufacturers can apply it using the same equipment as is used for a traditional top coat.
Can someone please provide a detailed explanation for this question. Thanks!
[Reveal] Spoiler: OA
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Re: Paint on a new airliner is usually applied in two stages: f [#permalink]
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25 Feb 2013, 08:43
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Hi,
In this questions you want to strengthen the idea that the new coating will be beneficial financially in the long run even though it is much more expensive initially. So the answer will provide some financial benefit that could outweigh the initial extra cost.
Paint on a new airliner is usually applied in two stages: first, a coat of primer, and then a top coat. A new process requires no primer, but instead uses two layers of the same newly developed coating, with each layer of the new coating having the same thickness and weight as a traditional top coat. Using the new process instead of the old process increases the price of a new aircraft considerably.
Which of the following, if true, most strongly indicates that it is in an airline's long-term economic interest to purchase new airliners painted using the new process rather than the old process?
(A) Although most new airliners are still painted using the old process, aircraft manufacturers now offer a purchaser of any new airliner the option of having it painted using the new process instead.
No financial benefit
(B) A layer of primer on an airliner weighs more than a layer of the new coating would by an amount large enough to make a difference to that airliner's load-bearing capacity.
The plane will be able to carry more with the new coating and therefore make more money per flight. CORRECT
(C) A single layer of the new coating provides the aluminum skin of the airliner with less protection against corrosion than does a layer of primer of the usual thickness.
This only addresses a single layer. There will be two layers. So this is irrelevant.
(D) Unlike the old process, the new process was originally invented for use on spacecraft, which are subject to extremes of temperature to which airliners are never exposed.
(E) Because the new coating has a viscosity similar to that of a traditional top coat, aircraft manufacturers can apply it using the same equipment as is used for a traditional top coat.
This is saying that you won't have to buy new equipment to put on the coating BUT it does not address any new financial benefit to offset the costs.
Let me know if you need more advice on this questions. Happy Studies.
HG.
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25 Feb 2013, 09:51
Thanks for the wonderful explanation Kudos!
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Re: Paint on a new airliner is usually applied in two stages: f [#permalink]
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26 Feb 2013, 09:03
fozzzy wrote:
Paint on a new airliner is usually applied in two stages: first, a coat of primer, and then a top coat. A new process requires no primer, but instead uses two layers of the same newly developed coating, with each layer of the new coating having the same thickness and weight as a traditional top coat. Using the new process instead of the old process increases the price of a new aircraft considerably.
Which of the following, if true, most strongly indicates that it is in an airline's long-term economic interest to purchase new airliners painted using the new process rather than the old process?
(A) Although most new airliners are still painted using the old process, aircraft manufacturers now offer a purchaser of any new airliner the option of having it painted using the new process instead.
(B) A layer of primer on an airliner weighs more than a layer of the new coating would by an amount large enough to make a difference to that airliner's load-bearing capacity.
(C) A single layer of the new coating provides the aluminum skin of the airliner with less protection against corrosion than does a layer of primer of the usual thickness.
(D) Unlike the old process, the new process was originally invented for use on spacecraft, which are subject to extremes of temperature to which airliners are never exposed.
(E) Because the new coating has a viscosity similar to that of a traditional top coat, aircraft manufacturers can apply it using the same equipment as is used for a traditional top coat.
Can someone please provide a detailed explanation for this question. Thanks!
Hi fozzzy,
This is "Evaluate a Plan" type question. The keywords in the question is "long term" and "economic interest", so the correct answer choice will provide a strong economic incentive for the airliner to use the new painting process.
Stimulus says that the new process increases the price of the new aircraft considerably. So, there should be something that could offset this price and increase the revenue in long term
.
(A) Although most new airliners are still painted using the old process, aircraft manufacturers now offer a purchaser of any new airliner the option of having it painted using the new process instead.
The aircraft manufacturer offers an option, but this will not offset the price of new painting process.
(B) A layer of primer on an airliner weighs more than a layer of the new coating would by an amount large enough to make a difference to that airliner's load-bearing capacity.
The key here is "load-bearing capacity". If the airliner's load-baring capacity increases then it can carry more luggage/ passengers etc, this in turn would generate greater revenue for the airliner. This choice most strongly indicates that switching to the aircraft with the new paint will be in the long-term economic interest for the airliner
(C) A single layer of the new coating provides the aluminum skin of the airliner with less protection against corrosion than does a layer of primer of the usual thickness.
This choice discusses about a "single layer" but the new process will include "two layers" of coating. So, this choice is will not help us determine whether the new painting process will be in the economic interest of the airline.
(D) Unlike the old process, the new process was originally invented for use on spacecraft, which are subject to extremes of temperature to which airliners are never exposed.
The stimulus discusses aircraft; the performance of new painting process on spacecraft makes a wrong comparison. It could be possible that the aircraft will never face extreme temperature.
(E) Because the new coating has a viscosity similar to that of a traditional top coat, aircraft manufacturers can apply it using the same equipment as is used for a traditional top coat.
Even if the aircraft manufacturers are using the same equipment, it doesn't mean that they will sell the aircraft cheap enough so that the airline can have an economic advantage.
Hope that helps,
Vercules
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26 Feb 2013, 09:06
Do you have any tips on how to tackle Evaluate a plan question stems effectively?
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Re: Paint on a new airliner is usually applied in two stages: f [#permalink]
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03 Mar 2013, 19:28
fozzzy wrote:
Do you have any tips on how to tackle Evaluate a plan question stems effectively?
Hi,
e-GMAT had recently written an article on the approach to handle Evaluate questions. You may find it useful. Here's the link:
a-primer-on-variance-analysis-147036.html#p1180642
Thanks,
Chiranjeev
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Re: Paint on a new airliner is usually applied in two stages: [#permalink]
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10 May 2013, 02:56
Topic is locked as it is duplicate post.
Re: Paint on a new airliner is usually applied in two stages: [#permalink] 10 May 2013, 02:56
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Display posts from previous: Sort by | 3,028 | 13,164 | {"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} | 3.984375 | 4 | CC-MAIN-2017-22 | longest | en | 0.909167 |
https://www.thestreet.com/story/11645025/1/execuspeak-dictionary-hurdle-rate.html | 1,529,505,149,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00516.warc.gz | 916,721,657 | 28,808 | NEW YORK ( ExecuSpeak Dictionary) -- Is it a good deal? The fast answer is this: If it exceeds the hurdle rate. But there's more to it than just a number.
The hurdle rate is the measure used to determine if an investment is considered to be worth spending the money. It's used to compare and contrast a multitude of investment and project choices.
Most explanations indicate that the calculation is designed to make the decision (this investment vs. that investment vs. no investment) a quantitative and financial decision. Presumably, the hurdle rate considers risk. Thus projects to reduce costs are likely to have a lower hurdle than projects associated with new products since new products are more risky.
It is essential to understand that the hurdle rate varies. The decision makers have determined, in advance, which measure will be considered their "hurdle rate." The preferred measure to define the hurdle rate at one company may be return on investment (ROI). Another company prefers internal rate of return (IRR). Others might select net present value (NPV). Sometimes the hurdle rate is referred to as the "required rate of return." That's the first part.
The second element of the hurdle rate information is the actual number or the metric. There are lots of ways to select, or justify, the actual value once NPV or ROI or IRR is selected. The choice of interest rates (whether the formula uses 2% or 3% or 3.5%) selected for the cost of capital and the alternative investment (U.S. Treasuries or money market) are decisions. But even then the actual metric will vary since one venture capital (VC) firm may set their hurdle rate at an ROI of 30% on a project in cleantech while another venture capital firm may set the bar at 35%. Or a large corporation might use NPV to compare and contrast investments across divisions and subsidiaries.
The choices made in selecting the composition of the formula -- the interest rate selected, the measurement selected (NPV or ROI or IRR), the way the calculation is used and discussed -- all provide insight into the investing group's priorities and values.
Beyond this technical discussion of the metric, the number and the calculation, the hurdle rate is not about the number. It's a filter. It's a means of separating the wheat from the chaff to organize a potentially overwhelming set of investment choices and options. It's a management tool to organize a broad array of investment options.
With the exception of the interest rate or cost of money element, it doesn't take strategy into consideration. But it aids in the discussion. Effectively, it's a way of getting the financials out of the way so the strategy discussion can begin.
More from ExecuSpeak Dictionary
--By Carol Heiberger
Carol Heiberger is the author of ExecuSpeak Dictionary. Experience includes positions with the Ford Motor Company, Bell Atlantic, consulting, and as COO of a start-up CATV/ISP. She has expertise in strategic planning, marketing and finance. She has taught degree-seeking graduate students and knowledge-seeking adults of all ages and walks of life. She earned her MBA from Wharton.
Contact: Carol@execuspeakdictionary.com | 653 | 3,178 | {"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} | 2.71875 | 3 | CC-MAIN-2018-26 | latest | en | 0.937824 |
https://stats.stackexchange.com/questions/103741/how-to-compute-pca-scores-from-eigendecomposition-of-the-covariance-matrix | 1,722,725,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00607.warc.gz | 457,483,248 | 40,934 | How to compute PCA scores from eigendecomposition of the covariance matrix?
Given a data matrix $\mathbf X$ of $12 \times 7$ size with samples in rows and variables in columns, I have calculated centered data $\mathbf X_c$ by subtracting column means, and then computed covariance matrix as $\frac{1}{N-1} \mathbf X_c^\top \mathbf X_c$.
The dimensions of this covariance matrix are $7 \times 7$. After that I have calculated the eigenvalue decomposition, obtaining eigenvector matrix $\mathbf V$ with eigenvectors in columns.
Now I want to know about projections (i.e. principal component scores in PCA): is it $\mathbf{V}^\top \mathbf{X}_c$ or $\mathbf{X}_c \mathbf{V}$?
mean_matrix = X - repmat(mean(X),size(X,1),1);
covariance = (transpose(mean_matrix) * mean_matrix)/(12-1);
[V,D] = eig(covariance);
[e,i] = sort(diag(D), 'descend');
V_sorted = V(:,i);
• If you want the PCA, why not call [COEFF,SCORE] = princomp(X) (see here) directly? Commented Jun 17, 2014 at 22:45
• Also, why bother with the long road of standardizing the covariance matrix, when simply using the correlation matrix will get you there in one step? Commented Jun 18, 2014 at 4:15
• it is just self understanding,that why i preferred to do it myself Commented Jun 18, 2014 at 5:10
• @Dato, in my previous answer to you there is clearly written: "Raw component scores ... are computed directly as X*V". All necessary formulas for PCA and factor analysis were given by me there. Commented Jun 18, 2014 at 5:38
The projection is given by $\mathbf{X}_c \mathbf{V}$; principal component scores are columns of this matrix.
Your first formula cannot be computed, as $\mathbf{V}^\top$ is of $7 \times 7$ size and $\mathbf{X}_c$ has $12$ rows; the dimensions do not match, and the matrix product is not defined. Instead, $\mathbf{V}^\top \mathbf{X}_c^\top$ would be another correct formula, but it is simply a transpose of the first equation: $\mathbf{V}^\top \mathbf{X}^\top_c = (\mathbf{X}_c\mathbf{V})^\top$. Here PC scores are in rows. | 578 | 2,012 | {"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} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.8448 |
http://www.chegg.com/homework-help/suppose-point-pinhole-camera-15-m-tall-tree-75-m-away-detec-chapter-19-problem-36gp-solution-9780321595485-exc | 1,469,684,495,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00097-ip-10-185-27-174.ec2.internal.warc.gz | 371,193,627 | 19,564 | View more editions
# TEXTBOOK SOLUTIONS FOR College Physics A Strategic Approach with MasteringPhysics 2nd Edition
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Chapter: Problem:
Suppose you point a pinhole camera at a 15-m-tall tree that is 75 m away.
a. If the detector is 22 cm behind the pinhole, what will be the size of the tree’s image on the detector?
b. If you would like the image to be larger, should you get closer to the tree or farther from the tree? Explain.
c. If you had time, you could make the image larger by rebuilding the camera, changing the length or the pinhole size. What one change would give a larger image?
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College Physics A Strategic Approach with MasteringPhysics | 2nd Edition
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Alternate ISBN: 9780321596079, 9780321596291, 9780321596307, 9780321696304, 9780321737014, 9780321738592, 9780321777836 | 297 | 1,098 | {"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-2016-30 | latest | en | 0.818638 |
https://permanentkisses.com/understanding-what-is-work/ | 1,726,791,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00193.warc.gz | 397,983,805 | 12,239 | # Understanding What is Work
Physics refers to the science of the forces that act on matter to cause it to move. In layman’s terms, work is energy that is transferred from a source to an end with the help of a dynamic force or motive power. The definition of “work” is very broad, but we can give some examples that will better illustrate the concept. From this discussion of work it should be evident that it can be a source of power such as in rocket engines or the movement of particles in the earth’s crust. It can also be a force, for example the force of gravity that keeps everything on earth’s surface up and running.
In physics, work is any energy that is transferred from a source to an end through the application of a particular force or motive power. In its most simple form, it can be represented as the resultant of a certain displacement and force applied to an object. Any source of power can be used to cause an object to move, whether it is a constant speed force such as gravity or a specific momentary force like heat coming out of a hot object. In general, one finds that the work cannot be created nor destroyed. It is only a change from zero to one.
One way to put this process into perspective is through the example of a spring. A spring is designed so that when it is placed at the ends it remains at the same distance apart from the starting point, always moving in the same direction. But this spring could not exist if it did not have a motion of its own. In order for this to happen, a combination of three things would need to be the exact values: the kinetic energy of the spring, the potential energy needed to overcome gravity and the total time taken to go from the starting point to the end. Now if we take the total time, we find that it takes a great amount of time for a spring to reach its destination! This is because a spring cannot move entirely in one direction for a very long time.
Thus, the only thing that makes a spring move is a change in direction through a force that causes the spring to alter its position from zero to a higher value. If you take a baseball and place it on a table, you will see that there is an enormous change in displacement when the ball is thrown in different directions. A particle that has been carefully chosen so that it has a particular directional distribution, will also exhibit similar dispersion. These particles are referred to as a work flow.
The relationship between the force and the power that cause an object to move can be expressed in terms of a force and power law. Force and power are always equal, only in different units. A force is any kind of forward-directed effort, which pushes an object from a straight line to a curved path. Power, on the other hand, is any kind of backward-directed effort that pushes an object in a curved direction. In simple terms, force equals distance times the power of the force, while power is equal to distance times the force times the square of the acceleration.
The concept of a force or pressure exerted on an object depends on the property that governs the orientation of the object along the x-axis. The formula for the force is actually the derivative of the product of the force acting on the object and the time for the object to move with a constant speed. In the field of mathematics this is called a closed system, where a constant force acts uniformly on an arbitrarily chosen system. This type of law is used to solve problems involving the Law of Conservation of Energy, which states that energy cannot be changed from a point in zero velocity to a point in a greater velocity. | 739 | 3,630 | {"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-2024-38 | latest | en | 0.957173 |
http://blog.csdn.net/qq_26122039/article/details/52346505 | 1,521,381,795,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645775.16/warc/CC-MAIN-20180318130245-20180318150245-00666.warc.gz | 44,943,236 | 14,755 | # Codeforces Round #143 (Div. 2)-D. Magic Box
D. Magic Box
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point(0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
The numbers are located on the box like that:
• number a1 is written on the face that lies on the ZOX plane;
• a2 is written on the face, parallel to the plane from the previous point;
• a3 is written on the face that lies on the XOY plane;
• a4 is written on the face, parallel to the plane from the previous point;
• a5 is written on the face that lies on the YOZ plane;
• a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers xy and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1y1z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
input
2 2 2
1 1 1
1 2 3 4 5 6
output
12
input
0 0 10
3 2 3
1 2 3 4 5 6
output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6(on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
int main(){
// freopen("in.txt", "r", stdin);
int a[7];
int x, y, z, x1, y1, z1;
scanf("%d%d%d%d%d%d", &x, &y, &z, &x1, &y1, &z1);
for(int i = 1; i <= 6; i++)
scanf("%d", a+i);
int s = 0;
if(y > y1)
s += a[2];
if(y < 0)
s += a[1];
if(z > z1)
s += a[4];
if(z < 0)
s += a[3];
if(x > x1)
s += a[6];
if(x < 0)
s += a[5];
cout << s << endl;
return 0;
}
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