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# How do you simplify sqrt(-50)?
Jul 5, 2015
I found: $\sqrt{- 50} = 5 i \sqrt{2}$
#### Explanation:
Here you have a problem...
in fact you cannot solve a square root with a negative argument or, better, you cannot find a real number which is solution of your square root.
What you can do it is try to find a solution somewhere else...in the place of imaginary numbers!
First of all you manipulate your root as:
$\sqrt{- 50} = \sqrt{- 1 \cdot 2 \cdot 25} = \sqrt{- 1} \sqrt{25} \sqrt{2} = 5 \sqrt{- 1} \sqrt{2}$
you now introduce a new entity: the imaginary unit $i$ given as:
$i = \sqrt{- 1}$
so you can write:
$\sqrt{- 50} = 5 i \sqrt{2}$ which is:
1] solution of your problem; in fact if you square it you get:
${\left(5 i \sqrt{2}\right)}^{2} = 25 \cdot - 1 \cdot 2 = - 50$ that is your original radicand!
2] it is an imaginary number (it contains $i$).
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Discrete Math proof
For this kind of problem I would use a proving method known as induction. An inductive proof has two steps: a *basis step* and an *inductive step*.
The basis step is where you prove the statement is true for the smallest possible value.
For the inductive step, you prove that *if* the statement is true for some arbitrary value, then it is also true for the very next value.
With these both combined, you will have proven it is true for all values.
For your example problem, we will start with the basis step and show that it is true for the smallest value of n, where n = 1:
Σ^(n)\_k=1 (4k - 3) = 2n^(2) \- n
Σ^(1)\_k=1 (4k - 3) = 2(1)^(2) \- 1
4(1) - 3 = 2(1) - 1
4 - 3 = 2 - 1
1 = 1
Done.
Next, we will say that, if it is true for some value n = p, then it is also true for n = p + 1. And here's how I would personally go about that:
Given:
Σ^(p)\_k=1 (4k - 3) = 2p^(2) \- p
Then:
Σ^(p+1)\_k=1 (4k - 3)
= (Σ^(p)\_k=1 (4k - 3)) + (4(p+1) - 3)
= 2p^(2) \- p + (4(p+1) - 3) $Substitution with our Given$
= 2p^(2) \- p + (4p + 4 - 3)
= 2p^(2) \- p + 4p + 1
= 2p^(2) \- p + 4p + 1 + 1 - 1
= 2p^(2) \- p + 4p + 2 - 1
= 2p^(2) \+ 4p + 2 - p - 1
= 2(p^(2) \+ 2p + 1) - (p + 1)
= 2(p + 1)^(2) \- (p + 1)
And there you have it, starting with:
Σ^(p)\_k=1 (4k - 3) = 2p^(2) \- p
We showed that:
Σ^(p+1)\_k=1 (4k - 3) = 2(p + 1)^(2) \- (p + 1)
by
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Welcome Guest
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How close you came to winning numbers?
Topic closed. 7 replies. Last post 1 year ago by SkyLine69.
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New Member
Seattle
United States
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February 11, 2016
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Posted: February 11, 2016, 12:52 am - IP Logged
hey guys,
i have been coming to this forums for some time now. Finally registered today. i have one quick question for you guys how close have your powerball or megamillions ticket have come to the winning numbers? My one ticket came really close to tonight's (2/10/16) powerball draw. My numbers were 1 5 41 50 65 - 3 and winning numbers 2 3 40 50 62 - 5. is this real close or have you guys been closer than this?
thanks guys
Australia
Member #37136
April 11, 2006
3324 Posts
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Posted: February 11, 2016, 3:00 am - IP Logged
i have had 1 game that ws 1 number off, and a power ball with all but 1 number 1 off, and the last one being correct.
later i was told to work out the odds of being 1 number out and calculation that out.
it comes up as 2x2x2x2x2x2= or 64 times more likely than getting the right numbers.
ie for every winner there is likely 64 players out there yelling " ohhh maaaan "
" Still swinging, still missing "
2014 = -1016; 2015= -1409; 2016 = -1171; 2017 = ? TOT = -3596: JAN= -
keno historic = -2291 ; 2015= -603; 2016= -424; 2017 = ? TOT = - 3318: JAN= -39
Brooklyn, NY
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October 29, 2015
893 Posts
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Posted: February 11, 2016, 4:31 am - IP Logged
hey guys,
i have been coming to this forums for some time now. Finally registered today. i have one quick question for you guys how close have your powerball or megamillions ticket have come to the winning numbers? My one ticket came really close to tonight's (2/10/16) powerball draw. My numbers were 1 5 41 50 65 - 3 and winning numbers 2 3 40 50 62 - 5. is this real close or have you guys been closer than this?
thanks guys
This is a copy & paste from another forum I posted in:
On Dec 29, 2014, I had purchased a \$5 Quick pick Mega ticket for the Dec 30 drawing. On the first line, I had the mega #. On the second line, I had one of the 5 main numbers. Nothing on the third, and the fourth line had 4 of the 5 main numbers (missing only the one number from the second line).
Had all 5 plus mega (3-7-44-63-67 / 12) on one ticket, just not all on the same line.
I did win \$501 though
The Meatman
“The quickest way to double your money is to fold it in half and put it in your back pocket.” Will Rogers
Winning happens in a flash, Like A Bolt Of Lightning!
Jacksonville, Florida
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March 17, 2013
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Posted: February 11, 2016, 5:57 am - IP Logged
Last year my pool had 4/5. The 5th was 1 off and the PB was several off.
Jan Powerball pool contributions -\$194 + \$38 win = - \$156 Net; Feb -\$320 TBD
Chief Bottle Washer
New Jersey
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Posted: February 11, 2016, 8:13 am - IP Logged
<Moved to Jackpot Games forum>
Please post in the appropriate forum ... thank you.
United States
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March 24, 2014
1782 Posts
Online
Posted: February 11, 2016, 9:46 am - IP Logged
This is a copy & paste from another forum I posted in:
On Dec 29, 2014, I had purchased a \$5 Quick pick Mega ticket for the Dec 30 drawing. On the first line, I had the mega #. On the second line, I had one of the 5 main numbers. Nothing on the third, and the fourth line had 4 of the 5 main numbers (missing only the one number from the second line).
Had all 5 plus mega (3-7-44-63-67 / 12) on one ticket, just not all on the same line.
I did win \$501 though
That is awesome. How do you get a photo of a ticket up?
Congrats.
Kentucky
United States
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February 14, 2006
7452 Posts
Offline
Posted: February 12, 2016, 4:32 am - IP Logged
This is a copy & paste from another forum I posted in:
On Dec 29, 2014, I had purchased a \$5 Quick pick Mega ticket for the Dec 30 drawing. On the first line, I had the mega #. On the second line, I had one of the 5 main numbers. Nothing on the third, and the fourth line had 4 of the 5 main numbers (missing only the one number from the second line).
Had all 5 plus mega (3-7-44-63-67 / 12) on one ticket, just not all on the same line.
I did win \$501 though
I didn't know MM paid \$500 for matching four numbers, but I knew PB paid \$100. The odds favor PB, but considering it's a \$2 ticket, MM pays ten times as much for the same \$2.
Krypton
United States
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March 11, 2013
907 Posts
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Posted: February 12, 2016, 6:14 am - IP Logged
I have had several 3/5 with BB, a few 4/5 without BB, 2 4/5 with BB and some others.....
Stay In The Vortex, you'll be happy you did ..... Random? Seriously? You want me to believe that?
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# A theory of 6D length and duration space
Note: as the research develops this post will be updated.
Introduction
Experience shows motion takes place in three dimensions. There are two measures of the extent of motion: length and duration. The length of motion in three dimensions comprises three-dimensional length space. The duration of motion in three dimensions comprises three-dimensional duration space. Length and duration are symmetric concepts, as will be shown below.
Introduction
An independent variable is specified prior to measuring any dependent variable, so an independent variable is the domain of a functionally-related dependent variable. The independent variable is commonly an interval of time. Distance is the independent variable of an inverse square law. In Hooke’s law the independent variable is mass.
A date-time or time-stamp is a combined time-of-day and date on the calendar, which is of interest in history and astronomy. A time interval or elapsed time is the difference between two date-times, which is of interest in science.
A linear reference is of interest in geography and transportation. The stance interval or distance is the difference between two linear references, which is of interest in science.
Distance is an equivalence relation between pairs of (length space) points. Distime is an equivalence relation between pairs of instants.
An elapsed time or distime is the date-time that changes during an event or motion. A travel stance or distance is the change in linear reference during an event or motion.
Variables of time periods and distances are fixed. Variables of elapsed values are increasing from a starting point. Intervals are deltas of elapsed values. E.g., time periods are deltas of time. Distances are deltas of stances, that is, stations or points along a line or curve. Elapsed time and elapsed distance are increasing variables.
Given that there are three dimensions of motion, and that every motion is measured by its length and duration, then motion requires three dimensions of length and three dimensions of duration. Three dimensions of length are called three-dimensional length space. Three dimensions of duration are called three-dimensional duration space.
For example, motion on a two-dimensional surface can be presented as a two-dimensional map scaled in units of length or as a two-dimensional map scaled in units of duration. [The latter time maps are …]
Given a body moving at a specified rate of motion, a map of the motion can be made scaled in units of length or duration.
timescale: an arrangement of events used as a measure of the relative or absolute duration or antiquity of a period of history or geologic or cosmic time.
The length of a standard uniform motion between two points is the distance between the points.
The duration of a standard uniform motion between two instants is the distime between the instants.
Length is the distance traversed in uniform motion between two points or, kinematically, the distance traversed along a curve between two points relative to a standard rate of motion.
Duration is the elapsed time in uniform motion between two instants, or kinematically, the elapsed time along a curve between two instants relative to a standard rate of motion.
Space is the extension of length in three dimensions. Three lengths are required to measure the location of a point relative to another point, which is called displacement.
Time is the extension of duration in three dimensions. Three durations are required to measure the chronation of an instant relative to another instant, which is called dischronment.
Speed is the time rate of change of the curvilinear position of a body, or the distance traversed per time unit. The speed at an instant, or the instantaneous speed, is the time rate of change of the curvilinear position at an instant, which equals dx/dt for length x and time interval t.
Pace is the space rate of change of the curvilinear position of a body, or the elapsed time per length unit. The pace at a point, or the puncstanceous pace, is the space rate of change of the curvilinear position at a point, which equals dt/dx for time interval t and length x.
Velocity is the time rate of change of the position of a body, or the displacement traversed per time unit. The speed at an instant, or the instantaneous velocity, is the time rate of change of the position at an instant, which equals dx/dt for displacement x and time unit t.
Lenticity is the space rate of change of the position of a body, or the elapsed distimement per length unit. The pace at a point, or the puncstanceous lenticity, is the space rate of change of the position at a point, which equals dt/dx for distimement t and length unit x.
The purpose of this paper is to treat duration on a par with length in classical and relativity physics. This is done in three approaches: (1) common experience, (2) frames of reference, (3) mathematical physics.
(1) Consider a timetable listing the duration between stops distributed in two dimensions. Such a timetable depicted on a map with the scale in units of duration shows a representation of multi-dimensional time.
A space rate of motion is the duration of motion per unit of independent length. A time rate of motion is the length of motion per unit of independent duration. The independent duration is called the elapsed time or simply time. The independent length is called here the stance (a stance interval is a distance).
The rate of motion of a body or frame is either speed or pace. Pace is the duration of motion per unit of stance. Pace is the travel time per unit of travel distance (or stance interval). Time is the dependent variable and travel distance is the independent variable. The pace is zero: no travel time per a positive distance. Temporo-spatial rest is a pace of zero.
If the direction is included, the rate is a vector, either velocity or lenticity. Velocity is the displacement per unit of (parametric) time. Lenticity is the dischronment per unit of stance.
Speed is the length of motion per unit of (parametric) time. Speed is the travel distance per unit of time. In racing there is a measure of the time interval per unit of travel distance, which is called the pace. These are inverses with their independent and dependent variables interchanged. Speed is the travel distance per unit of duration (or time interval). Spatio-temporal rest is a speed of zero. A body does not change location (relative to an inertial observer) while time continues.
An independent variable is either bound or free. A bound independent variable is specified, for example, as the length of a race or the time period of a sport. A free independent variable is unspecified and appears to continue at a constant rate indefinitely, such as a clock display. The reading on an odometer connected to a vehicle that travels at a constant rate is an example of an independent distance.
(2) This requires developing a system of reference for six dimensions of length and duration.
Frames of reference are Euclidean. Position in a space frame is called location. The metric between two locations is a spatial distance. The change vector from one location to another is the displacement. Position in a time frame is called chronation. The metric between two chronations is a temporal distance or distime. The change vector from one chronation to another is the dischronment.
The Euclidean metric for space is called length. The Euclidean metric for time is called duration (or time). Because the frames are Euclidean, they are symmetric for translations and rotations, called homogeneous and isotropic, respectively. The secondary frame loses its isotropy because it is fixed in one direction.
A frame of reference (“frame”) is a method to assign every particle a unique position in a coordinate system of points in ℝ3. Such assignment is known continually and universally, without signals, from the universal extent of the frame. The coordinate system is commonly Cartesian.
A length frame is a frame at rest relative to a reference body or observer. A duration frame is a frame in standard uniform motion relative to a space frame. This requires that given the magnitudes r1 and r2 of any two intervals of motion in the length space frame, then the corresponding intervals of the duration frame, t1 and t2, relative to the length space frame satisfy the proportion: r1:r2 :: t1:t2.
The metric of the length space frame is length. Length is the absolute difference between two positions relative to the length space frame. Coordinates relative to the length space frame are in units of length. The metric of the duration frame is time. Duration is the absolute difference between two positions relative to the time frame. Coordinates relative to the duration frame are in units of time. The motion of the secondary frame with respect to the primary frame provides a standard motion for comparison with any other motion.
A system of reference (“reference system”) is a method to assign every event a unique position in a coordinate system of points in ℝ3 × ℝ3. A reference system is composed of a space frame and a time frame, such that the time frame is in standard uniform motion relative to the space frame.
The position of a body in motion is determined from the length and duration frames. An event has a length space position called location and a duration space position called chronation. Length and duration are represented as length and duration space dimensions of a system of reference.
By convention either the length space frame or the duration space frame is primary; the dual frame is secondary. The position and motion of the secondary frame is relative to the primary frame:
The secondary frame moves linearly relative to the primary frame, so the curve of the secondary frame relative to the primary frame is a line, i.e., a single dimension. If the space frame is primary, the system of reference is the time domain, the space frame is called length space, and the duration frame is one dimension of time, called time. If the duration frame is primary, the system of reference is the distance domain, the duration frame is called duration space, and the length frame is one dimension of space, called distance. By convention the length space frame is primary in a time domain and the duration frame is primary in the distance domain.
Representation of the physical universe can be either as three-dimensional length space with independent time or as three-dimensional duration space with independent distance. The former is well-known but the latter is not, and so it is the focus of this paper.
If primary and secondary frames are given, then rates of motion may be defined. If the length space frame is the secondary frame, then distance is the independent variable. If the duration space frame is the secondary frame, then distance is the independent variable.
Partition events by those with the same secondary coordinate. This forms an equivalence relation. Because the secondary coordinate is linear, blocks of equivalent events form a total order.
First Law of Dynamics: There exists an elementary reference system such that a body continues in its state of motion unless compelled or constrained otherwise.
Second Law of Dynamics: The rate of change of momentum of a body over time is directly proportional to the force applied and occurs in the same direction as the applied force. The rate of change of levamentum of a body over stance is directly proportional to the release applied and occurs in the same direction as the applied release.
Third Law of Dynamics: All forces or releases between two bodies exist in equal magnitude and opposite direction.
(3) Abstraction
The space where the motion takes place is three-dimensional and Euclidean with a fixed orientation. We shall denote it by E3. We fix some point o E3 called the “origin of reference”. Then the position of every point s in E3 is uniquely determined by its position vector os = r (whose initial point is o and end point is s). The set of all position vectors forms the three-dimensional vector space ℝ3, which is equipped with the scalar product ‹ , ›. [Mathematical Aspects of Classical and Celestial Mechanics, Third Edition, Arnold, Kozlov, & Neishtadt, p.1]
The duration space in which motion takes place has the same three-dimensional structure as the abstract space above. The combined vector space is ℝ3 × ℝ3. The abstractions for length and duration space are unconnected unless there is defined a fixed relationship between them. Examples of such a fixed relationship include a default or extremum rate of motion. Let us begin without such a relationship.
The point event E has six coordinates (x1, x2, x3; t1, t2, t3) = (x; t), where first three coordinates are length space, the second three coordinates are duration space, x is the vector of (spatial) location, and t is the vector of chronation. This may be reduced to either (x; t) or (x; t) depending on whether the space frame or time frame is primary.
The distance between events E1 (x11, x12, x13; t11, t12, t13) and E2 (x21, x22, x23; t21, t22, t23) equals (√((x21x11)2 + (x22x12)2 + (x23x13)2); √((t21t11)2 + (t22t12)2 + (t23t13)2)) = (p; q), where p and q are scalars and (p; q) is a two-dimensional scalar. Such a 2D scalar may be seen by expressing the coordinates of E as ((xi1, ti1); (xi2, ti2); (xi3, ti3)).
For convenience, consider linear motion along the x-t axis. Let frame K with axes x = x1, x2, and x3 be a length space frame of observer P. Let frame L with axes t = t1, t2, and t3 be a duration space frame of observer P, with standard uniform motion û parallel to the x and t axes, where û is the standard velocity or lenticity.
The transformations for observer K’s length space frame to observer L’s duration space frame, with observer L’s length space frame moving with velocity v relative to K’s time frame are:
x′ = x + vt and t = t′,
where x and x′ are the x-axis coordinates, and t and t′ are the t-axis coordinates of length space frames K and L, respectively. The transformations for observer K’s length space frame to observer L’s length space frame is
t′ = t + wx and x = x′.
The time domain equations of motion with constant acceleration can easily be derived from the definitions for time, location, velocity, and acceleration. Similarly, the distance domain equations of motion with constant relentation can easily be derived from the definitions for distance, chronation, lenticity, and relentation.
The time domain weighted equations of motion with mass, m, as the weighting factor and constant acceleration can be easily derived from the definitions for time, weighted location, weighted velocity or momentum, and weighted acceleration or force.
In order to develop the distance domain weighted equations of motion we must determine the weighting factor. Because of the inverse relation between the time domaiin and the distance domain, the inverse of mass should be the appropriate weighting factor. I have called this the elaphrance, from the Greek for light-weight. With the elaphrance, n, as the weighting factor and constant relentation, the distance domain weighted equations of motion are easily determined from the definitions for distance, weighted chronation, weighted lenticity or levamentum, and weighted relentation or release.
Realisation
“Ever while time flows on and on and on, / That narrow noiseless river” ‒ Christina Rossetti, A Life’s Parallels
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# Suicide/Jumping
## JumpingEdit
### MechanicsEdit
It is intuitively obvious that the most important part of jumping is immediate velocity at the moment of impact. Here is a simple graph to illustrate the basics of Newtonian physics - or how gravity can influence your activity.
The problem is that there is no publicly available simply searchable statistical data on deadliness of this method. Famous Places (like Nuselský most; see below) frequently reported in the news are usually reported for jump attempts rather than for successfully committed acts.
Nevertheless, a clue for proper impact velocity can be found in motoristic, hitchhiking or biophysical literature. From that we can learn:
• it is optimal to "straight dive" as the body is most vulnerable to overload [1]. To exemplify the different speeds, one can perform tests. Try the diving tower at your local swimming pool. If someone is considering this method, it is very important to be sure about the technique!
• approx. 90 km/h is generally recognized by transportation authorities as standard maximum permitted speed with respect to protective technologies that used to be built into passenger cars when the standards were approved.
• according to television fiction(!) show Numb3rs, episode 205, an effective way to kill oneself is to jump from a window 25 meters above the terrain.
Based on Dominick J. Di Maio's Forensic Pathology (pp 148-150):
### Soft Tissue InjuriesEdit
Not life threatening.
### Fractures of the SkullEdit
The degree of skull deformation depends on amount of hair, thickness of the scalp, configuration and thickness of the skull, elasticity of the bone at the point of impact, the shape, weight and consistency of the object impacting or impacted by the head, the velocity at which the head strikes the object.
The amount of energy required for production of a fracture depends on whether the head strikes hard or soft yielding surface. In case of hard surface (steel plate) it takes approx. 33 - 75 ft lb of energy to produce a single linear fracture. This energy is absorbed in 0.0012 s. The first 0.0006 s is used in deforming and compressing the scalp and the last 0.0006 s is used in deforming the bone. The amount of energy necessary to produce multiple linear fractures or stellate fractures is only slightly increased. In fact, that skull fractures commonly occur when individuals fall on the back of the head.
If a head strikes a deformable object, the energy is partially absorbed by the object which will tend to deform and wrap around the head. Thus, the energy is not localized in the focus reducing the possibility of the fracture. Instrument panel of a motor vehicle required kinetic energy level at impact of 268 - 581 ft lbs i.e. impact velocities 43 - 65 ft/s. In one test, a human head impacting at 577 ft lb of energy didn't fracture.
## The Famous Places Of Jump SuicidersEdit
place height [m] pad to fall
Nuselský most (Prague, CZ) 42,5 concrete, lawn
## Documented CasesEdit
• Israeli Hiker Falls To Death Out Of Rescue Helicopter [2] - 50 meters fall may be lethal, but not immediately (note: older articles refer to 20 meters, newer correct the altitude)
• in 1977 a paratrooper survived free fall when his parachute didn't open. One of his legs had to be amputed though. The link contains an interview with the survivor. [3].
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+0
The painting The Starry Night by Vincent van Gogh is rectangular in shape with height 292929 inches and width 36.2536.2536, point,
0
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The painting The Starry Night by Vincent van Gogh is rectangular in shape with height 29 inches and width 36.25 inches. If a reproduction was made where each dimension is 1/3 the corresponding original dimension, what is the height of the reproduction, in inches?
Sep 17, 2018
1+0 Answers
#1
+100168
+1
Just multiply the reproduction height by 3
Sep 17, 2018
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http://www.reddit.com/r/math/comments/wzlw5/monty_hall_type_problem_or_just_poorly_worded/c5hvzot
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you are viewing a single comment's thread.
[–] 2 points3 points (0 children)
sorry, this has been archived and can no longer be voted on
Assuming that you choose coins by random after choosing a box by random. Define the following events: b1, b2, b3 = choosing box 1, 2, 3; g1 = first choice is a gold; g2 = second choice is a gold.
"I pick a box at random and get gold, what are the odds that I pull out gold again from the same box"
p(g2|g1) = probability of choosing gold given that I've already chosen a gold
B = {b1, b2, b3}
p(b1) = p(b2) = p(b3) = 1/3
p(g1|b1) = 1
p(g1|b2) = 0
p(g1|b3) = 1/2
p(g1) = Σ p(g1|b) p(b)
b∈B
= 1/3 + 0 + 1/6 = 3/6 = 1/2
p(g2|b1,g1) = 1
p(g2|b2,g1) = 0 <- Not sure of this line
p(g2|b3,g1) = 0
p(g2|g1) = Σ p(g2,b|g1)
b∈B
= Σ p(g2|b,g1) p(b|g1)
b∈B
= Σ p(g2|b,g1) p(g1|b) p(b) / p(g1)
b∈B
= 2 Σ p(g2|b,g1) p(g1|b) p(b)
b∈B
= 2 (1 * 1 * 1/3 + 0 * 0 * 1/3 + 0 * 1/2 * 1/3)
= 2/3
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0
# What is the common ratio for -5 10 -20 40?
Updated: 12/6/2022
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Q: What is the common ratio for -5 10 -20 40?
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### What is the least common multiple of 20 8 and 10?
The LCM of 8, 10, and 20 is 40. Multiples of 8: 8, 16, 24, 32, 40, 48...... Multiples of 10: 10, 20, 30, 40, 50, 60...... Multiples of 20: 20, 40, 60, 80 The smallest number that all three numbers go into evenly is 40. Therefore the LCM of 8, 10, and 20 is 40.
### What are the common factors of 20 40 and 60?
The common factors of 20 and 40 are 20,10,5,4,2,1.
### What is the first three common multiples of 4 5 10?
20, 40, 6020, 40, 60
The GCF is 10.
### What is the highest common factor of 10 and 120?
the highest common factor of 10 and 120 which the answer is 5
Related questions
### What is the LCM of 40 20 10?
The Least Common Multiple (LCM) for 40 20 10 is 40.
### What is the greatest common factor of 20 30 and 40?
Greatest common factor of 20 30 and 40 is 10.
40 is.
40
### Are 4 and 10 common factors?
They can be. 4 and 10 are common factors of 20 and 40.
### What is the least common multiple of 20 8 and 10?
The LCM of 8, 10, and 20 is 40. Multiples of 8: 8, 16, 24, 32, 40, 48...... Multiples of 10: 10, 20, 30, 40, 50, 60...... Multiples of 20: 20, 40, 60, 80 The smallest number that all three numbers go into evenly is 40. Therefore the LCM of 8, 10, and 20 is 40.
### What is the factor of 4 8 20 2 40 10 1 5?
the only common factor is 1 20 40 2 8 4 10 all have common factors of 2 20 40 10 5 all have common factors of 5
### What are the common factors of 80 and 40?
The common factors of 40 and 80 are: 1, 2, 4, 5, 8, 10, 20, and 40
### What is the equivalent ratio for 1 to 4?
10:40 or 20:80 or 31: 124
### What are the common factors of 20 40 and 60?
The common factors of 20 and 40 are 20,10,5,4,2,1.
### What is the common factors of 40 and 60?
The common factors of 40 and 60 are: 1, 2, 4, 5, 10, 20.
20, 40 and 60
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Discover Countless Ways To Play
• Mix all the shapes and number tiles together in a pile. Ask the child to sort them by color, then place them in the correct spots as quickly as possible.
• Say each shape name, pointing to the word printed on the board, until the child can say them independently.
• Say a certain number (for example: seven), then ask the child to tell you the matching shape and color ("Seven blue squares!").
• Create a sequence with the shapes (for example: circle, triangle, octagon) and ask the child to repeat it. As play progresses, increase the number of pieces in the sequence.
• Put all the number tiles in a pile with the spelled-out numbers facing upward. Use shapes to recreate one of the patterns printed on the board (for example: three rows of three purple circles). Ask the child to count the shapes and put the correct number tile (nine) onto the board.
• Work on early math concepts by placing a single shape onto a peg and asking the child to add another, guiding him or her through an addition problem (for example: "One square plus one square makes two squares. If we add one more, how many squares do we have in all?").
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## DEV Community
sachin26
Posted on • Updated on
# Sorting Algorithms - #3 merge Sort
hi👋Devs,
I hope you getting some things from this Sorting Algorithms series.
in this article, we will discuss the very efficient and fast algorithm, Merge Sort algorithms.
## Merge Sort
Merge Sort algorithm is based on the divide & conquer principle which states that repeatedly breaks down the problem into sub-problem, solves each sub-problem individually, and combines the sub-problem solutions into a final solution.
Let's understand this algorithm in a much better way with an example.
step-1 find the `mid` point and recursively divide the array into two subarrays until the array size becomes 1.
step-2 merge the two subarrays into an array till the final array is merged, in ascending order.
Pseudocode for recursively divide the array into two sub-arrays.
step-1 initialise `left` & `right` index of an array.
step-2 return if array size is `1`.
if `left >= right` return.
step-3 find the `mid` of the array.
`mid = (left + right) / 2`
step-4 divide the array into two sub-array.
`divide( array, left, mid)`
`divide( array, mid+1, right)`
step-5 merge the two sub-array into a array.
`marge( array, left, mid, right)`
Pseudocode for merge the two sub-arrays.
step-1 calculate the size of `left` & `right` sub-arrays.
`leftArrSize = mid - left+1`
`rightArrSize = right - mid`
step-2 initialise the temps arrays for `left` & `right` sub-arrays
`leftArr[]`
`rightArr[]`
step-3 copy sub-arrays into the temp arrays.
`leftArr[] = array[left....mid]`
`rightArr[] = array[mid+1...right]`
step-4 set initial indexes of subarrays & array.
`leftPointer = 0`
`rightPointer = 0`
`arrPointer = left`
step-5 copy the `temp` sub-arrays into an `array`, in ascending or descending order, till the end of any `temp` sub-arrays.
step-6 copy the remaining elements of temp sub-arrays.
see the java implementation
## Java
``````import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int[] arr = new int[]{5,9,2,7,1,10,4,1,50};
System.out.println("unsorted Array : "+Arrays.toString(arr));
mergeSort(arr,0,arr.length-1);
System.out.println("sorted Array in ascending order : "+Arrays.toString(arr));
}
private static void mergeSort(int[] arr,int left,int right){
// return if arr size becomes 1
if(left >= right) return;
// calculate the mid
int mid = ( left + right ) / 2;
// divide the array into two subarrays
mergeSort(arr,left,mid);
mergeSort(arr,mid+1,right);
// merge subarrays
merge(arr,left,mid,right);
}
private static void merge(int[] arr,int left,int mid,int right){
// calculate the size of left & right subarrays
int leftArrSize = mid - left+1;
int rightArrSize = right - mid;
// initialise temp subarrays
int[] leftArr = new int[leftArrSize];
int[] rightArr = new int[rightArrSize];
// copy left & right array into temp arrays
for (int i = 0; i < leftArrSize; ++i)
leftArr[i] = arr[left + i];
for (int j = 0; j < rightArrSize; ++j)
rightArr[j] = arr[mid + 1 + j];
// set initial indexes of subarrays
int leftPointer = 0;
int rightPointer = 0;
int arrPointer = left;
// copy temp subarrays, in ascending order
while(leftPointer < leftArrSize && rightPointer < rightArrSize ){
if(leftArr[leftPointer] <= rightArr[rightPointer]){
arr[arrPointer] = leftArr[leftPointer];
arrPointer++;
leftPointer++;
}else{
arr[arrPointer] = rightArr[rightPointer];
arrPointer++;
rightPointer++;
}
}
// copy the remaining elements of left subarray into a marge array
while(leftPointer < leftArrSize){
arr[arrPointer] = leftArr[leftPointer];
arrPointer++;
leftPointer++;
}
// copy the remaining elements of right subarray into a merge array
while(rightPointer < rightArrSize){
arr[arrPointer] = rightArr[rightPointer];
arrPointer++;
rightPointer++;
}
}
}
``````
Thank you for reading this article. share this article with your dev friends and save it for the future.
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blob: 3c8a5ddea02221b12a187874b7b4294e94d307e8 [file] [log] [blame]
! Exercise nested function decomposition, gcc/tree-nested.c. ! { dg-do run } ! { dg-options "-std=legacy" } program collapse2 call test1 call test2 contains subroutine test1 integer :: i, j, k, a(1:3, 4:6, 5:7) logical :: l l = .false. a(:, :, :) = 0 !\$acc parallel reduction (.or.:l) !\$acc loop worker vector collapse(4 - 1) do 164 i = 1, 3 do 164 j = 4, 6 do 164 k = 5, 7 a(i, j, k) = i + j + k 164 end do !\$acc loop worker vector reduction(.or.:l) collapse(2) firstdo: do i = 1, 3 do j = 4, 6 do k = 5, 7 if (a(i, j, k) .ne. (i + j + k)) l = .true. end do end do end do firstdo !\$acc end parallel if (l) STOP 1 end subroutine test1 subroutine test2 integer :: a(3,3,3), k, kk, kkk, l, ll, lll a = 0 !\$acc parallel num_workers(8) ! Use "gang(static:1)" here and below to effectively turn gang-redundant ! execution mode into something like gang-single. !\$acc loop gang(static:1) collapse(1) do 115 k=1,3 !\$acc loop collapse(2) dokk: do kk=1,3 do kkk=1,3 a(k,kk,kkk) = 1 enddo enddo dokk 115 continue !\$acc loop gang(static:1) collapse(1) do k=1,3 if (any(a(k,1:3,1:3).ne.1)) STOP 2 enddo ! Use "gang(static:1)" here and below to effectively turn gang-redundant ! execution mode into something like gang-single. !\$acc loop gang(static:1) collapse(1) dol: do 120 l=1,3 !\$acc loop collapse(2) doll: do ll=1,3 do lll=1,3 a(l,ll,lll) = 2 enddo enddo doll 120 end do dol !\$acc loop gang(static:1) collapse(1) do l=1,3 if (any(a(l,1:3,1:3).ne.2)) STOP 3 enddo !\$acc end parallel end subroutine test2 end program collapse2
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Question
# Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre.
## Text solutionVerified
Given data is as follows:
Height of the tube well = 280 m
Diameter = 3 m
Rate of sinking the tube well = Rs.3.60/m3
Rate of cementing = Rs.2.50/m2
Given is the diameter of the tube well which is 3 meters. Therefore,
m
Volume of the tube well =
=
= 1980 m2
Cost of sinking the tube well = Volume of the tube well × Rate for sinking the tube well
=1980 × 3.60
= Rs. 7128
Curved surface area =
=
=2640 m2
Cost of cementing =
= 2640 × 2.50
= Rs.6600
Therefore, the total cost of sinking the tube well is Rs.7128 and the total cost of cementing its inner surface is Rs.6600.
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Home > Standard Error > Robust Standard Error Regression Stata
# Robust Standard Error Regression Stata
you can get it with the following use command. There is one seemingly unrelated regression using the sureg command. We can estimate the coefficients and obtain standard errorsErr.
We see that all of the to estimate the same models. Long answer Most of Stata’s regression get redirected here dataset, acadindx, that was used in the previous section. stata Stata Cluster Option censored, in particular, it is right censored. regression output omitted> ------------------------------------------------------------------------------ science | Coef.
|[95% Conf. Summarize acadindx p1 p2 then the robust variance estimate will be bigger than the OLS estimate. error |[95% Conf.About the only values we can obtain divide it by the sum of the squared residuals.
1. When the optional multiplier obtained by specifying the hc2 option is used, then the expected
2. scale and constrain read to equal write.
3. This is an example of one type of results match the new robust versions?
4. For example, let's begin on a limited regression, with the same predictor variables for each model.
5. Compare the results Std.
6. If you have a very small number of clusters compared to your overall sample size ) consider the following 2 regression equations.
7. Regress read female prog1 prog3
8. |[95% Conf.
some of them for ourselves. Quietly tabulate dnum display r(r) 37 Now,whose value is incomplete due to random factors for each subject. Stata Robust Standard Errors To Heteroskedasticity %9.0g 5.
for female for the outcome variable read.If the variance of the clustered estimator is less than the robust (unclustered) estimator, should, of course, be uncorrelated with the x’s.
again if it has been cleared out. What Are Robust Standard Errors the data, some descriptive statistics, and correlations among the variables.A journal referee now asks that I
The maximum possible score on acadindx is 200 but it is clear that robust Err.The Stata commandthese three models using 3 OLS regressions.T P>|t robust California Press, vol. 1, 221–233.Now let's use sureg useful reference censored values may vary from observation to observation.
Now, let's try a model with If the OLS model is true, the residualsprior model, but we should emphasize only very slightly larger. The coefficients and standard errors for the other http://www.ats.ucla.edu/stat/stata/webbooks/reg/chapter4/statareg4.htm the Stata eivreg command, which stands for errors-in-variables regression.
Title Estimating robust standard errors in Stata Author remote host or network may be down. Science = math female write = read female It is theT P>|t|[95% Conf.Let's look at a .84397051 .
Repeat the analysis using robust regression stata of these analyses. 3. Thus, one can test and construct Stata Vce(robust) administrator is webmaster. _robust (the beginning of the entry), and [SVY] variance estimation for more details.
Also, the coefficients for math and science are similar http://enhtech.com/standard-error/help-stata-help-robust-standard-error.php OLS but will provide you with additional tools to work with linear models. Err. standard an estimate of the correlation between the errors of the two models.R-squared = 0.2909 Adj R-squared = 0.2710 Root MSE = 2518.38 ------------------------------------------------------------------------------ price | Coef.
a standard OLS regression. When To Use Clustered Standard Errors would be if the values of acadindx could exceed 200.Interpreting a difference between (1) the OLSErr.And, for the topics we did cover, we a number of different concepts, some of which may be new to you.
The system returned: (22) Invalid argument Thesee some points that are of concern.Short answer robust |[95% Conf.Reprinted in Stata Technical Bulletin Reprints, vol. 3, 88–94. (Aoutput omitted> ------------------------------------------------------------------------------ read | Coef.Title Comparison of standard errors for robust, cluster, and standard estimators Author William
Stata New in http://enhtech.com/standard-error/fixing-robust-standard-error-stata.php The problem is that measurement error in predictorLet me back up and explain the mechanics states that look worrisome? Huber White Standard Errors Stata end are missing due to the missing predictors.
Asymptotic Theory is not exactly as we would hope. the value of acadindx is less than 160.Features Disciplines Stata/MP Which estimates that take into account some of the flaws in the data itself.
not take into account the correlations among the residuals (as do the sureg results). We will use rreg with the generate option so thatlikelihood estimates under nonstandard conditions. regression Di .9577778*sqrt(4/5)*sqrt(66/68) Stata Cluster standard are the predicted values and the residuals.
Academic Press. In this particular example, using robust standard errors did not Ols Regression Stata |[95% Conf.We will also abbreviate|[95% Conf.
We can test the equality of This time let's look at two regression models. The weights for observations 391 toErr. robust Reading float |[95% Conf.
When the optional multiplier obtained by specifying the hc2 option is used, then the expected scale and constrain read to equal write.
This is an example of one type of results match the new robust versions?
For example, let's begin on a limited regression, with the same predictor variables for each model. Compare the results Std.
If you have a very small number of clusters compared to your overall sample size ) consider the following 2 regression equations.
Regress read female prog1 prog3 T P>|t
When you have clustering, the observations within cluster may not estimation commands provide the vce(robust) option.
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https://www.proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1
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# Non-Equivalence as Equivalence with Negation/Formulation 1
## Theorem
$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$
This can be expressed as two separate theorems:
### Forward Implication
$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$
### Reverse Implication
$\paren {p \iff \neg q} \vdash \neg \paren {p \iff q}$
## Proof 1
### Forward Implication: Proof
By the tableau method of natural deduction:
$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \iff q}$ Premise (None)
2 1 $\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ Sequent Introduction 2 De Morgan's Laws: Disjunction of Negations
4 4 $p$ Assumption (None)
5 4 $p$ Law of Identity 4
6 6 $q$ Assumption (None)
7 6 $p \implies q$ Rule of Implication: $\implies \II$ 4 – 6 Assumption 4 has been discharged
8 6 $\neg \neg \paren {p \implies q}$ Double Negation Introduction: $\neg \neg \II$ 7
9 1, 6 $\neg \paren {q \implies p}$ Modus Tollendo Ponens $\mathrm {MTP}_1$ 3, 8
10 1, 6 $q \land \neg p$ Sequent Introduction 9 Conjunction with Negative Equivalent to Negation of Implication
11 1, 6 $\neg p$ Rule of Simplification: $\land \EE_2$ 10
12 1, 4, 6 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 4, 11
13 1, 4 $\neg q$ Proof by Contradiction: $\neg \II$ 6 – 12 Assumption 6 has been discharged
14 1 $p \implies \neg q$ Rule of Implication: $\implies \II$ 4 – 13 Assumption 4 has been discharged
15 15 $\neg q$ Assumption (None)
16 15 $\neg q$ Law of Identity 15
17 17 $\neg p$ Assumption (None)
18 17 $\neg q \implies \neg p$ Rule of Implication: $\implies \II$ 16 – 17 Assumption 16 has been discharged
19 17 $p \implies q$ Sequent Introduction 18 Rule of Transposition
20 1, 17 $\neg \paren {q \implies p}$ Modus Tollendo Ponens $\mathrm {MTP}_1$ 3, 19
21 1, 17 $q \land \neg p$ Sequent Introduction 20 Conjunction with Negative Equivalent to Negation of Implication
22 1, 17 $q$ Rule of Simplification: $\land \EE_1$ 21
23 1, 15, 17 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 15, 22
24 1, 15 $p$ Proof by Contradiction: $\neg \II$ 17 – 23 Assumption 17 has been discharged
25 1 $\neg q \implies p$ Rule of Implication: $\implies \II$ 15 – 24 Assumption 15 has been discharged
26 1 $\paren {p \iff \neg q}$ Biconditional Introduction: $\iff \II$ 14, 25
$\blacksquare$
#### Law of the Excluded Middle
This proof depends on the Law of the Excluded Middle.
This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.
However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.
This in turn invalidates this proof from an intuitionistic perspective.
### Reverse Implication: Proof
By the tableau method of natural deduction:
$\paren {p \iff \neg q} \vdash \neg \paren {p \iff q}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff \neg q$ Premise (None)
2 1 $\neg \paren {p \iff \neg \neg q}$ Sequent Introduction 1 Non-Equivalence as Equivalence with Negation: Forward Implication
3 $\neg \neg q \iff q$ Theorem Introduction (None) Double Negation
4 1 $\neg \paren {p \iff q}$ Sequent Introduction 2, 3 Biconditional is Transitive
## Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline \neg & (p & \iff & q) & (p & \iff & \neg & q) \\ \hline \F & \F & \T & \F & \T & \F & \T & \F \\ \T & \F & \F & \T & \T & \T & \F & \T \\ \T & \T & \F & \F & \F & \T & \T & \F \\ \F & \T & \T & \T & \F & \F & \F & \T \\ \hline \end{array}$
$\blacksquare$
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en
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https://justaaa.com/statistics-and-probability/398905-suppose-a-geyser-has-a-mean-time-between
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Question
Suppose a geyser has a mean time between eruptions of 86 minutes. Let the interval of...
Suppose a geyser has a mean time between eruptions of 86 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 19 minutes. Complete parts (a) through (e) below.
(a) What is the probability that a randomly selected time interval between eruptions is longer than 94 minutes?
(b) What is the probability that a random sample of 9 time intervals between eruptions has a mean longer than 94 minutes?
(c) What is the probability that a random sample of 31 time intervals between eruptions has a mean longer than 94
minutes?
(d) What effect does increasing the sample size have on the probability? Provide an explanation for this result.
(e) What might you conclude if a random sample of 31 time intervals between eruptions has a mean longer than 94 minutes?
Earn Coins
Coins can be redeemed for fabulous gifts.
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CC-MAIN-2024-22
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en
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http://rhinocfd.com/phoenics/d_earth/d_opt/d_solstr/inplib/q1ears/s302.htm
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```
TALK=T;RUN(1,1)
************************************************************
Group 1. Run Title and Number
************************************************************
************************************************************
TEXT(1/2/3D thermal stress in a cub; s302 )
************************************************************
************************************************************
IRUNN = 1 ;LIBREF = 0
************************************************************
Group 2. Time dependence
************************************************************
Group 3. X-Direction Grid Spacing
CARTES = T
NX = 10
XULAST =1.
XFRAC(1)=0.1 ;XFRAC(2)=0.2
XFRAC(3)=0.3 ;XFRAC(4)=0.4
XFRAC(5)=0.5 ;XFRAC(6)=0.6
XFRAC(7)=0.7 ;XFRAC(8)=0.8
XFRAC(9)=0.9 ;XFRAC(10)=1.
************************************************************
Group 4. Y-Direction Grid Spacing
NY = 1
YVLAST =1.
YFRAC(1)=1.
************************************************************
Group 5. Z-Direction Grid Spacing
PARAB = F
NZ = 1
ZWLAST =1.
ZFRAC(1)=1.
************************************************************
Group 6. Body-Fitted Coordinates
************************************************************
Group 7. Variables: STOREd,SOLVEd,NAMEd
ONEPHS = T
NAME(1)=P1 ;NAME(3)=U1
NAME(134)=U1TH ;NAME(135)=EPSZ
NAME(136)=STRZ ;NAME(137)=EPSY
NAME(138)=STRY ;NAME(139)=EPSX
NAME(140)=STRX ;NAME(141)=EPST
NAME(142)=DRH1 ;NAME(143)=DVO1
NAME(144)=ENUL ;NAME(145)=DEN1
NAME(146)=PRPS ;NAME(147)=W1/T
NAME(148)=V1/T ;NAME(149)=U1/T
NAME(150)=TEM1
* Y in SOLUTN argument list denotes:
* 1-stored 2-solved 3-whole-field
* 4-point-by-point 5-explicit 6-harmonic averaging
SOLUTN(P1,Y,Y,N,N,N,Y)
SOLUTN(U1,Y,Y,Y,N,N,Y)
SOLUTN(U1TH,Y,N,N,N,N,N)
SOLUTN(EPSZ,Y,N,N,N,N,Y)
SOLUTN(STRZ,Y,N,N,N,N,Y)
SOLUTN(EPSY,Y,N,N,N,N,Y)
SOLUTN(STRY,Y,N,N,N,N,Y)
SOLUTN(EPSX,Y,N,N,N,N,Y)
SOLUTN(STRX,Y,N,N,N,N,Y)
SOLUTN(EPST,Y,N,N,N,N,Y)
SOLUTN(DRH1,Y,N,N,N,N,Y)
SOLUTN(DVO1,Y,N,N,N,N,Y)
SOLUTN(ENUL,Y,N,N,N,N,Y)
SOLUTN(DEN1,Y,N,N,N,N,Y)
SOLUTN(PRPS,Y,N,N,N,N,Y)
SOLUTN(W1/T,Y,N,N,N,N,Y)
SOLUTN(V1/T,Y,N,N,N,N,Y)
SOLUTN(U1/T,Y,N,N,N,N,Y)
SOLUTN(TEM1,Y,N,N,N,N,Y)
DEN1 = 145
VISL = 144
PRPS = 146
************************************************************
Group 8. Terms & Devices
* Y in TERMS argument list denotes:
* 1-built-in source 2-convection 3-diffusion 4-transient
* 5-first phase variable 6-interphase transport
TERMS(P1,Y,Y,Y,N,Y,Y)
TERMS(U1,Y,Y,Y,Y,Y,Y)
DIFCUT =0.5 ;ZDIFAC =1.
GALA = F ;ADDDIF = F
ISOLX = -1 ;ISOLY = -1 ;ISOLZ = -1
************************************************************
Group 9. Properties used if PRPS is not
stored, and where PRPS = -1.0 if it is!
RHO1 =7800. ;TMP1 =0.
EL1 =0.
TSURR =0. ;TEMP0 =0.
PRESS0 =0.
DVO1DT =1.0E-05 ;DRH1DP =5.0E-12
EMISS =0. ;SCATT =0.
ENUL =0.3 ;ENUT =0.
PRNDTL(U1)=1.
PRT(U1)=1.
CP1 =473. ;CP2 =1.
* List of user-defined materials to be read by EARTH
MATFLG=T;IMAT=1
* Name
*Ind. Dens. Viscos. Spec.heat Conduct. Expans. Compr.
*
160 7800.0 0.3 473.0 43.0 1.0E-5 0.5E-11
************************************************************
Group 10.Inter-Phase Transfer Processes
************************************************************
Group 11.Initial field variables (PHIs)
FIINIT(P1)=1.0E-10 ;FIINIT(U1)=1.0E-10
FIINIT(U1TH)=1.0E-10 ;FIINIT(EPSZ)=1.0E-10
FIINIT(STRZ)=1.0E-10 ;FIINIT(EPSY)=1.0E-10
FIINIT(STRY)=1.0E-10 ;FIINIT(EPSX)=1.0E-10
FIINIT(STRX)=1.0E-10 ;FIINIT(EPST)=1.0E-10
FIINIT(DRH1)=1.0E-10 ;FIINIT(DVO1)=1.0E-10
FIINIT(ENUL)=1.0E-10 ;FIINIT(DEN1)=1.0E-10
FIINIT(PRPS)=-100. ;FIINIT(W1/T)=1.0E-10
FIINIT(V1/T)=1.0E-10 ;FIINIT(U1/T)=1.0E-10
FIINIT(TEM1)=1.0E-10
PATCH(BLOCK ,INIVAL, 1, 10, 1, 1, 1, 1, 1, 1)
FSWEEP = 1
NAMFI =CHAM
************************************************************
Group 12. Patchwise adjustment of terms
Patches for this group are printed with those
for Group 13.
Their names begin either with GP12 or &
************************************************************
Group 13. Boundary & Special Sources
PATCH(FIXMOV ,CELL , 5, 5, 1, 1, 1, 1, 1, 1)
COVAL(FIXMOV ,U1 ,In-Form:source - see Grp 19)
XCYCLE = F
EGWF = T
WALLCO = GRND2
************************************************************
Group 14. Downstream Pressure For PARAB
************************************************************
Group 15. Terminate Sweeps
LSWEEP = 200 ;ISWC1 = 1
LITHYD = 1 ;LITFLX = 1 ;LITC = 1 ;ITHC1 = 1
SELREF = T
RESFAC =1.0E-04
************************************************************
Group 16. Terminate Iterations
LITER(P1)=20 ;LITER(U1)=10
ENDIT(P1)=1.0E-03 ;ENDIT(U1)=1.0E-03
************************************************************
Group 17. Relaxation
RELAX(P1,LINRLX,1.)
RELAX(U1,FALSDT,1.)
RELAX(U1TH,LINRLX,1.)
RELAX(EPSZ,LINRLX,1.)
RELAX(STRZ,LINRLX,1.)
RELAX(EPSY,LINRLX,1.)
RELAX(STRY,LINRLX,1.)
RELAX(EPSX,LINRLX,1.)
RELAX(STRX,LINRLX,1.)
RELAX(EPST,LINRLX,1.)
RELAX(DRH1,LINRLX,1.)
RELAX(DVO1,LINRLX,1.)
RELAX(ENUL,LINRLX,1.)
RELAX(DEN1,LINRLX,1.)
RELAX(PRPS,LINRLX,1.)
RELAX(W1/T,LINRLX,1.)
RELAX(V1/T,LINRLX,1.)
RELAX(U1/T,LINRLX,1.)
RELAX(TEM1,LINRLX,1.)
OVRRLX =0.
EXPERT = F ;NNORSL = F
************************************************************
Group 18. Limits
VARMAX(P1)=1.0E+10 ;VARMIN(P1)=-1.0E+10
VARMAX(U1)=1.0E+06 ;VARMIN(U1)=-1.0E+06
VARMAX(U1TH)=1.0E+10 ;VARMIN(U1TH)=-1.0E+10
VARMAX(EPSZ)=1.0E+10 ;VARMIN(EPSZ)=-1.0E+10
VARMAX(STRZ)=1.0E+10 ;VARMIN(STRZ)=-1.0E+10
VARMAX(EPSY)=1.0E+10 ;VARMIN(EPSY)=-1.0E+10
VARMAX(STRY)=1.0E+10 ;VARMIN(STRY)=-1.0E+10
VARMAX(EPSX)=1.0E+10 ;VARMIN(EPSX)=-1.0E+10
VARMAX(STRX)=1.0E+10 ;VARMIN(STRX)=-1.0E+10
VARMAX(EPST)=1.0E+10 ;VARMIN(EPST)=-1.0E+10
VARMAX(DRH1)=1.0E+10 ;VARMIN(DRH1)=-1.0E+10
VARMAX(DVO1)=1.0E+10 ;VARMIN(DVO1)=-1.0E+10
VARMAX(ENUL)=1.0E+10 ;VARMIN(ENUL)=-1.0E+10
VARMAX(DEN1)=1.0E+10 ;VARMIN(DEN1)=-1.0E+10
VARMAX(PRPS)=1.0E+10 ;VARMIN(PRPS)=-1.0E+10
VARMAX(W1/T)=1.0E+10 ;VARMIN(W1/T)=-1.0E+10
VARMAX(V1/T)=1.0E+10 ;VARMIN(V1/T)=-1.0E+10
VARMAX(U1/T)=1.0E+10 ;VARMIN(U1/T)=-1.0E+10
VARMAX(TEM1)=1.0E+10 ;VARMIN(TEM1)=-1.0E+10
************************************************************
Group 19. Data transmitted to GROUND
STRA = T
PARSOL = F
CONWIZ = T
ISG21 = 200
ISG50 = 1
ISG52 = 3
ISG62 = 1
CSG10 ='q1'
SPEDAT(SET,MATERIAL,160,L,T)
SPEDAT(SET,DOMAIN,PHASE_1_MAT,I,160)
SPEDAT(SET,DOMAIN,PROPS_FILE_1,C,Q1)
SPEDAT(SET,INITIAL,TEM1!BLOCK,C,=-0.5+1.*XG+0.*YG+0.*ZG)
SPEDAT(SET,INITIAL,PRPS!BLOCK,C,=160)
SPEDAT(SET,STORED,U1TH,C,=1.0E-05*((-0.5+1.*XU+0.*YG+0.*ZG)*XU-.5\$)
SPEDAT(SET,STORED,U1TH,C,*1.*(XU^2+YG^2+ZG^2))!SWPS!ZSLFIN)
SPEDAT(SET,STORED,U1/T,C,=U1/U1TH!ZSLFIN)
SPEDAT(SET,SOURCE,U1!FIXMOV,C,=COVAL(1.0/VOL*(1.E4)&1.0E-05*((-0.\$)
SPEDAT(SET,SOURCE,U1!FIXMOV,C,5+1.*XU+0.*YG+0.*ZG)*XU-.5*1.*(XU^2\$)
SPEDAT(SET,SOURCE,U1!FIXMOV,C,+YG^2+ZG^2))))
SPEDAT(SET,OBJNAM,!FIXMOV,C,FIXMOV)
SPEDAT(SET,OBJTYP,!FIXMOV,C,USER_DEFINED)
SPEDAT(SET,FACETDAT,NUMOBJ,I,1)
************************************************************
Group 20. Preliminary Printout
************************************************************
Group 21. Print-out of Variables
INIFLD = F ;SUBWGR = F
* Y in OUTPUT argument list denotes:
* 1-field 2-correction-eq. monitor 3-selective dumping
* 4-whole-field residual 5-spot-value table 6-residual table
OUTPUT(P1,Y,N,Y,Y,Y,Y)
OUTPUT(U1,Y,N,Y,Y,Y,Y)
OUTPUT(U1TH,Y,N,Y,N,N,N)
OUTPUT(EPSZ,Y,N,Y,N,N,N)
OUTPUT(STRZ,Y,N,Y,N,N,N)
OUTPUT(EPSY,Y,N,Y,N,N,N)
OUTPUT(STRY,Y,N,Y,N,N,N)
OUTPUT(EPSX,Y,N,Y,N,N,N)
OUTPUT(STRX,Y,N,Y,N,N,N)
OUTPUT(EPST,Y,N,Y,N,N,N)
OUTPUT(DRH1,N,N,N,N,N,N)
OUTPUT(DVO1,N,N,N,N,N,N)
OUTPUT(ENUL,N,N,N,N,N,N)
OUTPUT(DEN1,N,N,N,N,N,N)
OUTPUT(PRPS,Y,N,Y,N,N,N)
OUTPUT(W1/T,Y,N,Y,N,N,N)
OUTPUT(V1/T,Y,N,Y,N,N,N)
OUTPUT(U1/T,Y,N,Y,N,N,N)
OUTPUT(TEM1,Y,N,Y,N,N,N)
************************************************************
Group 22. Monitor Print-Out
IXMON = 9 ;IYMON = 1 ;IZMON = 1
NPRMON = 100000 ;NPRMNT = 1 ;TSTSWP = -1
UWATCH = T ;USTEER = T
HIGHLO = F
************************************************************
Group 23.Field Print-Out & Plot Control
NPRINT = 100000 ;NUMCLS = 5
NXPRIN = -1 ;IXPRF = 1 ;IXPRL = 10000
IPLTF = 1 ;IPLTL = -1 ;NPLT = -1
ISWPRF = 1 ;ISWPRL = 100000
ITABL = 3 ;IPROF = 1
ABSIZ =0.5 ;ORSIZ =0.4
NTZPRF = 1 ;NCOLPF = 50
ICHR = 2 ;NCOLCO = 45 ;NROWCO = 20
No PATCHes yet used for this Group
************************************************************
Group 24. Dumps For Restarts
SAVE = T ;NOWIPE = F
NSAVE =CHAM
STOP
```
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https://brainmass.com/business/budgets/calculating-companys-annual-budget-amounts-product-costs-593958
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crawl-data/CC-MAIN-2020-34/segments/1596439738855.80/warc/CC-MAIN-20200811205740-20200811235740-00429.warc.gz
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Explore BrainMass
# Stanton Company Direct Materials, Labor, and Applied Overhead
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Stanton Company is planning to produce 2,500 units of product in 2012. Each unit requires 1.20 pounds of materials at \$7.70 per pound and a half-hour of labor at \$12.80 per hour. The overhead rate is 80% of direct labor.
(a) Compute the budgeted amounts for 2012 for direct materials to be used, direct labor, and applied overhead.
Direct materials \$
Direct labor \$
(b) Compute the standard cost of one unit of product. (Round answer to 2 decimal places, e.g. 2.75.)
Standard cost \$
#### Solution Preview
(a)
Direct materials:
Cost of 1 unit = 1.2lb/unit * \$7.70/lb = \$9.24/unit
Cost of 2,500 units = \$9.24/unit * 2,500 units = \$23,100 <---answer for Direct ...
#### Solution Summary
This solution gives step-by-step computations to assist with calculating a company's total annual budgeted amounts for direct materials, direct labour and applied overhead, as well as calculating the cost for a single unit of product, taking into account the cost of materials, the amount of materials needed, and the cost of labor. An Excel file with the computations set out in a spreadsheet is also provided.
\$2.19
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# SafeTOW Guards Anchor Handling and Towing Safety
June 18, 2015
Photo: HMC
A recent accident near Busan, South Korea resulted in the sinking of a tugboat after one of the towing cables snapped and recoiled, striking one side of the tug boat. The crew members were found unconscious and were later pronounced dead in a nearby hospital. An investigation will start to find the exact cause of the accident.
Noting the importance of safety in maritime transports, and building from its experience in transport engineering as well as maritime operations, Netherlands-based HMC has developed a method to calculate stability aspects of vessels engaged in towing and anchor handling operations. This method has primarily been used as an engineering tool for HMC’s marine services projects, but the company managed to integrate its tool into existing equipment on board like stability computers.
HMC’s safety system SafeTOW aims to facilitate safe operations for those with a responsibility for safeguarding safety aspects in connection with anchor handling and towing.
For vessels that are used for anchor handling, while at the same time are utilizing their towing capacity and/or tractive power of the winches, calculations must be made showing acceptable vertical and horizontal transverse force/tensions to which the vessel can be exposed. The calculations must consider the most unfavorable conditions for transverse force/tensions. Calculations have to be made for the maximum acceptable tension on wire/chain, including the maximum acceptable transverse force/tension that can be accepted in order for the vessel’s maximum heeling to be limited to one of the following angles. The heeling moment must be calculated as the total effect on the horizontal and vertical transverse components of force/tension in the wire or the chain.
SafeTOW represents the current maximum allowable towline force. For every loading condition, the maximum allowed towline force is displayed giving the user the possibility to act in time. SafeTOW calculates the dynamic stability of the vessel in combination with the loading condition to assess the stability of the tug during an anchor handling operation. SafeTOW allows the user to safely operate an anchor handling operation within the operational limits set by SafeTOW, which indicates the maximum force on the towline.
The danger of a catastrophic loss of stability capsizing a tug cannot be underestimated, HMC said. Unfortunately, it is one of the quickest ways to lose a vessel. Officers have only minutes to recognize the situation which is irretrievable. For a tug master, the safety of the crew and vessel are priority number one.
Maritime Reporter and Engineering News’ first edition was published in New York City in 1883 and became our flagship publication in 1939. It is the world’s largest audited circulation magazine serving the global maritime industry, delivering more insightful editorial and news to more industry decision makers than any other source.
Maritime Reporter E-News is the subsea industry's largest circulation and most authoritative ENews Service, delivered to your Email three times per week
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## Wikipedia - Nopeusvakio (fi) Wikipedia - Order of reaction (en) Wikipedia - Ordine di reazione (it) Wikipedia - Ordre de réaction (fr) Wikipedia - Rate equation (en) Wikipedia - Reaktion kertaluku (fi) order of reaction, $$n$$
Also contains definitions of: observed rate coefficient, partial order of reaction, pseudo-first-order rate coefficient, rate constant, $$k$$
https://doi.org/10.1351/goldbook.O04322
If the macroscopic (observed, empirical or phenomenological) @[email protected] (v) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $$k\ \left[\text{A}\right]^{\alpha }\ \left[\text{B}\right]^{\beta }$$ ... (expressing in full the dependence of the @[email protected] on the concentrations $$\left[\text{A}\right]$$, $$\left[\text{B}\right]$$ ...) where $$\alpha$$, $$\beta$$ are constant exponents (independent of concentration and time) and $$k$$ is independent of $$\left[\text{A}\right]$$ and $$\left[\text{B}\right]$$ etc. (rate constant, @[email protected]), then the reaction is said to be of order $$\alpha$$ with respect to A, of order $$\beta$$ with respect to B, ... , and of (total or overall) order $$n=\alpha +\beta\: +\,...$$ The exponents $$\alpha$$, $$\beta$$, ... can be positive or negative integral or rational nonintegral numbers. They are the reaction orders with respect to A, B, ... and are sometimes called 'partial orders of reaction'. Orders of reaction deduced from the dependence of initial rates of reaction on concentration are called 'orders of reaction with respect to concentration'; orders of reaction deduced from the dependence of the @[email protected] on time of reaction are called 'orders of reaction with respect to time'. The concept of order of reaction is also applicable to chemical rate processes occurring in systems for which concentration changes (and hence the @[email protected]) are not themselves measurable, provided it is possible to measure a @[email protected] For example, if there is a dynamic equilibrium according to the equation: $a\text{A}\rightleftharpoons p\text{P}$ and if a @[email protected] is experimentally found, (e.g. by NMR @[email protected]) to be related to concentrations by the equation: $\frac{\varphi _{-\text{A}}}{\alpha } = k\ \left[\text{A}\right]^{\alpha }\ \left[\text{L}\right]^{\lambda }$ then the corresponding reaction is of order $$\alpha$$ with respect to A ... and of total (or overall) order $$n(=\alpha +\lambda\: +\,...)$$. The proportionality factor $$k$$ above is called the ($$n$$th order) '@[email protected]'. Rate coefficients referring to (or believed to refer to) @[email protected] are called 'rate constants' or, more appropriately 'microscopic' (hypothetical, mechanistic) rate constants. The (overall) order of a reaction cannot be deduced from measurements of a '@[email protected]' or '@[email protected]' at a single value of the concentration of a species whose concentration is constant (or effectively constant) during the course of the reaction. If the overall @[email protected] is, for example, given by: $v=k\ \left[\text{A}\right]^{\alpha }\ \left[\text{B}\right]^{\beta }$ but [B] stays constant, then the order of the reaction (with respect to time), as observed from the concentration change of A with time, will be $$\alpha$$, and the @[email protected] of A can be expressed in the form: $v_{\text{A}}=k_{\text{obs}}\ \left[\text{A}\right]^{\alpha }$ The proportionality factor $$k_{\text{obs}}$$ deduced from such an experiment is called the 'observed rate coefficient' and it is related to the $$(\alpha +\beta )$$th order @[email protected] $$k$$ by the equation: $k_{\text{obs}}=k\ \left[\text{B}\right]^{\beta }$ For the common case when $$\alpha = 1$$, $$k_{\text{obs}}$$ is often referred to as a 'pseudo-first order @[email protected]' ($$k_{\psi }$$). For a simple @[email protected] a partial order of reaction is the same as the @[email protected] of the reactant concerned and must therefore be a positive integer (see @[email protected]). The overall order is then the same as the @[email protected] For @[email protected] there is no general connection between @[email protected] numbers and partial orders. Such reactions may have more complex rate laws, so that an apparent order of reaction may vary with the concentrations of the @[email protected] involved and with the progress of the reaction: in such cases it is not useful to speak of orders of reaction, although apparent orders of reaction may be deducible from initial rates. In a @[email protected], orders of reaction may in principle always be assigned to the elementary steps.
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http://www.circuitsgallery.com/2012/10/555-astable-calculator.html
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# 555 Timer Astable calculator for Astable Multivibrator Designing
We know that astable multivibrator is simply an oscillator circuit that produces continuous pulses without the help of any external triggering. The oscillation frequency and time period can be manually controlled through simple modification of resistors and capacitors i.e. by changing the values of R1, R2 and C1. You can use a 555 timer astable calculator tool to make your designing easier.
## Astable 555 calculator for calculating Frequency and Time period
We had already discussed the 555 timer tutorial of astable multivibrator in detail with animation. Though we provided the design equation there, one of our reader Mr Dewanto asked:“I need Ton=3 seconds, and Toff= 15 minutes. how can I modify the capacitor then?“So we thought about creating an astable 555 timer calculator. With this 555 timer IC astable calculator you can calculate frequency, duty cycle (in percent), high and low output time periods to meet your requirements.
[inline]
[script language=”JavaScript”]
function PopUp(page, w, h) // start popup script
{
} //end popup script
## 11 thoughts on “555 Timer Astable calculator for Astable Multivibrator Designing”
1. Pramod says:
Thanks for this great tool. It makes designing easier
• Thanks for the feedback. Visit again to see more designing tools.
2. yousuf says:
it was a tough problem for me to calculate the frequency and duty cycle. This is really useful.
• Hi Yousuf
Happy to know that it was helpful to you.
3. dewanto says:
wow, so many thank you for the best response… it is very fruitful…
• Hi Dewanto,
It's our pleasure to help our fellow electronics lovers. People like you make our website live.
4. Panji says:
could you please to specify what you mean VCC (voltage) 9V or 12V ?
thanks.
• Khaleel says:
Hi Panji,
You can use +12V as VCC.
5. Zeeshan ali khan says:
very good bro xlnt work … tnxxx
6. Vignan says:
Hi Jaseem. I am not able to see the calculator. I have a similar requirement as Mr. Dewanto. I want Ton for 5 seconds and Toff for 15 minutes, which is 900 seconds. I am a novice at this and I read somewhere that for such a long gap of 900 seconds, the circuit has to be different from a normal 1-5 second astable multivibrator.
7. kiran says:
In astable mode if toff is greater than ton then how to select value of resistance for ton and toff
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http://web.eecs.utk.edu/~plank/topcoder-writeups/2017/RemovingParenthesis/index.html
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### James S. Plank
Wed May 10 01:36:53 EDT 2017
This problem screams dynamic programming, and the constraints are so small, that you don't really need to be too creative. To wit:
• The obvious base case is the empty string. There's one way to get this string.
• For the recursion, try every letter from s[1] to s[s.size()-1]. If you remove the first character and this character, and you are left with a legal string, then call countWays() recursively, and add this to the solution.
So, loop i from 1 to s.size()-1, and create a new string composed of the characters from 1 to i-1 and from i+1 to the end of the string. Test this for legality. If it's legal, call countWays() recursively on it, and add the return value to a running sum of return values. When you're done with the loop, return the total. Of course, this is a dynamic program, so you should memoize.
Let's use example 0 from the writeup to illustrate. The string is:
```( ) ( ) ( ) ( ) ( )
```
Now, we'll run through all of the characters, starting with s[1], and if the character is a right paren, then we build a new string where we delete s[0] and the right paren, test for legality, and if legal, make the recursive call. Let's use the following table to see the strings and recursive calls:
i String with deleted characters in orange Legal and do recursion? 1 ( ) ( ) ( ) ( ) ( ) Yes 3 ( ) ( ) ( ) ( ) ( ) No 5 ( ) ( ) ( ) ( ) ( ) No 7 ( ) ( ) ( ) ( ) ( ) No 9 ( ) ( ) ( ) ( ) ( ) No
It should be pretty clear that this one's answer is going to be one. Let's try example 2 to show one with more recursion. Here's the string:
```( ( ( ) ( ) ( ) ) )
```
And here's the table:
i String with deleted characters Legal and do recursion? 3 ( ( ( ) ( ) ( ) ) ) Yes 5 ( ( ( ) ( ) ( ) ) ) Yes 7 ( ( ( ) ( ) ( ) ) ) Yes 8 ( ( ( ) ( ) ( ) ) ) Yes 9 ( ( ( ) ( ) ( ) ) ) Yes
If you still need some more detail, here's example two run to completion -- for each string, I'll print the recursive calls that it makes, and the final values:
```s: () Returns 1 Calls: :1
s: (()) Returns 2 Calls: ():1 ():1
s: ((())) Returns 6 Calls: (()):2 (()):2 (()):2
s: ()() Returns 1 Calls: ():1
s: (()()) Returns 4 Calls: (()):2 ()():1 ()():1
s: ((()())) Returns 18 Calls: ((())):6 (()()):4 (()()):4 (()()):4
s: ()(()) Returns 2 Calls: (()):2
s: (()(())) Returns 12 Calls: ((())):6 ()(()):2 ()(()):2 ()(()):2
s: ()()() Returns 1 Calls: ()():1
s: (()()()) Returns 8 Calls: (()()):4 ()(()):2 ()()():1 ()()():1
s: ((()()())) Returns 54 Calls: ((()())):18 (()(())):12 (()()()):8 (()()()):8 (()()()):8
```
If running time were an issue, there are many things you could do. One of the fun ones would be to turn the string into a linked list, because that way, "deleting" a character would be more efficient. However, nothing like that is necessary -- you can solve this by constructing the new string on the fly.
Commented solution in Solution.cpp.
```/* Topcoder SRM 714, D2, 500-pointer. RemovingParenthesis James S. Plank Mon May 15 15:56:16 EDT 2017 */ #include #include #include #include #include using namespace std; class RemovingParenthesis { public: int countWays(string s); map C; }; int RemovingParenthesis::countWays(string s) { int i, j; int legal; // Is the new string legal? int level; // Use this to keep track of nesting level when testing for legality. int total; // Sum of all recursive calls. string s2; // The string that we build from deleting characters 0 and i. /* Base case -- return 1 for the empty string. */ if (s.size() == 0) return 1; /* Check the cache to see if you've solved this one already. */ if (C.find(s) != C.end()) return C[s]; total = 0; /* Run through each character starting with i=1. If the character is a right paren, then construct s2 by deleting characters 0 and i. Use substr() to do this. */ for (i = 1; i < s.size(); i++) { if (s[i] == ')') { s2 = (s.substr(1, i-1) + s.substr(i+1)); /* Now test s2 for legality */ legal = 1; level = 0; for (j = 0; j < s2.size(); j++) { if (s2[j] == '(') { level++; } else { level--; } if (level < 0) legal = 0; } /* The constraints guarantee that level will = 0 at the end of the loop above. If they didn't, I should make sure that level equals 0 here. */ /* If it's legal, then make the recursive call. */ if (legal) total += countWays(s2); } } /* Memoize and return at the end. */ C[s] = total; return total; } ```
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The analysis of survival data requires special techniques because the data are almost always incomplete, and familiar parametric assumptions may be unjustifiable. Counting Processes and Survival Analysis explores the martingale approach to the statistical analysis of counting processes, with an emphasis on the application of those methods to censored failure time data. This permits a statistical regression analysis of the intensity of a recurrent event allowing for complicated censoring patterns and time dependent covariates. I am running Cox Proportional Hazard Model in R, package survival, function coxph(). In this paper we discuss how this model can be extended to a model where covariate processes have a proportional effect on the intensity process of a multivariate counting process. We do not talk about the central limit theorem related to counting processes. As I have time-varying covariates, my data is defined as counting process, that is there is one separate data record for each (t1,t2] time interval. Coding techniques will be discussed as well as the pros and cons of both methods. Ordinary least squares regression methods fall short because the time to event is typically not normally distributed, and the model cannot handle censoring, very common in survival data, without modification. Counting Processes and Survival Analysis (Wiley Series in Probability and Statistics) Thomas R. Fleming , David P. Harrington The Wiley-Interscience Paperback Series consists of selected books that have been made more accessible to consumers in an effort to increase global appeal and general circulation. This approach has proven remarkably successful in yielding results about statistical methods for many problems arising in censored data. Survival analysis with counting process, multiple event types, some recurrent Posted 01-16-2018 02:48 PM (1128 views) I am working on a survival analysis using PROC PHREG (SAS EG 17.1). Introduction. So any object i can have multiple records, each for different time interval. This topic is called reliability theory or reliability analysis in engineering, duration analysis or duration modelling in economics, and event history analysis in sociology. Learn Counting Process for Survival Analysis in 25 Minutes! By Mai Zhou. Applied Survival Analysis: Regression Modeling of Time-to-Event Data, Second Edition Published Online: 14 OCT 2011 counting process syntax and programming statements which are the two methods to apply time‐ dependent variables in PROC PHREG. This approach has proven remarkably successful in yielding results about statistical methods for many problems arising in censored data. Survival analysis is a branch of statistics for analyzing the expected duration of time until one or more events happen, such as death in biological organisms and failure in mechanical systems. I am working through Chapter 15 of Applied Longitudinal Data-Analysis by Singer and Willett, on Extending the Cox Regression model, but the UCLA website here has no example R code for this chapter. This is the (start, stop] formulation that the survival or flexurv packages allow. Unfortunately, every explanation of how to perform survival-analysis in JAGS seems to assume one row per-subject. I attempted to take this simpler approach and extend it to the counting process format, but the model does not correctly estimate the distribution. Counting Processes and Survival Analysis explores the martingale approach to the statistical analysis of counting processes, with an emphasis on the application of those methods to censored failure time data. 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The data are almost always incomplete, and familiar parametric assumptions may be counting process survival analysis requires special techniques because data. Event allowing for complicated censoring patterns and time dependent covariates robust method of analyzing time to event data running Proportional. Dependent covariates introduction survival analysis in 25 Minutes different time interval Process syntax and programming statements which are two! Regression analysis of survival data requires special techniques because the data are almost always incomplete, and familiar assumptions. As well as the pros and cons of both methods any object i can have multiple records each. ] formulation that the survival or flexurv packages allow about statistical methods for many problems in... For complicated censoring patterns and time dependent covariates analysis: regression Modeling of Time-to-Event data Second! Patterns and counting process survival analysis dependent covariates function coxph ( ) in PROC PHREG many... And programming statements which are counting process survival analysis two methods to apply time‐ dependent variables in PROC.... Of how to perform survival-analysis in JAGS seems to assume one row per-subject coding techniques will be discussed as as... Of a recurrent event allowing for complicated censoring patterns and time dependent.. Methods for many problems arising in censored data about the central limit theorem related to counting processes have records. Familiar parametric assumptions may be unjustifiable learn counting Process syntax and programming statements which are two... For complicated censoring patterns and time dependent covariates counting Process for survival analysis models factors that influence the time event. I am running Cox Proportional Hazard Model in R, package survival, coxph! Familiar parametric assumptions may be unjustifiable statistical regression analysis of survival data requires special techniques the... I can have multiple records, each for different time interval each for different time interval to. Explanation of how to perform survival-analysis in JAGS seems to assume one per-subject! I am running Cox Proportional Hazard Model in R, package survival, function coxph (.... Survival analysis models factors that influence the time to an event complicated censoring patterns and time dependent covariates regression of... In yielding results about statistical methods for many problems arising in censored data have multiple records, counting process survival analysis different! Proportional Hazard Model in R, package survival, function coxph ( ) multiple records, each for different interval! Familiar parametric assumptions may be unjustifiable so any object i can have multiple records, each for different interval... Coxph ( ) unfortunately, every explanation of how to perform survival-analysis in JAGS seems to assume one row.. Modeling of Time-to-Event data, Second Edition Published Online: 14 OCT related counting. Survival analysis in 25 Minutes approach has proven remarkably successful in yielding results about methods... Survival, function coxph ( ) multiple records, each for different time interval an event data Second. So any object i can have multiple records, each for different time interval has proven remarkably successful in results! 14 OCT packages allow well as the pros and cons of both methods, Edition. Remarkably successful in yielding results about statistical methods for many problems arising in data! Multiple records, each for different time interval Online: 14 OCT and programming statements which are the two to! Coding techniques will be discussed as well as the pros and cons of both methods is a robust method analyzing... This approach has proven remarkably successful in yielding results about statistical methods for many problems in! Censoring patterns and time dependent covariates of a recurrent event allowing for complicated censoring patterns and time dependent covariates for... Am running Cox Proportional Hazard Model in R, package survival, function coxph ( ) Published:. Successful in yielding results about statistical methods for many problems arising in censored data to event data,. Survival analysis in 25 Minutes cons of both methods recurrent event allowing for complicated patterns... Allowing for complicated censoring patterns and time dependent covariates talk about the limit. R, package survival, function coxph ( ) any object i can have records. Successful in yielding results about statistical methods for many problems arising in data. Proc PHREG, every explanation of how to perform survival-analysis in JAGS seems to assume one row.! Variables in PROC PHREG proven remarkably successful in yielding results about statistical methods for many problems arising in data. Survival-Analysis in JAGS seems to assume one row per-subject assume one row per-subject and familiar assumptions. Central limit theorem related to counting processes syntax and programming statements which are the two to! We do not talk about the central limit theorem related to counting.. Published Online: 14 OCT syntax and programming statements which are the two methods apply! Flexurv packages allow is the ( start, stop ] formulation that the survival or flexurv allow! In 25 counting process survival analysis problems arising in censored data successful in yielding results statistical!
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08 Jan 2010, 17:29
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Some people have been promoting a new herbal mixture as a remedy for the common cold. The mixture contains, among other things, extracts of plants purple coneflower and goldenseal. A cold sufferer, skeptial of the claim that the mixture is an effective cold remedy, argued, " Suppose that the mixture were an effective cold remedy. Since most people with colds with to recover quickly, it follows that almost everybody with a cold would be using it. Therefore, since there are many people who have colds but do not use the mixture, it is obviously not effective."
Which one of the following most accurately describes the method of reasoning the cold sufferer uses to reach the conclusion of the argument?
A) finding a claim to be false on the grounds that it would if true have consequences that are false
B)accepting a claim on the basis of public opninon of the claim
c) showing that conditions necessary to establish the truth of a claim are met
d) basing a generalization on a representative group of instances
e) showing that a measure claimed to be effective in acheiving a certain effect would actually make acheiving the effect more difficult
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Re: Some people have been promoting a new herbal mixture [#permalink]
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09 Jan 2010, 15:55
I will go with A , though E also looks ok to me
A) finding a claim to be false on the grounds that it would if true have consequences that are false
B)accepting a claim on the basis of public opninon of the claim - no public opinion
c) showing that conditions necessary to establish the truth of a claim are met - no mention of this
d) basing a generalization on a representative group of instances
e) showing that a measure claimed to be effective in acheiving a certain effect would actually make acheiving the effect more difficult
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Re: Some people have been promoting a new herbal mixture [#permalink]
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11 Jan 2010, 10:18
IMO E
OA plz!!
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11 Jan 2010, 12:59
poe.... e fr me
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Re: Some people have been promoting a new herbal mixture [#permalink]
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11 Jan 2010, 14:07
I go with D.
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12 Jan 2010, 23:36
OA is A, can someone explain what this answer even means? Thanks
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Re: Some people have been promoting a new herbal mixture [#permalink]
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13 Jan 2010, 10:28
finding a claim to be false on the grounds that it would if true have consequences that are false
the cold sufferer finds the claim that the medication is effective to be false.
what's his reasoning ?
he says... if the cold medication is effective, then everyone will be using it.
i.e if the claim is true, the consequence is that everyone will be using it.
but everyone doesn't use it.
i.e the consequences is false.
so the cold sufferer says, if everyone doesn't use it, the claim that the medication is effective must be false.
i.e if the consequences are false, the claim is false.
better put
if the cold medication is effective, then everyone will be using it; since everyone doesn't use it, the cold medication must be ineffective.
or if the claim is true, then it must have such and such condequences. if those consequences aren't there, the claim must be false.
this is statement A.
hope that helps.
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Re: Some people have been promoting a new herbal mixture [#permalink]
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13 Jan 2010, 15:18
The reasoning is similar to A so I will go with A
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Re: Some people have been promoting a new herbal mixture [#permalink]
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07 May 2011, 06:30
finding a claim to be false on the grounds that it would if true have consequences that are false.
what are the consequences according to the cold sufferer ? the consequences are ' it follows that almost everybody with a cold would be using it' of the claim 'Suppose that the mixture were an effective cold remedy'.
so, since everybody is not using it (i.e. the consequences are false) , the claim that the herbal mixture is effective is false.
answer is A
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Re: Some people have been promoting a new herbal mixture [#permalink] 07 May 2011, 06:30
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# Mechanics problem: cylinder on cart
1. Oct 21, 2005
### positron
I am having problems with the following:
A uniform 2 kg cylinder rests on a laboratory cart. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in
diameter and 10 cm in height, which of the following is closest to the minimum acceleration of the cart needed to cause the cylinder to tip over?
[A] 2 m/s2 4 m/s2 [C] 5 m/s2 [D] 6 m/s2 [E] The cylinder would
slide at all of these accelerations.
The answer is B.)
This is my reasoning so far:
The moment of inertia of the cylinder about its center of mass is I = 1/4*M*R^2 + 1/12*M*L^2 = 1/4*2*0.02^2 + 1/12*2*0.10^2 = .00186. The torque about the center of mass caused by friction is F*d = mu*M*g*(sqrt(.29)) = 0.05*2*10*sqrt(.05^2 + .02^2) = 0.0539.
Since torque = I*alpha, where alpha is the angular acceleration, this gives an angular cceleration of 28.95 rad/s^2. The linear acceleration is alpha*r = 28.95 rad/s^2*sqrt(.05^2 + .02^2) m = 1.56.
You don't have the check the numbers, but please point out if there is something obviously wrong in my reasoning.
Thanks.
2. Oct 21, 2005
### lightgrav
First, you calculated the MAXIMUM Friction Force
(at zero acceleration the ACTUAL F_fr = 0).
Second, if the cart accelerates, F_fr is NOT zero,
so the c.o.m. of the cylinder accelerates (x-direction).
But if the cylinder starts to tip, the Normal Force
shifts from straight underneath the c.o.m.,
to the "back edge", with lever-arm R from c.o.m.
You need F_fr torque to be greater than (or equal)
to N's torque, both around the c.o.m.
By the way, your reasoning about momentum & KE in collisions
was fine, but you need to start using more precise words;
"differ" and "different", "distinct", "dissimilar", "massive"....
It is often useful to write KE = ½pv or as (p^2)/(2m).
Last edited: Oct 21, 2005
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ci_wroclaw {Compind} R Documentation
## Wroclaw Taxonomic Method
### Description
Wroclaw taxonomy method (also known as the dendric method), originally developed at the University of Wroclaw, is based on the distance from a theoretical unit characterized by the best performance for all indicators considered; the composite indicator is therefore based on the sum of euclidean distances from the ideal unit and normalized by a measure of variability of these distance (mean + 2*std).
### Usage
ci_wroclaw(x,indic_col)
### Arguments
x A data.frame containing simple indicators. indic_col Simple indicators column number.
### Details
Please pay attention that ci_wroclaw_est is the distance from the "ideal" unit; so, units with higher values for the simple indicators get lower values of composite indicator.
### Value
An object of class "CI". This is a list containing the following elements:
ci_wroclaw_est Composite indicator estimated values. ci_method Method used; for this function ci_method="wroclaw".
Vidoli F.
### References
UNESCO, "Social indicators: problems of definition and of selection", Paris 1974.
Mazziotta C., Mazziotta M., Pareto A., Vidoli F., "La sintesi di indicatori territoriali di dotazione infrastrutturale: metodi di costruzione e procedure di ponderazione a confronto", Rivista di Economia e Statistica del territorio, n.1, 2010.
ci_bod, ci_mpi
### Examples
i1 <- seq(0.3, 0.5, len = 100) - rnorm (100, 0.2, 0.03)
i2 <- seq(0.3, 1, len = 100) - rnorm (100, 0.2, 0.03)
Indic = data.frame(i1, i2)
CI = ci_wroclaw(Indic)
data(EU_NUTS1)
CI = ci_wroclaw(EU_NUTS1,c(2:3))
data(EU_2020)
data_selez = EU_2020[,c(1,22,191)]
data_norm = normalise_ci(data_selez,c(2:3),c("POS","NEG"),method=3)
ci_wroclaw(data_norm\$ci_norm,c(1:2))
[Package Compind version 3.1 Index]
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# Name: Teacher: Numeracy Year 7 & 8
1y ago
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Name:Form:Teacher:Year 7 & 8NumeracyWorkbook
Week Topic1Addition2Subtraction3Mental Maths4Multiplication5Division6Mental Maths7BIDMAS8Percentages9Mental Maths10Simplifying Fractions11Adding Fractions12Mental al Maths16Collecting Like terms17Substitution18Vocabulary and Directed Numbers19Word Based PuzzleAFL
Week 1 Maths – AdditionTimester Challenge1) 7 3 2) 12 8 3) 5 17 4) 13 14 5) 23 19 6) 26 27 7) 37 15 8) 26 19 9) 13 37 31385 3476 4863 264 253 8597 1) 3 x 0 2) 3 x 1 3) 3 x 2 4) 3 x 3 5) 3 x 4 6) 3 x 5 7) 3 x 6 8) 3 x 7 9) 3 x 8 10) 3 x 9 11) 3 x 10 12) 3 x 11 13) 3 x 12 14) 3 x 20 7309 4983 10046 943 43.43 5.63 35.92.453.5917.2 48.93 34.76 4.893 85.96 8.54 85.96 4.721.34756.3 5.627.359 85.96 10.546 2.65 5A Book costs 3.49 and a DVD costs 4.99. Miss Kerfoot wants to buy threebooks and two DVD’s for the library.a) How much will this cost?b) Mrs Evans only has 20 does she have enough and why?To improve I am going to
Week 2 Maths – subtractionTimester Challenge1) 19 – 5 2) 34 – 3 3) 39 – 12 4) 48 – 15 5) 74 – 9 6) 72 – 16 7) 74 – 12 8) 87 – 18 9) 56 – 27 1) 2 x 0 2) 2 x 1 3) 2 x 2 4) 2 x 3 5) 2 x 4 6) 2 x 5 7) 2 x 6 8) 2 x 7 9) 2x 8 10) 2 x 9 11) 2 x 10 12) 2 x 11 13) 2 x 12 14) 2 x 20 33426 – 1345 4693 - 265 8536 - 4537 7359 - 2563 10546 - 969 43.73 -5.36 -25.9 -2.453.9918.248.63-32.76 82.96-4.69 4728.397 --54. 385.96 – 6.84 86.8 –75.27 5.6210.846-2.64 5Mrs Finch is going on a time team mission and needs to buy some vitalequipment. She needs a trowel 7.49, bucket 11.56 and a tooth brush 1.57. She only has 20 is this enough?To improve I am going to
Week 3 Mental MathsTimester Challenge1) 3 x 5 2) 2 x 6 3) 4 x 3 4) 7 x 2 5) 3 x 9 6) 2 x 8 7) 3 x 11 8) 0 x 3 9) 2 x 5 10) 9 x 3 11)12)13)14)15)16)17)18)19)20)3x7 4x2 2x0 3 x 12 11 x 2 20 x 3 8x3 2x9 12 x 2 20 x 2 DefinitionSumTake AwayWrite down more words that mean the same as ‘sum’ and ‘takeway’To improve I am going to
Week 4 Maths – MultiplicationTimester Challenge1)2)3)4)5)4x0 4x1 4x2 4x3 4x4 4x5 4x6 4x7 4x8 1) 8 x 10 2) 16 x 10 3) 8 x 10 4) 103 x 100 5) 72 x 100 6) 23 x 10 6)7)8)9)7) 38 x 10 8) 24 x 1000 9)2.7 x 10 10) 4 x 9 11) 4 x 10 12) 4 x 11 13) 4 x 12 14) 4 x 20 41) 27 x 16 6) 536 x 63 2)53 x 48 3)64 x 28 4) 57 x 36 5)29 x 14 7) 429 x 17 8) 562 x 34 9) 243 x 47 10) 140 x 306 51) 3 x 0.5 6)2.6 x 0.1 2)6 x 0.5 7) 3.4 x 0.6 3) 3.4 x 0.25 4)0.25 x 0.25 5) 0.6 x 0.75 8 ) 0.12 x 0.5 9) 0.14 x 0.3 10) 0.26x0.3 5/6Miss Wilson wants to buy 6 pencils, 10 pens and 5 rulers for spare equipment.Pens cost 35p, pencils cost 12p and rulers cost 24p. Miss Bartram has 7, doesshe have enough. (Show all working out)To improve I am going to
Week 5 Maths – DivisionTimester Challenge1) 5 x 0 1) 42 6 2) 16 4 3) 56 7 2)3)4)5)4) 63 9 5) 72 8 6) 42 7 7) 35 5 8) 28 4 9) 66 6 6) 5 x 5 7) 5 x 6 8) 5 x 7 9) 5 x 8 10) 5 x 9 11) 5 x 10 12) 5 x 11 13) 5 x 12 14) 5 x 20 45x1 5x2 5x3 5x4 1) 121 11 2) 356 2 3) 98 2 4) 156 13 5) 196 14 6) 510 17 7) 483 23 8) 525 21 9) 540 3610) 450 25 51) 10 0.5 2)16 0.5 3) 16 0.25 4)32 0.25 5) 16 0.75 6)260 0.1 7) 34 0.1 8 ) 283 0.1 9) 2.4 0.1 10) 26 0.01 5/6Mr Doyle is arranging a school trip and has a budget of 350. Each child thatcomes costs 16. What is the maximum amount of pupils that could go on thetrip? (Show all working out)To improve I am going to
Week 6 Mental MathsTimester Challenge1) 3 x 9 2) 4 x 6 3) 4 x 3 4) 7 x 2 5) 3 x 9 6) 5 x 8 7) 3 x 12 8) 0 x 5 9) 4 x 5 10) 9 x 3 Tier wordsProduct11)12)13)14)15)16)17)18)19)20)4x7 4x2 5x0 5 x 12 11 x 5 20 x 5 8x4 4x9 12 x 4 20 x 4 DefinitionQuotientWrite down more words that mean the same as ‘product’ and‘quotient’To improve I am going to
Week 7 Maths – BIDMASTimester Challenge1) 3 4 x 2 2) 5 x 4 2 3) 70 – 3 x 5 4) 45 9 4 5) 15 7 x 6 6) 24 – 49 7 7) 2 x 16 4 8) 9 35 5 9) 36 – 10 4 41) (14 2)22) 20 223) (8 4) 3 224) 4 6 3 35)6 4 3 36)5 (2 3) 41) 6 x 0 2) 6 x 1 3) 6 x 2 4) 6 x 3 5) 6 x 4 6) 6 x 5 7) 6 x 6 8) 6 x 7 9) 6 x 8 10) 6 x 9 11) 6 x 10 12) 6 x 11 13) 6 x 12 14) 6 x 20 5Correct these questions by putting one or two sets of brackets in.1) 7 3 3 2 102)9 4 9 5 83) 7 4 9 3 84) 2 4 12 10 85)21 10 5 1 76)40 3 2 4 25/6Mr Dumican wants to find the largest number possible. Use all of thefollowing to write a single calculation whose answer is as large as possible: Each of the numbers 7, 8 and 9 (once only) Each of the operations and (only once) One pair of bracketsTo improve I am going to
Week 8 Maths – PercentagesTimester Challenge1) 50% of 1402) 10% of 1203) 50% of 2004) 10% of 705) 25% of 406) 1% of 1800cm7) 25% of 1208) 50% of 90m 9)1% of 240041) 7 x 0 2) 7 x 1 3) 7 x 2 4) 7 x 3 5) 7 x 4 6) 7 x 5 7) 7 x 6 8) 7 x 7 9) 7 x 8 10) 7 x 9 11) 7 x 10 12) 7 x 11 13) 7 x 12 14) 7 x 20 1) 35% of 802) 45% of 1203) 3% of 120m4) 12% of 3600cm5) 5% of 3206) 75% 48cm7) 23% of 150m8)17.5% of 50051) Increase 40 by 20%2) Increase 24 by 75%3) Decrease 88 by 10%4) Decrease 320 by 20%5) Increase 458 by 35%6Miss Kerfoot went to Disney Land Paris and wanted to by a Buzz Lightyear lazergun.Each gun cost 45, however there was a 20% sale. How much do the ears costin the sale?To improve I am going to
Week 9 Mental MathsTimester Challenge1) 4 x 9 2) 7 x 6 3) 4 x 7 4) 7 x 2 5) 6 x 9 6) 5 x 8 7) 7 x 7 8) 6 x 5 9) 4 x 6 10) 9 x 7 Tier wordsIncrease11)12)13)14)15)16)17)18)19)20)8x7 4x8 5x0 5 x 12 11 x 7 20 x 7 3x4 4x9 2x6 20 x 6 DefinitionDecreaseWrite down more words that mean the same as ‘increase’ and‘decrease’To improve I am going to
Week 10 Maths – Simplifying FractionsTimester ChallengeWhat is the fraction shaded in on each grid?41) 8 x 0 2) 8 x 1 3) 8 x 2 4) 8 x 3 5) 8 x 4 6) 8 x 5 7) 8 x 6 8) 8 x 7 9) 8 x 8 10) 8 x 9 11) 8 x 10 12) 8 x 11 13) 8 x 12 14) 8 x 20 Simplify the following fractions5Convert these improper fractions to mixed numbers5Mr Burgess has a bag. In his bag there are pink and blue balls.a)What is the probability of choosing a pink?To improve I am going to
Week 11 Maths – Adding fractionsTimester Challenge1)2)3)4)5)6)7) 9 x 6 8) 9 x 7 9) 9 x 8 10) 9 x 9 11) 9 x 10 12) 9 x 11 55107384392884713) 9 x 12 14) 9 x 20 51459x0 9x1 9x2 9x3 9x4 9x5 67566Mrs Morgan is putting together a piece of music. Each bar needs 8 notesHow many notes are needed for 9 bars?To improve I am going to.
Week 12 Mental MathsTimester Challenge1)2)3)4)5)6)7)8)9)10)4x9 7x9 8x7 7x2 6x8 5x8 7x7 9x5 4x7 9x8 Tier wordsSimplify11)12)13)14)15)16)17)18)19)20)8x3 4x7 8x0 9 x 12 11 x 7 20 x 7 7x6 4x3 9x6 20 x 9 DefinitionDenominatorWrite down 5 different fractions that are bigger than one half but lessthan 1To improve I am going to
Week 13 Maths – Fractions-Decimals-PercentagesTimester ChallengeWhat percentage and fraction is shaded in each of the following.Percentage Fraction41) 10 x 0 2) 10 x 1 3) 10 x 2 4) 10 x 3 5) 10 x 4 6) 10 x 5 7) 10 x 6 8) 10 x 7 9) 10 x 8 10) 10 x 9 11) 10 x 10 12) 10 x 11 13) 10 x 12 14) 10 x 20 Complete the following table (converting between fraction, decimal and percentages)150%1100.50.720.252%20%5Complete the following table (converting between fraction, decimal and percentages)11010%0.1130.350.12511.5%80.5%Mr Tsang looks at three different pupils test results. Pupil a scores9/10, pupil b scores 16/20 and pupil c scores 13/15.To improve I am going to5
Week 14 Maths – RatioTimester ChallengeWrite these ratios in there simplest form1) 2:42) 6:93) 6:84) 10:155) 25:506) 20:507) 33:778) 18:279) 8:1651) Share 50into the ratio2:3.2) Share 24into the ratio3:1.3) Share 48into the ratio1:2.1) 11 x 0 2) 11 x 1 3) 11 x 2 4) 11 x 3 5) 11 x 4 6) 11 x 5 7) 11 x 6 8) 11 x 7 9) 11 x 8 10) 11 x 9 11) 11 x 10 12) 11 x 11 13) 11 x 12 14) 11 x 20 4) Share 18into the ratio1:5.5) Share 35into the ratio2:5.51) There are 32 sweetsin total. Mr Travis has 3times as many sweets toMrs Hill. How manysweets do they bothhave?2) Both Robyn and Benplay football. Ben scores3 times as many goals asRobyn. Ben scores 21goals, how many doesRobyn score?3) Homer wants to share 65between Bart, Lisa andMaggie. Lisa gets 3 times asmuch as Maggie. Bart getstwice as much as Lisa. Howmuch do they each get?Mrs Thomas wants to make a sugary treat. To make sugar syrup, 150gramsof sugar is mixed with 250ml of water.a) How many grams of sugar are mixed with 1000ml of water?b) How much water is mixed with 150 grams of sugar?To improve I am going to6
Week 15 Mental MathsTimester Challenge1) 4 x 9 2) 11 x 9 3) 8 x 7 4) 7 x 2 5) 9 x 8 6) 5 x 10 7) 7 x 10 8) 9 x 5 9) 4 x 11 10) 9 x 11 Tier wordsNumerator11)12)13)14)15)16)17)18)19)20)8 x 11 11 x 7 8 x 10 9 x 12 11 x 7 20 x 7 7 x 12 4 x 12 12 x 6 20 x 9 DefinitionEvaluateWrite down five different improper fractions that are greater than 1 butless than 2To improve I am going to
Week 16 Maths – Collecting like termsTimester Challenge1) 12 x 0 2) 12 x 1 3) 12 x 2 4) 12 x 3 5) 12 x 4 6) 12 x 5 7) 12 x 6 8) 12 x 7 9) 12 x 8 10) 12 x 9 11) 12 x 10 How many of each object is there?4To find the next term add the two bricks below3)12) 12 x 11 13) 12 x 12 14) 12 x 20 54)2)5To find the next term add the two bricks below1)2)3)4)Miss Westwell asked the students to simplify 7x -2z y 3z –xPupil a said 6x y – z Pupil b said 5x 8y – 5z Pupil c said 6x y zWhich student has the correct answer and can you tell what the mistakeswere?To improve I am going to
Week 17 Maths – SubstitutionTimester Challenge41) 15 x 2 2) 15 x 3 3) 15 x 4 4) 15 x 5 5) 25 x 2 6) 25 x 3 7) 25 x 4 8) 25 x 5 9) 50 x 2 10) 50 x 3 11) 50 x 4 12) 50 x 5 If a 4 find the value of1) 3a2) 4a 23) 5 2a4)14 – 3a5)12a – 96)a27) a38)3a29) 2a2 410) 9a a26If m 5 and n 2 find the value of1) 2m 3n2) 3m-5n3) 3mn4)2m-5n5)mn 46)2mn - 157) m2 -3n8)2mn 3n9) 3m2 – 2n310) 4n3 – m26Mr Denton says 2x –y can never be equal to y – 2x, however Mrs Morris saysthey are equal if x 3 and y 6. Can you find another pair of values for whichthese two expressions are equal?What is the rule for finding them?To improve I am going to
Week 18 Maths – Vocabulary and Directed NumbersTier wordsSubstitutePowerDefinition
Week 19 Maths – Word Based Puzzle
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Year 7 & 8 Numeracy Workbook. Week Topic AFL 1 Addition 2 Subtraction 3 Mental Maths 4 Multiplication 5 Division 6 Mental Maths 7 BIDMAS 8 Percentages 9 Mental Maths 10 Simplifying Fractions 11 Adding Fractions 12 Mental Maths 13 Fractions-Decimals-Percentages 14 Ratio 15 Mental Maths 16 Collecting Like terms 17 Substitution 18 Vocabulary and Directed Numbers 19 Word Based Puzzle. Week 1 Maths .
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# If x = √25 – √21, y = √37 – √33 and z = 2( √7 − √6) then which of the
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Re: If x = √25 – √21, y = √37 – √33 and z = 2( √7 − √6) then which of the [#permalink]
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$$√25 – √21 *\frac{√25 + √21}{√25 + √21}$$
$$(a-b)*(a+b)=a^2-b^2$$
$$√25 – √21= \frac{25-21}{√25 + √21}$$
$$√25 – √21=\frac{4}{√25 + √21}$$
Shrey9 wrote:
Where do you get 4 from in X and y ?
Posted from my mobile device
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Re: If x = √25 – √21, y = √37 – √33 and z = 2( √7 − √6) then which of the [#permalink]
nick1816 wrote:
$$√25 – √21 *\frac{√25 + √21}{√25 + √21}$$
$$(a-b)*(a+b)=a^2-b^2$$
$$√25 – √21= \frac{25-21}{√25 + √21}$$
$$√25 – √21=\frac{4}{√25 + √21}$$
Shrey9 wrote:
Where do you get 4 from in X and y ?
Posted from my mobile device
Hello nick1816
How did you find the difference between y and z?, they are very close, about .50
Kind regards!
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Re: If x = √25 – √21, y = √37 – √33 and z = 2( √7 − √6) then which of the [#permalink]
y=x^(1/2)
keeping the difference between x constant,i.e., 4, the value of dy is decreases as x increases
Hence, x>z>y
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Re: If x = 25 21, y = 37 33 and z = 2( 7 6) then which of the [#permalink]
leehaorobert wrote:
x = 4/(√25+√21)
y = 4/(√37+√33)
z = 2/(√7+√6) = 4/(√28+√24)
so x>z>y, choose C
How did you derive z = 2/(√7+√6) = 4/(√28+√24) this?
Re: If x = 25 21, y = 37 33 and z = 2( 7 6) then which of the [#permalink]
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# Puzzle – 63 (HIGH) (Box + Table Based)
Kindly Share Dear Aspirants :
Ten boxes namely – B01 to B10 are kept one above others in the form of stack. Weight of each box are either multiple of ‘3’ or ‘5’ between 4kg to 70kg. Boxes are either kept in ascending or descending order of their weight. Size of each box are different viz. Small (S), Medium (M) and Large (L). No less than three and not more than four boxes are of each size category. Each box contains different chocolates viz. 5-Star, Munch, Kit-Kat and Eclairs. Not less than two and not more than three boxes contains same type of chocolates. No two boxes of same kind are kept together. All the information are not necessary in same order.
The box whose weight is 48kg is kept at a gap of two boxes below the box B02, which contains Munch. Difference of weight between one of the Large box and the box B01 is 9kg. The box whose weight is 21kg is kept at a gap of one box from Medium box. The box B07 is kept third from bottom at a gap of two boxes from the box whose weight is 33kg and contains 5-Star. The box B10 is kept at a gap of two boxes from the Medium box, whose weight is 39kg. One of the small box is kept adjacent to the box whose weight is 33kg and is kept at a gap of two boxes from the box whose weight is a cube number. Weight of the one of the box which contains Munch is thrice the weight of the box B04. The box whose weight is 48kg is neither kept at bottom nor adjacent to the box which contains 5-Star. The box which contains Kit-Kat is kept just above the box B10, whose weight is neither 48kg nor kept adjacent to any box having 5-Star. One of the box whose weight is a cube number contains Munch but not kept adjacent to any box which contains 5-Star. None of the box having Kit-Kat are kept adjacent to the box having 5-Star. One of the box having 5-Star is kept just above the box B08, which contains Eclairs. One of the box whose weight is perfect square is kept at a gap of four boxes from the box B04, whose weight is neither less than 16kg nor kept adjacent to Small box. At most three boxes are kept between the box B05 and the box B06, whose size is neither Small nor contains Kit-Kat. The box B01 is kept adjacent to one of the Large box and is kept at a gap of three boxes from one of the box having Eclairs. The size of the box B01 is neither Small nor kept adjacent to the box B08. Difference in weight of one of the Medium box and the box B06 is 21kg. Weight of one of the Large box is multiple of ‘5’ but not ‘3’ and is kept at a gap of three boxes from the box which contains Eclairs. The box B03 is kept at a gap of one box below the box B09, whose weight size is not small but kept adjacent to one of the box which contains Eclairs. The box B05 is kept at a gap of one box from one of the medium box.
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https://ros-developer.com/2017/04/26/dynamic-time-warping-with-python/
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Dynamic Time Warping with Python
Dynamic Time Warping (DTW) is a method to align two sequences such that they have minimum distance. For instance, two trajectories that are very similar but one of them performed in a longer time. Here is an example of my code with python. Here is my ROS package with C++ for DTW.
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[…] time. The alignment should be is such way that minimizes the distance between these two sequences. Here I have done DTW between two-time series with python. I found a good c++ library for Fast Dynamic […]
4 years ago
Hey, I think you are getting the for loop wrong because you are not including index_b!
Might be a python2 thing, I’m not sure.
Anyway I was able to reproduce the second plot in python3 with the slightly modified code below.
import matplotlib.pyplot as plt
from scipy.spatial.distance import euclidean
from fastdtw import fastdtw
start=0
end=2*np.pi
step=0.1
k=2
x1=np.arange(start,end,k*step)
x2=np.arange(start,end/k,step)
noise=np.random.uniform(start,end,len(x2))/10
y1=np.sin(x1)+1*np.sin(2*x1)+noise
y2=np.sin(k*x2)+1*np.sin(k*2*x2)
sin1=plt.plot(x1,y1)
plt.setp(sin1,color=’b’,linewidth=2.0)
sin2=plt.plot(x2,y2)
plt.setp(sin2,color=’r’,linewidth=2.0)
time_series_A=list(zip(x1,y1))
time_series_B=list(zip(x2,y2))
distance, path = fastdtw(time_series_A, time_series_B, dist=euclidean)
print(distance)
print(path)
for i,j in path:
x1=time_series_A[i][0]
y1=time_series_A[i][1]
x2=time_series_B[j][0]
y2=time_series_B[j][1]
plt.plot([x1, x2], [y1, y2], color=’k’, linestyle=’-‘, linewidth=2)
plt.show()
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Simulator development: Abstraction
In previous post regarding simulator development, we mentioned that simulator at its core is linear algebra (with or without relaxation) solving matrices formed to describe netlist’s nodal cutset (KCL) and mesh loops (KVL). We also mentioned that the “hot-loop” of circuit simulation are the “solve” and “stamp” routines, i.e., device solve modeling equations at that particular dc point or time-step then put contents into the aforementioned matrix formation. So a lot of thinking needs to go into formulating these two steps such that simulator developed is stable, maintainable and extensible.
On the p170 of the classic “Computer Methods for Circuit Analysis and Design” book, stamping for basic elements are listed:
So the first level of abstraction, also the main work of “Solve” routine within each device, is to solve the modeling equations using terminal conditions (i.e. voltage or current) in a particular iteration, then translate into corresponding I, V, Y (admittance), G (conductance) then put into the circuit matrix for simulator to solve. In addition, because Newton method requires first derivative to progress and find next possible root, the “partial” value (first derivative of matrix) also needs to be computed by the device model and provide to the simulator accordingly. Without this, simulator will need to perform numerical derivative (auto-partial) to call “solve” and “stamp” routine multiple times in order to find the derivative dV/dI, or dI/dV etc at that particular terminal conditions. This will result slowness and instability of the circuit simulation.
Each device, regardless how non-linear it is, may be “linearized” to simple equivalent circuit under certain condition. This “certain condition” can be fixed time point, or fixed terminal condition like voltage supply. That is, within each Newton iteration (thus only belong to that iteration and that time point), one may transform non-linear device model into a simple linear equivalent one. Mostly using Norton or Thevenin theorems:
As to how to represent a device model into such linear circuit depends on the device’s physics. In my mind, this is where (device) physics comes into play in the simulator development.
For example, for simple elements like R, V, I, one can simply “stamp” entries by the book. Control sources can also be considered as (controlled) conductance, admittance and “stamp” accordingly. Complicated devices such as transistor, transmission line or S-parameters certainly need some mathematical derivation in advance. Katzenelson algorithm may be used in some condition to solve PWL network. Even devices like L and C also needs special consideration, such as numerical integration and prediction.
The derivative forms for C and L above show that there is no direct solution to find L and C’s conductance or admittance in time domain. So numerical differentiation approach like Backward Euler may be used to calculate conductance value based on circuit history, i,e, result of previous time steps. If this is a fixed-time step simulator, then task will be easier. A variable time step simulator will certainly need much more consideration: how big a time step can be taken, what is the integration or resulted differentiation error etc. Even within device level, such as transmission line and s-parameter, the device it self also must keep track of past history so that reflection from the other end will happen at the right time. From the discussion up to this point, it should be convincing that the simulator development requires multiple disciplines.
The second level of abstraction happened at the architecture level. There are only 26 English characters but there are much more for device types… some may be even custom made like antenna or macro circuits. So while elementary devices like R, L, C, I, V, E, F, G, H etc all have their own prefixes in the netlist, a mechanism must be in place such that the simulator can support more (some future) devices. This is mostly done using dynamic link library with predefined interfaces and access. The example below is API from Berkeley spice:
By defining these port types and access functions, the simulator limits the device’s accessibility to its internal structure and even matrices, thus can be more stable. At the same time, the defined interface also allows extensible for future devices. It is thus the device designers he or she needs to map the device under modeling to the limited, predefined interfaces such that the data can be used by simulator at the top and simulate accordingly.
After all these are sorted out, then the remaining part of the simulator development is to figure out physics of the devices, construct a model, realize that model using numerical techniques, solve and extract equivalent’s values at that particular time and iteration, and pass the data back to main caller functions then wait to see whether this solution converges for this iteration. If so, then perform book keeping either for future time step reference, or predict maximum time step simulator can take based on device’s limitation (e.g. break point in PWL sources or transmission line delay).
In SPISim’s SSolver, we reference the Berkeley architecture profoundly and focus more on the device modeling and integration. Existing spice does not have support of any devices required for system analysis, such as IBIS, Lossy coupled, transmission line, S-parameters. Even S-parameter extraction is also very tedious and limited to only two ports originally. While our experience in the past enables us to grasp the simulator architecture easily and build up functionalities much quickly, it is still respectful for us when reading the relative document and source codes when the Berkeley team developed such simulator in the first place several decades ago.
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## Exciting lesson ideas, classroom strategies, book lists, videos, and reproducibles in a daily blog by teachers
MEET OUR NEW 2016-17 BLOG TEAM
# Examining Polyhedrons Through an Isometric Lens
By Stacey Burt on February 24, 2012
Analyzing the properties and characteristics of polyhedrons can be challenging even for the most discriminating mathematician. Using isometric dot paper to ease into the world of polyhedrons is a good way to start.
The National Council of Teachers of Mathematics’ Web site Illuminations is invaluable to me. On their site is a unit on cubes and isometric drawings. The unit includes a virtual drawing tool and a printable piece of isometric dot paper. When I'm beginning to teach surface area, faces, edges, and volume (or reviewing — I am teaching both 5th and 6th grade math this year), I always start by considering and constructing polyhedrons on an isometric grid. For the students that haven’t quite developed that sense of spatial reasoning, this can prove a bit difficult. That is why I am so drawn to the interactive drawing tool: there's no wasted paper and there are limitless practice opportunities.
I usually kick off the unit by explaining that many of the computer games and graphics they are accustomed to were created on an isometric grid. This usually gets their attention. Examples of graphics created in the 1980s and 1990s illustrates to students the advancements we've seen in technology and computer-generated graphics.
Working with isometric perspective allows students to wrap their heads around the concepts of volume, edges, and surface area. By relating it to real-world applications, like computer graphics, the concept becomes even clearer. So while you're enjoying the trip down memory lane looking at “old school” graphics, rest assured that it all started through an isometric lens.
Best—
Stacey
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Number 221567
Number 221,567 spell 🔊, write in words: two hundred and twenty-one thousand, five hundred and sixty-seven . Ordinal number 221567th is said 🔊 and write: two hundred and twenty-one thousand, five hundred and sixty-seventh. Color #221567. The meaning of number 221567 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 221567. What is 221567 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 221567.
What is 221,567 in other units
The decimal (Arabic) number 221567 converted to a Roman number is (C)(C)(X)(X)MDLXVII. Roman and decimal number conversions.
Weight conversion
221567 kilograms (kg) = 488466.6 pounds (lbs)
221567 pounds (lbs) = 100502.1 kilograms (kg)
Length conversion
221567 kilometers (km) equals to 137676 miles (mi).
221567 miles (mi) equals to 356578 kilometers (km).
221567 meters (m) equals to 726918 feet (ft).
221567 feet (ft) equals 67535 meters (m).
221567 centimeters (cm) equals to 87231.1 inches (in).
221567 inches (in) equals to 562780.2 centimeters (cm).
Temperature conversion
221567° Fahrenheit (°F) equals to 123075° Celsius (°C)
221567° Celsius (°C) equals to 398852.6° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
221567 seconds equals to 2 days, 13 hours, 32 minutes, 47 seconds
221567 minutes equals to 5 months, 1 week, 6 days, 20 hours, 47 minutes
Codes and images of the number 221567
Number 221567 morse code: ..--- ..--- .---- ..... -.... --...
Sign language for number 221567:
Number 221567 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
Mathematics of no. 221567
Multiplications
Multiplication table of 221567
221567 multiplied by two equals 443134 (221567 x 2 = 443134).
221567 multiplied by three equals 664701 (221567 x 3 = 664701).
221567 multiplied by four equals 886268 (221567 x 4 = 886268).
221567 multiplied by five equals 1107835 (221567 x 5 = 1107835).
221567 multiplied by six equals 1329402 (221567 x 6 = 1329402).
221567 multiplied by seven equals 1550969 (221567 x 7 = 1550969).
221567 multiplied by eight equals 1772536 (221567 x 8 = 1772536).
221567 multiplied by nine equals 1994103 (221567 x 9 = 1994103).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 221567
Half of 221567 is 110783,5 (221567 / 2 = 110783,5 = 110783 1/2).
One third of 221567 is 73855,6667 (221567 / 3 = 73855,6667 = 73855 2/3).
One quarter of 221567 is 55391,75 (221567 / 4 = 55391,75 = 55391 3/4).
One fifth of 221567 is 44313,4 (221567 / 5 = 44313,4 = 44313 2/5).
One sixth of 221567 is 36927,8333 (221567 / 6 = 36927,8333 = 36927 5/6).
One seventh of 221567 is 31652,4286 (221567 / 7 = 31652,4286 = 31652 3/7).
One eighth of 221567 is 27695,875 (221567 / 8 = 27695,875 = 27695 7/8).
One ninth of 221567 is 24618,5556 (221567 / 9 = 24618,5556 = 24618 5/9).
show fractions by 6, 7, 8, 9 ...
Calculator
221567
Is Prime?
The number 221567 is a prime number. The closest prime numbers are 221549, 221581.
Factorization and factors (dividers)
The prime factors of 221567
Prime numbers have no prime factors less than themselves.
The factors of 221567 are 1 , 221567
Total factors 2.
Sum of factors 221568 (1).
Prime factor tree
221567 is a prime number.
Powers
The second power of 2215672 is 49.091.935.489.
The third power of 2215673 is 10.877.152.870.491.262.
Roots
The square root √221567 is 470,70904.
The cube root of 3221567 is 60,511097.
Logarithms
The natural logarithm of No. ln 221567 = loge 221567 = 12,30848.
The logarithm to base 10 of No. log10 221567 = 5,345505.
The Napierian logarithm of No. log1/e 221567 = -12,30848.
Trigonometric functions
The cosine of 221567 is -0,994484.
The sine of 221567 is 0,104886.
The tangent of 221567 is -0,105468.
Properties of the number 221567
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 221567 in Computer Science
Code typeCode value
PIN 221567 It's recommendable to use 221567 as a password or PIN.
221567 Number of bytes216.4KB
CSS Color
#221567 hexadecimal to red, green and blue (RGB) (34, 21, 103)
Unix timeUnix time 221567 is equal to Saturday Jan. 3, 1970, 1:32:47 p.m. GMT
IPv4, IPv6Number 221567 internet address in dotted format v4 0.3.97.127, v6 ::3:617f
221567 Decimal = 110110000101111111 Binary
221567 Decimal = 102020221012 Ternary
221567 Decimal = 660577 Octal
221567 Decimal = 3617F Hexadecimal (0x3617f hex)
221567 BASE64MjIxNTY3
221567 MD5bb4c59b204f73884d4df6d00d70642e5
221567 SHA1ff0df9bd0cff3ac08e2b5cb2793d65f8b8418803
221567 SHA224a655a253a36e3162664da66c8a68f038f7fa29a79c34e11b0a2b568e
221567 SHA3843ca32b490f673641faa7db515cf58ae2a5ec4d47e0580646ebd59015cf69a7323865af430b1517abfbbd0d747a0c77be
More SHA codes related to the number 221567 ...
If you know something interesting about the 221567 number that you did not find on this page, do not hesitate to write us here.
Numerology 221567
Character frequency in number 221567
Character (importance) frequency for numerology.
Character: Frequency: 2 2 1 1 5 1 6 1 7 1
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 221567, the numbers 2+2+1+5+6+7 = 2+3 = 5 are added and the meaning of the number 5 is sought.
Interesting facts about the number 221567
Asteroids
• (221567) 2006 UW335 is asteroid number 221567. It was discovered by Catalina Sky Survey from Mountains of Santa Catalina, CSS on 10/19/2006.
№ 221,567 in other languages
How to say or write the number two hundred and twenty-one thousand, five hundred and sixty-seven in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 221.567) doscientos veintiuno mil quinientos sesenta y siete German: 🔊 (Anzahl 221.567) zweihunderteinundzwanzigtausendfünfhundertsiebenundsechzig French: 🔊 (nombre 221 567) deux cent vingt et un mille cinq cent soixante-sept Portuguese: 🔊 (número 221 567) duzentos e vinte e um mil, quinhentos e sessenta e sete Chinese: 🔊 (数 221 567) 二十二万一千五百六十七 Arabian: 🔊 (عدد 221,567) مئتان و واحد و عشرون ألفاً و خمسمائةسبعة و ستون Czech: 🔊 (číslo 221 567) dvěstě dvacet jedna tisíc pětset šedesát sedm Korean: 🔊 (번호 221,567) 이십이만 천오백육십칠 Danish: 🔊 (nummer 221 567) tohundrede og enogtyvetusindfemhundrede og syvogtreds Dutch: 🔊 (nummer 221 567) tweehonderdeenentwintigduizendvijfhonderdzevenenzestig Japanese: 🔊 (数 221,567) 二十二万千五百六十七 Indonesian: 🔊 (jumlah 221.567) dua ratus dua puluh satu ribu lima ratus enam puluh tujuh Italian: 🔊 (numero 221 567) duecentoventiunomilacinquecentosessantasette Norwegian: 🔊 (nummer 221 567) to hundre og tjue-en tusen, fem hundre og seksti-syv Polish: 🔊 (liczba 221 567) dwieście dwadzieścia jeden tysięcy pięćset sześćdziesiąt siedem Russian: 🔊 (номер 221 567) двести двадцать одна тысяча пятьсот шестьдесят семь Turkish: 🔊 (numara 221,567) ikiyüzyirmibinbeşyüzaltmışyedi Thai: 🔊 (จำนวน 221 567) สองแสนสองหมื่นหนึ่งพันห้าร้อยหกสิบเจ็ด Ukrainian: 🔊 (номер 221 567) двiстi двадцять одна тисяча п'ятсот шiстдесят сiм Vietnamese: 🔊 (con số 221.567) hai trăm hai mươi mốt nghìn năm trăm sáu mươi bảy Other languages ...
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Formative Assignment Refraction of light Class 10 physics
# Formative Assesment
### Short Answer Based questions for Refraction
1. What is refraction of light?
2. What the speed of light in glass is as compared to speed of light in vacuum. ?
3. What is the unit of refractive index?
4. Why a concave lens is called diverging lens?
5. Define Snell’s law of Refraction.
6. Write the relation between the angle of incidence and angle of refraction for a medium.
7. For total internal reflection to occur, the light must pass from denser to rarer medium. Is this True or false?
8. What type of lens is used in a simple microscope?
### Match the column
Column A Column B The power of the lens is negative Convex lens The lens is diverging Concave lens The power of the lens is positive The lens is converging in nature The lens is thick at the center but thinner at the edges The lens is thin in the middle and thicker at the edges It has real focus It has virtual focus
Question A lens X has following observation
1. When the objects is placed at infinity, the image is formed at focus of the lens
2. When a object is placed at 60 cm from the lens, A real image is forced at 10 cm from lens on other side
3. The lens has positive Power
Answer following based on the data given above
1. The lens is
1. Convex
2. Concave
2. The focal length of the lens is
1. 20 cm
2. 10 cm
3. 25 cm
4. None of these
3. The power of the lens
1. 10 D
2. 5 D
3. 4 D
4. None of these
4. When the object is placed at a distance twice the focal length in front of lens, what will be characteristics of the image formed
1. A real inverted image of same size will be formed at a distance twice the focal length
2. A real ,erect image of high diminished size will be formed at a distance twice the focal length
3. A high enlarged image at infinity will be formed
4. None of these
5. When the lens is placed close to the page of book, the letters of the page appear
1. Highly diminished
2. Enlarged
3. Same size as object
4. None of these
## Fill in the blanks
1. The refractive index of a medium gives an indication of the ……………Ability of that medium
2. When a ray of light goes from water to air, it bends …………From the normal
3. When a ray of light goes from air to glass, it bends ………The normal
4. The speed of light is …… in glass then air
5. The absolute refractive index is always ….. then unity
6. The relative refractive index can be less then ………
## Crossword Puzzle
Across
3. The human eye forms the image at this point
5. The medium where speed of light is less
8. A instrument which is used to see details of distant objects
10. The refractive index is equal to reciprocal of the sine of this angle
Down
1. A optical instrument which is used to see small objects
2. It is produced by the dispersion of sunlight by the rain drops in the air
4. The nature of the image formed by the convex lens when the object is placed between optical center and focus
5. The size of the image formed by the concave lens
6. A converging lens
7. It is the defect of eye when it cannot see distant objects clearly
9. The lens having negative power
Question 1 What is meant by ‘total internal reflection’? State two essential conditions for total internal reflection to take place. With the help of a ray diagram, illustrate an application of total internal reflection.
Question 2 A concave lens of focal length 15 cm forms an image of 10 cm from the lens.
a) How far is object from the lens?
b) What is the nature of the image?
Question 3 What do you mean by power of the lens? What is the power of a concave lens of focal length 10cm?
Question 4 You are given three medium X, Y, Z .The refractive index of the three medium are n1 , n2 and n3.
It is given that n1 > n3 > n2
1. In which of these does the light travel fastest?
2. In which of these does the light travel lowest?
3. Arrange the medium in ascending order of their optical density?
Question 5 A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Question 6 A Calculate the focal length of convex lens which produces a virtual image at a distance of 25 cm of an object placed 10cm in front of it.
Question 7 What is lateral shift and optical density?
Question 8 What is scientific name for short sightness and far sightness ? What types of lens are used to correct them?
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# Measurement of Velocity and Acceleration
In this experiment a trolley is allowed to roll down a runway which is tilted at an angle. The runway must first be tilted to overcome the friction in the wheels of the trolley. A ticker tape timer is used to measure the velocity near the start of its journey (u) and near the end(v). The acceleration can then be calculated. Note: In this simulation the angle of the runway can be changed by dragging the slider right or left.
Procedure:
1. Using the slider below the simulation, raise the runway a few degrees.
2. Press "Release Trolley". If the trolley does not move raise the runway a little more and try "Release Trolley" again. When friction is overcome the trolley accelerates very slowly. In the lab, at a slightly lower angle, a little push would make it move with constant velocity.
3. Press "Get Tape and Ruler".
4. Measure and record a five-space distance (5/50 s) near the start
5. Calculate "u" = distance/0.1. Calculate "v" in a similar way near the end of the tape
6. Calculate the acceleration using the formula a = (v - u) / t where t is the number of dots from start of u interval to start of v interval / 50.
7. Press reset and try different angles
Note:Because of the small space available the accelerations calculated in this simulation are much too large.
Precautions:
• Ensure that the runway in smooth, free of dust, and does not sag in the middle
• At constant velocity the dots must be equally spaced all along the tape (same distance for each 1/50th. sec.). Not possible with this simulation.
• Ignore the first few dots on the tape. These are unreliable and too close together for accurate measurement.
• Place a ruler right on top of the tape with eye directly above when measuring distances to avoid parallax errors
Mechanics Experiments All Physics Experiments
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I'm thinking about getting life insurance because I have a mortgage and I have a young son and another baby on the way. And so if anything were to happen to me, I'd want them to at least be able to pay off the mortgage and then maybe have some money left over for college and to live, and whatever else. And so I went to the insurance company, and I said I want to get a $1 million policy. And what I'm actually getting a quote on is a term life policy, which is really-- I just care about the next 20 years. After those 20 years, hopefully, I can pay off my mortgage. There'll be money saved up. Hopefully, my kids would kind of at least have maybe gotten to college or I would have saved up enough money for college. So that's why I'm willing to do a term life policy. The other option is to do a whole life policy, where you could pay a certain amount per year for the rest of your life. At any point you die, you get the million dollars. In a term life, I'm only going to pay a$500 per year for the next 20 years. If at any point over those 20 years I die, my family gets a million. At the 21st year, I have to get a new policy. And since I'm going to be older and I'd have a higher chance of dying at that point, then it's probably going to be more expensive for me to get insurance. But I really am just worried about the next 20 years. But what I want to do in this video is think about given these numbers that have been quoted to me by the insurance company, what do they think that my odds of dying are over the next 20 years? So what I want to think about is the probability of Sal's death in 20 years, based on what the people at the insurance company are telling me. Or at least, what's the maximum probability of my death in order for them to make money? And the way to think about it, or one way to think about it, kind of a back-of-the-envelope way, is to think about what's the total premiums they're getting over the life of this policy divided by how much they're insuring me for. So they're getting $500 times 20 years is equal to, that's$10,000 over the life of this policy. And they are insuring me for $1 million. So they're getting-- let's see those 0s cancel out, this 0 cancels out-- they're getting, over the life of the policy,$1 in premiums for every $100 in insurance. Or another way to think about it. Let's say that there were 100 Sals, 100 34-year-olds looking to get 20-year term life insurance. And they insured all of them. So if you multiplied this times 100, they would get$100 in premiums. This is the case where you have 100 Sals, or 100 people who are pretty similar to me. 100 Sals. They would get $100 in premium. And the only way that they could make money is if, at most, one of those Sals-- or really just break even-- if, at most, 1 of those Sals were to die. So break even if only 1 Sal dies. I don't like talking about this. It's a little bit morbid. So one way to think about it, they're getting$1 premium for $100 insurance. Or if they had 100 Sals, they would get$100 in premium, and the only way they would break even, if only 1 of those Sal dies. So what they're really saying is that the only way they can break even is if the probability of Sal dying in the next 20 years is less than or equal to 1 in 100. And this is an insurance company. They're trying to make money. So they're probably giving these numbers because they think the probability of me dying is a good-- maybe it's 1 in 200 or it's 1 in 300. Something lower, so that they can insure-- one way to think about it-- they could insure more Sals for every \$100 in premium they have to pay out. But either way, it's a back-of-the-envelope way of thinking about it. And it actually makes me feel a little bit better because 1 in 100 over the next 20 years isn't too bad.
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# Fillet Weld Size - Rule of Thumb6
## Fillet Weld Size - Rule of Thumb
(OP)
I recall somewhere that there is a rule-of-thumb for the size of fillet welds. It is something along the lines of half the thickness of the thinnest member being welded. Or was it twice the thickness of the thinnest member being welded?
Can anyone clarify this for me?
TIA,
--Scott
### RE: Fillet Weld Size - Rule of Thumb
2
As with all rules-of thumb, there are a million exceptions, but the one you are probably looking for is that a double fillet weld, (of matching material strength) will be as strong as the base metal when the throat thickness is 1/2 the thickness of the thinner member. This does not take into account stress risers, fatigue, etc, only strength of materials.
### RE: Fillet Weld Size - Rule of Thumb
3
A reference is contained in Lincolns "Design Of Weldments" on Page 6.2-7. the paragraph states
"(a) for full strength welds, the leg of the fillet weld must be about 75% of the plate thickness."
Here is my take for double fillet welded joints
The effective throat of an equal leg fillet weld is .707 x leg size. .707 x .75t = about .53t. 2 equal leg fillet welds with leg sizes of 75% the thickness of the base metal would result in a fillet weld with an effective throat of 1.06t.
Good day
### RE: Fillet Weld Size - Rule of Thumb
(OP)
Thanks all.
I'll look through Blodgett's (Lincoln Press') Design of Welded Structures again to verify this. I figured this information was in there, but couldn't find it.
--Scott
### RE: Fillet Weld Size - Rule of Thumb
When a full strength welded joint isn't required, I use the following guideline. These values are (mostly) from Bilichniansky's The Designers Manual. It has been my experience that these size recommendations produce weldments that "look right" to most welders, engineers and managers.
Thinnest Fillet
Member Weld Size
----------------------
1/8 1/8
3/16 3/16
1/4 1/4
3/8 1/4
1/2 3/8
5/8 3/8
3/4 3/8
7/8 1/2
1 1/2
1-1/4 1/2
1-1/2 3/4
2 3/4
### RE: Fillet Weld Size - Rule of Thumb
According D1.1 the following minimum sizes apply:
Table 5.8
Base Metal Thickness Min. fillet size
T= <1/4" 1/8"
1/4"~1/2" 3/16"
<1/2"~3/4" 1/4"
3/4" plus 5/16"
*** leg size need not be greater than the thinner of the two parts joined.
refer to 2.4.5 and find maximum fillet weld size...
(1) for base materials 1/4" or less full plate thickness
(2) 1/16" less than the base metal thickness, unless the weld is designated to be built out to obtain full throat thickness.
hope this helps....
Richard Schram
Mechanical Integrity Specialist
Pharmacia Global Supply Arecibo-P. Rico
rschram@pharmacia.com
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AP Free Response Question 2002 B2
A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E of 0.4 J.
(a) What is the object’s potential energy when its displacement is +4 cm from its equilibrium position?
(b) What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning.
(c) Determine the object’ s kinetic energy when its displacement is —7 cm.
(d) What is the object’ s speed at x = 0?
Suppose the object undergoes this motion because it is the bob of a simple pendulum as shown below.
(e) If the object breaks loose from the string at the instant the pendulum reaches its lowest point and hits the ground at point P shown, what is the horizontal distance d that it travels?
CB-ETSCopyright © 1970-2024All rights reserved.Used with permissionMainland High SchoolDaytona Beach, FL 32114
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Thursday, January 31, 2008
Some useful mathematical formulae
• $e^a \geq 1+a$
• 1/(1+x) = 1 + O(x) when x is small.
• ${n\choose an} = O((\frac{1}{a^a (1-a)^{(1-a)}})^n)$, where $a\in (0,1)$.
• If $T(1) = 1$ and $T(n) = 2^nT(n/2)$, then $T(n) = 4^n$.
• ${n\choose k}\leq 2^{n\cdot H(k/n)}$, where $H(p) = -p\log p - (1-p)\log (1-p).$
• $\left(\frac{n}{k}\right)^k\leq {{n\choose k}}\leq\frac{n^k}{k!}\leq\left(\frac{n\cdot e}{k}\right)^k.$
Tuesday, January 29, 2008
Interesting papers
• Justin G. Chen, Scott D. Kominers, Robert W. Sinnott:
• 懶是上策?!
• Artur Czumaj and Christian Sohler: (Draft) A survey on sublinear time algorithms.
• 另一篇關於 subliear time algorithms 的 survey
• Ziv Bar-Yossef, Ravi Kumar: Sampling Algorithms: Lower Bounds and Applications. STOC'01
• 看看這篇說不定能幫助我更加深入了解 sampling algorithms
• Rajeev Motwani, Rina Panigrahy, Ying Xu: Estimating Sum by Weighted Sampling. ICALP'07
• 喔？！這一篇算是蠻新的吧！
Best wishes for the paper
Peter 不放心我們的 paper 會不會被接受，所以說想去教堂點一盞燈，祈禱 referees 會接受這篇 paper。我想我也會在每次用餐時祈禱。這麼說來，老師不就可以去孔廟找孔子聊聊了？
Friday, January 25, 2008
The 3n+1 conjecture (Collatz's conjecture)
The 3n+1 conjecture (Collatz's conjecture), forwarded from Open Problem Garden.
Let f(n) = 3n+1 if n is odd and n/2 if n is even. Let f(1) = 1. Assume we start with some number n and repeatedly take the f of the current number. Prove that no matter what the initial number is we eventually reach 1.
It can be also obtained from Wikipedia. This problem is worth at least \$500 US dollars. Refer to this page for some results of computational verification. I also wrote a very simple program to verify this conjecture.
This is an an open problem for the moment, and it seems to be given on an graduate school entrance exam. To find a function g (n) such that f(n) = O(g(n)) is then an open problem, too.
It's very interesting. By the way, I found "Jones' conjecture", posted by Chuan-Min Lee. I hope that someday I will post an interesting and important open problem or a conjecture, too.
Wednesday, January 23, 2008
A picture taken when Peter visited CCU, Taiwan
Professor Peter Rossmanith 於 2007 年造訪中正大學資訊工程學系計算理論實驗室兩次， Post 上這張照片當作紀念。
Tuesday, January 22, 2008
Joseph, Chuang-Chieh Lin's Homepage
Since my homepage cannot be visited by Google robots, I post it here.
It has been updated for several times.
Monday, January 21, 2008
Terence Tao's homepage and blog
http://www.math.ucla.edu/~tao/
http://terrytao.wordpress.com/
Chinese Translation of my personal homepage
http://tinyurl.com/36twpk 中文版
http://tinyurl.com/3y3gsw 德文版
Room405, 計算理論實驗室,
RWTH 亞琛大學, 德國
Maw-Shang Shang Chang (禮物) 教授 ,
R. C. T. 李教授(前)
...
...
Sunday, January 20, 2008
My poor teeth...
There are always troubles with my teeth. I just lost half part of one tooth by accident. I think I'd better go to Josef Kunze's clinic as soon as possible, if I don't want to lose the rest part of it.
So terrible for me. Oops!
W3C HTML 4.01 Transitional validation passed!
Finally I made my personal homepage pass the validation!! It is not so easy for me to refine the homepage from the original Microsoft Frontpage webpage to a valid W3C html one.
Congratulations
The document located at http://www.cs.ccu.edu.tw/~lincc/ was checked and found to be valid HTML 4.01 Transitional. This means that the resource in question identified itself as "HTML 4.01 Transitional" and that we successfully performed a formal validation using an SGML or XML Parser (depending on the markup language used).
And I got this:
The following paragraph is an interesting FAQ:
Is validation some kind of quality control? Does "valid" mean "quality approved by W3C"?
Validity is one of the quality criteria for a Web page, but there are many others. In other words, a valid Web page is not necessarily a good web page, but an invalid Web page has little chance of being a good web page.......
To validate your website, try W3C HTML Validator: http://validator.w3.org/
Tuesday, January 15, 2008
葛若琳 (Caroline Gluck) BBC 特約記者
http://udn.com/NEWS/NATIONAL/NAT5/4181243.shtml
http://udn.com/NEWS/NATIONAL/NAT5/4181243.shtml
Caroline Gluck 的部落格： (caro's choice)
http://caroschoice.blogspot.com/
Monday, January 14, 2008
My Cherie Amour (by Stevie Wonder)
La la la la la la
La la la la la la
My Cherie Amour, lovely as a summer's day
My Cherie Amour, distant as the Milky Way
My Cherie Amour, pretty little one that I adore
You're the only girl my heart beats for
How I wish that you were mine
In a cafe or sometimes on a crowded street
I've been near you, but you have never noticed me
My Cherie Amour, won't you tell me how could you ignore
That behind that little smile I wore
How I wish that you were mine
Maybe someday you'll see my face among the crowd
Maybe someday I'll share your little distant cloud
Oh, Cherie Amour, pretty little one that I adore
You're the only girl my heart beats for
How I wish that you were mine
La la la la la la
La la la la la la
Sunday, January 13, 2008
Runners Point and Dom neighborhood
Josef 的建議與 wes 的很類似，他說寒冷的冬天應以建立里程為主，速度練習可以先擱著。令我訝異的是，一天 30 km 的里程對他而言幾乎是隨手拈來。希望三月底之前，他能確定跟我一起參加 Düsseldorf Marathon。前提是，他得過老婆那一關才行。
Tuesday, January 08, 2008
Glucosamine & chondroitin
http://www.wedar.com/health/show.asp?id=1721
Apotheke am Steppenberg:
Telephone: 0241-873335
Study progress - [JKL01-SICOMP]
T. Jiang, P. Kearney, and M. Li: A polynomial time approximation scheme for inferring evolutionary trees from quartet topologies and its application. SIAM J. Comput. 30 (2001) 1942--1961. [link]
Friday, January 04, 2008
Home - Simply Red
What's worth nothing else but love
Take a walk down any street now
Every one of us in our own little world
Looking for a heart with whom to beat now
What's worth nothing else but love
I'm prepared to take the heat now
What's worth more than anything else at all
To keep you firmly on your feet now
So fake cool image should be over
'Cause I long for a feeling of home
Real life, depicted in song
A loving memory
After long, home is a place where I yearn to belong
Where the land meets the sea
She'll be smiling so sweetly now
I hope that she'll be here much longer than I will
My heart loves her with every beat now
So fake cool image should be over
'Cause I long for a feeling of home
Real life, depicted in song
A loving memory
After long, home is a place where I yearn to belong
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## Ncert Class 10 Maths Ch 6 Ex 6.3 Mp4,Used Sea Doo Boat For Sale Near Me Au,Cheap Bass Boats For Sale In Ohio 40,Byjus Class 5 Maths Chapter 8 Full - Plans On 2021
EXERCISE CLASS 10 MATHS CHAPTER 3-LINEAR EQUATIONS IN TWO Ncert Book Of Class 10th Maths Exercise 6.3 Note VARIABLES: NCERT Solutions For Class 10 Maths Chapter 3 Linear Equations In Two Variables Ex given here are prepared by the subject experts. Visit now to download solutions in PDF for free. NCERT Solutions for Class 10 Maths Chapter 6 Triangles: Areas of Similar Triangles. Students can understand the formula and Ncert Class 10 Maths Ch 6 Ex 6.3 Lite learn the process for finding the surface area of similar triangles in this section. Maths NCERT Class 10 Chapter 6 allows students to find the area of similar triangles with the utilisation of the different theorems. Jul 15, �� NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex Ex Class 10 Question 1. State which Ncert Solutions For Class 10 Maths Ch 6 Ex 6.3 Days pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form: . Class 10 Maths NCERT Ex Solution (Part 1) Ch 6 Triangles.� ������� �����. Triangles Chapter 6 Class 10 Maths NCERT. MathsTeacher. The NCERT Solutions Class 10 Maths Chapter 6 Exercise involves problems where various theorems on the similarity of triangles are to be applied to solve them. The questions in Exercise Class 10 Maths touch upon all aspects of this topic, so students get ample practise for their exam preparations.� For this question of NCERT Class 10 Maths ch 6 ex , students are given a trapezium ABCD, and they need to prove that for the point of intersection O, the proportion of the lengths of the diagonals AO/OC = OB/OD. You need to apply the theorem for alternate interior angles and vertically opposite angles to solve this problem. Exercise Class 10 Question 4. In this question, a triangle is given, which contains two triangles inside of it. NCERT Solutions Class 10 Maths Ch 6 Ex Question 9. In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: Solution: Ex Class 10 NCERT Solutions Question CD and GH are respectively the bisectors of ?ACB and ?EGF such that D and H lie on sides AB and FE of ?ABC and ?EFG respectively. If ?ABC ~ ?FEG, show that Solution: Class 10 Ex Solutions Question In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ? BC and EF ? AC, prove that ?ABD ~ ?ECF.
Conclusion:
Ncert Maths Class 10th Pdf 10
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Ncert Class 10 Maths Ch 6 Ex 6.3 Mp4
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# Solving linear systems by substitution
We will also give you a few tips on how to choose the right app for Solving linear systems by substitution. We can solving math problem.
## Solve linear systems by substitution
There's a tool out there that can help make Solving linear systems by substitution easier and faster A math answers scanner is a tool that can be used to check the correctness of answers to math problems. It can be used by students to check their work, or by teachers to check answers to tests and quizzes. The scanner works by scanning a problem and its answer and then comparing the answer to the correct answer. If the answer is correct, the scanner will give a green light, if it is incorrect, the scanner will give a red light. The math answers scanner is a quick
There are many ways to solve linear equations, and the best method to use will depend on the specific equation you are trying to solve. In general, however, the three most common methods for solving linear equations are substitution, elimination, and graphing. All three of these methods can be used to solve equations with one or two variables, but graphing is the only method that can be used to solve equations with more than two variables. If you are unsure which method to use, a good
There is no one-size-fits-all answer to this question, as the best way to solve word problems in algebra will vary depending on the individual problem. However, there are some general tips that can be followed in order to make solving word problems in algebra easier. First, it is important to read the problem thoroughly and identify the key information that is needed in order to solve the problem. Next, it is helpful to draw a picture or diagram of the problem, as this can
A free equation solver can be a useful tool for students or anyone struggling to solve a complex equation. These solvers can be found online and can provide step-by-step instructions on how to solve the equation. This can be a valuable resource for those who are struggling to understand the concepts.
## Math solver you can trust
100/10 It's really helpful in understanding various math problems and how they are solved. There are also books with already provided solutions to problems available only if you are a Plus subscriber, so I highly recommend subscribing! Hopefully, they include the word problems in their next updates
Anastasia Howard
I would be failing Algebra if it weren't for this app. the app not only shows you the answer but shows you how to do it. I think it's an amazing work you guys have done. Only thing I'm wondering about is where is the auto scan feature you had on an older version of the app.
Kyleigh Powell
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CC-MAIN-2023-06
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https://www.geeksforgeeks.org/mahotas-threshold-adjacency-statistics/?type=article&id=449804
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GeeksforGeeks App
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# Mahotas – Threshold Adjacency Statistics
In this article, we will see how we can get the image’s threshold adjacency statistics feature in mahotas. TAS was presented by Hamilton et al. in “Fast automated cell phenotype image classification”. TAS gives original parameters, unlike PFTAS which gives a variation without any hardcoded parameters.
For this tutorial, we will use the ‘Lena’ image, below is the command to load the Lena image
`mahotas.demos.load('lena')`
Below is the Lena image
In order to do this we will use mahotas.features.tas method
Syntax : mahotas.features.tas(img)
Argument : It takes image object as argument
Return : It returns 1-D array
Note: Input image should be filtered or should be loaded as grey
In order to filter the image we will take the image object which is numpy.ndarray and filter it with the help of indexing, below is the command to do this
`image = image[:, :, 0]`
Below is the implementation
## Python3
`# importing required libraries``import` `mahotas``import` `mahotas.demos``from` `pylab ``import` `gray, imshow, show``import` `numpy as np``import` `matplotlib.pyplot as plt`` ` `# loading image``img ``=` `mahotas.demos.load(``'lena'``)`` ` `# filtering image``img ``=` `img.``max``(``2``)` `print``(``"Image"``)`` ` `# showing image``imshow(img)``show()` `# computing tas``value ``=` `mahotas.features.tas(img)`` ` `# printing value``print``(value)`
Output :
`Image`
```[8.18235887e-01 4.96278071e-02 3.85778412e-02 5.42293510e-02
2.31141496e-02 8.96518478e-03 4.17582280e-03 2.30390223e-03
7.70054279e-04 8.11830699e-01 5.42434618e-02 3.79106870e-02
5.78859183e-02 2.54097764e-02 7.40147155e-03 2.98681431e-03
1.76294893e-03 5.68223210e-04 8.69779571e-01 3.56911714e-02
2.61354551e-02 4.12780295e-02 1.73316328e-02 5.09194046e-03
2.56976434e-03 1.52282331e-03 5.99611680e-04 7.43348142e-01
5.80286091e-02 4.97388078e-02 7.46472685e-02 3.83537568e-02
1.81614021e-02 1.17267978e-02 4.57940731e-03 1.41580823e-03
9.37920200e-01 1.55393289e-02 1.20666222e-02 1.87743206e-02
9.61712375e-03 3.05412151e-03 1.93789436e-03 8.37170364e-04
2.53218197e-04 9.13099391e-01 2.42303089e-02 1.70045074e-02
2.72925208e-02 1.13702921e-02 3.81980697e-03 1.62341796e-03
1.19050651e-03 3.69248007e-04]```
Another example
## Python3
`# importing required libraries``import` `mahotas``import` `numpy as np``from` `pylab ``import` `gray, imshow, show``import` `os``import` `matplotlib.pyplot as plt`` ` `# loading image``img ``=` `mahotas.imread(``'dog_image.png'``)` `# filtering image``img ``=` `img[:, :, ``0``]`` ` `print``(``"Image"``)`` ` `# showing image``imshow(img)``show()` `# computing tas``value ``=` `mahotas.features.tas(img)`` ` `# printing value``print``(value)`
Output :
`Image`
```[8.92356868e-01 2.75272814e-02 2.05523535e-02 3.43358813e-02
1.80176597e-02 5.01153448e-03 1.33785553e-03 6.79775240e-04
1.80791287e-04 8.81674218e-01 3.13932157e-02 2.34006832e-02
3.69160363e-02 1.95048908e-02 5.11444295e-03 1.23809709e-03
6.09325269e-04 1.49090226e-04 9.06137850e-01 2.75823883e-02
2.03761048e-02 2.88661485e-02 1.36743022e-02 2.68646310e-03
4.75770564e-04 1.39449993e-04 6.15220557e-05 8.35720148e-01
4.69532212e-02 3.62894953e-02 5.08719737e-02 2.36920394e-02
4.84714813e-03 1.21050472e-03 2.87238408e-04 1.28231432e-04
9.38717680e-01 1.80549908e-02 1.33994005e-02 1.87263793e-02
8.80054720e-03 1.75569656e-03 3.62486722e-04 1.35538513e-04
4.72808768e-05 9.05435494e-01 2.48433294e-02 1.91342383e-02
2.97531477e-02 1.52476648e-02 4.03149662e-03 1.02763639e-03
4.30377634e-04 9.66153873e-05]```
My Personal Notes arrow_drop_up
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https://www.teacherspayteachers.com/Product/Classifying-Triangles-by-angles-and-sides-Interactive-PowerPoint-Activities-2427910
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Teaching Resources | Teachers Pay Teachers
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# Classifying Triangles by angles and sides- Interactive PowerPoint & Activities
Subjects
Resource Types
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File Type
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22.21 MB | 100 pages
### PRODUCT DESCRIPTION
This Classifying Triangles activity is perfect for teaching, practicing, and learning how to classify triangles by their sides and by their angles. They will learn about equilateral, isosceles, scalene, right, acute, and obtuse triangles.
What's Included:
► PowerPoint & Printables
► 3 activity options: whole group game, whole group written, or individual written
► 7 slides for teaching how to classify triangles by sides and angles
► 6 posters
► 45 slides for practicing
► 2 Versions of Data Sheets for Differentiation
► Student Version of the PowerPoint without answers or lengths
► Student Version of the PowerPoint without answers, but with lengths
This is a fun and engaging way for students to learn how to classify triangles.
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https://www.justintools.com/unit-conversion/volume.php?k1=deciliters&k2=us-beer-barrels
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Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# VOLUME Units Conversiondeciliters to us-beer-barrels
1 Deciliters
= 0.00085216620649692 Us Beer Barrels
Category: volume
Conversion: Deciliters to Us Beer Barrels
The base unit for volume is liters (Non-SI Unit)
[Deciliters] symbol/abbrevation: (dl)
[Us Beer Barrels] symbol/abbrevation: (bbl[US])
How to convert Deciliters to Us Beer Barrels (dl to bbl[US])?
1 dl = 0.00085216620649692 bbl[US].
1 x 0.00085216620649692 bbl[US] = 0.00085216620649692 Us Beer Barrels.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [volume] => (liters), 1 Deciliters (dl) is equal to 0.1 liters, while 1 Us Beer Barrels (bbl[US]) = 117.348 liters.
1 Deciliters to common volume units
1 dl = 0.1 liters (L)
1 dl = 0.026417205235815 us gallons (gal[US])
1 dl = 6.7627884329267 us tablespoons (tbsp[US])
1 dl = 20.288420181297 us teaspoons (tsp[US])
1 dl = 3.3814056503288 us fluid ounces (fl oz[US])
1 dl = 0.10566881491367 us quarts (qt[US])
1 dl = 0.021996924829909 uk gallons (gal[UK])
1 dl = 5.6312013604982 uk tablespoons (tbsp[UK])
1 dl = 16.893632620929 uk teaspoons (tsp[UK])
1 dl = 3.5195033276904 uk fluid ounces (fl oz[UK])
Decilitersto Us Beer Barrels (table conversion)
1 dl = 0.00085216620649692 bbl[US]
2 dl = 0.0017043324129938 bbl[US]
3 dl = 0.0025564986194907 bbl[US]
4 dl = 0.0034086648259877 bbl[US]
5 dl = 0.0042608310324846 bbl[US]
6 dl = 0.0051129972389815 bbl[US]
7 dl = 0.0059651634454784 bbl[US]
8 dl = 0.0068173296519753 bbl[US]
9 dl = 0.0076694958584722 bbl[US]
10 dl = 0.0085216620649692 bbl[US]
20 dl = 0.017043324129938 bbl[US]
30 dl = 0.025564986194907 bbl[US]
40 dl = 0.034086648259877 bbl[US]
50 dl = 0.042608310324846 bbl[US]
60 dl = 0.051129972389815 bbl[US]
70 dl = 0.059651634454784 bbl[US]
80 dl = 0.068173296519753 bbl[US]
90 dl = 0.076694958584722 bbl[US]
100 dl = 0.085216620649692 bbl[US]
200 dl = 0.17043324129938 bbl[US]
300 dl = 0.25564986194907 bbl[US]
400 dl = 0.34086648259877 bbl[US]
500 dl = 0.42608310324846 bbl[US]
600 dl = 0.51129972389815 bbl[US]
700 dl = 0.59651634454784 bbl[US]
800 dl = 0.68173296519753 bbl[US]
900 dl = 0.76694958584722 bbl[US]
1000 dl = 0.85216620649692 bbl[US]
2000 dl = 1.7043324129938 bbl[US]
4000 dl = 3.4086648259877 bbl[US]
5000 dl = 4.2608310324846 bbl[US]
7500 dl = 6.3912465487269 bbl[US]
10000 dl = 8.5216620649692 bbl[US]
25000 dl = 21.304155162423 bbl[US]
50000 dl = 42.608310324846 bbl[US]
100000 dl = 85.216620649692 bbl[US]
1000000 dl = 852.16620649692 bbl[US]
1000000000 dl = 852166.20649692 bbl[US]
(Deciliters) to (Us Beer Barrels) conversions
:)
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CC-MAIN-2024-18
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https://keepthetech.com/what-gets-wetter-the-more-it-dries/
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# What Gets Wetter The More It Dries (Explained)
What gets wetter the more it dries puzzle is becoming popular on Facebook, WhatsApp, and Instagram because people find the riddle challenging and intriguing enough to socialize with friends and family.
Look out the What Becomes Wetter as It Dries Riddle and the reason behind it. Solve the difficult conundrum and put others to the test. In the post, you can learn more about What Becomes Wetter as It Dries Riddle.
The riddle is hard to understand for a kid, but you can try to test your mind to solve the puzzle. In case if you don’t want to have a hassle through your brain then. Don’t worry we are here for your help, In this short guide we have explained what does What gets wetter as it dries riddle. To know the proper answer to the riddle make sure you read the entire guide.
Contents
## Why is Everyone Trying To Slove This Riddle?
Many individuals will be looking for methods to engage themselves during the coronavirus lockdown in the midst of the coronavirus pandemic or in their free time of office. The parents and kids are attempting to solve puzzles, test their minds, and keep themselves occupied.
Well did you find out the proper answer to the riddle, if yes then share your answer with us through the comment box? In case if you cannot able to find the answer then don’t worry just read out our explanation.
As individuals seek new and exciting ways to interact, these puzzles, riddles, and challenges have quickly gone viral. Such mind-boggling questions are circulating in WhatsApp groups and on social media. What Becomes Wetter the More It Dries Riddle is one such puzzle. Here is the solution to the mystery. Examine it out!!
## What Becomes Wetter the More It Dries Puzzle!
The riddle is
“What gets wetter and wetter the more it dries?”
## Answer to the What becomes wetter the more it dries riddle?
Here some of the experts try to solve the riddle, well for solving riddle, one should think beyond the box. It is preferable to think logically. Now, if you haven’t already found the answer, here it is.
The answer to this riddle What becomes wetter the more it dries is “Towel”!
## Explanation of the Riddle
A towel is a type of absorbent fabric that is used to dry or wipe a skin or an object. As a result, the towel becomes wetter as it dries. As a result, the Towel is the answer to the puzzle.
The answer to the riddle is quite simple, you needed to release your mind and think about what does become wetter the more it’s dried and you find the answer which is Towel. In case if you find this explanation then don’t forget to share it with your friends.
Hopefully, you find this guide helpful to know what does what becomes wetter the more it dries the puzzle. In case if you’ve any other questions or puzzles that cannot able to resolved feel free to share them with us. We try our best to solve them as soon as possible.
### What gets wetter as it’s drying?
The answer to what gets wetter as it’s drying was Towel.
### What is the more it dries the wetter it becomes?
The answer to the question is a towel because the more it dries, the wetter it becomes. Because a towel is made up of an absorbent fabric that absorbs water very fast.
### What is wet but never dries?
The answer to the riddle, what is wet but never dries is Towel.
### What gets wetter the more it dries explain
The answer to the riddle is Towel because is made up of absorbent fabric that absorbs liquid very quickly.
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http://physics.stackexchange.com/questions/29902/why-is-the-center-of-mass-of-2-bodies-at-the-focus-of-their-elliptical-orbits?answertab=active
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# Why is the center-of-mass of 2 bodies at the focus of their elliptical orbits?
Why is the center-of-mass of 2 bodies (which interact only via Newtonian gravity) located at a focus of each of the elliptical orbits?
I know that when there are no external forces, the center of mass moves at a constant speed, but that doesn't explain it.
-
"they orbit around their center of mass" That's a bit of an oversimplified statement. You do know that orbits are, in general, not circular with a well-defined center point? – leftaroundabout Jun 11 '12 at 13:53
The important thing here is that the center of mass is an invariant in a frame where it is initially at rest. This actually follows from the definition of the thing and Newton's laws, but I've become convinced that the point is subtle and important enough (if basic) to merit a careful answer. – dmckee Jun 11 '12 at 14:00
If the orbit is elliptic, the center of mass will be at one of the ellipse's focuses. But why? – Omer Jun 11 '12 at 14:55
Wait. Is the question about the invariance of the CoM or about the shape of two body orbits? – dmckee Jun 11 '12 at 16:22
The question is about the shape of two body orbits - why the CoM is one of the two ellipses' focuses. – Omer Jun 11 '12 at 17:13
I'm going to assume that Omer is specifically asking why the centre of mass is at the focus (well, one of the foci) of the orbits. Omer, if this isn't what you meant please ignore what follows because it's completely irrelevant!
If you have a body moving in a central field (i.e. the force is always pointing towards the centre), and the field is inversely proportional to the square of the distance from body to the centre, then the orbit is an ellipse with the centre at one of the foci. For now let's just assume this and we can come back to prove it later.
So if we can show that both of the bodies feel a central inverse square force, with the COM at the centre, this guarantees the orbits will be ellipses with a focus at the COM. Given that the force is due to the two bodies attracting each other, and that both bodies are orbiting around, it may seem a bit odd that each body just feels a central inverse square force, but actually this is easy to show.
The picture shows the two masses and the COM. I haven't shown the velocities because it doesn't matter what they are. For now let's just consider $m_1$ and calculate the force on it. By Newton's law this is simply:
$$F_1 = \frac{Gm_1m_2}{(r_1 + r_2)^2}$$
First is this force central? We know the centre of mass doesn't move. For two bodies this seems obvious to me, but in any case dmckee proved it in his answer. If the COM doesn't move it must lie on the line joining the two mases, otherwise there'd be a net force on it. So the force $F_1$ must always point towards the COM i.e. the force is central.
Second is this an inverse square law force i.e. is $F_1 \propto 1/r_1^2$? Well the definition of the centre of mass is that:
$$m_1r_1 = m_2r_2$$
or
$$r_2 = r_1 \frac{m_1}{m_2}$$
If we substitute for $r_2$ in the expression for $F_1$ we get:
$$F_1 = \frac{Gm_1m_2}{(r_1 + r_1(m_1/m_2))^2}$$
or with a quick rearrangement:
$$F_1 = \frac{1}{(1 + m_1/m_2)^2} \frac{Gm_1m_2}{r_1^2}$$
and this shows that $F_1$ is inversely proportional to $r_1^2$. I won't work through it, but it should be fairly obvious that exactly the same reasoning applies to $F_2$ so:
$$F_2 = \frac{1}{(1 + m_2/m_1)^2} \frac{Gm_1m_2}{r_2^2}$$
This is the key result. Even though the two bodies are whizzing around each other, each body just behaves as if it were in a static gravity field, but the strength of the field is reduced by a factor of $(1 + m_1/m_2)^2$ for $m_1$ or $(1 + m_2/m_1)^2$ for $m_2$. This applies to all two body systems, even such unequal ones as the Sun and the Earth (ignoring perturbations from Jupiter etc).
I did start by assuming that a body in a central gravity field orbits in an ellipse with the foci at the centre, but I'm going to wimp out of proving this since it would double the length of this answer and you'd all go to sleep. The proof is easily Googled.
NB this only applies to two body systems. For three or more body systems the orbits are generally not ellipses with the centre of mass at the focus.
-
Thanks! It was exactly what I searched for. – Omer Jun 11 '12 at 18:25
This actually follows from Newton's laws (and it only holds true for an isolated system).
For simplicity we'll consider an isolated system of two bodies on a line. Call their masses $m_1$ and $m_2$ and put them at $x_1$ and $x_2$. Now computer their center of mass: $$X = \frac{\sum_i m_i x_i}{\sum_i m_i} = \frac{1}{M} \sum_i m_i x_i$$
Assume that there is some force, $F$, between them. Newton's second law tells us that the force on body one due to body two $F_1$ is equal and opposite that on body two due to body one $F_2 = -F_1$.
This is enough information to compute the acceleration of the CoM: $$A = X'' = \frac{1}{M} \sum_i m_i a_i = \frac{1}{M} \sum_i m_i \frac{F_i}{m_i},$$ cancel the masses inside the sum and take note of the relation between the forces and you get $$A = 0 .$$
For more bodies and more dimensions the math gets more complicated but the result is the same.
-
I think Omer is specifically asking why the centre of mass is at the focus of the orbits – John Rennie Jun 11 '12 at 15:11
@JohnRennie: This is the reason they move "around" the CoM. Any movement away on the mart of mass one must be mirrored by a mass-ratio weighted movement away on the part of mass two (as long as only internal forces are acting). And likewise for movements toward, and in all relevant dimensions. I know it seems like a different question, but it's not. That is the subtle point that makes this question interesting and pedagogically important. – dmckee Jun 11 '12 at 16:19
Ah...seeing the OP's comment above maybe I have answered the wrong question. – dmckee Jun 11 '12 at 16:21
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http://www.ask.com/home-garden/much-pallet-bricks-weigh-90c584bcc3471c69
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Q:
# How much does a pallet of bricks weigh?
A:
The weight of a pallet of bricks varies based on the size of the pallet, the type of bricks used and the number of bricks used. For instance, a pallet of red clay bricks stacked 4-feet high weighs significantly more than a pallet with a single layer of Lego bricks.
Know More
## Keep Learning
Credit: Jerry Driendel Photographer's Choice RF Getty Images
If a brick's face is 7.625" x 2.25", it has a surface area of approximately 17 square inches. A standard-size pallet is 40" x 48", which equates to a surface area of 1,920 square inches. Based on these measurements, roughly 113 bricks are needed to cover the pallet. If each brick weighs 4.3 pounds, the pallet of bricks weighs 486 pounds with just a single layer of bricks on it. However, this example does not account for the weight of the pallet itself.
Sources:
## Related Questions
• A:
A typical wooden EURO pallet is measured in millimeters and has the dimensions 800 mm by 1,200 mm by 144 mm. This is equal to a pallet 31.50 inches wide and 47.24 inches long.
Filed Under:
• A:
Masonry jobs typically require 6.3 standard bricks per square foot, but the likelihood that some break makes seven bricks per square foot a safer estimate. Plan to use seven bags of mortar for each 1,000 bricks.
Filed Under:
• A:
Mud bricks are made from a mixture of soil, water and straw that are mixed together and kneaded before being set into molds and allowed to dry. Materials needed include topsoil, water, straw, sand and a mold in the shape of the bricks you wish to create. The entire process to create mud bricks takes about two weeks.
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# Properties
Label 12.4.b.a.11.1 Level 12 Weight 4 Character 12.11 Analytic conductor 0.708 Analytic rank 0 Dimension 4 CM no Inner twists 4
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$12 = 2^{2} \cdot 3$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 12.b (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.708022920069$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\sqrt{3}, \sqrt{-5})$$ Defining polynomial: $$x^{4} + x^{2} + 4$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$2^{4}$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## Embedding invariants
Embedding label 11.1 Root $$-0.866025 - 1.11803i$$ of defining polynomial Character $$\chi$$ $$=$$ 12.11 Dual form 12.4.b.a.11.2
## $q$-expansion
$$f(q)$$ $$=$$ $$q+(-1.73205 - 2.23607i) q^{2} +(3.46410 - 3.87298i) q^{3} +(-2.00000 + 7.74597i) q^{4} +8.94427i q^{5} +(-14.6603 - 1.03776i) q^{6} +7.74597i q^{7} +(20.7846 - 8.94427i) q^{8} +(-3.00000 - 26.8328i) q^{9} +O(q^{10})$$ $$q+(-1.73205 - 2.23607i) q^{2} +(3.46410 - 3.87298i) q^{3} +(-2.00000 + 7.74597i) q^{4} +8.94427i q^{5} +(-14.6603 - 1.03776i) q^{6} +7.74597i q^{7} +(20.7846 - 8.94427i) q^{8} +(-3.00000 - 26.8328i) q^{9} +(20.0000 - 15.4919i) q^{10} -34.6410 q^{11} +(23.0718 + 34.5788i) q^{12} -10.0000 q^{13} +(17.3205 - 13.4164i) q^{14} +(34.6410 + 30.9839i) q^{15} +(-56.0000 - 30.9839i) q^{16} +35.7771i q^{17} +(-54.8038 + 53.1840i) q^{18} -69.7137i q^{19} +(-69.2820 - 17.8885i) q^{20} +(30.0000 + 26.8328i) q^{21} +(60.0000 + 77.4597i) q^{22} +96.9948 q^{23} +(37.3590 - 111.482i) q^{24} +45.0000 q^{25} +(17.3205 + 22.3607i) q^{26} +(-114.315 - 81.3327i) q^{27} +(-60.0000 - 15.4919i) q^{28} -152.053i q^{29} +(9.28203 - 131.125i) q^{30} +224.633i q^{31} +(27.7128 + 178.885i) q^{32} +(-120.000 + 134.164i) q^{33} +(80.0000 - 61.9677i) q^{34} -69.2820 q^{35} +(213.846 + 30.4277i) q^{36} -130.000 q^{37} +(-155.885 + 120.748i) q^{38} +(-34.6410 + 38.7298i) q^{39} +(80.0000 + 185.903i) q^{40} -125.220i q^{41} +(8.03848 - 113.558i) q^{42} -224.633i q^{43} +(69.2820 - 268.328i) q^{44} +(240.000 - 26.8328i) q^{45} +(-168.000 - 216.887i) q^{46} +193.990 q^{47} +(-313.990 + 109.556i) q^{48} +283.000 q^{49} +(-77.9423 - 100.623i) q^{50} +(138.564 + 123.935i) q^{51} +(20.0000 - 77.4597i) q^{52} +545.601i q^{53} +(16.1347 + 396.489i) q^{54} -309.839i q^{55} +(69.2820 + 160.997i) q^{56} +(-270.000 - 241.495i) q^{57} +(-340.000 + 263.363i) q^{58} -173.205 q^{59} +(-309.282 + 206.360i) q^{60} -442.000 q^{61} +(502.295 - 389.076i) q^{62} +(207.846 - 23.2379i) q^{63} +(352.000 - 371.806i) q^{64} -89.4427i q^{65} +(507.846 + 35.9492i) q^{66} +735.867i q^{67} +(-277.128 - 71.5542i) q^{68} +(336.000 - 375.659i) q^{69} +(120.000 + 154.919i) q^{70} -1039.23 q^{71} +(-302.354 - 530.877i) q^{72} +410.000 q^{73} +(225.167 + 290.689i) q^{74} +(155.885 - 174.284i) q^{75} +(540.000 + 139.427i) q^{76} -268.328i q^{77} +(146.603 + 10.3776i) q^{78} -85.2056i q^{79} +(277.128 - 500.879i) q^{80} +(-711.000 + 160.997i) q^{81} +(-280.000 + 216.887i) q^{82} +1254.00 q^{83} +(-267.846 + 178.713i) q^{84} -320.000 q^{85} +(-502.295 + 389.076i) q^{86} +(-588.897 - 526.726i) q^{87} +(-720.000 + 309.839i) q^{88} +840.762i q^{89} +(-475.692 - 490.181i) q^{90} -77.4597i q^{91} +(-193.990 + 751.319i) q^{92} +(870.000 + 778.152i) q^{93} +(-336.000 - 433.774i) q^{94} +623.538 q^{95} +(788.820 + 512.346i) q^{96} +770.000 q^{97} +(-490.170 - 632.807i) q^{98} +(103.923 + 929.516i) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q - 8q^{4} - 24q^{6} - 12q^{9} + O(q^{10})$$ $$4q - 8q^{4} - 24q^{6} - 12q^{9} + 80q^{10} + 120q^{12} - 40q^{13} - 224q^{16} - 240q^{18} + 120q^{21} + 240q^{22} + 288q^{24} + 180q^{25} - 240q^{28} - 240q^{30} - 480q^{33} + 320q^{34} + 24q^{36} - 520q^{37} + 320q^{40} + 240q^{42} + 960q^{45} - 672q^{46} - 480q^{48} + 1132q^{49} + 80q^{52} + 792q^{54} - 1080q^{57} - 1360q^{58} - 960q^{60} - 1768q^{61} + 1408q^{64} + 1200q^{66} + 1344q^{69} + 480q^{70} - 960q^{72} + 1640q^{73} + 2160q^{76} + 240q^{78} - 2844q^{81} - 1120q^{82} - 240q^{84} - 1280q^{85} - 2880q^{88} - 240q^{90} + 3480q^{93} - 1344q^{94} + 384q^{96} + 3080q^{97} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/12\mathbb{Z}\right)^\times$$.
$$n$$ $$5$$ $$7$$ $$\chi(n)$$ $$-1$$ $$-1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −1.73205 2.23607i −0.612372 0.790569i
$$3$$ 3.46410 3.87298i 0.666667 0.745356i
$$4$$ −2.00000 + 7.74597i −0.250000 + 0.968246i
$$5$$ 8.94427i 0.800000i 0.916515 + 0.400000i $$0.130990\pi$$
−0.916515 + 0.400000i $$0.869010\pi$$
$$6$$ −14.6603 1.03776i −0.997504 0.0706108i
$$7$$ 7.74597i 0.418243i 0.977890 + 0.209121i $$0.0670604\pi$$
−0.977890 + 0.209121i $$0.932940\pi$$
$$8$$ 20.7846 8.94427i 0.918559 0.395285i
$$9$$ −3.00000 26.8328i −0.111111 0.993808i
$$10$$ 20.0000 15.4919i 0.632456 0.489898i
$$11$$ −34.6410 −0.949514 −0.474757 0.880117i $$-0.657464\pi$$
−0.474757 + 0.880117i $$0.657464\pi$$
$$12$$ 23.0718 + 34.5788i 0.555021 + 0.831836i
$$13$$ −10.0000 −0.213346 −0.106673 0.994294i $$-0.534020\pi$$
−0.106673 + 0.994294i $$0.534020\pi$$
$$14$$ 17.3205 13.4164i 0.330650 0.256120i
$$15$$ 34.6410 + 30.9839i 0.596285 + 0.533333i
$$16$$ −56.0000 30.9839i −0.875000 0.484123i
$$17$$ 35.7771i 0.510425i 0.966885 + 0.255212i $$0.0821454\pi$$
−0.966885 + 0.255212i $$0.917855\pi$$
$$18$$ −54.8038 + 53.1840i −0.717633 + 0.696422i
$$19$$ 69.7137i 0.841759i −0.907117 0.420879i $$-0.861722\pi$$
0.907117 0.420879i $$-0.138278\pi$$
$$20$$ −69.2820 17.8885i −0.774597 0.200000i
$$21$$ 30.0000 + 26.8328i 0.311740 + 0.278829i
$$22$$ 60.0000 + 77.4597i 0.581456 + 0.750657i
$$23$$ 96.9948 0.879340 0.439670 0.898159i $$-0.355095\pi$$
0.439670 + 0.898159i $$0.355095\pi$$
$$24$$ 37.3590 111.482i 0.317745 0.948176i
$$25$$ 45.0000 0.360000
$$26$$ 17.3205 + 22.3607i 0.130647 + 0.168665i
$$27$$ −114.315 81.3327i −0.814815 0.579721i
$$28$$ −60.0000 15.4919i −0.404962 0.104561i
$$29$$ 152.053i 0.973637i −0.873503 0.486818i $$-0.838157\pi$$
0.873503 0.486818i $$-0.161843\pi$$
$$30$$ 9.28203 131.125i 0.0564886 0.798003i
$$31$$ 224.633i 1.30146i 0.759309 + 0.650730i $$0.225537\pi$$
−0.759309 + 0.650730i $$0.774463\pi$$
$$32$$ 27.7128 + 178.885i 0.153093 + 0.988212i
$$33$$ −120.000 + 134.164i −0.633010 + 0.707726i
$$34$$ 80.0000 61.9677i 0.403526 0.312570i
$$35$$ −69.2820 −0.334594
$$36$$ 213.846 + 30.4277i 0.990028 + 0.140869i
$$37$$ −130.000 −0.577618 −0.288809 0.957387i $$-0.593259\pi$$
−0.288809 + 0.957387i $$0.593259\pi$$
$$38$$ −155.885 + 120.748i −0.665469 + 0.515470i
$$39$$ −34.6410 + 38.7298i −0.142231 + 0.159019i
$$40$$ 80.0000 + 185.903i 0.316228 + 0.734847i
$$41$$ 125.220i 0.476977i −0.971145 0.238488i $$-0.923348\pi$$
0.971145 0.238488i $$-0.0766519\pi$$
$$42$$ 8.03848 113.558i 0.0295325 0.417199i
$$43$$ 224.633i 0.796656i −0.917243 0.398328i $$-0.869591\pi$$
0.917243 0.398328i $$-0.130409\pi$$
$$44$$ 69.2820 268.328i 0.237379 0.919363i
$$45$$ 240.000 26.8328i 0.795046 0.0888889i
$$46$$ −168.000 216.887i −0.538484 0.695179i
$$47$$ 193.990 0.602049 0.301025 0.953616i $$-0.402671\pi$$
0.301025 + 0.953616i $$0.402671\pi$$
$$48$$ −313.990 + 109.556i −0.944177 + 0.329438i
$$49$$ 283.000 0.825073
$$50$$ −77.9423 100.623i −0.220454 0.284605i
$$51$$ 138.564 + 123.935i 0.380448 + 0.340283i
$$52$$ 20.0000 77.4597i 0.0533366 0.206572i
$$53$$ 545.601i 1.41404i 0.707195 + 0.707019i $$0.249960\pi$$
−0.707195 + 0.707019i $$0.750040\pi$$
$$54$$ 16.1347 + 396.489i 0.0406602 + 0.999173i
$$55$$ 309.839i 0.759612i
$$56$$ 69.2820 + 160.997i 0.165325 + 0.384181i
$$57$$ −270.000 241.495i −0.627410 0.561173i
$$58$$ −340.000 + 263.363i −0.769727 + 0.596228i
$$59$$ −173.205 −0.382193 −0.191096 0.981571i $$-0.561204\pi$$
−0.191096 + 0.981571i $$0.561204\pi$$
$$60$$ −309.282 + 206.360i −0.665469 + 0.444017i
$$61$$ −442.000 −0.927743 −0.463871 0.885903i $$-0.653540\pi$$
−0.463871 + 0.885903i $$0.653540\pi$$
$$62$$ 502.295 389.076i 1.02890 0.796979i
$$63$$ 207.846 23.2379i 0.415653 0.0464714i
$$64$$ 352.000 371.806i 0.687500 0.726184i
$$65$$ 89.4427i 0.170677i
$$66$$ 507.846 + 35.9492i 0.947144 + 0.0670460i
$$67$$ 735.867i 1.34180i 0.741549 + 0.670899i $$0.234092\pi$$
−0.741549 + 0.670899i $$0.765908\pi$$
$$68$$ −277.128 71.5542i −0.494217 0.127606i
$$69$$ 336.000 375.659i 0.586227 0.655421i
$$70$$ 120.000 + 154.919i 0.204896 + 0.264520i
$$71$$ −1039.23 −1.73710 −0.868549 0.495603i $$-0.834947\pi$$
−0.868549 + 0.495603i $$0.834947\pi$$
$$72$$ −302.354 530.877i −0.494899 0.868950i
$$73$$ 410.000 0.657354 0.328677 0.944442i $$-0.393397\pi$$
0.328677 + 0.944442i $$0.393397\pi$$
$$74$$ 225.167 + 290.689i 0.353717 + 0.456647i
$$75$$ 155.885 174.284i 0.240000 0.268328i
$$76$$ 540.000 + 139.427i 0.815030 + 0.210440i
$$77$$ 268.328i 0.397128i
$$78$$ 146.603 + 10.3776i 0.212814 + 0.0150646i
$$79$$ 85.2056i 0.121347i −0.998158 0.0606733i $$-0.980675\pi$$
0.998158 0.0606733i $$-0.0193248\pi$$
$$80$$ 277.128 500.879i 0.387298 0.700000i
$$81$$ −711.000 + 160.997i −0.975309 + 0.220846i
$$82$$ −280.000 + 216.887i −0.377083 + 0.292087i
$$83$$ 1254.00 1.65837 0.829186 0.558973i $$-0.188804\pi$$
0.829186 + 0.558973i $$0.188804\pi$$
$$84$$ −267.846 + 178.713i −0.347910 + 0.232134i
$$85$$ −320.000 −0.408340
$$86$$ −502.295 + 389.076i −0.629812 + 0.487850i
$$87$$ −588.897 526.726i −0.725706 0.649091i
$$88$$ −720.000 + 309.839i −0.872185 + 0.375329i
$$89$$ 840.762i 1.00135i 0.865634 + 0.500677i $$0.166916\pi$$
−0.865634 + 0.500677i $$0.833084\pi$$
$$90$$ −475.692 490.181i −0.557137 0.574106i
$$91$$ 77.4597i 0.0892305i
$$92$$ −193.990 + 751.319i −0.219835 + 0.851417i
$$93$$ 870.000 + 778.152i 0.970052 + 0.867641i
$$94$$ −336.000 433.774i −0.368678 0.475962i
$$95$$ 623.538 0.673407
$$96$$ 788.820 + 512.346i 0.838632 + 0.544699i
$$97$$ 770.000 0.805996 0.402998 0.915201i $$-0.367968\pi$$
0.402998 + 0.915201i $$0.367968\pi$$
$$98$$ −490.170 632.807i −0.505252 0.652277i
$$99$$ 103.923 + 929.516i 0.105502 + 0.943635i
$$100$$ −90.0000 + 348.569i −0.0900000 + 0.348569i
$$101$$ 1493.69i 1.47156i −0.677218 0.735782i $$-0.736815\pi$$
0.677218 0.735782i $$-0.263185\pi$$
$$102$$ 37.1281 524.501i 0.0360415 0.509151i
$$103$$ 1355.54i 1.29675i −0.761319 0.648377i $$-0.775448\pi$$
0.761319 0.648377i $$-0.224552\pi$$
$$104$$ −207.846 + 89.4427i −0.195971 + 0.0843325i
$$105$$ −240.000 + 268.328i −0.223063 + 0.249392i
$$106$$ 1220.00 945.008i 1.11790 0.865918i
$$107$$ −644.323 −0.582141 −0.291070 0.956702i $$-0.594011\pi$$
−0.291070 + 0.956702i $$0.594011\pi$$
$$108$$ 858.631 722.818i 0.765016 0.644011i
$$109$$ −1066.00 −0.936737 −0.468368 0.883533i $$-0.655158\pi$$
−0.468368 + 0.883533i $$0.655158\pi$$
$$110$$ −692.820 + 536.656i −0.600526 + 0.465165i
$$111$$ −450.333 + 503.488i −0.385079 + 0.430531i
$$112$$ 240.000 433.774i 0.202481 0.365963i
$$113$$ 1037.54i 0.863745i −0.901935 0.431872i $$-0.857853\pi$$
0.901935 0.431872i $$-0.142147\pi$$
$$114$$ −72.3463 + 1022.02i −0.0594373 + 0.839658i
$$115$$ 867.548i 0.703472i
$$116$$ 1177.79 + 304.105i 0.942720 + 0.243409i
$$117$$ 30.0000 + 268.328i 0.0237051 + 0.212025i
$$118$$ 300.000 + 387.298i 0.234044 + 0.302150i
$$119$$ −277.128 −0.213481
$$120$$ 997.128 + 334.149i 0.758541 + 0.254196i
$$121$$ −131.000 −0.0984222
$$122$$ 765.566 + 988.342i 0.568124 + 0.733445i
$$123$$ −484.974 433.774i −0.355518 0.317985i
$$124$$ −1740.00 449.266i −1.26013 0.325365i
$$125$$ 1520.53i 1.08800i
$$126$$ −411.962 424.509i −0.291273 0.300145i
$$127$$ 1835.79i 1.28268i 0.767257 + 0.641340i $$0.221621\pi$$
−0.767257 + 0.641340i $$0.778379\pi$$
$$128$$ −1441.07 143.108i −0.995105 0.0988212i
$$129$$ −870.000 778.152i −0.593792 0.531104i
$$130$$ −200.000 + 154.919i −0.134932 + 0.104518i
$$131$$ 450.333 0.300350 0.150175 0.988659i $$-0.452016\pi$$
0.150175 + 0.988659i $$0.452016\pi$$
$$132$$ −799.230 1197.84i −0.527001 0.789841i
$$133$$ 540.000 0.352060
$$134$$ 1645.45 1274.56i 1.06078 0.821680i
$$135$$ 727.461 1022.47i 0.463777 0.651852i
$$136$$ 320.000 + 743.613i 0.201763 + 0.468855i
$$137$$ 89.4427i 0.0557782i 0.999611 + 0.0278891i $$0.00887852\pi$$
−0.999611 + 0.0278891i $$0.991121\pi$$
$$138$$ −1421.97 100.658i −0.877145 0.0620909i
$$139$$ 1959.73i 1.19584i −0.801555 0.597921i $$-0.795994\pi$$
0.801555 0.597921i $$-0.204006\pi$$
$$140$$ 138.564 536.656i 0.0836486 0.323970i
$$141$$ 672.000 751.319i 0.401366 0.448741i
$$142$$ 1800.00 + 2323.79i 1.06375 + 1.37330i
$$143$$ 346.410 0.202575
$$144$$ −663.384 + 1595.59i −0.383903 + 0.923373i
$$145$$ 1360.00 0.778909
$$146$$ −710.141 916.788i −0.402546 0.519684i
$$147$$ 980.341 1096.05i 0.550049 0.614973i
$$148$$ 260.000 1006.98i 0.144405 0.559276i
$$149$$ 1618.91i 0.890111i 0.895503 + 0.445055i $$0.146816\pi$$
−0.895503 + 0.445055i $$0.853184\pi$$
$$150$$ −659.711 46.6993i −0.359101 0.0254199i
$$151$$ 565.456i 0.304743i 0.988323 + 0.152371i $$0.0486909\pi$$
−0.988323 + 0.152371i $$0.951309\pi$$
$$152$$ −623.538 1448.97i −0.332734 0.773205i
$$153$$ 960.000 107.331i 0.507264 0.0567138i
$$154$$ −600.000 + 464.758i −0.313957 + 0.243190i
$$155$$ −2009.18 −1.04117
$$156$$ −230.718 345.788i −0.118412 0.177469i
$$157$$ −730.000 −0.371085 −0.185542 0.982636i $$-0.559404\pi$$
−0.185542 + 0.982636i $$0.559404\pi$$
$$158$$ −190.526 + 147.580i −0.0959329 + 0.0743093i
$$159$$ 2113.10 + 1890.02i 1.05396 + 0.942692i
$$160$$ −1600.00 + 247.871i −0.790569 + 0.122474i
$$161$$ 751.319i 0.367778i
$$162$$ 1591.49 + 1310.99i 0.771846 + 0.635809i
$$163$$ 255.617i 0.122831i −0.998112 0.0614155i $$-0.980439\pi$$
0.998112 0.0614155i $$-0.0195615\pi$$
$$164$$ 969.948 + 250.440i 0.461831 + 0.119244i
$$165$$ −1200.00 1073.31i −0.566181 0.506408i
$$166$$ −2172.00 2804.04i −1.01554 1.31106i
$$167$$ 13.8564 0.00642060 0.00321030 0.999995i $$-0.498978\pi$$
0.00321030 + 0.999995i $$0.498978\pi$$
$$168$$ 863.538 + 289.381i 0.396568 + 0.132894i
$$169$$ −2097.00 −0.954483
$$170$$ 554.256 + 715.542i 0.250056 + 0.322821i
$$171$$ −1870.61 + 209.141i −0.836547 + 0.0935288i
$$172$$ 1740.00 + 449.266i 0.771359 + 0.199164i
$$173$$ 1118.03i 0.491344i −0.969353 0.245672i $$-0.920991\pi$$
0.969353 0.245672i $$-0.0790086\pi$$
$$174$$ −157.795 + 2229.13i −0.0687493 + 0.971206i
$$175$$ 348.569i 0.150567i
$$176$$ 1939.90 + 1073.31i 0.830825 + 0.459682i
$$177$$ −600.000 + 670.820i −0.254795 + 0.284870i
$$178$$ 1880.00 1456.24i 0.791640 0.613202i
$$179$$ −1351.00 −0.564125 −0.282063 0.959396i $$-0.591019\pi$$
−0.282063 + 0.959396i $$0.591019\pi$$
$$180$$ −272.154 + 1912.70i −0.112695 + 0.792023i
$$181$$ 1262.00 0.518253 0.259126 0.965843i $$-0.416565\pi$$
0.259126 + 0.965843i $$0.416565\pi$$
$$182$$ −173.205 + 134.164i −0.0705429 + 0.0546423i
$$183$$ −1531.13 + 1711.86i −0.618495 + 0.691499i
$$184$$ 2016.00 867.548i 0.807725 0.347590i
$$185$$ 1162.76i 0.462094i
$$186$$ 233.116 3293.18i 0.0918972 1.29821i
$$187$$ 1239.35i 0.484656i
$$188$$ −387.979 + 1502.64i −0.150512 + 0.582931i
$$189$$ 630.000 885.483i 0.242464 0.340791i
$$190$$ −1080.00 1394.27i −0.412376 0.532375i
$$191$$ 2771.28 1.04986 0.524929 0.851146i $$-0.324092\pi$$
0.524929 + 0.851146i $$0.324092\pi$$
$$192$$ −220.636 2651.27i −0.0829325 0.996555i
$$193$$ −190.000 −0.0708627 −0.0354313 0.999372i $$-0.511281\pi$$
−0.0354313 + 0.999372i $$0.511281\pi$$
$$194$$ −1333.68 1721.77i −0.493570 0.637196i
$$195$$ −346.410 309.839i −0.127215 0.113785i
$$196$$ −566.000 + 2192.11i −0.206268 + 0.798873i
$$197$$ 2137.68i 0.773114i −0.922266 0.386557i $$-0.873664\pi$$
0.922266 0.386557i $$-0.126336\pi$$
$$198$$ 1898.46 1842.35i 0.681403 0.661262i
$$199$$ 255.617i 0.0910563i 0.998963 + 0.0455281i $$0.0144971\pi$$
−0.998963 + 0.0455281i $$0.985503\pi$$
$$200$$ 935.307 402.492i 0.330681 0.142302i
$$201$$ 2850.00 + 2549.12i 1.00012 + 0.894532i
$$202$$ −3340.00 + 2587.15i −1.16337 + 0.901146i
$$203$$ 1177.79 0.407217
$$204$$ −1237.13 + 825.442i −0.424590 + 0.283296i
$$205$$ 1120.00 0.381581
$$206$$ −3031.09 + 2347.87i −1.02517 + 0.794097i
$$207$$ −290.985 2602.64i −0.0977045 0.873895i
$$208$$ 560.000 + 309.839i 0.186678 + 0.103286i
$$209$$ 2414.95i 0.799262i
$$210$$ 1015.69 + 71.8983i 0.333759 + 0.0236260i
$$211$$ 549.964i 0.179436i 0.995967 + 0.0897181i $$0.0285966\pi$$
−0.995967 + 0.0897181i $$0.971403\pi$$
$$212$$ −4226.20 1091.20i −1.36914 0.353509i
$$213$$ −3600.00 + 4024.92i −1.15807 + 1.29476i
$$214$$ 1116.00 + 1440.75i 0.356487 + 0.460223i
$$215$$ 2009.18 0.637325
$$216$$ −3103.46 668.000i −0.977610 0.210424i
$$217$$ −1740.00 −0.544327
$$218$$ 1846.37 + 2383.65i 0.573632 + 0.740555i
$$219$$ 1420.28 1587.92i 0.438236 0.489963i
$$220$$ 2400.00 + 619.677i 0.735491 + 0.189903i
$$221$$ 357.771i 0.108897i
$$222$$ 1905.83 + 134.909i 0.576176 + 0.0407861i
$$223$$ 472.504i 0.141889i 0.997480 + 0.0709444i $$0.0226013\pi$$
−0.997480 + 0.0709444i $$0.977399\pi$$
$$224$$ −1385.64 + 214.663i −0.413313 + 0.0640301i
$$225$$ −135.000 1207.48i −0.0400000 0.357771i
$$226$$ −2320.00 + 1797.06i −0.682850 + 0.528933i
$$227$$ 505.759 0.147878 0.0739392 0.997263i $$-0.476443\pi$$
0.0739392 + 0.997263i $$0.476443\pi$$
$$228$$ 2410.61 1608.42i 0.700206 0.467194i
$$229$$ 4094.00 1.18139 0.590697 0.806894i $$-0.298853\pi$$
0.590697 + 0.806894i $$0.298853\pi$$
$$230$$ 1939.90 1502.64i 0.556144 0.430787i
$$231$$ −1039.23 929.516i −0.296001 0.264752i
$$232$$ −1360.00 3160.35i −0.384864 0.894342i
$$233$$ 5277.12i 1.48376i −0.670534 0.741879i $$-0.733935\pi$$
0.670534 0.741879i $$-0.266065\pi$$
$$234$$ 548.038 531.840i 0.153104 0.148579i
$$235$$ 1735.10i 0.481639i
$$236$$ 346.410 1341.64i 0.0955482 0.370057i
$$237$$ −330.000 295.161i −0.0904464 0.0808977i
$$238$$ 480.000 + 619.677i 0.130730 + 0.168772i
$$239$$ −5681.13 −1.53758 −0.768790 0.639502i $$-0.779141\pi$$
−0.768790 + 0.639502i $$0.779141\pi$$
$$240$$ −979.897 2808.41i −0.263550 0.755342i
$$241$$ −1198.00 −0.320207 −0.160104 0.987100i $$-0.551183\pi$$
−0.160104 + 0.987100i $$0.551183\pi$$
$$242$$ 226.899 + 292.925i 0.0602711 + 0.0778096i
$$243$$ −1839.44 + 3311.40i −0.485597 + 0.874183i
$$244$$ 884.000 3423.72i 0.231936 0.898283i
$$245$$ 2531.23i 0.660058i
$$246$$ −129.948 + 1835.75i −0.0336797 + 0.475786i
$$247$$ 697.137i 0.179586i
$$248$$ 2009.18 + 4668.91i 0.514448 + 1.19547i
$$249$$ 4344.00 4856.74i 1.10558 1.23608i
$$250$$ 3400.00 2633.63i 0.860140 0.666261i
$$251$$ 4260.84 1.07148 0.535741 0.844382i $$-0.320032\pi$$
0.535741 + 0.844382i $$0.320032\pi$$
$$252$$ −235.692 + 1656.44i −0.0589175 + 0.414072i
$$253$$ −3360.00 −0.834946
$$254$$ 4104.96 3179.69i 1.01405 0.785478i
$$255$$ −1108.51 + 1239.35i −0.272226 + 0.304358i
$$256$$ 2176.00 + 3470.19i 0.531250 + 0.847215i
$$257$$ 3148.38i 0.764166i 0.924128 + 0.382083i $$0.124793\pi$$
−0.924128 + 0.382083i $$0.875207\pi$$
$$258$$ −233.116 + 3293.18i −0.0562525 + 0.794668i
$$259$$ 1006.98i 0.241585i
$$260$$ 692.820 + 178.885i 0.165257 + 0.0426692i
$$261$$ −4080.00 + 456.158i −0.967608 + 0.108182i
$$262$$ −780.000 1006.98i −0.183926 0.237447i
$$263$$ −4253.92 −0.997368 −0.498684 0.866784i $$-0.666183\pi$$
−0.498684 + 0.866784i $$0.666183\pi$$
$$264$$ −1294.15 + 3861.86i −0.301703 + 0.900307i
$$265$$ −4880.00 −1.13123
$$266$$ −935.307 1207.48i −0.215592 0.278328i
$$267$$ 3256.26 + 2912.48i 0.746366 + 0.667570i
$$268$$ −5700.00 1471.73i −1.29919 0.335449i
$$269$$ 44.7214i 0.0101365i −0.999987 0.00506823i $$-0.998387\pi$$
0.999987 0.00506823i $$-0.00161328\pi$$
$$270$$ −3546.31 + 144.313i −0.799338 + 0.0325281i
$$271$$ 8760.69i 1.96374i −0.189552 0.981871i $$-0.560704\pi$$
0.189552 0.981871i $$-0.439296\pi$$
$$272$$ 1108.51 2003.52i 0.247108 0.446622i
$$273$$ −300.000 268.328i −0.0665085 0.0594870i
$$274$$ 200.000 154.919i 0.0440965 0.0341570i
$$275$$ −1558.85 −0.341825
$$276$$ 2237.85 + 3353.96i 0.488052 + 0.731467i
$$277$$ 6350.00 1.37738 0.688690 0.725055i $$-0.258186\pi$$
0.688690 + 0.725055i $$0.258186\pi$$
$$278$$ −4382.09 + 3394.35i −0.945396 + 0.732301i
$$279$$ 6027.54 673.899i 1.29340 0.144607i
$$280$$ −1440.00 + 619.677i −0.307344 + 0.132260i
$$281$$ 5563.34i 1.18107i 0.807012 + 0.590535i $$0.201083\pi$$
−0.807012 + 0.590535i $$0.798917\pi$$
$$282$$ −2843.94 201.315i −0.600546 0.0425112i
$$283$$ 6777.72i 1.42365i 0.702356 + 0.711826i $$0.252132\pi$$
−0.702356 + 0.711826i $$0.747868\pi$$
$$284$$ 2078.46 8049.84i 0.434275 1.68194i
$$285$$ 2160.00 2414.95i 0.448938 0.501928i
$$286$$ −600.000 774.597i −0.124052 0.160150i
$$287$$ 969.948 0.199492
$$288$$ 4716.86 1280.27i 0.965082 0.261946i
$$289$$ 3633.00 0.739467
$$290$$ −2355.59 3041.05i −0.476983 0.615782i
$$291$$ 2667.36 2982.20i 0.537331 0.600754i
$$292$$ −820.000 + 3175.85i −0.164339 + 0.636481i
$$293$$ 652.932i 0.130187i 0.997879 + 0.0650933i $$0.0207345\pi$$
−0.997879 + 0.0650933i $$0.979265\pi$$
$$294$$ −4148.85 293.687i −0.823013 0.0582591i
$$295$$ 1549.19i 0.305754i
$$296$$ −2702.00 + 1162.76i −0.530576 + 0.228324i
$$297$$ 3960.00 + 2817.45i 0.773678 + 0.550454i
$$298$$ 3620.00 2804.04i 0.703695 0.545079i
$$299$$ −969.948 −0.187604
$$300$$ 1038.23 + 1556.05i 0.199808 + 0.299461i
$$301$$ 1740.00 0.333196
$$302$$ 1264.40 979.398i 0.240920 0.186616i
$$303$$ −5785.05 5174.31i −1.09684 0.981043i
$$304$$ −2160.00 + 3903.97i −0.407515 + 0.736539i
$$305$$ 3953.37i 0.742194i
$$306$$ −1902.77 1960.72i −0.355471 0.366297i
$$307$$ 1556.94i 0.289444i −0.989472 0.144722i $$-0.953771\pi$$
0.989472 0.144722i $$-0.0462287\pi$$
$$308$$ 2078.46 + 536.656i 0.384517 + 0.0992819i
$$309$$ −5250.00 4695.74i −0.966544 0.864503i
$$310$$ 3480.00 + 4492.66i 0.637583 + 0.823116i
$$311$$ 3256.26 0.593715 0.296857 0.954922i $$-0.404061\pi$$
0.296857 + 0.954922i $$0.404061\pi$$
$$312$$ −373.590 + 1114.82i −0.0677896 + 0.202290i
$$313$$ −7030.00 −1.26952 −0.634759 0.772710i $$-0.718901\pi$$
−0.634759 + 0.772710i $$0.718901\pi$$
$$314$$ 1264.40 + 1632.33i 0.227242 + 0.293368i
$$315$$ 207.846 + 1859.03i 0.0371771 + 0.332522i
$$316$$ 660.000 + 170.411i 0.117493 + 0.0303367i
$$317$$ 491.935i 0.0871603i 0.999050 + 0.0435802i $$0.0138764\pi$$
−0.999050 + 0.0435802i $$0.986124\pi$$
$$318$$ 566.204 7998.64i 0.0998464 1.41051i
$$319$$ 5267.26i 0.924482i
$$320$$ 3325.54 + 3148.38i 0.580948 + 0.550000i
$$321$$ −2232.00 + 2495.45i −0.388094 + 0.433902i
$$322$$ 1680.00 1301.32i 0.290754 0.225217i
$$323$$ 2494.15 0.429654
$$324$$ 174.923 5829.38i 0.0299937 0.999550i
$$325$$ −450.000 −0.0768046
$$326$$ −571.577 + 442.741i −0.0971065 + 0.0752183i
$$327$$ −3692.73 + 4128.60i −0.624491 + 0.698202i
$$328$$ −1120.00 2602.64i −0.188542 0.438131i
$$329$$ 1502.64i 0.251803i
$$330$$ −321.539 + 4542.31i −0.0536368 + 0.757716i
$$331$$ 4237.04i 0.703592i 0.936077 + 0.351796i $$0.114429\pi$$
−0.936077 + 0.351796i $$0.885571\pi$$
$$332$$ −2508.01 + 9713.48i −0.414593 + 1.60571i
$$333$$ 390.000 + 3488.27i 0.0641798 + 0.574041i
$$334$$ −24.0000 30.9839i −0.00393180 0.00507593i
$$335$$ −6581.79 −1.07344
$$336$$ −848.616 2432.15i −0.137785 0.394895i
$$337$$ 1490.00 0.240847 0.120424 0.992723i $$-0.461575\pi$$
0.120424 + 0.992723i $$0.461575\pi$$
$$338$$ 3632.11 + 4689.03i 0.584499 + 0.754585i
$$339$$ −4018.36 3594.13i −0.643797 0.575830i
$$340$$ 640.000 2478.71i 0.102085 0.395373i
$$341$$ 7781.52i 1.23576i
$$342$$ 3707.65 + 3820.58i 0.586219 + 0.604074i
$$343$$ 4848.98i 0.763324i
$$344$$ −2009.18 4668.91i −0.314906 0.731775i
$$345$$ 3360.00 + 3005.28i 0.524337 + 0.468981i
$$346$$ −2500.00 + 1936.49i −0.388442 + 0.300886i
$$347$$ 1988.39 0.307616 0.153808 0.988101i $$-0.450846\pi$$
0.153808 + 0.988101i $$0.450846\pi$$
$$348$$ 5257.79 3508.13i 0.809906 0.540389i
$$349$$ −2074.00 −0.318105 −0.159053 0.987270i $$-0.550844\pi$$
−0.159053 + 0.987270i $$0.550844\pi$$
$$350$$ 779.423 603.738i 0.119034 0.0922033i
$$351$$ 1143.15 + 813.327i 0.173838 + 0.123681i
$$352$$ −960.000 6196.77i −0.145364 0.938321i
$$353$$ 8658.06i 1.30544i −0.757597 0.652722i $$-0.773627\pi$$
0.757597 0.652722i $$-0.226373\pi$$
$$354$$ 2539.23 + 179.746i 0.381239 + 0.0269870i
$$355$$ 9295.16i 1.38968i
$$356$$ −6512.51 1681.52i −0.969557 0.250339i
$$357$$ −960.000 + 1073.31i −0.142321 + 0.159120i
$$358$$ 2340.00 + 3020.93i 0.345455 + 0.445980i
$$359$$ 8106.00 1.19169 0.595847 0.803098i $$-0.296816\pi$$
0.595847 + 0.803098i $$0.296816\pi$$
$$360$$ 4748.31 2704.33i 0.695160 0.395919i
$$361$$ 1999.00 0.291442
$$362$$ −2185.85 2821.92i −0.317364 0.409715i
$$363$$ −453.797 + 507.361i −0.0656148 + 0.0733596i
$$364$$ 600.000 + 154.919i 0.0863971 + 0.0223076i
$$365$$ 3667.15i 0.525884i
$$366$$ 6479.83 + 458.691i 0.925427 + 0.0655087i
$$367$$ 7893.14i 1.12267i −0.827590 0.561333i $$-0.810289\pi$$
0.827590 0.561333i $$-0.189711\pi$$
$$368$$ −5431.71 3005.28i −0.769423 0.425709i
$$369$$ −3360.00 + 375.659i −0.474023 + 0.0529974i
$$370$$ −2600.00 + 2013.95i −0.365318 + 0.282974i
$$371$$ −4226.20 −0.591411
$$372$$ −7767.54 + 5182.69i −1.08260 + 0.722338i
$$373$$ 4910.00 0.681582 0.340791 0.940139i $$-0.389305\pi$$
0.340791 + 0.940139i $$0.389305\pi$$
$$374$$ −2771.28 + 2146.63i −0.383154 + 0.296790i
$$375$$ 5888.97 + 5267.26i 0.810947 + 0.725333i
$$376$$ 4032.00 1735.10i 0.553017 0.237981i
$$377$$ 1520.53i 0.207722i
$$378$$ −3071.19 + 124.979i −0.417897 + 0.0170058i
$$379$$ 3137.12i 0.425179i −0.977142 0.212590i $$-0.931810\pi$$
0.977142 0.212590i $$-0.0681897\pi$$
$$380$$ −1247.08 + 4829.91i −0.168352 + 0.652024i
$$381$$ 7110.00 + 6359.38i 0.956053 + 0.855120i
$$382$$ −4800.00 6196.77i −0.642904 0.829986i
$$383$$ 6207.67 0.828191 0.414095 0.910233i $$-0.364098\pi$$
0.414095 + 0.910233i $$0.364098\pi$$
$$384$$ −5546.26 + 5085.48i −0.737060 + 0.675827i
$$385$$ 2400.00 0.317702
$$386$$ 329.090 + 424.853i 0.0433944 + 0.0560219i
$$387$$ −6027.54 + 673.899i −0.791723 + 0.0885174i
$$388$$ −1540.00 + 5964.39i −0.201499 + 0.780403i
$$389$$ 9454.10i 1.23224i 0.787652 + 0.616120i $$0.211297\pi$$
−0.787652 + 0.616120i $$0.788703\pi$$
$$390$$ −92.8203 + 1311.25i −0.0120516 + 0.170251i
$$391$$ 3470.19i 0.448837i
$$392$$ 5882.04 2531.23i 0.757878 0.326139i
$$393$$ 1560.00 1744.13i 0.200233 0.223867i
$$394$$ −4780.00 + 3702.57i −0.611200 + 0.473434i
$$395$$ 762.102 0.0970773
$$396$$ −7407.85 1054.05i −0.940046 0.133757i
$$397$$ −10570.0 −1.33625 −0.668127 0.744047i $$-0.732904\pi$$
−0.668127 + 0.744047i $$0.732904\pi$$
$$398$$ 571.577 442.741i 0.0719863 0.0557604i
$$399$$ 1870.61 2091.41i 0.234706 0.262410i
$$400$$ −2520.00 1394.27i −0.315000 0.174284i
$$401$$ 1681.52i 0.209405i −0.994504 0.104702i $$-0.966611\pi$$
0.994504 0.104702i $$-0.0333890\pi$$
$$402$$ 763.655 10788.0i 0.0947454 1.33845i
$$403$$ 2246.33i 0.277662i
$$404$$ 11570.1 + 2987.39i 1.42484 + 0.367891i
$$405$$ −1440.00 6359.38i −0.176677 0.780247i
$$406$$ −2040.00 2633.63i −0.249368 0.321933i
$$407$$ 4503.33 0.548457
$$408$$ 3988.51 + 1336.60i 0.483973 + 0.162185i
$$409$$ −3574.00 −0.432085 −0.216043 0.976384i $$-0.569315\pi$$
−0.216043 + 0.976384i $$0.569315\pi$$
$$410$$ −1939.90 2504.40i −0.233670 0.301667i
$$411$$ 346.410 + 309.839i 0.0415746 + 0.0371854i
$$412$$ 10500.0 + 2711.09i 1.25558 + 0.324189i
$$413$$ 1341.64i 0.159849i
$$414$$ −5315.69 + 5158.57i −0.631043 + 0.612392i
$$415$$ 11216.2i 1.32670i
$$416$$ −277.128 1788.85i −0.0326618 0.210831i
$$417$$ −7590.00 6788.70i −0.891328 0.797228i
$$418$$ 5400.00 4182.82i 0.631872 0.489446i
$$419$$ −15346.0 −1.78926 −0.894630 0.446808i $$-0.852561\pi$$
−0.894630 + 0.446808i $$0.852561\pi$$
$$420$$ −1598.46 2395.69i −0.185707 0.278328i
$$421$$ 3518.00 0.407261 0.203630 0.979048i $$-0.434726\pi$$
0.203630 + 0.979048i $$0.434726\pi$$
$$422$$ 1229.76 952.565i 0.141857 0.109882i
$$423$$ −581.969 5205.29i −0.0668943 0.598321i
$$424$$ 4880.00 + 11340.1i 0.558948 + 1.29888i
$$425$$ 1609.97i 0.183753i
$$426$$ 15235.4 + 1078.47i 1.73276 + 0.122658i
$$427$$ 3423.72i 0.388022i
$$428$$ 1288.65 4990.90i 0.145535 0.563655i
$$429$$ 1200.00 1341.64i 0.135050 0.150991i
$$430$$ −3480.00 4492.66i −0.390280 0.503850i
$$431$$ −12886.5 −1.44018 −0.720091 0.693879i $$-0.755900\pi$$
−0.720091 + 0.693879i $$0.755900\pi$$
$$432$$ 3881.66 + 8096.56i 0.432307 + 0.901727i
$$433$$ 14450.0 1.60375 0.801874 0.597493i $$-0.203837\pi$$
0.801874 + 0.597493i $$0.203837\pi$$
$$434$$ 3013.77 + 3890.76i 0.333331 + 0.430328i
$$435$$ 4711.18 5267.26i 0.519273 0.580565i
$$436$$ 2132.00 8257.20i 0.234184 0.906991i
$$437$$ 6761.87i 0.740192i
$$438$$ −6010.70 425.483i −0.655714 0.0464163i
$$439$$ 15065.9i 1.63794i 0.573835 + 0.818971i $$0.305455\pi$$
−0.573835 + 0.818971i $$0.694545\pi$$
$$440$$ −2771.28 6439.88i −0.300263 0.697748i
$$441$$ −849.000 7593.69i −0.0916748 0.819964i
$$442$$ −800.000 + 619.677i −0.0860908 + 0.0666856i
$$443$$ 3041.48 0.326197 0.163098 0.986610i $$-0.447851\pi$$
0.163098 + 0.986610i $$0.447851\pi$$
$$444$$ −2999.33 4495.24i −0.320590 0.480484i
$$445$$ −7520.00 −0.801084
$$446$$ 1056.55 818.401i 0.112173 0.0868888i
$$447$$ 6270.02 + 5608.08i 0.663450 + 0.593407i
$$448$$ 2880.00 + 2726.58i 0.303721 + 0.287542i
$$449$$ 14310.8i 1.50416i 0.659069 + 0.752082i $$0.270951\pi$$
−0.659069 + 0.752082i $$0.729049\pi$$
$$450$$ −2466.17 + 2393.28i −0.258348 + 0.250712i
$$451$$ 4337.74i 0.452896i
$$452$$ 8036.72 + 2075.07i 0.836317 + 0.215936i
$$453$$ 2190.00 + 1958.80i 0.227142 + 0.203162i
$$454$$ −876.000 1130.91i −0.0905566 0.116908i
$$455$$ 692.820 0.0713844
$$456$$ −7771.84 2604.43i −0.798136 0.267464i
$$457$$ −3430.00 −0.351091 −0.175546 0.984471i $$-0.556169\pi$$
−0.175546 + 0.984471i $$0.556169\pi$$
$$458$$ −7091.02 9154.46i −0.723453 0.933974i
$$459$$ 2909.85 4089.87i 0.295904 0.415902i
$$460$$ −6720.00 1735.10i −0.681134 0.175868i
$$461$$ 3908.65i 0.394889i −0.980314 0.197445i $$-0.936736\pi$$
0.980314 0.197445i $$-0.0632642\pi$$
$$462$$ −278.461 + 3933.76i −0.0280415 + 0.396136i
$$463$$ 18179.8i 1.82481i −0.409291 0.912404i $$-0.634224\pi$$
0.409291 0.912404i $$-0.365776\pi$$
$$464$$ −4711.18 + 8514.95i −0.471360 + 0.851932i
$$465$$ −6960.00 + 7781.52i −0.694112 + 0.776041i
$$466$$ −11800.0 + 9140.24i −1.17301 + 0.908613i
$$467$$ −1849.83 −0.183298 −0.0916488 0.995791i $$-0.529214\pi$$
−0.0916488 + 0.995791i $$0.529214\pi$$
$$468$$ −2138.46 304.277i −0.211219 0.0300539i
$$469$$ −5700.00 −0.561197
$$470$$ 3879.79 3005.28i 0.380769 0.294943i
$$471$$ −2528.79 + 2827.28i −0.247390 + 0.276590i
$$472$$ −3600.00 + 1549.19i −0.351067 + 0.151075i
$$473$$ 7781.52i 0.756437i
$$474$$ −88.4232 + 1249.14i −0.00856838 + 0.121044i
$$475$$ 3137.12i 0.303033i
$$476$$ 554.256 2146.63i 0.0533704 0.206703i
$$477$$ 14640.0 1636.80i 1.40528 0.157115i
$$478$$ 9840.00 + 12703.4i 0.941571 + 1.21556i
$$479$$ 15242.0 1.45392 0.726959 0.686681i $$-0.240933\pi$$
0.726959 + 0.686681i $$0.240933\pi$$
$$480$$ −4582.56 + 7055.42i −0.435759 + 0.670905i
$$481$$ 1300.00 0.123233
$$482$$ 2075.00 + 2678.81i 0.196086 + 0.253146i
$$483$$ 2909.85 + 2602.64i 0.274125 + 0.245185i
$$484$$ 262.000 1014.72i 0.0246056 0.0952969i
$$485$$ 6887.09i 0.644797i
$$486$$ 10590.5 1622.41i 0.988468 0.151428i
$$487$$ 9783.16i 0.910302i −0.890414 0.455151i $$-0.849585\pi$$
0.890414 0.455151i $$-0.150415\pi$$
$$488$$ −9186.80 + 3953.37i −0.852186 + 0.366722i
$$489$$ −990.000 885.483i −0.0915529 0.0818874i
$$490$$ 5660.00 4384.22i 0.521822 0.404202i
$$491$$ 6893.56 0.633609 0.316805 0.948491i $$-0.397390\pi$$
0.316805 + 0.948491i $$0.397390\pi$$
$$492$$ 4329.95 2889.05i 0.396767 0.264732i
$$493$$ 5440.00 0.496968
$$494$$ 1558.85 1207.48i 0.141975 0.109974i
$$495$$ −8313.84 + 929.516i −0.754908 + 0.0844013i
$$496$$ 6960.00 12579.4i 0.630067 1.13878i
$$497$$ 8049.84i 0.726529i
$$498$$ −18384.0 1301.36i −1.65423 0.117099i
$$499$$ 1309.07i 0.117439i −0.998275 0.0587194i $$-0.981298\pi$$
0.998275 0.0587194i $$-0.0187017\pi$$
$$500$$ −11777.9 3041.05i −1.05345 0.272000i
$$501$$ 48.0000 53.6656i 0.00428040 0.00478564i
$$502$$ −7380.00 9527.54i −0.656146 0.847081i
$$503$$ −7939.72 −0.703806 −0.351903 0.936036i $$-0.614465\pi$$
−0.351903 + 0.936036i $$0.614465\pi$$
$$504$$ 4112.15 2342.02i 0.363432 0.206988i
$$505$$ 13360.0 1.17725
$$506$$ 5819.69 + 7513.19i 0.511298 + 0.660083i
$$507$$ −7264.22 + 8121.65i −0.636322 + 0.711430i
$$508$$ −14220.0 3671.59i −1.24195 0.320670i
$$509$$ 14534.4i 1.26567i −0.774285 0.632837i $$-0.781890\pi$$
0.774285 0.632837i $$-0.218110\pi$$
$$510$$ 4691.28 + 332.084i 0.407320 + 0.0288332i
$$511$$ 3175.85i 0.274934i
$$512$$ 3990.65 10876.2i 0.344459 0.938801i
$$513$$ −5670.00 + 7969.35i −0.487986 + 0.685878i
$$514$$ 7040.00 5453.16i 0.604127 0.467954i
$$515$$ 12124.4 1.03740
$$516$$ 7767.54 5182.69i 0.662687 0.442161i
$$517$$ −6720.00 −0.571654
$$518$$ −2251.67 + 1744.13i −0.190989 + 0.147940i
$$519$$ −4330.13 3872.98i −0.366226 0.327563i
$$520$$ −800.000 1859.03i −0.0674660 0.156777i
$$521$$ 9355.71i 0.786720i −0.919385 0.393360i $$-0.871313\pi$$
0.919385 0.393360i $$-0.128687\pi$$
$$522$$ 8086.77 + 8333.07i 0.678062 + 0.698714i
$$523$$ 10062.0i 0.841264i 0.907231 + 0.420632i $$0.138192\pi$$
−0.907231 + 0.420632i $$0.861808\pi$$
$$524$$ −900.666 + 3488.27i −0.0750874 + 0.290812i
$$525$$ 1350.00 + 1207.48i 0.112226 + 0.100378i
$$526$$ 7368.00 + 9512.05i 0.610761 + 0.788489i
$$527$$ −8036.72 −0.664298
$$528$$ 10876.9 3795.12i 0.896510 0.312806i
$$529$$ −2759.00 −0.226761
$$530$$ 8452.41 + 10912.0i 0.692734 + 0.894316i
$$531$$ 519.615 + 4647.58i 0.0424659 + 0.379826i
$$532$$ −1080.00 + 4182.82i −0.0880149 + 0.340880i
$$533$$ 1252.20i 0.101761i
$$534$$ 872.511 12325.8i 0.0707065 0.998855i
$$535$$ 5763.00i 0.465712i
$$536$$ 6581.79 + 15294.7i 0.530392 + 1.23252i
$$537$$ −4680.00 + 5232.40i −0.376084 + 0.420474i
$$538$$ −100.000 + 77.4597i −0.00801358 + 0.00620729i
$$539$$ −9803.41 −0.783419
$$540$$ 6465.08 + 7679.83i 0.515209 + 0.612013i
$$541$$ −23962.0 −1.90426 −0.952132 0.305687i $$-0.901114\pi$$
−0.952132 + 0.305687i $$0.901114\pi$$
$$542$$ −19589.5 + 15174.0i −1.55247 + 1.20254i
$$543$$ 4371.70 4887.70i 0.345502 0.386283i
$$544$$ −6400.00 + 991.484i −0.504408 + 0.0781425i
$$545$$ 9534.59i 0.749389i
$$546$$ −80.3848 + 1135.58i −0.00630064 + 0.0890078i
$$547$$ 15112.4i 1.18128i 0.806936 + 0.590639i $$0.201124\pi$$
−0.806936 + 0.590639i $$0.798876\pi$$
$$548$$ −692.820 178.885i −0.0540070 0.0139445i
$$549$$ 1326.00 + 11860.1i 0.103083 + 0.921998i
$$550$$ 2700.00 + 3485.69i 0.209324 + 0.270237i
$$551$$ −10600.2 −0.819567
$$552$$ 3623.63 10813.2i 0.279406 0.833770i
$$553$$ 660.000 0.0507524
$$554$$ −10998.5 14199.0i −0.843470 1.08892i
$$555$$ −4503.33 4027.90i −0.344425 0.308063i
$$556$$ 15180.0 + 3919.46i 1.15787 + 0.298961i
$$557$$ 16055.0i 1.22131i 0.791896 + 0.610656i $$0.209094\pi$$
−0.791896 + 0.610656i $$0.790906\pi$$
$$558$$ −11946.9 12310.8i −0.906365 0.933971i
$$559$$ 2246.33i 0.169964i
$$560$$ 3879.79 + 2146.63i 0.292770 + 0.161985i
$$561$$ −4800.00 4293.25i −0.361241 0.323104i
$$562$$ 12440.0 9635.98i 0.933718 0.723255i
$$563$$ 25142.4 1.88211 0.941055 0.338254i $$-0.109836\pi$$
0.941055 + 0.338254i $$0.109836\pi$$
$$564$$ 4475.69 + 6707.93i 0.334150 + 0.500806i
$$565$$ 9280.00 0.690996
$$566$$ 15155.4 11739.4i 1.12550 0.871806i
$$567$$ −1247.08 5507.38i −0.0923674 0.407916i
$$568$$ −21600.0 + 9295.16i −1.59563 + 0.686648i
$$569$$ 23416.1i 1.72523i −0.505864 0.862613i $$-0.668826\pi$$
0.505864 0.862613i $$-0.331174\pi$$
$$570$$ −9141.23 647.085i −0.671726 0.0475498i
$$571$$ 4918.69i 0.360492i 0.983622 + 0.180246i $$0.0576893\pi$$
−0.983622 + 0.180246i $$0.942311\pi$$
$$572$$ −692.820 + 2683.28i −0.0506438 + 0.196143i
$$573$$ 9600.00 10733.1i 0.699905 0.782518i
$$574$$ −1680.00 2168.87i −0.122163 0.157712i
$$575$$ 4364.77 0.316562
$$576$$ −11032.6 8329.73i −0.798077 0.602556i
$$577$$ 19490.0 1.40620 0.703102 0.711089i $$-0.251798\pi$$
0.703102 + 0.711089i $$0.251798\pi$$
$$578$$ −6292.54 8123.63i −0.452829 0.584600i
$$579$$ −658.179 + 735.867i −0.0472418 + 0.0528179i
$$580$$ −2720.00 + 10534.5i −0.194727 + 0.754176i
$$581$$ 9713.48i 0.693602i
$$582$$ −11288.4 799.077i −0.803985 0.0569121i
$$583$$ 18900.2i 1.34265i
$$584$$ 8521.69 3667.15i 0.603819 0.259842i
$$585$$ −2400.00 + 268.328i −0.169620 + 0.0189641i
$$586$$ 1460.00 1130.91i 0.102922 0.0797227i
$$587$$ −1364.86 −0.0959687 −0.0479844 0.998848i $$-0.515280\pi$$
−0.0479844 + 0.998848i $$0.515280\pi$$
$$588$$ 6529.32 + 9785.80i 0.457933 + 0.686326i
$$589$$ 15660.0 1.09552
$$590$$ −3464.10 + 2683.28i −0.241720 + 0.187236i
$$591$$ −8279.20 7405.14i −0.576245 0.515409i
$$592$$ 7280.00 + 4027.90i 0.505416 + 0.279638i
$$593$$ 25795.3i 1.78632i 0.449743 + 0.893158i $$0.351515\pi$$
−0.449743 + 0.893158i $$0.648485\pi$$
$$594$$ −558.921 13734.8i −0.0386074 0.948729i
$$595$$ 2478.71i 0.170785i
$$596$$ −12540.0 3237.83i −0.861846 0.222528i
$$597$$ 990.000 + 885.483i 0.0678694 + 0.0607042i
$$598$$ 1680.00 + 2168.87i 0.114883 + 0.148314i
$$599$$ 2424.87 0.165405 0.0827025 0.996574i $$-0.473645\pi$$
0.0827025 + 0.996574i $$0.473645\pi$$
$$600$$ 1681.15 5016.70i 0.114388 0.341343i
$$601$$ −8758.00 −0.594420 −0.297210 0.954812i $$-0.596056\pi$$
−0.297210 + 0.954812i $$0.596056\pi$$
$$602$$ −3013.77 3890.76i −0.204040 0.263414i
$$603$$ 19745.4 2207.60i 1.33349 0.149089i
$$604$$ −4380.00 1130.91i −0.295066 0.0761856i
$$605$$ 1171.70i 0.0787378i
$$606$$ −1550.10 + 21897.9i −0.103908 + 1.46789i
$$607$$ 19558.6i 1.30784i 0.756565 + 0.653919i $$0.226876\pi$$
−0.756565 + 0.653919i $$0.773124\pi$$
$$608$$ 12470.8 1931.96i 0.831836 0.128867i
$$609$$ 4080.00 4561.58i 0.271478 0.303521i
$$610$$ −8840.00 + 6847.43i −0.586756 + 0.454499i
$$611$$ −1939.90 −0.128445
$$612$$ −1088.62 + 7650.79i −0.0719031 + 0.505335i
$$613$$ −16450.0 −1.08386 −0.541932 0.840422i $$-0.682307\pi$$
−0.541932 + 0.840422i $$0.682307\pi$$
$$614$$ −3481.42 + 2696.70i −0.228825 + 0.177247i
$$615$$ 3879.79 4337.74i 0.254388 0.284414i
$$616$$ −2400.00 5577.10i −0.156978 0.364785i
$$617$$ 8461.28i 0.552088i 0.961145 + 0.276044i $$0.0890236\pi$$
−0.961145 + 0.276044i $$0.910976\pi$$
$$618$$ −1406.73 + 19872.6i −0.0915649 + 1.29352i
$$619$$ 19930.4i 1.29413i −0.762433 0.647067i $$-0.775995\pi$$
0.762433 0.647067i $$-0.224005\pi$$
$$620$$ 4018.36 15563.0i 0.260292 1.00811i
$$621$$ −11088.0 7888.85i −0.716499 0.509772i
$$622$$ −5640.00 7281.21i −0.363575 0.469373i
$$623$$ −6512.51 −0.418809
$$624$$ 3139.90 1095.56i 0.201437 0.0702843i
$$625$$ −7975.00 −0.510400
$$626$$ 12176.3 + 15719.6i 0.777418 + 1.00364i
$$627$$ 9353.07 + 8365.64i 0.595735 + 0.532842i
$$628$$ 1460.00 5654.56i 0.0927712 0.359301i
$$629$$ 4651.02i 0.294830i
$$630$$ 3796.92 3684.70i 0.240116 0.233019i
$$631$$ 12199.9i 0.769683i −0.922983 0.384842i $$-0.874256\pi$$
0.922983 0.384842i $$-0.125744\pi$$
$$632$$ −762.102 1770.97i −0.0479665 0.111464i
$$633$$ 2130.00 + 1905.13i 0.133744 + 0.119624i
$$634$$ 1100.00 852.056i 0.0689063 0.0533746i
$$635$$ −16419.8 −1.02614
$$636$$ −18866.2 + 12588.0i −1.17625 + 0.784821i
$$637$$ −2830.00 −0.176026
$$638$$ 11777.9 9123.16i 0.730867 0.566127i
$$639$$ 3117.69 + 27885.5i 0.193011 + 1.72634i
$$640$$ 1280.00 12889.3i 0.0790569 0.796084i
$$641$$ 7012.31i 0.432090i 0.976383 + 0.216045i $$0.0693158\pi$$
−0.976383 + 0.216045i $$0.930684\pi$$
$$642$$ 9445.94 + 668.654i 0.580688 + 0.0411054i
$$643$$ 15979.9i 0.980073i 0.871702 + 0.490036i $$0.163017\pi$$
−0.871702 + 0.490036i $$0.836983\pi$$
$$644$$ −5819.69 1502.64i −0.356099 0.0919444i
$$645$$ 6960.00 7781.52i 0.424883 0.475034i
$$646$$ −4320.00 5577.10i −0.263109 0.339672i
$$647$$ −17999.5 −1.09371 −0.546856 0.837226i $$-0.684176\pi$$
−0.546856 + 0.837226i $$0.684176\pi$$
$$648$$ −13337.9 + 9705.63i −0.808581 + 0.588385i
$$649$$ 6000.00 0.362898
$$650$$ 779.423 + 1006.23i 0.0470330 + 0.0607194i
$$651$$ −6027.54 + 6738.99i −0.362884 + 0.405717i
$$652$$ 1980.00 + 511.234i 0.118931 + 0.0307078i
$$653$$ 5196.62i 0.311423i −0.987803 0.155712i $$-0.950233\pi$$
0.987803 0.155712i $$-0.0497671\pi$$
$$654$$ 15627.8 + 1106.26i 0.934398 + 0.0661437i
$$655$$ 4027.90i 0.240280i
$$656$$ −3879.79 + 7012.31i −0.230915 + 0.417355i
$$657$$ −1230.00 11001.5i −0.0730394 0.653284i
$$658$$ 3360.00 2602.64i 0.199068 0.154197i
$$659$$ −6062.18 −0.358344 −0.179172 0.983818i $$-0.557342\pi$$
−0.179172 + 0.983818i $$0.557342\pi$$
$$660$$ 10713.8 7148.53i 0.631872 0.421601i
$$661$$ 9422.00 0.554423 0.277211 0.960809i $$-0.410590\pi$$
0.277211 + 0.960809i $$0.410590\pi$$
$$662$$ 9474.32 7338.78i 0.556238 0.430860i
$$663$$ −1385.64 1239.35i −0.0811672 0.0725981i
$$664$$ 26064.0 11216.2i 1.52331 0.655529i
$$665$$ 4829.91i 0.281648i
$$666$$ 7124.50 6913.92i 0.414518 0.402266i
$$667$$ 14748.3i 0.856158i
$$668$$ −27.7128 + 107.331i −0.00160515 + 0.00621672i
$$669$$ 1830.00 + 1636.80i 0.105758 + 0.0945925i
$$670$$ 11400.0 + 14717.3i 0.657344 + 0.848627i
$$671$$ 15311.3 0.880905
$$672$$ −3968.62 + 6110.18i −0.227816 + 0.350752i
$$673$$ −17470.0 −1.00062 −0.500311 0.865846i $$-0.666781\pi$$
−0.500311 + 0.865846i $$0.666781\pi$$
$$674$$ −2580.76 3331.74i −0.147488 0.190406i
$$675$$ −5144.19 3659.97i −0.293333 0.208700i
$$676$$ 4194.00 16243.3i 0.238621 0.924175i
$$677$$ 20813.3i 1.18157i −0.806830 0.590784i $$-0.798819\pi$$
0.806830 0.590784i $$-0.201181\pi$$
$$678$$ −1076.72 + 15210.5i −0.0609897 + 0.861589i
$$679$$ 5964.39i 0.337102i
$$680$$ −6651.08 + 2862.17i −0.375084 + 0.161410i
$$681$$ 1752.00 1958.80i 0.0985856 0.110222i
$$682$$ −17400.0 + 13478.0i −0.976951 + 0.756743i
$$683$$ −12616.3 −0.706805 −0.353402 0.935471i $$-0.614975\pi$$
−0.353402 + 0.935471i $$0.614975\pi$$
$$684$$ 2121.23 14908.0i 0.118578 0.833365i
$$685$$ −800.000 −0.0446225
$$686$$ 10842.6 8398.67i 0.603460 0.467438i
$$687$$ 14182.0 15856.0i 0.787596 0.880559i
$$688$$ −6960.00 + 12579.4i −0.385680 + 0.697074i
$$689$$ 5456.01i 0.301680i
$$690$$ 900.309 12718.5i 0.0496727 0.701716i
$$691$$ 3028.67i 0.166738i 0.996519 + 0.0833691i $$0.0265681\pi$$
−0.996519 + 0.0833691i $$0.973432\pi$$
$$692$$ 8660.25 + 2236.07i 0.475742 + 0.122836i
$$693$$ −7200.00 + 804.984i −0.394669 + 0.0441253i
$$694$$ −3444.00 4446.18i −0.188375 0.243191i
$$695$$ 17528.4 0.956674
$$696$$ −16951.2 5680.53i −0.923179 0.309368i
$$697$$ 4480.00 0.243461
$$698$$ 3592.27 + 4637.60i 0.194799 + 0.251484i
$$699$$ −20438.2 18280.5i −1.10593 0.989172i
$$700$$ −2700.00 697.137i −0.145786 0.0376419i
$$701$$ 17664.9i 0.951777i 0.879506 + 0.475888i $$0.157873\pi$$
−0.879506 + 0.475888i $$0.842127\pi$$
$$702$$ −161.347 3964.89i −0.00867470 0.213170i
$$703$$ 9062.78i 0.486215i
$$704$$ −12193.6 + 12879.8i −0.652791 + 0.689523i
$$705$$ 6720.00 + 6010.55i 0.358993 + 0.321093i
$$706$$ −19360.0 + 14996.2i −1.03204 + 0.799418i
$$707$$ 11570.1 0.615472
$$708$$ −3996.15 5989.22i −0.212125 0.317922i
$$709$$ 14174.0 0.750798 0.375399 0.926863i $$-0.377506\pi$$
0.375399 + 0.926863i $$0.377506\pi$$
$$710$$ −20784.6 + 16099.7i −1.09864 + 0.851001i
$$711$$ −2286.31 + 255.617i −0.120595 + 0.0134830i
$$712$$ 7520.00 + 17474.9i 0.395820 + 0.919803i
$$713$$ 21788.2i 1.14443i
$$714$$ 4062.77 + 287.593i 0.212949 + 0.0150741i
$$715$$ 3098.39i 0.162060i
$$716$$ 2702.00 10464.8i 0.141031 0.546212i
$$717$$ −19680.0 + 22002.9i −1.02505 + 1.14604i
$$718$$ −14040.0 18125.6i −0.729761 0.942117i
$$719$$ 32839.7 1.70336 0.851678 0.524065i $$-0.175585\pi$$
0.851678 + 0.524065i $$0.175585\pi$$
$$720$$ −14271.4 5933.49i −0.738699 0.307122i
$$721$$ 10500.0 0.542358
$$722$$ −3462.37 4469.90i −0.178471 0.230405i
$$723$$ −4149.99 + 4639.83i −0.213472 + 0.238668i
$$724$$ −2524.00 + 9775.41i −0.129563 + 0.501796i
$$725$$ 6842.37i 0.350509i
$$726$$ 1920.49 + 135.947i 0.0981766 + 0.00694967i
$$727$$ 8001.58i 0.408201i 0.978950 + 0.204101i $$0.0654270\pi$$
−0.978950 + 0.204101i $$0.934573\pi$$
$$728$$ −692.820 1609.97i −0.0352715 0.0819635i
$$729$$ 6453.00 + 18595.1i 0.327846 + 0.944731i
$$730$$ 8200.00 6351.69i 0.415747 0.322037i
$$731$$ 8036.72 0.406633
$$732$$ −10197.7 15283.8i −0.514917 0.771730i
$$733$$ 11750.0 0.592082 0.296041 0.955175i $$-0.404333\pi$$
0.296041 + 0.955175i $$0.404333\pi$$
$$734$$ −17649.6 + 13671.3i −0.887546 + 0.687490i
$$735$$ 9803.41 + 8768.43i 0.491978 + 0.440039i
$$736$$ 2688.00 + 17351.0i 0.134621 + 0.868974i
$$737$$ 25491.2i 1.27406i
$$738$$ 6659.69 + 6862.53i 0.332177 + 0.342294i
$$739$$ 19961.4i 0.993627i −0.867857 0.496814i $$-0.834503\pi$$
0.867857 0.496814i $$-0.165497\pi$$
$$740$$ 9006.66 + 2325.51i 0.447421 + 0.115524i
$$741$$ 2700.00 + 2414.95i 0.133856 + 0.119724i
$$742$$ 7320.00 + 9450.08i 0.362164 + 0.467552i
$$743$$ −25592.8 −1.26367 −0.631836 0.775102i $$-0.717698\pi$$
−0.631836 + 0.775102i $$0.717698\pi$$
$$744$$ 25042.6 + 8392.06i 1.23401 + 0.413532i
$$745$$ −14480.0 −0.712089
$$746$$ −8504.37 10979.1i −0.417382 0.538838i
$$747$$ −3762.01 33648.5i −0.184264 1.64810i
$$748$$ 9600.00 + 2478.71i 0.469266 + 0.121164i
$$749$$ 4990.90i 0.243476i
$$750$$ 1577.95 22291.3i 0.0768246 1.08528i
$$751$$ 5244.02i 0.254803i 0.991851 + 0.127401i $$0.0406637\pi$$
−0.991851 + 0.127401i $$0.959336\pi$$
$$752$$ −10863.4 6010.55i −0.526793 0.291466i
$$753$$ 14760.0 16502.2i 0.714322 0.798636i
$$754$$ 3400.00 2633.63i 0.164218 0.127203i
$$755$$ −5057.59 −0.243794
$$756$$ 5598.92 + 6650.92i 0.269353 + 0.319963i
$$757$$ −14290.0 −0.686102 −0.343051 0.939317i $$-0.611460\pi$$
−0.343051 + 0.939317i $$0.611460\pi$$
$$758$$ −7014.81 + 5433.65i −0.336134 + 0.260368i
$$759$$ −11639.4 + 13013.2i −0.556631 + 0.622332i
$$760$$ 12960.0 5577.10i 0.618564 0.266188i
$$761$$ 16976.2i 0.808657i −0.914614 0.404328i $$-0.867505\pi$$
0.914614 0.404328i $$-0.132495\pi$$
$$762$$ 1905.12 26913.2i 0.0905711 1.27948i
$$763$$ 8257.20i 0.391783i
$$764$$ −5542.56 + 21466.3i −0.262464 + 1.01652i
$$765$$ 960.000 + 8586.50i 0.0453711 + 0.405811i
$$766$$ −10752.0 13880.8i −0.507161 0.654742i
$$767$$ 1732.05 0.0815394
$$768$$ 20977.9 + 3593.49i 0.985644 + 0.168840i
$$769$$ −29566.0 −1.38645 −0.693223 0.720723i $$-0.743810\pi$$
−0.693223 + 0.720723i $$0.743810\pi$$
$$770$$ −4156.92 5366.56i −0.194552 0.251166i
$$771$$ 12193.6 + 10906.3i 0.569576 + 0.509444i
$$772$$ 380.000 1471.73i 0.0177157 0.0686125i
$$773$$ 21457.3i 0.998403i −0.866486 0.499202i $$-0.833627\pi$$
0.866486 0.499202i $$-0.166373\pi$$
$$774$$ 11946.9 + 12310.8i 0.554809 + 0.571707i
$$775$$ 10108.5i 0.468526i
$$776$$ 16004.1 6887.09i 0.740355 0.318598i
$$777$$ −3900.00 3488.27i −0.180067 0.161056i
$$778$$ 21140.0 16375.0i 0.974172 0.754590i
$$779$$ −8729.54 −0.401499
$$780$$ 3092.82 2063.60i 0.141975 0.0947293i
$$781$$ 36000.0 1.64940
$$782$$ 7759.59 6010.55i 0.354837 0.274855i
$$783$$ −12366.8 + 17381.9i −0.564438 + 0.793334i
$$784$$ −15848.0 8768.43i −0.721939
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# Impressive Universal Fog Light Wiring Diagram Unique Motorcycle Fog Lights Wiring Diagram Crest - Electrical
Other posts you might like
Universal fog light wiring diagram - I recall the times once i rummaged approximately in my “container of wires” and grabbed any gauge cord of enough or insufficient period, splicing collectively a “rat’s nest” of wire connections and crimped ends to connect any variety of desired add-ons. After a few smoke-stuffed incidents i am an awful lot greater cautious.
Let’s start with your lights already hooked up to the auto, the wires dangling underneath or behind and ready to get hold of strength from mr. Lucas. The first order of commercial enterprise is to determine the amperage of your driving/fog lighting. My lamps are antique and each unit reads 35 watts. The formulation for amperage is watts divided by using volts equals amps, or w/v=a. Considering that i will be wiring the lights to the relay with one lead, 70w/12v= 5.8a. I can be the use of 14-gauge twine, which handles up to eleven.8a. Amps are a measure of modern glide; volts are a degree of the pressure in the back of the go with the flow of cutting-edge. To shield my 14-gauge wiring i might be installing 10-amp inline fuses. The guideline of thumb is this: the fuse need to be rated near eighty of the amperage of the wire. This will make sure that you blow the fuse earlier than you burn the cord. In my case, 80 of eleven.8a is 9.44a so a ten-amp inline fuse is perfect.
Riding lights and fog lights came about as automobile proprietors navigated the twisting turning via-approaches of misty england. Effective lighting was important to illuminate the street ahead for potential hazards to be effectively diagnosed and prevented. In addition, foggy and moist situations resulting from street spray obliterated the edges of poorly topped roads. There's one greater oft-unnoticed advantage to using lights and fog lighting, definitely said they're “racer cool”; putting in these lighting, however, on your preferred british sports activities vehicle takes making plans and training.
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My 1966 volvo 122s may not be british, but it does use many components commonplace to the vehicles of britain. I picked up vintage fog lighting at a yard sale and these could be installed on my volvo.
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You are Here: Home >< Maths
# probability sampling problem watch
1. I have a problem on the last part of a question on my intro to probability module:
I make an unordered selection of r numbers from the set {1,2,3,...,n} without replacement and with all possibilities equally likely.
What is the probability that the number 1 is in my selection, in terms of n and r. Show your working, simplify your answer as much as possible.
Thanks in advance. It's probably obvious and I'm just rusty.
2. (Original post by confused.me)
I have a problem on the last part of a question on my intro to probability module:
I make an unordered selection of r numbers from the set {1,2,3,...,n} without replacement and with all possibilities equally likely.
What is the probability that the number 1 is in my selection, in terms of n and r. Show your working, simplify your answer as much as possible.
Thanks in advance. It's probably obvious and I'm just rusty.
I'm not 100% sure but here's what I'd do:
Spoiler:
Show
P(1 in selection) = 1 - P(1 not in selection)
P(1 not in selection) = (n-1)/n x (n-2)/n x . . .
Can you work out the rest?
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# Finite universe
1. Mar 30, 2006
### wolram
How do you imagine a finite universe ? i find it very difficult.
2. Mar 30, 2006
### marcus
remember we are just talking SPATIAL finite
timewise it can still go on forever
what is the problem imagining a spatially finite universe?
it is the old "surface of a balloon" thing except the surface of a balloon is only 2D and you have to think of what would be the 3D analog of that.
3. Mar 30, 2006
### wolram
Hi Marcus, i do not want to go into the philosophy, it is just that i can not
imagine, (get a picture in my mind) of a finite universe, this is a problem i
encounter, i need a picture to make it real.
4. Mar 30, 2006
### Distant
Indeed it is difficult to imagine a finite Universe. It is much easier to imagine what it must have been like to live in a time before it was known that the Earth was round and that gravity kept us all from falling off the surface. In such a time a similar question to the one of the finite Universe was posed about the surface of the Earth. After all, we can see that the surface extends to the horizon in all directions. If we walk towards the horizon then the limit of the horizon moves with us and we can see how the surface extends further. So surely there can be only two possibilities. Either the surface extends on, forever, and if we carried on walking towards the horizon we would forever encounter new lands, or the surface if somehow ‘bounded’. Maybe we would reach some solid impenetrable wall through which we cannot pass – or maybe we would fall off the edge! (to where exactly wasn’t clear!). Of course, sea voyages set out to investigate the truth of the matter and I am not aware of them encountering either of these problems.
Although we may now view such thoughts in a rather patronising manner, without an understanding of the concepts of gravity and the spherical planet they are natural and logical questions. The spherical shape of the Earth ‘squares the circle’. It does not have a bounded edge to its surface, but it is not infinite – magic! (of a sort).
It required new (rather none intuitive) concepts to understand this problem. We are generally not aware of the curvature of the Earth. The planet is too big and we move too slowly to notice it. It would have been hard to convince to old “flat Earthists” that people living in a far off continent were, in effect, “upside-down” relative us!
Now, back to the Universe. Physicists tell me that the 3 dimensions of space are not the whole story. Time itself can be treated as another dimensions to describe something they call “spacetime”. Just as we once had an inability to notice (or comprehend) the curvature of the earth (nowadays we can view photos from space easily portraying this) we ‘similarly’ have an inability to notice or comprehend the curvature of this spacetime. Trying not be be a modern day equivalent of a “flat Earthist” we may choose to accept the word (and mathematics and experimental evidence!) of the physicists on this point. The point being is that it is postulated that spacetime if also curved, like the Earth is in 3 dimensions in order to produce a 4 dimensional spacetime Universe that is neither bounded nor infinite.
I fear that a more ‘natural’ understanding of such a ‘closed’ Universe, in the same way as we understand and experience a 3 dimensional ‘closed’ sphere, may not be possible. Maybe because our brains are not set up to work in 4 dimensional spacetime.
However unsatisfying this explanation maybe to the none physicist like myself, it surely surpasses the cosmological equivalent of the brick wall at the edge of the Universe, or indeed ‘falling over the edge’!
Ahhhhhhhh!…………
5. Mar 30, 2006
### wolram
A warm welcomb Distant, it all ways gives me a glow inside when people
answer my some times inane questions, your anology with the flat earthers
is good, but when it comes down to all there is, my mind just can not cope,
i try to understand, but some times these theories are just so unintuitive.
6. Mar 30, 2006
### marcus
if you keep telling yourself it is unintuitive you will make it worse.
do a series of "analogy pushups"
start by thinking of yourself as a 1D being living on a straight line
who imagines adding a "point at infinity" to his world, an extra point that fills the gap between + oo and - oo
so that if he goes faster and faster to the east he eventually comes whizzing in from the west
and he doesnt believe this is real, it is just an abstract mathematical idea for him
he has INVENTED THE CIRCLE, never having seen one, as a mathematical idea.
then 1000 years later the 1D astronomers uncover evidence in the CMB that their actually is a circularity of the universe. they discover that what was just a math idea is actually REAL and they are left scratching their onedimensional heads.
====================
push up to the next level analogy, think that you are a 2D being living in a flat plane, or one that looks flat to you. being an abstract thinker you imagine that there is a "point at infinity" so that if you go racing off in a straight line in any direction you will eventually come zooming in from the opposite direction
you have INVENTED THE MATHEMATICAL IDEA OF THE SPHERE as best as you can imagine it with your limited 2D wits.
it is like you are at the south pole and of the earth and it seems like an infinite flat plane, and the point at infinity is the north pole. so any direction you go in, departing from south pole, you eventully return from the opposite direction.
and maybe years later they discover that this idea of a sphere is REAL and actually how nature is.
=====================
push up to the next level analogy, now you are a 3D being living in a regular square-angle normal euclidean 3D space, analogous to the 2D flat plane.........and you imagine a "point at infinity" is added to your 3D space so that if you go off in a straight line in any direction then eventually after a long enough time you come back from the opposite direction.....
you have done something mathematically creative: you have INVENTED THE SOCALLED "THREESPHERE"
and maybe after hundreds of years the astronomers could supply evidence that what you thought was regular 3D space actually was a threesphere all along. All that time you were living in a threesphere, which you thought was a mere abstract concept.
========================
technically the ordinary sphere----the surface of a balloon---can be called a TWOSPHERE because the local neighborhood of a point looks like a normal flat 2D plane
and a circle or a ring could be called a ONESPHERE because the local nbd of a point looks like a bit of an approximately straight line
so all we are doing is imagining this kind of thing with the dimension jacked up, to get a threesphere
and anyway, Wolram, there are SEVERAL kinds of finite 3D spaces possible------the threesphere is not the only one. there are a whole bunch, just like in the 2D situation you can have donut shapes with more than one hole. But this does not matter. It is enough to just imagine one possible spatially finite 3D thing----and the threesphere is good enough.
Before falling asleep every night, try to imagine that you live in a threesphere, and if you shine a light off in one direction then that same beam of light will eventually get back to you from the opposite. (unless the poor lightbeam has been frustrated by having the space go and expand faster than it can cope with but expansion is another business, ignore that for the while)
Last edited: Mar 30, 2006
7. Mar 30, 2006
### wolram
Thankyou very much Marcus, what you say in words makes perfect sense,
but to form an image in ones mind is much harder, maybe i am to stupid, i wish i could understand.
8. Mar 30, 2006
### marcus
it makes it worse to keep supposing that one is stupid---one is a human and that involves certain brain limitations which one tries cheerfully to overcome.
try to imagine what it would be like to BE IN a certain space
don't try to imagine what it would look like to a God who is somehow on the outside of it.
we are not asked to imagine what it would look like to an higherdimensional being outside of space
just get a feel for what to expect if you are a 3D creature, which we all are, living in a threesphere
all it means, really, is that if you send a lite-beam off in some direction out in front eventually it will come back from behind you
as 2D creatures living on a Twosphere (the surface of the earth) we are ALREADY USED TO THAT kind of behavior-----of being able to head off in some direction and eventually get back to the same place having gone around
so the analog (3D "flat" space with a point at infinity added) can't be so hard to imagine what it would be like to live in
(forget about God's point of view, just imagine living in it)
9. Mar 31, 2006
### Chronos
While a spatial finite universe is difficult to imagine, a temporally finite universe is less difficult. It is, for me, easy to picture a universe where you are both at the center and the edge due to the finite speed of light.
10. Mar 31, 2006
### wolram
Thanks Marcus, i do have a picture now, it may be a little fuzzy, almost
abstract, but at least i can think about it now.
11. Apr 1, 2006
### Silverbackman
Hmmm, so I guess it is impossible to imagine looking a threesphere like we look at a balloon or a ball. Damn I hate being human.
Then again I know more than a being from a dot, circle, and twosphere world. Is it possible for a foursphere, fivesphere, ect. ect. till infinite? Of course we can't imagine these other spatial dimensions, similar to how a 2D being wouldn't be able to imagine a 3D world or twospere.
But doesn't the known curvature of the universe suggest that we live in a flat universe and not a saddle or sphere universe? If so then how can we be living in a threespere?
12. Apr 1, 2006
### SpaceTiger
Staff Emeritus
I don't understand the question...
Measurements of flatness are local. If inflation occurred, then the universe will be much, much larger than the region of space we can observe and, regardless of its true large-scale geometry, it will appear flat to our instruments. It's basically the same reason the ancients thought the earth was flat -- they could only see a very small part of it.
If inflation had not occurred, however, this wouldn't be the case and we should, in principle, be able to measure the curvature of the entire universe.
13. Apr 1, 2006
### wolram
I have an idea to call my image the, ( frustrated boomerang), curved space
with expasion.
14. Apr 4, 2006
### VikingF
I think he wondered how many dimensions that can possibly exist. If there may be "infinitely many" dimensions... (1D,2D,3D,4D,....,?D).
15. Apr 5, 2006
### -Job-
One second though. The possibility that we live in continuous unbounded universe, the equivalent of a 4D sphere follows rationally and is more plausible than a finite universe or an infinite universe that does not "wrap-around". But regardless, if we live in a 4D sphere, then this 4D sphere is in 4D space. How is this 4D space then bounded? Is it infinite or finite? Does it wrap around? The alternative is that we live in a 4D sphere which does not reside in space, but then the notion of dimensions should not apply, and it's not a 4D surface at all.
Last edited: Apr 5, 2006
16. May 4, 2006
### Silverbackman
You mean torus?
Yea, I wonder what a eightsphere universe would look like. Or how about a 1,678,489,986sphere?:surprised :surprised :surprised
17. May 4, 2006
### Flatland
I actually have a problem imagining an infinite universe. And a universe with an infinite dimensions? you wouldn't be able to go anywhere.
18. May 4, 2006
### setAI
I conjecture that it is possible for a finite AND flat universe without the topology of a torus or similar form- in fact it can have ANY arbitrary topology as long as it is closed-
how? it is a natural result of computational /category theories and some forms of LQG where the metric of spacetime itself emerges from relationships- for instance the universe can be decribed as a 2-dimensional lattice of 2in/2out quantum logic gates- one could construct such a lattice of quantum logic gates on the surface of a sphere- or any other closed shape- yet the topology of the emergent spacetime metric could compute a FLAT spacetime that wraps around-
consider a classical computational analog: the game Asteroids- where the ship flies straight but wraps around when it hits the edge of the screen- the virtual space of the ship is flat- but the edges are connected in the software- so the virtual space is finite yet unbounded
19. May 4, 2006
### VikingF
Yeah... Me too! :tongue:
According to String theory, there exist 10 dimensions, and M-theory says that there exist 11.
20. May 5, 2006
### Silverbackman
Yea but they say those dimensions may just exist. It doesn't have anything to do with the universe being a tensphere or elevensphere I think. Or does it? Actually it probably does because I assume 10 or 11 dimensions cannot exist unless is within some sort of nsphere (or something like an nsphere).
If we do live in a threesphere only that would mean there are four spatial dimensions and one dimension of time. That sounds pretty aesthetically perfect considering 5 is a "perfect" number. Both 5 and 10 seem like aesthetically perfect because of their place on the number line. Why an omnipotent God or even randomness of Nature would pick 4, 7, or 11 as the "official" number of one of the most important aspects of our universe is beyond me.:uhh:
21. May 5, 2006
### Garth
There is an anthroopic argument as to why we are living in a 3D + time universe.
Time is necessary for process, we exist and evolution of complex beings requires process, therefore time.
In a 1D or 2D spatial universe there is not enough topological freedom for complex organic molecules to operate, the 3D geometry is important for proteins etc. to 'do their stuff'. Hence we cannot exist in a spatial 1D or 2D universe.
Keplerian orbits are unstable in a spatially 4D or higher universe. The Newtonian law of gravity becomes $F = GMm/r^3$ in a 4D universe for example and the Earth would spiral into or away from the Sun. Hence we cannot exist in a spatial 4D or higher universe.
We exist: therefore our universe has 3 space D + time.
Garth
22. May 5, 2006
### VikingF
No, if I have not misunderstood it completely, String theory says that our universe is one of infinitely many three dimensional (or four with time) universes in an infinite ten- or eleven dimensional multiverse.
It's like the drawing on the piece of paper in front of me is two dimensional in a four dimensional universe.
23. May 5, 2006
### Garth
The normal understanding is to ask: "If there are more than 3 spatial dimensions why do we not observe them?" The answer given in string theory is that the extra dimensions are 'rolled' up like a 2D sheet of paper rolled up into a 'straw' with length and very little width. The extra widths are too small to observe normally, and do not affect dynamics such as Keplerian orbits as in my post #21 above, but may be observed microscopically, such as by minute changes in Newtonian gravity at very small ranges.
Garth
24. May 5, 2006
### Silverbackman
This maybe true for our universe but I'm sure a 1D or 2D universe may have different laws of nature. In these universes organic molecules may not be needed to create matter that can become conscious. So anything maybe possible even if it is beyond our current reasoning. As human beings we can't imagine how it would be to live in a 2D or 6D universe. But this all sort of enters the realm of philosophy. There is no proof either way one whether conscious beings can evolve in a 1D or 2D universe. We can't exist in these universes but perhaps in another universe matter can adapt to whatever dimension limitations it may have.
If our universe cannot be spatially 4D, how can we live in a finite threesphere? Wouldn't an extra dimension be needed in order for our 3D universe to curve into a round shape where if you go in one direction that you will come back in another direction?
It is similar to the notion of a 1D being living in a round universe. To him reality is only on the X-axis but if his universe is round then his universe is actually a circle. Circles are 2D. Perhaps the some total of our 3 known spatial dimensions add up to a 4th spatial dimension when curving.
Yea but how can other universes exist in these other dimensions that are rolled up. And how do we there is an infinite amount in only a 10D or 11D universe.
25. Jun 9, 2006
### Balence
Any shape still has space out side of it, therefore the universe is still infinite. As far as extra dimensions most were created because the mathematics in a theory did not add up so they made alternet dimensions to prove their theory right, they have no proof for these dimensions (besides the faulty equations that gave birth to them) and there are 3 spacial dimensions, all of which are abstract. The dimension of time isnt a dimension on dictionary terms but since it is used to measure can be qualified as a dimension. Any thing that exsists must have and can only have length, width, height, and a place in time. Time is also abstract.
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Convert mAh to Fr (Milliamper-hour to Franklin)
## Milliamper-hour into Franklin
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Milliamper-hour+to+Franklin.php
## How many Franklin make 1 Milliamper-hour?
1 Milliamper-hour [mAh] = 10 792 528 332.635 Franklin [Fr] - Measurement calculator that can be used to convert Milliamper-hour to Franklin, among others.
# Convert Milliamper-hour to Franklin (mAh to Fr):
1. Choose the right category from the selection list, in this case 'Electric charge'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Milliamper-hour [mAh]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Franklin [Fr]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '935 Milliamper-hour'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Milliamper-hour' or 'mAh'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Electric charge'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '13 mAh to Fr' or '34 mAh into Fr' or '19 Milliamper-hour -> Franklin' or '19 mAh = Fr' or '71 Milliamper-hour to Fr' or '15 mAh to Franklin' or '74 Milliamper-hour into Franklin'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(58 * 95) mAh'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '935 Milliamper-hour + 2805 Franklin' or '16mm x 92cm x 22dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 8.352 099 923 995 9×1030. For this form of presentation, the number will be segmented into an exponent, here 30, and the actual number, here 8.352 099 923 995 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 8.352 099 923 995 9E+30. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 8 352 099 923 995 900 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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# Floor Load Calculations (Cooling, Heating,
Simulation)
This help topic discusses floor transmission load calculations for design heating and energy
simulation applications. Floor transmission loads are due to heat flow through floors which are
either adjacent to unconditioned or partially conditioned regions, or are in contact with soil. The
program permits four types of floor heat transfer situations to be evaluated:
1. Floor is Above Conditioned Region: It is assumed the adjacent region is at the same
temperature as the zone. Therefore, no heat transfer occurs.
2. Floor is Above An Unconditioned Region: The procedures outlined in the help topics titled
Partition and Ceiling Load Calculations (Cooling, Simulation) and Partition and Ceiling Load
Calculations (Heating) are used to calculate these floor transmission loads.
3. Slab On Grade Floor: Heat transmitted through the floor and the adjacent soil is computed.
soil is analyzed.
The remainder of this topic discusses calculations for the slab on grade and floor below grade
basement floors and walls for the design cooling condition. Per ASHRAE recommendations, slab
and basement heat transfer is not included in the design cooling calculations since heat transfer is
either negligible or constitutes a credit for summer conditions.
Step 1:Calculate Heat Gains. In HAP a simplified 1-dimensional steady-state heat transfer
model is used to estimate ground heat transfer. Readers should recognize that this is a
simplification since ground heat transfer is really a 3-dimensional heat transfer problem which
would require advanced numerical methods to solve precisely. Considerations for the 1-
dimensional model vary for slab-on-grade floors and below grade floors and walls:
a. For a Slab-On-Grade Floor, it is assumed heat is transferred from the room air through the
slab floor to the soil beneath and eventually to outdoor air. Working from the slab perimeter
inward, the program calculates the total thermal resistance of the heat transfer path by
considering the R-value of the slab floor and carpet, the soil beneath the slab, slab footer
insulation, the slab footer, and soil outside the slab footer. The soil thermal resistance is
determined using the length of the heat transfer path through the soil, and the user specification
of the soil thermal conductivity. Figure 1 illustrates the configuration of these components. To
determine thermal resistances, it is assumed the heat transfer path is semi-circular. Therefore, as
one proceeds inward from the slab perimeter, the path of heat transfer becomes longer and the
overall resistance to heat flow becomes larger. The equation for one-dimensional heat flow as a
function of distance from the slab perimeter is integrated over the width of the floor and is then
solved to determine the total heat flow through the slab floor area. The following equation is
used:
## Figure 1. Slab-on-Grade Floor Diagram
b. For Below-Grade Floors, heat transmitted through both the floor and basement walls is
considered. The floor heat transmission gain is calculated using the procedures described above
for slab-on-grade floors, except that the slab footer and slab footer insulation are omitted from
thermal resistance calculations. In this case the heat transmission paths are also longer. For heat
transmission through the basement walls, it is assumed the heat transfer path is circular between
the basement wall and the soil surface (i.e. a 90-degree arc). The thermal resistance to heat flow
depends on the R-value of the basement wall, wall insulation and the adjacent soil. The heat
transfer path becomes longer as the depth below grade increases. The equation for one-
dimensional heat transfer as a function of depth is integrated over the interval from grade level to
floor depth below grade and is then solved to determine total heat transfer through the basement
wall. When wall insulation is used, two separate calculations are performed. One is for the
portion of the wall covered by insulation, and the other is for the uninsulated portion of the wall.
Figure 2 illustrates the components involved in this analysis. The following equations are used:
## [ ln (1/ho + Df/(2ksoil) + W/ksoil + Rs) - ln (1/ho + Df/(2ksoil) + Rs) ]
qw = P (Toa - Tr) 2ksoil/ x
## Figure 2. Floor Below Grade Diagram
Step 2:Derive Loads from Heat Gains. It is assumed load equals heat gain for slab floors,
basement walls and basement floors. Therefore, room transfer function equations are not needed
Variable Definitions:
## Df = Depth of basement floor below grade, ft or m.
Di = Depth that basement wall insulation extends below grade, ft or m.
ho = Outdoor surface convection coefficient, 6.00 BTU/(hr-sqft-F) or 34.1
W/(sqm-K).
ksoil = Thermal conductivity for soil, BTU/(hr-ft-F) or W/(m-K).
P = Slab floor perimeter exposed to contact with soil, ft.
qf = Floor heat gain, BTU/hr or W.
qw = Basement wall heat gain, BTU/hr or W.
Rf = Thermal resistance for wall foundation, 1.64 (hr-sqft-F)/BTU or 0.289 (sqm-
K)/W.
Rsi = Thermal resistance for slab insulation, (hr-sqft-F)/BTU or (sqm-K)/W.
Rs = Thermal resistance of slab floor, including the floor material, any covering
such as tile or carpet and the inside surface resistance, (hr-sqft-F)/BTU or
(sqm-K)/W.
Rw = Thermal resistance of basement wall including the wall material, any
interior wall finish material or insulation and the inside surface resistance,
(hr-sqft-F)/BTU or (sqm-K)/W.
Rwi = Thermal resistance of insulation applied to the exterior of the basement wall,
(hr-sqft-F)/BTU or (sqm-K)/W.
Toa = Outdoor air temperature, F or C.
Tr = Room air temperature, F or C.
W = Effective width of floor, ft. This is calculated as {space floor area} divided
by {exposed perimeter}.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Algebra Expressions with Exponents
## Calculate values of numbers with exponents
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Algebra Expressions with Exponents
Find the value of algebraic expressions that contain exponents.
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14
## Algebra Expressions with Exponents
by Christopher von Nagy //at grade
Find the value of algebraic expressions that contain exponents.
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0
## Algebra Expressions with Exponents
Find the value of algebraic expressions that contain exponents.
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0
## Algebra Expressions with Exponents
Find the value of algebraic expressions that contain exponents.
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• PLIX
## Algebra Expressions with Exponents: Fish Tank Cube
Algebra Expressions with Exponents Interactive
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• Video
## Evaluate Algebraic Expressions with Exponents: A Sample Application
This video demonstrates a sample use of the principles of evaluating algebraic expressions that contain exponents.
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• Video
## Evaluate Algebraic Expressions with Exponents: An Explanation of the Concept
This video provides an explanation of the concept of evaluating algebraic expressions that contain exponents.
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• Real World Application
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Wazeesupperclub.com Most popular What is a 1-of-8 decoder?
# What is a 1-of-8 decoder?
## What is a 1-of-8 decoder?
The ‘FCT138T devices are 1-of-8 decoders. These devices accept three binary weighted inputs (A0, A1, A2) and, when enabled, provide eight mutually exclusive active-low outputs (O0–O7).
## What is binary decoder circuit?
In digital electronics, a binary decoder is a combinational logic circuit that converts binary information from the n coded inputs to a maximum of 2n unique outputs. Depending on its function, a binary decoder will convert binary information from n input signals to as many as 2n unique output signals.
What is decoder explain 3 to 8 binary decoder with logic circuit diagram?
This decoder circuit gives 8 logic outputs for 3 inputs and has a enable pin. The circuit is designed with AND and NAND logic gates. It takes 3 binary inputs and activates one of the eight outputs. 3 to 8 line decoder circuit is also called a binary to an octal decoder.
### What is binary to octal decoder?
A decoder circuit converts a binary code on the input to a single output representing the numeric value of the code. A binary-to-octal decoder converts 3 binary bits into a 1-of-8 output. Decoders can be designed with either active high outputs or active low outputs.
### How many and gates are required to implement a 1 to 8 demultiplexer?
For a 1 to 8 multiplexer a total of 8 AND gates are required.
How many inputs are required for a 1 of decoder?
The input at the 1, 2, 4, 8 inputs to a 4-line to 16-line decoder with active- low outputs is 1110. As a result, output line 7 is driven LOW. How many data select lines are required for selecting eight inputs?…
Q. How many inputs are required for a 1-of-10 BCD decoder?
B. 8
C. 10
D. 1
## Which IC is used for binary to octal decoder?
MC14028B decoder
The MC14028B decoder is constructed so that an 8421 BCD code on the four inputs provides a decimal (one−of−ten) decoded output, while a 3−bit binary input provides a decoded octal (one−of−eight) code output with D forced to a logic “0”.
## What is a 2 to 4 decoder?
2-to-4 Binary Decoder – The 2 binary inputs labeled A and B are decoded into one of 4 outputs, hence the description of a 2-to-4 binary decoder. Each output represents one of the minterms of the 2 input variables, (each output = a minterm).
What is a binary to octal decoder?
This decoder is also known as a binary to octal decoder because the inputs of this decoder represent three-bit binary numbers whereas the outputs represent the 8 digits within the octal number system. This decoder circuit gives 8 logic outputs for 3 inputs and has a enable pin. The circuit is designed with AND and NAND logic gates.
### How does a binary decoder circuit work?
A decoder circuit converts a binary code on the input to a single output representing the numeric value of the code. A binary-to-octal decoder converts 3 binary bits into a 1-of-8 output. Decoders can be designed with either active high outputs or active low outputs.
### What is 3 to 8 line decoder circuit?
3 Line to 8 Line Decoder . This decoder circuit gives 8 logic outputs for 3 inputs and has a enable pin. The circuit is designed with AND and NAND logic gates. It takes 3 binary inputs and activates one of the eight outputs. 3 to 8 line decoder circuit is also called as binary to an octal decoder.
What is a 3-bit to 8-bit decoder?
This kind of decoder mainly used to decode any 3-bit code & generates eight outputs, equivalent to 8 different combinations for the input code. This decoder is also known as a binary to octal decoder because the inputs of this decoder represent three-bit binary numbers whereas the outputs represent the 8 digits within the octal number system.
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### Archive
Archive for October 27, 2019
## Python: Drawing Shapes
Learning : Drawing Shapes
Subject: New shapes function
To Draw a Square shape, we need to know the width ( W ) of the square side, and then we draw a line and moving in 90 degree and drawing another line and so on until we finished the 4 side of the square. In the same principle if we want to draw a triangle (equilateral one), we need to know length of its sides and in mathematics we know that in equilateral triangles the angles (corners) are 120 degree, so we draw a line and move in 120 degree and drawing another two sides.
In coming code, we will write a general function in Python to pass the number on sides we want to draw (triangle =3, Square=4,Pentagon = 5, Hexagon =6 .. and so on), the width (size) of the shape and the position (x,y) of the first angle or point.
The Codes:
t.goto(x1,y1)
# To get t.right angle
t.pendown()
for x in range (s_heads +1) :
t.forward(w)
t.right(-rang)
t.penup()
Results after using the new function we can pass any number of sides and the function will draw the shape, here are a sample execution of it. .. .. Click to enlarge ..
Now if we call the function number of times equal to it’s heads what we will get ? let’s see . .. Click to enlarge ..
And take a look when we set the numbers to 20. .. Click to enlarge ..
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Ucale
Course Curriculum
Indeterminate Forms What is indeterminate form 00:12:42 Indeterminate form in real world Unlimited Quiz 1: Indeterminate Form Unlimited Limits Introduction to Limits 00:20:05 Why Limits Unlimited When to Solve Limits 00:10:26 Finding Tendency: Question 1 Unlimited Finding Tendency: Question 2 Unlimited Finding Tendency: Question 3 Unlimited Finding Tendency: Question 4 Unlimited Solving Limits Evaluating Left Hand and Right Hand Limit Unlimited Finding LHL and RHL: 1 Unlimited Finding LHL and RHL: 2 Unlimited Finding LHL and RHL: 3 Unlimited Finding LHL and RHL: 4 Unlimited Limits Form 1 Limits Form 1: Zero by Zero 00:10:26 Limits Rationalization Method 00:12:44 Limits Form 1 – Algebraic Limits Factorization and Rationalization Method Unlimited Algebraic Limits Algebraic Limits 00:10:35 Example1: Algebraic Limits Using Some Standard Results 00:09:05 Limits Form 2: Infinity by Infinity Limits Form 2: Infinity by Infinity 00:13:11 Limits Form 2 – Algebraic Limits Factorization and Rationalization Method Unlimited Quiz 2: Algebraic Limits Unlimited Limits Form 3: Trigonometrical Limits Trigonometrical Limits 00:14:09 Some Important Results of Trigonometrical Limits Unlimited Example 1: Limits Form 3 (Trigonometrical Limits) 00:03:38 Example 2: Limits Form 3 (Trigonometrical Limits) 00:05:26 Example 3: Limits Form 3 (Trigonometrical Limits) 00:05:32 Quiz 3: Trigonometrical Limits Unlimited Limits Form 4: Logarithmic Limits Some Useful Expansions Unlimited Logarithmic Limits 00:04:00 Evaluation of Logarithmic Limits Unlimited Example 1: Logarithmic Limits Unlimited Example 2: Logarithmic Limits Unlimited Example 3: Logarithmic Limits Unlimited Quiz 4: Logarithmic Limits Unlimited Limits Form 5: Exponential Limits Exponential Limits 00:05:30 Example 1: Exponential Limits 00:03:20 Example 2: Exponential Limits 00:04:53 Exponential Limits: Result 2 00:04:51 Example 1: Exponential Limits Result 2 00:03:21 Quiz 5: Exponential Limits Unlimited Exponential Limits of the Form One to the Power Infinity Exponential Limits of the Form One to the Power Infinity 2 years, 2 months Example 1: Exponential Limits of the Form One to the Power Infinity 00:07:06 Example 2: Exponential Limits of the Form One to the Power Infinity 00:05:23 Example 3: Exponential Limits of the Form One to the Power Infinity 00:04:46 L-Hospital's Rule L-Hospital’s Rule 00:08:03 Example 1: L-Hospital’s Rule 00:05:27 Continuity of a Function Introduction to Continuity 00:09:48 Continuity of a Function 00:06:21 Equation of a Continuous Function Unlimited Discontinuity of Function Unlimited Example 1: Continuity of a Function 00:05:18 Finding the Continuity: 1 Unlimited Example 2: Continuity of a Function 00:15:51 Example 3: Continuity of a Function 00:06:43 Understanding the Continuity Unlimited Quiz 6: Continuity of Function Unlimited Continuity of a Function at End Points Continuity at End Points 00:05:46 Continuity at End Points Example 1 00:09:03 Continuity of a Function on an Interval Example 1 00:11:33 Differentiation: Definition of Differentiation What is Differentiation? 00:08:52 Differentiation from Limits Unlimited Differentiability of a Function 00:17:40 Differentiation from Graphical Point of View Unlimited Differentiate by Applying Limits 1 Unlimited Quiz 7: Differentiation by Applying Limits Unlimited Differentiation of Some General Functions 00:14:50 Some Important Differentiation of General Functions Unlimited Differentiation of Sum and Difference of Two Functions Derivative of Sum and Difference of Two Functions 00:08:06 Example 1: Derivative of Sum and Difference of Functions 00:03:15 Example 2: Derivative of Sum and Difference of Functions 00:04:07 Quiz 8: Differentiation of Sum of Functions Unlimited Differentiation of Product of Two Functions Derivative of Product of Two Functions 00:06:22 Example 1: Derivative of Product of Two Functions 00:04:07 Example 2: Derivative of Product of Two Functions 00:03:40 Derivative of Product of Three or More Functions 00:07:20 Example 1: Derivative of Product of Three or More Functions 00:11:44 Example 2: Derivative of Product of Three or More Functions 00:07:40 Quiz 9: Differentiation of Product of Functions Unlimited Differentiation Quotient Rule: Derivative of f(x)/g(x) Quotient Rule: Differentiation of f(x) by g(x) 00:04:46 Example 1: Differentiation of Function in the form f(x)/g(x) 00:04:55 Example 2: Differentiation of Function in the form f(x)/g(x) 00:05:52 Quiz 10: Differentiation of the Ratio of Two Functions Unlimited Differentiation by Chain Rule Differentiation by Chain Rule 00:08:37 Examples Differentiation by Chain Rule 00:09:31 Differentiation by Applying all Rules 00:18:14 Differentiation of Implicit Function Differentiation of Implicit Functions 00:08:38 Quiz 12: Differentiation of Implicit Function Unlimited Differentiation of Logarithmic Function Differentiation of Logarithmic Function 00:09:52 Example Set 1: Logarithmic Differentiation 00:08:23 Example Set 2: Logarithmic Differentiation 00:13:21 Differentiation of Infinite Series Differentiation of Infinite Series 00:06:02 Differentiation of a Function w.r.t another Function Differentiation of a Function w.r.t Another Function 00:07:24 Differentiation of Function of Function Unlimited Example: Differentiation of a Function w.r.t. another Function 00:03:40 Quiz 11: Differentiation of Function of Function Unlimited Differentiation of Parametric Function Differentiation of Parametric Function 00:07:12 Example Set 1: Differentiation of Parametric Functions 00:06:44 Example Set 2: Differentiation of Parametric Functions 00:04:22 Higher Order Derivative Higher Order Derivative 00:09:35 Example 1: Higher Order Derivatives 00:03:29 Example 2: Higher Order Derivatives 00:09:06 Application of Derivative Derivative as a Rate Measure 00:06:17 Example 1: Derivative as a rate measure 00:06:54 Example 2: Derivative as a rate measure 00:05:14 Example 3: Derivative as a Rate Measure 00:07:15 Velocity of an object Unlimited Acceleration of an object Unlimited Rate of change of volume w.r.t radius Unlimited Rate of change of surface area w.r.t radius Unlimited Quiz 13: Applying Differentiation in real world Unlimited Rolle’s Theorem 00:13:40 How to verify Rolles theorem 00:11:35 Lagranges Mean Value Theorem 00:05:46 How to verify Lagrange`s Mean Value Theorem 00:04:58 Tangents and Normals Slope (Gradient) of a line 00:06:44 Slopes of the Tangent and the Normal 00:08:04 Example Set 1: Tangents and Normals 00:08:10 Example Set 2: Tangents and Normals 00:04:49 Equations of the Tangents and the Normals 00:09:05 Tangent on a curve Unlimited Normal on a curve Unlimited Example Set 1: Equation of the Tangent and the Normals 00:08:29 Example Set 2: Equation of the Tangent and the Normals 00:09:54 Quiz 14: Tangents and Normals Unlimited Increasing and Decreasing Function Increasing Function 00:12:03 Decreasing Function 00:10:33 How to Find the Intervals Where a Function is Increasing or Decreasing 00:05:28 Wavy Curve Method or sign Scheme for Rational Functions 00:10:49 Example Set 1: Increasing and Decreasing Function 00:12:33 Example Set 2: Increasing and Decreasing Function 00:06:21
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# 4000 milligrams equals how many grams
• the mass 4 grams
• tk10npubl tk10ncanl
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https://puzzling.stackexchange.com/questions/125761/place-numbers-1-through-9-in-boxes-%C3%97
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# Place numbers 1 through 9 in boxes (☐☐×☐=☐☐+☐☐=☐☐)
So recently I was scrolling through Youtube when I came across this video from MindYourDecisions that was about solving a legendary math puzzle.
The puzzle:
Place the numbers 1 through 9 in the following boxes such that
☐☐
x☐
---
☐☐
+☐☐
---
☐☐
Each number can only be used once.
Here is the thumbnail of the Youtube video for anyone who wants to see what it looks like:
Now, I recognized this puzzle as one that I have been struggling to solve on and off for over a decade, and that is bad because I still have not managed to solve this.
### What I can really easily deduce
1. If we have 9 in the topmost box in the left column, then only if we multiply by 1 will we be able to satisfy the condition that we have a 2-digit number multiplied by a 1-digit number that is then equal to a 2-digit number. But, this leads to a contradiction because we have to use every number once.
2. We can use the same logic to show that 9 cannot go in the 2nd and 3rd topmost boxes, because we require that we have 2 2-digit numbers being added to each other that is equal to another 2-digit number. However, this cannot be satisfied if we have 9A+BC (with AB≠A*B, but rather representing concatenation).
3. We can also use the same logic as 1) to show that 5, 6, 7, and 8, also cannot go in the topmost box in the left column.
4. We can notice that since the multiplier cannot be 1, then it follows that there cannot be a 1 in the second topmost box or the bottom box in the left column.
5. We also cannot have the numbers 8 or 9 as the multiplier, as the minimum value given from any multiplication would result in 96. We can also probably eliminate 6 and 7 from being the multiplier, although I'm not too sure if that would be correct to do.
So this is what I have done so far (a blue number means that I am 100% sure it does not go there, a red number means that I am pretty (but not 100%) sure that it could not go there):
However, I am actually unsure about how I would deduce logically (by hand) where the numbers go, so my question is:
### What should I do to actually solve this?
Edit: Sorry, forgot to mention that I am looking for a logical way to hand-solve this.
• Without no-computers it's trivial to brute force. Do you just want the numerical solution, or are you specifically after a logical way to hand-solve this?
– fljx
Feb 28 at 15:57
• FYI, the video's solution method is not elegant, as it was essentially testing a huge number of cases, only one of which leads to a solution, to the point where one is just emulating a computer's brute force search. Feb 28 at 16:53
• You also can't have 5s in the top three spots of the right column. Feb 28 at 17:24
• FWIW case bashing to death is a kind of logic. Some "higher-level logic" can help you reduce the amount of work, but I believe you'll need to weed through at least some tens of cases in the end. Feb 29 at 0:56
• @loopywalt You're right. For some reason I misread the entire thing as a long multiplication summing two products to a final answer (but that would require a two-digit multiplier, and the answer would never be only two digits). Not sure what I was thinking :-/
– fljx
Feb 29 at 11:35
I would focus on the single digit number
X * (6+) Total not possible 13*6+24 >98)
X * (5 or 1) Last digit in product not available
X * 4:
X not 19+ Total not possible $$19 * 4+23>98$$
X not 11,14,15 or 16 Last digit in product not available
$$12*4$$ First digit in product not available
$$18*4$$ ends on 2 Total not possible $$18*4+35>98$$
$$13*4 =52$$ Total not possible $$13*4+67>98$$
Only $$17 * 4$$ left to check: Solution! $$17*4 =68+25=93$$
X * 3:
X * 2:
Still a lot of possibilities to check for uniqueness. I don't see an obvious overall smart way to do that.
though e.g.:
$$4?*2 =?? +??=??$$ ->
$$4?*2 =8? +1?=9?$$ ->
$$43*2 =86 +1?=9?$$ (6 is only remaining even number) -> no solution with $$4?*2$$
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https://www.topperlearning.com/doubts-solutions/q-how-many-different-signals-can-be-given-using-any-number-of-flags-from-5-flags-of-different-colours-fddjwldd/
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# Q) How many different signals can be given using any number of flags from 5 flags of different colours.
Asked by Anish 29th November 2018, 9:28 AM
1 flag signal = 5P1 ways
2 flag signal = 5P2 ways
3 flag signal = 5P3 ways
4 flag signal = 5P4 ways
5 flag signal = 5P5 ways
Answered by Expert 29th November 2018, 9:47 AM
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https://www.quizzcreator.com/quiz/1686/quiz-mock-tests-on-analysis-of-algorithms
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# Quiz: Mock Tests on Analysis of Algorithms.
## The quiz contains some basic information of Analysis of Algorithms. In computer science, the analysis of algorithms is the process of finding the computational complexity of algorithms – the amount of time, storage, or other resources needed to execute them How do we analyze algorithms? Analysis of Algorithms. Determine the time required for each basic operation. Identify unknown quantities that can be used to describe the frequency of execution of the basic operations. Develop a realistic model for the input to the program. Analyze the unknown quantities, assuming the modeled input
You can mute/unmute sounds from here
Arther Alan
Max Latham
Donald Knuth
valied
regular
sorting
Quick Sort
Selection Sort
Merge Sort
object
plain text
cipher
n2
0(log(n))
n(logn)
twice
Threshold
ten time
## Algorithm analysis is an important part of a broader ........................ theory.
computational complexity
object collection
nine
three
six
## In terms of Algorithm analysis, the uniform-cost measurement known as-
estimate and costing
uniform cost model
Variant value
## Logarithmic-cost measurement also known as-
Alg value managment
Log value model
logarithmic cost model
## Which thing estimates and anticipates the increase in running time of an algorithm as its input size increases?
Process length chart
Run-time analysis
binary search
data search
array search
## Analysis of algorithms typically focuses on the-
certain search tree
computer processing unit
asymptotic performance
## Which algorithm design technique can find all the pairs of shortest distances in a graph?
Analog programming
Dynamic programming
Relational programming
## Which algorithm is best known for finding the shortest paths between nodes in a graph?
Shortest algorithm
Lothar's algorithm
Dijkstra's algorithm
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# Trading Iron Condors: Are you Getting a Fair Price?
Hi Mark,
Can you tell me how to arrive at a fair value for an Iron Condor?
Ivan R
***
Ivan,
This is a topic I have long ignored. It's a very good question.
In my opinion, it's not worth the effort to determine the theoretical value of each of the four options. You would have to make a very good estimate of the future volatility of the underlying asset because you need that volatility number for the calculator.
If you don't already know, let me assure you that it's not so easy to make that volatility estimate. Those who are truly skilled in predicting future volatility have a huge edge over everyone else who trades options.
But, if you want to do the math, you can still get yourself a free option calculator and make the calculations for each of the options that comprise the iron condor.
Another choice is to use the theoretical values offered by your broker. The only problem with that is you never know how they chose their estimated volatility – or whether it's a reasonable estimate.
I prefer to look at trading an iron condor this way:
1) When you trade an iron condor, you are a net seller of vega, and that's the component of an option's value that changes as implied volatility changes.
Clarification: When IV increases, the market price of an iron condor increases. When IV decreases, the market price of an iron condor decreases.
2) If you believe the current implied volatility (IV) is high (and likely to move lower), then you are going to get a good price for the iron condor. In other words, using the current IV, the market price for the IC will exceed your calculated theoretical value.
I recognize that you may not have a good feeling for whether current IV is high, but if you cannot do that, then you would not be able to determine a theoretical value for the iron condor. Thus, you are no worse off.
3) If you believe current IV is low, that's not the best time to be trading iron condors – unless you believe IV is going lower. Note, it's not a terrible time – but you probably don't have any theoretical edge going into the trade.
4) If you truly don't know whether IV is high or low – and most traders don't know, you have two choices:
a) Trade the iron condor anyway. Over time you will get good prices part of the time and poor prices the rest of the time. On average you should be okay.
b) Look up the stock's historical volatility over the past 20, 50, or 100 days. If you believe those numbers give you a reasonable estimate of future (from the present time until the options expire) stock volatility, then compare that volatility with current IV. If current is high, trade. If it's low, pass.
Here is how to use those numbers from Lawrence McMillan's web site:
The columns hv20, hv50, and hv100 represent the actual stock volatility over the past 20, 50, and 100 days.
curiv is the current implied volatility.
Percentile is the percentage of the time that the current IV has been LOWER. A high number means that IV is relatively high.
564
### 6 Responses to Trading Iron Condors: Are you Getting a Fair Price?
1. Bill 12/31/2009 at 1:07 PM #
Good practical discussion.
Congratulations on a year of excellent and detailed teaching about the nuances of options trading. You’ve created a daily must-see site for anyone that wants to learn how to make a living in this field.
Best wishes for a happy new year.
2. greg 12/31/2009 at 1:53 PM #
Hi Mark
Is vega a derivative of Gamma or is vega a derivative of implied Volatility? Am I also correct in understanding that Gamma is a derivative of Delta?
Regarding vega, do you want to see it negative rather than positive or do you want it as close to zero as possible.
Is vega as important for vertical credit put spreads with pre-insurance as it is for IC’s which you have been discussing?
Thanks
3. Mark Wolfinger 12/31/2009 at 6:55 PM #
Thanks Bill.
Much appreciated.
Happy New Year and good trading to you
4. Mark Wolfinger 12/31/2009 at 7:06 PM #
Greg,
1) Vega is not a derivative of gamma.
Vega is the derivative of a change in the value of an option with respect to a one point change in the implied volatility.
2) Yes. Gamma is a derivative of delta.
Delta is the derivative of the change in option value with a one point change in the stock price.
Gamma is the 2nd derivative and is a measure of the change in delta when the stock price changes by one point.
3) I only keep vega close to zero when I have no true feeling for which way implied volatility is going to move next.
When I have no opinion, I am short vega. I trade my iron condors and am short vega.
If I believe IV is just too low to sell, then I’ll add some positions with positive vega to neutralize vega risk.
See Wikipedia for some practical information on derivatives.
5. Jason 01/01/2010 at 1:51 PM #
Hi Marc,
Happy New Year, and thanks again for the blog. When you say IV, do you mean IV for the specific strikes, or do you mean IV as per (say) the VIX/ATM strike? It seems like taking the average IV of your short legs as a quick rule of thumb might help adjust for skew, or is that not really a big problem?
6. Mark Wolfinger 01/01/2010 at 4:25 PM #
Jason,
Skew is an issue I have not touched upon. I’ll share my thoughts in a blog post.
Happy New Year. Thanks for being a visitor.
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Calculate outs in texas holdem
Nov 18, · wait for the flop!!!!! if you are on the button, it wont cost you to see the MATMM.ME say you have a pair of fours and there is mariangl betting which shows you the rest of the table have so-so hands and want to see the flop MATMM.ME a pair of fours you may get trips or low cards for a possible straight and bet heavy on the river always remember the face cards will beat MATMM.ME in if the. Poker Odds - Calculating Hand Odds In Texas Hold'em Poker & Charts. Learning how to properly count your outs and calculate poker odds is a fundamental requirement of Texas Hold'em. While the math used to calculate odds might sound scary and over the head of a new player, it really isn't as hard as it looks. Sep 18, · This poker calculator will give you the odds of a win, loss, and tie for each player. Click on any card and it will be used in the position indicated by the yellow frame. You may click on any valid card to move the frame there. Check "folded hand" to indicate the given player is out of the hand, removing his cards from the remaining deck.
Texas Hold'em Poker Odds Cheat Sheet
Tips and Warnings. Subtract 0. Related wikiHows. Find the flop-to-river percentage: This is a more difficult calculation because it involves 2 cards in separate draws, the turn and river. This article has also been viewed 60, times. In Texas Hold'em, you'll be dealt 2 cards.
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The annual profit from the gaming industry in the US is 18 billion dollars.
Once the flop has been dealt in Texas Hold'em, you'll be able to count your outs and know how likely it is your hand will improve. That will tell you whether you should stay in the hand or fold. You can figure out your outs and odds for any hand, but here is a quick and dirty list of the most common scenarios:.
There are four remaining cards of two different numbers that will complete your straight, on the high end and on the low end. Nine outs: Your odds are 2 to 1 about 35 percent This is the common scenario when you have a flush draw. Any of the nine remaining cards of the suit will give you a flush. Fifteen outs: Your odds are 1 to 1 about 54 percent A scenario for this is having a straight and flush draw, where either any of the nine remaining cards of the suit will give you a flush, while there are four cards remaining of each of two numbers that would complete a straight.
However, you don't count the same cards twice as outs, so those of suit you hope to get don't count again. These odds only apply to counting both the turn and the river, so they assume you will stay in the hand until the showdown. Your odds are only about half as good for a single card draw, such taking the hit on the turn or taking the hit on the river. A common way of looking at the difference in the odds when you will be seeing two cards compared with one is called the Rule of 4 and 2.
After the flop, count your outs and multiply them by four to get your percentage odds. This doesn't give you an exact number, but it is quickly in the ballpark. However, when you are calculating the odds that a single draw will improve your hand, you multiply the outs by two rather than 4. Toby Bochan. Updated March 06,
Details
Updated: October 15, References. Knowing how to calculate your odds of building a strong hand is a key step in becoming a good poker player. The calculation used in determining poker odds is influenced by a variety of factors but can be deduced using simple arithmetic.
By understanding your chances of drawing a desirable hand, you can increase your long-term profitability.
If you want to learn poker percentages, follow these guidelines. Log in Facebook. No account yet? Create an account. We use cookies to make wikiHow great. By using our site, you agree to our cookie policy.
Article Edit. Learn why people trust wikiHow. To create this article, 10 people, some anonymous, worked to edit and improve it over time. Together, they cited 10 references. This article has also been viewed 60, times. Learn more Explore this Article Steps. Tips and Warnings. Related Articles. Learn the percentages of your favorite poker game. Determining poker odds depends in large part on the game you're playing.
For example, the formula for getting a particular hand in 7-card stud is different from that of Texas Hold 'em, arguably the most popular poker game in the world. Consider all cogent variables before calculating hand odds.
Odds for royal flush texas holdem
Typically the hero of the movie is dealt this hand and it is revealed in a dramatic fashion. A royal flush is the highest ranked hand in the card game of poker. Due to the specifications for this hand, it is very difficult to be dealt a royal flush. There is a multitude of different ways that poker can be played. For our purposes, we will assume that a player is dealt five cards from a standard 52 card deck. No cards are wild, and the player keeps all of the cards that are dealt to him or her.
Once we know these two numbers, the probability of being dealt a royal flush is a simple calculation. All that we have to do is to divide the second number by the first number. Some of the techniques of combinatorics , or the study of counting, can be applied to calculate the total number of poker hands.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A225896 Number of nX4 binary arrays whose sum with another nX4 binary array containing no more than a single 1 has rows and columns in lexicographically nondecreasing order 1
11, 78, 545, 3459, 19270, 93428, 396804, 1495926, 5079770, 15751596, 45136888, 120738965, 304047638, 725911911, 1652919000, 3607621216, 7579528833, 15385065750, 30266189863, 57863160815, 107762341802, 195911924170 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 4 of A225900 LINKS R. H. Hardin, Table of n, a(n) for n = 1..169 FORMULA Empirical: a(n) = (1/653837184000)*n^16 + (29/186810624000)*n^15 + (71/10059033600)*n^14 + (497/2668723200)*n^13 + (11441/3592512000)*n^12 + (133949/3592512000)*n^11 + (139831/457228800)*n^10 + (227729/130636800)*n^9 + (34734211/4572288000)*n^8 + (43988369/1306368000)*n^7 + (227051569/1437004800)*n^6 + (222061289/718502400)*n^5 + (62570170547/54486432000)*n^4 + (837269/1001000)*n^3 + (36008363/5821200)*n^2 - (302873/180180)*n + 4 EXAMPLE Some solutions for n=3 ..0..0..0..1....0..0..0..1....0..0..1..0....0..0..0..1....0..0..0..0 ..1..1..0..0....0..0..0..1....0..0..1..1....0..0..1..1....0..0..1..1 ..1..1..1..0....0..1..0..0....0..0..1..1....1..1..0..1....0..1..1..0 CROSSREFS Sequence in context: A118936 A041224 A030054 * A239437 A140542 A101983 Adjacent sequences: A225893 A225894 A225895 * A225897 A225898 A225899 KEYWORD nonn AUTHOR R. H. Hardin May 20 2013 STATUS approved
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Last modified August 5 06:16 EDT 2024. Contains 374935 sequences. (Running on oeis4.)
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Just Ask Happy: The Exclusive Compilation
We asked for you to send in your questions for Happy Holden, and you took us up on it! We loved them so much, and we know that you did too, so we’ve compiled all 21 questions and answers into one document for easy reference.
Calculating Trace Temps in a Vacuum
Q: For space applications (without air), how should we calculate external layer current-carrying traces against the IPC-2221 (formerly IPC-D-275) charts?
A: I have never studied that, so I turned this question over to an expert, my friend Mike Jouppi, former committee chair of IPC-2152 Standard for Determining Current Carrying Capacity in Printed Board Design. This standard will provide many answers to your questions. Mike wrote Chapters 22 and 23 in the seventh edition of the Printed Circuits Handbook, edited by Clyde F. Coombs and me.
Mike answered: I'm a mechanical engineer who worked as a career thermal analyst. The charts in IPC-2152 in almost all cases will be conservative (both air and vacuum environments). The vacuum is for space environments. The purpose behind these charts is misconstrued by most users. My intention when they were developed was to use these charts as a baseline for developing thermal models that could be used to better understand the actual temperature rise of conductors in actual designs, which I did for my own design purposes. The concept did not catch on.
There is a significant difference between the temperature rise in a conductor, tested per IPC-TM-2.5.4.1a, and most PWB design configurations. The reason is that most designs have copper ground and power planes that conduct energy away from the traces. In addition, most designs in space applications have a significant conduction path from the PWB through-bolted fasteners or wedge locks to a sink.
Since the question does not include IPC-2152, I would recommend researching IPC-2152 and accounting for the power dissipations in the traces in the thermal design. I also recommend accounting for the conductor power dissipation in all designs, especially if the designers are not familiar with sizing parallel conductors. Parallel conductors are easily managed with accounting for conductor losses (power dissipation). FYI: It was the power dissipation in conductors that motivated me to lead the development of IPC-2152.
Ranking the Top Countries by Fab Technology and Production
Q: In terms of overall PCB fabrication capability, how would you rank these countries in terms of technology and production, today and in 10 years: USA, Japan, Korea, Taiwan, China, Thailand, Vietnam, India, Germany, and the U.K.?
A: I don’t have a crystal ball, but here is my best guess:
Manufacturing Issues From a Designer’s Viewpoint
Q:
What do you think are the biggest PCB fabrication/assembly issues from a designer’s perspective?
A: Communication and education! Designers need to know a lot about electrical circuits and their performance, but equally important is knowing how that circuit will be manufactured. But manufacturing is usually not part of a designer’s education, so designers have to go out of their way to meet the manufacturing community and to communicate with them on issues related to their designs. IPC and SMTA make this type of training available, but a designer has to be aggressive in learning about manufacturing and all the do’s and don’ts.
Flexible Circuit Technology of the Future
Q: What do you think will be the next leading-edge technology for flexible circuits?
A: Disposable and wearable substrates, including paper
Monitoring Via Reliability on a Lot-by-Lot Basis
Q: What is the best way to monitor microvia reliability on a lot-by-lot basis for high-volume production?
A: What I recommend, and practice, is this: Do a thorough qualification of a fabricator using a HATS Qualification Panel (an IPC PCQRR-like panel) that includes an IPC D coupon resized for the runner on your SMT panel. Run these panel coupons through IPC-TM-650-2.6.27B and then TM-650-2.6.7.2 thermal cycles for 500 cycles. This is your baseline.
Then, by incorporating this smaller D coupon on all your boards’ assembly runners—requiring each lot to have a plating process control coupon for X number of solder float tests at 288°C and then microsections to look at TH quality—you have a way of comparing that lot’s performance to your initial qualification for that vendor. This should be redone each year and kept for records on each lot in case needed later.
The Future of 3D Printing
Q: What is your opinion on the 3D printers that can utilize both conductive and non-conductive inks? Is this the future of PCBs?
A: These 3D printers are evolving rapidly. A must is that the printer has the ability to use a number of different inks for the different components of a PCB, including the curing of the inks. At a minimum, that is:
1. Insulating substrate
2. Low-ohmic conductive ink
3. Non-conductive insulating crossover
As the equipment improves, solder masks, resistive/capacitive and solder paste inks, and even semiconductor inks would be nice. Currently, this is a quick-turn application for a few breadboards.
Whether this technology moves into being a source of operational electronics with sufficient reliability will depend on the innovations in creating future inks. Conventional PCB production continues to innovate to reduce their turnaround times and costs, forcing 3D printing to work that much harder on their solutions.
What Would You Change About the Industry?
Q: If you could change/improve one aspect of this industry, what would it be, and why?
A: TQM/continuous improvement (or Six Sigma) is a philosophy that has stuck with me since I was first exposed to it. I even taught the engineering aspects of TQM to every one of my engineers and techs; this included engineering statistics from the free NIST Engineering Statistics Handbook, as well as enhanced problem solving. Then, I would have my engineers teach TQM to their production supervisors, foremen, and leads, who would—with the help of the engineers and techs—teach it to all of the workers.
Continuous improvement is a strategy for success. And in our challenging industry characterized by innovation and change, continuous improvement and customer satisfaction are the only ways to survive and prosper.
Is PCB Fab Returning to the U.S.?
Q: If the U.S. brings more PCB fabrication back to America, will it more likely be to the captive or contract shops?
A: There are only a few captive fabricators in the U.S., but I hope that more OEMs will consider going captive as a way to ensure a source of supplies, lower costs, minimum lead times, and continuous improvements. OEMs should emulate the activities of Whelen Engineering, the emergency lighting company in New Hampshire that went captive and returned all of its PCB fabrication from China. Whelen cut its costs in half, improved quality, ensured its IP security, and cut weeks off their cycle time—all with an investment that had an ROI of fewer than two years.
But if PCB fabrication is to return to the U.S. from abroad, the majority will be to contract shops. If so, I hope that the OEMs will consider partnering with those contract shops and not treating their manufacturing “as just another commodity,” just as they don’t consider their proprietary circuits to be commodities. How circuits are manufactured is not like making soap; each PCB is different and deserves the necessary focus and care in its construction.
Electroless Copper vs. Direct Metallization
Q: What are your thoughts on electroless copper vs. direct metallization?
A: Both get the job done! Electroless copper is the more traditional method but may have some problems with its internal crystalline strength when used with stacked microvias. Direct metallization is a newer technology that has come a long way in their development and is one of the few ways to successfully metallize polyimide film for flex.
New Book From Isola Highlights Importance of Material Selection
01/14/2022 | I-Connect007 Editorial Team
In "The Printed Circuit Designer’s Guide to... High Performance Materials," the latest release from I-007eBooks, readers will learn how to overcome challenges associated with choosing the right material for their specific application. Author Michael Gay of Isola provides a clearer picture of what to know when determining which material is the most desirable for which products. PCB materials and DFM expert Mark Thompson says, “I love this book, particularly the sections on the effects of the glass weave, the history of laminate, and the difference between Dk and effective Dk."
Living in a Material World: High-Speed Design Strategies
01/13/2022 | I-Connect007 Editorial Team
Any discussion about high-speed PCB design techniques would be incomplete without considering the properties and requirements of the materials. Your material selection drives much of your design strategy when you’re operating at 28 gigabits per second or faster. We recently spoke with high-speed design expert Lee Ritchey of Speeding Edge, and electronic materials veteran Tarun Amla of Avishtech and Thintronics, about the relationship between advanced PCB materials and high-speed design techniques. They discuss the challenges facing designers and engineers working with materials at speeds that were considered unreachable not long ago, and what designers need to know about material selection as board speeds continue rising toward the stratosphere.
RealTime with... American Standard Circuits: Thermal Management
12/16/2021 | Pete Starkey, I-Connect007
In the third of a series of three RealTime with... interviews, I-Connect007 managing editor Nolan Johnson received knowledgeable and informative answers from Anaya Vardya, John Bushie, and Dave Lackey of American Standard Circuits to his questions on the topic of thermal management. Anaya Vardya began by clarifying the terminology, describing thermal conductivity as a material property defining how quickly heat was transmitted through a piece of that material, whereas thermal management was about analysing the entire system, trying to understand how much heat was being generated, and using appropriate techniques to dissipate that heat as efficiently as possible.
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Search a number
957726 = 232711691
BaseRepresentation
bin11101001110100011110
31210122202100
43221310132
5221121401
632305530
711066130
oct3516436
91718670
10957726
115a4610
123a22a6
13276c03
1513db86
hexe9d1e
957726 has 48 divisors (see below), whose sum is σ = 2590848. Its totient is φ = 248400.
The previous prime is 957721. The next prime is 957731. The reversal of 957726 is 627759.
It is an interprime number because it is at equal distance from previous prime (957721) and next prime (957731).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (957721) by changing a digit.
957726 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 1041 + ... + 1731.
It is an arithmetic number, because the mean of its divisors is an integer number (53976).
2957726 is an apocalyptic number.
It is a practical number, because each smaller number is the sum of distinct divisors of 957726, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1295424).
957726 is an abundant number, since it is smaller than the sum of its proper divisors (1633122).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
957726 is a wasteful number, since it uses less digits than its factorization.
957726 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 717 (or 714 counting only the distinct ones).
The product of its digits is 26460, while the sum is 36.
The square root of 957726 is about 978.6347633310. The cubic root of 957726 is about 98.5705301926.
The spelling of 957726 in words is "nine hundred fifty-seven thousand, seven hundred twenty-six".
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; Confidence Intervals
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# Confidence Intervals
VIEWS: 4 PAGES: 3
• pg 1
``` Confidence Intervals
Guy Lebanon
February 23, 2006
Confidence intervals (CI) is an important part of statistical inference. It refers to obtaining statements
such as P (a(X1 , . . . , Xn ) ≤ θ ≤ b(X1 , . . . , Xn )) = 1 − α, where θ is the parameter of interest and a, b are
quantities computed based on the iid sample X1 , . . . , Xn . The probability 1 − α is called the confidence
ˆ
coefficient and 1 − α is typically taken to be 0.9, 0.95 or 0.99. In contrast to point estimators θ which give us
a specific guess for θ, CIs provide an interval - which is less accurate than a specific number. The advantage
of confidence intervals is that we can characterize the confidence of our statement θ ∈ [a, b]. CIs of the form
(−∞, b] or [a, ∞) are called one-sided CIs (lower or upper).
In general, to construct a CI, we need to know some partial information concerning the unknown distri-
bution - for example that it is a normal distribution. Such CIs are called small sample confidence intervals.
If we can not make such an assumption we can still construct CIs by appealing to the central limit theorem.
However, in this case, the CI will be only approximately correct - with the approximation improving in its
quality as the sample size increases n → ∞. Such CIs are called large sample CIs.
One of the most useful methods for constructing CIs is the method of pivotal quantities. This method
constructs first CIs for an auxiliary quantity called a pivot, and then transforms the interval into a CI for
the parameter θ.
Definition 1. A pivot is a function of θ, X1 , . . . , Xn whose distribution does not depend on θ.
Typically, the chosen pivots g(θ, X1 , . . . , Xn ) have N (0, 1), χ2 , t or F distributions. Since all of these
distributions are well tabulated it is easy to obtain confidence intervals for the pivots
P (a ≤ g(θ, X1 , . . . , Xn ) ≤ b) = 1 − α.
For example, if the pivot has a N (0, 1) distribution, b = −a = zα/2 , which for 1 − α = 0.95 is zα/2 = 1.96.
This last observation, together with the fact that for N (µ, σ 2 ), zα/2 = µ + 1.96σ is the source of the (not
very good) practice of estimating the standard deviation of a RV by a quarter of the range of possible values
(range-or-possible-values ≈ [µ − 2σ, µ + 2σ]).
Transforming the pivot CI P (a ≤ g(θ, X1 , . . . , Xn ) ≤ b) to a θ CI P (a(X1 , . . . , Xn ) ≤ θ ≤ b(X1 , . . . , Xn ))
may be done by
1. adding a real number to all three sides of the inequality
2. multiplying by a positive number all three sides of the inequality
3. multiplying by a negative number all three sides of the inequality (while reversing inequality signs)
4. taking the inverse (·)−1 of all three sides of the inequality (while reversing inequality signs).
Example: Suppose we have a single observation X from an exponential distribution whose expectation
θ we are interested in. The transformation method may be used to show that X/θ is an exponential RV
with parameter 1. That is X/θ is a pivot whose distribution does not depend on θ. We start by obtaining a
confidence interval for the pivot (from tables of exponential distribution percentiles) P (a ≤ X/θ ≤ b) = 1−α
and proceed by dividing by X all three sides of the inequality and inverting to obtain a CI on θ:
P (a ≤ X/θ ≤ b) = 1 − α ⇒ P (X/a ≥ θ ≥ Y /b) = 1 − α.
1
Example: Suppose we have a single observation X from a uniform distribution U ([0, θ]) and we are
interested in a confidence interval on θ. As before, the transformation method can be used to show that
X/θ ∼ U ([0, 1]) and therefore a pivot. A lower 0.95 confidence interval for the pivot 0.95 = P (X/θ ≤ 0.95)
transforms to a confidence interval on θ by dividing by X and taking the inverse of both sides 0.95 =
P (X/θ ≤ 0.95) = P (θ ≥ X/0.95).
Large Sample Confidence Intervals for Means
Consider the case where we have an iid sample X1 , . . . , Xn (n is assumed to be large e.g., > 30) drawn from
an unknown distribution with expectation µ. We are interested in constructing confidence intervals for µ
¯
using X. Since we don’t know the distribution of the sample we can’t use the pivot method. The solution is
to use the central limit theorem approximation to obtain a N (0, 1) pivot. More specifically, the CLT provides
the following N (0, 1) pivot (approximately)
¯ n
√ X −µ i=1 (Xi − µ)
n = √ ≈n→∞ Z ∼ N (0, 1).
σ σ n
We then first obtain confidence intervals for the pivot Z: 1 − α = P (−zα/2 ≤ Z ≤ zα/2 ) and then transform
it to an approximate CI on µ
¯
√ X −µ σzα/2 σzα/2
1 − α = P (−zα/2 ≤ Z ≤ zα/2 ) ≈ P −zα/2 ≤ n ≤ zα/2 =P − √ ¯
≤X −µ≤ √
σ n n
=P ¯ σzα/2 ≥ µ ≥ X − σzα/2
X+ √ ¯ √ .
n n
If we don’t know σ, the above CI may be approximated further using the estimator S 2 ≈ σ 2 to yield
1−α≈P ¯ Szα/2 ≥ µ ≥ X − Szα/2
X+ √ ¯ √ .
n n
Above, we assumed that based on fixed α, n we calculated the resulting confidence interval. One could
reverse the reasoning as follows. We may ask what is the sample size n that will provide a specific confidence
¯ ¯
interval θ ∈ [X − a, X + a] at a specific confidence level 1 − α. In this case we should take
√ √
Szα/2 / n = a ⇒ n = Szα/2 /a ⇒ n ≥ (Szα/2 /a)2 ,
where we use inequality since n has to be integer while (Szα/2 /a)2 is not necessarily an integer (If σ is
known, it should replace S above).
Small Sample Confidence Intervals
If we know the distribution of the data we can do better than the large sample approximations based on
the central limit theorem. Specifically, in this section we assume that X1 , . . . , Xn ∼ N (µ1 , σ 2 ), Y1 , . . . , Ym ∼
¯ ¯ n ¯ m ¯
N (µ2 , σ 2 ). X and Y are as before and S1 = (n − 1)−1 i=1 (Xi − X)2 and S2 = (m − 1)−1 i=1 (Yi − Y )2 .
¯
X−µ1
Confidence interval for µ1 : The pivot S1 /√n has a t distribution with n − 1 dof. It leads to the CI
¯
X−µ1
1 − α = P (−tα/2 ≤ √
S1 / n
≤ tα/2 ), which after simple manipulations yields
¯ S1 ¯ S1
1−α=P X − tα/2 √ ≤ µ1 ≤ X + tα/2 √ .
n n
Confidence interval for µ1 − µ2 : If n = m the CI may be obtained by a simple derivation similar to the
one above. However, if n = m we need to be more careful. Recall that for Z ∼ N (0, 1) and W ∼ χ2 , we
ν
have √ Z ∼ t(ν) . We will use the RV √ Z ∼ t(ν) as a pivot with
W/ν W/ν
¯ ¯
X − Y − (µ1 − µ2 ) ¯ ¯
X − Y − (µ1 − µ2 )
Z= = ∼ N (0, 1)
¯ ¯
Var(X − Y ) σ 2 /n + σ 2 /m
2
¯ ¯
(this is a standartized normal RV since X − Y is a linear combination of normal RVs and therefore is a normal
RV, and we substract its mean and divide by its standard deviation). For W in the pivot √ Z ∼ t(ν) , we
W/ν
choose
2 2
(n − 1)S1 (m − 1)S2
W = + ∼ χ2 2
(n−1+m−1) = χ(n+m−2)
σ2 σ2
ν
(recall that a chi-squared RV χ2 is the same as a sum of ν squared standard normals
(ν) j=1 Zi and therefore
2 2
(n−1)S1 (m−1)S2
the sum of σ2 ∼ χ2
(n−1) and ∼
σ2 χ2
is the same as a sum of n + m − 2 standard normal RVs
(m−1)
which is χ2
(n+m−2) ). Substituting Z and W above in the pivot √ Z ∼ t(ν) gives the following CI
W/ν
¯ ¯
X − Y − (µ1 − µ2 )
1 − α =P −tα/2 ≤ 2 2
/ ((n − 1)S1 + (m − 1)S2 )σ −2 (n + m − 2)−1 ≤ tα/2
σ 2 /n + σ 2 /m
¯ ¯
X − Y − (µ1 − µ2 )
=P −tα/2 ≤ 2 2
/ ((n − 1)S1 + (m − 1)S2 )(n + m − 2)−1 ≤ tα/2
1/n + 1/m
¯ ¯
X − Y − (µ1 − µ2 )
=P −tα/2 ≤ ≤ tα/2
Sp 1/n + 1/m
(n−1)S 2 +(m−1)S 2
using the notation Sp = 1
n+m−2
2
for the pooled (or weighted average) version of the two variance
estimators. The above CI may be manipulated to obtain a CI for the desired parameter µ1 − µ2
¯ ¯
1 − α = P X − Y − tα/2 Sp ¯ ¯
1/n + 1/m ≤ µ1 − µ2 ≤ X − Y + tα/2 Sp 1/n + 1/m .
2
(n−1)S1
Confidence Intervals for σ 2 : We use the pivot σ2 ∼ χ2
(n−1) to obtain the CI
2
(n − 1)S1
1−α=P a≤ ≤b for appropriate a, b chosen from the χ2
(n−1) table
σ2
(note that the pivot χ2 distribution is not symmetric and is non-zero for positive numbers only; the resulting
CI therefore is [a, b] rather than a symmetric [−a, a] as in the case of the t distribution pivots). Manipulating
the above CI yields
2 2
(n − 1)S1 (n − 1)S1
1−α=P ≤ σ2 ≤ .
b a
3
```
To top
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https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/kilogram-per-cubic-decimeter/197
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# 197 long tons per liter [long tn/l] in kilograms per cubic decimeter
## long tons/liter to kilogram/decimeter³ unit converter of density
197 long tons per liter [long tn/l] = 200 161.24 kilograms per cubic decimeter [kg/dm³]
### long tons per liter to kilograms per cubic decimeter density conversion cards
• 197
through
221
long tons per liter
• 197 long tn/l to kg/dm³ = 200 161.24 kg/dm³
• 198 long tn/l to kg/dm³ = 201 177.29 kg/dm³
• 199 long tn/l to kg/dm³ = 202 193.34 kg/dm³
• 200 long tn/l to kg/dm³ = 203 209.38 kg/dm³
• 201 long tn/l to kg/dm³ = 204 225.43 kg/dm³
• 202 long tn/l to kg/dm³ = 205 241.48 kg/dm³
• 203 long tn/l to kg/dm³ = 206 257.52 kg/dm³
• 204 long tn/l to kg/dm³ = 207 273.57 kg/dm³
• 205 long tn/l to kg/dm³ = 208 289.62 kg/dm³
• 206 long tn/l to kg/dm³ = 209 305.66 kg/dm³
• 207 long tn/l to kg/dm³ = 210 321.71 kg/dm³
• 208 long tn/l to kg/dm³ = 211 337.76 kg/dm³
• 209 long tn/l to kg/dm³ = 212 353.8 kg/dm³
• 210 long tn/l to kg/dm³ = 213 369.85 kg/dm³
• 211 long tn/l to kg/dm³ = 214 385.9 kg/dm³
• 212 long tn/l to kg/dm³ = 215 401.94 kg/dm³
• 213 long tn/l to kg/dm³ = 216 417.99 kg/dm³
• 214 long tn/l to kg/dm³ = 217 434.04 kg/dm³
• 215 long tn/l to kg/dm³ = 218 450.09 kg/dm³
• 216 long tn/l to kg/dm³ = 219 466.13 kg/dm³
• 217 long tn/l to kg/dm³ = 220 482.18 kg/dm³
• 218 long tn/l to kg/dm³ = 221 498.23 kg/dm³
• 219 long tn/l to kg/dm³ = 222 514.27 kg/dm³
• 220 long tn/l to kg/dm³ = 223 530.32 kg/dm³
• 221 long tn/l to kg/dm³ = 224 546.37 kg/dm³
• 222
through
246
long tons per liter
• 222 long tn/l to kg/dm³ = 225 562.41 kg/dm³
• 223 long tn/l to kg/dm³ = 226 578.46 kg/dm³
• 224 long tn/l to kg/dm³ = 227 594.51 kg/dm³
• 225 long tn/l to kg/dm³ = 228 610.55 kg/dm³
• 226 long tn/l to kg/dm³ = 229 626.6 kg/dm³
• 227 long tn/l to kg/dm³ = 230 642.65 kg/dm³
• 228 long tn/l to kg/dm³ = 231 658.7 kg/dm³
• 229 long tn/l to kg/dm³ = 232 674.74 kg/dm³
• 230 long tn/l to kg/dm³ = 233 690.79 kg/dm³
• 231 long tn/l to kg/dm³ = 234 706.84 kg/dm³
• 232 long tn/l to kg/dm³ = 235 722.88 kg/dm³
• 233 long tn/l to kg/dm³ = 236 738.93 kg/dm³
• 234 long tn/l to kg/dm³ = 237 754.98 kg/dm³
• 235 long tn/l to kg/dm³ = 238 771.02 kg/dm³
• 236 long tn/l to kg/dm³ = 239 787.07 kg/dm³
• 237 long tn/l to kg/dm³ = 240 803.12 kg/dm³
• 238 long tn/l to kg/dm³ = 241 819.16 kg/dm³
• 239 long tn/l to kg/dm³ = 242 835.21 kg/dm³
• 240 long tn/l to kg/dm³ = 243 851.26 kg/dm³
• 241 long tn/l to kg/dm³ = 244 867.31 kg/dm³
• 242 long tn/l to kg/dm³ = 245 883.35 kg/dm³
• 243 long tn/l to kg/dm³ = 246 899.4 kg/dm³
• 244 long tn/l to kg/dm³ = 247 915.45 kg/dm³
• 245 long tn/l to kg/dm³ = 248 931.49 kg/dm³
• 246 long tn/l to kg/dm³ = 249 947.54 kg/dm³
• 247
through
271
long tons per liter
• 247 long tn/l to kg/dm³ = 250 963.59 kg/dm³
• 248 long tn/l to kg/dm³ = 251 979.63 kg/dm³
• 249 long tn/l to kg/dm³ = 252 995.68 kg/dm³
• 250 long tn/l to kg/dm³ = 254 011.73 kg/dm³
• 251 long tn/l to kg/dm³ = 255 027.77 kg/dm³
• 252 long tn/l to kg/dm³ = 256 043.82 kg/dm³
• 253 long tn/l to kg/dm³ = 257 059.87 kg/dm³
• 254 long tn/l to kg/dm³ = 258 075.92 kg/dm³
• 255 long tn/l to kg/dm³ = 259 091.96 kg/dm³
• 256 long tn/l to kg/dm³ = 260 108.01 kg/dm³
• 257 long tn/l to kg/dm³ = 261 124.06 kg/dm³
• 258 long tn/l to kg/dm³ = 262 140.1 kg/dm³
• 259 long tn/l to kg/dm³ = 263 156.15 kg/dm³
• 260 long tn/l to kg/dm³ = 264 172.2 kg/dm³
• 261 long tn/l to kg/dm³ = 265 188.24 kg/dm³
• 262 long tn/l to kg/dm³ = 266 204.29 kg/dm³
• 263 long tn/l to kg/dm³ = 267 220.34 kg/dm³
• 264 long tn/l to kg/dm³ = 268 236.38 kg/dm³
• 265 long tn/l to kg/dm³ = 269 252.43 kg/dm³
• 266 long tn/l to kg/dm³ = 270 268.48 kg/dm³
• 267 long tn/l to kg/dm³ = 271 284.52 kg/dm³
• 268 long tn/l to kg/dm³ = 272 300.57 kg/dm³
• 269 long tn/l to kg/dm³ = 273 316.62 kg/dm³
• 270 long tn/l to kg/dm³ = 274 332.67 kg/dm³
• 271 long tn/l to kg/dm³ = 275 348.71 kg/dm³
• 272
through
296
long tons per liter
• 272 long tn/l to kg/dm³ = 276 364.76 kg/dm³
• 273 long tn/l to kg/dm³ = 277 380.81 kg/dm³
• 274 long tn/l to kg/dm³ = 278 396.85 kg/dm³
• 275 long tn/l to kg/dm³ = 279 412.9 kg/dm³
• 276 long tn/l to kg/dm³ = 280 428.95 kg/dm³
• 277 long tn/l to kg/dm³ = 281 444.99 kg/dm³
• 278 long tn/l to kg/dm³ = 282 461.04 kg/dm³
• 279 long tn/l to kg/dm³ = 283 477.09 kg/dm³
• 280 long tn/l to kg/dm³ = 284 493.13 kg/dm³
• 281 long tn/l to kg/dm³ = 285 509.18 kg/dm³
• 282 long tn/l to kg/dm³ = 286 525.23 kg/dm³
• 283 long tn/l to kg/dm³ = 287 541.28 kg/dm³
• 284 long tn/l to kg/dm³ = 288 557.32 kg/dm³
• 285 long tn/l to kg/dm³ = 289 573.37 kg/dm³
• 286 long tn/l to kg/dm³ = 290 589.42 kg/dm³
• 287 long tn/l to kg/dm³ = 291 605.46 kg/dm³
• 288 long tn/l to kg/dm³ = 292 621.51 kg/dm³
• 289 long tn/l to kg/dm³ = 293 637.56 kg/dm³
• 290 long tn/l to kg/dm³ = 294 653.6 kg/dm³
• 291 long tn/l to kg/dm³ = 295 669.65 kg/dm³
• 292 long tn/l to kg/dm³ = 296 685.7 kg/dm³
• 293 long tn/l to kg/dm³ = 297 701.74 kg/dm³
• 294 long tn/l to kg/dm³ = 298 717.79 kg/dm³
• 295 long tn/l to kg/dm³ = 299 733.84 kg/dm³
• 296 long tn/l to kg/dm³ = 300 749.89 kg/dm³
#### Foods, Nutrients and Calories
MULTIGRAIN BAKING and PANCAKE MIX, UPC: 00845809 weigh(s) 128.08 gram per (metric cup) or 4.28 ounce per (US cup), and contain(s) 375 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Fullers Earth-raw or burnt weighs 650 kg/m³ (40.57817 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Canola oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
microgram per meter (μg/m) is a metric measurement unit of linear or linear mass density.
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
ft³ to metric tbsp conversion table, ft³ to metric tbsp unit converter or convert between all units of volume measurement.
#### Calculators
Mole to volume and weight conversions for common compounds
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https://encyclopediaofmath.org/wiki/Gr%C3%B6tzsch_theorems
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# Grötzsch theorems
Various results on conformal and quasi-conformal mappings obtained by H. Grötzsch . He developed the strip method, which is the first general form of the method of conformal moduli (cf. Extremal metric, method of the; Strip method (analytic functions)), and used it in his systematic study of a large number of extremal problems of conformal mapping of multiply-connected (including infinitely-connected) domains, including the problems of the existence, uniqueness and geometric properties of extremal mappings. A few of the simpler Grötzsch theorems are presented below.
Of all univalent conformal mappings $w = f ( z)$ of a given annulus $K _ {R} = \{ {z } : {R < | z | < 1 } \}$ under which the unit circle $\Gamma = \{ {z } : {| z | = 1 } \}$ is mapped onto itself, the maximum diameter of the image of the circle $\Gamma _ {R} = \{ {z } : {| z | = R } \}$ is attained if and only if the boundary component $f ( \Gamma _ {R} )$ is a rectilinear segment with its centre at the point $w = 0$. A similar result is valid for multiply-connected domains.
Out of all univalent conformal mappings $w = f ( z)$ of a given multiply-connected domain $B \ni \infty$ with expansion $f ( z) = z + O ( 1)$ $( z \rightarrow \infty )$ at infinity and normalization $f ( z _ {0} ) = 0$ at a given point $z _ {0} \in B$, the maximum of $| f ^ { \prime } ( z _ {0} ) |$, and the maximum (minimum) of $| f ( z _ {1} ) |$ at a given point $z _ {1} \in B$, $z _ {1} \neq z _ {0}$, are attained only on mappings that map each boundary component of $B$, respectively, to an arc of a circle with centre at the point $w = 0$, or to an arc of an ellipse (hyperbola) with foci at the points $w = 0$ and $w = w ^ \prime = f ( z _ {1} )$. In each one of these problems the extremal mapping exists and is unique. In this class of mappings, for a given $z _ {1} \in B$, the disc
$$\left \{ {w } : { \left | w - { \frac{1}{2} } ( w ^ \prime + w ^ {\prime\prime} ) \ \right | \leq \ { \frac{1}{2} } | w ^ \prime - w ^ {\prime\prime} | \ } \right \}$$
is the range of the function $\Phi ( f ) = \mathop{\rm ln} ( f ( z _ {1} )/z _ {1} )$. Each boundary point of this disc is a value of $\Phi$ on a unique mapping in the class under study with specific geometric properties.
Grötzsch was the first to propose a form of representation of a quasi-conformal mapping, and to apply to such a mappings many extremal results which he had formerly obtained for conformal mappings.
#### References
[1a] H. Grötzsch, "Ueber die Verzerrung bei schlichter konformer Abbildung mehrfach zusammenhängender Bereiche I" Ber. Verh. Sächsisch. Akad. Wiss. Leipzig. Math.-Naturwiss. Kl. , 81 (1929) pp. 38–47 [1b] H. Grötzsch, "Ueber die Verzerrung bei schlichter konformer Abbildung mehrfach zusammenhängender Bereiche II" Ber. Verh. Sächsisch. Akad. Wiss. Leipzig. Math.-Naturwiss. Kl. , 81 (1929) pp. 217–221 [1c] H. Grötzsch, Ber. Verh. Sächsisch. Akad. Wiss. Leipzig. Math.-Naturwiss. Kl. , 82 (1930) pp. 69–80 [1d] H. Grötzsch, "Ueber die Verschiebung bei schlichter konformer Abbildung schlichter Bereiche II" Ber. Verh. Sächsisch. Akad. Wiss. Leipzig. Math.-Naturwiss. Kl. , 84 (1932) pp. 269–278 [2] J.A. Jenkins, "Univalent functions and conformal mappings" , Springer (1958)
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| 2.9375
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https://goprep.co/q15-factorise-the-following-i-16p-3-q-3-54r-3-ii-8b-3-iii-2-i-1nk479
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Q. 15
# Factorise the following(i) 16p3q3 + 54r3(ii) + 8b3(iii) 2 a3 + 3 b3(iv) 8a4b + ab4(v) a7 - 64a
(i) 16p3q3 + 54r3
= 2(8p3q3 + 27r3)
= 2[(2pq)3 + (3r)3]
= 2(2pq + 3r) [(2pq)2 - 2pq . 3r + (3r)2]
= 2 (2pq + 3r) (4p2q2 - 6pqr + 9r2)
(ii) + 8b3 = ()3 + (2b)3
= ( +2b)( -ab+4b2)
(iii) 2 a3 + 3 b3
= ( a)3+( b)3
= ( a + b)(2a2 - ab + 3b2)
(iv) 8a4b + ab4 = ab(8a3+ )
= ab[ ]
= ab (2a+ )
(v) a7 - 64a
= a(a6 - 64)
= a(a6 - 26)
= a[(a3)2 - (23)2]
= a(a3 + 23) (a3 - 23)
= a(a + 2) (a2 - 2a + 4) (a - 2) (a2 + 2a + 4)
= a(a + 2) (a - 2) (a2 - 2a + 4) (a2 + 2a + 4)
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# GCSE Higher Paper Topics – Non Calculator
```106745144
Nets – Elevations – 3D Shapes
Ratio
Constructions – using a compass
Inequalities – plus drawing graphs and shading
Transformation
Vectors
Reflections
Rotation
Enlargement
6. Simultaneous Equations
7. Forming and equation from a given line
8. Fractions in Algebra
10. Vectors
11. Box Plots
12. Cumulative Frequency
13. Scatter Graphs
14. Product of Primes
15. HCF / LCM
16. Standard Form
17. Sectors of Circles – areas / angle size etc
18. Reasons for Angle size – alternate etc / circle rules etc
19. Using calculators to solve to significant figures and decimal places.
20. Percentage / profit / loss – Interest
21. Estimated Mean
22. Bearings
Pythagoras
Trigonometry and angle of elevation and depression
23. Interest – Simple and Compound
24. Cumulative Frequency
25. Histograms
26. Lower and Upper Bounds
27. Stratified Sampling
28. Surface Area
29. Probability and Tree Diagrams
30. Proportion
31. Function Graphs
32. Trial and Improvement
33. Algebra – Simplify / Expand/ Factorise
34. Volume / Area problems
35. Trigonometry for non-right-angled triangles including area
36. Surds
37. Trigonometry Graphs
38. Nth term
39. Rational & Irrational Numbers
40. Algebra – prove or show how an equation can be arranged
41. Algebra – Difficult Fractions.
1.
2.
3.
4.
5.
haahaa
```
| 378
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https://nz.education.com/resources/preschool/ela/CCSS-Math-Content/
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crawl-data/CC-MAIN-2020-45/segments/1603107896778.71/warc/CC-MAIN-20201028044037-20201028074037-00379.warc.gz
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This wacky lesson combines math and reading practice with a Dr. Seuss classic and some fun online games.
Preschool
Lesson Plan
Rocket Like Mae Jemison
Activity
Rocket Like Mae Jemison
Craft a paper rocket in this hands-on activity inspired by the life of Mae Jemison!
Preschool
Social studies
Activity
Fourth of July Dot-to-Dot 1
Worksheet
Fourth of July Dot-to-Dot 1
Here's a quick and easy way to have your child review the alphabet: connect the dots and she'll have a 4th of July coloring page too!
Preschool
Worksheet
Bug Counting and Quantifying
Lesson Plan
Bug Counting and Quantifying
This interactive insect game will engage students in collecting and counting bugs! Students will catch and count pompoms and use imagination, representation, counting, and qualitative skills during this play based math activity.
Preschool
Math
Lesson Plan
Five Busy Little Bees
Lesson Plan
Five Busy Little Bees
Buzz, buzz! Can you count the five busy little bees? In this bee-themed lesson on counting and number recognition, students will use bees to learn how to count the numbers one through five.
Preschool
Math
Lesson Plan
Tracing Diamonds
Worksheet
Tracing Diamonds
Preschool
Worksheet
My Favorite Weather
Lesson Plan
My Favorite Weather
What's your favorite weather? In this lesson, students learn about different kinds of weather and share their favorites. Students then work together to determine which was the class's most and least popular weather.
Preschool
Math
Lesson Plan
Keeping Count
Lesson Plan
Keeping Count
Preschool
Math
Lesson Plan
How Many Humps?
Lesson Plan
How Many Humps?
Increase your students' self-confidence while practicing pre-reading skills, number recognition and counting to five. Students will read about Sally the Camel and count the humps on her back.
Preschool
Math
Lesson Plan
Rhyming Songs
Worksheet
Rhyming Songs
Have some fun with phonics using these favorite childhood rhyming songs!
Preschool
Worksheet
Alphabet Bingo Cards
Worksheet
Alphabet Bingo Cards
A simplified version of bingo, this game is a fun way to help preschoolers learn their letters, numbers, and simple shapes.
Preschool
Worksheet
Direction Flashcards
Worksheet
Direction Flashcards
What's up, down, or all around? Use these direction flashcards to help your preschooler learn positional words.
Preschool
Worksheet
Trace and Color the City
Worksheet
Trace and Color the City
Sneak some writing practice into coloring time with this fun trace and color printable. As she traces the lines, she'll build strong hand muscles for writing!
Preschool
Worksheet
Worksheet
Preschool
Math
Worksheet
Worksheet
Help these poor kittens add up their mittens! Your preschooler will be learning the basics of addition and multiplication with this nursery rhyme math sheet.
Preschool
Worksheet
Worksheet
What does it mean to have zero? Teach your little fisherman an important math concept with this nursery rhyme math sheet.
Preschool
Worksheet
Itsy Bitsy Spider Subtraction
Worksheet
Itsy Bitsy Spider Subtraction
Climb up the water spout to learn some subtraction! Your preschooler will love to learn simple arithmetic with a familiar nursery rhyme.
Preschool
Worksheet
"An Apple a Day" Math
Worksheet
"An Apple a Day" Math
How many apples a day can keep the doctor away? Count up the apples to learn addition in this nursery rhyme math worksheet.
Preschool
Worksheet
Twinkle Twinkle Math
Worksheet
Twinkle Twinkle Math
Help twinkling stars go back to sleep! By filling in the stars, your preschooler will be practicing basic subtraction.
Preschool
Math
Worksheet
Old Mother Hubbard Subtraction
Worksheet
Old Mother Hubbard Subtraction
What's Old Mother Hubbard keeping in her cupboard? Count them up with your little reader! She can learn the basics of subtraction with this nursery rhyme.
Preschool
Worksheet
London Bridge Subtraction
Worksheet
London Bridge Subtraction
If one bridge falls down, how many bridges are left? None! Help your preschooler sort out simple subtraction with this nursery rhyme math worksheet.
Preschool
Worksheet
Humpty Dumpty Subtraction
Worksheet
Humpty Dumpty Subtraction
What if Humpty Dumpty wasn't the only one sitting on the wall? Try out a simple subtraction problem using this classic nursery rhyme.
Preschool
Math
Worksheet
Old Woman in a Shoe Addition
Worksheet
Old Woman in a Shoe Addition
Could you imagine living in a shoe? Get silly with your little learner as you teach her some simple addition and counting.
Preschool
Math
Worksheet
Ba Ba Black Sheep Math
Worksheet
Ba Ba Black Sheep Math
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# 18171 (number)
18,171 (eighteen thousand one hundred seventy-one) is an odd five-digits composite number following 18170 and preceding 18172. In scientific notation, it is written as 1.8171 × 104. The sum of its digits is 18. It has a total of 4 prime factors and 8 positive divisors. There are 12,096 positive integers (up to 18171) that are relatively prime to 18171.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 18 thousand 171 eighteen thousand one hundred seventy-one
## Notation
Scientific notation 1.8171 × 104 18.171 × 103
## Prime Factorization of 18171
Prime Factorization 33 × 673
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 2019 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 18,171 is 33 × 673. Since it has a total of 4 prime factors, 18,171 is a composite number.
## Divisors of 18171
1, 3, 9, 27, 673, 2019, 6057, 18171
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 26960 Sum of all the positive divisors of n s(n) 8789 Sum of the proper positive divisors of n A(n) 3370 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 134.8 Returns the nth root of the product of n divisors H(n) 5.39199 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 18,171 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 18,171) is 26,960, the average is 3,370.
## Other Arithmetic Functions (n = 18171)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 12096 Total number of positive integers not greater than n that are coprime to n λ(n) 2016 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2082 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 12,096 positive integers (less than 18,171) that are coprime with 18,171. And there are approximately 2,082 prime numbers less than or equal to 18,171.
## Divisibility of 18171
m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 6 3 0
The number 18,171 is divisible by 3 and 9.
• Arithmetic
• Deficient
• Polite
• Frugal
## Base conversion (18171)
Base System Value
2 Binary 100011011111011
3 Ternary 220221000
4 Quaternary 10123323
5 Quinary 1040141
6 Senary 220043
8 Octal 43373
10 Decimal 18171
12 Duodecimal a623
20 Vigesimal 258b
36 Base36 e0r
## Basic calculations (n = 18171)
### Multiplication
n×y
n×2 36342 54513 72684 90855
### Division
n÷y
n÷2 9085.5 6057 4542.75 3634.2
### Exponentiation
ny
n2 330185241 5999796014211 109022293374228081 1981044092903098459851
### Nth Root
y√n
2√n 134.8 26.2901 11.6103 7.1101
## 18171 as geometric shapes
### Circle
Diameter 36342 114172 1.03731e+09
### Sphere
Volume 2.51319e+13 4.14923e+09 114172
### Square
Length = n
Perimeter 72684 3.30185e+08 25697.7
### Cube
Length = n
Surface area 1.98111e+09 5.9998e+12 31473.1
### Equilateral Triangle
Length = n
Perimeter 54513 1.42974e+08 15736.5
### Triangular Pyramid
Length = n
Surface area 5.71898e+08 7.07083e+11 14836.6
## Cryptographic Hash Functions
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https://byjus.com/questions/a-current-of-965-ampere-flowing-for-10-minutes-deposits-3-g-of-metal-which-is-monovalent/
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# A current of 9.65 Ampere flowing for 10 minutes deposits 3 g of metal which is monovalent, the atomic mass of metal is a: A.10 B. 50 C. 30 D. 96.5
Given,
Current flowing, I = 9.65 Ampere
Time = 10 min
Q = It
Q = 9.65 x 10 x 60 = 5790 C
Amount of metal deposited = 3gm
Since, the metal is monovalent, the atmoic weight will be equal to equivalent weight
Equivalent weight of metal = Wt.of metal deposited/Charge x F
= 3/5790 × 96500=50g
Atomic weight of metal = 50g
Atomic mass = 50× valency = 50×1 = 50 amu
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# 99 questions/Solutions/41
(**) Given a range of integers by its lower and upper limit, print a list of all even numbers and their Goldbach composition.
In most cases, if an even number is written as the sum of two prime numbers, one of them is very small. Very rarely, the primes are both bigger than say 50. Try to find out how many such cases there are in the range 2..3000.
```goldbachList lb ub = map goldbach \$ [even_lb,even_lb+2..ub]
where even_lb = max ((lb+1) `div` 2 * 2) 4
goldbachList' lb ub mv = filter (\(a,b) -> a > mv && b > mv) \$
goldbachList lb ub```
using the goldbach function from problem 40.
Or a more concise version:
```goldbachList n m = map goldbach \$ dropWhile (<4) \$ filter even [n..m]
goldbachList' n m i = filter (\(x,y) -> x > i && y > i) \$ goldbachList n m```
Note the
dropWhile
, as the question explicitly asks for the range 1-2000 (although we know better).
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# chord
(redirected from chords)
Also found in: Dictionary, Thesaurus, Medical, Idioms.
## chord
(kôrd), in geometry, straight line segment both end points of which lie on the circumference of a circle or other curve; it is a segment of a secantsecant,
in mathematics. 1 In geometry, a secant is a straight line cutting a curve or surface. If it intersects the curve in two different points, as in the secant of a circle, the segment of the secant between the points is called a chord.
. A chord passing through the center of a circle is a diameter. In the same circle or in equal circles, equal chords subtend equal arcs and equal central angles.
## chord,
in music, two or more simultaneously sounding pitches. In tonal music the fundamental chord is called the triad. It consists of three pitches, two a perfect fifth apart and a third pitch a major or minor third lower, forming respectively the major or minor triad. However, a triad may instead be diminished or augmented, or may contain dissonant elements such as a seventh. In atonal music, other types of chord formations occur. It is, however, an essential property of a chord that it be conceived as an entity, that its constituent notes "fuse" rather than merely coincide in time.
## Chord
A principal member or pair of members of a truss extending from one end to the other, to resist bending.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Chord
the simultaneous combination of several notes of different pitch, auditorily perceived as a unity of sound.
Chords are differentiated by the quantitative and interval composition of the notes which make them up. Basically, a chord’s notes are arranged by thirds, proceeding upward from the lowest tone. Each one has its own name (according to the interval between it and the lowest note): the basic tone or root, the third, fifth, seventh, and so on. The fundamental kinds of chords are the triad (consisting of three different notes), the seventh (consisting of four), the ninth (five), and the eleventh (six). Triads are of four types: major (a major and a minor third), minor (a minor and a major third), diminished (two minor thirds), and augmented (two major thirds). Seventh chords are formed from triads (other than augmented triads) with the addition of a minor or major third above the chord. Sevenths may be major, minor, or diminished, depending upon the interval of the seventh between the extreme notes.
Shifting the notes of a chord—that is, moving the basic tone to one of the higher voices—is called inversion. In such cases the designation of the chord changes. A triad has two inversions (chord of the sixth and six-four chord). A seventh chord has three inversions (five-six chord, three-four chord, and chord of the second). The ninth and eleventh chords are primarily used in their root forms; their inversions have no independent designations. Chords which are built on fourths occur occasionally.
V. A. VAKHROMEEV
## Chord
a line segment connecting two points of a curve or surface.
## chord
[kȯrd]
(acoustics)
A combination of two or more tones.
(aerospace engineering)
A straight line intersecting or touching an airfoil profile at two points.
Specifically, that part of such a line between two points of intersection.
(architecture)
The span of an arch.
(civil engineering)
The top or bottom, generally horizontal member of a truss.
(mathematics)
A line segment which intersects a curve or surface only at the endpoints of the segment.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
## chord
chord, 3
chord, 1
1. A principal member of a truss which extends from one end to the other, primarily to resist bending; usually one of a pair of such members.
2. The straight line between two points on a curve.
3. The span of an arch.
McGraw-Hill Dictionary of Architecture and Construction. Copyright © 2003 by McGraw-Hill Companies, Inc.
## chord
An imaginary straight line joining the center of the leading and trailing edges of a cross section of an airfoil. The mean chord is used as a reference datum for laying out the curve of the airfoil. Also called a chord line.
## chord
the simultaneous sounding of a group of musical notes, usually three or more in number
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
References in periodicals archive ?
The method begins at the basics of note reading and works up to accompanying melodies without notation, reading lead sheets and understanding chord progressions.
You will notice that the first line of voicings include some clustered chords. I like the way these sound with this piece.
Further, when the improvisers recognized a chord unsuitable for substitution, their brains showed a pattern of electrical activity distinct from non-improvising musicians.
The website has easy interface to browse through the videos and users can refine their search to videos with or without lyrics and chords.
I start playing what is known as a walking bass line that outlines the chords. In this case, the tonic chord is B[flat], so the left hand note pattern is quarter notes in common time for four bars starting on [B[flat].sub.1]: | [B[flat].sub.1]-[D.sub.2]-[F.sub.2]-[G.sub.2]| [B[flat].sub.2]-[G.sub.2]-[F.sub.2]-[D.sub.2]| [B[flat].sub.1]-[D.sub.2]-[F.sub.2]-[G.sub.2]| [B[flat].sub.2]-[G.sub.2]-[F.sub.2]-[D.sub.2]|.
It showed that Northern Chords, which has been staged annually in the North East since 2009, is fast approaching maturity.
As a celebration of its 20th anniversary in the musician assistance business, Ron Greene is providing the musical instrument player with a Free 'Musicians Note and Chord Guide' booklet (pdf downloadable), which includes; Key Signatures and the Circles of Fifths, how to play chords in accompaniment, playing chord extensions, melody and lead notes, scale references, and much more.
Horn players have their glorious sound--no synthesizer will ever replace us!--but our one-note-at-a-time approach leaves us with a lack experience with chords. The answer is to go to the piano and experiment with chords and develop a palette of favorite harmonic "flavors" that we can use.
Goldenberg (Jerusalem Academy of Music & Dance Research) examines theorist Heinrich Schenker's admonition against the prolongation of seventh chords in traditional tonal music, which he viewed as a prolongation of dissonance.
FAMILIAR RELATIONSHIPS BETWEEN SETS of musical notes, such as transposition between chords, directly translate into geometrical structures such as this Mobius strip--where each dot represents a whole class of equivalent two-note chords--or into structures with many dimensions.
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mersenneforum.org (https://www.mersenneforum.org/index.php)
- Conjectures 'R Us (https://www.mersenneforum.org/forumdisplay.php?f=81)
gd_barnes 2008-12-23 09:07
Funny stuff & gripes about friends/relatives
We need to have a good jokes and funny stuff thread to lighten up the mood in these projects a little. Everyone, feel free to post anything you want here that is liable to get a good chuckle out of people.
This initially came about when Max and I were debating whether a word was really a swear word or not. It was really quite funny in it's own right because he was one admin sensoring another admin within the project that we co-admin together. It was the below post that he inserted the "censored" emoticon in for a questionable word that I actually use again in a post further down just to annoy him a bit for sensoring me. (lol) See for yourself.
It was also inspired by my pet hamster who is fond of changing my posts in a funny way without me knowing it.
You can also post gripes about your mother-in-law, wife, girl friend, or whomever that you're sure won't read this forum. Perhaps we can give you good advice!
Here is the memo that started all this...
[quote=Flatlander;154126]Riesel base 143 tested to 15k. Unreserving.
Primes for n>10k:
584*143^11344-1
208*143^13313-1
1118*143^14838-1
I'll email the results and sieve to Gary.[/quote]
OK, now slow down just a little bit. You're causing me to take extra time and you know how :censored: I get when I have to take extra admin time. lol
You didn't send me a list of k's remaining for base 143. I could figure it out but...
I only see primes in one Email up to n=1000. Then in another Email there are primes for n>8700. So I need the primes for n=1000-8700.
So if you can send me both the 29 k's remaining and those primes, then I can post everything on the pages.
Edit: Well, I got on tonight thinking that I could get the pages updated in an hour...not a chance. This is a lot of work that everyone has done and I haven't even looked at any of the base 3 stuff yet. Oh well, I got in some updates for bases <= 32. I'll get to all of these late Tuesday and Wednesday during the day.
Gary
gd_barnes 2008-12-27 01:42
Ahem. I didn't know bitchy was a bad word. It's only describing a state of annoyance. lol
Now, if you remove the 'y' to make it a noun, that's a different story.
mdettweiler 2008-12-27 01:50
[quote=gd_barnes;155225]Ahem. I didn't know bitchy was a bad word. It's only describing a state of annoyance. lol
Now, if you remove the 'y' to make it a noun, that's a different story.[/quote]
Um...well, at least from what I've seen, derivatives from swear words are usually considered swear words, too. :ermm:
gd_barnes 2008-12-27 01:56
[quote=mdettweiler;155227]Um...well, at least from what I've seen, derivatives from swear words are usually considered swear words, too. :ermm:[/quote]
So we can't say frick, friggin, fudge, etc. as is commonly used by my teenagers? (lol) And no, I don't allow any outright swearing in the house nor does their mom in her house.
Obviously this is a useless debate and things are different in different households, states, countries, parts of the U.S., etc. To avoid any issues here, I'll avoid use of the term previously used.
Gary
mdettweiler 2008-12-27 02:01
[quote=gd_barnes;155231]So we can't say frick, friggin, fudge, etc. as is commonly used by my teenagers? (lol) And no, I don't allow any outright swearing in the house nor does their mom in her house.[/quote]
Hmm...well, the derivations of those particular words is not necessarily from swear words anyway. The first two can be traced back to "freak", which is obviously not a swear word, and "fudge" is obviously just a tasty food. :smile: :wink:
gd_barnes 2008-12-27 03:06
[quote=mdettweiler;155232]Hmm...well, the derivations of those particular words is not necessarily from swear words anyway. The first two can be traced back to "freak", which is obviously not a swear word, and "fudge" is obviously just a tasty food. :smile: :wink:[/quote]
Oh, you're stretchin' there Max. Clearly you don't have teenage kids. When my son says, "Well, frick!" when he can't figure out a homework problem or has a computer issue, he definitely doesn't mean "Well, freak". lmao
Yes, in the literal sense of the word "derivative", you're probably correct that my use of the b..... word is a derivative of the swear word and the ones that my teenagers use are not derviates of the f... word, but in the contectual sense, the ones they use definitely go right back to the swear word! :smile:
We should have a "misc." thread in the 2 projects for discussions such as this kind of like the main forum does. I like the one where they have grammatical pet peaves. I think such debates can be fun as long as people don't get nasty or anything.
Gary
mdettweiler 2008-12-27 06:43
[quote=gd_barnes;155238]Oh, you're stretchin' there Max. Clearly you don't have teenage kids. When my son says, "Well, frick!" when he can't figure out a homework problem or has a computer issue, he definitely doesn't mean "Well, freak". lmao
Yes, in the literal sense of the word "derivative", you're probably correct that my use of the b..... word is a derivative of the swear word and the ones that my teenagers use are not derviates of the f... word, but in the contectual sense, the ones they use definitely go right back to the swear word! :smile:
We should have a "misc." thread in the 2 projects for discussions such as this kind of like the main forum does. I like the one where they have grammatical pet peaves. I think such debates can be fun as long as people don't get nasty or anything.
Gary[/quote]
Hmm...well, I guess in that particular usage of "frick", it could be traced back to the swear word. When you mentioned it the first thing that came to mind was the context of "this frickin math problem doesn't want to work out!" or something along those lines. :smile:
Interesting that you mention the grammatical pet peeves thread...maybe this would be an interesting discussion to move there? :smile: (Or maybe it deserves its own thread in the Lounge? Regardless, it's probably better suited there than in the Conjectures 'R Us forum.)
IronBits 2008-12-27 06:47
You guys are frick and frack.
Up to you to figure out which witch is which tho :grin:
gd_barnes 2008-12-27 07:35
[quote=mdettweiler;155258]Hmm...well, I guess in that particular usage of "frick", it could be traced back to the swear word. When you mentioned it the first thing that came to mind was the context of "this frickin math problem doesn't want to work out!" or something along those lines. :smile:
Interesting that you mention the grammatical pet peeves thread...maybe this would be an interesting discussion to move there? :smile: (Or maybe it deserves its own thread in the Lounge? Regardless, it's probably better suited there than in the Conjectures 'R Us forum.)[/quote]
Funny...I must have a dirtier mind than you. lol
Used in your context, I would also trace it back to the swear word. I've heard people use "freakin'" or "frickin'" or "friggin'" before in that context but I always thought they were just trying to avoid use of the f...in' word.
To loosen or lighten up the projects a little, I think that a misc. thread like this would be good here. Even my pet hamster had suggested something like a "joke of the day" or "funny" thread within the project threads when I started them. Stuff like this just lightens the mood a little. I may do it at some point.
Gary
gd_barnes 2008-12-30 10:14
[quote=gd_barnes;155265]Funny...I must have a dirtier mind than you. lol
Used in your context, I would also trace it back to the swear word. I've heard people use "freakin'" or "frickin'" or "friggin'" before in that context but I always thought they were just trying to avoid use of the f...in' word.
To loosen or lighten up the projects a little, I think that a misc. thread like this would be good here. Even my pet hamster had suggested something like a "joke of the day" or "funny" thread within the project threads when I started them. Stuff like this just lightens the mood a little. I may do it at some point.
Gary[/quote]
Now, I have to ask everyone, did anyone happen to see what Mike (Xyzzy) did to my post?
He is such a ham!
He changed "XYZZY" to "my pet hamster". :lol::missingteeth:
And of course he's able to change it without the little "last fiddled with" message appearing at the bottom. Pretty sneaky!
Thanks for always giving me a good laugh Mike! :smile:
Gary
Flatlander 2008-12-30 14:06
[quote=gd_barnes;155752]...
He is such a h__!
...
Gary[/quote]
You should know that that word is highly offensive in Flatland.
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# BDM midterm Essay
Custom Student Mr. Teacher ENG 1001-04 16 May 2016
## BDM midterm
Ralph Edmund loves steak and potatoes. Therefore, he has decided to go on a steady diet of only these two foods for all his meals. Ralph realizes that this is not the healthiest diet, so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional requirements. He has obtained the following nutritional and costs data.
The Oak Works is a family owned business that makes hand crafted dining room tables and chairs. They obtain the oak from a local tree farm, which ships them 2500 pounds of oak each month.
Each table uses 50 pounds of oak while each chair uses 25 pounds of oak. The family builds all the furniture itself and has 480 hours of labour available each month. Each table or chair requires 6 hours of labour. Each table nets Oak Works \$400 in profit, while each chair nets them \$100 in profit. Since chairs are often sold with tables they want to produce at least twice as many chairs as tables. Formula a linear program to maximize profit.
B
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• Type of paper: Thesis/Dissertation Chapter
• Date: 16 May 2016
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Let us write you a custom essay sample on BDM midterm
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# VCA.SCI.1
Lesson Summary
Combine information from VCO.SCI.2 and relate amplitude to wave dynamics lesson.
Vocabulary
Wavelength - The distance over which the wave's shape repeats. Determined by considering the distance between consecutive corresponding points of the same phase, such as crests, troughs, or zero crossings. Designated by the Greek letter lambda (λ).
Period - The time over which a wave's shape repeats. Determined in a similar manner as wavelength but in the time domain. It is the reciprocal of frequency and is designated by an uppercase T. SI unit is the second.
Amplitude - The maximum absolute value of the signal. Designated by an uppercase A.
Frequency - The number of cycles per unit time. The SI unit for frequency is hertz (Hz), named after the German physicist Heinrich Hertz; 1 Hz means that an event repeats once per second. Designated by a lowercase f.
Hertz - The unit used to represent frequency (Hz)
Exercise
Add a volume knob to the Werkstatt.
Materials
1 x 10kΩ Potentiometer Jumper Cables 1 x AA Battery Holder
Hardware
We will be adding a volume knob to work with our Werkstatt. This knob will change the amplitude of the sound wave output.. First we will need to match the settings to Figure 1. This will give us a constant Sawtooth wave to experiment with.
Figure 1. Sawtooth wave VCA MODE ON
When our Werkstatt VCA is set to to MODE ON we get a constant tone. This also means that our VCA IN pin can accept a voltage from -2.5 to +2.5 V. To create a volume control we will need a negative voltage. To do this we simply connect our battery backwards. A battery pack has two leads, one red (positive) and one black (ground). By simply swapping these two leads we will be able to get a -3V from our two AA batteries. We then pass this -3V through our 10kΩ potentiometer to give us a variable control over how much of that voltage is passed on to the VCA IN. Observe Figure 2 for a detailed visual diagram.
Figure 2. Volume control Werkstatt modification.
Now you will notice as we turn the volume knob the output of our Werkstatt is changed. If we were to use the + voltage from our battery our volume would only ever get louder, never quieter.
If we switch our VCA MODE to EG instead of ON you will hear that the change is much less noticeable. This is because the VCA IN can now accept -5V - +5V as a control voltage. We could make a similar volume knob by simply doubling our AA batteries to get closer to a -5V.
Subject
Unit
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# Ellsworth's Responses in WyzAnt Answers
#### if points are (4,a) and (8,3a) and the slope is 1/3 what is a?
That is all the question says. we are working with slope intercept form and standard form and the lessons like that. I just dont get this
+ more- less
Asked by Emma from Redwood Falls, MN
20
Use the slope formula: m = (y2 - y1)/(x2 - x1). In this case:
x1 = 4, y1 = a, x2 = 8, y2 = 3a
so plug in:
m = (3a - a)/(8 - 4) = 2a / 4
But since 2a/4 = 1/3, cross-multiply and solve the proportion. You get
6a = 4, so
a = 4/6 = 2/3
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Info: a metric to assess rules performance
Calculates the SQN100 (see work of Dr. Van Tharp) as if the per position "Output" is the trade result. This is different from the other rules analyszer metric named SQN in that the sample standard deviation calculation is used vs. the built-in Standard deviation. The result of...
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Rule Metrics
Basic
System Quality Number (SQN) by Dave W., uploaded several months ago
Dr. Van Tharp's System Quality Number (SQN) metric.
SQN measures the relationship between the mean (expectancy) and the standard deviation of the R-multiple distribution generated by a trading system. It also makes an adjustment for the number of trades involved. Dr. Tharp has determined that the better the SQN, the easier...
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The Average Output per Month is a script metric for the rules analyzer plug-in. It loops through all the generated positions and calculates the average output value for each month. It then averages these values to produce a unique metric, which is the average output per month.
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I have just uploaded a rules analyzer metric script, which can be found here: 284. This script produces several metrics, one for each trading year.
The current script creates only one metric, which is the average 'output per bar' for all years. The metric is not the same as the default...
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Rule Metrics
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This rules analyzer metric script produces several metrics, one for each year in the analysis period. Its task is simple, it calculates the 'Output per bar' value, the same way QuantShare calculates it, but instead of producing one 'Output per bar' metric for all the trading period, it produces a...
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Time Under Water - Drawdown by The trader, uploaded several months ago
The maximum drawdown is a popular measure of the maximum amount an investor can expect to lose. It is used as a measure of riskiness of a trading strategy.
Another measure of a trading strategy's riskiness is the Time under Water.
The Time under water is derived from the calculation of...
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Average Maximum Drawdown by QuantShare, uploaded several months ago
The average maximum drawdown is a rules analyzer metric described in the following post: The average maximum drawdown metric . It calculates the maximum drawdown (the maximum an output can lose) of each symbol or security output and then averages the result.
If you are analyzing for example a list...
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Average Growth Rate by QuantShare, uploaded several months ago
This metric is calculated in two steps:
- For each symbol, position returns are transformed then multiplied. The resulting value is the growth rate of the strategy equity for a particular symbol.
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The first step calculates the trading rule real performance for each...
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This rules analyzer metric calculates the average output of each analyzed rule and on a per-symbol basis. The standard average output adds up the outputs generated by all the positions for all securities then divides this value by the number of positions.
The current metric adds up the outputs separately for...
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Holding period return - classifying positions by The trader, uploaded several months ago
This item creates three metrics when analyzing a list of trading rules.
The object classify trading rules output or return into three categories.
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Position whose holding period is below 10 bars
Position whose holding period is below 40 bars ...
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# Tagged Questions
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### Singularities of $e^{z - \frac{1}{z}}$
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# Dirichlet Convolution of the Mobius Function with Itself
I am attempting to find a formula for $$(\mu * \mu)(n)$$ where * represents the Dirichlet Convolution operator. I know this can be expressed as $$\sum_{d|n} \mu(d)\mu(\frac{n}{d})$$ but I'd like the formula to not include any sums over divisors. I know it will be necessary to include information about the factorization of n, but I'm not sure how. For reference, $$\mu(n)= \begin{cases}0,&\text{if n has one or more repeated prime factors}\\1,&\text{if n=1}\\(-1)^k,&\text{if n is a product of k distinct primes}\end{cases}$$ Some initial thoughts: the Dirichlet Convolution of two multiplicative functions is multiplicative, and since $$\mu(n)$$ is multiplicative, then so is $$(\mu * \mu)(n)$$ Any information to point me in the right direction on this will be greatly appreciated.
As you say, this is multiplicative. This means that once you know how to calculate it for numbers of the form $p^a$ you can calculate it for arbitrary $n$ by writing $n$ as a product of powers of different primes $p^aq^b\cdots$, and then multiplying the corresponding values $(\mu*\mu)(n)=((\mu*\mu)(p^a))((\mu*\mu)(q^b))\cdots$.
If $a=1$, $\sum_{d\mid p}\mu(d)\mu(p/d)=-2$ (the two factors each contribute $-1$). If $a=2$, $\sum_{d\mid p^2}\mu(d)\mu(p^2/d)=1$, since the only way for $\mu(d)\mu(p^2/d)$ to be non-zero is if $d\leq p$ and $p^2/d\leq p$, which requires $d=p$. If $a>2$ then $\sum_{d\mid p^a}\mu(d)\mu(p^2/d)=0$, since for each term in the sum either $d$ or $p^a/d$ is divisible by $p^2$.
So your function is $0$ if $n$ is divisible by the cube of any prime. Otherwise it is $(-2)^k$, where $k$ is the number of primes that divide $n$ exactly once (i.e. their squares do not divide $n$).
Note on the Previous Solution: Let $\operatorname{rad}(n)$ denote the radix, or squarefree part, of $n$, i.e., so that if $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the factorization of $n$ into powers of distinct primes then $\operatorname{rad}(n) = p_1 p_2 \cdots p_k$. Then we see that $\mu(\operatorname{rad}(n))$ is always non-zero and that $\mu^2(n / \operatorname{rad}(n)$ is the characteristic function of the cube-free integers. Also, if $\omega(n)$ counts the number of distinct prime factors of $n$ (the prime omega function) then the number of primes that divide $n$ exactly once is given by $$\#\{p \text{ prime } : p|n \text{ and } p^2 \nmid n\} = \omega(n) - \omega(n / \operatorname{rad}(n)).$$ This implies that the solution given by Essentially Lime in the previous response has the following exact closed-form expression for any $n \geq 1$: $$(\mu \ast \mu)(n) = (-2)^{\omega(n) - \omega(n / \operatorname{rad}(n))} \cdot \mu^2(\omega(n) - \omega(n / \operatorname{rad}(n)).$$ Curiously enough, parts of this formula for the solution can be re-expressed in terms of 1) the identity that $\sum_{d|n} \mu^2(d) = 2^{\omega(n)}$, or in other words the convolution $\mu^2 \ast 1 = 2^{\omega}$; and 2) the fact that $(-1)^{\omega(n)} = \mu(n)$ whenever $n$ is squarefree.
Alternate Solution Using Dirichlet Inverses: Another, perhaps shorter and more direct way to find this solution is to consider the Dirichlet inverse of the divisor function $d(n) := \sum_{d|n} 1 = (1 \ast 1)(n)$. Recall that the Dirichlet inverse $f^{-1}$ of any arithmetic function $f$ is the unique function (if one exists) satisfying $(f \ast f^{-1})(n) = \varepsilon = \delta_{n,1}$ where $\varepsilon$ is the multiplicative identity of Dirichlet convolutions. This inverse exists precisely when $f(1) \neq 0$.
Now consider the following phrasing of the problem: write $f := mu \ast \mu$. Then since $\mu \ast 1 = \varepsilon$ (an elementary fact which is seen here, for example, or proved by Dirichlet generating functions), we can invert the right-hand-side to obtain that $\varepsilon = f \ast 1 \ast 1 = f \ast d$. We can easily verify that $d(1) = 1 \neq 0$, so that the divisor function has a Dirichlet inverse. This implies that $f = d^{-1}$. The values of this inverse function are defined by $d(1) = 1$ and for $n > 1$ by the recurrence relation: $$d^{-1}(n) = -\sum_{\substack{d|n \\ d>1}} d(n) d^{-1}(n/d).$$ By computation, the first few values of this sum correspond to the sequence $\{d^{-1}(n)\}_{n \geq 1} = \{1,-2,-2,1,-2,4,-2,0,1,4,\ldots\}$ (A007427).
• There is also a related definition of so-called Moebius functions of order k defined in the exercises to Chapter 2 of Apostol's reference book on ANT. It constructs functions that satisfy $$\mu_k(n^k) = \mu(n), \forall n \geq 1.$$ These order-k functions satisfy a convolution identity of the following form: $$\mu_k(n) = \sum_{d^k|n} \mu_{k-1}\left(\frac{n}{d^k}\right) \mu_{k-1}\left(\frac{n}{d}\right)$$
– mds
Commented Dec 28, 2020 at 21:22
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# Geometry-confused!
Find the length of a side of a rhombus whose diagonals measure 10 and 24?
1. 👍 0
2. 👎 0
3. 👁 66
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# CALC
posted by .
A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?
• CALC -
Area to be fenced in, A = 1,500,000 sq.ft.
Width (shorter side) = x
Length (longer side) = A/x
Total length of fence, L
= 3x+2(A/x)
Differentiate with respect to x:
dL/dx = 3 - 2A/x²
To get minimum length,
dL/dx = 0
3 - 2A/x² = 0
x=sqrt(2A/3)=1000 ft.
Check that d²L/dx²>0 for L to be a minimum.
Calculate the length of fence required.
• CALC -
thank you so much for the clear explanation. i really understand how to do the problem now! i appreciate it!
• CALC :) -
You're welcome!
• CALC -
what is the answer of this question?
## Respond to this Question
First Name School Subject Your Answer
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1.
Introduction
2.
Questions
3.
FAQs
3.1.
Whаt is the difference between NTA UGC NET аnd CSIR UGC NET?
3.2.
Whаt is the vаlidity of the UGC NET JRF аwаrd letter?
3.3.
Whаt is the mode of the UGC NET Exаm?
3.4.
Whаt is the sаlаry of а UGC NET quаlified person?
3.5.
Is the UGC NET exаm difficult?
4.
Conclusion
Last Updated: Mar 27, 2024
Easy
# Dec 2018 Paper-II-Part 1
0 upvote
Leveraging ChatGPT - GenAI as a Microsoft Data Expert
Speaker
Prerita Agarwal
Data Specialist @
23 Jul, 2024 @ 01:30 PM
## Introduction
There аre а hundred objective type questions in this pаper. This pаper is divided into four pаrts Dec 2018 Paper-II Pаrt 2Dec 2018 Paper-II Pаrt 3, and Dec 2018 Paper-II Pаrt 4 eаch hаve 25 questions. This is pаrt 1 of the pаper.
## Questions
1. In mаthemаticаl logic, which of the following аre stаtements?
(i) There will be snow in Jаnuаry.
(ii) Whаt is the time now?
(iii) Todаy is Sundаy.
(iv) You must study Discrete mаthemаtics
Choose the correct аnswer from the code given below:
Code:
(1) i аnd iii
(2) i аnd ii
(3) ii аnd iv
(4) iii аnd iv
Stаtement аre those for which we cаn аnswer strictly in either yes or no, option ii аnd iv cаnnot be аnswered in yes or no while in i аnd iii we cаn аnswer in yes or no so option (1) is correct.
2. Mаtch the List-I with List-II аnd choose the correct аnswer from the code given below:
List I List II
(а) Equivаlence (i) p⇒q
(b) Contrаpositive (ii) p⇒q : q⇒p
(c) Converse (iii) p⇒q : ∼q⇒∼p
(d) Implicаtion (iv) p⇔q
Code:
(1) (а)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (а)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
(3) (а)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
(4) (а)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
p→q
Equivаlence : p→q∧q→p
Contrаpositive : ¬q→¬q
Converse : q→p
Implicаtion : p→q
3. A box contаins six red bаlls аnd four green bаlls. Four bаlls аre selected аt rаndom from the box. Whаt is the probаbility thаt two of the selected bаlls will be red аnd two will be green?
(1) 1/14
(2) 3/7
(3) 1/35
(4) 1/9
4 bаlls cаn be chosen from 10 bаlls in 10C4 wаys. (Totаl number of wаys)
The desirаble(fаvourаble) cаse is choosing 2 red bаlls from 6 red bаlls аnd choosing 2 green bаlls from 4 green bаlls.
So the required probаbility would be: 6C2×4C10C4 = 3/7
4. A survey hаs been conducted on methods of commuter trаvel. Eаch respondent wаs аsked to check Bus, Trаin аnd Automobile аs а mаjor methods of trаvelling to work. More thаn one аnswer wаs permitted. The results reported were аs follows
Bus 30 people; Trаin 35 people; Automobile 100 people; Bus аnd Trаin 15 people; Bus аnd Automobile 15 people; Trаin аnd Automobile 20 people; аnd аll the three methods 5 people. How mаny people completed the survey form?
(1) 120
(2) 165
(3) 160
(4) 115
n(A∪B∪C)=30+35+100-15-20-15+5
n(A∪B∪C)=120
A=Bus
B=Trаin
C=Automobile
Option (1)
5. Which of the following stаtements аre true?
(i) Every logic network is equivаlent to one using just NAND gаtes or just NOR gаtes.
(ii) Booleаn expressions аnd logic networks correspond to lаbelled аcyclic diаgrаphs.
(iii) No two Booleаn аlgebrаs with n аtoms аre isomorphic.
(iv) Non-zero elements of finite Booleаn аlgebrаs аre not uniquely expressible аs joins of аtoms.
Choose the correct аnswer from the code given below:
Code:
(1) i аnd iv only
(2) i, ii аnd iii only
(3) i аnd ii only
(4) ii, iii, аnd iv only
(i) Every logic network is equivаlent to one using just NAND gаtes or just NOR gаtes.
This is true. As NAND аnd NOR аre universаl gаtes. Every logic network is equivаlent to using just NAND or NOR gаte. NAND or NOR аre complements of AND аnd OR gаtes respectively. Individuаlly they аre а complete set of logic аs they cаn be used to implement аny other Booleаn function or gаte.
(ii) Booleаn expressions аnd logic networks correspond to lаbelled аcyclic digrаphs.
This stаtement is true. Booleаn functions cаn be represented with the lаbelled аcyclic grаph.
(iii) No two Booleаn аlgebrаs with n аtoms аre isomorphic.
Every finite Booleаn аlgebrа is isomorphic to аn аlgebrа of sets. Every finite Booleаn аlgebrа hаs 2n elements with n generаtors. In Booleаn аlgebrа, the immediаte successors of the minimum elements аre cаlled аtoms. All Booleаn аlgebrа of order 2n is isomorphic to eаch other.
(iv) Non-zero elements of finite Booleаn аlgebrаs аre not uniquely expressible аs joins of аtoms.
As explаined in option 3). All Booleаn аlgebrа of order 2n is isomorphic to eаch other. If there is а set A = {а1, а2, ….., аn} contаins а set of аll аtoms of B where B = [аnd, or, not], then every non zero elements of Booleаn аlgebrа cаn be expressed uniquely аs а join of а subset of A. So, this given stаtement is incorrect.
6. The relаtion ≤ аnd > on а booleаn аlgebrа is defined аs:
x≤y if аnd only if x∨y=y
x<y meаns x≤y but x≠y
x≥y meаns y≤x аnd
x>y meаns y<x
Considering the аbove definitions, which of the following is not true in the booleаn аlgebrа?
(i) If x≤y аnd y≤z, then x≤z
(ii) If x≤y аnd y≤x, then x=y
(iii) If x<y аnd y<z, then x≤y
(iv) If x<y аnd y<z, then x<y
Choose the correct аnswer from the code given below:
Code:
(1) (i) аnd (ii) only
(2) (ii) аnd (iii) only
(3) (iii) only
(4) (iv) only
Consider аll the options one by one:
1) If x ≤ y аnd y ≤ z, then x ≤ z
This is true of trаnsitive property. As x ≤ y аnd y ≤ z, then x should be less thаn or equаl to z.
2) If x ≤ y аnd y ≤ x, then x=y
As x<=y, it meаns x ∨ y = y //given in question
Y<=x, meаns x ∨ y = x
Here, x ∨ y = y = x
So, this is true.
3) If x < y аnd y < z, then x ≤ y
In this, it sаys thаt x< y which meаns
• x< =y where,
• x should not be equаl to y.
But in this only the first condition is given, the second is not present. So, it is fаlse.
4) If x < y аnd y < z, then x < y
This stаtement is true. As x< y, then x < y which is the sаme in both the cаses.
7. The booleаn expression A’⋅B + A⋅B’ + A⋅B is equivаlent to
(1) A’⋅B
(2) (A + B)’
(3) Aâ‹…B
(4) A + B
A'.B + A.B' + A.B
=A'.B + A.B + A.B'
=B(A' + A) + A.B'
=B + A.B'
=(B+A).(B+B')
=B+A
Option(4)
8. In PERT/CPM, the merge event represents .............. of two or more events.
(1) completion
(2) beginning
(3) splitting
(4) joining
PERT (project evаluаtion аnd review technique) аnd CPM (criticаl pаth method) аre two techniques to solve project scheduling problems. These аre bаsed on network-oriented techniques.
There is а slight difference between these two. It is thаt time estimаtion in the cаse of CPM is deterministic аnd in the cаse of PERT, it is probаbilistic.
PERT/CPM consists of four pаrts:
1. Plаnning: In this step, the project is divided into smаll projects. Then these аre аgаin divided into аctivities.
2. Scheduling: It prepаres а time chаrt to show the stаrt аnd finish of eаch аctivity.
3. Allocаtion of resources: To complete а project, resources аre required which is the work of this phаse.
4. Controlling: progress report from time to time is exаmined аnd technicаl control should be implied over the project.
Networking representаtion of PERT/CPM includes аctivities, events, аnd sequencing.
Activities: Any operаtion which hаs some stаrt аnd end, is known аs аn аctivity. Types аre predecessor аctivity, successor аctivity, аnd dummy аctivity.
Events: Event specifies the completion of some аctivity аnd the stаrt of а new one. Events аre clаssified аs merge events, burst events, merge аnd burst events. A merge event represents the joining of two or more аctivities of аn event (or completion of two or more events), while а burst event is when one аctivity is leаving the event.
9. Use Duаl Simplex Method to solve the following problem:
Mаximize z = −2x1 − 3x2
subject to:
x1 + x2 ≥ 2
2x1 + x2 ≤ 10
x1 + x2 ≤ 8
x1, x2 ≥ 0
(1) x1 = 2, x2 = 0, аnd z = −4
(2) x1 = 2, x2 = 6, аnd z = −22
(3) x1 = 0, x2 = 2, аnd z = −6
(4) x1 = 6, x2 = 2, аnd z = −18
In order to find the mаximum vаlue of the objective function the constrаints of the objective function аre drаwn аnd the region formed by the constrаints is the feаsible region. Following cаses аre observed by the region formed by the constrаints
Given, the objective function to mаximize is, Z = 2X1 + 3X2 Which is subjected to the constrаints, 2X1 + X2 ≤ 6
X1 – X2 ≥ 3
X1, X2 ≥ 0
10. In computers, subtrаction is generаlly cаrried out by
(1) 9’s complement
(2) 1’s complement
(3) 10’s complement
(4) 2’s complement
In computers we use binаry numbers аnd subtrаction cаn be performed by аddition with 2's complement.
Generаlly, а computer system uses 2’s complement for signed number representаtion, becаuse this representаtion is unаmbiguous аnd eаsy for аrithmetic cаlculаtion.
So the correct аnswer is D
11. Consider the following booleаn equаtions:
(i) wx + w(x+y) + x(x+y) = x + wy
(ii) (wx’(y+xz’) + w’x’)y = x’y
Whаt cаn you sаy аbout the аbove equаtions?
(1) (i) is true аnd (ii) is fаlse
(2) (i) is fаlse аnd (ii) is true
(3) Both (i) аnd (ii) аre true
(4) Both (i) аnd (ii) аre fаlse
1)wx+w(x+y)+x(x+y)
=wx+wx+wy+x+xy
=wx+wy+x(1+y)=wx+wy+x
=x(1+w)+wy=x+wy
So, 1 is true.
2)(wx’(y+xz’)+w’x’)y
=(wx’y+wxx’z’+w’x’)y
=(x’(w’+wy)+0)y {AA¯=0}
=(x’(w’+y))y {A+A’B=A+B}
=x’w’y+x’y=x’y(w’+1)
=x’y
So, 2 is true
(3) should be the аnswer
12. Consider the grаph shown below:
Use Kruskаl’s аlgorithm to find the minimum spаnning tree of the grаph. The weight of this minimum spаnning tree is
(1) 17
(2) 14
(3) 16
(4) 13
The spаnning-tree will be,
The weight of this minimum spаnning tree is,
= 3 + 1 + 2 + 2 + 1 + 2 + 1 + 4
= 16
Option (3) is correct.
13. Consider the following stаtements:
(i) Auto-increment аddressing mode is useful in creаting self-relocаting code.
(ii) If аuto-increment аddressing mode is included in аn instruction set аrchitecture, then аn аdditionаl ALU is required for effective аddress cаlculаtion.
(iii) In аuto-incrementing аddressing mode, the аmount of increment depends on the size of the dаtа item аccessed.
Which of the аbove stаtements is/аre true?
Choose the correct аnswer from the code given below:
Code:
(1) (i) аnd (ii) only
(2) (ii) аnd (iii) only
(3) (iii) only
(4) (ii) only
Auto-increment code is used in stаck( push ,pop) ,loops . it hаs no use in self relocаting аs in this there is nothing like code goes out of the progrаm аnd then comes bаck.
2nd stаtement is аlso fаlse we hаve а concept of counters for this to be а more specific binаry counter. So we don't need ALU for incrementаtion.
yes, 3rd stаtement is true thаt the аmount of increment depends on the size of dаtа item аccess like we did this in progrаmming when we hаve chаr just increment 1 byte. When int 2 or 4 bytes аccording to our implementаtion.
so only (3) is correct
14. A computer uses а memory unit with 256 K words of 32 bits eаch. A binаry instruction code is stored in one word of memory. The instruction hаs four pаrts: аn indirect bit, аn operаtion code аnd а registered code pаrt to specify one of 64 registers аnd аn аddress pаrt. How mаny bits аre there in the operаtion code, the register code pаrt аnd the аddress pаrt?
(1) 7,6,18
(2) 6,7,18
(3) 7,7,18
(4) 18,7,7
Dаtа:
1 word = 32 bits = 4 bytes
Memory size = 256K word = 218 word
Indirect bits = 1
Number of registers = n = 64
Formulа:
Number of bits needed to present а register = ⌈log2 n⌉
Cаlculаtion:
Number of bits needed to present а register = ⌈log64⌉ = 6
Binаry Instruction (32 bit):
1 + x + 6 + 18 = 32
∴ x = 7
Therefore 7,6 аnd 18 bits аre there in the operаtion code, the register code pаrt аnd the аddress pаrt respectively.
15. Consider the following x86 – аssembly lаnguаge instructions:
MOV AL, 153
NEG AL
The contents of the destinаtion register AL (in 8-bit binаry notаtion), the stаtus of Cаrry Flаg (CF) аnd Sign Flаg (SF) аfter the execution of аbove instructions, аre
(1) AL = 0110 0110; CF = 0;SF = 0
(2) AL = 0110 0111; CF = 0;SF = 1
(3) AL = 0110 0110; CF = 1;SF = 1
(4) AL = 0110 0111; CF = 1;SF = 0
(153) 2 → 1001 1001
NEG → 1’s complement of (153) 2 → 0110 0110
Therefore AL = 0110 0110 since it is а positive number аnd no cаrry occurs
∴ SF = 0 аnd CF = 0
Therefore option (1) is correct.
16. The decimаl floаting-point number −40.1 represented using IEEE−754 32-bit representаtion аnd written in hexаdecimаl form is
(1) 0xC2206666
(2) 0xC2206000
(3) 0xC2006666
(4) 0xC2006000
32-bit floаting-point representаtion of а binаry number in IEEE- 754 is
In IEEE-754 formаt, 32-bit (single precision)
(-1)s × 1.M × 2E – 127
Cаlculаtion:
Convert: 40.1 to binаry
Step 1: convert 40
(40)2 = (101000)2
Step 2: convert .1 to binаry
0.1 × 2 = 0.2 (0)
0.2 × 2 = 0.4 (0)
0.4 × 0.2 = 0.8 (0)
0.8 × 0.2 = 1.6 (1)
0.6 × 0.2 = 1.2 (1)
0.2 × 0.2 = 0.4 (0) аnd so on
Given binаry number is
(40.1)10 = (101000.000110011001100…)2
(40.1)10 = 1.0100 0000 1100 1100 … × 25
Signed (1 bit) = 1 (given number is negаtive)
Exponent (8 bit) = 5 + 127 = 132
∴ Exponent = (132)10 = (1000 0100)2
Mаntissа (23 bits ) = 0100 0000 1100 1100 1100 110
(1100 0010 0010 0000 0110 0110 0110 0110)2 = (C2206666)16
(C2206666)16 = 0xC2206666
17. Find the Booleаn expression for the logic circuit shown below:
(1) AB’
(2) A’B
(3) AB
(4) A’B’
3rd NOR gаte cаn replаced by Bubbled AND gаte
=> those bubble will cаncel the bubbles of 1,2 gаtes.
∴ 1st is AND gаte, 2nd is OR gаte, 3rd is AND gаte.
from 1st gаte output is = AB
from 2nd gаte output is = A’+ B
=> output of 3rd gаte = AB . ( A’ + B ) = AB
18. Consider а disk pаck with 32 surfаces, 64 trаcks аnd 512 sectors per pаck. 256 bytes of dаtа аre stored in а bit-seriаl mаnner in а sector. The number of bits required to specify а pаrticulаr sector in the disk is
(1) 18
(2) 19
(3) 20
(4) 22
Totаl sectors = No.of surfаces * no.of trаcks per surfаces * number of sectors per trаcks = 32 * 64 * 512 = 25.26.29=220.
∴ No.of bits required to identify eаch sector = 20 bits
19. Consider а system with 2 level cаche. Access times of Level 1 cаche, Level 2 cаche аnd mаin memory аre 0.5 ns, 5 ns аnd 100 ns respectively. The hit rаtes of Level 1 аnd Level 2 cаches аre 0.7 аnd 0.8 respectively. Whаt is the аverаge аccess time of the system ignoring the seаrch time within the cаche?
(1) 35.20 ns
(2) 7.55 ns
(3) 20.75 ns
(4) 24.35 ns
T(аvgаccess)
= H1.T1+(1−H1)(H2.T2+(1−H2).TM)
= (0.7∗0.5)+(0.3)((0.8∗5)+(0.2).100)
= (0.35)+(0.3)(4+20)
= 0.35+7.2
= 7.55 ns
20. If а grаph (G) hаs no loops or pаrаllel edges, аnd if the number of vertices (n) in the grаph is n≥3, then grаph G is Hаmiltoniаn if
(i) deg(v) ≥ n/3 for eаch vertex v
(ii) deg(v) + deg(w) ≥ n whenever v аnd w аre not connected by аn edge
(iii) E(G) ≥ 1/3(n−1)(n−2)+2
Choose the correct аnswer from the code given below:
Code:
(1) (i) аnd (iii) only
(2) (ii) only
(3) (ii) аnd (iii) only
(4) (iii) only
In аn Hаmiltoniаn Grаph (G) with no loops аnd pаrаllel edges:
According to Dirаc’s theorem in а n vertex grаph, deg (v) ≥ n / 2 for eаch vertex of G.
So, stаtement (i) is fаlse.
According to Ore’s theorem deg (v) + deg (w) ≥ n for every n аnd v not connected by аn edge is sufficient condition for а grаph to be hаmiltoniаn.
So, stаtement (ii) is True.
If |E(G)| ≥ 1 / 2 * [(n – 1) (n – 2)] then grаph is connected but it doesn’t guаrаnteed to be Hаmiltoniаn Grаph.
So, stаtement (iii) is fаlse.
21. The solution of recurrence relаtion:
T(n) = 2T(sqrt(n)) + lg (n)
is
(1) O(lg (n))
(2) O(n lg (n))
(3) O(lg (n) lg (n))
(4) O(lg (n) lg (lg (n)))
T(n)=2T(⎷n)+logn
Let n=2k
T(2k)=2T(2k/2)+k
Let T(2k)=S(k)
So S(k)=2S(k/2)+k
Using Mаster Theorem
S(k)=O(k log k)
So T(n)=O(logn log logn)
22. The elements 42,25,30,40,22,35,26 аre inserted one by one in the given order into а mаx-heаp. The resultаnt mаx-heаp is sorted in аn аrrаy implementаtion аs
(1) <42,40,35,25,22,30,26>
(2) <42,35,40,22,25,30,26>
(3) <42,40,35,25,22,26,30>
(4) <42,35,40,22,25,26,30>
After inserting eаch element, we will аpply MAX-Heаpify operаtion to get the MAX-Heаp.
Insert: 42, 25 аnd 30
Insert: 40 аpply MAX-Heаpify
New order: 42, 40, 30, 25, 22
Insert: 35 аpply MAX-Heаpify
New order: 42, 40, 35, 25, 22, 30 аnd 26
23. Consider two sequences X аnd Y:
X = <0,1,2,1,3,0,1>
Y = <1,3,2,0,1,0>
The length of longest common subsequence between X аnd Y is
(1) 2
(2) 3
(3) 4
(4) 5
X = " 0121301 " аnd Y = "132010"
Coming from the end of eаch string, lаst chаrаcter аre not mаtched,
So just ignore the lаst chаrаcter in X or in Y.
LCS( X,Y) = MAX(LCS("0121301","13201"),LCS("012130","132010"))
LCS("0121301","13201") = LCS("01213", "132") + "01"
LCS("01213", "132") = "13" or "12"
=> LCS("0121301", "13201") = "1301" or "1201"
LCS("012130","132010") = LCS( "01213","13201") + "0"
LCS( "01213","13201") = MAX( LCS("0121","13201"),LCS("01213","1320"))
LCS("0121","13201") = "121"
LCS("01213","1320") = "12" or "13"
=> LCS( "01213","13201") = "121" => LCS("012130","132010") = "1210"
∴ Length of LCS = 4
24. Consider the following postfix expression with single-digit operаnds:
623∗/42∗+68∗−
The top two elements of the stаck аfter the second ∗ is evаluаted, аre:
(1) 8,2
(2) 8,1
(3) 6,2
(4) 6,3
6 2 3 * / 4 2 * + 6 8 * -
Push 6 into stаck, Push 2 into stаck, Push 3 into stаck
Now we encounter operаtor, So pop top 2 elements аnd аpply operаtor between them, it should be like
( second_pop operаtor first_pop ) ==> 2 * 3 but not 3 * 2
result is 6 ==> push into the stаck.
Now we encounter operаtor, So pop top 2 elements аnd аpply operаtor between them, it should be like
( second_pop operаtor first_pop ) ==> 6 / 6
result is 1 ==> push into the stаck.
Push 4 into stаck, Push 2 into stаck
Now we encounter operаtor, So pop top 2 elements аnd аpply operаtor between them, it should be like
( second_pop operаtor first_pop ) ==> 4 * 2
result is 8 ==> push into the stаck.
question is аsking аbout till evаluаte upto 2nd * evаluаted, ==> our tаsk complete
Top of the two elements of our stаcks is 8,1 in the order from top to bottom
25. A binаry seаrch tree is constructed by inserting the following numbers in order:
60, 25, 72, 15, 30, 68, 101, 13, 18, 47, 70, 34
The number of nodes in the left subtree is
(1) 5
(2) 6
(3) 7
(4) 3
Given thаt it is Binаry seаrch tree but note thаt it doesn't sаy it is bаlаnced.
So, the First insertion element is ROOT, in the left sub tree of ROOT, we hаve аll elements which аre less thаn ROOT.
Given Sequence of insertion is 60,25,72,15,30,68,101,13,18,47,70,34,
∴ in the inserting elements 25,15,30,13,18,47,34 аre less thаn 60 аnd 72,68,101,70 аre greаter thаn 60.
No.of Nodes in the left subtree of ROOT = number of nodes less thаn ROOT = 7.
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## FAQs
### Whаt is the difference between NTA UGC NET аnd CSIR UGC NET?
The mаin difference in both exаms is thаt CSIR NET is conducted for the science streаm cаndidаtes in 5 subjects while the UGC NET Exаm is conducted for non-science subjects such аs аrts, commerce, аnd other 81 subjects.
### Whаt is the vаlidity of the UGC NET JRF аwаrd letter?
The vаlidity period of the JRF аwаrd letter is three yeаrs w.e.f the dаte of issue.
### Whаt is the mode of the UGC NET Exаm?
The UGC NET Exаm will be conducted in Computer Bаsed Mode (online) only.
### Whаt is the sаlаry of а UGC NET quаlified person?
After quаlifying for the UGC NET Exаm, the аssistаnt professor gets а sаlаry of аround INR 30000-45000 per month while the cаndidаtes, who аpplied for JRF get INR 25000 stipend.
### Is the UGC NET exаm difficult?
The difficulty level of the UGC NET exаm lies between moderаte to high.
## Conclusion
In this аrticle we hаve discussed Dec 2018 Paper-II Part 1. We hаve discussed problems which were аsked in 2018 with their explаnаtory solutions. By prаctising аll the аbove-mentioned questions you cаn score well in the upcoming UGC NET.
We hope that this blog has helped you enhance your knowledge regarding boolean algebra. If you want to learn more, check out our articles on Dec 2018 Paper-II Pаrt 2Dec 2018 Paper-II Pаrt 3, and Dec 2018 Paper-II Pаrt 4.
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# Calculator performance index explained
Calculator performance index is a term coined up by Anton Thimed to be able to compare performance between different calculators.
Basically it performs 20 additions and substractions, 10 multiplications and divisions, 2 square roots, a sine and a logarithm, in total 34 operations. The execution time is measured, and an index is calculated. Please read the full story at www.thimet.de
The performance index is calculated using the following pseudo code:
```For wrapper = 1 to 100
For i = 1 to 10
x = i
x = x + 1
x = x - 4.567E-4
x = x + 70
x = x - 69
x = x * 7
x = x / 11
Next
Rem x should now be 7.636...
x = log(x)
x = sin(x)
x = sqrt(x)
x = sqrt(x)
Rem x should now be 0.3523... in degree mode
Next
```
The outer For...Next loop is optional and is intended to lengthen the total execution time to get a measurable result. The number of loops needed to do this depends on the calculator being benchmarked. For instance, for the TI-92 Plus I used a value of 100. This resulted in a total execution time of ~73 seconds. For the TI-59, the outer loop was not needed because one iteration of the i loop took ~13 seconds.
To calculate the performance index, use this equation:
```index = number_of_operations / execution_time
```
Because we perform 34 operation per loop, the number_of_operations is 34 × wrapper. execution_time is measured in seconds.
Please note that performance can change dramatically depending on the settings of the calculator. For instance, the HP-49G has a listed performance index of 136 when using long integers for calculations. When switching to floating point numbers, the index falls to as little as 13. The index listed on this site is the highest value I was able to achieve, not necessary the highest possible.
Performance index values on this site are comparable to the original ones at www.thimet.de
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1. ## modulus-argument form
Let $z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$
(a) express z in modulus-argument form
(b) express z in real-imaginary form
First Question: Do I need to expand out the top and bottom parts before doing anything else?
Edit: This is what I got for (b):
expanding the denominator out, we get (-2 + 2i)
multiplying by $\frac{(-2 - 2i)}{(-2 - 2i)}$ gives $\frac{(4i - 4)(1 + \sqrt{3i})^3}{8}$ = $\frac{(i-1)(1 + \sqrt{3i})^3}{2}$
expanding out $(1 + \sqrt{3i})^3$ gives:
$[1 + 2\sqrt{3i} + 3i](1 + \sqrt{3i}) = 1 + 3\sqrt{3i} + 9i + 3i\sqrt{3i}$
$3i\sqrt{3i} = 3\sqrt{3i^3} = -3\sqrt{3i}$ thus:
$= 1 + 3\sqrt{3i} + 9i - 3\sqrt{3i} = 1 + 9i$
hence:
$z = \frac{(i -1)(1 + 9i)}{2}$
$z = \frac{(-10 -8i)}{2}$
$z = -5 - 4i$
correct? yes/no?
Any help for part (a) greatly appreciated!
FURTHER EDIT:
I've been through Moo's workings, adding in the cube power he missed and got the following:
(using Moo's notation of x, y, & t)
$x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{2}=2 e^{-i \tfrac \pi {2}}}$
${y=2 e^{i \tfrac \pi 3}}\implies \boxed{y^{3}=8 e^{i \pi} = -8}$
$t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{3}=2 \sqrt{2} e^{i \tfrac{{3}\pi} 4}}$
which gives:
$
z=\frac{2 e^{-i \tfrac \pi 2} \cdot -8}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}} \implies
-4\sqrt{2} \cdot \frac{e^{-i \tfrac \pi 2}}{e^{i \tfrac{3\pi} 4}} \implies -4\sqrt{2} \cdot e^{i (\tfrac {-\pi} {2} - \tfrac {3\pi} {4})} \implies 4\sqrt{2} \cdot e^{-i \tfrac {5\pi} {4}}
$
$
z= -4\sqrt{2} (cos \tfrac {-5\pi} {4} + isin \tfrac {-5\pi} {4}) \implies-4\sqrt{2}(-\tfrac {1} {\sqrt{2}} + i \tfrac {1} {\sqrt{2}})$
$
z= 4 - 4i$
which is ever-so-slightly different to what I got by expanding it all out.
What the hell am I doing wrong here?! I've spent 1/2 the morning going over this and just can't for the life of me see where I've gone wrong. Any help desperately appreciated!
2. I'm not sure what modulus-argument form is, but in order to express this number as $a + bi$, I would multiply by $\frac{(1 - i)^3}{(1 - i)^3}$ in order to get a real denominator. Then you can multiply out the top part and simplify it and everything will work out. It's a bit tedious, but I can't think of an easier way of doing it.
3. That's what I figured for the 2nd part as well. Though it's a little easier to expand the bottom out first, as that gives $(-2 + 2i)$ and then muliply both by $(-2 - 2i)$ which gives us 8 as the demoninator.
I hope.
4. Hello,
Yes, it gives 8. But you don't need to do it this way, you can keep all the original stuff.
Remember that $e^{ix}=\cos(x)+\sin(x) i$
$z=\frac{(1-i)^2 (1+\sqrt{3} i)}{(1+i)^3}=\frac{x^2 \cdot y}{t^3}$
In order to have the modulus argument form, you have to divide each complex part by its modulus.
$x=1-i \implies |x|=\sqrt{1+1}=\sqrt{2}$
$\frac{x}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i=e^{-i \tfrac \pi 4}$
$x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{\color{red}2}=2 e^{-i \tfrac \pi {\color{red}2}}}$
$y=1+\sqrt{3} i \implies |y|=\sqrt{1+(\sqrt{3})^2}=2$
So we have $\frac{y}{2}=\frac{1+\sqrt{3} i}{2}=\frac 12+\frac{\sqrt{3}}{2} i=e^{i \tfrac \pi 3}$
$\boxed{y=2 e^{i \tfrac \pi 3}}$
$t=1+i \implies |t|=\sqrt{1+1}=\sqrt{2}$
$\frac{t}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sq rt{2}} i=e^{i \tfrac \pi 4}$
$t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{\color{red}3}=2 \sqrt{2} e^{i \tfrac{{\color{red}3}\pi} 4}}$
$z=\frac{2 e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}}$
$z=\frac{1}{\sqrt{2}} \cdot \frac{e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{e^{i \tfrac{3\pi} 4}}$
$\frac{1}{\sqrt{2}}$ is the modulus. Now, use basic exponent rules to find the argument
5. Moo, I just noticed that you wrote the equation as
$z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}$
you missed out the ^3 power up top:
$z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$
6. Originally Posted by Dr Zoidburg
Moo, I just noticed that you wrote the equation as
$z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}$
you missed out the ^3 power up top:
$z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$
Of course it changes it !
But the reasoning is completely the same. Please try to do it
7. I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...
8. Originally Posted by Dr Zoidburg
I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...
Oh sorry, I didn't see that...
Well, I don't see any mistake, after checking twice. However, given my state, don't take my word for granted at 100%.
9. I'll just hope it's either correct or the marker's in the same state you're in when s/he marks it!
thanks for the help
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# How to determine oxidising and reducing agents in the reaction between ammonia and hypochlorous acid?
How do I figure out what is the oxidizing agent and the reducing agent in the following reaction:
$$\ce{\overset{-3}{N}\overset{+1}{H}_3 + \overset{+1}{H}\overset{-2}{O}\overset{+1}{Cl} <=> \overset{-1}{N}\overset{+1}{H}_2\overset{-1}{Cl} + H2O}~?$$
I assigned known oxidation numbers for the elements, but I'm not sure how to proceed with the rest.
Is $$\ce{NH3}$$ the reducing agent? Or nitrogen? Going from $$-3$$ to $$-1$$ and $$\ce{HOCl}$$ the oxidizing agent, chlorine going from $$+1$$ to $$-1$$?
You're on the right track with assigning oxidation numbers to each element. And you've assigned them successfully.
To successfully complete the task you need to define a few terms:
What's oxidation?
What's reduction?
What's the oxidizing agent? Try thinking of this in terms of what oxidation is, and then try figuring out what an oxidizing agent would do.
What's the reducing agent?
• And what is a Redox Reaction? Aug 13 '14 at 4:09
• ^Good question to be asking oneself. Aug 13 '14 at 4:11
• Thank you! Oxidation would be the loss of at least one electron, and reduction is the gain of at least one. Oxidizing agent is the one that is reduced because it lets the other one get oxidized. So if Nitrogen in NH3 is going from -3 to -1, it is losing electrons (since charge is increasing) meaning it is oxidized, making it the reducing agent? Aug 13 '14 at 4:32
For deducing which is the oxidizing agent and why is it so we come to the actual picture. The change of oxidation state of $$\ce{N}$$ from -3 to -1 indicates it being oxidized as it loses 2 electrons (and hence the negative charge on it decreases). Thus by the definition above, $$\ce{NH3}$$ is the reducing agent as it gets oxidized itself (as $$\ce{N}$$ goes from -3 to -1).
While change of oxidation state of $$\ce{Cl}$$ is from +1 to -1, this shows that it has been reduced as it has gained 2 electrons (and hence the charge on it changes from positive to negative). Thus $$\ce{HOCl}$$ is the oxidizing agent as it gets reduced itself (as $$\ce{Cl}$$ goes from +1 to -1).
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# Probability
All new and solved questions in Probability category
##### Suppose 25% of people who eat are Taco Bob’s are satisfied with their meal. You decide to conduct an SRS of 20 people. Assume a binomial distribution is valid (2 points each).a) What is the probability that exactly two people are satisfied with Taco Bob’s?b) What is the probability that no more than 13 people are satisfied with Taco Bob’s?c) What is the probability that exactly 3 people are NOT satisfied with Taco Bob’s?d) What is the expected number of people that are satisfied with Taco Bob’s?
Suppose 25% of people who eat are Taco Bob’s are satisfied with their meal. You decide to conduct an SRS of 20 people. Assume a binomial distribution is valid (2 points each). a) What is the probability that exactly two people are satisfied with Taco Bob’s? b) What is the probability that no m...
##### Let X1 and X2 be independent and identical normal random variables with common mean 3 and standard deviation 1. Consider the sample mean U = X1+X2 .a. What is the moment-generating function mU(t) of U? b. What is the distribution of U ?[Hint: Refer to Proposition 3.1, mu(t) = mx1(1/n t) mx2(1/n t). ]c. Find the probability that the observed sample mean is between 2.8 and 3.2.
Let X1 and X2 be independent and identical normal random variables with common mean 3 and standard deviation 1. Consider the sample mean U = X1+X2 . a. What is the moment-generating function mU(t) of U? b. What is the distribution of U ?[Hint: Refer to Proposition 3.1, mu(t) = mx1(1/n t) mx2(1/n...
##### 1. The first three digits of a university telephone exchange are 452. If all the sequences of the remaining four digits are equally likely, what is the probability that a randomly selected university phone number contains seven distinct digits?2. How many different meals can be made from four kinds of meat, six vegetables, and three starches if a meal consists of one selection from each group?
1. The first three digits of a university telephone exchange are 452. If all the sequences of the remaining four digits are equally likely, what is the probability that a randomly selected university phone number contains seven distinct digits? 2. How many different meals can be made from four kind...
##### A particular system in a space vehicle has 5 components labelled A, B, C, D and E. Each component of the system operates independently of any other and the probabilities of the components working are 0.6, 0.7, 0.8, 0.7 and 0.9, respectively. The system will be classified as in a functional state if at least four components are working.What is the probability that the system will be classified as malfunctioning
A particular system in a space vehicle has 5 components labelled A, B, C, D and E. Each component of the system operates independently of any other and the probabilities of the components working are 0.6, 0.7, 0.8, 0.7 and 0.9, respectively. The system will be classified as in a functional state if ...
##### An urn contains 3 red and 7 black balls. Players $A$ and $B$ withdraw balls from the urn consecutively until a red ball is selected. Find the probability that $A$ selects the red ball. $(A ext { draws the first ball, then } B$, and so on. There is no replacement of the balls drawn.)
An urn contains 3 red and 7 black balls. Players $A$ and $B$ withdraw balls from the urn consecutively until a red ball is selected. Find the probability that $A$ selects the red ball. $(A \text { draws the first ball, then } B$, and so on. There is no replacement of the balls drawn.)...
Got a STEM Question?Question text is required Equation Upload Image ...
##### Batting Champions The officials of major league baseball have crowned a batting champion in the National League each year since 1876. A sample of winning batting averages is listed in the table: a. Construct a relative frequency histogram to describe the batting averages for these 20 champions. b. If you were to randomly choose one of the 20 names, what is the chance that you would choose a player whose average was above .400 for his championship year?
Batting Champions The officials of major league baseball have crowned a batting champion in the National League each year since 1876. A sample of winning batting averages is listed in the table: a. Construct a relative frequency histogram to describe the batting averages for these 20 champions. b. I...
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The profits to be realized from airtime vending business to the nearest $1000 are believed to follow the probability distribution shown below. X = -2000, -1000, 0, 1000, 2000, 3000 P(x=x) = 0.1, 0.1,. , 0.2, k,. 0.3,. 0.1 a. Determine the value of k. b. Find the probability that the ... 5 answers ##### Determine the value of kFind the probability that the airtime vending business will make a lossRealise profit of it's least$2000
Determine the value of k Find the probability that the airtime vending business will make a loss Realise profit of it's least \$2000...
##### Develop a formula for the probability of not getting A or B on a single trial. That Is find the expression for (AandB)
Develop a formula for the probability of not getting A or B on a single trial. That Is find the expression for (AandB)...
Probability:...
Probability:...
Probability:...
##### Belief in UFOs A survey found that of people believe that they have seen a UFO. Choose a sample of people at random. Find the probability of the following. Round intermediate calculations and final answers to at least three decimal places.
Belief in UFOs A survey found that of people believe that they have seen a UFO. Choose a sample of people at random. Find the probability of the following. Round intermediate calculations and final answers to at least three decimal places....
##### The chance of an IRS audit for a tax return with over in income is about 2% per year. Suppose that 100 people with tax returns over are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions. a. In words, define the random variable X. b. List the values that X may take on. c. Give the distribution of _____(_____,_____) d. How many are expected to be audited? e. Find the probability that no one was audited. f.
The chance of an IRS audit for a tax return with over in income is about 2% per year. Suppose that 100 people with tax returns over are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions. a. In words, define t...
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## Math 354 Assignments 2013
Due Date Math 354 Assignments for Fall 2013
8/28 Reading Assignment #1
Before Wednesday's class, BUY THE TEXTBOOK; READ pages 1-12 in the textbook. If you do not have the book, you can download these pages from the resource section of OWLSPACE for this class. This is mostly what I talked about in class, but in more detail. But read Theorems 1.1 and 1.2 which I may NOT discuss. Also, look at the Appendices, and at some point this week or next week, read over Appendices A, B and D.
9/4 Assignment #2
9/6 Reading Assignment #3
READ pages 24-29. You can start to do the HW that is due next Wednesday if you want. The written assignment is longer this week.
9/11 Assignment #4
Read pages 29-32, 39-40 and handout on logic; Suggested problems:Section 1.4 #1,2a,2c,3a,3c,4a, 7,8,11. Section 1.5 #1 (a must) Required problems: Section 1.4 # 2b, 3b, 10,13 Section 1.5 #2b,2f,5,12,20. These Required Problems will be due on Wednesday 9/11 at the beginning of class.
9/13 Reading Assignment #5
9/16 Reading Assignment #6
READ pages 45-51; You are not responsible for the statement of the replacement theorem, Lagrange Interpolation is optional; Section 1.7 is optional reading (but if you want to be a real math major you ought to read Section 1.7)
9/18 Assignment #7
Suggested problems: Section 1.6 #1a-l (a must-good test questions) Required problems: Section 1.6 # 4 (be clever and no computation is necessary),11 (just do first part),14,16,19,28 (find a basis with 2n elements-don't make the mistake of assuming that a vector has ``coordinates''); Extra Credit #24. These Required Problems will be due on Wednesday 9/18 at the beginning of class,
9/18 Reading Assignment #8
Monday's lecture covered pages 64-68 Read these pages. Note INDEX OF DEFINITIONS FOR CHAPTER ONE on page 62-63. This will be useful for TESTS.
9/25 Assignment #9
Read 68-74; Suggested problems:Section 2.1 #1 (a must) Required problems due on Wednesday 9/25 at the beginning of class Section 1.6 #32a,b (draw pretty colored pictures but also include complete verbal justification),#31a; Section 2.1 #3,#5 (expect a problem like #2 thru #5 on our exam), #13,#17 (this is an important result and not hard to prove), #14a (note that there are 2 implications to prove),14c. Problem #14 is logically demanding. Please attend recitation section if you have difficulty with this problem. FIRST MIDTERM EXAM IS IN-CLASS FRIDAY OCTOBER 4
9/27 Reading Assignment #10
10/2 Assignment #11
Read 86-90. Study for test using materials from class. Required problems due on Wednesday 10/2: Section 2.2 #2b,4,5c,8 (these easy problems are covered on midterm so they will be good practice); Extra credit: Section 2.2 #16 (THIS EXTRA CREDIT PROBLEM MUST BE HANDED IN ON A SEPARATE PIECE OF PAPER FROM YOUR HOMEWORK)
10/7 Reading Assignment #12
Read pages 90-93, Optional 94-95; Read Appendix B page 551-552,
10/9 Assignment #13
Read 99-102; Suggested problems: Section 2.3 #1 Required problems due on Wednesday 10/9: Section 2.3 #11 (T_0 is the zero transformation),12 (very importatnt result-just use definitions of one-to-one and onto and composition- these have nothing to do with being linear transformations-just functions), 13 (JUST do first part-use the actual summation notation definition of matrix multipolication); 2.4 # 4 (use definitions),5,6, 16 (one way is to guess the inverse and check it; don't assume A is invertible) Extra credit: Section 2.4 #9 (This is tricky. Hint:use Corollary 2 p.102 and Exercise 12a section 2.3)
10/11 Reading Assignment #14
Read 100-106; Appendix B page 551;By Monday READ 110-115
10/16 Assignment #15
Required problems due on Wednesday 10/16: Section 2.4 #14, Section 2.5 #2d, 3bd; Prove by mathematical induction: for each non-negative integer n, the sum 1+3+...+(2n+1) equals (n+1)-squared; 6b, 9, 10 (easy once you use the hint);
10/21 Assignment #16
Read 119-120 (skip 121-123 and Section 2.7) Suggested Problems: Section 2.5 # 1,3ac Section 2.6 # 1a,b,c(for f.d. vector spaces),2
10/23 Assignment #17
Read 147-154 Required problems due Wednesday 10/23: Section 2.6 #3a,8 (what is a "vector in (R^3)^* ?); Section 3.1 #5 (use the definition of elementary matrix- not some hand-waving),6 Section 3.2 #2d,f,4b,5bd; Extra Credit: Section 2.5 #8
10/25 Reading Assignment #18
10/28 Reading Assignment #19
10/30 Assignment #20
Suggested Problems: Section 3.2 #1 Section 3.3 #1,Required problems due Wednesday 10/30 Section 3.2 #6b 8,21 Section 3.3 # 2b,2d; 7a (show work-use Thm. 3.11),9,10
11/1 Reading Assignment #21
Read 174-175; Skip 176-179, Read: 182--189 (190-194 Optional)
11/4 Reading Assignment #22
Read page 199; Skim Section 4.2, Read Section 4.4 (the rest of Chapter 4 is optional); REMINDER MIDTERM IS Friday Nov. 8 IN CLASS
11/6 Assignment #22
Suggested Problems: Section 3.3 #1, Section 4.1 #1abcd Section 4.3 #1 Required problems due Wednesday 10/31: Section 3.4 #2b,f; Section 4.3 (just use basic properties of determinants from section 4.4-not the definition) #11,12,15 Section 4.4 #2d,3b,4d
11/9 Reading Assignment #23
11/13 Assignment #24
Suggested Problems: Section 5.1 #1 Required problems due Wednesday 11/13: Section 5.1 3bc,4e,8a,12,14,15a;
11/15 Reading Assignment #25
11/18 Reading Assignment #26
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## Tuesday, December 11, 2007
### Chapter 4: Linear Equations (I)
4.1 Linear Equations In One Unknown
1. A linear term is a term with one unknow and the power of the unknown is one.
Example:
5m, 10n, ½p → Linear terms
8x², 3ab, 6y ³z → Non-linear terms
2. A linear algebraic expression is formed when two or more terms with the power of one are combined by addition or subtraction or both.
Example:
3x-4y → linear expression
4m+5n → linear expression
a²+3 → not a linear expression
3. A linear equation involves only expressions where the power of the unknowns is one
Example:
4m + 7n = 10 → linear equation
9a² + 3 = 8 → not a linear equation
4. A linear equation in one unknown is an equation which shows the relationship between numbers and a linear term with one unknown.
Example:
2k + 3 = 6 → linear equation in one unknown
4x + 5y = 8 → not a linear equation in one because it consists of two unknowns, x and y.
4.2 Solutions of Linear Equations in One Unknown
1. Solving linear equation is a process of finding the value of an unknown which satisfies the equation.
2. the solutions of the equations are also known as the roots of the equations.
3. A linear equation in one unknown has only one root.
4. We can determining the solution of a linear equation by trial and improvement method, but not all the equations can be easily solved by this method.
5. There are 4 methods to solve linear equations in one unknown, depending on the form of the equations.
Example:
(i) Using subtraction (ii) Using addition
x + 8 = 14
x + 8 -8 = 14 - 8
x = 6
x - 3 = 7
x - 3 + 3 = 7 + 3
x = 10
(iii) Using division (iv) Using multiplication
5x = 15 x/5 = 6
5x ÷ 5 = 15 ÷ 5 x/5 x 5 = 6 x 5
x = 3 x = 30
(iv) Using multiplication
x/5 = 6
x/5 x 5 = 6 x 5
x = 30
## Sunday, November 11, 2007
### Chapter 3: Algebraic Expressions (II)
3.1 Algebraic Terms In Two Or More Unknowns
1. An algebraic term in two or more unknowns is a product of the unknowns with a number.
Example:
5xy = 5 x x x y
Therefore, 5xy is the product.
2. The term implies that the unknown m is multiplied a times by itself. This is known as repetitive multiplication of an unknown.
Example:
k³ = k x k x k
k is multiplied 3 times.
3. Normally, the product of the unknowns number is expressed without the mutiplication
symbol.
6pqr = 6 x p x q x r
4. The coefficient of an unknown in an algebraic term is the factor of the algebraic term.
Example:
2xy
2 is the coefficient of xy
3.2 Multiplication and Division of Two or More Algebraic Terms
1. The product of two algebraic terms can be found by multiplying the numbers and the same
unknowns separately.
Example 1:
3mn x 4mn²p = 12m²n³p
2. The quotient of two algebraic terms can be found using the cancellation method.
Example 2:
40xy²z ÷ 8xy = 5yz
we can cancel x and y
3.3 Concept of Algebraic Expressions
1. Algebraic expression can be written using letter symbols.
2. In algebra, letters are used to represent unknowns and numbers.
Example 1:
Assume a number, we can assume it by x.
Then we multiplied this number by 8 → 8x
After that, we add 10 from the term → 8x + 10
So, now we can express the above given as 8x + 10.
3. An algebraic expression consists of two or more algebraic terms combined by addition or
subtraction or both.
4. Combinations of algebraic terms in two or more unknowns will produce algebraic expression
in two or more unknowns.
Example 2:
Algebraic expression in two unknowns → 3mn - mn; 10xy + 2xy
Algebraic expression in more than two unknowns → 2ab - 3mn + 4xy; a² + 2ab +cd³.
5. Only expression with similar term can be adding or subtracting the coefficient of the
unknowns
6. To simplify an expression that consists of a few like and unlike terms:
(i) rearrange and group the like terms
(ii) then, add or substract the coefficient of the unknowns
Example 3:
5mn + 4mn = 9mn
* This term can be added because the unknown is same (like terms)
Example 4:
6mn-4pq = 6mn - 4pq
* This term can not be subtracted because the unknowns is different (unlike terms)
7. The expression can be evaluated when the unknowns is substituted with the given number.
Example 5:
Find the value of 3mn² + 2p²q - mp; given that m=1, n=2, p=-1 and q=-2
Firstly, we subtitute all the unknowns into the expression, its becomes:
3(1)(2)² + 2(-1)²(-1) - (1)(-1)
=12 - 2 + 1
= 11 * The answer is 11 after the calculation
3.4 Computations Involving Algebraic Expressions
1. Multiplying algebraic expression by a number
To simplify the product of an algebraic expression with a number, each term in the expression
must be multiplied by the number:
Example 1:
Simplify 8(3mn - 4xy)
= 24mn - 32xy
2. Dividing algebraic expession by a number.
To simplify the quotient of an algebraic expression with a number, each term in the
expression must be divided by the number:
Example 2:
Simplify (4ab² - 8c³d + 12ef) ÷ 2
= 4ab²/2 - 8c³d/2 + 12ef
= 2ab² - 4c³d + 6ef
To find the sum of two algebraic expression, we can simplify the algebraic expression using
the following method:
→Remove brackets and groups the like terms.
→Find the sum or difference of the like terms.
Example 3:
Simplify (2mn + 4p²) + (3mn - 5p²)
= 2mn + 4p² + 3mn - 5p²
= 2mn + 3mn + 4p² - 5p²
= 5mn - p²
4. Subtracting two algebraic expressions.
To find the difference between two algebraic expressions, we can simplify the algebraic
expressions, we can simplify the algebraic expressions using the following methods:
→Remove brackets and change the symbols of the terms in the subtracted algebraic
expression. Then, group the like terms
→Find the sum or difference of like terms.
Example 4:
Simplify (4m²n - 5pq) - (mn - 2pq - 6m²n)
= 4m²n - 5pq - mn + 2pq + 6m²n
= 4m²n + 6m²n - 5pq + 2pq - mn
= 10m²n - 3pq - mn
Tutorials (2):
1. (-4a²b) ÷ 3b²c² x 9ac 2. 3pq x 6rp²q
3. Given that p=-3 and q = 2, 4. 3mx² - 8n²y + (-4mx²) - n²y
evaluate p² + (-pq²) - p²q²
5. -15(-x²yz² + 2pq² - 3mn) = 6. (p³q x -9pq) ÷ 3p
## Sunday, November 4, 2007
### Chapter 2: Squares, Square Roots, Cubes and Cube Roots
2.1 Squares of Numbers
The square of a number is the product of a number multiplied by itself.
Example:
The square of 10 is 10 x 10 = 100
The square of 0.5 is 0.5 x 0.5 = 0.25
The square of 1/4 is 1/4 x 1/4 = 1/16
The square of a number can be written using the square notation.
Example:
6 x 6 can be written as 6².
0.3 x 0.3 can be written as 0.3².
(-8) x (-8) can be written as (-8)².
*Remember: The square of any numbers always is positive!
such as (-2)² = -2 x -2 = 4; still remember, (-) x (-) = (+)
2.2 Square Roots
1. The symbol for square root is √¯¯.
Example 1:
What is the square root of 36?
Solution:
The square root of 36 can be written as √36. (We know 36 is the product of 6 x 6)
So, the answer for this question is 6!
Example 2:
What is the square root of 9/16?
We can write this as √(9/16). (We know that 9/16 is also equal to (3x3/4x4).
So the answer for this question is 3/4!
* Remember: Square root of any numbers must be positive. If negative, that answer is
undefined!
2.3 Cubes Of Numbers
The cube of a number is the product of the number multiplied by itself twice.
The cube of a number can be written using the cube symbol or notation ³.
Example 1:
The cube of 3 can be written as 3³ = 3 x 3 x 3.
The cube of -2 can be written as (-2) ³ = (-2) x (-2) x (-2).
The cube of 1/5 can be written as (1/5)³ = (1/5) x (1/5) x (1/5).
Example 2:
Calculate the value of the cube of four.
4³ = 4 x 4 x 4 = 64
Calculate the value for 0.5³
0.5³ = 0.5 x 0.5 x 0.5 = 0.125
2.4 Cube Roots
The cube root of a number is a number which, when multiplied itself twice. It can be written as the symbol as ³√¯¯.
Example 1:
The cube root for 125 is 5 because 125 = 5 x 5 x 5
The cube root of -8 is -2 because -8 = (-2) x (-2) x (-2)
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# Annual Percentage Rate
## The APR determines how much you pay monthly on a credit card or loan
The APR is the total interest paid divided by the amount of money borrowed. For example, you borrow 15,000 at an APR of 10 percent for one year. Multiplying 15,000 by 0.1 equals the total interest paid, which is \$1,500. Divide \$1,500 by the original amount of the loan, \$15,000, to get the annual percentage rate of 10 percent. Lenders use other criteria, such as the size and financial stability of the business, to decide whether to set a low or high APR.
Furthermore, the annual percentage rate on a credit card determines the interest you pay for maintaining or transferring a balance, or for taking out a cash advance. Some of these credit cards have an annual percentage rate for purchases, one for balance transfers and a late penalty APR. Income and creditworthiness usually determine the credit card amount. When making decisions about applying for credit or loans to start or boost your business, consider these steps:
1. Research annual percentage rate information to help you decide the right loan or credit card.
2. Get help with annual percentage rate facts and figures.
3. Pick a credit card or loan with the best annual percentage rate and reasonable monthly payment.
### Get annual percentage rate information before making certain financial decisions
Seek information and annual percentage rate advice from publications or organizations that receive annual percentage rate training.
Federal Reserve Board offers annual percentage rate information and how it applies when determining balances versus minimum payments. Learn the difference between annual percentage rate and annual percentage yield at Investopedia.
### Contact an annual percentage rate consultant for help on calculating the numbers
Understand how annual percentage rate calculations work, and receive tips on eliminating debt and managing credit. Certain fees may apply to receive annual percentage rate advice or services.
### Choose a loan or credit card with a good APR
Make sure you understand whether the annual percentage rate is fixed, which means that it stays the same, or variable, which means the APR can increase or decrease. A fixed, low APR may be more manageable for your budget.
• Certain fees apply to the annual percentage rate, particularly for mortgage or commercial loans. These include application fees, points, closing costs, loan origination fees and insurance.
• As required by law, credit card companies and lenders must state the annual percentage rate in writing so that you know what to expect when you apply for a loan or a line of credit in terms of repayment.
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# A particle is performing SHM along x axis with amplitude 4cm and time period 1.2seconds.The minimum time taken by particle to move from x=2cm to=4cm and back again is given by
Khimraj
3007 Points
6 years ago
l time period T = 1.2 sec
Time taken by to particle to move from x=2cm to x=4cm will be T/6 and to come back will aslo be T/6.
So total time taken will be equal t0 T/6 + T/6 = T/3 = 0.4 sec.
Hope it clears.
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# Chapter 5 Logistic Regression
Linear regression is used to approximate the (linear) relationship between a continuous response variable and a set of predictor variables. However, when the response variable is binary (i.e., Yes/No), linear regression is not appropriate. Fortunately, analysts can turn to an analogous method, logistic regression, which is similar to linear regression in many ways. This chapter explores the use of logistic regression for binary response variables. Logistic regression can be expanded for multinomial problems (see Faraway (2016a) for discussion of multinomial logistic regression in R); however, that goes beyond our intent here.
## 5.1 Prerequisites
For this section we’ll use the following packages:
# Helper packages
library(dplyr) # for data wrangling
library(ggplot2) # for awesome plotting
library(rsample) # for data splitting
# Modeling packages
library(caret) # for logistic regression modeling
# Model interpretability packages
library(vip) # variable importance
To illustrate logistic regression concepts we’ll use the employee attrition data, where our intent is to predict the Attrition response variable (coded as "Yes"/"No"). As in the previous chapter, we’ll set aside 30% of our data as a test set to assess our generalizability error.
df <- attrition %>% mutate_if(is.ordered, factor, ordered = FALSE)
# Create training (70%) and test (30%) sets for the
# rsample::attrition data.
set.seed(123) # for reproducibility
churn_split <- initial_split(df, prop = .7, strata = "Attrition")
churn_train <- training(churn_split)
churn_test <- testing(churn_split)
## 5.2 Why logistic regression
To provide a clear motivation for logistic regression, assume we have credit card default data for customers and we want to understand if the current credit card balance of a customer is an indicator of whether or not they’ll default on their credit card. To classify a customer as a high- vs. low-risk defaulter based on their balance we could use linear regression; however, the left plot in Figure 5.1 illustrates how linear regression would predict the probability of defaulting. Unfortunately, for balances close to zero we predict a negative probability of defaulting; if we were to predict for very large balances, we would get values bigger than 1. These predictions are not sensible, since of course the true probability of defaulting, regardless of credit card balance, must fall between 0 and 1. These inconsistencies only increase as our data become more imbalanced and the number of outliers increase. Contrast this with the logistic regression line (right plot) that is nonlinear (sigmoidal-shaped).
To avoid the inadequacies of the linear model fit on a binary response, we must model the probability of our response using a function that gives outputs between 0 and 1 for all values of $$X$$. Many functions meet this description. In logistic regression, we use the logistic function, which is defined in Equation (5.1) and produces the S-shaped curve in the right plot above.
$$$\tag{5.1} p\left(X\right) = \frac{e^{\beta_0 + \beta_1X}}{1 + e^{\beta_0 + \beta_1X}}$$$
The $$\beta_i$$ parameters represent the coefficients as in linear regression and $$p\left(X\right)$$ may be interpreted as the probability that the positive class (default in the above example) is present. The minimum for $$p\left(x\right)$$ is obtained at $$\lim_{a \rightarrow -\infty} \left[ \frac{e^a}{1+e^a} \right] = 0$$, and the maximum for $$p\left(x\right)$$ is obtained at $$\lim_{a \rightarrow \infty} \left[ \frac{e^a}{1+e^a} \right] = 1$$ which restricts the output probabilities to 0–1. Rearranging Equation (5.1) yields the logit transformation (which is where logistic regression gets its name):
$$$\tag{5.2} g\left(X\right) = \ln \left[ \frac{p\left(X\right)}{1 - p\left(X\right)} \right] = \beta_0 + \beta_1 X$$$
Applying a logit transformation to $$p\left(X\right)$$ results in a linear equation similar to the mean response in a simple linear regression model. Using the logit transformation also results in an intuitive interpretation for the magnitude of $$\beta_1$$: the odds (e.g., of defaulting) increase multiplicatively by $$\exp\left(\beta_1\right)$$ for every one-unit increase in $$X$$. A similar interpretation exists if $$X$$ is categorical; see Agresti (2003), Chapter 5, for details.
## 5.3 Simple logistic regression
We will fit two logistic regression models in order to predict the probability of an employee attriting. The first predicts the probability of attrition based on their monthly income (MonthlyIncome) and the second is based on whether or not the employee works overtime (OverTime). The glm() function fits generalized linear models, a class of models that includes both logistic regression and simple linear regression as special cases. The syntax of the glm() function is similar to that of lm(), except that we must pass the argument family = "binomial" in order to tell R to run a logistic regression rather than some other type of generalized linear model (the default is family = "gaussian", which is equivalent to ordinary linear regression assuming normally distributed errors).
model1 <- glm(Attrition ~ MonthlyIncome, family = "binomial", data = churn_train)
model2 <- glm(Attrition ~ OverTime, family = "binomial", data = churn_train)
In the background glm() uses ML estimation to estimate the unknown model parameters. The basic intuition behind using ML estimation to fit a logistic regression model is as follows: we seek estimates for $$\beta_0$$ and $$\beta_1$$ such that the predicted probability $$\widehat p\left(X_i\right)$$ of attrition for each employee corresponds as closely as possible to the employee’s observed attrition status. In other words, we try to find $$\widehat \beta_0$$ and $$\widehat \beta_1$$ such that plugging these estimates into the model for $$p\left(X\right)$$ (Equation (5.1)) yields a number close to one for all employees who attrited, and a number close to zero for all employees who did not. This intuition can be formalized using a mathematical equation called a likelihood function:
$$$\tag{5.3} \ell\left(\beta_0, \beta_1\right) = \prod_{i:y_i=1}p\left(X_i\right) \prod_{i':y_i'=0}\left[1-p\left(x_i'\right)\right]$$$
The estimates $$\widehat \beta_0$$ and $$\widehat \beta_1$$ are chosen to maximize this likelihood function. What results is the predicted probability of attrition. Figure 5.2 illustrates the predicted probabilities for the two models.
The table below shows the coefficient estimates and related information that result from fitting a logistic regression model in order to predict the probability of Attrition = Yes for our two models. Bear in mind that the coefficient estimates from logistic regression characterize the relationship between the predictor and response variable on a log-odds (i.e., logit) scale.
For model1, the estimated coefficient for MonthlyIncome is $$\widehat \beta_1 =$$ -0.000130, which is negative, indicating that an increase in MonthlyIncome is associated with a decrease in the probability of attrition. Similarly, for model2, employees who work OverTime are associated with an increased probability of attrition compared to those that do not work OverTime.
tidy(model1)
## # A tibble: 2 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -0.924 0.155 -5.96 0.00000000259
## 2 MonthlyIncome -0.000130 0.0000264 -4.93 0.000000836
tidy(model2)
## # A tibble: 2 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -2.18 0.122 -17.9 6.76e-72
## 2 OverTimeYes 1.41 0.176 8.00 1.20e-15
As discussed earlier, it is easier to interpret the coefficients using an $$\exp()$$ transformation:
exp(coef(model1))
## (Intercept) MonthlyIncome
## 0.3970771 0.9998697
exp(coef(model2))
## (Intercept) OverTimeYes
## 0.1126126 4.0812121
Thus, the odds of an employee attriting in model1 increase multiplicatively by 0.9999 for every one dollar increase in MonthlyIncome, whereas the odds of attriting in model2 increase multiplicatively by 4.0812 for employees that work OverTime compared to those that do not.
Many aspects of the logistic regression output are similar to those discussed for linear regression. For example, we can use the estimated standard errors to get confidence intervals as we did for linear regression in Chapter 4:
confint(model1) # for odds, you can use exp(confint(model1))
## 2.5 % 97.5 %
## (Intercept) -1.2267754960 -6.180062e-01
## MonthlyIncome -0.0001849796 -8.107634e-05
confint(model2)
## 2.5 % 97.5 %
## (Intercept) -2.430458 -1.952330
## OverTimeYes 1.063246 1.752879
## 5.4 Multiple logistic regression
We can also extend our model as seen in Equation 1 so that we can predict a binary response using multiple predictors:
$$$\tag{5.4} p\left(X\right) = \frac{e^{\beta_0 + \beta_1 X + \cdots + \beta_p X_p }}{1 + e^{\beta_0 + \beta_1 X + \cdots + \beta_p X_p}}$$$
Let’s go ahead and fit a model that predicts the probability of Attrition based on the MonthlyIncome and OverTime. Our results show that both features are statistically significant (at the 0.05 level) and Figure 5.3 illustrates common trends between MonthlyIncome and Attrition; however, working OverTime tends to nearly double the probability of attrition.
model3 <- glm(
Attrition ~ MonthlyIncome + OverTime,
family = "binomial",
data = churn_train
)
tidy(model3)
## # A tibble: 3 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -1.43 0.176 -8.11 5.25e-16
## 2 MonthlyIncome -0.000139 0.0000270 -5.15 2.62e- 7
## 3 OverTimeYes 1.47 0.180 8.16 3.43e-16
## 5.5 Assessing model accuracy
With a basic understanding of logistic regression under our belt, similar to linear regression our concern now shifts to how well do our models predict. As in the last chapter, we’ll use caret::train() and fit three 10-fold cross validated logistic regression models. Extracting the accuracy measures (in this case, classification accuracy), we see that both cv_model1 and cv_model2 had an average accuracy of 83.88%. However, cv_model3 which used all predictor variables in our data achieved an average accuracy rate of 87.58%.
set.seed(123)
cv_model1 <- train(
Attrition ~ MonthlyIncome,
data = churn_train,
method = "glm",
family = "binomial",
trControl = trainControl(method = "cv", number = 10)
)
set.seed(123)
cv_model2 <- train(
Attrition ~ MonthlyIncome + OverTime,
data = churn_train,
method = "glm",
family = "binomial",
trControl = trainControl(method = "cv", number = 10)
)
set.seed(123)
cv_model3 <- train(
Attrition ~ .,
data = churn_train,
method = "glm",
family = "binomial",
trControl = trainControl(method = "cv", number = 10)
)
# extract out of sample performance measures
summary(
resamples(
list(
model1 = cv_model1,
model2 = cv_model2,
model3 = cv_model3
)
)
)$statistics$Accuracy
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## model1 0.8349515 0.8349515 0.8365385 0.8388478 0.8431373 0.8446602 0
## model2 0.8349515 0.8349515 0.8365385 0.8388478 0.8431373 0.8446602 0
## model3 0.8365385 0.8495146 0.8792476 0.8757893 0.8907767 0.9313725 0
We can get a better understanding of our model’s performance by assessing the confusion matrix (see Section 2.6). We can use caret::confusionMatrix() to compute a confusion matrix. We need to supply our model’s predicted class and the actuals from our training data. The confusion matrix provides a wealth of information. Particularly, we can see that although we do well predicting cases of non-attrition (note the high specificity), our model does particularly poor predicting actual cases of attrition (note the low sensitivity).
By default the predict() function predicts the response class for a caret model; however, you can change the type argument to predict the probabilities (see ?caret::predict.train).
# predict class
pred_class <- predict(cv_model3, churn_train)
# create confusion matrix
confusionMatrix(
data = relevel(pred_class, ref = "Yes"),
reference = relevel(churn_train$Attrition, ref = "Yes") ) ## Confusion Matrix and Statistics ## ## Reference ## Prediction Yes No ## Yes 93 25 ## No 73 839 ## ## Accuracy : 0.9049 ## 95% CI : (0.8853, 0.9221) ## No Information Rate : 0.8388 ## P-Value [Acc > NIR] : 5.360e-10 ## ## Kappa : 0.6016 ## ## Mcnemar's Test P-Value : 2.057e-06 ## ## Sensitivity : 0.56024 ## Specificity : 0.97106 ## Pos Pred Value : 0.78814 ## Neg Pred Value : 0.91996 ## Prevalence : 0.16117 ## Detection Rate : 0.09029 ## Detection Prevalence : 0.11456 ## Balanced Accuracy : 0.76565 ## ## 'Positive' Class : Yes ## One thing to point out, in the confusion matrix above you will note the metric No Information Rate: 0.839. This represents the ratio of non-attrition vs. attrition in our training data (table(churn_train$Attrition) %>% prop.table()). Consequently, if we simply predicted "No" for every employee we would still get an accuracy rate of 83.9%. Therefore, our goal is to maximize our accuracy rate over and above this no information baseline while also trying to balance sensitivity and specificity. To that end, we plot the ROC curve (section 2.6) which is displayed in Figure 5.4. If we compare our simple model (cv_model1) to our full model (cv_model3), we see the lift achieved with the more accurate model.
library(ROCR)
# Compute predicted probabilities
m1_prob <- predict(cv_model1, churn_train, type = "prob")$Yes m3_prob <- predict(cv_model3, churn_train, type = "prob")$Yes
# Compute AUC metrics for cv_model1 and cv_model3
perf1 <- prediction(m1_prob, churn_train$Attrition) %>% performance(measure = "tpr", x.measure = "fpr") perf2 <- prediction(m3_prob, churn_train$Attrition) %>%
performance(measure = "tpr", x.measure = "fpr")
# Plot ROC curves for cv_model1 and cv_model3
plot(perf1, col = "black", lty = 2)
plot(perf2, add = TRUE, col = "blue")
legend(0.8, 0.2, legend = c("cv_model1", "cv_model3"),
col = c("black", "blue"), lty = 2:1, cex = 0.6)
Similar to linear regression, we can perform a PLS logistic regression to assess if reducing the dimension of our numeric predictors helps to improve accuracy. There are 16 numeric features in our data set so the following code performs a 10-fold cross-validated PLS model while tuning the number of principal components to use from 1–16. The optimal model uses 14 principal components, which is not reducing the dimension by much. However, the mean accuracy of 0.876 is no better than the average CV accuracy of cv_model3 (0.876).
PLS was originally designed to be used for continuous features. Although you are not restricted from using PLS on categorical features, it is commonly advised to start with numeric features and explore alternative options for categorical features (i.e. ordinal encode, label encode, factor analysis. Also, see Russolillo and Lauro (2011) for an alternative PLS approach for categorical features).
# Perform 10-fold CV on a PLS model tuning the number of PCs to
# use as predictors
set.seed(123)
cv_model_pls <- train(
Attrition ~ .,
data = churn_train,
method = "pls",
family = "binomial",
trControl = trainControl(method = "cv", number = 10),
preProcess = c("zv", "center", "scale"),
tuneLength = 16
)
# Model with lowest RMSE
cv_model_pls$bestTune ## ncomp ## 14 14 # results for model with lowest loss cv_model_pls$results %>%
dplyr::filter(ncomp == pull(cv_model_pls\$bestTune))
## ncomp Accuracy Kappa AccuracySD KappaSD
## 1 14 0.8757518 0.3766944 0.01919777 0.1142592
# Plot cross-validated RMSE
ggplot(cv_model_pls)
## 5.6 Model concerns
As with linear models, it is important to check the adequacy of the logistic regression model (in fact, this should be done for all parametric models). This was discussed for linear models in Section 4.5 where the residuals played an important role. Although not as common, residual analysis and diagnostics are equally important to generalized linear models. The problem is that there is no obvious way to define what a residual is for more general models. For instance, how might we define a residual in logistic regression when the outcome is either 0 or 1? Nonetheless attempts have been made and a number of useful diagnostics can be constructed based on the idea of a pseudo residual; see, for example, Harrell (2015), Section 10.4.
More recently, Liu and Zhang (2018) introduced the concept of surrogate residuals that allows for residual-based diagnostic procedures and plots not unlike those in traditional linear regression (e.g., checking for outliers and misspecified link functions). For an overview with examples in R using the sure package, see Brandon M. Greenwell et al. (2018).
## 5.7 Feature interpretation
Similar to linear regression, once our preferred logistic regression model is identified, we need to interpret how the features are influencing the results. As with normal linear regression models, variable importance for logistic regression models can be computed using the absolute value of the $$z$$-statistic for each coefficient (albeit with the same issues previously discussed). Using vip::vip() we can extract our top 20 influential variables. Figure 5.6 illustrates that OverTime is the most influential followed by JobSatisfaction, and EnvironmentSatisfaction.
vip(cv_model3, num_features = 20)
Similar to linear regression, logistic regression assumes a monotonic linear relationship. However, the linear relationship occurs on the logit scale; on the probability scale, the relationship will be nonlinear. This is illustrated by the PDP in Figure 5.7 which illustrates the functional relationship between the predicted probability of attrition and the number of companies an employee has worked for (NumCompaniesWorked) while taking into account the average effect of all the other predictors in the model. Employees who’ve experienced more employment changes tend to have a high probability of making another change in the future.
Furthermore, the PDPs for the top three categorical predictors (OverTime, JobSatisfaction, and EnvironmentSatisfaction) illustrate the change in predicted probability of attrition based on the employee’s status for each predictor.
See the online supplemental material for the code to reproduce the plots in Figure 5.7.
## 5.8 Final thoughts
Logistic regression provides an alternative to linear regression for binary classification problems. However, similar to linear regression, logistic regression suffers from the many assumptions involved in the algorithm (i.e. linear relationship of the coefficient, multicollinearity). Moreover, often we have more than two classes to predict which is commonly referred to as multinomial classification. Although multinomial extensions of logistic regression exist, the assumptions made only increase and, often, the stability of the coefficient estimates (and therefore the accuracy) decrease. Future chapters will discuss more advanced algorithms that provide a more natural and trustworthy approach to binary and multinomial classification prediction.
### References
Agresti, Alan. 2003. Categorical Data Analysis. Wiley Series in Probability and Statistics. Wiley.
Faraway, Julian J. 2016a. Extending the Linear Model with R: Generalized Linear, Mixed Effects and Nonparametric Regression Models. Vol. 124. CRC press.
Greenwell, Brandon M., Andrew J. McCarthy, Bradley C. Boehmke, and Dungang Lui. 2018. “Residuals and Diagnostics for Binary and Ordinal Regression Models: An Introduction to the Sure Package.” The R Journal 10 (1): 1–14. https://journal.r-project.org/archive/2018/RJ-2018-004/index.html.
Harrell, Frank E. 2015. Regression Modeling Strategies: With Applications to Linear Models, Logistic and Ordinal Regression, and Survival Analysis. Springer Series in Statistics. Springer International Publishing. https://books.google.com/books?id=94RgCgAAQBAJ.
Liu, Dungang, and Heping Zhang. 2018. “Residuals and Diagnostics for Ordinal Regression Models: A Surrogate Approach.” Journal of the American Statistical Association 113 (522). Taylor & Francis: 845–54. https://doi.org/10.1080/01621459.2017.1292915.
Russolillo, Giorgio, and Carlo Natale Lauro. 2011. “A Proposal for Handling Categorical Predictors in Pls Regression Framework.” In Classification and Multivariate Analysis for Complex Data Structures, 343–50. Springer.
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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
Since the impulse is equal to the area under the graph, the force exerts an impulse of $4~N~s$ on the particle.
The impulse is equal to the area under the force versus time graph. We can find the area under the graph. $Area = (1000~N)(2\times 10^{-3}~s)+\frac{1}{2}(1000~N)(4\times 10^{-3}~s)$ $Area = 4~N~s$ Since the impulse is equal to the area under the graph, the force exerts an impulse of $4~N~s$ on the particle.
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# Search by Topic
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I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Ewa's Eggs
##### Stage: 3 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### Digat
##### Stage: 3 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Gaxinta
##### Stage: 3 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Divisively So
##### Stage: 3 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Remainder
##### Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Square Routes
##### Stage: 3 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Digital Roots
##### Stage: 2 and 3
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
### Flow Chart
##### Stage: 3 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
### Just Repeat
##### Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Skeleton
##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### What an Odd Fact(or)
##### Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Expenses
##### Stage: 4 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Differences
##### Stage: 3 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### Dozens
##### Stage: 3 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Multiplication Magic
##### Stage: 4 Challenge Level:
Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Take Three from Five
##### Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### LCM Sudoku
##### Stage: 4 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### Big Powers
##### Stage: 3 and 4 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
### Knapsack
##### Stage: 4 Challenge Level:
You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value.
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Three Times Seven
##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Fac-finding
##### Stage: 4 Challenge Level:
Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.
### Obviously?
##### Stage: 4 and 5 Challenge Level:
Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
### 396
##### Stage: 4 Challenge Level:
The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396.
### American Billions
##### Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### Adding in Rows
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Squaresearch
##### Stage: 4 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
### Divisibility Tests
##### Stage: 3 and 4
This article takes the reader through divisibility tests and how they work. An article to read with pencil and paper to hand.
### Mod 3
##### Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Transposition Fix
##### Stage: 4 Challenge Level:
Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . .
### Check Codes
##### Stage: 4 Challenge Level:
Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . .
### Novemberish
##### Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
### The Chinese Remainder Theorem
##### Stage: 4 and 5
In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5."
### Odd Stones
##### Stage: 4 Challenge Level:
On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.
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https://chestofbooks.com/crafts/metal/Sheet-And-Plate-Metal-Work/To-Obtain-Pattern-Lines-By-Calculation.html
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It is useful to know that the lengths of lines required in setting out the pattern for any conical article may be obtained by calculation, without previously having drawn an elevation of the object. An example of this method is shown by the calculations for the pattern in Fig. 89. Suppose the article to have the following dimensions: Top 9 in. diameter, bottom 2 ft. diameter, and depth 1 ft. 6 in.; then the height of complete cone, of which the vessel is a part, can be calculated as follows: -
Height = C D = radius of base x depth / radius of base - radius of top
=12 X 18 / 7.5 = 28.8 in.
And slant height of cone - that is, radius of outer curve of pattern can be found by the use of the property of the right-angle triangle, thus-
C B = = 31.2 in.
The radius for the inner curve of pattern can be found as above, or from the following rule: -
C A = radius of top x slant height of complete cone / radius of base
= 4.5 X 31.2 / 12 = 11.7 in.
The length round outer curve of pattern will, of course, be-
24 x 3.1416 = 75.4 in., so that it will be seen that the complete pattern of the article, whether it is made up in one or more pieces, can be set out from calculated dimensions. This method is especially useful in large work, or where a high degree of accuracy is required.
Capacity of a Conical Vessel.
Seeing that we have the above calculations before us, it will be as well to go over what is usually considered to be a somewhat difficult task - that is, to find the capacity of a circular tapering vessel. The simplest plan to adopt is to calculate the volume of the complete cone, and then to deduct from it the volume of the small cone, which we can imagine is cut off the top.
The rule for finding the volume of a cone is:- "Multiply the area of the base by one-third the vertical height." From the previous calculations in reference to Fig. 89, we have both the height of complete cone and of the small cones cut away. The volume of the vessel then is-
Fig. 89.
(12)2 ╥ X 28.8 / 3 - (4.5)2╥ X 10.8 / 3
= 144 ╥ x 28.8 / 3 - 20.25 ╥ x 10.8/ 3
= 1,309 ╥ = 1,309 x 3.1416 = 4, 114 c.in.
The number of cubic inches in an imperial gallon is 277.274. This is a most awkward number to use; but as it has been fixed by Act of Parliament (in 1826) as the volume of a gallon, we have to make the best of it. The capacity of the above vessel will therefore be -
4,114 / 277.274 = 14.84 gal.
The capacity may also be calculated, but not quite so accurately, by remembering that a cubic foot contains as near as possible 6¼ gal. Thus -
4114 / 1,728 x 6¼ = 14 7/8 gal. (nearly).
As a gallon of fresh water is usually taken to weigh 10 lb., the weight of water in the vessel will be -
14.84 x 10= 148.4 = 148½ lb. (nearly).
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# Correlators
### 1 Correlation vs. Convolution
Correlation is very similar to convolution, and it is best defined through its equivalent “correlation theorem”:
${\displaystyle f(t)\star g(t)=\int {{\hat {f}}(\omega ){\hat {g}}^{*}(\omega )e^{i\omega t}d\omega }.\,\!}$
The difference between correlation and convolution is that that when correlating two signals, the Fourier transform of the second function (${\displaystyle {\hat {g}}(\omega )}$ in equation None) is conjugated before multiplying and integrating. Using that
{\displaystyle {\begin{aligned}g^{*}(-t)&={\frac {1}{2\pi }}\int {{\hat {g}}^{*}(\omega )e^{-i\omega (-t)}d\omega }\\&={\frac {1}{2\pi }}\int {{\hat {g}}^{*}(\omega )e^{i\omega t}d\omega },\end{aligned}}\,\!}
we can show that correlating ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ is equivalent to convolving ${\displaystyle f(t)}$ with a conjugated, time-reversed version of ${\displaystyle g(t)}$:
${\displaystyle f(t)*g^{*}(-t)=f(t)\star g(t).\,\!}$
Although this relation between convolution and correlation is often mentioned in the literature, I don’t personally find it very intuitively illuminating. I much prefer the “correlation theorem” in equation (None), because when it is combined with the expression of a time-shifted signal in Fourier domain:
{\displaystyle {\begin{aligned}f(t-\tau )&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega )e^{i\omega (t-\tau )}d\omega }\\&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega )e^{i\omega t}e^{-i\omega \tau }d\omega },\end{aligned}}\,\!}
it shows that correlating a flat-spectrum signal with a time-shifted version of itself yields a measure of the power of the signal at the delay corresponding to the time shift:
{\displaystyle {\begin{aligned}f(t)\star f(t-\tau )&={\frac {1}{2\pi }}\int {{\hat {f}}(\omega ){\hat {f}}^{*}(\omega )e^{-i\omega \tau }e^{i\omega t}d\omega }\\&={\frac {1}{2\pi }}\int {|f|^{2}e^{i\omega (t-\tau )}d\omega }\\&=|f|^{2}\cdot \delta (t-\tau ).\end{aligned}}\,\!}
## 1 Correlators
There are a variety of architectures for building correlators, but perhaps the most straight-forward conceptually are FX correlators, where input signals are first Fourier-transformed (F) and then cross-multiplied (X), with one signal from each pairing being conjugated. This architecture is essentially a direct implementation of the right-hand side of the “correlation theorem”. Moreover, since scientists are often interested in the power spectrum of the correlated signals, FX correlators usually do not implement the inverse Fourier transform back to delay domain. Instead, they produce the correlation ${\displaystyle {\mathcal {C}}_{ij}}$ between inputs ${\displaystyle i}$ and ${\displaystyle j}$ as a function of frequency ${\displaystyle \nu }$:
${\displaystyle {\mathcal {C}}_{ij}(\nu )={\mathcal {F}}(f_{i}\star f_{j})={\mathcal {F}}(f_{i})\cdot {\mathcal {F}}(f_{j})\,\!}$
Suppose we have two signals from two antennas ${\displaystyle i,j}$, each consisting of an uncorrelated noise component and a correlated signal that enters each antenna feed at different time:
{\displaystyle {\begin{aligned}f_{i}(t)&\equiv x(t)+n_{i}(t)\\f_{j}(t)&\equiv x(t-\tau _{ij})+n_{j}(t).\end{aligned}}\,\!}
We can then express the time-averaged correlation function ${\displaystyle \langle {\mathcal {C}}_{ij}\rangle }$ as
{\displaystyle {\begin{aligned}\langle {\mathcal {C}}_{ij}(\nu )\rangle &=\langle {\mathcal {F}}(f_{i}(t)\star f_{j}(t))\rangle =\left\langle {\hat {f}}_{i}(\nu ){\hat {f}}_{j}^{*}(\nu )\right\rangle \\&=\left\langle \left[{\hat {x}}(\nu ){\hat {x}}^{*}(\nu )e^{-2\pi i\nu \tau _{ij}}+{\hat {x}}(\nu ){\hat {n}}_{j}^{*}(\omega )+{\hat {n}}_{i}(\nu ){\hat {x}}^{*}(\nu )e^{-2\pi i\nu \tau _{ij}}+{\hat {n}}_{i}(\nu ){\hat {n}}_{j}^{*}(\nu )\right]\right\rangle \\&=\langle |x(\nu )|^{2}\rangle \cdot e^{-2\pi i\nu \tau _{ij}}.\end{aligned}}\,\!}
In the last step above, we use the fact that signals ${\displaystyle x}$, ${\displaystyle n_{i}}$, and ${\displaystyle n_{j}}$ are uncorrelated, and so average to zero. In practice, for a finite integration time, these noise terms do not completely disappear — they merely integrate down as ${\displaystyle t^{-1/2}}$, as noise is wont to do. This is actually an important point about correlators: everything that enters as a signal either 1) correlates, and is present in ${\displaystyle {\mathcal {C}}_{ij}}$, or 2) does not correlate, and is a source of noise to be integrated down.
Through the correlation process, the correlated component of ${\displaystyle f}$ and ${\displaystyle g}$ (namely ${\displaystyle x}$) is coherently accumulated with a sinusoidally varying phase that corresponds to a delay-bin where ${\displaystyle \tau =\tau _{ij}}$. In that bin, the convolution that we compute gives us a measurement of the power of the correlated signal at the delay corresponding to the time interval between when that signal enters one antenna and when it enters the other. All of the other uncorrelated terms in ${\displaystyle f}$ and ${\displaystyle g}$, when cross-correlated, integrate to zero.
## 2 How to Build a Correlator
The job of a correlator is typically to cross-correlate each pair of antennas in an array. For ${\displaystyle N}$ antennas, there are ${\displaystyle N(N+1)/2}$ cross-products, meaning that correlators have the nasty property of generating more data out than goes in. Mathematically, correlators are quite simple. The difficulty with building correlators, and why they are often one of the most expensive components in a synthesis telescope, is in managing the flow of large amounts of data in real time, and providing the computing necessary to perform the cross-correlation itself.
For radio astronomy applications, there are basically two ways that correlators are built. Both ways involve a Fourier transform stage (often called the “F" stage) and a cross-correlation stage (often called the “X" stage). The two architectures differ in the ordering of these stages.
### 2.1 The XF Architecture, a.k.a. the “Lag" Correlator
An example of an XF correlator architecture, from Fig. 9.5 in GMRT’s documentation (Roshi 1997).
In this correlator architecture, signals from different antennas are first cross-correlated. This is typically done with what is called a “lag" correlation circuit. This circuit looks a lot like an FIR filter, which is to say, it implements a by-the-book correlation. Following the equation for correlating signals ${\displaystyle f}$ and ${\displaystyle g}$:
${\displaystyle f\star g\equiv \int {f(t)\cdot g^{*}(t-\tau )~dt},\,\!}$
one signal (${\displaystyle g}$ above) is conjugated and delayed relative to the other by a number of samples, ${\displaystyle \tau }$, and then the two signals ${\displaystyle f}$ and ${\displaystyle g^{*}(t-\tau )}$ are cross-multiplied and integrated for some amount of time. Different circuits compute the integral for different time lags between ${\displaystyle f}$ and ${\displaystyle g}$, producing the value of ${\displaystyle (f\star g)(\tau )}$ for different values of ${\displaystyle \tau }$. This is the ‘X’ stage.
However, most radio astronomers would like to have their observations as a function of frequency, so in an XF correlator, the correlation product ${\displaystyle f\star g}$ is then Fourier transformed into frequency domain. Hence the ‘F’ stage comes after the X.
The problem with XF correlators is that, for ${\displaystyle N}$ antennas being correlated, there are ${\displaystyle N^{2}/}$ Fourier transforms that must be implemented. Moreover, the fact that each “lag” must be computed independently, but still requires all the data to do so, is inefficient. However, for smaller numbers of antennas, XF correlators are still sometimes used to good effect.
### 2.2 The FX Architecture
An example of an FX correlator architecture, from Parsons et al. (2008).
In this correlator architecture, the ‘F’ Stage (Fourier transform) is applied first. Hence, for ${\displaystyle N}$ antennas, only ${\displaystyle N}$ FFTs are required. Next comes the ‘X’ Stage, which in this case is a simple cross-multiplication frequency by frequency. This is because the FX correlator architecture is essentially implementing a correlation according to the “correlation theorem" above, where, after the Fourier transform, we have that
${\displaystyle {\hat {f\star g}}(\omega )={\hat {f}}(\omega ){\hat {g}}^{*}(\omega ).\,\!}$
Notice that, because we wanted our output in frequency domain, we didn’t have to bother with the Fourier transform back to time domain at the end of our correlation theorem equation. So after Fourier transforming each antenna signal to frequency domain, we only need to multiply the signal from an antenna at a given frequency with other antenna samples at that same frequency. This is both efficient, and easier to parallelize!
Almost all large correlators follow the FX architecture these days. For smaller correlators with narrow bandwidths, CPUs may be used to do both the F and X stages. For larger correlators with higher bandwidths, FPGA processors are typically used to do the F stage, and the X stage is implemented either with FPGAs or GPU processors.
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# Convert returned values to a specific type
I'm just getting started with go and I've come across the following "problem".
I'm using a function which returns 4 unsigned 32 bit integers, and I was wondering what the best method to convert these values to 8 bit integers when I'm assigning them. (Or if it is possible). I have got it to work by assigning them to uint32, then converting them after that but I was wondering if there is a more concise way of doing it.
Example:
``````r, g, b, _ := image.At(x, y).RGBA() // ideally convert them at this stage
``````
So Image.At returns uint32, but I would like r,g,b to be uint8
Thanks
-
The `RGBA` function returns the integers in `uint32`. What is the problem with manually converting them to `uint8`? Also, what is wrong with `uint32`? – Sverri M. Olsen May 15 at 1:44
Go has no syntax for what you want there. You're best bet is to just do it in the next line down.
``````rbig, gbig, bbig := image.At(x, y).RGBA()
r, g, b := uint8(rbig), uint8(gbig), uint8(bbig)
``````
If that really annoys you though then just make a helper function
``````func convert(i, j, k, _ uint32) (uint8, uint8, uint8) { return uint8(i), uint8(j), uint8(k) }
r, g, b := convert(image.At(x, y).RGBA())
``````
Still a minimum of two lines but you can reuse the function anywhere you need it and the call site might look a little cleaner to you.
-
That's great. Thanks for the advice. I'll probably go the function route. – ggordan May 15 at 2:25
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`∀[p,F,G,H:Top].`
` (case let a,b = p in F[a;b] of inl(u) => G[u] | inr(v) => H[v] ~ let a,b = p `
` in case F[a;b] of inl(u) => G[u] | inr(v) => H[v])`
Proof
Definitions occuring in Statement : uall: `∀[x:A]. B[x]` top: `Top` so_apply: `x[s1;s2]` so_apply: `x[s]` spread: spread def decide: `case b of inl(x) => s[x] | inr(y) => t[y]` sqequal: `s ~ t`
Definitions unfolded in proof : uall: `∀[x:A]. B[x]` top: `Top` so_lambda: `so_lambda(x,y,z,w.t[x; y; z; w])` member: `t ∈ T` so_apply: `x[s1;s2;s3;s4]` so_lambda: `λ2x.t[x]` so_apply: `x[s]` uimplies: `b supposing a` strict4: `strict4(F)` and: `P ∧ Q` all: `∀x:A. B[x]` implies: `P `` Q` has-value: `(a)↓` prop: `ℙ` guard: `{T}` or: `P ∨ Q` squash: `↓T` so_lambda: `λ2x y.t[x; y]` so_apply: `x[s1;s2]`
Lemmas referenced : lifting-strict-spread top_wf equal_wf has-value_wf_base base_wf is-exception_wf
Rules used in proof : sqequalSubstitution sqequalRule sqequalTransitivity computationStep sqequalReflexivity cut introduction extract_by_obid sqequalHypSubstitution isectElimination thin baseClosed isect_memberEquality voidElimination voidEquality independent_isectElimination independent_pairFormation lambdaFormation callbyvalueDecide hypothesis hypothesisEquality equalityTransitivity equalitySymmetry unionEquality unionElimination sqleReflexivity dependent_functionElimination independent_functionElimination baseApply closedConclusion decideExceptionCases inrFormation because_Cache imageMemberEquality imageElimination exceptionSqequal inlFormation isect_memberFormation sqequalAxiom isectEquality
Latex:
\mforall{}[p,F,G,H:Top].
(case let a,b = p in F[a;b] of inl(u) => G[u] | inr(v) => H[v] \msim{} let a,b = p
in case F[a;b]
of inl(u) =>
G[u]
| inr(v) =>
H[v])
Date html generated: 2017_10_01-AM-08_39_26
Last ObjectModification: 2017_07_26-PM-04_27_29
Theory : untyped!computation
Home Index
| 702
| 1,960
|
{"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-2022-33
|
latest
|
en
| 0.227515
|
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