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https://www.jiskha.com/display.cgi?id=1348163200 | 1,503,469,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00193.warc.gz | 923,461,408 | 3,925 | # Physics
posted by .
A solid of density 5000 kg/m^3 weighs 0.5 kgf in air.It is completely immersed in water of density 1000 kg/m^3. Calculate the apparent weight of the solid in water.
• Physics -
volume=.5kg/5000kg/m3= 100ml
mwass water displaced=100g
apparent mass=origianal-displaced=400g
• Physics -
We have,
Density of solid, d = 5000 kg/m3
Weight in air, W = 0.5 kgf = 0.5g N
Density of water, q = 1000 kg/m3
Let, V be the volume of the solid.
So, W = Vdg
=> 0.5g = V × 5000 × g
=> V = 10-4 m-3
So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf
Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf
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https://practicepaper.in/gate-me/thermodynamics | 1,638,360,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00585.warc.gz | 538,226,586 | 24,862 | # Thermodynamics
Question 1
An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at 500 kPa and 300 K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, and 2 kg/s of cold air stream is leaving the device from outlet 3 at 100 kPa and 240 K.
Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).
A 1.2 B 4.3 C 2.2 D 3.8
GATE ME 2021 SET-2 Second Law, Carnot Cycle and Entropy
Question 1 Explanation:
\begin{aligned} \left(\frac{d S}{d t}\right)_{C . V} &=\dot{S}_{i}+\dot{S}_{g e n}-\dot{S}_{\theta} \\ \dot{S}_{\text {gen }} &=\dot{S}_{e}-\dot{S}_{i} \\ &=\dot{m}_{2} s_{2}+\dot{m}_{3} s_{3}-\dot{m}_{1} s_{1} \\ &=3\left(s_{2}-s_{1}\right)+2\left(s_{3}-s_{1}\right) \\ &=3 \times 0.587+2(0.237) \\ &=2.235 \mathrm{~kW} / \mathrm{K} \simeq 2.2 \mathrm{~kW} / \mathrm{K} \end{aligned}
Question 2
Consider the open feed water heater (FWH) shown in the figure given below:
Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be kg/s (round off to one decimal place).
A 25.2 B 45.6 C 62.3 D 18.4
GATE ME 2021 SET-2 Power System
Question 2 Explanation:
\begin{aligned} 100 h_{5}+(x-100) h_{2} &=x h_{6} \\ 100 \times 226.7+(x-100) 2624&=708.6 x \\ 22670+2624 x-262400 &=708.6 x \\ 2624 x-708.6 x &=239730 \\ 1915.4 x &=239730 \\ x &=125.159 \simeq 125.2 \mathrm{~kg} / \mathrm{s} \end{aligned}
Mass flow rate at state $2(x-100)=25.2 \mathrm{~kg} / \mathrm{s}$
Question 3
Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s, temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant $c_p=1.005$ kJ/kg.K. The stagnation temperature at this point is ______K (round off to two decimal places).
A 374.71 B 352.24 C 874.65 D 458.32
GATE ME 2021 SET-2 Thermodynamic System and Processes
Question 3 Explanation:
\begin{aligned} M_{1}&=\frac{V_{1}}{\sqrt{\gamma R T_{1}}}=\frac{300}{\sqrt{1.4 \times 287 \times 330}}=0.823 \\ \frac{T_{o1}}{T_{1}}&=1+\frac{\gamma-1}{2} M_{1}^{2}=1+\frac{1.4-1}{2}(0.823)^{2} \\ \frac{T_{o1}}{T_{1}}&=1.154 \\ T_{01}&=374.7037 \mathrm{~K} \end{aligned}
Question 4
In the vicinity of the triple point, the equation of liquid-vapour boundary in the $P-T$ phase diagram for ammonia is $\ln P=24.38-3063/T$, where $P$ is pressure (in Pa) and $T$ is temperature (in K). Similarly, the solid-vapour boundary is given by $\ln P=27.92-3754/T$. The temperature at the triple point is ________K (round off to one decimal place).
A 195.2 B 25.6 C 254.6 D 125.5
GATE ME 2021 SET-1 Pure Substances
Question 4 Explanation:
Liquid vapour, $\ln P=24.38-\frac{3063}{T}$
Solid vapour,$\ln P=27.92-\frac{3754}{T}$
At triple point temperature of solid, liquid and vapour is same.
$\therefore$equating
$24.38-\frac{3063}{T}=27.92-\frac{3754}{T}$
Multiplying by T on both sides
\begin{aligned} 24.38 T-3063 &=27.92 T-3754 \\ 3.54 T &=691 \\ T &=195.197 \mathrm{~K} \end{aligned}
Question 5
Consider a steam power plant operating on an ideal reheat Rankine cycle. The work input to the pump is 20 kJ/kg. The work output from the high pressure turbine is 750 kJ/kg. The work output from the low pressure turbine is 1500 kJ/kg. The thermal efficiency of the cycle is 50 %. The enthalpy of saturated liquid and saturated vapour at condenser pressure are 200 kJ/kg and 2600 kJ/kg, respectively. The quality of steam at the exit of the low pressure turbine is ________ % (round off to the nearest integer).
A 45 B 68 C 98 D 93
GATE ME 2021 SET-1 Power System
Question 5 Explanation:
\begin{aligned} h_{f} &=200 \mathrm{~kJ} / \mathrm{kg} \\ h_{g} &=2600 \mathrm{~kJ} / \mathrm{kg} \\ w_{p} &=20 \mathrm{~kJ} / \mathrm{kg}=h_{6}-h_{5} \\ h_{1}-h_{2} &=750 \mathrm{~kJ} / \mathrm{kg} \\ h_{3}-h_{4} &=1500 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=0.5=\frac{W_{\mathrm{NET}}}{Q_{s}}=\frac{W_{T}-W_{P}}{Q_{s}}\\ 0.5 &=\frac{750+1500-20}{Q_{S}} \\ Q_{S} &=4460 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=1-\frac{Q_{R}}{Q_{S}} \\ \frac{Q_{R}}{Q_{S}} &=0.5 \\ Q_{R} &=2230 \mathrm{~kJ} / \mathrm{kg} \\ Q_{R} &=h_{4}-h_{5} \\ 2230 &=h_{4}-200 \\ h_{4} &=2430 \mathrm{~kJ} / \mathrm{kg} \\ h_{4} &=h_{f}+x\left(h_{g}-h_{f}\right) \\ 2430 &=200+x(2600-200) \\ x &=0.9291 \\ x &=93 \% \end{aligned}
Question 6
The fundamental thermodynamic relation for a rubber band is given by $dU=TdS+\tau dL$, where $T$ is the absolute temperature, $S$ is the entropy, $\tau$ is the tension in the rubber band, and $L$ is the length of the rubber band. Which one of the following relations is CORRECT:
A $\tau =\left ( \frac{\partial U}{\partial S}\right )_L$ B $\left ( \frac{\partial T}{\partial L}\right )_S =\left ( \frac{\partial \tau}{\partial S}\right )_L$ C $\left ( \frac{\partial T}{\partial S}\right )_L =\left ( \frac{\partial \tau}{\partial L}\right )_S$ D $T=\left ( \frac{\partial U}{\partial S}\right )_ \tau$
GATE ME 2021 SET-1 Second Law, Carnot Cycle and Entropy
Question 6 Explanation:
\begin{aligned} d U&=T d s+\tau d L \\ \left(\frac{\partial T}{\partial L}\right)_{S}&=\left(\frac{\partial \tau}{\partial S}\right)_{L}\\ \text { Comparing with } M=\left(\frac{\partial z}{\partial x}\right)_{y}& \\ T&=\left(\frac{\partial U}{\partial S}\right)_{L}\\ \text { Comparing with } N=\left(\frac{\partial z}{\partial y}\right)_{x} \\ \tau&=\left(\frac{\partial U}{\partial L}\right)_{S} \\ M&=\left(\frac{\partial z}{\partial x}\right)_{y} \\ N&=\left(\frac{\partial z}{\partial y}\right)_{x} \\ d z&=M d x+N d y \end{aligned}
If z is exact different
$\left(\frac{\partial M}{\partial y}\right)_{x}=\left(\frac{\partial N}{\partial x}\right)_{y}$
Question 7
A rigid insulated tank is initially evacuated. It is connected through a valve to a supply line that carries air at a constant pressure and temperature of 250 kPa and 400 K respectively. Now the valve is opened and air is allowed to flow into the tank until the pressure inside the tank reaches to 250 kPa at which point the valve is closed. Assume that the air behaves as a perfect gas with constant properties $(c_p=1.005\; kJ/kg.K, c_v=0.718\; kJ/kg.K, R=0.287 kJ/kg.K)$. Final temperature of the air inside the tank is _______K (round off to one decimal place).
A 512 B 248 C 688 D 560
GATE ME 2021 SET-1 Thermodynamic System and Processes
Question 7 Explanation:
$\begin{array}{l} T_{2}=\frac{C_{P}}{C_{V}} T_{1} \\ T_{2}=\left(\frac{1.005}{0.718}\right) \times 400 \\ T_{2}=559.888 \mathrm{~K} \approx 560 \mathrm{~K} \end{array}$
Question 8
In which of the following pairs of cycles, both cycles have at least one isothermal process?
A Diesel cycle and Otto cycle B Carnot cycle and Stirling cycle C Brayton cycle and Rankine cycle D Bell-Coleman cycle and Vapour compression refrigeration cycle
GATE ME 2021 SET-1 Second Law, Carnot Cycle and Entropy
Question 8 Explanation:
1.Brayton
$\begin{array}{ll} 1 \text { to } 2: & S=C \\ 2 \text { to } 3: & P=C \\ 3 \text { to } 4: & S=C \\ 4 \text { to } 1: & P=C \end{array}$
2. VCRS
1 to 2 : Isentropic compression
2 to 3: constant pressure (P=C)
3 to 4: Isenthalpic expansion $\left(h_{3}=h_{4}\right)$
4 to 1: Constant pressure (P=C)
3. Carnot
1 to 2 : lsentropic compression
2 to 3 : Isothermal heat addition
3 to 4 : lsentropic expansion
4 to 1 : Isothermal heat rejection
4. Bell Coleman
1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure
5. Rankine
1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure
6. Diesel
1 to 2 : lsentropic compression
2 to 3 : Constant pressure (P = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume
7. Otto
1 to 2 : lsentropic compression
2 to 3 : Constant pressure (V = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume (V = C)
Question 9
Keeping all other parameters identical, the Compression Ratio (CR) of an air standard diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle $r_c = 2$.
The difference between the new and the old efficiency values, in percentage,
$(\eta _{new}|_{ CR = 21})-(\eta _{old}|_{CR = 15})=$ _______ %. (round off to one decimal place)
A 4.8 B 2.4 C 6.2 D 2.8
GATE ME 2020 SET-2 Availability and Irreversibility
Question 9 Explanation:
\begin{aligned} \eta_{d, r=21} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=54.87 \% \\ \eta_{d, r=15} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=50.08 \% \\ \eta_{d, r=21}-\eta_{d, r=15} &=4.8 \% \end{aligned}
Question 10
In a steam power plant, superheated steam at 10 MPa and 500$^{\circ}C$, is expanded isentropically in a turbine until it becomes a saturated vapour. It is then reheated at constant pressure to 500$^{\circ}C$. The steam is next expanded isentropically in another turbine until it reaches the condenser pressure of 20 kPa. Relevant properties of steam are given in the following two tables. The work done by both the turbines together is ______ kJ/kg (roundoff to the nearest integer).
A 1513 B 1245 C 832 D 1825
GATE ME 2020 SET-2 First Law, Heat, Work and Energy
Question 10 Explanation:
Given data: $h_{1}=3373.6 \mathrm{kJ} / \mathrm{kg}, h_{3}=3478.4 \mathrm{kJ} / \mathrm{kg}, h_{2}=2778.1 \mathrm{kJ} / \mathrm{kg}, s_{1}=s_{2}$ (as from table )
\begin{aligned} s_{3} &=s_{4} \\ s_{3} &=7.7621=0.8319+x+(7.9085-0.8319) \\ x_{4} &=0.9793 \\ h_{4} &=h_{f}+x_{4} \times\left(h_{g}-h_{f}\right)=2560.91 \mathrm{kJ} / \mathrm{kg} \\ W_{T} &=\left(h_{1}-h_{2}\right)+\left(h_{3}-h_{4}\right)=1512.95 \mathrm{kJ} / \mathrm{kg} \end{aligned}
There are 10 questions to complete. | 3,637 | 10,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 38, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-49 | longest | en | 0.701907 |
https://informesia.com/1804/in-the-above-example-if-the-average-weight-is-decreased-by-1kg-then-find-the-weight-of-new-person | 1,685,467,627,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00024.warc.gz | 373,701,723 | 14,747 | 39 views
In the above example, if the average weight is decreased by 1Kg , then find the weight of new person.
## 1 Answer
Best Answer
Solution —
Let the average weight of a group = x
∴ Total weight of 30 persons =30x
Let the weight of new person = y
∴From the quetions,
30x - 40 + y = 30(x - 1)
⇒ 30x - 40 + y = 30x - 30
⇒ y = 40-30
∴ y = 10 kg
Hence, The age of new person is 10kg
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42 views | 242 | 688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-23 | latest | en | 0.759757 |
https://www.physicsforums.com/threads/light-speed-relative-to-accelerating-const-vel-source.251542/ | 1,480,781,653,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00134-ip-10-31-129-80.ec2.internal.warc.gz | 1,009,857,750 | 16,784 | # Light speed relative to accelerating & const vel source
1. Aug 22, 2008
### brightonb
Two observers, A & B, are moving apart at constant velocity V. At distance D, B sends a pulse to A which arrives T seconds later. If B were instead accelerating at a rate such that he attains velocity V just when he is at distance D, and sends a pulse at this instant (while still accelerating but at velocity V), would the pulse still arrive T seconds later?
Thanks for any responses, Manny
2. Jul 16, 2009
### JCOX
Let me start by saying I am not a physics major and am starting to try and learn this stuff so I would appreciate any skepticism in my response.
You can’t say T sec. later. You are using T as a time without any particular Frame of reference. T for the B will not be T for A. Mainly due to the fact that B is accelerating.
3. Jul 16, 2009
### vin300
Yes, definitely, the time depends only on the distance.
4. Jul 16, 2009
### JCOX
Ah ... are you taking into account G.R... I mean time is not the same for someone accelerating compared to someone not accelerating ... Am I correct??
5. Jul 16, 2009
### JCOX
What I mean is WHO is holding the clock ... you?
6. Jul 16, 2009
### JesseM
The question is not really well-defined unless you specify what frame (coordinate system) you're using to define time and distance. If you are using A's inertial rest frame, for example, then it would make no difference whether B was accelerating when he emitted the signal, for example. But if you use an accelerating frame in which B is at rest, the answer will probably be different than it would have been if B had been moving inertially and you used B's inertial rest frame.
7. Jul 16, 2009
### vin300
The time measured by A would be the same but to find that measured in any other frame in relative motion divide the time measured by A bygamma, that is 1/sqrt.(1-v^2/c^2) v is the relative velocity that gives a lesser value
8. Jul 16, 2009
### JesseM
That's not right, the time dilation formula only works for events which occur at the same location in frame where they are separated by a time T (for example, ticks of a clock which is at rest in that frame), then in another frame the same pair of events will be separated by a larger time T/sqrt(1 - v^2/c^2). In this example we are dealing with the time between the signal being emitted by B and the signal being received by A, which don't occur at the same location in A's frame; if the time between them in A's frame is T and the distance between them in A's frame is X then you would use the temporal part of the Lorentz contraction equation, T' = (T - vX/c^2)/sqrt(1 - v^2/c^2), to find the time T' between the events in some other inertial frame moving at speed v relative to A's frame.
9. Jul 19, 2009
### brightonb
To clarify this problem, what I am actually asking is whether a signal from 2 transmitters will arrive simultaneously at the observer given that one transmitter is accelerating & the other moving at constant velocity relative to the observer. The signals are sent as the 2 transmitters pass each other (i.e. from the same position).
10. Jul 19, 2009
### ZikZak
In that case, since the signals travel at c, and they are being sent from the same spacetime event, then they arrive simultaneously at the receiver. It makes no difference whether the emitter is accelerating, at rest, moving, running in circles, or wearing its underwear on its head. Photons travel at c.
11. Jul 19, 2009
### diazona
Man, I'm going to have some strange dreams tonight... | 881 | 3,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-50 | longest | en | 0.941828 |
http://nrich.maths.org/7224/index | 1,503,530,672,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00531.warc.gz | 302,187,905 | 9,550 | ### Whole Number Dynamics I
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Whole Number Dynamics II
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics III
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
# Difference Dynamics Discussion
##### Stage: 5
Published January 2011,February 2011.
Before you read this article you are advised to try the Difference Dynamics investigation. Start with a finite sequence, for example (25, 42, 6). The differences between your numbers give you three new numbers (17, 36, 19). Repeat this operation to give a sequence of sequences. In this example the sequence of sequences continues: (19, 17, 2), (2, 15, 17), (13, 2, 15), (11, 13, 2), (2, 11, 9) , (9, 2, 7), (7, 5, 2), (2, 3, 5), (1, 2, 3), (1, 1, 2), (0,1,1) etc. You can produce chains of sequences like this for sequences of any length. The rule for generating these sequences is that you take the moduli of the differences between adjacent terms, taking the difference between the last term and the first term, to get the next sequence in the chain and repeat this process to get a sequence of sequences.
Read no further. You can try this for yourself, starting with a sequence of any length. Carry out your own investigation, make some conjectures about the behaviour of the sequences that you observe, at least for sequences of two, three and four terms, and try to prove your conjectures. When you tire of your own thoughts and you want to know what others have done with these ideas then read on.
This investigation is well known and has a long history. The problem is attributed to the Italian Enrico Ducci in the late 1800's, and the first proof of the theorem that is proved in this article seems to have appeared in 1937. The usual proof involves, among other things, expanding a sum of matrices by the binomial theorem over the field $\mathbb{Z}_2$. However, as the problem can be tackled by school students, it seems desirable to have elementary proofs of the conjectures they might make. The purpose of this article is to provide such proofs.
A little experimentation leads to sequences that repeat in a cyclic pattern or to the sequence of sequences stopping with the zero sequence $(0, 0, ...0)$ which we call a fixed point of this iteration. This behaviour is typical of dynamical systems which are very important in mathematics.
There are many situations in which we want to measure something that changes with time, either discretely (say every week) or continuously. For example, we may count the number of elephants in Africa each year or record the temperature continuously at some spot in London. A dynamical system is a mathematical model of such a system, and dynamical systems are often used to predict population growth, weather, and so on. The essential feature of this type of system is that there is a set of all the possibilities for the measurements, together with some mathematical rule which tells us how our measurement will change with time. An important feature in any such model are the fixed points in the model: if the measurement is $X_n$ and the rule is $X_{n+1}=f(X_n)$ for some function $f$, then a fixed point is a measurement $X$ such that $f(X) = X$. In this case, if some $X_n$ takes the value $X$ then $X_{n+1}=X_n=X$ and the population (for example) will remain at this fixed value for ever. A more general situation is when the sequence of measurements oscillates like for example, $A,B,A,B,A,...$ or $A,B,C,A,B,C,A,...$ giving cycles of periods two and three. Measurements in nature are often seen to follow cyclic patterns.
In this Difference Dynamics investigation, many sequences converge to repeating cycles such as$(k, k, 0), (0, k, k), (k, 0, k), (k, k, 0)$ and some sequences end with the zero sequence $(0, 0, 0, ... 0)$ making it a fixed point for this iteration. So which sequences of sequences end with the zero sequence? Do all other sequences end in cycles?
As a first step let us prove that if this process does not produce a sequence of zeros then it must lead to a repeating cyclic pattern. Notice that after the first sequence all the numbers in the sequences are positive so without loss of generality we'll consider sequences with positive terms. Then taking differences in this iterative process makes sure that the largest number in the sequence stays the same or gets smaller.
The proof depends on the same reasoning as you can use to prove that if your school is big enough then there are at least two people in the school who have the same birthday. As there are only 366 possible birthdays, if there are more than 366 people in your school then at least two people must share a birthday.
The sequence we used as an example started with (25, 42, 6) so no number in any of the subsequent sequences can be bigger than 42. It follows that there are at most 43 choices (0 to 42) for each number in the sequences making at most $43^3$ different possible sequences. There are not an infinite number of possibilities, so just as with the birthday example, sooner or later the iteration must repeat a sequence that has appeared earlier and repeat the earlier pattern. The same argument applies for any finite starting sequence of any length and so we have proved that these sequences always end in the zero sequence or in cycles.
You will no doubt have noticed that all sequences of length 2 and 4 seem to end with the zero sequence and all non-constant sequences of length 3 end in cycles and you may have conjectured that there is a significant difference for sequences of odd and even length. This is not so as you will have discovered if you tried sequences of length 6. If you pursued the investigation far enough you may have noticed the result in the theorem below for which we give a proof that uses nothing more than school mathematics.
Let $(a_1,\ldots,a_n)$ be a sequence of integers, and let $\theta$ be the operation $$\theta: (a_1,\ldots,a_n) \mapsto (|a_1-a_2|,\ldots,|a_{n-1}-a_n|, |a_n-a_1|).$$ Throughout we write $\mathbf{a} = (a_1,\ldots,a_n)$, and so on, and $\mathbf{0}=(0,\ldots,0)$. Clearly, we need only consider sequences with non-negative elements (since this will be so after just one application of $\theta$), so all sequences in this discussion will be assumed to have non-negative integral elements. For a positive integer $q$, $\theta^q$ denotes the $q$-th iterate of $\theta$.
First, suppose that $\theta(\mathbf{a}) =\mathbf{b}$. We show that $\mathbf{a}$ is a constant sequence if and only if $\mathbf{b}$ is the zero sequence. Clearly if $\mathbf{a}$ is a constant sequence then $\mathbf{b}$ is the zero sequence. Now suppose that $\mathbf{b}$ is the zero sequence so we have $|a_j - a_{j+1}|= b_j = 0$ for all $j$. Thus $a_1 = a_2 = ... = a_n$ so that $\mathbf{a}$ is a constant sequence. Can $\mathbf{b}$ be a constant sequence without being the zero sequence? Now if $n$ is even and $\theta(\mathbf{a}) =\mathbf{b}$ it is possible for $\mathbf{b}$ to be a constant non-zero sequence, for example if $\mathbf{a} =(1,2)$ then $\mathbf{b}= (1,1)$. However if $n$ is odd, this is not possible as we show below.
Theorem Every sequence $(a_1,\ldots,a_n)$ eventually maps to $(0,\ldots,0)$ under repeated applications of $\theta$ if and only if $n=1, 2, 4, 8, 16 ...$, that is $n=2^m$ for some integer $m$.
We divide our proof into three cases: (i) $n$ is odd, (ii) $n$ has an odd factor, and (iii) $n=2^m$ for some $m$.
Section (i) $n$ is odd
We suppose that $n$ is odd. Suppose $\theta(\mathbf{a})$ is constant, say $(b,b,\ldots,b)$. Then, for $j=1,\ldots,n$, we have $|a_j-a_{j+1}|=b$ (where we put $a_{n+1}=a_1$). Thus we can write $a_j-a_{j+1} = \varepsilon_jb$, where $\varepsilon_j = \pm 1$, and $b(\varepsilon_1+\cdots +\varepsilon_n)=0$. Since the sum of the $\varepsilon_j$ cannot be zero (because $n$ is odd), we see that $b=0$. Thus $\theta(\mathbf{a})=\mathbf{0}$. Hence, from above, $\mathbf{a}$ is constant. These facts show that only constant sequences map to constant sequences under $\theta$ or, equivalently, that non-constant sequences map to non-constant sequences. Thus if $n$ is odd, and if we start with a sequence that is not constant, then it will never reach $\mathbf{0}$.
Section (ii) $n$ has an odd factor
We shall give the argument for $n=6$ (which has $3$ as an odd factor), and this clearly generalises to any $n$ with an odd factor. We know from Section (i) for any non-constant sequence $(a_1,a_2,a_3)$ we have $\theta^p(a_1,a_2,a_3) \neq (0,0,0)$ for any integer $p$. Now if $\theta^p(a_1,a_2,a_3) = (b_1,b_2,b_3)$, then $$\theta^p(a_1,a_2,a_3,a_1,a_2,a_3) = (b_1,b_2,b_3,b_1,b_2,b_3),$$ and it follows from this that $\theta^p(a_1,a_2,a_3,a_1,a_2,a_3) \neq (0,0,0,0,0,0)$ for any $p$. In general, if $n$ has an odd factor then there is a sequence $\mathbf{a}$ which will never reach $\mathbf{0}$. We have now shown that if all sequences eventually reach $\mathbf{0}$, then $n$ cannot have an odd factor, so that $n=2^m$ for some $m$.
Section (iii) $n$ is a power of $2$
We now assume that $n=2^m$. Our proof uses the following lemma.
Lemma For any starting sequence $\mathbf{a}$, $\theta^n(\mathbf{a})$ has even components.
Let us assume this lemma for the moment, and show how it enables us to complete the proof of the theorem. Take any $\mathbf{a}$ and choose an integer $p$ such that each component of $\mathbf{a}$ is strictly less than $2^p$. Now, by our Lemma, we can write $\theta^n(\mathbf{a}) = 2\mathbf{a}_1$, where $\mathbf{a}_1$ is some sequence. In the same way, we can write $\theta^n(\mathbf{a_1}) = 2\mathbf{a}_2$, and so on, until we reach $\theta^n(\mathbf{a}_{p-1}) = 2\mathbf{a}_p$. It is now clear that $\theta^{np}(\mathbf{a}) = 2^p\mathbf{a}_p.$ Now, for any $\mathbf{x}$, the maximum component of $\theta(\mathbf{x})$ is at most the maximum component of $\mathbf{x}$. Because the maximum component of $\theta^{np}(\mathbf{a})$ is strictly less than $2^p$, we must have $\mathbf{a}^p = \mathbf{0}$; thus $\theta^{np}(\mathbf{a}) =\mathbf{0}$ and so if the lemma holds we have proved the theorem.
It remains to prove the lemma and we do this by considering a function $\rho$ that is closely related to $\theta$. The function $\rho$ acts on infinite sequences of non-negative integers as follows: $$\rho(a_1,a_2,a_3,\ldots) = (|a_1-a_2|,|a_2-a_3|, |a_3-a_4|,\ldots).$$
Notice that the action of $\theta$ on sequences of length $n$ can be obtained from $\rho$ by taking the sequence $(a_1,a_2,\ldots)$ to be periodic of period $n$ so this proof works for all values of $n$. Now $$\rho^2(a_1,a_2,a_3,\ldots) = (\big||a_1-a_2|-|a_2-a_3|\big|,\ldots).$$ The key observation now is that if $q$ is an integer then $q$ and $|q|$ have the same parity (both odd or both even). Thus $\big||a_1-a_2|-|a_2-a_3|\big|$ has the same parity as $a_1-2a_2+a_3$, which has the same parity as $a_1+a_3$. If we now use the notation $(a_1,a_2,\ldots)\sim (b_1,b_2,\ldots)$ to mean that $a_j$ and $b_j$ have the same parity for all $j$, then we see that $$\rho^2(a_1,a_2,a_3,\ldots) \sim (a_1+a_3,a_2+a_4,a_3+a_5,\ldots).$$ If we now apply $\rho^2$ to the sequence on the right we find that $$\rho^4(a_1,a_2,a_3,\ldots) \sim (a_1+a_5,a_2+a_6,a_3+a_7,a_4+a_8,a_5+a_9,\ldots).$$ Next, we apply $\rho^4$ to the sequence on the right: this gives $$\rho^8(a_1,a_2,a_3,\ldots) \sim (a_1+a_9,a_2+a_{10},\ldots).$$ Quite generally, this shows that $$\rho^{2^p}(a_1,a_2,a_3,\ldots) \sim (a_1+a_{1+2^p},a_2+a_{2+2^p},a_3+a_{3+2^p},\ldots).$$ Now suppose that the original sequence $(a_1,a_2,\ldots)$ is periodic with period $2^m$. Then $a_j=a_{j+2^m}$ for all $j$, so that $$\rho^{2^m}(a_1,a_2,a_3,\ldots) \sim (2a_1,2a_2,2a_3,\ldots)\sim (0,0,0,\ldots).$$ Clearly this implies that $\theta^{2^m}(a_1,\ldots,a_{2^m})$ has all components even, and the lemma (and hence the theorem) is proved.
References
(1) Ciamberlini, C. and Maregoni, A., Su una interessante curiosit\`a numerica, {\sl Period. Mat. Ser. 4} 17 (1937), 25-30.
(2) Honsberger, R., {\sl Ingenuity in Mathematics}, Random House, New York, 1970.
(3) Lotan, M., A problem in difference sets, {\sl Amer. Math. Monthly} 56 (1949), 535-541.
(4) Miller, R., A game with $n$ numbers, {\sl Amer. Math. Monthly} 85 (1978), 183-185.
(5) Zvengrowski, P., Iterated absolute differences, {\sl Math. Mag.} 52 (1979), 36-37. | 3,826 | 12,754 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-34 | latest | en | 0.923564 |
https://circlecoder.com/k-th-smallest-prime-fraction/ | 1,679,558,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00273.warc.gz | 209,875,235 | 9,424 | # K-Th Smallest Prime Fraction Problem
## Description
LeetCode Problem 786.
You are given a sorted integer array arr containing 1 and prime numbers, where all the integers of arr are unique. You are also given an integer k.
For every i and j where 0 <= i < j < arr.length, we consider the fraction arr[i] / arr[j]. Return the k^th smallest fraction considered. Return your answer as an array of integers of size 2, where answer[0] == arr[i] and answer[1] == arr[j].
Example 1:
``````1
2
3
4
5
Input: arr = [1,2,3,5], k = 3
Output: [2,5]
Explanation: The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.
The third fraction is 2/5.
``````
Example 2:
``````1
2
Input: arr = [1,7], k = 1
Output: [1,7]
``````
Constraints:
• 2 <= arr.length <= 1000
• 1 <= arr[i] <= 3 * 10^4
• arr[0] == 1
• arr[i] is a prime number for i > 0.
• All the numbers of arr are unique and sorted in strictly increasing order.
• 1 <= k <= arr.length * (arr.length - 1) / 2
## Sample C++ Code using Priority Queue
``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
class Solution {
public:
vector<int> kthSmallestPrimeFraction(vector<int>& A, int K) {
priority_queue<pair<double, pair<int,int>>> pq;
for (int i = 0; i < A.size(); i++)
pq.push({-1.0*A[i]/A.back(), {i,A.size()-1}});
while (--K > 0) {
pair<int,int> cur = pq.top().second;
pq.pop();
cur.second--;
pq.push({-1.0*A[cur.first]/A[cur.second], {cur.first, cur.second}});
}
return {A[pq.top().second.first], A[pq.top().second.second]};
}
};
`````` | 520 | 1,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-14 | latest | en | 0.574471 |
https://olete.in/learn-2656-30-Aptitude-Aptitude%20Areas%20Mcq%20Set%203 | 1,726,591,919,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00697.warc.gz | 406,494,983 | 11,711 | # Aptitude/Aptitude Areas Mcq Set 3 Sample Test,Sample questions
1.76 min
2.80 min
3.90 min
4.96 min
1.65%
2.70%
3.80%
4.60%
1.14m
2.24m
3. 28m
4.40m
1.15.16cm
2.16.17cm
3.17.18cm
4.18.19m
1.320 tiles
2.640 tiles
3.160 tiles
4.6.4 tiles
1.0.525cm
2.4.2cm
3.80%
4.. 60%
1.1520 m2
2. 2420 m2
3.2480 m2
4.2520 m2
1.14m
2.21m
3.28m
4.None of these
1.15500Sq.m
2.15400Sq.m
3.15200Sq.m
4.15300Sq.m
1.4 4/13 m
2.6 4/11m
3. 12 4/11m
4.12 8/11m
1.125cm, 35cm
2.126cm,32cm
3.132cm,26cm
4.135cm,25cm
1.2dm
2.6dm
3.18dm
4.72dm
1.1.233m
2.1.225m
3.1.227m
4.1.229m
1.34
2.40
3.68
4.88
1. 5000 m/hr
2.5200m/hr
3.5400 m/hr
4.5600 m/hr
1.4m
2.4.1cm
3. 4.2cm
4.4.3m
1.32
2.36
3.42
4.46
1.12km/hr
2.10km/hrC.
3.14km/hr
4.None of these
1.2times
2. 3times
3.1/3 times
4.9times
1. 25%
2.40.5%
3.55%
4.56.25%
1.14.64cm
2.16.36m
3. 17.60m
4.18.40m
1. 6cm
2.3.2 cm
3.3 cm
4.3.5 cm
1. 110m
2. 112m
3.120cm
4.135m
1.2R cm2
2.R2 cm2
3.1/2 R2 cm 2
4.2 R2 cm 2
1.2.5m
2.3m
3.5.5m
4.4m
1.17.5cm
2. 21cm
3.28cm
4.None of these
1.25% increase
2. 50% increase
3.50 decrease
4.75% decrease
## Question: ```The product of the area of three adjacent faces of a rectangular box is equal to ```
1.the volume of the box
2.Twice the volume of the box
3.. the square of the volume of the box
4.Cube root of the volume of the box
1.10
2.100
3.1000
4.10,000
1.125 Sq.cm
2.236 Sq.cm
3. 361 Sq.cm
4.None of these
1.26hours
2.42hours
3. 52hours
4.65hours
1.484Sq.cm
2.538 2/7 Sq.cm
3.616 Sq.cm
4. 644 Sq.cm
1.1/9
2.2/9
3.7/9
4. 8/9
## More MCQS
##### Olete Team
Online Exam TestTop Tutorials are Core Java,Hibernate ,Spring,Sturts.The content on Online Exam Testwebsite is done by expert team not only with the help of books but along with the strong professional knowledge in all context like coding,designing, marketing,etc! | 925 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.612283 |
https://www.coursehero.com/file/5638775/EE3113-A2009-L4/ | 1,516,797,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00008.warc.gz | 855,437,614 | 69,708 | EE3113_A2009_L4
# EE3113_A2009_L4 - ECE 3113 INTRODUCTION TO RF CIRCUIT...
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ECE 3113 INTRODUCTION TO RF CIRCUIT DESIGN Lecture Notes for A-term 2009 LECTURE 4 Prof. R. Ludwig Department of Electrical and Computer Engineering Worcester Polytechnic Institute Worcester, MA 01609 copyright © 2009, R. Ludwig Copyright, 1998 © R. Ludwig
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ECE3113_L4 2 Objectives • What is the difference between a lumped and distributed system? • What are the characteristics of a Transmission Line? • What sets the voltages and currents apart from low frequency lines?
ECE3113_L4 3 General transmission line equations • Detailed analysis of a differential section Note: Analysis applies to all types of transmission lines such as coax cable, two-wire, microstrip, etc.
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ECE3113_L4 4 Kirchhoff’s laws on a microscopic level • Over a differential section we can again use basic circuit theory • Model takes into account line losses and dielectric losses • Ideal line involves only L and C
ECE3113_L4 5 Advantages versus disadvantages of electric circuit representation • Clear intuitive
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• Spring '09
• Ludwig
• Transmission line, Prof. R. Ludwig, Prof. R. Ludwig Department of Electrical and Computer Engineering Worcester Polytechnic Institute Worcester, R. Ludwig, General Transmission Line, differential transmission line
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EE3113_A2009_L4 - ECE 3113 INTRODUCTION TO RF CIRCUIT...
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Ask a homework question - tutors are online | 469 | 1,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-05 | latest | en | 0.798829 |
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Generating Random Numbers [02/24/2001] Would you know of any congruential models for the generation of random numbers? Getting Complex with Euclid, Sturm, and Newton [02/20/2016] An adult seeks a method for determining a polynomial's complex roots. Doctor Vogler offers some clarifications, iterative techniques — and caveats. Greatest Common Factor [03/28/1997] How do you find the greatest common factor? How a CPU Adds Decimal Numbers [10/22/2000] How does an 8-bit CPU add up two four-digit decimal numbers, and why does it take four operations to do so? How to Solve Equations with No Analytic Solution Method [10/26/2005] A discussion of solving equations that can't be solved analytically by using iterative estimation methods including bisection, false position, and Newton's method. The equation x(e^x) = 3 is used as an example. The Hungarian Job Assignment [03/10/2011] A company owner writes in for help cost-efficiently assigning tasks to different employees when each one commands her own fee for every job. Invoking a little graph theory, Doctor Jacques introduces the Hungarian algorithm and walks through an application to an example assignment. An Introduction to Basic Diophantine Equations [08/27/2007] A birdcage contains both 2-legged and 1-legged birds, and there are a total of 11 legs in the cage. Use a Diophantine equation to find all possible combinations of birds. Inverses in the Field GF(2^8) [11/07/2000] How can I get the multiplicative inverse of a byte in the polynomial field GF(2^8)? An Iterative Method of Calculating Pi [06/09/2004] I recently saw a method of calculating pi that involves an iterative function, P(n + 1) = P(n) + sin(P(n)) where P(n) is the approximation of pi at the nth iteration. 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Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1875 Tribol 00664º 15206695 C++17 0.000 8/14/19, 2:33:27 PM
1038 Snack 21812º 15191815 C++17 0.000 8/13/19, 1:10:34 PM
1019 Time Conversion 24098º 15191412 C++17 0.000 8/13/19, 12:33:14 PM
1018 Banknotes 22667º 15191227 C++17 0.000 8/13/19, 12:10:45 PM
1016 Distance 24573º 15154213 C++17 0.000 8/9/19, 3:22:24 PM
1015 Distance Between Two Points 27303º 15154046 C++17 0.000 8/9/19, 3:05:22 PM
1017 Fuel Spent 25191º 15153018 C++17 0.000 8/9/19, 1:43:53 PM
1014 Consumption 27152º 15153042 C++17 0.000 8/9/19, 10:51:38 AM
1238 Combiner 03756º 15139058 C++17 0.192 8/8/19, 11:30:42 AM
1168 LED 07630º 15129687 C++17 0.020 8/7/19, 2:45:09 PM
1332 One-Two-Three 02319º 12349109 C++17 0.000 11/24/18, 3:41:34 PM
1099 Sum of Consecutive Odd... 07116º 12348642 C++17 0.000 11/24/18, 2:34:14 PM
1068 Parenthesis Balance I 01919º 12348516 C++17 0.032 11/24/18, 2:14:41 PM
1115 Quadrant 08813º 12348316 C++17 0.000 11/24/18, 1:46:18 PM
1013 The Greatest 22497º 12345453 C++17 0.000 11/24/18, 12:46:48 AM
1012 Area 24752º 12345370 C++17 0.000 11/24/18, 12:34:37 AM
1011 Sphere 22847º 12345247 C++17 0.000 11/24/18, 12:12:35 AM
1010 Simple Calculate 26868º 12345225 C++17 0.000 11/24/18, 12:07:38 AM
1009 Salary with Bonus 25982º 12345185 C++17 0.000 11/23/18, 11:57:54 PM
1007 Difference 34325º 12345115 C++17 0.000 11/23/18, 11:41:54 PM
1006 Average 2 31073º 12344488 C++17 0.000 11/23/18, 9:35:19 PM
1005 Average 1 32338º 12344463 C++17 0.000 11/23/18, 9:31:19 PM
1004 Simple Product 39700º 12344327 C++17 0.000 11/23/18, 9:15:06 PM
1003 Simple Sum 40959º 12344268 C++17 0.000 11/23/18, 9:02:39 PM
1002 Area of a Circle 36767º 12344212 C++17 0.000 11/23/18, 8:52:05 PM
1001 Extremely Basic 52645º 12343830 C++17 0.000 11/23/18, 7:53:11 PM
2670 Máquina de Café 00821º 12345074 C++17 0.000 11/22/18, 9:56:32 PM
2518 FNDI's Staircase 00267º 12336397 C++17 0.000 11/22/18, 9:50:24 PM
1173 Array fill I 10027º 12032456 C++17 0.000 10/20/18, 9:57:08 PM
1059 Even Numbers 15369º 11630693 C++17 0.000 9/18/18, 5:27:46 PM
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https://jp.mathworks.com/matlabcentral/cody/problems/58-tic-tac-toe-ftw/solutions/1729551 | 1,606,497,944,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00484.warc.gz | 340,724,006 | 17,167 | Cody
Problem 58. Tic Tac Toe FTW
Solution 1729551
Submitted on 18 Feb 2019 by Daniel Bergman
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
a = [ 1 0 0 0 -1 0 -1 0 1]; b = [0]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
1 0 0 0 -1 0 -1 0 1 3 3 rows and wins follow: 0
2 Pass
a = [ 1 0 0 0 1 -1 1 -1 -1]; b = [2 7]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
1 0 0 0 1 -1 1 -1 -1 3 3 rows and wins follow: 1 8 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 7
3 Pass
a = [ 1 0 0 -1 1 -1 1 -1 0]; b = [7 9]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
1 0 0 -1 1 -1 1 -1 0 3 3 rows and wins follow: 7 8 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 7 9
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Start Hunting! | 407 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-50 | latest | en | 0.724747 |
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Hi, I am trying to project a point x in 3D Space onto a point x' on the surface of a cone. x' should be the point on the cone's surface closest to the original point x such that the normalized vector x-x' is the cone's surface normal in the point x'. The Cone is parameterize in two distinct ways... (1) - the cone's apex a e R³ - the cone's normal n e R³ - the cone's opening semi-angle alpha (2) - a point p e R³ on the cone's axis but not the apex - a normal n e R³ pointing from p towards the axis - the ortohgonal distance s from from p to the cone's surface - the opening semi-angle alpha I have thought quite a lot about this problem but have not come up with a satisfying result. Does anybody know a solution and is willing to help out? Thanks a lot...
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The closest point is going to be on the plane defined by x, the cone's apex, and a point on the cone's axis other than the apex. When you intersect this plane with the cone, you should get two lines. The problem then simplifies to finding the closest point on each line, and choosing the smaller distance of the two solutions.
You can either do the problem in the original space, or you can transform everything to a space where the cone's apex is the origin, and its axis is aligned to X/Y/Z. Then once you find a solution you reverse the transformation. If you're confident with your 3D linear algebra then you shouldn't need to do this, but it can simplify many aspects of the problem.
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Quote:
Original post by ZipsterThe closest point is going to be on the plane defined by x, the cone's apex, and a point on the cone's axis other than the apex. When you intersect this plane with the cone, you should get two lines. The problem then simplifies to finding the closest point on each line, and choosing the smaller distance of the two solutions.You can either do the problem in the original space, or you can transform everything to a space where the cone's apex is the origin, and its axis is aligned to X/Y/Z. Then once you find a solution you reverse the transformation. If you're confident with your 3D linear algebra then you shouldn't need to do this, but it can simplify many aspects of the problem.
Thank's a lot for your thoughts on this one Zipster! I came up with a different solution now but your idea sure helped me to get to this point now!
My solution (if anyone is interested)...
Vector3f Cone::getNormalOfProjection( Vector3f p_ ){ //choose new coord system with the the base vectors... //a. cone normal //b. a perpendicular vector u pointing towards p_ //c. a third vector, perpendicular to normal and u Vector3f u = ( p_ - m_apex ); u = u - ( m_normal * ( u * m_normal ) ); u.normalize(); Vector3f v = m_normal.crossProduct( u ); v.normalize(); //transformation matrix of the new coord system Matrix3f m = makeMatrix3f( m_normal[0], m_normal[1], m_normal[2], u[0], u[1], u[2], v[0], v[1], v[2] ); //p_ transformed in alternate coord system //(translated, such that the apex is centered in the origin) Vector3f p = m * ( p_ - m_apex ); //definition of point x which is the point of intersection //of the line defined by p_ and the searched normal //and the axis of the cone Vector3f x; if( p[0] < 0 ) x = makeVector3f( p[0] - tan( m_alpha ) * p[1], 0.0f, 0.0f ); else x = makeVector3f( p[0] + tan( m_alpha ) * p[1], 0.0f, 0.0f ); //transform x back into world coord system x = ( m.transpose() * x ) + m_apex; //the normal of the projected point is p_ - x Vector3f n = p_ - x; n.normalize(); return n;}
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× | 1,069 | 4,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-43 | latest | en | 0.925374 |
https://www.biostars.org/p/12227/ | 1,560,890,959,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998817.58/warc/CC-MAIN-20190618203528-20190618225528-00097.warc.gz | 667,065,428 | 8,330 | Question: How To Simulate The Null Distribution For Tajima'S D?
3
7.8 years ago by
David W4.7k
New Zealand
David W4.7k wrote:
I know there are a few populations genetics/molecular evolution types in the Biostar community, so hopefully someone has already run into this problem.
I've calculated Tajima's D statistic for a couple of DNA alignments using the R package PEGAS, which gives a p-value using a Beta distribution. It happens that the Beta distribution doesn't perfectly model the distribution of D under neutral evolution, so many authors simulate data under neutrality to make a null distribution. Unfortunately, all the methods for these papers tend to say about this procedure is "we used `ms` to simulate the null...".
Does anyone know how to do this properly. My understanding was you simulated data with a matching number of segregating sites, so it you had 100 samples and 20 segregating sites you'd do
``````ms 100 1000 -s 20
``````
giving 1000 simulations. However when I calculate D from those simulations I get a skewed distribution with a mean near to 1 (I would have though it would b 0) but a p-value pretty close to the one using the Beta distribution. It's entirely possible I messed up somewhere in the calculations, (you can see my function here) but I thought I'd make sure I'm actually simulating the distribution correctly before I pick that apart.
population sequence • 4.3k views
written 7.8 years ago by David W4.7k
Yeah! Love population genetics questions!
4
7.8 years ago by
Jarretinha3.3k
São Paulo, Brazil
Jarretinha3.3k wrote:
Aha! Found the problem! After checking you code carefully for mistakes (found none) I've noted that your interpretation (and expectation) of the test could be you problem. Tajima's D isn't normally distributed and is naturally skewed. By using a fixed number of segregating sites, 4Nu/2Nu will vary between samples to enforce the restriction. So, you can obtain a positively skewed distribution even in a neutral setup. I'm not aware of the overall sensitivity of Tajima's D in this situation. You can try to generate a much larger number of samples and verify if the skeweness still persists. If PEGAS is correct, it should decrease.
Cheers Jarretinha, looks like I might need to do some digging into pop. gen. or just use Arlequin and move on :)
3
7.8 years ago by
Casey Bergman18k
Athens, GA, USA
Casey Bergman18k wrote:
You are on the right track and Jarretinha provides additional (correct) insight into the problem. If you need to check whether your solution in R is giving good results, you can run a test alignment through your script and also through DNAsp, which provides methods to compute the confidence interval on Tajima's D via standard neutral coalescent simulations.
3
7.8 years ago by
David W4.7k
New Zealand
David W4.7k wrote:
(not sure if this should be an answer, a comment or an edit to the original question - let me know if I'm doing it wrong...)
So, it turns out there was something funny going on in the way I was calculating D_ from my simulated data (I suspect the polymorphism counting step, but having dissected that yet). Thankfully `ms` comes with another utility `sample_stats` that calculates some of the standard diviersity indices including _D for each replicate so
``````ms 20 100 -s 20 | sample_stat > sims.tsv
``````
Creates a text file with D_ in the sixth column. To call this in R and get the distribution of _D I've used
``````cmd <- paste("ms", n, nreps, "-s", S, "| sample_stats")
sims <- system(cmd, intern=TRUE)
get.D <- function(x) unlist(strsplit(x, "\t"))[6]
Ds <- as.numeric(unlist(lapply(sims, get.D)))
``````
Which gives a result close to that of Arlequin (haven't checked DNAsp).
Thanks Casey and Jarretinha - both helpful answers, Jarrenthinha get's the Big Green Tick since getting to the bottom of what I was actually expecting from the simulations is what helped me get this answer. | 956 | 3,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-26 | longest | en | 0.939723 |
https://electronics.stackexchange.com/questions/71614/guessing-power-line-voltage-from-the-number-of-disc-insulator?noredirect=1 | 1,685,915,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00740.warc.gz | 255,801,887 | 41,911 | # Guessing power line voltage from the number of disc insulator
Is there a good way to guess the power line voltage based on the number of insulator discs that attach the line to the transmission tower?
Wikipedia seems to suggest that there is a "Typical number of disc insulator units for standard line voltages".
Is this a good ball park upper bound to the voltage on the power line?
Is there a better way of guessing line voltages?
Is it possible to further refine this guess based on the type of insulator discs?
• The total length of the insulator gives you some idea of maximum voltage it could withstand, since the limit is often the arc distance thru air, not what the material is made from. In any case, the arc distance is still there, so the voltage can't exceed that minus some derating factor. Jun 4, 2013 at 15:56
Is there a good way to guess the power line voltage based on the number of insulator discs that attach the line to the transmission tower?
We usually use as a rule of thumb for determining the number of porcelain/glass suspension disc insulators: 10kV for each disc (standard 5-1/4' x 10").
For typical system voltages in North America, this would be:
69kV: 4-6 discs; 115kV: 7-9 discs; 138kV: 8-10 discs; 230kV: 12 discs; 345kV: 18 discs; 500kV: 24 discs.
You may check Electric Power Generation, Transmission & Distribution by Grigsby, et al.
In general, probably not.
25KV Low Voltage Silicone Rubber Composite Tension Suspension Electric Insulator for Railway
Low Voltage 220kV Polymer Composite Power Line Insulator
In a specific country, with a single supplier of high-voltage transmission tower systems, for a small group of towers built at similar times, the answer may be yes.
• Your right this is a very location specific question. Most of california seems to have the same sort of high voltage distribution lines. aka the large metal transmission lines with what seem to be ceramic disc insulators, it seems like there might be a standard for modern ceramic disc insultors Jun 4, 2013 at 15:19
• The 220kV one is much larger (2240mm vs 480mm). Jun 4, 2013 at 19:07
See attached Table 4-6 from the USDA-RUS Bulletin 1724e300.
Usually, it is possible to say what voltage the insulator can withstand, because it is a function of its length. Typically these discs are set with the same distance so it CAN be possible but only as some approximation. The purpose of these discs is not however being markings of voltage, but (among others) they function to halt arc on the insulator surface.
The insulator length provides more information, but this is only design data. The line which was designed for 400 kV can be used as 110 kV, so it says nothing. The same information might be taken from distance between phase conductors, tower height (the distance between conductor and earth), distance from line to its neighbourhood and so on.
Taking a look on the line will not even show you if it is operational.
According to My research I have found that yes and no that you can and can’t base the line voltages off of the insulator number of discs.
Here is why you could NOT:
High voltages act differently in cold, hot, moist, or and dry areas. In Utah there are power lines with 12 ceramic discs on the power poles holding or suspending the power lines. The power lines only have 138 kV flowing through the power lines. But the number of insulating ceramic discs suggests that a little more than 161 kV is flowing through the power lines. So the answer for this is no, you couldn’t base voltages flowing through power lines on the number of ceramic discs on the power lines.
The reason why you could:
The main and only reason why you could base the voltage flowing through a power line off of the number of insulating discs is if you wanted to be over safe. Or what I call over-protective. Which means basically you are estimating higher than needed to make things even safer. | 905 | 3,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.914369 |
https://justaaa.com/accounting/36356-cramerica-has-8-coupon-bond-issue-with-8-years-to | 1,685,347,102,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00290.warc.gz | 391,986,961 | 9,845 | Question
# CraMerica has 8% coupon bond issue with 8 years to maturity. Each of these bonds make...
CraMerica has 8% coupon bond issue with 8 years to maturity. Each of these bonds make semi-annual interest payments. These bonds have a yield to maturity of 10%. Suddenly, the yield to maturity on these bonds fall to 8%. What is the percentage change in the price of these bonds. Assume a par value of \$1,000. Enter your answer to two decimal plances with no percentage sign. That is, like this: 13.61
Face Value = \$1,000
Annual Coupon Rate = 8%
Semiannual Coupon Rate = 4%
Semiannual Coupon = 4% * \$1,000
Semiannual Coupon = \$40
Time to Maturity = 8 years
Semiannual Period = 16
If interest rate is 10%:
Annual Interest Rate = 10%
Semiannual Interest Rate = 5%
Price of Bond = \$40 * PVIFA(5%, 16) + \$1,000 * PVIF(5%, 16)
Price of Bond = \$40 * (1 - (1/1.05)^16) / 0.05 + \$1,000 / 1.05^16
Price of Bond = \$891.62
If interest rate is 8%:
Annual Interest Rate = 8%
Semiannual Interest Rate = 4%
Price of Bond = \$40 * PVIFA(4%, 16) + \$1,000 * PVIF(4%, 16)
Price of Bond = \$40 * (1 - (1/1.04)^16) / 0.04 + \$1,000 / 1.04^16
Price of Bond = \$1,000.00
Percentage Change in Price = (\$1,000.00 - \$891.62) / \$891.62
Percentage Change in Price = 0.1216 or 12.16
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Most questions answered within 1 hours. | 456 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-23 | latest | en | 0.827791 |
https://nicoguaro.github.io/posts/bem_corrected/ | 1,709,422,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00894.warc.gz | 413,055,174 | 9,465 | # Coming back to the Boundary element method
During October (2017) I wrote a program per day for some well-known numerical methods in both Python and Julia. It was intended to be an exercise to learn some Julia. You can see a summary here. I succeeded in 30 of the challenges, except for the BEM (Boundary Element Method), where I could not figure out what was wrong that day. The original post is here.
Thomas Klimpel found the mistake and wrote an email where he described my mistakes. So, I am creating a new post with a correct implementation of the BEM.
## The Boundary Element Method
We want so solve the equation
\begin{equation*} \nabla^2 u = -f(x, y)\quad \forall (x, y) \in \Omega\, , \end{equation*}
with
\begin{equation*} u(x) = g(x, y), \quad \forall (x, y)\in \partial \Omega \, . \end{equation*}
For this method, we need to use an integral representation of the equation, that, in this case, is
\begin{equation*} u(\boldsymbol{\xi}) = \int_{S} [u(\mathbf{x}) F(\mathbf{x}, \boldsymbol{\xi}) - q(\mathbf{x})G(\mathbf{x}, \boldsymbol{\xi})]\mathrm{d}S(\mathbf{x}) + \int_{V} f(\mathbf{x}) G(\mathbf{x}, \boldsymbol{\xi}) \mathrm{d}V(\mathbf{x}) \end{equation*}
with
\begin{equation*} G(\mathbf{x}, \boldsymbol{\xi})= -\frac{1}{2\pi}\ln|\mathbf{x} - \xi| \end{equation*}
and
\begin{equation*} F(\mathbf{x}, \boldsymbol{\xi}) = -\frac{1}{2\pi |\mathbf{x} - \boldsymbol{\xi}|^2} (\mathbf{x} - \boldsymbol{\xi})\cdot\hat{\mathbf{n}} \end{equation*}
Then, we can form a system of equations
\begin{equation*} [G]\{q\} = [F]\{u\}\, , \end{equation*}
that we obtain by discretization of the boundary. If we take constant variables over the discretization, the integral can be computed analytically by
\begin{equation*} G_{nm} = -\frac{1}{2\pi}\left[r \sin\theta\left(\ln|r| - 1\right) + \theta r\cos\theta\right]^{\theta_B, r_B}_{\theta_A, r_A} \end{equation*}
and
\begin{equation*} F_{nm} = \left[\frac{\theta}{2\pi}\right]^{\theta_B}_{\theta_A} \end{equation*}
for points $n$ and $m$ in different elements, and the subindices $A,B$ refer to the endpoints of the evaluation element. We should be careful evaluating this expression since both $r_A$ and $r_B$ can be (close to) zero and make it explode. Also, here it was the main problem since I forgot to compute the angles with respect to elements that are, in general, not aligned with horizontal or vertical axes.
For diagonal terms the integrals evaluate to
\begin{equation*} G_{nn} = -\frac{L}{2\pi}\left(\ln\left\vert\frac{L}{2}\right\vert - 1\right) \end{equation*}
and
\begin{equation*} F_{nn} = - \frac{1}{2} \end{equation*}
with $L$ the size of the element.
Following is the code. Keep in mind that this code works for purely Dirichlet problems. For mixed Dirichlet-Neumann the influence matrices would need rearranging to separate known and unknowns in opposite sides of the equation.
You can download the files for this project here. It includes a YML file to create a conda environment with the dependencies listed. For example, it uses version 3.0 of Meshio.
import numpy as np
from numpy import log, arctan2, pi, mean, arctan
from numpy.linalg import norm, solve
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import meshio
def assem(coords, elems):
"""Assembly matrices for the BEM problem
Parameters
----------
coords : ndarray, float
Coordinates for the nodes.
elems : ndarray, int
Connectivity for the elements.
Returns
-------
Gmat : ndarray, float
Influence matrix for the flow.
Fmat : ndarray, float
Influence matrix for primary variable.
"""
nelems = elems.shape[0]
Gmat = np.zeros((nelems, nelems))
Fmat = np.zeros((nelems, nelems))
for ev_cont, elem1 in enumerate(elems):
for col_cont, elem2 in enumerate(elems):
pt_col = mean(coords[elem2], axis=0)
if ev_cont == col_cont:
L = norm(coords[elem1[1]] - coords[elem1[0]])
Gmat[ev_cont, ev_cont] = - L/(2*pi)*(log(L/2) - 1)
Fmat[ev_cont, ev_cont] = - 0.5
else:
Gij, Fij = influence_coeff(elem1, coords, pt_col)
Gmat[ev_cont, col_cont] = Gij
Fmat[ev_cont, col_cont] = Fij
return Gmat, Fmat
def influence_coeff(elem, coords, pt_col):
"""Compute influence coefficients
Parameters
----------
elems : ndarray, int
Connectivity for the elements.
coords : ndarray, float
Coordinates for the nodes.
pt_col : ndarray
Coordinates of the colocation point.
Returns
-------
G_coeff : float
Influence coefficient for flows.
F_coeff : float
Influence coefficient for primary variable.
"""
dcos = coords[elem[1]] - coords[elem[0]]
dcos = dcos / norm(dcos)
rotmat = np.array([[dcos[1], -dcos[0]],
[dcos[0], dcos[1]]])
r_A = rotmat.dot(coords[elem[0]] - pt_col)
r_B = rotmat.dot(coords[elem[1]] - pt_col)
theta_A = arctan2(r_A[1], r_A[0])
theta_B = arctan2(r_B[1], r_B[0])
if norm(r_A) <= 1e-6:
G_coeff = r_B[1]*(log(norm(r_B)) - 1) + theta_B*r_B[0]
elif norm(r_B) <= 1e-6:
G_coeff = -(r_A[1]*(log(norm(r_A)) - 1) + theta_A*r_A[0])
else:
G_coeff = r_B[1]*(log(norm(r_B)) - 1) + theta_B*r_B[0] -\
(r_A[1]*(log(norm(r_A)) - 1) + theta_A*r_A[0])
F_coeff = theta_B - theta_A
return -G_coeff/(2*pi), F_coeff/(2*pi)
def eval_sol(ev_coords, coords, elems, u_boundary, q_boundary):
"""Evaluate the solution in a set of points
Parameters
----------
ev_coords : ndarray, float
Coordinates of the evaluation points.
coords : ndarray, float
Coordinates for the nodes.
elems : ndarray, int
Connectivity for the elements.
u_boundary : ndarray, float
Primary variable in the nodes.
q_boundary : [type]
Flows in the nodes.
Returns
-------
solution : ndarray, float
Solution evaluated in the given points.
"""
npts = ev_coords.shape[0]
solution = np.zeros(npts)
for k in range(npts):
for ev_cont, elem in enumerate(elems):
pt_col = ev_coords[k]
G, F = influence_coeff(elem, coords, pt_col)
solution[k] += u_boundary[ev_cont]*F - q_boundary[ev_cont]*G
return solution
#%% Simulation
elems = mesh.cells["line"]
bound_nodes = list(set(elems.flatten()))
coords = mesh.points[bound_nodes, :2]
x, y = coords.T
x_m, y_m = 0.5*(coords[elems[:, 0]] + coords[elems[:, 1]]).T
theta = np.arctan2(y_m, x_m)
u_boundary = 3*np.cos(6*theta)
#%% Assembly
Gmat, Fmat = assem(coords, elems)
#%% Solution
q_boundary = solve(Gmat, Fmat.dot(u_boundary))
#%% Evaluation
ev_coords = mesh.points[:, :2]
ev_x, ev_y = ev_coords.T
solution = eval_sol(ev_coords, coords, elems, u_boundary, q_boundary)
#%% Visualization
tris = mesh.cells["triangle"]
fig = plt.figure()
ax.plot_trisurf(ev_x, ev_y, solution, cmap="RdYlBu", lw=0.3,
edgecolor="#3c3c3c")
plt.xticks([])
plt.yticks([])
ax.set_zticks([])
plt.savefig("bem_solution.png", bbox_inches="tight", transparent=True,
dpi=300)
The result in this case is the following. | 2,046 | 6,654 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-10 | latest | en | 0.76246 |
https://www.jiskha.com/similar?question=Determine+the+%23+of+moles+of+Al2S3+that+can+be+prepared+by+the+reaction+of+0.400+moles+of+Aluminum+with+0.700+moles+of+Sulfur%2C+and+Also+find+the+%23+of+moles+of+reactant+in+excess | 1,561,467,576,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999838.23/warc/CC-MAIN-20190625112522-20190625134522-00302.warc.gz | 771,603,873 | 19,366 | # Determine the # of moles of Al2S3 that can be prepared by the reaction of 0.400 moles of Aluminum with 0.700 moles of Sulfur, and Also find the # of moles of reactant in excess
43,347 questions
1. ## Chemistry
Determine the # of moles of Al2S3 that can be prepared by the reaction of 0.400 moles of Aluminum with 0.700 moles of Sulfur, and Also find the # of moles of reactant in excess
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2. ## Chemistry
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3. ## Chemistry
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4. ## Chemistry
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5. ## chemistry
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6. ## CHEMISTRY
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7. ## Chemistry
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8. ## Chemistry
In the reaction Al2S3(s) + 6HCl(aq) --> 2AlCl3(aq) + 3H2S(g), how many moles of HCl are used for each mole of AlCl3 formed? A.0 B.2 C.3 D.1 E.1.5 where do you start and how do you figure it out?
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9. ## Chemistry
Did I do these right? Can someone let me know if I did them correctly? I had someone help me with the problems but know I want to make sure their the right answers? 1. Methane burns in oxygen according to the following equation: CH4 + 2O2 ----> CO2 + 2H2O
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10. ## chemistry
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11. ## Chemistry
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12. ## chemistry
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asked by anna on March 31, 2008
13. ## chemistry
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14. ## Chemistry
Hello, I really don't understand how to find the grams of moles in some of these reactions. If someone could help I would appreciate it. 1) Balance the following chemical reaction and determine the number of moles of HI produced when 2.33 moles of H2(g)
asked by Anonymous on February 19, 2017
15. ## Chemistry II
The enthalpy of neutralizatino for the reaction of a strong acid with a strong base is -56 kJ/mol of water produced. How much energy will be released when 230.0 mL of 0.400 M HCl is mixedwith 150.5 mL of 0.500 M NaOH? How much water is produced in the
asked by Jayd on May 20, 2007
16. ## chem
Calculate the mole fraction of total ions in an aqueous solution prepared by dissolving .400 moles of MgCl2 in 850.0g of water. mole fraction =amount solute(in mols) /total amount of solute and solvent (in mols) so wouldn't it be .400/ .400+ 47.2 which is
asked by natash on May 11, 2008
17. ## Chemistry
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18. ## Chemistry
Following question I need help with part b) Please verify my answer for part a) too if possible. Consider the following reaction for silver tarnishing: 3Ag2S(s) + 2Al(s) -> 6Ag(s) + Al2S3(s) a. State the oxidation number for each element in the reaction.
asked by Dexter on June 14, 2011
19. ## chemistry
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20. ## Chemistry
The combustion reaction of propane is described by the reaction. C3H8 + 5O2 → 4H2O + 3CO2. How many moles of O2 are required to generate 3 moles of CO2? 2.5 moles 3 moles 4 moles 5 moles
asked by Wesley on June 3, 2016
21. ## Chemistry
At 1200 K a gaseous mixture contains 0.30 moles of carbon monoxide, 0.10 moles of hydrogen, 0.20 moles of water vapor and 0.059 moles of methane in a 1.00 liter container. Determine whether or not equilibrium has been reached for the following reaction:
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22. ## chemistry
Is this equation correctly balanced? If not why. Al2S3+H2O----> Al2S3+3H2O---->Al2(SO)3+3H2
asked by What? on October 22, 2012
23. ## Chemistry
Based on the balanced equation Al2S3 + 6H2O → 2Al(OH)3 + 3H2S calculate the number of excess reagent units remaining when 4 Al2S3 formula units and 30 H2O molecules react? Molar Mass (g/mol) Al2S3 150.16 H2O 18.015 Al(OH)3 75.361 H2S 34.082 Avogadro's
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24. ## chemistry
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25. ## Chemistry
A mass of 37.5 grams of mercury oxide is decomposed by heating. The reaction is 2HgO > 2Hg + O2. A. How many moles of mercury oxide is decomposed? B. How many moles of oxygen are prepared? C. How many grams of oxygen are prepared?
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26. ## chemistry
Given this data in a study on how the rate of a reaction was affected by the concentration of the reactants Initial Rate, RUN #(A, M) (B, M) (C, M) (mol L-1 s-1 1 0.200 0.100 0.600 5.00 2 0.200 0.400 0.400 80.0 3 0.600 0.100 0.200 15.0 4 0.200 0.100 0.200
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27. ## chemistry
6. Determine Keq for the reaction: 2 SO2 (g) + O2(g) 2 SO3(g), given that 1.00 x 10-2 moles of SO2 and 2.00 x 10-2 moles of O2 were placed in a 2.00L reaction chamber. The chamber contained 7.5 x 10-3 moles of SO3 when equilibrium was established at 727oC.
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28. ## chemistry
The first step of the synthesis is described by the reaction below. When 1.750 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4 , the theoretical yield of FeC2O42H2O is ? grams.. I do not understand why my answer is not correct. How do you find
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29. ## chemistry
Balance the following chemical reaction and determine the number of moles of HI produced when 2.33 moles of H2(g) are consumed in the following chemical reaction: ___H2(g) + __I2(g) -> ___HI (g)
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30. ## chem
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asked by jack on February 14, 2018
31. ## chemistry
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asked by j on September 30, 2009
32. ## chemistry
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is s2(g)+C(s)-->CS2(g) Kc=9.40 at 900k How many grams of CS2(g) can be prepared by heating 14.5 moles of S2(g) with excess carbon in a 8.15 L reaction vessel held at 900 K
asked by kitor on June 10, 2013
33. ## chemistry
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2 +C =CS2 Kc= 9.40 at 900k How many grams of CS2(g) can be prepared by heating 14.3 moles of S2(g) with excess carbon in a 8.55 L reaction vessel held at 900 K until
asked by Paul on January 19, 2013
34. ## Chemistry
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is: S2(g)+C(s) CS2(g) Kc=9.40 at 900K How many grams of CS2(g) can be prepared by heating 16.8 moles of S2(g) with excess carbon in a 9.80 L reaction vessel held at 900 K
asked by Anonymous on March 25, 2013
35. ## chemistry
Ammonia gas can be prepared by the reaction CaO(s) + 2NH4Cl(s) -> 2NH3(g) + H2O(g) + CaCl2(s) If 112 g CaO reacts with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?
asked by Jazmin on January 20, 2011
36. ## chemistry
Ammonia gas can be prepared by the reaction CaO(s) + 2NH4Cl(s) -> 2NH3(g) + H2O(g) + CaCl2(s) If 112 g CaO reacts with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?
asked by Jazmin on January 20, 2011
37. ## chemistry
Ammonia gas can be prepared by the reaction CaO(s) + 2NH4Cl(s) -> 2NH3(g) + H2O(g) + CaCl2(s) If 112 g CaO reacts with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?
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38. ## chemistry
Ammonia gas can be prepared by the reaction CaO(s) + 2NH4Cl(s) -> 2NH3(g) + H2O(g) + CaCl2(s) If 112 g CaO reacts with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?
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39. ## chemistry
what type of reaction is Al+S8=Al2S3
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40. ## chemistry
Strong base is dissolved in 665 mL of 0.400 M weak acid (Ka = 3.69 × 10-5) to make a buffer with a pH of 3.94. Assume that the volume remains constant when the base is added. HA(aq) + OH-(aq) -> H2O(l) + A-(aq) Calculate the pKa value of the acid and
asked by craig on July 8, 2012
41. ## Chemistry
The Cu2+ ions in this experiment are produced by the reaction of 1.0g of copper. Information already found out .016 mol Cu2+ produced Overall reaction is Cu2+(aq)+Cl-(aq)+Cu(s)=2 CuCl(s) Maximum mass of CuCl that can be prepared from the reaction sequence
asked by Julie on October 23, 2012
42. ## Gen Chm II
Determine the pressure-based Keq for the reaction: 2SO2(g) + O2(g)2SO3(g) given that 1.00×10−2 moles of SO2 and 2.00×10−2 moles of O2 were initially placed in a 2.00 L reaction chamber. The chamber contained 7.5×10−3 moles of SO3 when equilibrium
asked by James on October 7, 2013
43. ## Chemistry
can you please please help me with these questions I don't know how to do any of them. I know how to calculate moles and grams and mole ratio,etc but I don't understand the limiting excess reactant and yield stuff. Thanks! 1.In the following chemical
asked by molly on October 2, 2016
44. ## Chemistry
can you please please help me with these questions I don't know how to do any of them. I know how to calculate moles and grams and mole ratio,etc but I don't understand the limiting excess reactant and yield stuff. Thanks! 1.In the following chemical
asked by molly on October 2, 2016
45. ## Chemistry 130
Will someone help me by letting me know if I did these correct, or if there the correct answers? Thanks! 2 (10). If two moles of hydrogen gas (H2) react in the following equation, how many moles of H2O will be formed? 2H2 + O2 „³ 2H2O a. 2 moles b. 4
asked by Jessie on September 8, 2009
46. ## CHEMISTRY!
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation for this is S2 (g) + C (s) CS2 (g) Kc = 9.40 at 900 K How many grams of CS2 (g) can be prepared by heating 11.7 moles of S2 (g) with excess carbon in a 5.95 L reaction
asked by cal on October 14, 2015
47. ## chem
Determine the number of moles of MgO that form from the reaction of 4.4 moles of O2 in the following unbalanced reaction: Mg(s) + O2(g) → MgO(s)
asked by a on November 17, 2009
48. ## College Chemistry
Consider the reaction. 4Fe(s) + 3O2(g) ¨ 2Fe2O3(s) If 0.500 mol of Fe reacts with 0.400 mol of O2, what are the limiting reactant and the theoretical yield in moles of Fe2O3 for the reaction, respectively?
asked by Lisa on March 26, 2012
49. ## Chemistry
For the reaction 2H2 + 1O2 à 2H2O, there are 8.08 grams of H2 available. 2. Use the ratio of stoichiometric coefficients in the reaction above to determine how many moles of H2O can be produced from the H2 available. Then, use molar mass to convert the
asked by Feather on September 13, 2014
50. ## chemistry
Am doing a lab report on the determination of the product of a redox reaction: the reaction of bromate and hydroxylammonium ions. here is the lab Experiment: In this experiment you will determine the equation of a redox reaction by the use of experimental
asked by khela on October 26, 2010
51. ## Chem
Review example 5.10 and exercise 5.10. Suppose 1.58 moles of N2 and 0.420 moles of O2 occupy 22.41 L at a total pressure of 2.00 atm. Use XA = to determine the (decimal) mole fractions of both components. Then, use XA = to determine the partial pressures
asked by Feather on October 11, 2014
52. ## Chem
How many grams of CS2 can be prepared by heating 11.0 moles of S2 with excess carbon in 5.20 L reaction vessel held at 900 kelvin until equilibrium is attained?
asked by Mer on February 9, 2014
53. ## Chemistry
How many grams of CS2(g) can be prepared by heating 12.9 moles of S2(g) with excess carbon in a 7.00 L reaction vessel held at 900 K until equilibrium is attained? Stoichiometry is 1 to 1
asked by Moni on February 21, 2013
54. ## chemistry
3. Use the ratio of stoichiometric coefficients in the reaction above to determine how many moles of O2 will react with the H2 available. Then, use molar mass to convert the moles of O2 into grams. For the reaction 2H2 + 1O2 =2H2O, there are 8.08 grams of
asked by feather on September 13, 2014
55. ## Chem
Hydrochloric acid can be prepared using the reaction described by the chemical equation: 2 NaCl(s) + H2SO4(l) ----> 2 HCl(g) + Na2 SO4(s). How many grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl? A) 4.00 g B) 2.49 g C) 146 g D) 284
asked by Jessica on April 11, 2007
56. ## Chemistry
7.0 mL of 0.64 M HBr reacts with 38 mL of 0.39 M KOH. 1) What are the moles of acid before the reaction? 2) What are the moles of base before the reaction? 3) What is the limiting reagent? 4) How many moles of excess reagent are there after the reaction?
asked by Fontaine on December 6, 2016
57. ## Chemistry
In the gas-phase reaction given below, the [I2] changes from 0.400 M at 0.00 min to 0.200 M at 3.00 min. I2 + Cl2 --> 2 ICl Calculate the average reaction rate in moles I2 consumed per liter per minute. mol/(L·min)
asked by Lionel on April 27, 2017
58. ## Chemistry
Thanks for the help earlier My question Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si + 2Cl2 yields SiCl4 In one reaction 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction? Ok
asked by Kate on July 3, 2010
59. ## Chemistry
2NaOH+2Al+2H2O --> 2NaAlO2+3H2 here is the reaction. If i am trying to find the mass of hydrogen produced and i am trying to convert .95 moles Al to moles H, which number from the reaction do I use for H when converting from moles to moles? Do I 6 because
asked by Erica on March 9, 2012
60. ## chemistry
5.00g aluminum sulfide and 10.0g water react according to the reaction: Al2S3 + 6 H2O -> 2 Al(OH)3 + 3 H2S What is maximum mass of H2S that can be formed? What mass of excess reactant remains when the reaction is complete?
asked by Jessi on October 12, 2014
61. ## Chemistry - check answer
A 20.0 L reaction vessel contains 0.016 moles CO2, 1.05 moles of CO, and 2.00 moles of C. The mixture is approaching the following equilibrium: CO2(g) + C(s) 2CO(g) a) Calculate the reaction quotient Q. b) If value of K is 1.17, predict which direction the
asked by Amanda on April 20, 2014
62. ## Chemistry
If 0.110 mol of liquid H2O and 40.5 g of solid Al2S3 are reacted stoichiometrically according to the balanced equation, how many moles of solid Al(OH)3 are produced?
asked by Julie on May 21, 2009
63. ## Chemistry
A reaction mixture of 4.0 mL of 0.002 M SCN- and 5.0 mL of 0.002 M Fe3+ is diluted to 10.0 mL with 0.1 M HNO3 to form the blood-red FeNCS2+ complex. The equilibrium molar concentration of the FeNCS2+ determined from a standardization curve is 1.5x10^-4
asked by Anonymous on March 18, 2013
64. ## chemistry
A reaction mixture of 4.0 mL of 0.002 M SCN- and 5.0 mL of 0.002 M Fe3+ is diluted to 10.0 mL with 0.1 M HNO3 to form the blood-red FeNCS2+ complex. The equilibrium molar concentration of the FeNCS2+ determined from a standardization curve is 1.5x10^-4
asked by Anonymous on October 13, 2015
65. ## Chemistry
Iron metal reacts with oxygen gas to produce iron(III) oxide. If you have 12.0 moles of iron, for complete reaction you need: (HINT you need the balanced chemical reaction!) A) 9.0 moles of O2 and produce 3.0 moles of Fe2O3 B) 9.0 moles of O2 and produce
asked by Petey on April 9, 2016
66. ## CHEmistry
The balanced reaction equation for combustion of Mg is: 2 Mg + O2 --> 2 MgO. In the reaction vessel you have 1.0 moles of Mg and 0.30 moles of O2. What is the limiting reactant? How many moles of MgO can you possibley make?
asked by Amanda on October 9, 2011
67. ## Math
determine whether the number sentence is true. In each case explain how you could answer without calculating. Check your answers by doing the indicated calculations. A. 50 x 432= (50 x 400) + (50 x 32) B. 50 x 368 = (50 x 400) - (50 x 32) C. 50 x 800= ( 50
68. ## Chem
S2 + C ----> CS2 How many grams of CS2 can be prepared by heating 11.0 moles of S2 with excess carbon in 5.20 L reaction vessel held at 900 kelvin until equilibrium is attained? Kc= 9.40
asked by Help on February 10, 2014
69. ## chemistry
Consider the following reaction: 2 NO(g) + O2(g) ---> 2 NO2(g) + energy if the number of moles of oxygen is increased to two moles, the a. reaction speeds up b. reaction slows down c. temperature increases d. temperature decreases I'm not really sure what
asked by Aria on February 21, 2012
70. ## chemistry 30
Consider the following reaction: 2NO(g)+O2(g)-->2 NO2(g)+ energy if the number of moles of oxygen is increased to two moles, the a. reaction speeds up b. reaction slows down c. temperature increases d. temperature decreases
asked by Tucker on December 30, 2011
71. ## CHemistry
The first step of the synthesis is described by the reaction below. When 1.200 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is ? grams im not sure what to do. do u convert 1.0M to grams first? by
asked by Jason on April 22, 2007
72. ## CHEMISTRY
******* Please check. Also check sig figs. Ion Concentrations 1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and
asked by Anonymous on October 25, 2013
73. ## chemistry
Asprin, C9H8O4, is prepared by reacting salicylic acid, C7H6O3, with acetic anhydride, C4H6O3, in the reaction C7H6O3(s) + C4H6O3(l) -> C9H8O4(s) + C2H4O2(l) A student is told to prepare 45.0 g of aspirin. She is also told to use a 55.0% excess of the
asked by Chris on October 18, 2006
74. ## chemistry
Given the following reaction: 4 KO2 + 2 H2O + 4 CO2 4 KHCO3 + 3 O2 How many moles of H2O are needed to react with 12.15 moles of CO2? moles A chemical reaction occurs between solutions of LiCl and silver nitrate. This reaction produces a precipitate that
asked by Erika on October 22, 2010
75. ## chemistry
Given the reaction: H2SO4 = 2NaOH -> NaSO4 = 2H20. Determine the # of moles of sulfuric acid id needed to completely react with 2.3 L of 0.5 M of NaOH. Then determine the concentration of the solution if 3.0 L of it are used instead.
asked by sarah on November 4, 2012
76. ## chemistry
Given the reaction: H2SO4 = 2NaOH -> NaSO4 = 2H20. Determine the # of moles of sulfuric acid id needed to completely react with 2.3 L of 0.5 M of NaOH. Then determine the concentration of the solution if 3.0 L of it are used instead.
asked by sarah on November 4, 2012
77. ## chemistry
a 500ml sample of o2 gas at 24 degrees celsius was prepared by decomposing a 3% aqueous solution of hydrogen peroxide, in this presence of a small amount of manganese catalyst by the reaction 2H2O2>2H2O+O2. The oxygen thus prepared was collected by
asked by Michelle on April 16, 2012
78. ## chemistry- electrochemical cell
What makes this an oxidation-reaction? 3Ag2S+2Al(s) -> Al2S3+6 Ag(s)? Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reason. What is oxidized in the reaction? What
asked by princess on March 25, 2012
79. ## CHEM! (SCIENCE) NEED HELP
How many moles of water will form if the reaction has 9 moles of Cu? how many moles of NO will form if the reaction has 9 moles of Cu? 3Cu+8HNO3->3Cu(NO3)2+4H2O+2NO
asked by tyneisha on January 24, 2013
80. ## AP CHEM-- check work?
How many moles of ATP must be converted to ADP by the reaction ATP(aq) + H2O ---> ADP(aq) + HPO4-2(aq) + 2H+(aq) delta-G*= -31 kJ to bring about a nonspontaneous biochemical reaction in which delta-G*= +372 kJ? I did my work out like this..: 372 kJ x 1
asked by janelle on April 15, 2007
81. ## chem
In a chemical reaction, A and B reacts to form D. C is the intermediate product of this reaction. Percent yield of the reaction A+B->C is 50% and the percent yield of the reaction C->D is 90%. How many moles of D can be produced when 14 moles of A and B
asked by bekah on October 5, 2014
82. ## Chemistry
5.2 grams of magnesium metal are place in 400.0 mL of 1.0 M hydrochloric acid. The balanced equation for the reaction is: Mg + 2HCl = MgCl2 + H2. How many moles of H2 gas were produced?
asked by Heidi on June 9, 2014
83. ## Chemistry
Given the balanced equation representing a reaction : 2Fe+ 3Cu^2+ --> 2Fe^3+ +3Cu When the iron atoms lose six moles of electrons, how many moles of electrons are gained by the copper ions? (1) 12 moles (2) 2 moles (3) 3 moles (4) 6 moles how would i even
asked by Anonymous on February 16, 2014
84. ## chemistry
Calculate the PH value of 1.00 L of a solution prepared by mixture of 1.00 mol HCl, 2.00 moles of NaOH and 1.50 moles of NH4Cl Kb (NH3) = 1.8 * 10 ^-5
asked by wallace on January 13, 2019
85. ## chemistry
In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 10 mL 5.0 M acetone + 10 mL 1.5 M HCl + 10 mL 0.005 M I2 + 20 mL H2O A student found that it took 400 seconds for the color of the I2 to
asked by leticia on March 5, 2015
86. ## Chem 2
Need help.... calculate the theorectical yield of the compounds to be prepared in the experiment. The metal ion in both cases is the limiting reagent.Find nuber of moles of Cu(II) in the sample of CuSO4 x 5H2O that you used. The equal the number of moles
asked by Debbie on April 2, 2012
87. ## chemistry
When 1.24g of an organic compound with the formula CxHyOz is burned in excess oxygen, 1.76g of carbon dioxide and 1.08g of water vapor are obtained. What is the empirical formula of the compound? Determine the moles of CO2 in 1.76grams, that will give you
asked by kevin on August 23, 2005
88. ## Chemistry
In a reaction chamber, 3.0mol of aluminum is mixed with 5.3mol Cl2 and reacts. The reaction is described by the following balanced chemical equation. 2Al + 3Cl2 ==> 2AlCl3 a)Identify the limiting reagent for the reaction. b)Calculate the number of moles or
asked by Jack on April 5, 2010
89. ## Chemistry
A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 265 mL of solution. How many moles of ammonium chloride are present in the resulting solution? When thinking about the amount of solute present in a solution,
asked by Onye on March 3, 2018
90. ## chemistry
A 0.286g sample of Zn was used to produce H2(g)by the reaction. Zn(s)+ 2HCl(aq)> ZnCl2(aq) + H2(g) The barometric pressure was recorded as 761.6mm Hg and room temperature was 20degreesC. When the water levels of the eudiometer tube and graduated cylinder
asked by joann on December 3, 2012
91. ## chemistry
2 Al + 3 S --> Al2S3 In an experiment 10.0 g Al is combined with 15.0 g S. If the reaction continues until one reactant is totally consumed, which reactant would be in excess and how many grams would be unreacted?
asked by Sheree' on November 4, 2012
92. ## Chemistry 2
calculate the theorectical yield of the compounds to be prepared in the experiment. The metal ion in both cases is the limiting reagent.Find nuber of moles of Cu(II) in the sample of CuSO4 x 5H2O that you used. The equal the number of moles of [Cu(NH3)4]x
asked by Debbie on April 7, 2012
93. ## chemistry
Propane fuel is used in BBQs. Use the following reaction to determine how many moles of oxygen gas are needed to react with 1.6 moles of propane gas. C3H8(g)+5O2(g) --------> 4H2O(g)+ 3CO2(g)
asked by stacy on March 30, 2015
94. ## Chemistry
Consider the following reaction: CO2 (g) + H2 (g) CO (g) + H2O (g) Calculate the value of the equilibrium constant, Kc, for the above system, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapour were present in a
asked by kevin on February 25, 2016
95. ## Chemistry
Oxygen gas may be prepared by decomposing hydrogen peroxide, h2o2, using manganese (IV) oxide catalyst: 2H2O2(liquid) ----> 2H2O(liquid) + O2(gas) Calculate the amount of oxygen that will be produced from 1.35 moles of H2O2 Determine the mass of oxygen gas
asked by kai on December 5, 2014
96. ## science
Oxygen gas may be prepared by decomposing hydrogen peroxide, h2o2, using manganese (IV) oxide catalyst: 2H2O2(liquid) ----> 2H2O(liquid) + O2(gas) Calculate the amount of oxygen that will be produced from 1.35 moles of H2O2 Determine the mass of oxygen gas
asked by Nat on December 8, 2014
97. ## CHEM! (SCIENCE) NEED HELP
Determine the moles of aluminum acetate,AL(C2H3O2)3,that can be made if you do this reaction with 12 moles of acetic acid (C2H3O2H) C2H3O2H+AL(OH)3->AL(C2H3O2)3+3H2O
asked by tyneisha on January 24, 2013
98. ## Chemistry
In the reaction 2A+B2 to 2AB if 0.5 moles of A and an excess of B are combined and the reaction goes to %100 completion, then how many moles of C are formed?
asked by Alanna on November 23, 2014
99. ## Chemistry Need Help
Given the chemical equation, Cl2 + F2 2ClF, how many moles of ClF are produced in the reaction, if 1.50 moles Cl2 are made to react with 1.75 moles F2 ? I got the answer 3.50 moles but it was wrong. Can anyone help me please.
asked by Matthew on February 27, 2015
100. ## chemistry
I am given the following reaction: 2NH3(g) -----> N2(g) + 3H2(g) 6.4 mols of ammonia gas has been put into a 1.7 L flask and has been permitted to reach equilibrium in accordance to the reaction listed above. If the equilibrium mixture has 4.2 mols of
asked by Eric on May 26, 2008 | 9,182 | 28,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-26 | latest | en | 0.91924 |
https://derivativedribble.wordpress.com/2020/03/03/a-note-on-uncertainty/ | 1,669,724,768,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00079.warc.gz | 240,132,148 | 24,272 | # A Note on Uncertainty
Shannon’s equation is often used as a measure of uncertainty, and that’s not unreasonable, and he provides a mathematical argument as to why it works as a measure of uncertainty in his seminal paper on information theory.
But, I’ve introduced a different way of thinking about uncertainty rooted in partial information, that is quite elegant, since as you’re given partial information about a system, your uncertainty approaches zero as a matter of arithmetic.
However, this model doesn’t take into account the probability distribution of a system, and instead looks only to the number of states the system can be in. Your uncertainty is simply the log of the number of states of the system.
We can also take into account the distribution of the states of the system by instead calculating the expected number of states, which will be some portion of the total number of possible states.
So if a system is always stuck in some particular state, then your uncertainty is much lower than a uniform distribution of states, and so is the expected number of unique states over any number of observations. Calculating the expected number of states is trivial, and you simply calculate the portion of the total number that would have been observed given a uniform distribution, with a maximum of one, for each possible state of the system. This will always produce an expected number of states that is less than or equal to the actual total number of states. Your uncertainty is then simply the logarithm of the expected number of states of the system.
So for example, if a system can be in 10 states, and after 100 observations, State A occurred 8 times, then that state would contribute 8/10 = .8 to the total expected number of states. In contrast, if State B occurred 12 times, then that state would contribute 1 to the total, since we cap the contribution at 1. If instead all states occur an equal number of times, then each state would contribute exactly 1 to the expected number of states, for a total of 10 states. As a result, if the distribution of states is uniform, then you end up simply taking the logarithm of the number of states.
The information content of a message is measured using the same method in the previous article linked to above, which is to simply measure the change in uncertainty that results from receipt of the message. | 478 | 2,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-49 | latest | en | 0.945565 |
https://quantumcomputing.stackexchange.com/questions/10024/how-to-deal-with-the-relative-phase-from-the-evaluation-of-the-not-interacting-q | 1,660,215,482,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00454.warc.gz | 470,319,006 | 65,835 | # How to deal with the relative phase from the evaluation of the not interacting qubit?
Every qubit is a realization of some two-level quantum system with $$\left| 0 \right\rangle$$ and $$\left| 1 \right\rangle$$ states. These states have their energies $$E_0$$ and $$E_1$$ respectively. By solving the Schrödinger equation for the two-level quantum system one can show time evaluation of the $$\left| 0 \right\rangle$$, $$\left| 1 \right\rangle$$ and $$\left| + \right\rangle$$ states:
$$\left| \psi_0(t) \right\rangle = e^{-i \frac{E_0}{\hbar} t} \left| 0 \right\rangle \\ \left| \psi_1(t) \right\rangle = e^{-i \frac{E_1}{\hbar} t} \left| 1 \right\rangle \\ \left| \psi_+(t) \right\rangle = \frac{1}{\sqrt{2}}\left( e^{-i \frac{E_0}{\hbar} t} \left| 0 \right\rangle + e^{-i \frac{E_1}{\hbar} t} \left| 1 \right\rangle \right)= \frac{e^{-i \frac{E_0}{\hbar} t}}{\sqrt{2}} \left(\left| 0 \right\rangle + e^{-i \frac{E_1 - E_0}{\hbar} t} \left| 1 \right\rangle \right)$$
My question is about the relative phase $$\varphi = e^{-i \frac{E_1 - E_0}{\hbar} t}$$. This $$\phi$$ acts as a phase gate and changes the quantum state. Is this relative phase important and how in real quantum computers we should deal with it? One problem that I see is the following circuit:
After the measurement, the state should be $$\left| 0 \right\rangle$$, because $$HH = I$$. But If $$t$$ is such that the $$\varphi = \pi$$ then we will measure $$\left| 1 \right\rangle$$. We can forget about it if $$\varphi(t) << \pi$$ in the whole computational time. Otherwise, we should make sure that the time between gates should be $$2\pi$$.
Is this a problem and if yes how IBM, Google or etc. deal with this problem when constructing the circuits? How I understand the frequency $$(E_1 - E_0)/\hbar$$ value is important and for example, the ibmq_armonk qubit's frequency is equal to ~4.97428 GHz (qiskit_textbook), so should we take into account this value when we play with ibmq_armonk by using OpenPulse?
The could be a problem, but it depends on how you're realising your qubits. Some realisations are configured so that $$E_0=E_1$$, and then there's no problem.
There is (at least from the theoretical perspective) a simple fix: if you're supposed to be waiting a time $$t$$, then, instead:
• wait time $$t/2$$
• apply bit flip
• wait time $$t/2$$
• apply bit flip.
This exchanges the roles of 0 and 1 for half the time so that they end up with equal phases.
Alternatively, you just take this into account and apply compensating phase gates inside your circuit.
• Thanks for the answer. Can you mention an example of a qubit realization that has $E_0 = E_1$? Feb 28, 2020 at 11:38
• IIRC, something like the hyperfine ground states of a Rubidium atom (e.g. trapped in an optical lattice) are a very good approximation to that because the timescales of the gates are much shorter than the timescale due to $1/(E_0-E_1)$. Feb 28, 2020 at 11:58
• You may wish to save your answer tick in case someone can answer your specific question about the ibm hardware, which is not something I kno anything about. Feb 28, 2020 at 11:59
• Thanks a lot. I will keep the answer tick for a while, but still, your answer was very helpful. Feb 28, 2020 at 12:11 | 979 | 3,230 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | longest | en | 0.771276 |
https://www.physicsforums.com/threads/permutable-matrices.134675/ | 1,481,200,357,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00365-ip-10-31-129-80.ec2.internal.warc.gz | 1,006,350,685 | 14,597 | # Permutable matrices
1. Oct 3, 2006
### esmeco
I have this question as homework from my Algebra class:
A square matrix X is called exchangeable with A if AX=XA.Determine the set of permutable matrices with .
My question is,how do I find that set?I know that a matrix to be permutable all rows and columns must be the same and that a square matrix is composed by the same number of rows and columns.
Thanks in advance for the help!
2. Oct 4, 2006
### HallsofIvy
Staff Emeritus
?? You defined "exchangeable" with A and then asked for "permutable" with A?? Then you defined "permutable" matrix without any reference to a matrix A?? What am I missing?
If you want to find all matrices that are "exchangeable" with A (standard terminology: "that commute with A"), then look at
$$\left[\begin{array}{cc}a & b \\ c & d \end{array}\right]\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\0 & 1\end{array}\right]\left[\begin{array}{cc}a & b \\c & d\end{array}\right]$$
If I understand your definition of "permutable" correctly: "all rows and columns must be the same", then all 2 by 2 permutable matrices are of the form
$$\left[\begin{array}{cc}a & a \\ a & a\end{array}\right]$$
and the only "permutable" matrix that is "exchangeable" with A is
$$\left[\begin{array}{cc}0 & 0 \\ 0 & 0 \end {array} \right]$$
3. Oct 4, 2006
### esmeco
...
Sorry for the mistypeing! When I said "exchengeable I meant to say permutable,so it would be like:
"A square matrix X is called permutable with A if AX=XA..." | 466 | 1,542 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.85912 |
http://radiotec.ru/article/15836 | 1,571,550,198,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986703625.46/warc/CC-MAIN-20191020053545-20191020081045-00408.warc.gz | 165,952,827 | 9,092 | Publishing house Radiotekhnika "Publishing house Radiotekhnika":scientific and technical literature.Books and journals of publishing houses: IPRZHR, RS-PRESS, SCIENCE-PRESS Тел.: +7 (495) 625-9241
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# Parallel implementation for solving optimization problems by means of databases
Keywords:
V.I. Munerman – Ph. D. (Eng.), Associate Professor, Smolensk State University. E-mail: vimoon@gmail.com T.A. Samoylova – Ph. D. (Eng.), Associate Professor, Smolensk State University. E-mail: tatsam@hotbox.ru
The article describes a way to solve a wide range of optimization problems, whose common name would be «the shortest path problems». In practice, this class presents various problems. The weights of the edges of the graph can be elements of different Algebraic Systems. The choice of an Algebraic System is defined by the semantics of the domain, to which belongs the problem under consideration. Various algorithms for solving these problems are considered. The algorithm based on the computation of the transitive closure matrix of the weights of graph G, which is calculated as (Z is the null matrix), is selected. This algorithm is based on matrix multiplication and is therefore easily parallelized. But most of the advantages of that algorithm are lost in the case when G is a sparse matrix, in which most of the elements are neutral. The technology of storage and processing of sparse matrices is selected. That technology is based on the fact that in-memory computing system matrix is stored as a set of tuples of the form (i, j, w), where i, j are indexes of the element of the matrix, w is the weight of the edges between vertices i and j (the value of w is different from the additive neutral element of the Algebraic System, which belong to the weights of the edges). If you use such a representation of sparse matrices, they can be stored and processed in relational databases. In this case, the weight matrix of the graph can be given by a table with the scheme G(i, j, w). The query: SELECT Gk.i, G.j, Q(G.w×Gk.w) FROM Gk INNER JOIN G ON Gk.j = G.i GROUP BY G k.i, G.j; consequently performs all powers of table G, resulting in a sequence of tables with the scheme of Gk(i, j, w) (k = 1, …, K). Q is a function that implements a group operation, such as Sum, Min, Max and the like. The program in the programming language Tranzact-SQL, which allows solving the problem under consideration, is shown. The basic operation used in solving the problem of the search of the shortest paths is JOIN operation. This operation has a large computational complexity. The time of its implementation on Big Data is large. A method is provided which allows speeding up the solution of the problem by a parallel implementing JOIN operation. This possibility is based on the principle of symmetric horizontal distribution of database tables. It is shown that the heuristic algorithm boustrophedon which distributes pair fragments tables Gk and G between calculators constituting firmware computing system can also be implemented by means of databases. The program in the programming language Tranzact-SQL, which implements the algorithm, is shown. The following conclusions are made: when solving optimization problems, which belong to a class of problems of the shortest path, it is possible to use methods based on databases; these methods are applied most effectively by using algorithms in the matrix when the weights matrix is spars; the modern data manipulation language can effectively implement the matrix algorithms and heuristic algorithms such as the boustrophedon algorithm; the approach based on the use of the symmetric horizontal distribution of data principle can effectively solve optimization problems on graphs, by constructing a virtual software-hardware- complexes for parallel implementation of mass data processing, specifically for each task.
References:
1. Fletcher P., Hoyle H., Patty C.W. Foundations of Discrete Mathematics. Boston: PWS-KENTPub. Co. 1991. 781 p.
2. Lidl R., Pilz G. Applied abstract algebra, 2nd edition. Undergraduate Texts in Mathematics. Springer. 1998. 488 p.
3. Gross J.L., Yellen J. Handbook of Graph Theory, Discrete Mathematics and Its Applications. CRCPress. 2004. 779 p.
4. Aho A., Ullman J.D., Hopcroft J.E. Data Structures and Algorithms. AddisonWesley. 1983.
5. Agarwal R.C., Gustavson F.G., Zubair M. A High-Performance Matrix-Multiplication Algorithm on a Distributed-Memory Parallel Computer Using Overlapped Communication // IBM Journal of Research and Development. 1994. № 38(6). P. 673−681.
6. Gupta H., Sadayappan P. Communication Efficient Matrix-Multiplication on Hypercubes. Technical Report // Stanford InfoLab. Publication Note: Sixth Annual ACM Symposium on Parallel Algorithms and Architectures. 1994. (SPAA 1994). ExtendedversioninParallelComputing. 1996. № 22. P. 75−99.
7. Pissanetzky Sergio Sparse Matrix Technology. AcademicPress. 1984. 321 p.
8. Munerman V.I. The Object-Oriented Model of Mass Data Processing // Highly available systems. 2011. V. 7. № 4. P. 72−74.
9. Munerman V.I. A multidimesional matrix model of mass data processing // Highly available systems. 2012. V. 8. № 1. P. 19−22.
10. Levin N.A., Munerman V.I. Models of big data processing in massively parallel systems // Highly available systems. 2013. V. 9. № 2. P. 36−39.
11. Munerman V.I. Implementation of big data processing on symmetric multiprocessing systems // Highly available systems. 2013. V. 9. № 3. P. 35−43.
12. Munerman V.I. The experience of massive data processing in the cloud using Windows Azure (as an example) // Highly available systems. 2014. V. 10. № 2. P. 3−8.
© Издательство «РАДИОТЕХНИКА», 2004-2017 Тел.: (495) 625-9241 Designed by [SWAP]Studio | 1,391 | 5,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | latest | en | 0.908557 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=Task/SurfaceArea | 1,498,510,171,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320865.14/warc/CC-MAIN-20170626203042-20170626223042-00089.warc.gz | 594,584,042 | 21,962 | Surface Area - Maple Programming Help
Surface Area
Description Determine the surface area of a surface that is not necessarily a surface of revolution.
Surface Area
Surface
>
${x}{}{y}$ (1)
Domain: $\left\{u\left(x\right)\le y\le v\left(x\right),a\le x\le b\right\}$
$u\left(x\right)$
> ${{x}}^{{2}}$
${{x}}^{{2}}$ (2)
$v\left(x\right)$
> ${x}$
${x}$ (3)
$a$
> ${0}$
${0}$ (4)
$b$
> ${1.}$
${1.}$ (5)
Inert integral:
> $\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{SurfaceArea}\right]\left(,{x}=..,{y}=..,\mathrm{output}=\mathrm{integral}\right)$
${{∫}}_{{0}}^{{1.}}{{∫}}_{{{x}}^{{2}}}^{{x}}\sqrt{{1}{+}{{y}}^{{2}}{+}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}$ (6)
Value
> $\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{SurfaceArea}\right]\left(,{x}=..,{y}=..\right)$
${0.2031164292}$ (7)
Commands Used | 374 | 946 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-26 | latest | en | 0.358898 |
https://isabelle.in.tum.de/repos/isabelle/file/db9043a8e372/doc-src/Ref/simplifier.tex | 1,685,927,412,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00323.warc.gz | 378,828,240 | 9,372 | doc-src/Ref/simplifier.tex
author clasohm Tue, 16 Nov 1993 14:13:11 +0100 changeset 122 db9043a8e372 parent 104 d8205bb279a7 child 133 2322fd6d57a1 permissions -rw-r--r--
replaced \un by \union in "Simplification sets"
%% $Id$
\chapter{Simplification} \label{simp-chap}
\index{simplification|(}
This chapter describes Isabelle's generic simplification package, which
provides a suite of simplification tactics. This rewriting package is less
general than its predecessor --- it works only for the equality relation,
not arbitrary preorders --- but it is fast and flexible. It performs
conditional and unconditional rewriting and uses contextual information
(local assumptions''). It provides a few general hooks, which can
provide automatic case splits during rewriting, for example. The
simplifier is set up for many of Isabelle's logics: {\tt FOL}, {\tt ZF},
{\tt LCF} and {\tt HOL}.
\section{Simplification sets}
\index{simplification sets}
The simplification tactics are controlled by {\bf simpsets}. These consist
of five components: rewrite rules, congruence rules, the subgoaler, the
solver and the looper. Normally, the simplifier is set up with sensible
defaults, so that most simplifier calls specify only rewrite rules.
Sophisticated usage of the other components can be highly effective, but
most users should never worry about them.
\subsection{Rewrite rules}\index{rewrite rules}
Rewrite rules are theorems like $Suc(\Var{m}) + \Var{n} = \Var{m} + Suc(\Var{n})$, $\Var{P}\conj\Var{P} \bimp \Var{P}$, or $\Var{A} \union \Var{B} \equiv \{x.x\in A \disj x\in B\}$. {\bf Conditional} rewrites such as
$\Var{m}<\Var{n} \Imp \Var{m}/\Var{n} = 0$ are permitted; the conditions
rewrite rules, while \ttindex{delsimps} deletes rewrite rules from a
simpset.
Each simpset contains a (user-definable) function for extracting equalities
from arbitrary theorems. For example $\neg(x\in \{\})$ could be turned
into $x\in \{\} \equiv False$. This function can be set with
\ttindex{setmksimps} but only the definer of a logic should need to do
this. Exceptionally, one may want to install a selective version of
\ttindex{mksimps} in order to filter out looping rewrite rules arising from
local assumptions (see below).
Internally, all rewrite rules are translated into meta-equalities:
theorems with conclusion $lhs \equiv rhs$. To this end every simpset contains
a function of type \verb$thm -> thm list$ to extract a list
of meta-equalities from a given theorem.
\begin{warn}\index{rewrite rules}
The left-hand side of a rewrite rule must look like a first-order term:
after eta-contraction, none of its unknowns should have arguments. Hence
${\Var{i}+(\Var{j}+\Var{k})} = {(\Var{i}+\Var{j})+\Var{k}}$ and $\neg(\forall x.\Var{P}(x)) \bimp (\exists x.\neg\Var{P}(x))$ are acceptable, whereas
$\Var{f}(\Var{x})\in {\tt range}(\Var{f}) = True$ is not. However, you can
replace the offending subterms by new variables and conditions: $\Var{y} = \Var{f}(\Var{x}) \Imp \Var{y}\in {\tt range}(\Var{f}) = True$ is again
acceptable.
\end{warn}
\subsection {Congruence rules}\index{congruence rules}
Congruence rules are meta-equalities of the form
$\List{\dots} \Imp f(\Var{x@1},\ldots,\Var{x@n}) \equiv f(\Var{y@1},\ldots,\Var{y@n}).$
They control the simplification of the arguments of certain constants. For
example, some arguments can be simplified under additional assumptions:
$\List{\Var{P@1} \bimp \Var{Q@1};\; \Var{Q@1} \Imp \Var{P@2} \bimp \Var{Q@2}} \Imp (\Var{P@1} \imp \Var{P@2}) \equiv (\Var{Q@1} \imp \Var{Q@2})$
This rule assumes $Q@1$ and any rewrite rules it implies, while
simplifying~$P@2$. Such local'' assumptions are effective for rewriting
formulae such as $x=0\imp y+x=y$. The next example makes similar use of
such contextual information in bounded quantifiers:
$\List{\Var{A}=\Var{B};\; \Forall x. x\in \Var{B} \Imp \Var{P}(x) = \Var{Q}(x)} \Imp (\forall x\in \Var{A}.\Var{P}(x)) = (\forall x\in \Var{B}.\Var{Q}(x))$
This congruence rule supplies contextual information for simplifying the
arms of a conditional expressions:
$\List{\Var{p}=\Var{q};~ \Var{q} \Imp \Var{a}=\Var{c};~ \neg\Var{q} \Imp \Var{b}=\Var{d}} \Imp if(\Var{p},\Var{a},\Var{b}) \equiv if(\Var{q},\Var{c},\Var{d})$
A congruence rule can also suppress simplification of certain arguments.
Here is an alternative congruence rule for conditional expressions:
$\Var{p}=\Var{q} \Imp if(\Var{p},\Var{a},\Var{b}) \equiv if(\Var{q},\Var{a},\Var{b})$
Only the first argument is simplified; the others remain unchanged.
This can make simplification much faster, but may require an extra case split
to prove the goal.
must be a meta-equality, as in the examples above. It is more
natural to derive the rules with object-logic equality, for example
$\List{\Var{P@1} \bimp \Var{Q@1};\; \Var{Q@1} \Imp \Var{P@2} \bimp \Var{Q@2}} \Imp (\Var{P@1} \imp \Var{P@2}) \bimp (\Var{Q@1} \imp \Var{Q@2}),$
So each object-logic should provide an alternative combinator
\ttindex{addcongs} that expects object-equalities and translates them into
meta-equalities.
\subsection{The subgoaler} \index{subgoaler}
The subgoaler is the tactic used to solve subgoals arising out of
conditional rewrite rules or congruence rules. The default should be
simplification itself. Occasionally this strategy needs to be changed. For
example, if the premise of a conditional rule is an instance of its
conclusion, as in $Suc(\Var{m}) < \Var{n} \Imp \Var{m} < \Var{n}$, the
default strategy could loop.
The subgoaler can be set explicitly with \ttindex{setsubgoaler}. For
example, the subgoaler
\begin{ttbox}
fun subgoal_tac ss = resolve_tac (prems_of_ss ss) ORELSE'
asm_simp_tac ss;
\end{ttbox}
tries to solve the subgoal with one of the premises and calls
simplification only if that fails; here {\tt prems_of_ss} extracts the
current premises from a simpset.
\subsection{The solver} \index{solver}
The solver attempts to solve a subgoal after simplification, say by
recognizing it as a tautology. It can be set with \ttindex{setsolver}.
Occasionally, simplification on its own is not enough to reduce the subgoal
to a tautology; additional means, like \verb$fast_tac$, may be necessary.
\begin{warn}
Rewriting does not instantiate unknowns. Trying to rewrite $a\in \Var{A}$ with the rule $\Var{x}\in \{\Var{x}\}$ leads nowhere. The
solver, however, is an arbitrary tactic and may instantiate unknowns as
it pleases. This is the only way the simplifier can handle a conditional
rewrite rule whose condition contains extra variables.
\end{warn}
\begin{warn}
If you want to supply your own subgoaler or solver, read on. The subgoaler
is also used to solve the premises of congruence rules, which are usually
of the form $s = \Var{x}$, where $s$ needs to be simplified and $\Var{x}$
needs to be instantiated with the result. Hence the subgoaler should call
the simplifier at some point. The simplifier will then call the solver,
which must therefore be prepared to solve goals of the form $t = \Var{x}$,
usually by reflexivity. In particular, reflexivity should be tried before
any of the fancy tactics like {\tt fast_tac}. It may even happen that, due
to simplification, the subgoal is no longer an equality. For example $False \bimp \Var{Q}$ could be rewritten to $\neg\Var{Q}$, in which case the
solver must also try resolving with the theorem $\neg False$.
If the simplifier aborts with the message {\tt Failed congruence proof!},
it is due to the subgoaler or solver who failed to prove a premise of a
congruence rule.
\end{warn}
\subsection{The looper} \index{looper}
The looper is a tactic that is applied after simplification, in case the
solver failed to solve the simplified goal. If the looper succeeds, the
simplification process is started all over again. Each of the subgoals
generated by the looper is attacked in turn, in reverse order. A
typical looper is case splitting: the expansion of a conditional. Another
possibility is to apply an elimination rule on the assumptions. More
adventurous loopers could start an induction. The looper is set with
\ttindex{setloop}.
\begin{figure}
\indexbold{*SIMPLIFIER}
\begin{ttbox}
setsubgoaler setsolver setloop setmksimps;
signature SIMPLIFIER =
sig
type simpset
val simp_tac: simpset -> int -> tactic
val asm_simp_tac: simpset -> int -> tactic
val asm_full_simp_tac: simpset -> int -> tactic
val addeqcongs: simpset * thm list -> simpset
val addsimps: simpset * thm list -> simpset
val delsimps: simpset * thm list -> simpset
val empty_ss: simpset
val merge_ss: simpset * simpset -> simpset
val setsubgoaler: simpset * (simpset -> int -> tactic) -> simpset
val setsolver: simpset * (thm list -> int -> tactic) -> simpset
val setloop: simpset * (int -> tactic) -> simpset
val setmksimps: simpset * (thm -> thm list) -> simpset
val prems_of_ss: simpset -> thm list
val rep_ss: simpset -> \{simps: thm list, congs: thm list\}
end;
\end{ttbox}
\caption{The signature \ttindex{SIMPLIFIER}} \label{SIMPLIFIER}
\end{figure}
\section{The simplification tactics} \label{simp-tactics}
\index{simplification!tactics|bold}
\index{tactics!simplification|bold}
The actual simplification work is performed by the following tactics. The
rewriting strategy is strictly bottom up, except for congruence rules, which
are applied while descending into a term. Conditions in conditional rewrite
rules are solved recursively before the rewrite rule is applied.
There are three basic simplification tactics:
\begin{description}
\item[\ttindexbold{simp_tac} $ss$ $i$] simplifies subgoal~$i$ using the rules
in~$ss$. It may solve the subgoal completely if it has become trivial,
using the solver.
\item[\ttindexbold{asm_simp_tac}] is like \verb$simp_tac$, but also uses
\item[\ttindexbold{asm_full_simp_tac}] is like \verb$asm_simp_tac$, but also
simplifies the assumptions one by one, using each assumption in the
simplification of the following ones.
\end{description}
Using the simplifier effectively may take a bit of experimentation. The
tactics can be traced with the ML command \verb$trace_simp := true$. To
remind yourself of what is in a simpset, use the function \verb$rep_ss$ to
return its simplification and congruence rules.
\section{Example: using the simplifier}
\index{simplification!example}
Assume we are working within {\tt FOL} and that
\begin{description}
\item[\tt Nat.thy] is a theory including the constants $0$, $Suc$ and $+$,
\item[\tt add_0] is the rewrite rule $0+n = n$,
\item[\tt add_Suc] is the rewrite rule $Suc(m)+n = Suc(m+n)$,
\item[\tt induct] is the induction rule
$\List{P(0); \Forall x. P(x)\Imp P(Suc(x))} \Imp P(n)$.
\item[\tt FOL_ss] is a basic simpset for {\tt FOL}.\footnote
{These examples reside on the file {\tt FOL/ex/nat.ML}.}
\end{description}
We create a simpset for natural numbers by extending~{\tt FOL_ss}:
\begin{ttbox}
\end{ttbox}
Proofs by induction typically involve simplification:
\begin{ttbox}
goal Nat.thy "m+0 = m";
{\out Level 0}
{\out m + 0 = m}
{\out 1. m + 0 = m}
\ttbreak
by (res_inst_tac [("n","m")] induct 1);
{\out Level 1}
{\out m + 0 = m}
{\out 1. 0 + 0 = 0}
{\out 2. !!x. x + 0 = x ==> Suc(x) + 0 = Suc(x)}
\end{ttbox}
Simplification solves the first subgoal:
\begin{ttbox}
{\out Level 2}
{\out m + 0 = m}
{\out 1. !!x. x + 0 = x ==> Suc(x) + 0 = Suc(x)}
\end{ttbox}
The remaining subgoal requires \ttindex{asm_simp_tac} in order to use the
induction hypothesis as a rewrite rule:
\begin{ttbox}
{\out Level 3}
{\out m + 0 = m}
{\out No subgoals!}
\end{ttbox}
\medskip
The next proof is similar.
\begin{ttbox}
goal Nat.thy "m+Suc(n) = Suc(m+n)";
{\out Level 0}
{\out m + Suc(n) = Suc(m + n)}
{\out 1. m + Suc(n) = Suc(m + n)}
\ttbreak
by (res_inst_tac [("n","m")] induct 1);
{\out Level 1}
{\out m + Suc(n) = Suc(m + n)}
{\out 1. 0 + Suc(n) = Suc(0 + n)}
{\out 2. !!x. x + Suc(n) = Suc(x + n) ==> Suc(x) + Suc(n) = Suc(Suc(x) + n)}
\ttbreak
{\out Level 2}
{\out m + Suc(n) = Suc(m + n)}
{\out 1. !!x. x + Suc(n) = Suc(x + n) ==> Suc(x) + Suc(n) = Suc(Suc(x) + n)}
\end{ttbox}
Switching tracing on illustrates how the simplifier solves the remaining
subgoal:
\begin{ttbox}
trace_simp := true;
{\out Rewriting:}
{\out Suc(x) + Suc(n) == Suc(x + Suc(n))}
{\out Rewriting:}
{\out x + Suc(n) == Suc(x + n)}
{\out Rewriting:}
{\out Suc(x) + n == Suc(x + n)}
{\out Rewriting:}
{\out Suc(Suc(x + n)) = Suc(Suc(x + n)) == True}
{\out Level 3}
{\out m + Suc(n) = Suc(m + n)}
{\out No subgoals!}
\end{ttbox}
As usual, many variations are possible. At Level~1 we could have solved
both subgoals at once using the tactical \ttindex{ALLGOALS}:
\begin{ttbox}
{\out Level 2}
{\out m + Suc(n) = Suc(m + n)}
{\out No subgoals!}
\end{ttbox}
\medskip
Here is a conjecture to be proved for an arbitrary function~$f$ satisfying
the law $f(Suc(n)) = Suc(f(n))$:\footnote{The previous
simplifier required congruence rules for such function variables in
order to simplify their arguments. The present simplifier can be given
congruence rules to realize non-standard simplification of a function's
arguments, but this is seldom necessary.}
\begin{ttbox}
val [prem] = goal Nat.thy
"(!!n. f(Suc(n)) = Suc(f(n))) ==> f(i+j) = i+f(j)";
{\out Level 0}
{\out f(i + j) = i + f(j)}
{\out 1. f(i + j) = i + f(j)}
{\out val prem = "f(Suc(?n)) = Suc(f(?n)) [!!n. f(Suc(n)) = Suc(f(n))]" : thm}
\ttbreak
by (res_inst_tac [("n","i")] induct 1);
{\out Level 1}
{\out f(i + j) = i + f(j)}
{\out 1. f(0 + j) = 0 + f(j)}
{\out 2. !!x. f(x + j) = x + f(j) ==> f(Suc(x) + j) = Suc(x) + f(j)}
\end{ttbox}
We simplify each subgoal in turn. The first one is trivial:
\begin{ttbox}
{\out Level 2}
{\out f(i + j) = i + f(j)}
{\out 1. !!x. f(x + j) = x + f(j) ==> f(Suc(x) + j) = Suc(x) + f(j)}
\end{ttbox}
The remaining subgoal requires rewriting by the premise, so we add it to
\begin{ttbox}
{\out Level 3}
{\out f(i + j) = i + f(j)}
{\out No subgoals!}
\end{ttbox}
No documentation is available on setting up the simplifier for new logics.
Please consult {\tt FOL/simpdata.ML} to see how this is done, and {\tt
FOL/simpdata.ML} for a fairly sophisticated translation of formulae into
rewrite rules.
%%\section{Setting up the simplifier} \label{SimpFun-input}
%%Should be written!
\index{simplification|)} | 4,438 | 14,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-23 | latest | en | 0.799158 |
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# Under the politician’s ambitious new health plan, all citizens are req
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Under the politician’s ambitious new health plan, all citizens are req [#permalink]
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28 Mar 2017, 11:00
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Question Stats:
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Source: McGraw Hill GMAT
Under the politician’s ambitious new health plan, all citizens are required either to purchase health insurance with vouchers provided by the government or they may acquire health insurance through their employers.
A. or they may acquire health insurance
B. or they must acquire it
C. or they acquire it
D. or acquiring it
E. or to acquire it
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Under the politician’s ambitious new health plan, all citizens are req [#permalink]
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29 Mar 2017, 15:59
1
Source: McGraw Hill GMAT
Under the politician’s ambitious new health plan, all citizens are required either to purchase health insurance with vouchers provided by the government or they may acquire health insurance through their employers.
A. or they may acquire health insurance
B. or they must acquire it
C. or they acquire it
D. or acquiring it
E. or to acquire it
I'm happy to respond.
This is another question I solved in less than 5 seconds. The structure "either . . . or" is a parallel marker. We have "either to purchase," so we need another infinitive after "or." Only (E) works.
If the difficulty of real GMAT is like a Formula One race tournament, then then difficulty level of this question is like riding a pedal tricycle. Much as riding a tricycle will not prepare you for Formula One racing, this question will fall abysmally short of preparing you for the GMAT.
Does all this make sense?
Mike
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Re: Under the politician’s ambitious new health plan, all citizens are req [#permalink]
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30 Mar 2017, 20:02
Under the politician’s ambitious new health plan, all citizens are required either to purchase health insurance with vouchers provided by the government or they may acquire health insurance through their employers.
A. or they may acquire health insurance (Not parallel)
B. or they must acquire it (Same as A)
C. or they acquire it (Same as A)
D. or acquiring it (Same as A)
E. or to acquire it (Parallel and correct)
Kudos please if you like my explanation
_________________
Kindly hit kudos if my post helps!
Re: Under the politician’s ambitious new health plan, all citizens are req &nbs [#permalink] 30 Mar 2017, 20:02
Display posts from previous: Sort by | 1,085 | 4,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-04 | latest | en | 0.895134 |
https://allinternetradio.com/qa/question-what-is-1m-ohm.html | 1,627,229,646,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00113.warc.gz | 106,873,154 | 9,726 | Question: What Is 1m Ohm?
How many ohms is 4 megohms?
We assume you are converting between megaohm and ohm.
You can view more details on each measurement unit: megaohms or ohms The SI derived unit for electric resistance is the ohm.
1 megaohms is equal to 1000000 ohm.
Note that rounding errors may occur, so always check the results..
What is resistance measured in?
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
How many ohms is an open circuit?
Open-When the term open is used in relation to an electrical circuit it means current cannot flow or no continuity. Infinity ohms-This is what an ohmmeter reads when placed on an open circuit. On an analog meter infinity ohms is when the needle doesn’t move at all and on a digital meter infinity ohms is 1 .
Is higher ohms better?
Ohms just means number of volts required for 1 amp of current. Higher Ohms means more damping power the amp has over your headphones = better quality. Lower Ohms means easier to drive BUT also more sensitive to amp quality!
Why is the Omega symbol used for ohms?
In physics, Omega denotes ohm, a unit defining the electrical resistance. In chemistry, the symbol is in association with oxygen-18, one of the oxygen isotopes.
What is current measured in?
ampereAn ampere (AM-pir), or amp, is the international unit used for measuring current. It expresses the quantity of electrons (sometimes called “electrical charge”) flowing past a point in a circuit over a given time.
What M ohm means?
One megaohm is equal to 1,000,000 ohms, which is the resistance between two points of a conductor with one ampere of current at one volt. The megaohm is a multiple of the ohm, which is the SI derived unit for electrical resistance. In the metric system, “mega” is the prefix for 106.
What is a bad ohm reading?
A measurement of Zero, or very close to zero (less than .5 OHM) indicates a very low resistance to current flow. Applying voltage to this low level of resistance will result in extremely high current flow.
How many ohms are in a Milliohm?
10-3 ohmsDefinition: Milliohm So 1 milliohm = 10-3 ohms. The definition of a ohm is as follows: The ohm (symbol: Ω) is the SI unit of electrical impedance or, in the direct current case, electrical resistance, named after Georg Ohm.
Is kilo ohms bigger than ohms?
One kiloohm is equal to 1,000 ohms, which is the resistance between two points of a conductor with one ampere of current at one volt. The kiloohm is a multiple of the ohm, which is the SI derived unit for electrical resistance.
How much is a ohm?
Ohm defines the unit of resistance of “1 Ohm” as the resistance between two points in a conductor where the application of 1 volt will push 1 ampere, or 6.241×10^18 electrons. This value is usually represented in schematics with the greek letter “Ω”, which is called omega, and pronounced “ohm”.
How many ohms is a dead short?
Very low resistance — about 2 ohms or less — indicates a short circuit.
How many watts is 2 ohms?
550 wattsWith a 2 ohm speaker, the maximum output power will be 550 watts.
What does K mean in ohms?
Definition: Kiloohm The SI prefix “kilo” represents a factor of 103, or in exponential notation, 1E3. So 1 kiloohm = 103 ohms. The definition of a ohm is as follows: The ohm (symbol: Ω) is the SI unit of electrical impedance or, in the direct current case, electrical resistance, named after Georg Ohm.
How do you convert kg to ohms?
The answer is 0.001. We assume you are converting between kiloohm and ohm. You can view more details on each measurement unit: kiloOhms or Ohms The SI derived unit for electric resistance is the ohm. 1 kiloOhms is equal to 1000 ohm.
How much is 2 mega ohms?
Megohm to Ohm Conversion TableMegohmOhm1 megohm1000000 ohm2 megohm2000000 ohm3 megohm3000000 ohm5 megohm5000000 ohm7 more rows
How do you convert ohms?
Example. Calculate the resistance in ohms of a resistor when the voltage is 5 volts and the power is 2 watts. The resistance R is equal to square of 5 volts divided by 2 watts, which is equal to 12.5 ohms.
How many ohms is a Gigaohm?
1000000000 ohmYou can view more details on each measurement unit: gigaohm or ohm The SI derived unit for electric resistance is the ohm. 1 gigaohm is equal to 1000000000 ohm.
Does higher ohms mean more resistance?
Ohms are units of resistance. The lower the resistance of your atomizer, the larger the amount of current flowing through it. If you increase resistance, the atomizer receives less current. … Also, the relationship between voltage (Volts) and resistance (Ohms) is important.
What does Ohm’s law mean?
Ohm’s Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit. To students of electronics, Ohm’s Law (E = IR) is as fundamentally important as Einstein’s Relativity equation (E = mc²) is to physicists.
How do you convert ohms to Megaohms?
How to Convert Ohms to Megaohms. The electrical resistance in megaohms is equal to the ohms divided by 1,000,000.
Is mega ohms more than kilohms?
How many kiloOhms in 1 megaOhms? The answer is 1000. We assume you are converting between kiloohm and megaohm. You can view more details on each measurement unit: kiloOhms or megaOhms The SI derived unit for electric resistance is the ohm.
How do you write kilo ohms?
One thousand (103) ohms. Symbol: kΩ.
How do I calculate resistance?
The resistance R in ohms (Ω) is equal to the voltage V in volts (V) divided by the current I in amps (A): Since the current is set by the values of the voltage and resistance, the Ohm’s law formula can show that if you increase the voltage, the current will also increase.
How many ohms does a 9 volt battery have?
Though the internal resistance may be or appear low, around 0.1Ω for an AA alkaline battery, and about 1Ω to 2Ω for a 9-volt alkaline battery, it can cause a noticeable drop in output voltage if a low-resistance load is attached to it. | 1,570 | 6,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-31 | latest | en | 0.938116 |
https://istopdeath.com/determine-if-true-3e6/ | 1,675,055,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00391.warc.gz | 345,023,789 | 13,554 | # Determine if True 3e<=6
The left side is greater than the right side , which means that the given statement is false.
False
Determine if True 3e<=6 | 41 | 150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.777602 |
https://homeworktutors.net/category/uncategorized/ | 1,653,361,209,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00107.warc.gz | 367,249,008 | 16,609 | Stanford University System of Linear Equation Problems
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Question Description I need an explanation for this Algebra question to help me study. For the discussion board assignment in each unit, you will complete the problem associated with the letter that you have been assigned by the instructor. Post the entire example/word problem you have been assigned from the textbook. Then fully explain how […]
GCCCD Climate Change and Genomic Variation Questions
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Question Description I’m working on a Biology question and need guidance to help me study. 1) Climate change is probably the most serious threat to human existence on Planet Earth. The main cause of climate change is global warming. First, indicate the one molecule that alone is most responsible for the earth’s warming. Second, describe […]
Walden University Accounting Budgeting Discussion
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Question Description I’m working on a Accounting question and need guidance to help me study. Describe budgeting; it’s objective, and its impact on human behavior. Based on your reading and outside research, please communicate your own understanding of the requirements.
Chamberlain University Special Factoring Strategies Project
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Description Required Resources Read/review the following resources for this activity: OpenStax Textbook Readings Lesson in Canvas Assignments in Knewton Factoring Trinomials with a Leading Coefficient of 1 Factoring Trinomials with a Leading Coefficient Other than 1 Factoring Special Products Choosing a Factoring Strategy Solving Quadratic Equations by Factoring Solving Polynomial Equations by Factoring Initial Post […]
GCCCD Climate Change and Genomic Variation Questions
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Question Description I don’t know how to handle this Biology question and need guidance. 1) Climate change is probably the most serious threat to human existence on Planet Earth. The main cause of climate change is global warming. First, indicate the one molecule that alone is most responsible for the earth’s warming. Second, describe at […]
University of Phoenix Cash Management Strategies Discussion Questions
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Question Description I’m stuck on a Accounting question and need an explanation. Unfortunately, most firms lack liquidity. To make sure there is sufficient cash flow to meet day-to-day financial needs, financial managers adopt short-term financing methods to raise needed capital to meet debt obligations in time while maximizing profitability through investment with cash surpluses. Respond […]
Columbia Southern University Learning Tools Journal
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Description Introduce yourself to your professor with your name, location, current employment, and future goals. Part II: Throughout this course, there are many different learning aids available to assist you with understanding the material. Watch the Introduction to Learning Tools video here, and practice using the tools within your Unit I Homework Assignment. You may […]
CHC BIO 200 Organisms To Genetics Questions
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Question Description I need support with this Biology question so I can learn better. Must have prior knowledge of the following topics: – Nature & Process Of Science – Genetics -Evolution -Phylogenetics – MIcrobes -Early Land Plants -Gymnosperms -Angiosperms -Fungi -Protostomes -Deuetrostomes -Matter & Energy -Nitrogen
ASU Not for Profit Accounting Voluntary Health and Welfare Organization Report
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Description Your city has a voluntary health and welfare organization (VHWD) that provides musical opportunities for inner-city youth. It does not use fund accounting, but it does identify all revenues by their net asset class. The following transactions have occurred in the past year: The VHWD received gift pledges from donors in the amount of […]
AU MAT 120 Systems of Linear Equations and Inequalities Discussion
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Description For the discussion board assignment in each unit, you will complete the problem associated with the letter that you have been assigned by the instructor. Post the entire example/word problem you have been assigned from the textbook. Then fully explain how you would go about finding the solution. Please explain all steps in a […] | 999 | 4,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-21 | latest | en | 0.882745 |
http://assorted-experience.blogspot.com/2009/09/how-d40-handles-multi-dimensional.html | 1,532,213,256,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00199.warc.gz | 27,961,001 | 24,219 | ### How the D40 handles a multi dimensional problem
There are three dimensions to exposure:
1. Aperture
2. Shutter speed
3. Sensitivity
In film days adjusting sensitivity on the fly was not practical, so you basically had a two dimensional plane once you loaded film.
The life of the camera computer was easy when you set the camera to A or S modes: all it had to do was adjust the free variable to maintain correct exposure. In M mode the camera merely informed you of exposure letting you roam freely in this 2D space.
The D40 can be setup exactly this way if you select fixed ISO. If you select auto ISO, however, the camera now is handed two free parameters (for A and S) and one free parameter (for M). How does the D40 deal with this?
For A mode the computer minimizes ISO and maximizes shutter speed. You can set a lower limit for Shutter speed and upper limit for ISO and the camera will drop shutter speed until it hits the lower limit and the start to bump ISO.
For S mode the computer minimizes ISO. It will maximize the aperture until it reaches the lens' limit. The it bumps ISO.
This all sounds sensible, but what about M mode? This is funny. In M mode you no longer have the freedom to mess with exposure - the camera runs loose with ISO changing it to give you correct exposure regardless of your A and S combination. Eventually it hits the camera ISO limits and starts to show you over- and under-exposure.
M mode with auto-ISO will enable you to play with particular shutter (motion capturing) and aperture (DOF) combinations for your subject that would normally be inaccessible to you (cumbersome with manual ISO, impractical with film).
### Python: Multiprocessing: passing multiple arguments to a function
Write a wrapper function to unpack the arguments before calling the real function. Lambda won't work, for some strange un-Pythonic reason.
import multiprocessing as mp def myfun(a,b): print a + b def mf_wrap(args): return myfun(*args) p = mp.Pool(4) fl = [(a,b) for a in range(3) for b in range(2)] #mf_wrap = lambda args: myfun(*args) -> this sucker, though more pythonic and compact, won't work p.map(mf_wrap, fl)
### Flowing text in inkscape (Poster making)
You can flow text into arbitrary shapes in inkscape. (From a hint here).
You simply create a text box, type your text into it, create a frame with some drawing tool, select both the text box and the frame (click and shift) and then go to text->flow into frame.
UPDATE: | 564 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-30 | latest | en | 0.868014 |
https://www.hackmath.net/en/example/6228 | 1,558,434,531,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00173.warc.gz | 822,877,171 | 6,853 | # Three 43
Three brothers inherited a cash amount of 62,000 and they divided it among themselves in the ratio of 5:4:1. How much more is the largest share than the smallest share?
Result
x = 24800
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator.
## Next similar examples:
1. Three workers
Three workers got paid for work. First earn € 50, the second € 100 and third € 60. For further work should receive together € 735. How do you divide that amount in order to be distributed in the same proportion as the first job?
2. Babysitting
The amount that Susan charges per hour for babysitting are directly proportional to the number of children she is watching. She charges \$4.50 for 3 kids. How much would she charge to babysit 5 kids?
3. Peppers
In the box are yellow (a), green (b) and red (c) peppers. Their amount is in a ratio 2:4:1 . Most are yellow peppers and green the least. Calculate the number of peppers each type if the total number of peppers is 70.
4. Division of money
Calculate how many euros have Matthew, Miriam, Lucy, Michael, Janka when together have 2,700 euros and the amounts are at a ratio of 1:5:6:7:8.
5. Moneys in triple ratio
Milan, John and Lili have a total 344 euros. Their amounts are in the ratio 1:2:5. Determine how much each of them has?
6. Points
Gryffindor won 437 points. How many points obtained by each of the faculties if they were split at a ratio of 5: 7: 3: 4?
7. Building
At the building, we divided 240 boards into two piles in a 5: 3 ratio. How many were fewer boards in the lower pile?
8. 1.5 divided
1.5 divided by 1 = w divided by 4
9. Divide money
Divide 1200 USD at a ratio of 1:2:3:4:5:6:9:10
10. Land area
A land area of Asia and Africa are in a 3: 2 ratio, the European and African are is 1:3. What are the proportions of Asia, Africa, and Europe?
11. Donuts
Find how many donuts each student will receive if you share 126 donuts in a ratio of 1:5:8
12. Pamela
Lara Pamela spends her salary of 3000 for food, clothing, recreation and savings, which are in the ratio of 48:20:15:37, respectively. How much does he spend for savings?
13. The farmer
The farmer had 140 sheep. For the next year, she decided to change the number of sheep in ratio 10: 7. How many sheep will he have then?
14. Image scale
The actual image dimensions are 60 cm x 80cm and has a reduced size 3 cm x 4 cm. At what scale the image was reduced?
15. Line segment
The 4 cm long line segment is enlarged in the ratio of 5/2. How many centimeters will measure the new line segment?
16. Hour salary
You work for 4 hours on a Saturday and 8 hours on Sunday. You also receive a \$50 bonus. You earn \$164. How much did you earn per hour?
17. Ages
John, Teresa, Daniel and Paul have summary 56 years. Their ages are in a ratio of 1:2:5:6. Determine how many years have each of them. | 837 | 3,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-22 | latest | en | 0.943637 |
https://rdrr.io/cran/treebalance/man/B2I.html | 1,721,514,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00236.warc.gz | 424,379,776 | 7,796 | # B2I: Calculation of the B2 index for rooted trees In treebalance: Computation of Tree (Im)Balance Indices
B2I R Documentation
## Calculation of the B2 index for rooted trees
### Description
This function calculates the B2 index B2(T) for a given rooted tree T. The tree must not necessarily be binary. B2(T) is defined as
B2(T)=-\sum_{x\in V_L(T)} p_x\cdot log(p_x)
in which V_L(T) denotes the leaf set of T, and in which
p_x=\prod_{v\in anc(x)} \frac{1}{|child(v)|}
denotes the probability of reaching leaf x when starting at the root and assuming equiprobable branching at each vertex v\in anc(x) with anc(x) denoting the set of ancestors of x excluding x. child(v) denotes the set of children of the inner vertex v.
The B2 index is a balance index.
For n=1 the function returns B2(T)=0 and a warning.
For details on the B2 index, see also Chapter 11 in "Tree balance indices: a comprehensive survey" (https://doi.org/10.1007/978-3-031-39800-1_11).
### Usage
B2I(tree, logbase = 2)
### Arguments
tree A rooted tree in phylo format. logbase The base that shall be used for the logarithm. For binary trees it is common to use base 2.
### Value
B2I returns the B2 index of the given tree.
### Author(s)
Sophie Kersting, Luise Kuehn
### References
K.-T. Shao and R. R. Sokal. Tree Balance. Systematic Zoology, 39(3):266, 1990.
doi: 10.2307/2992186.
P.-M. Agapow and A. Purvis. Power of Eight Tree Shape Statistics to Detect Nonrandom Diversification: A Comparison by Simulation of Two Models of Cladogenesis. Systematic Biology, 51(6):866-872, 2002.doi: 10.1080/10635150290102564.
URL https://doi.org/10.1080/10635150290102564.
M. Hayati, B. Shadgar, and L. Chindelevitch. A new resolution function to evaluate tree shape statistics. PLOS ONE, 14(11):e0224197, 2019. doi: 10.1371/journal.pone.0224197.
URL https://doi.org/10.1371/journal.pone.0224197.
M. Kirkpatrick and M. Slatkin. Searching for evolutionary patterns in the shape of a phylogenetic tree. Evolution, 47(4):1171-1181, 1993. doi: 10.1111/j.1558-5646.1993.tb02144.x.
### Examples
tree <- ape::read.tree(text="((((,),),(,)),(((,),),(,)));")
B2I(tree)
treebalance documentation built on May 29, 2024, 1:15 a.m. | 686 | 2,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.647638 |
moose.londonderry.org | 1,726,635,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00738.warc.gz | 370,896,440 | 24,172 | # Early Math Skills
Preschool Math
I’ve listed some websites below that you can use as resources. They have some great activities or games you can do with your child. Other skills that we focus on at school include: patterning, sorting by attributes, 1-to-1 correspondence, numeral recognition, simple addition and subtraction, naming and recognizing two- and three-dimensional shapes and providing opportunities and materials that build a foundation for understanding economic concepts (e.g., playing restaurant, running a toy store, identifying and exchanging money, etc.). We strongly encourage teaching children new skills through hands on activities and manipulatives (small objects such as counting bears, buttons, cubes, beans, etc.). Once they understand the concept and are consistently able to demonstrate the skill at a concrete level, we introduce activities which encourage the use of mental images or pictures of real objects. The highest level for our preschoolers would be functioning at an abstract level. This would involve very basic and simple skills such as 1+1=2 (without using fingers or thinking of objects).
Please take some time and check out a few of the websites listed below and let me know what you think. If you have some suggestions for math activities or websites that you would like to share with other families, please let me know. Remember to make working with math fun and interesting. Unless you have a learner who loves worksheet, provide hands on materials that your child can feel and touch. By doing so, you’re utilizing different parts of the brain and building a stronger experience to store in memory. So, remember what Mickey Mouse said, "Arithmetic is being able to count up to twenty without taking off your shoes."
Parent Resources
Articles Related to Preschool Math Skills
Understanding numbers and counting skills:
Why early math is just as important as early reading:
https://www.greatschools.org/gk/articles/early-math-equals-future-success/
Review important concepts and explore new topics—the options are endless with Education.com! Join for free today and browse 30,000+ worksheets, online games, lesson plans and more.
https://www.education.com/games/preschool/math/
Math 4 Children - Lots of Resources
Math 4 Children
Websites for Kids
Cool Math 4 Kids – Games for kids
https://math4children.com/
Edhelper.com- Math Activities for Kids
Ed Helper
Jumpstart Use the Free Online Learning Games Section- Recommended by USA Today
Jump Start
Math Foundations
http://www.topmarks.co.uk/Interactive.aspx?cat=1
Number Theme Preschool Activities and Crafts - Go down to: Online Story Time and Activities
Online number themes
Online Math Learning Includes other areas of learning too!
Online Math Learning | 566 | 2,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.936457 |
https://www.infocomm.ky/indicative-quote-definition/ | 1,718,780,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00254.warc.gz | 729,077,902 | 11,119 | # Indicative Quote Definition.
An indicative quote is a type of quote that is used to provide information about the current market price of a financial instrument, without necessarily committing to a trade. Indicative quotes are typically used by market makers to help them assess the level of interest in a particular instrument, and they are also often used by traders as a way to gauge market activity and potential trading opportunities.
#### How do you calculate indicative value?
The first step is to find the current spot price of the currency pair. You can do this by looking at a currency quote or using a currency converter.
Next, you will need to find the current forward price of the currency pair. The forward price is the price at which the currency pair can be bought or sold at a future date. Forward prices are usually quoted for delivery two business days from the current date.
Once you have both the spot and forward price, you can calculate the indicative value using the following formula:
Indicative Value = Spot Price x (1 + Forward Points/10000)
where Forward Points are the number of pips between the spot and forward price.
For example, if the spot price of EUR/USD is 1.1000 and the forward price is 1.1010, the indicative value would be:
Indicative Value = 1.1000 x (1 + 10/10000)
Indicative Value = 1.1001
### What is an indicative proposal?
An indicative proposal is a type of proposal that is used to provide indicative pricing information to a potential customer. It is not a binding proposal, and is not usually used to secure a deal. Instead, it is used to provide an indication of what the company is willing to offer, and to gauge the level of interest from the potential customer.
##### How is indicative price determined?
Indicative pricing is the price of a security or financial instrument as quoted by a market maker at which that market maker is willing to buy or sell the security or instrument. It is not a binding price, but rather a guidepost for further negotiation. What does far indicative clearing price mean? The far indicative clearing price is the price at which the market is expected to clear at the end of the day. This price is based on the current market conditions and is used by traders to gauge where the market is likely to head in the future. What does firm quote mean? A firm quote is a quote that is not subject to change. In the forex market, a firm quote is typically given for large transactions and is not negotiable. | 522 | 2,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-26 | latest | en | 0.940551 |
http://www.exploredatabase.com/2018/03/find-number-of-disk-blocks-in-given-hard-disk-dbms.html | 1,581,931,348,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141806.26/warc/CC-MAIN-20200217085334-20200217115334-00036.warc.gz | 194,988,523 | 20,683 | ## TOPICS (Click to Navigate)
` I request GUEST WRITERS for this web site. Anyone interested please contact me at saravu2k2013@gmail.com Thanks `
## Find the number of disk blocks in a given hard disk in DBMS data storage
Question:
Following are the characteristics of a hard disk; Sector size of 512 bytes, 16 sectors per track, 16384 tracks per surface, 4 double sided platter, and 4096 bytes per block. Find the number of blocks in that disk.
Solution:
Let us find the total capacity of given disk in bytes.
A track has 16 sectors each of which is 512 bytes in size.
Capacity of a track = Sector size * Number of sectors
= 512 * 16 = 8192 bytes.
There are 16384 tracks in a surface of a disk platter.
Capacity of one surface of disk platter = No. of tracks * Capacity of a track
= 16384 * 8192 = 134217728 bytes
Capacity of one double sided platter = 134217728 * 2 (2 sides per disk)
= 268435456 bytes
There are 4 double sided platters.
Capacity of the hard disk = Capacity of one disk platter * No. of disk platters
= 268435456 * 4
= 1073741824 bytes
It is given that the block size is 4096.
The number of blocks = (capacity of the hard disk) / (block size)
= 1073741824/4096 = 262144 blocks
The hard disk with the given characteristics will have 262144 blocks.
*************
find number of blocks in a hard disk drive
calculate the size of hard disk in bytes
find the capacity of a disk platter | 380 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-10 | longest | en | 0.824445 |
http://topengineeringcollegesintamilnadu.blogspot.com/2011/12/signals-and-systems-jntu-previous-years.html?widgetType=BlogArchive&widgetId=BlogArchive2&action=toggle&dir=open&toggle=DAILY-1266048000000&toggleopen=DAILY-1324108800000 | 1,582,503,013,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00053.warc.gz | 143,502,319 | 28,808 | ## December 17, 2011
### SIGNALS AND SYSTEMS JNTU previous years question papers
SIGNALS AND SYSTEMS JNTU previous years question papers
Time: 3 hours Max Marks: 75
All Questions carry equal marks
1. (a) Evaluate the following integrals:
i.
R8
&1
[u(t + 3) & 2f(t):u(t)]dt
ii.
5
R2
1
2
f(3t)dt
(b) A even function g(t) is described by
g(t) =
8<
:
2t
15 & 3t
&2
0 f t < 3
3 f t < 7
7 f t < 10
i. What is the val ue of g(t) at time t = 5
ii. What is the val ue of 1st derivative of g(t) at time t = 6. [8+7]
2. (a) Distinguish between Energy and Power signals.
(b) Derive the expression for Energy density spectrum function of a energy signal f(t) from fundamentals and interpret why it is called Energy density spectrum.[5+10]
3. (a) Explain the concept of generalizedFourier series representation of signal f(t).
(b) State the properties ofFourier series. [8+7]
4. (a) Explain the properties of the ROC of Z transforms.
(b) Z transform of a signal x(n) ifX(z) = 1+z&1
1+ 1
3z&1 .
Use long division method to determine the val ues of
i. x[0], x[1], and x[2], assuming the ROC to be jzj > 1
3
ii. x[0], x[-1], and x[-2] , assuming the ROC to be jzj < 1
3 . [7+8]
5. (a) A signal y(t) given by y(t) = C0 +
1P
n=1
CnCos(n!0t + fn). Find the autocorrelation and PSD of y(t).
(b) Explain the Graphical representation of convolution with an example. [8+7]
6. (a) Consider anLTI system with input and output related through the equation.
y(t) =
Rt
&f
e(t&f)x(f & 2)dfWhat is the impulse response h(t) for this system.
(b) Determine the response of this system when the input x(t) is as shown
(c) Consider the inter connection ofLTI system depicted in fgure 6c.
Here h(t) is an in part (a). Determine the output y(t) when input x(t) is again given fgure above, using the convolution integral. [5+5+5]
7. (a) Consider the signal x(t) = (sin 50 ft / ft)2 which to be sampled with a sampling frequency of !s = 150 f to obtain a signal g(t) withFourier transform G(j! ). Determine the maximum val ue of !0 for which it is guaranteed that
G(j!) = 75 X(j!) for j!j (b) The signal x(t) = u(t + T0) - u(t - T0) can undergo impulse train sampling without aliasing, provided that the sampling period T< 2T0 . Justify.[7+8]
8. (a) Explain the method of determining the inverse Laplace transforms using Partial fraction method, for the following cases
i. Simple and real roots
ii. Complex roots
iii. Multiple or repeated roots.
(b) Find the Laplace transform of the function
f(t) = A Sin !0t for 0 < t < T/2. [3+3+4+5] | 816 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-10 | latest | en | 0.824835 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_Languages/Book%3A_Think_Java_-_How_to_Think_Like_a_Computer_Scientist/04%3A_Objects_of_Arrays/4.02%3A_Shuffling_Decks | 1,627,653,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00121.warc.gz | 253,784,783 | 22,356 | # 4.2: Shuffling Decks
For most card games you need to be able to shuffle the deck; that is, put the cards in a random order. In Section 8.7 we saw how to generate random numbers, but it is not obvious how to use them to shuffle a deck.
One possibility is to model the way humans shuffle, which is usually dividing the deck in two halves and then choosing alternately from each one. Since humans usually don’t shuffle perfectly, after about seven iterations the order of the deck is pretty well randomized.
But a computer program would have the annoying property of doing a perfect shuffle every time, which is not very random. In fact, after eight perfect shuffles, you would find the deck back in the order you started in! (For more information, see https://en.Wikipedia.org/wiki/Faro_shuffle.)
A better shuffling algorithm is to traverse the deck one card at a time, and at each iteration choose two cards and swap them. Here is an outline of how this algorithm works. To sketch the program, we will use a combination of Java statements and English. This technique is sometimes called pseudocode.
for each index i {
// choose a random number between i and length - 1
// swap the ith card and the randomly-chosen card
}
The nice thing about pseudocode is that it often makes clear what methods you are going to need. In this case, we need a method that chooses a random integer between low and high, and a method that takes two indexes and swaps the cards at those positions. Methods like these are called helper methods, because they help you implement more complex algorithms.
And this process – writing pseudocode first and then writing methods to make it work – is called top-down development (see https://en.Wikipedia.org/wiki/Top-down_and_bottom-up_design).
One of the exercises at the end of the chapter asks you to write the helper methods randomInt and swapCards and use them to implement shuffle. | 407 | 1,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-31 | latest | en | 0.942679 |
https://nz.education.com/activity/article/Cause_Effect_Card_middle/ | 1,591,330,509,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00176.warc.gz | 461,034,886 | 30,710 | # Cause and Effect Card Game Activity
### What You Need:
• Pencils, pens or markers
• Sturdy paper (such as cardstock or construction)
• Scissors
### What You Do:
• Before diving in, review the principle of cause and effect with your child. Remind her that writers use cause and effect to show the relationship between two facts, concepts, or events, in which one is the result of the other (or others). Give her some examples to put the concept into context. For example, She got 100 percent on the test (effect)Because she studied hard (cause). He had A stomachache (effect) Because he ate junk food (cause).
• With your child, fold two sheets of sturdy paper down the middle lengthwise. Draw a line on the fold to split the paper into two columns, and label one column Effect and the other column Cause. Separately, write down 10 examples of cause and effect relationships, leaving enough space to cut each situation into a card. Some examples could include:
Effect Cause Aiden was not able to go to soccer today Because he had the flu. Mike spent all day in the kitchen Because all of his friends wanted him to make dinner for them.
• Cut up each piece of paper into cards, so that each cause and each effect has its own card. There should be 20 cards total. Shuffle the cards together, and then lay them out face down. Draw 7 cards each, and leave the rest in a stack between you and your child. Flip the card on the top of the stack, and put it face-up next to the stack. You and your kid should both be able to read it.
• Play the game following the rules below.
GAME RULES
• On their turn, players have the option of picking up the face-up card or choosing a new card from the stack. The object of each turn is to match a “cause” card with an “effect” card.
• Players must discard one card into the face-up stack after each turn. Players should always have 7 cards in their hands.
• When a player thinks she’s made a match, she must show the matching cards to her opponent. If the match makes sense, the match will be approved. If not, then she’ll have to try for a new match during her next turn.
• The game ends when one player has no cards left. The player with the most matches at this point wins.
To determine what makes a match, read the cause and effect together. If they make sense, it counts! Two cards that don’t represent the original cause and effect pair can be a match.
• Match: Aiden was not able to go to soccer today/because he needed to make dinner for all of his friends. This counts as a match, even if it's not the original cause and effect pair.
• Not a match:Mike spent all day in the kitchen/because he had the flu. This pair doesn’t count as a match because it doesn’t make sense.
This game offers a fun, hands-on way to practise this skill that will help in the classroom and beyond!
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0
### New Collection>
0Items
What could we do to improve Education.com? | 669 | 2,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.954063 |
https://number.academy/224229 | 1,660,776,142,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00125.warc.gz | 395,153,003 | 12,226 | # Number 224229
Number 224,229 spell 🔊, write in words: two hundred and twenty-four thousand, two hundred and twenty-nine . Ordinal number 224229th is said 🔊 and write: two hundred and twenty-four thousand, two hundred and twenty-ninth. Color #224229. The meaning of number 224229 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 224229. What is 224229 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 224229.
## What is 224,229 in other units
The decimal (Arabic) number 224229 converted to a Roman number is (C)(C)(X)(X)(IV)CCXXIX. Roman and decimal number conversions.
#### Weight conversion
224229 kilograms (kg) = 494335.3 pounds (lbs)
224229 pounds (lbs) = 101709.6 kilograms (kg)
#### Length conversion
224229 kilometers (km) equals to 139330 miles (mi).
224229 miles (mi) equals to 360862 kilometers (km).
224229 meters (m) equals to 735651 feet (ft).
224229 feet (ft) equals 68346 meters (m).
224229 centimeters (cm) equals to 88279.1 inches (in).
224229 inches (in) equals to 569541.7 centimeters (cm).
#### Temperature conversion
224229° Fahrenheit (°F) equals to 124553.9° Celsius (°C)
224229° Celsius (°C) equals to 403644.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
224229 seconds equals to 2 days, 14 hours, 17 minutes, 9 seconds
224229 minutes equals to 5 months, 2 weeks, 1 day, 17 hours, 9 minutes
### Codes and images of the number 224229
Number 224229 morse code: ..--- ..--- ....- ..--- ..--- ----.
Sign language for number 224229:
Number 224229 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. 224229
### Multiplications
#### Multiplication table of 224229
224229 multiplied by two equals 448458 (224229 x 2 = 448458).
224229 multiplied by three equals 672687 (224229 x 3 = 672687).
224229 multiplied by four equals 896916 (224229 x 4 = 896916).
224229 multiplied by five equals 1121145 (224229 x 5 = 1121145).
224229 multiplied by six equals 1345374 (224229 x 6 = 1345374).
224229 multiplied by seven equals 1569603 (224229 x 7 = 1569603).
224229 multiplied by eight equals 1793832 (224229 x 8 = 1793832).
224229 multiplied by nine equals 2018061 (224229 x 9 = 2018061).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 224229
Half of 224229 is 112114,5 (224229 / 2 = 112114,5 = 112114 1/2).
One third of 224229 is 74743 (224229 / 3 = 74743).
One quarter of 224229 is 56057,25 (224229 / 4 = 56057,25 = 56057 1/4).
One fifth of 224229 is 44845,8 (224229 / 5 = 44845,8 = 44845 4/5).
One sixth of 224229 is 37371,5 (224229 / 6 = 37371,5 = 37371 1/2).
One seventh of 224229 is 32032,7143 (224229 / 7 = 32032,7143 = 32032 5/7).
One eighth of 224229 is 28028,625 (224229 / 8 = 28028,625 = 28028 5/8).
One ninth of 224229 is 24914,3333 (224229 / 9 = 24914,3333 = 24914 1/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
224229
#### Is Prime?
The number 224229 is not a prime number. The closest prime numbers are 224221, 224233.
#### Factorization and factors (dividers)
The prime factors of 224229 are 3 * 41 * 1823
The factors of 224229 are 1 , 3 , 41 , 123 , 1823 , 5469 , 74743 , 224229
Total factors 8.
Sum of factors 306432 (82203).
#### Powers
The second power of 2242292 is 50.278.644.441.
The third power of 2242293 is 11.273.930.164.360.988.
#### Roots
The square root √224229 is 473,528246.
The cube root of 3224229 is 60,752468.
#### Logarithms
The natural logarithm of No. ln 224229 = loge 224229 = 12,320423.
The logarithm to base 10 of No. log10 224229 = 5,350692.
The Napierian logarithm of No. log1/e 224229 = -12,320423.
### Trigonometric functions
The cosine of 224229 is 0,568642.
The sine of 224229 is 0,822585.
The tangent of 224229 is 1,446579.
### Properties of the number 224229
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 224229 in Computer Science
Code typeCode value
PIN 224229 It's recommendable to use 224229 as a password or PIN.
224229 Number of bytes219.0KB
CSS Color
#224229 hexadecimal to red, green and blue (RGB) (34, 66, 41)
Unix timeUnix time 224229 is equal to Saturday Jan. 3, 1970, 2:17:09 p.m. GMT
IPv4, IPv6Number 224229 internet address in dotted format v4 0.3.107.229, v6 ::3:6be5
224229 Decimal = 110110101111100101 Binary
224229 Decimal = 102101120210 Ternary
224229 Decimal = 665745 Octal
224229 Decimal = 36BE5 Hexadecimal (0x36be5 hex)
224229 BASE64MjI0MjI5
224229 MD53751525100fa1880bc8bfe268fe1821a
224229 SHA16505a4d5f45fa82e47558585168062f597681016
224229 SHA224e072b21ac690c8daa534b1d65618ee9e05c3c129afa7f772a25b092d
224229 SHA256ab2765028d260e4e4fff21b312e9befac40265a85e917554cbfb0b225423938f
224229 SHA3841511942852797c903f72aac7b80f610b0e91342f72e25de6a8d660fd9610d7ed308ebd0e2d763878a2be1c671019ac53
More SHA codes related to the number 224229 ...
If you know something interesting about the 224229 number that you did not find on this page, do not hesitate to write us here.
## Numerology 224229
### Character frequency in number 224229
Character (importance) frequency for numerology.
Character: Frequency: 2 4 4 1 9 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 224229, the numbers 2+2+4+2+2+9 = 2+1 = 3 are added and the meaning of the number 3 is sought.
## Interesting facts about the number 224229
### Asteroids
• (224229) 2005 SH66 is asteroid number 224229. It was discovered by NEAT, Near Earth Asteroid Tracking from Palomar Mountains Observatory on 9/26/2005.
## № 224,229 in other languages
How to say or write the number two hundred and twenty-four thousand, two hundred and twenty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 224.229) doscientos veinticuatro mil doscientos veintinueve German: 🔊 (Anzahl 224.229) zweihundertvierundzwanzigtausendzweihundertneunundzwanzig French: 🔊 (nombre 224 229) deux cent vingt-quatre mille deux cent vingt-neuf Portuguese: 🔊 (número 224 229) duzentos e vinte e quatro mil, duzentos e vinte e nove Chinese: 🔊 (数 224 229) 二十二万四千二百二十九 Arabian: 🔊 (عدد 224,229) مئتان و أربعة و عشرون ألفاً و مئتان و تسعة و عشرون Czech: 🔊 (číslo 224 229) dvěstě dvacet čtyři tisíce dvěstě dvacet devět Korean: 🔊 (번호 224,229) 이십이만 사천이백이십구 Danish: 🔊 (nummer 224 229) tohundrede og fireogtyvetusindtohundrede og niogtyve Dutch: 🔊 (nummer 224 229) tweehonderdvierentwintigduizendtweehonderdnegenentwintig Japanese: 🔊 (数 224,229) 二十二万四千二百二十九 Indonesian: 🔊 (jumlah 224.229) dua ratus dua puluh empat ribu dua ratus dua puluh sembilan Italian: 🔊 (numero 224 229) duecentoventiquattromiladuecentoventinove Norwegian: 🔊 (nummer 224 229) to hundre og tjue-fire tusen, to hundre og tjue-ni Polish: 🔊 (liczba 224 229) dwieście dwadzieścia cztery tysiące dwieście dwadzieścia dziewięć Russian: 🔊 (номер 224 229) двести двадцать четыре тысячи двести двадцать девять Turkish: 🔊 (numara 224,229) ikiyüzyirmidörtbinikiyüzyirmidokuz Thai: 🔊 (จำนวน 224 229) สองแสนสองหมื่นสี่พันสองร้อยยี่สิบเก้า Ukrainian: 🔊 (номер 224 229) двiстi двадцять чотири тисячi двiстi двадцять дев'ять Vietnamese: 🔊 (con số 224.229) hai trăm hai mươi bốn nghìn hai trăm hai mươi chín Other languages ...
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If you know something interesting about the number 224229 or any natural number (positive integer) please write us here or on facebook. | 2,641 | 7,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.662197 |
https://forum.allaboutcircuits.com/threads/antenna-wattage.190679/ | 1,679,901,355,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00749.warc.gz | 325,396,698 | 21,596 | # Antenna wattage
#### Track99
Joined Jun 30, 2022
42
Hello my friends.
I have a transceiver system.
Inside the datasheet, under antenna data, it shows that the Power is 50 Watts Input.
1) What is this 50 Watts input power mean? Does it mean that at peak performance the antenna can withstand 50 Watts input without damaging itself?
2) What will happen if I use a antenna that is rated for 5 Watts?
#### Papabravo
Joined Feb 24, 2006
19,850
Hello my friends.
I have a transceiver system.
Inside the datasheet, under antenna data, it shows that the Power is 50 Watts Input.
1) What is this 50 Watts input power mean? Does it mean that at peak performance the antenna can withstand 50 Watts input without damaging itself?
2) What will happen if I use a antenna that is rated for 5 Watts?
Can you provide a link to the datasheet? I'm not sure I understand your description of what is in the datasheet. There is a quantity called ERP, which stands for Effective Radiated Power. Is there any possible connection?
#### ronsimpson
Joined Oct 7, 2019
2,617
On my antennas there is a "max power".
What is the transceiver? What is its power output?
On the receiving side these numbers do not matter. On transmitting it does matter.
#### vu2nan
Joined Sep 11, 2014
285
Antenna input power is the product of the square of the antenna current and the antenna resistance at the point where the current is measured.
Likewise, it is the ratio of the square of the antenna voltage to the antenna resistance at the point where the current is measured.
Consider an antenna with an impedance of 50 Ω at its feed point. Inputting 50 W will entail applying 50 V and driving 1A at its feed point.
An antenna with a maximum input power of 50 W means that it can safely withstand 50V, 1 A at its feed point without getting damaged.
Nandu.
Last edited:
#### crutschow
Joined Mar 14, 2008
31,526
So basically if you use an antenna beyond it's rated transmitter power rating, the dielectric isolating the connections may deteriorate or arc over.
#### Track99
Joined Jun 30, 2022
42
So basically if you use an antenna beyond it's rated transmitter power rating, the dielectric isolating the connections may deteriorate or arc over.
You answered my question my friend. Ty
#### Janis59
Joined Aug 21, 2017
1,576
N=V^2/R thus 50=V^2/50 or V^2=250 means V=12 V. Gueass the thjree times is it very much to antenna or not
N=i^2*R thus 50=i^2*50 0r i^2=1 what means i=1 Ampere. Guess again, is it very much or not for Your resonator cross section.
RE:""What will happen if I use a antenna that is rated for 5 Watts"" - It probably have sth made from very thin wire what may overheat. Sure 11 Volts will not cause any arcing. But smoke may flow out.
#### ronsimpson
Joined Oct 7, 2019
2,617
I have worked on antennas that we put 50,000 watts into. These antennas are out in the rain and snow. Life at the top of a tower is not like it is in your house. I had problems, when a lightning storm is close the elements would spark over. At 100mhz, with a charge cloud over head, and high humidity the electrons will jump much farther than you think.
#### Janis59
Joined Aug 21, 2017
1,576
50 000=V^2/50 thus V^2=2 50 00 00 or V=1100 Volts. Of course the least possibility and arcing begins.
50 000=i^2*50 thus i^2=1000 or i=33 Amperes. Impressively!
Had seen the tree leafs near the antenna park of transmitter radiating about 200 kW. The leafs was vibrating and releasing the saund so it was possible to hear from it what dictor speaks in the studio. Awful! | 942 | 3,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.914441 |
http://openstudy.com/updates/52dc258de4b003c643a028d1 | 1,448,754,591,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398454553.89/warc/CC-MAIN-20151124205414-00184-ip-10-71-132-137.ec2.internal.warc.gz | 170,690,423 | 9,970 | ## BlveXO one year ago Graphing an Equation Using the Slope-Intercept Form Manipulate the equation into slope-intercept form, y = mx + b. Identify and plot the y-intercept of the line. Remember, the y-intercept is b. Don’t forget: The sign goes with the number. Identify and use the slope of the line to find a second point. Remember, the slope is m. Don’t forget: Starting at the y-intercept, the numerator tells you the rise (count up if positive and down if negative), the denominator tells you the run (count right.) Draw a straight line through the two points to complete the graph.
1. BlveXO
This is my last assignment.
2. DontLikeMathButOhWell
So it's just telling you to graph any Equation in Slope int form?
3. BlveXO
Yeah, I believe so. | 190 | 752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2015-48 | longest | en | 0.834216 |
http://stackoverflow.com/questions/2481587/if-i-xor-2-numbers-do-i-only-get-identical-results-if-the-numbers-are-the-same?answertab=oldest | 1,432,626,374,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928817.29/warc/CC-MAIN-20150521113208-00038-ip-10-180-206-219.ec2.internal.warc.gz | 236,682,070 | 17,887 | # If I XOR 2 numbers, do I only get identical results if the numbers are the same?
For example, suppose I have `x XOR y = y XOR x = z`. Is it possible to have something like `a XOR b = z`?
-
Of course it is!!! – mjv Mar 20 '10 at 1:51
However, if you have x XOR y = z and x XOR w = z then w = y – George Phillips Mar 20 '10 at 1:58
Yes.
`z = y` because `x ^ y ^ x = y`
So it is entirely possible for a combination `a ^ b = y = z`.
In fact, for every `a` there exists a `b` such that `a ^ b = z`. To calculate that, `b = z ^ a`.
Be aware that XOR is commutative: this means that `x ^ y` = `y ^ x`.
-
Yes. As a degenerate proof, XORing a number with itself always results in 0.
-
It depends how many bits are in x, y, and z. In general, for every x, x XOR y = z for y = z XOR x.
-
XOR, will return true if both parameters are different, assuming that the parameters are Boolean values anyway. This is different from or, which will return true if either parameter is true, and NOR, which will return true only if both of them are false.
-
Long answer: XOR is a binary operation, it works on the individual bits and it's commutative.
It has the truth table:
``````A B Q
0 0 0
0 1 1
1 0 1
1 1 0
``````
As the number is made up of these bits then the result will be the same as long as for each bit position the two bits have the same result. For example take the 2 eight bit numbers 113 and 42
``````113 = 01110001
42 = 00101010
XOR = 01011011 = 91
``````
but if I swap the fourth bit from the left I get
``````97 = 01100001
58 = 00111010
XOR = 01011011 = 91
``````
So yes again...
-
nice clear demonstration of a concrete example – PP. Mar 20 '10 at 2:09 | 530 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2015-22 | latest | en | 0.854375 |
https://cpep.org/mathematics/1885165-a-playground-area-is-circular-with-a-diameter-of-32-feet-what-is-the-a.html | 1,716,813,896,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00440.warc.gz | 156,918,752 | 7,243 | 7 May, 04:17
A playground area is circular with a diameter of 32 feet. what is the area and circumference of the playground?
+1
1. 7 May, 04:40
0
Area:803.84ft^2
Circumference:100.48ft | 66 | 187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.906726 |
https://www.physicsforums.com/threads/volume-of-water-a-ship-must-displace-to-float.591259/ | 1,508,425,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00206.warc.gz | 963,717,125 | 15,462 | # Volume of water a ship must displace to float
1. Mar 28, 2012
### chaotiiic
1. The problem statement, all variables and given/known data
What volume of water must a ship that masses 2.00x10^5 kg displace to float?
2. Relevant equations
density = mass/volume
density of water = 1.00x10^3
3. The attempt at a solution
200,000/1000 = 200 m^3
im guessing that in order to float you must displace your own volume. ive read other answer you have to displace your own mass
2. Mar 28, 2012
### Nessdude14
Think of a free body diagram of a ship floating on water. There's a force from the weight of the ship, and since the ship isn't sinking or rising, the force of buoyancy from the water must equal that of the weight of the ship.
The equation for buoyant force is: Fb=pfVfg
In your case, pf is the density of water, and Vf is the volume of water displaced by the ship (your unknown variable.)
Find the volume of displaced water which makes this buoyant force equal to the weight of the ship, and you're done. Hope this helps.
3. Mar 28, 2012
### chaotiiic
so is it
g*(2.00x10^5) = (1.00x10^3)*V*g
g cancels
V = 200,000/1000 = 200m^3
4. Mar 28, 2012
### Nessdude14
Looks good. One thing you need to be sure of is your units on the water density. The density you used was in kg/m^3 which happens to be just what you needed for your problem to come to an answer of m^3. Always work through the units along with the numbers.
5. Mar 28, 2012
ok thanks | 430 | 1,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.936827 |
https://edu-answer.com/mathematics/question3099479 | 1,627,895,434,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00371.warc.gz | 234,974,180 | 16,715 | , 08.11.2019 12:31 Thomas7785
# Which graph represents the equation y = -1 2 x + 3 ? a) a b) b c) c d) d
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Find an equation for the linear function which has slope -10 and x-intercept -7
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The fraction of defective integrated circuits produced in a photolithography process is being studied. a random sample of 300 circuits is tested, revealing 17 defectives. (a) calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. round the answers to 4 decimal places. less-than-or-equal-to p less-than-or-equal-to (b) calculate a 95% upper confidence bound on the fraction of defective circuits. round the answer to 4 decimal places. p less-than-or-equal-to | 215 | 805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.836784 |
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/countif/how-to-use-a-countif-function-in-google-spreadsheets | 1,720,924,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00110.warc.gz | 684,120,585 | 19,248 | Get instant live expert help with Excel or Google Sheets
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The COUNTIF function in Google Sheets is used to count the number of times a value is found in a selected data range that meets the specified criteria. It is basically a combination of the IF function and COUNT function.
## Using the COUNTIF function in Google Sheets
The COUNTIF function is used where specified criterion is single. It can be divided into two parts for further clarification;
• IF portion of function supplies the criteria to meet for cells or data to count
• COUNT portion of function totals the number of cells that meet that specified criteria
### Basic formula for COUNTIF
The syntax for the COUNTIF function is:
`=COUNTIF(range, criterion)`
Where,
Range is group of cells that we need to search for. Criterion can be any of any of the followings;
• A Number, such as 100
• A text data or text string, such as “Grapes”
• A cell reference where data is located in sheet, such as A1
• An expression, such as “<=100”
There are some rules to follow depending upon data range contains number or text values while specifying the criterion in the COUNTIF function.
If the data range contains numbers:
• Comparison operators, such as > (greater than), >= (greater than equal to), < (less than), <> (not equal to) are used in expression to check the criterion with number. The expression must be enclosed in double quotation marks, such as “<=100”
• Equal sign (=) is not used in an expression to check equal values, such as equal to 100
• In expressions using comparison operators and cell references, the comparison operator is joined to the cell reference using the ampersand (&).
• Comparison operators must be enclosed in double quotation marks, whereas cell references are not enclosed in double quotation marks, such as “>=”&A1
If the data range contains numbers:
• Text string must be enclosed in double quotation marks, such as “Grapes”
• Text strings can contain “?” and “*” wildcard characters. To match one character “?” wildcard is used, and to match multiple contiguous characters “*” wildcard is used, such as “P?T” or “*es”
### How to enter the COUNTIF function in Google Sheets
When you enter any function in Google sheets, an auto-suggest box pops up to follow that function syntax instead of a dialog box in Excel. Like when we will enter COUNTIF function in Google Sheets auto-suggest box will popup; containing syntax, example, summary related to COUNTIF function and explanation of each part of function as shown below.
1. Keep your cursor in a cell where you want to show your calculations, like cell E1
2. Enter “=” sign and type COUNTIF function in this active cell E1, an auto-suggested box pops up
3. Select the range of cells which you want to text again your criterion
4. Then enter comma “,”
5. Enter criterion expression to apply at selected range
6. Finally, press the ENTER key on the keyboard to enter the closing bracket.
Suppose we have sales dataset of some inventory items and we want to use COUNTIF function in Google Sheets to perform the following calculation on this data set.
### Count of Sales Greater Than \$400
Now we want to test the sales range of B2:B9 again this criterion expression of “>400”. As data range contains numeric values, so we will place the express in double quotation marks, as shown below
You can see the COUNTIF function has tested the selected sales data and counted the values against specified criterion where sales figures are greater than \$400 and returned the result as 2
`=COUNTIF(B2:B9,">400")`
Now we will use comparison operator and cell reference in criterion expression to count sales that are greater than the value mentioned in cell reference, such as D2.
We will join comparison operator and cell reference by using an ampersand (&). Comparison operator will be placed in double quotation marks, whereas cell references will not be enclosed in double quotation marks, such as “>”&D2
`=COUNTIF(B2:B9,">"&D2)`
### Count of Sales Equal to
IF we want to count the sales figure found in a data range that is exactly equal to a specified value, say \$200, the COUNTIF function will be as follows;
`=COUNTIF(B2:B9,200)`
Here, we will not use the equal sign “=” as comparison operator in criterion expression and criterion numeric value will not be enclosed in double quotation marks.
### Count of Text value or string
We can count the number of times a text value or text string appears in data range in Google Sheets using COUNTIF functions;
`=COUNTIF(B2:B16,"Utensil")`
Criterion text value will be enclosed in double quotation marks, and cell reference of that value will be used in the formula without double quotation marks, like;
`=COUNTIF(B2:B16,D2)`
Here you can see that COUNTIF has returned the output as number of times (5) a text value is found in data range to test for.
### Count of Text value using Wildcard
You can count any instance of a text value in a data range using wildcards, like “*” and “?”. Wildcard search is not case sensitive and it will count any instance of text value, like
`=COUNTIF(A2:A9,"*Bottled*")`
As criterion text value can be found anywhere in a text string of selected range, then we will place “*” wildcard in start and end of this text value.
IF you have to count the occurrence of text value only at the end of a text string in the data range, then the formula would be;
`=COUNTIF(A2:A9,"*Soda")`
Still need some help with Excel formatting or have other questions about Excel? Connect with a live Excel expert here for some 1 on 1 help. Your first session is always free.
Related blogs | 1,358 | 5,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.824584 |
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## Problem Definition
Assume we have two trains moving on the same track. These trains are approaching each other head-on meaning that Train A is moving towards Train B & vice versa. The track these trains are on has a distance of 120 km. Train A has a speed of 26.6 km/hour and Train B has a speed of 19.7 km/hour.
Now on top of Train A, there is a little bird that is flying with a speed of 33.4 km/hour. The bird flies from Train A to Train B and back. He keeps doing this until both trains approach each other.
The question is: How much distance has the bird flown once both trains approached each other?
Do you have questions on this case? Ask our community!
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Do you have questions on this case? Ask our community! | 221 | 887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-39 | latest | en | 0.955579 |
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### New Book Reviews!
The Memory Principle - Computer Memory and Pigeonholes
Friday, 22 January 2016
Article Index
The Memory Principle - Computer Memory and Pigeonholes
The Memory Principle
## The Memory Principle
To understand the memory principle it we have to go back to my six-line Fortran program and think a little about how it might work. Well actually Fortran is a little advanced. What we really need is machine code and we might as well use Babbage’s.
His machine stored instructions on punched cards as a pattern of holes. Each card was a single instruction. Now consider a card that holds the instruction:
`Store 10 in B`
This looks superficially like the instruction to place a pigeon in a particular hole but it is more than this.
The holes punched in the card not only mean “Store 10 in B” they actually DO IT.
Imagine that resting on the card is a set of rods. Where there is a hole the rod goes through and so moves. The movement of rods into and out of holes is what sets the machine up to do a particular operation and it selects the stack of wheels that the number will be stored in.
In the case of “Store 10 in B” the holes that specify wheel stack B will let rods drop through that physically move it into position so that its, and only its, cogs engage with a gear wheel. The holes that specify the “Store” operation will set the machine up so that gear wheel is driven by the processor and the rods that specify 10 arrange that it will rotate exactly ten times.
I’m going to leave it to your imagination to work out how the last piece of machinery could be implemented but you should be able to see that it is possible. Also notice that the number stored is in fact added to the wheel stack rather than just transferred. It turns out in most computer implementations to be easier to add a value than store it. As long as the programmer remembers to clear, i.e. set to zero, the location before adding this is equivalent to a store operation.
This is the reason that some specialised memory locations where arithmetic was performed were, and still are, called “accumulators” - they sum up or accumulate all of the numbers stored in them.
If you want to work out the details of how a “Load A from B” operation might work I’m sure you will see that retrieving data from B is just a matter driving the cog wheel in the opposite direction until the value stored in B is reduced to zero. Of course the same cog wheel could be arranged to drive wheel stack A in the positive direction so adding what was stored in B to A. A “Subtract A from B” instruction could be…
Stop!
This is getting much to detailed.
I have no intention of building the Analytical Engine and we are in danger of missing the point.
What is important is that you can see that computer memory isn’t like a set of pigeonholes.
It has addresses and storage locations like a set of pigeonholes but in the case of computer memory the addresses aren’t just static labels.
Computer memory addresses actually do select the memory locations they address and this is very different to just having pigeonholes with labels. It is much more like having a bank of pigeonholes that are all covered by doors. In this case the instruction to store a pigeon in hole 1234 would automatically cause that pigeonhole door, and no other, to open. The pigeon would then fly into the waiting memory location.
The instruction to retrieve the pigeon from location 1234 would again cause just that door to spring open and the pigeon would fly out. I’m sure that you can imagine some wild and wonderful Heath Robinson or Rube Goldberg style mechanical machinery to automatically store and retrieve pigeons – and I wouldn’t dream of denying you this pleasure!
The key point is that in a computer memory the addresses are not just dead and static labels for locations they are active selectors.
A computer memory is a mechanism where presenting the address returns the contents of the memory or allows something to be stored there.
How does all this relate to today’s modern computer?
Surprisingly well, as it turns out.
I admit that we have got rid of the pigeons but the push rods are still there! In this case the push rods are digital signals that come out of the CPU’s electronics – to find out exactly how it works and some fascinating facts like why a Megabyte is exactly 1,048,576 bytes see How Memory Works.
Modern memory - no push rods and certainly no pigeons
There is a well-known demonstration of how computer memory works that isn’t as well known today as it used to be. It is an excellent demonstration and it’s fun so find yourself sixteen postcards or similar, a hole punch, a pair of scissors and some knitting needles..
Take your set of postcards and punch four holes in a row close to the top edge of each one. The holes have to line up so you can see through them all when you make a stack of the cards. If you are strong enough and have a good enough punch the easiest way of doing this is to punch all sixteen in one go. Next you have to number each card in binary writing the bits under each hole.
Take a pair of scissors and turn any hole with a zero below it into a U shaped slot (see the diagrams for clarification).
First punch holes in all 16 cards (only four shown)
Number the cards in binary and make each zero hole into a slot.
When you put the pack of cards back together you have a working computer memory.
If you don’t believe me I will now tell you how to automatically access a location. Shuffle the cards, because the selection process doesn’t depend on them being in any particular order.
Suppose you want the card corresponding to address 1010.
First take a pair of knitting needles and place them in the first and third holes counting from the right - i.e. the holes corresponding to the positions you want a zero in.
Now lift up the needles together and the cards left behind have to have zeros i.e. slots in both of those positions.
That is the cards that are left behind all have addresses like x0x0 where x is zero or one.
With this subset of cards repeat the procedure but now place a single needle in the fourth position from the right and lift out a subset of cards. These now correspond to the address 10x0 and so finally you can put the needle through the second position from the left and lift out the only card with the address 1010.
This is a completely general procedure and any card that you are looking for can be found in the same way. First use multiple needles to pull out cards with the required zeros and then use a single needle to reduce the deck to the cards with the required ones.
There is also a very simple procedure that can be used to put a shuffled deck of binary cards back into order - can you work out what it is?
Your next project is to automate this procedure using Lego or Meccano and build something that works as fast as a multicore processor!
The memory principle is that you store data at an address for later retrieval and presenting the address to the mechanism automatically retrieves the correct data. In other words the address physically selects the data returned it isn't just a passive label.
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# Find the missing number, it’s a challenge that can’t be missed!
Instanews challenges never disappoint, find out with today’s math puzzle. Take the opportunity to participate by applying your arithmetic skills and more. It’s about being skilled and knowing how to exploit the potential of the logic you’ve developed.
The ability to solve a math puzzle It is a great satisfaction, but it is not that simple, nor is it taken for granted. Because it’s true that important computing skills have to be put in place, but it’s not just about that. Indeed, as in all challenges, it is necessary to maintain focus and at the required time, to solve the puzzle by all the means one has, such as logic.
what is the math puzzle Today? It is a numerical sequence characterized by the presence of numbers and numbers that are partially separate from each other, but in fact share clearly visible elements. This is where you need to start solving today’s puzzle.
Above all, you will have to complete the game in a minute, which is quite a long time, considering the fact that I have already facilitated a lot for you with an explanation that practically suggests you the answer!
So are you close? Time is running out, as always in the next paragraph, I will give you the answer, but if you look at it before time runs out and without you understanding it yourself, I win it!
## Math puzzle, here’s the solution!
Before moving on to the correct answer and the usual explanation of the riddle, I suggest you make a good trip on our site, because you will find many riddles. Why don’t you try Logical puzzle in which you should discover the hidden puzzle? It’s a timed challenge, but with a mind-blowing investigative vibe. So, time is up, do you understand what the missing number is?
First, we know the values that already exist. We have three plugins that hide an extra paragraph. This step is invisible, but understandable, because as I always suggest, the puzzle should not be taken literally, but rather should be broken down and analyzed. Especially when it comes to math!
Also comes into play Logic. What could be the movement that allows you to find the correct value, which must correspond to the same operations that were performed in the previous numbers? Well, it is necessary not only to do addition, but also Square the second number.
So the solution:
3 + 4 (squared) = 3 + 16 = 19
5 + 6 (squared) = 5 + 36 = 41
1 + 3 (squared) = 1 + 9 = 10!
So, the correct answer is 10! Did you do it yourself? If you like it Try a quick puzzle in which you must find all the missing words, what are you waiting for? Challenges and puzzles await you and your friends to solve.
As usual, you win it! See you in the next puzzle. | 621 | 2,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-21 | latest | en | 0.957749 |
https://math.hecker.org/2013/12/08/linear-algebra-and-its-applications-review-exercise-2-20/ | 1,685,644,128,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00125.warc.gz | 439,470,227 | 21,325 | ## Linear Algebra and Its Applications, Review Exercise 2.20
Review exercise 2.20. Consider the set of all 5 by 5 permutation matrices. How many such matrices are there? Are the matrices linearly independent? Do the matrices span the set of all 5 by 5 matrices?
Answer: An example member of this set is
$\begin{bmatrix} 1&0&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&0&1 \\ 0&1&0&0&0 \\ 0&0&0&1&0 \end{bmatrix}$
Note that for any such permutation matrix each row has the value 1 in one column and the value 0 in all other columns. Also note that if a given row has a 1 in a given column then no other row can have a 1 in that column.
This allows us to calculate the number of permutation matrices as follows: For the first row we have five columns in which to put a 1. Having chosen a 1 column for the first row, we then have four choices for the column to contain a 1 in the second row (because we can’t put it in the same column as in the first row). Having chosen the 1 columns for the first two rows we have three choices for the 1 column in the third row, and having chosen the 1 columns for the first three rows we have two choices for the 1 column in the fourth row. Finally, having chosen the 1 columns for the first four rows there is only one choice for the 1 column in the fifth row.
The number of 5 by 5 permutation matrices is therefore
$5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 = 5!$
(In general the number of $n$ by $n$ permutation matrices is $n!$.)
The set of 5 by 5 matrices has dimension $5 \cdot 5 = 25$. Since the number of 5 by 5 permutation matrices is greater than 25 the set of permutation matrices cannot be linearly independent.
Does the set of 5 by 5 permutation matrices span the space of all 5 by 5 matrices? One way to address this question is to first look at the set of 3 by 3 permutation matrices, of which there are $3! = 6$ matrices. A linear combination of such matrices looks as follows:
$c_1 \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} + c_2 \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix}$
$+ c_3 \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} + c_4 \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix}$
$+ c_5 \begin{bmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{bmatrix} + c_6 \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix}$
$= \begin{bmatrix} c_1+c_2&c_3+c_4&c_5+c_6 \\ c_3+c_5&c_1+c_6&c_2+c_4 \\ c_4+c_6&c_2+c_5&c_1+c_3 \end{bmatrix}$
Notice that for each row in the resulting matrix the sum of the entries in the row is the same, namely $c_1+c_2+c_3+c_4+c_5+c_6$.
Now consider the set of 5 by 5 permutation matrices $P_i$ and consider forming a linear combination of those matrices; this produces a matrix
$A = c_1P_1 + c_2P_2 + \cdots + c_{120}P_{120}$
$= \sum_{i=1}^{120} c_iP_i$
for some arbitrary set of scalars $c_i$.
Consider an arbitrary row of $A$. Of the 120 5 by 5 permutation matrices there are 24 permutation matrices that have the value 1 in column 1, so the entry for column 1 in that row of $A$ will be the sum of the 24 scalars $c_i$ that multiply those matrices. Similarly, there is a different set of 24 permutation matrices that have a 1 in column 2, so the entry for column 2 in that row of $A$ will be the sum of the different 24 scalars $c_i$ that multiply those matrices. Continuing in this way for columns 3 through 5, we see that each scalar $c_i$ contributes to the value of one and only one column in that row of $A$, and that (as in the 3 by 3 case) the sum of the entries for all columns in that row is equal to the sum of the 120 scalars, or $\sum_{i = 1}^{120} c_i$.
We chose an arbitrary row, so this same argument applies to all rows of $A$: the sum of the entries for all columns in each row of $A$ is equal to the same value $\sum_{i = 1}^{120} c_i$.
There are matrices for which the sum of the entries in each row is not equal, for example the matrix
$B = \begin{bmatrix} 1&0&0&0&1 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \end{bmatrix}$
for which the sum of the entries in the first row is 2 and the sum of the entries in all other rows is 1. The matrix $B$ cannot be expressed as a linear combination of the 5 by 5 permutation matrices (otherwise the sum of the entries in each row of $B$ would be the same), and thus the set of 5 by 5 permutation matrices does not span the space of all 5 by 5 matrices.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
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# Correct Prices
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
You are the manager of a firm that produces output in two plants. The demand for your firm's product is P = 78 - 15Q, where Q = Q1 + Q2. The marginal cost associated with producing in the two plants are MC1 = 3Q1 and MC2 = 2Q2.
(1) How much output should be produced in palnt 1 in order to maximize profits?
(2) What price should be charged to maximize profits?
© BrainMass Inc. brainmass.com October 9, 2019, 10:22 pm ad1c9bdddf
https://brainmass.com/economics/pricing-output-decisions/218300
#### Solution Preview
(1)
For Profit Maximization MR=MC1=MC2.
First we need to calculate MR. Total Revenue = PQ = R = 78Q - 15Q^2
Thus Marginal Revenue = MR = 78 - ...
#### Solution Summary
The solution answers the question below and goes into quite a bit of detail regarding Profit Maximization. The answer is ideal for students looking for a detailed analysis of the question asked below. An excellent response to the question being asked.
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## Please can you help in this question
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# Evolution Hardy Weinberg Principle
## Hardy Weinberg principle: An Introduction
The principle or the equilibrium of discussion is named after G. H. Hardy and Wilhelm Weinberg. The genotype frequencies calculated following the Hardy–Weinberg rules can be used to test for population stratification and non-random mating. The Hardy-Weinberg principle studies the genotype frequencies in non-evolving populations.
For example, external forces like mutations are reported to introduce new alleles, in turn disrupting the equilibrium of allele frequencies. Natural selection and non-random, on the other hand, are believed to alter the gene frequencies resulting in disruption of the Hardy-Weinberg equilibrium. This disturbance mainly occurs because few alleles are reported to assist or harm the reproductive success of the individuals carrying them.
Genetic drift can also cause disruption in equilibrium, majorly in small populations. Gene flow that occurs when new alleles are introduced into a population due to breeding can also alter the Hardy-Weinberg equilibrium (HWE). These external forces are always present in nature; in that case, the HWE will always be disrupted. Hence, the Hardy-Weinberg equilibrium is hypothetical.
## Hardy Weinberg Equilibrium
The main assumptions for the Hardy Weinberg Principle are the following-
1. In order to maintain equilibrium, only sexual reproduction can occur.
2. Individuals of the population should randomly mate.
3. The size of the population should be indefinitely large, and there must be diploid entities.
4. The generations must not overlap, and the sex ratio should be equal.
5. Gene flow, selection, mutation, migration and other evolutionary influences must be absent.
The Hardy-Weinberg equation can be explained by considering a simple genetic locus containing two alleles, A and a. The equation can be written as-
p2 + 2pq + q2 = 1
where p is the frequency of the allele "A" and q is the frequency of allele "a". p2 denotes the frequency of the homozygous genotype AA, q2 indicates the frequency of the homozygous genotype aa, and 2pq represents Aa, which is the frequency of the heterozygous genotype. In studies related to population genetics, the Hardy-Weinberg equation is used to compute the difference between the observed genotype frequencies and the calculated frequencies given by the equation.
## Application of Hardy Weinberg Law
Applications of the Hardy Weinberg principle are mentioned below-
1. Population stratification and non-random mating can be studied from the Hardy-Weinberg genotype frequency tests.
2. As variations occur in genes due to mutation, genetic drift, migration, sexual selection and natural selection, the Hardy Weinberg principle acts as a statistical criterion for differentiating a non-evolving population from evolving populations.
3. The idea of evolution in a population can be obtained from the allele frequencies that are recorded and calculated following the Hardy-Weinberg principle.
4. The law can be used as a template to research the population genetics of diploid organisms. However, the law stands invalid for haploid organisms.
## Hardy Weinberg Law in Plant Breeding
Plant breeding deals with the alteration of the traits of plants to generate desired characteristics. In population genetics, the most frequently used mathematical model is the Hardy–Weinberg Equilibrium (HWE). The genotype and allele frequencies of future generations can be computed with the help of this principle. The equilibrium of earlier and current populations can also be interpreted from this principle.
Hence, the HWE has ecological significance. The facultative clonal plants are inherently problematic subjects for the application of the Hardy Weinberg principle. The problematic areas or assumptions that are not followed according to HWE are the generations overlapping in the case of clonal plants. Life spans of these plants are extreme i.e; some live for a short span and some for a long span. Hence, the study of generations cannot be done in such cases.
In the case of dioecious plants, the criteria of equal sex ratio are not maintained. In spite of these limitations, HWE can be used to obtain values such as expected heterozygosity or fixation index. In plants where clonality is not maintained, the Hardy-Wienberg principle can be used to calculate the genotype frequencies.
## Hardy Weinberg Population Genetics
The relationship among genotype frequencies, allele frequencies, and factors that are reported to alter these frequencies over time are examined with the help of population genetics. The Hardy-Weinberg principle is applicable to individual genes containing two alleles, a dominant and a recessive allele. A population with such a gene can be described in terms of its genotype numbers or genotype frequencies. The allele frequency of each genotype can be calculated by dividing the number of individuals with a particular genotype divided by the total number of genotypes in a population.
The HWE gives us an idea about the genetic composition and the inheritance of genes in living organisms. The idea of genetic variation can also be obtained. The study of HWE in population genetics helps us to better understand the contribution of genes to the incidence of diseases. Hardy-Weinberg law is essential to do a comparative study between the real variations in a population to the calculated one by the Hardy-Weinberg principle.
If there is a difference between the observed and predicted valu, then it indicates that the equilibrium in the population is disturbed. The prediction of the occurrence of a negative recessive gene in the heterozygous carriers can also be made by the study of HWE in population genetics.
## Key Features of Hardy Weinberg Principle
• The frequency of alleles in a population can be designated by p2 + q2 + 2pq = 1, where p2 is the frequency of homozygous dominant genotype, q2 is the frequency of recessive genotype, and 2pq is the frequency of heterozygous genotype.
• In presence of the disruptive forces like selection, mutation and genetic drift the HWE is not maintained.
• The Hardy Weinberg equilibrium is maintained if sexual reproduction and random mating should occur and the sex ratio should be equal.
## FAQs on Evolution Hardy Weinberg Principle
1. What is an allele frequency?
Allele frequency is calculated by dividing the number of times the allele of attention has appeared in a population by the total number of alleles at a particular genetic locus in the population. It is represented as a decimal.
2. What is population genetics?
Population genetics studies the overall gene pool and variations in the genetic composition of populations. The variations occur as a result of factors, such as natural selection. The concept is focused on the Hardy-Weinberg law.
3. What is random mating?
Random mating occurs in a panmictic population, i.e. when the probability of mating between individuals is not dependent on their genetic constitution within the population.
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Study materials | 1,420 | 7,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.876053 |
https://sphweb.bumc.bu.edu/otlt/MPH-Modules/PH717-QuantCore/PH717-Module7-T-tests/Module7-ttests3.html | 1,713,220,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00657.warc.gz | 494,219,613 | 4,888 | # P-Values
A test statistic enables us to determine a p-value, which is the probability (ranging from 0 to 1) of observing sample data as extreme (different) or more extreme if the null hypothesis were true. The smaller the p-value, the more incompatible the data are with the null hypothesis.
A p-value ≤ 0.05 is an arbitrary but commonly used criterion for determining whether an observed difference is "statistically significant" or not. While it does not take into account the possible effects of bias or confounding, a p-value of ≤ 0.05 suggests that there is a 5% probability or less that the observed differences were the result of sampling error (chance). Furthermore, while it does not indicate certainty, it suggests that the null hypothesis is probably not true, so we reject the null hypothesis and accept the alternative hypothesis if the p-value is less than or equal to 0.05. The 0.05 criterion is also called the "alpha level," indicating the probability of incorrectly rejecting the null hypothesis.
A p-value > 0.05 would be interpreted by many as "not statistically significant," meaning that there was not sufficiently strong evidence to reject the null hypothesis and conclude that the groups are different. This does not mean that the groups are the same. If the evidence for a difference is weak (not statistically significant), we fail to reject the null, but we never "accept the null," i.e., we cannot conclude that they are the same – only that there is insufficient evidence to conclude that they are different.
While commonly used, p-values have fallen into some disfavor recently because the 0.05 criterion tends to devolve into a hard and fast rule that distinguishes "significantly different" from "not significantly different."
"A P value of 0.05 does not mean that there is a 95% chance that a given hypothesis is correct. Instead, it signifies that if the null hypothesis is true, and all other assumptions made are valid, there is a 5% chance of obtaining a result at least as extreme as the one observed. And a P value cannot indicate the importance of a finding; for instance, a drug can have a statistically significant effect on patients' blood glucose levels without having a therapeutic effect."
[Monya Baker: Statisticians issue warning over misuse of P values. Nature, March 7,2016]
Consider two studies evaluating the same hypothesis. Both studies find a small difference between the comparison groups, but for one study the p-value =0.06, and the authors conclude that the groups are "not significantly different"; the second study finds p=0.04, and the authors conclude that the groups are significantly different. Which is correct? Perhaps one solution is to simply report the p-value and let the reader come to their own conclusion.
Cautions Regarding Interpretation of P-Values There is an unfortunate tendency for p-values to devolve into a conclusion of "significant" or "not significant" based on the p-value. If an effect is small and clinically unimportant, the p-value can be "significant" if the sample size is large. Conversely, an effect can be large, but fail to meet the p ≤ 0.05 criterion if the sample size is small. Therefore, p-values cannot determine clinical significance or relevance. When many possible associations are examined using a criterion of p ≤ 0.05, the probability of finding at least one that meets the this criterion increases in proportion to the number of associations that are tested. Statistical significance does not take into account the evaluation of bias and confounding. P-values do not imply causation. P-values do not indicate whether the null or alternative hypothesis is really true. P-values do not indicate the strength or direction of an effect, i.e., the "magnitude of effect." P-values do not provide a way of assessing the precision of an estimated difference, and do not provide a range of possible values for the measure of effect that are compatible with the observed data.
Many researchers and practitioners now prefer confidence intervals, because they focus on the estimated effect size and how precise the estimate is rather than "Is there an effect?"
Also note that the meaning of "significant" depends on the audience. To scientists it mean "statistically significant," i.e., that p ≤ 0.05, but to a lay audience significant means "important."
## What to Report
• Measure of effect: the magnitude of the difference between the groups, e.g., difference in means, risk ratio, risk difference, odds ratio, etc.
• P-value: The probability of observing differences this great or greater if the null hypothesis is true.
• Confidence interval: a measure of the precision of the measure of effect. The confidence interval estimates the range of values compatible with the evidence.
Many public health researchers and practitioners prefer confidence intervals, since p-values give less information and are often interpreted inappropriately. When reporting results one should provide all three of these. | 1,020 | 5,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.95736 |
https://documentation.aimms.com/language-reference/optimization-modeling-components/robust-optimization/solving-robust-optimization-models.html | 1,708,948,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00823.warc.gz | 216,178,226 | 6,848 | # Solving Robust Optimization Models
Solving robust optimization models
After you have specified all uncertain parameters, random parameters, chance constraints and adjustable variables that specify your robust optimization model, your original mathematical program can now be solved as a robust optimization model. It is also still possible to solve it as a deterministic model by just calling the `SOLVE` statement (see also The SOLVE Statement).
Generate robust counterpart
To solve a robust optimization model for a `MathematicalProgram` MP, the first step is to generate its robust counterpart. This can be accomplished by calling the GMP function
• `GenerateRobustCounterpart`(MP,UncertainParameters,UncertaintyConstraints[,Name])
The function returns an element into the set `AllGeneratedMathematicalPrograms`, i.e., the generated mathematical program representing the robust counterpart of the given robust optimization model.
Specifying uncertain data
Through the UncertainParameters and UncertaintyConstraints arguments you can specify the collection of uncertain and random parameters, as well as the uncertainty constraints that you want to take into account when generating the robust counterpart. Together, these completely determine the uncertain data which AIMMS will use to translate the uncertain matrix coefficients, chance constraints and adjustable variables into the generated mathematical program representing the robust counterpart.
Name argument
With the optional Name argument you can explicitly specify a name for the generated mathematical program. If you do not choose a name, AIMMS will use the name of the underlying `MathematicalProgram` as the name of the generated mathematical program as well. Please note, that AIMMS will also use this name as the default name for solving the deterministic model. Therefore, if you do not want the generated mathematical program of the deterministic model to be deleted, then you have to choose a non-default name.
Solving the robust counterpart
You can solve the generated mathematical program gmp representing the robust counterpart by calling the regular GMP procedure
The `GMP::Instance::Solve` method is discussed in full detail in Managing Generated Mathematical Program Instances. Alternatively, you can use any of the other available functions available to solve generated mathematical programs discussed in Implementing Advanced Algorithms for Mathematical Programs. Note that AIMMS will not allow you to use the GMP modification functions on any `gmp` generated by `GenerateRobustCounterpart`.
The resulting solution
The solution resulting from solving the robust counterpart will satisfy all non-chance constraints in your model for all realizations of the uncertain parameters that you passed to the `GenerateRobustCounterPart` function, and will satisfy all chance constraints with the given probabilities and approximations, given the random parameters taken into account. | 536 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.773864 |
https://www.scribd.com/document/2420611/Chapter08 | 1,529,612,856,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00372.warc.gz | 919,496,226 | 33,230 | # 8
Factor of safety and probability of failure
8.1 Introduction
How does one assess the acceptability of an engineering design? Relying on
judgement alone can lead to one of the two extremes illustrated in Figure 8.1. The first
case is economically unacceptable while the example illustrated in the lower drawing
violates all normal safety standards.
Figure 8.1: Rockbolting alternatives
involving individual judgement. (Drawings
based upon a cartoon in a brochure on
Mines of Western Australia.)
106 Chapter 8: Factor of safety and probability of failure
8.2 Sensitivity studies
The classical approach used in designing engineering structures is to consider the
relationship between the capacity C (strength or resisting force) of the element and
the demand D (stress or disturbing force). The Factor of Safety of the structure is
defined as F = C/D and failure is assumed to occur when F is less than 1.
Rather than base an engineering design decision on a single calculated factor of
safety, an approach which is frequently used to give a more rational assessment of the
risks associated with a particular design is to carry out a sensitivity study. This
involves a series of calculations in which each significant parameter is varied
systematically over its maximum credible range in order to determine its influence
upon the factor of safety.
This approach was used in the analysis of the Sau Mau Ping slope in Hong Kong
discussed in the previous chapter. It provided a useful means of exploring a range of
possibilities and reaching practical decisions on some difficult problems. On the
following pages this idea of sensitivity studies will be extended to the use of
probability theory and it will be shown that, even with very limited field data,
practical, useful information can be obtained from an analysis of probability of
failure.
8.3 An introduction to probability theory
A complete discussion on probability theory exceeds the scope of these notes and the
techniques discussed on the following pages are intended to introduce the reader to
the subject and to give an indication of the power of these techniques in engineering
decision making. A more detailed treatment of this subject will be found in a book by
geotechnical applications of probability theory entitled ‘Evaluating calculated risk in
reading for anyone with a serious interest in this subject. Pine (1992), Tyler et al
(1991), Hatzor and Goodman (1993) and Carter (1992) have published papers on the
application of probability theory to the analysis of problems encountered in
underground mining and civil engineering.
Most geotechnical engineers regard the subject of probability theory with doubt
and suspicion. At least part of the reason for this mistrust is associated with the
language which has been adopted by those who specialise in the field of probability
theory and risk assessment. The following definitions are given in an attempt to
dispel some of the mystery which tends to surround this subject.
Random variables: Parameters such as the angle of friction of rock joints, the
uniaxial compressive strength of rock specimens, the inclination and orientation of
discontinuities in a rock mass and the measured in situ stresses in the rock
surrounding an opening do not have a single fixed value but may assume any number
of values. There is no way of predicting exactly what the value of one of these
parameters will be at any given location. Hence these parameters are described as
random variables.
An introduction to probability theory 107
Probability distribution: A probability density function
(PDF) describes the relative likelihood that a random
variable will assume a particular value. A typical
probability density function is illustrated opposite. In
this case the random variable is continuously
distributed (i.e., it can take on all possible values). The
area under the PDF is always unity.
An alternative way of presenting the same
information is in the form of a cumulative distribution
function (CDF), which gives the probability that the
variable will have a value less than or equal to the
selected value. The CDF is the integral of the
corresponding probability density function, i.e., the
ordinate at x1 on the cumulative distribution is the area
under the probability density function to the left of x1.
Note the fx(x) is used for the ordinate of a PDF while
Fx(x) is used for a CDF.
One of the most common graphical representations of
a probability distribution is a histogram in which the
fraction of all observations falling within a specified
interval is plotted as a bar above that interval.
Data analysis: For many applications it is not necessary to use all of the information
contained in a distribution function and quantities summarised only by the dominant
features of the distribution may be adequate.
The sample mean or expected value or first moment indicates the centre of gravity of
a probability distribution. A typical application would be the analysis of a set of
results x x x
n 1 2
, ,........, from uniaxial strength tests carried out in the laboratory.
Assuming that there are n individual test values x
i
, the mean x is given by:
x
n
x
i
i
n
·
·
1
1
(8.1)
The sample variance s
2
or the second moment about the mean of a distribution is
defined as the mean of the square of the difference between the value of x
i
and the
mean value x . Hence:
s
n
x x
i
i
n
2
1
1
1
2
·
·
( ) 8.2)
Note that, theoretically, the denominator for calculation of variance of samples
should be n, not (n - 1). However, for a finite number of samples, it can be shown
that the correction factor n/(n-1), known as Bessel's correction, gives a better
estimate. For practical purposes the correction is only necessary when the sample
size is less than 30.
The standard deviation s is given by the positive square root of the variance s
2
. In
the case of the commonly used normal distribution, about 68% of the test values will
fall within an interval defined by the mean ± one standard deviation while
Cumulative distribution
function (CDF)
108 Chapter 8: Factor of safety and probability of failure
approximately 95% of all the test results will fall within the range defined by the
mean ± two standard deviations. A small standard deviation will indicate a tightly
clustered data set while a large standard deviation will be found for a data set in
which there is a large scatter about the mean.
The coefficient of variation (COV) is the ratio of the standard deviation to the mean,
i.e. COV = s/ x . COV is dimensionless and it is a particularly useful measure of
uncertainty. A small uncertainty would typically be represented by a COV = 0.05
while considerable uncertainty would be indicated by a COV = 0.25.
Normal distribution: The normal or Gaussian distribution is the most common type
of probability distribution function and the distributions of many random variables
conform to this distribution. It is generally used for probabilistic studies in
geotechnical engineering unless there are good reasons for selecting a different
distribution. Typically, variables which arise as a sum of a number of random effects,
none of which dominate the total, are normally distributed.
The problem of defining a normal distribution is to estimate the values of the
governing parameters which are the true mean ( µ ) and true standard deviation ( σ).
Generally, the best estimates for these values are given by the sample mean and
standard deviation, determined from a number of tests or observations. Hence, from
equations 8.1 and 8.2:
µ · x (8.3)
σ · s (8.4)
It is important to recognise that equations 8.3 and 8.4 give the most probable
values of µ and σ and not necessarily the true values.
Obviously, it is desirable to include as many samples as possible in any set of
observations but, in geotechnical engineering, there are serious practical and financial
limitations to the amount of data which can be collected. Consequently, it is often
necessary to make estimates on the basis of judgement, experience or from
comparisons with results published by others. These difficulties are often used as an
excuse for not using probabilistic tools in geotechnical engineering but, as will be
shown later in this chapter, useful results can still be obtained from very limited data.
Having estimated the mean µ and standard deviation σ, the probability density
function for a normal distribution is defined by:
π σ
]
]
]
]
,
`
.
|
σ
µ −
·
2
2
1
exp
) (
2
x
x f
x
(8.5)
for −∞ ≤ ≤ ∞ x .
As will be seen later, this range of −∞ ≤ ≤ ∞ x can cause problems when a normal
distribution is used as a basis for a Monte Carlo analysis in which the entire range of
values is randomly sampled. This can give rise to a few very small numbers
(sometimes negative) and very large numbers which, in certain analyses, can cause
numerical instability. In order to overcome this problem the normal distribution is
An introduction to probability theory 109
sometimes truncated so that only values falling within a specified range are
considered valid.
There is no closed form solution for the cumulative distribution function (CDF)
which must by found by numerical integration.
Other distributions: In addition to the commonly used normal distribution there are a
number of alternative distributions which are used in probability analyses. Some of
the most useful are:
• Beta distributions (Harr, 1987) are very versatile distributions which can be used
to replace almost any of the common distributions and which do not suffer from
the extreme value problems discussed above because the domain (range) is
bounded by specified values.
• Exponential distributions are sometimes used to define events such as the
occurrence of earthquakes or rockbursts or quantities such as the length of joints
in a rock mass.
• Lognormal distributions are useful when considering processes such as the
crushing of aggregates in which the final particle size results from a number of
collisions of particles of many sizes moving in different directions with different
velocities. Such multiplicative mechanisms tend to result in variables which are
lognormally distributed as opposed to the normally distributed variables resulting
• Weibul distributions are used to represent the lifetime of devices in reliability
studies or the outcome of tests such as point load tests on rock core in which a
few very high values may occur.
It is no longer necessary for the person starting out in the field of probability theory to
know and understand the mathematics involved in all of these probability
distributions since commercially available software programs can be used to carry out
many of the computations automatically. Note that the author is not advocating the
blind use of ‘black-box’ software and the reader should exercise extreme caution is
using such software without trying to understand exactly what the software is doing.
However there is no point in writing reports by hand if one is prepared to spend the
time learning how to use a good word-processor correctly and the same applies to
mathematical software.
One of the most useful software packages for probability analysis is a program
called BestFit
1
which has a built-in library of 18 probability distributions and which
can be used to fit any one of these distributions to a given set of data or it can be
allowed automatically to determine the ranking of the fit of all 18 distributions to the
data set. The results from such an analysis can be entered directly into a companion
program called @RISK which can be used for risk evaluations using the techniques
described below.
1
BestFit for Windows and its companion program @RISK for Microsoft Excel of Lotus 1-2-3 (for
Windows or Macintosh) are available from the Palisade Corporation, 31 Decker Road, Newfield, New
York 14867, USA. Fax number 1 607 277 8001.
110 Chapter 8: Factor of safety and probability of failure
Sampling techniques: Consider a problem in which the factor of safety depends upon
a number of random variables such as the cohesive strength c, the angle of friction φ
and the acceleration α due to earthquakes or large blasts. Assuming that the values of
these variables are distributed about their means in a manner which can be described
by one of the continuous distribution functions such as the normal distribution
described earlier, the problem is how to use this information to determine the
distribution of factor of safety values and the probability of failure.
The Monte Carlo method uses random or pseudo-random numbers to sample from
probability distributions and, if sufficiently large numbers of samples are generated
and used in a calculation such as that for a factor of safety, a distribution of values for
the end product will be generated. The term ‘Monte Carlo’ is believed to have been
introduced as a code word to describe this hit-and-miss technique used during secret
work on the development of the atomic bomb during World War II (Harr 1987).
Today, Monte Carlo techniques can be applied to a wide variety of problems
involving random behaviour and a number of algorithms are available for generating
random Monte Carlo samples from different types of input probability distributions.
With highly optimised software programs such as @RISK, problems involving
relatively large samples can be run efficiently on most desktop or portable computers.
The Latin Hypercube sampling technique (Imam et al (1980), Startzman and
Watterbarger (1985)) is a relatively recent development which gives comparable
results to the Monte Carlo technique but with fewer samples. The method is based
upon stratified sampling with random selection within each stratum. Typically an
analysis using 1000 samples obtained by the Latin Hypercube technique will produce
comparable results to an analysis using 5000 samples obtained using the Monte Carlo
method. Both techniques are incorporated in the program @RISK.
Note that both the Monte Carlo and the Latin Hypercube techniques require that
the distribution of all the input variables should either be known or that they be
assumed. When no information on the distribution is available it is usual to assume a
normal or a truncated normal distribution.
The Generalised Point Estimate Method, developed by Rosenbleuth (1981) and
discussed in detail by Harr (1987), can be used for rapid calculation of the mean and
standard deviation of a quantity such as a factor of safety which depends upon
random behaviour of input variables. Hoek (1989) discussed the application of this
technique to the analysis of surface crown pillar stability while Pine (1992) has
applied this technique to the analysis of slope stability and other mining problems.
To calculate a quantity such as a factor of safety, two point estimates are made at
one standard deviation on either side of the mean ( µ σ t ) from each distribution
representing a random variable. The factor of safety is calculated for every possible
combination of point estimates, producing 2
n
solutions where n is the number of
random variables involved. The mean and the standard deviation of the factor of
safety are then calculated from these 2
n
solutions.
While this technique does not provide a full distribution of the output variable, as do
the Monte Carlo and Latin Hypercube methods, it is very simple to use for problems
with relatively few random variables and is useful when general trends are being
investigated. When the probability distribution function for the output variable is
known, for example, from previous Monte Carlo analyses, the mean and standard
deviation values can be used to calculate the complete output distribution .
Probability of failure 111
8.4 Probability of failure
In the case of the Sau Mau Ping slope problem the factor of safety of the overall slope
with a tension crack is defined by:
1. Fixed dimensions:
Overall slope height H = 60 m
Overall slope angle ψ
f
= 50°
Failure plane angle ψ
p
= 35°
Unit weight of rock γ
r
= 2.6 tonnes/m
3
Unit weight of water γ
w
= 1.0 tonnes/m
3
2. Random variables Mean values
Friction angle on joint surface φ = 35°
Cohesive strength of joint surface c = 10 tonnes/m
2
Depth of tension crack z = 14 m
Depth of water in tension crack z
w
= z/2
Ratio of horizontal earthquake
to gravitational acceleration α = 0.08
Figure 8.2 illustrates the layout of a Microsoft Excel spreadsheet with plots of the
probability distribution functions of the random input variables and of the calculated
factor of safety. It is worth discussing each of the plots in detail to demonstrate the
reasoning behind the choice of the probability distribution functions.
1. Friction angle φ - A truncated normal distribution has been assumed for this
variable. The mean is assumed to be 35° which is the approximate centre of the
assumed shear strength range illustrated in Figure 7.8. The standard deviation of
5° implies that about 68% of the friction angle values defined by the distribution
will lie between 30° and 40°. The normal distribution is truncated by a minimum
value of 15° and a maximum value of 60° which have been arbitrarily chosen as
the extreme values represented by a smooth slickensided surface and a fresh,
rough tension fracture.
2. Cohesive strength c - Again using the assumed range of shear strength values
illustrated in Figure 7.8, a value of 10 tonnes/m
2
has been chosen as the mean
cohesive strength and the standard deviation has been set at 2 tonnes/m
2
on the
basis of this diagram. In order to allow for the wide range of possible cohesive
strengths the minimum and maximum values used to truncate the normal
distribution are 0 and 25 tonnes/m
2
respectively. Those with experience in the
interpretation of laboratory shear strength test results may argue that the friction
angle φ and the cohesive strength c are not independent variables as has been
assumed in this analysis. This is because the cohesive strength generally drops as
the friction angle rises and vice versa. The program @RISK allows the user to
define variables as dependent but, for the sake of simplicity, the friction angle φ
and the cohesive strength c have been kept independent for this analysis.
112 Chapter 8: Factor of safety and probability of failure
Figure 8.2: Spreadsheet for @RISK Latin Hypercube analysis of Sau Mau Ping slope with
distributions of random input variables and the probability density function for the
calculated factor of safety. The probability of failure, shown by the dark region for F<1, is
approximately 7% for the assumed conditions.
.
Probability of failure 113
3. Tension crack depth z - Equation 7.6, defining the tension crack depth, has been
derived by minimisation of equation 7.5. For the purposes of this analysis it has
been assumed that this value of z (14 m for the assumed conditions) represents the
mean tension crack depth. A truncated normal distribution is assumed to define
the possible range of tension crack depths and the standard deviation has been
arbitrarily chosen at 3 m. The minimum tension crack depth is zero but a value of
0.1 m has been chosen to avoid possible numerical problems. The maximum
tension crack depth is given by z H
p f
· − ( tan / tan ) 1 φ ψ = 24.75 m which occurs
when the vertical tension crack is located at the crest of the slope.
4. Water depth z
w
in tension crack - The water which would fill the tension crack in
this slope would come from direct surface run-off during heavy rains. In Hong
Kong the heaviest rains occur during typhoons and it is likely that the tension
crack would be completely filled during such events. The probability of
occurrence of typhoons has been defined by a truncated exponential distribution
where the mean water depth is assumed to be one half the tension crack depth.
The maximum water depth cannot exceed the tension crack depth z and, as
defined by the exponential distribution, this value would occur very rarely. The
minimum water depth is zero during dry conditions and this is assumed to be a
frequent occurrence. Note that the water depth z
w
is defined in terms of the
tension crack depth z which is itself a random variable. In calculating z
w
the
program @RISK first samples the truncated normal distribution defining z and
then combines this value with the information obtained from sampling the
truncated exponential distribution to calculate z
w
.
5. Ratio of horizontal earthquake acceleration to gravitational acceleration α - The
frequent occurrence of earthquakes of different magnitudes can be estimated by
means of an exponential distribution which suggests that large earthquakes are
very rare while small ones are very common. In the case of Hong Kong local
wisdom suggested a ‘design’ horizontal acceleration of 0.08g. In other words, this
level of acceleration could be anticipated at least once during the operating life of
a civil engineering structure. A rough rule of thumb suggests that the ‘maximum
credible’ acceleration is approximately twice the ‘design’ value. Based upon these
very crude guidelines, the distribution of values of α used in these calculations
was defined by a truncated exponential distribution with a mean value of α =
0.08, a maximum of 0.16 and a minimum of 0.
Using the distributions shown in Figure 8.2, the program @RISK was used, with
Latin Hypercube sampling to carry out 1,000 iterations on the factor of safety. The
resulting probability distribution was not a smooth curve, indicating that an
insufficient number of iterations had been performed for this combination of
variables. A second analysis was carried out using 10,000 iterations and the resulting
factor of safety distribution is plotted in the lower right hand corner of Figure 8.2.
Note that this distribution closely resembles a normal distribution.
From the statistical tables produced by the program @RISK it was determined that
the probability of failure for this slope is approximately 7%. This value is given by
the ration of the area under the distribution curve for F<1 (shown in black in Figure
8.2) divided by the total area under the distribution curve. This means that, for the
combination of slope geometry, shear strength, water pressure and earthquake
114 Chapter 8: Factor of safety and probability of failure
acceleration parameters assumed, 7 out of 100 similar slopes could be expected to fail
at some time during the life of the slope. Alternatively, a length of 70 m could be
expected to fail in every 1000 m of slope.
This is a small but not negligible risk of failure and it confirms the earlier
conclusion, discussed in Chapter 7, that this slope was not adequately stable for a
densely populated region such as Kowloon. Incidentally, a risk of this magnitude
would probably be acceptable in an open pit mine, with limited access of trained
miners, and even on a rural road. The decisions reached in Chapter 7 on the long
term stabilisation measures for this slope are considered appropriate and the type of
analysis described here could be used to evaluate the effectiveness of these remedial
measures. | 4,960 | 23,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-26 | latest | en | 0.929759 |
http://bbs.magnum.uk.net/?page=001-forum.ssjs&sub=usenet_recpuzzl&thread=105 | 1,720,943,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00480.warc.gz | 5,509,944 | 8,030 | • #### Word arithmetic : (CARL * R) + GINNOW = COINER
From James Dow Allen@21:1/5 to Carl G. on Sun Jun 28 02:17:21 2020
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles:
testit.
Dammit, I'm a coder not a mind-reader! What are you testing?
Given that
(a) Carl frequently posted word problems,
(b) He is an excellent COINER of all sorts of puzzles, and
(c) His surname might be Ginnow
... I'm going to guess that he's testing whether rec.puzzles
is still capable of designing word arithmetic puzzles, with
unique solutions. I think we are:
(CARL * R) + GINNOW = COINER
Since Carl may be trying to ATTRACT interest in rec.p,
I'll submit another, which may be rather easy.
(CARL * R) + GINNOW = ATTRACT
I hope others also post word arithmetic puzzles -- I'd like
to show off my own! :-)
But again, I'm not a mind-reader. Perhaps Carl is annoyed
that nobody has responded to some of his puzzles,
e.g. "Resistance is Futile":
On Monday, July 10, 2000 at 2:00:00 PM UTC+7, Carl G. wrote:
You have a set of ten very precise resistors (assume that they have an exact resistance value). The resistors can be used individually, in parallel, or in series. The resistance of two resistors in series is given by the sum of the two resistances ( = r1+r2). The resistance of two resistors in parallel is given by the reciprocal of the sum of the reciprocals of the two resistances (= 1 / (1/r1 + 1/r2)). Find the set of ten resistors that forms the largest contiguous range of integral resistances, starting at 1 (in ohms)?
For example: The set of ten resistors: 1, 2, 4, 8, 16, 32, 64, 128, 256,
and 512 ohms, could form all of the resistances from 1 to 1023 by only using them in series (just convert the desired resistance value into binary and
use the resistors that correspond to each non-zero bit). Can you get a larger range by also using your set in parallel?
Can you find a larger contiguous range of integral resistances if you don't have to start at 1 ohm?
Carl G.
"Doing things in series may have serious consequences, but doing things in parallel may have perilous consequences." - C. Ginnow
My excuse is that I was AWOL from rec,puzzles in 2000.
I'll to look into this puzzle tomorrow.
James Dow Allen
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Edward Murphy@21:1/5 to James Dow Allen on Sun Jun 28 10:08:27 2020
On 6/28/2020 2:17 AM, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles:
testit.
Dammit, I'm a coder not a mind-reader! What are you testing?
Given that
(a) Carl frequently posted word problems,
(b) He is an excellent COINER of all sorts of puzzles, and
(c) His surname might be Ginnow
... I'm going to guess that he's testing whether rec.puzzles
is still capable of designing word arithmetic puzzles, with
unique solutions. I think we are:
(CARL * R) + GINNOW = COINER
Partial solution:
This includes all ten digits (ACEGILNORW).
CARL * R <= 9999 * 9 = 89991, so:
C = G + 1
O < I
E > O
Also, if CARL * R <= 9999, then I < N, but also I = 9 and O = 0 (contradiction). So R > 1.
(L * R) + W = R mod 10
(L * R) = R - W (non-zero) mod 10
so L != 0, R != 0, and we can also rule out
(L = 2, R = 5)
and
(L = 5, R = 2)
Since Carl may be trying to ATTRACT interest in rec.p,
I'll submit another, which may be rather easy.
(CARL * R) + GINNOW = ATTRACT
Partial solution:
This includes all ten digits (ACGILNORTW).
Again, CARL * R is at most 5 digits, so:
?????
+ GINNOW
-------
ATTRACT
so A = 1, G = 9, T = 0.
CARL * R must be five digits, otherwise we would also get I = 9.
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ilan Mayer@21:1/5 to James Dow Allen on Sun Jun 28 10:39:49 2020
On Sunday, 28 June 2020 05:17:23 UTC-4, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles: >> testit.
Dammit, I'm a coder not a mind-reader! What are you testing?
Given that
(a) Carl frequently posted word problems,
(b) He is an excellent COINER of all sorts of puzzles, and
(c) His surname might be Ginnow
... I'm going to guess that he's testing whether rec.puzzles
is still capable of designing word arithmetic puzzles, with
unique solutions. I think we are:
(CARL * R) + GINNOW = COINER
Since Carl may be trying to ATTRACT interest in rec.p,
I'll submit another, which may be rather easy.
(CARL * R) + GINNOW = ATTRACT
I hope others also post word arithmetic puzzles -- I'd like
to show off my own! :-)
But again, I'm not a mind-reader. Perhaps Carl is annoyed
that nobody has responded to some of his puzzles,
e.g. "Resistance is Futile":
On Monday, July 10, 2000 at 2:00:00 PM UTC+7, Carl G. wrote:
You have a set of ten very precise resistors (assume that they have an exact
resistance value). The resistors can be used individually, in parallel, or in series. The resistance of two resistors in series is given by the sum of
the two resistances ( = r1+r2). The resistance of two resistors in parallel
is given by the reciprocal of the sum of the reciprocals of the two resistances (= 1 / (1/r1 + 1/r2)). Find the set of ten resistors that forms
the largest contiguous range of integral resistances, starting at 1 (in ohms)?
For example: The set of ten resistors: 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512 ohms, could form all of the resistances from 1 to 1023 by only using
them in series (just convert the desired resistance value into binary and use the resistors that correspond to each non-zero bit). Can you get a larger range by also using your set in parallel?
Can you find a larger contiguous range of integral resistances if you don't have to start at 1 ohm?
Carl G.
"Doing things in series may have serious consequences, but doing things in parallel may have perilous consequences." - C. Ginnow
My excuse is that I was AWOL from rec,puzzles in 2000.
I'll to look into this puzzle tomorrow.
James Dow Allen
SPOILER
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( CARL * R ) + GINNOW = COINER -> ( 7891 * 9 ) + 654420 = 725439
( CARL * R ) + GINNOW = ATTRACT -> ( 5136 * 3 ) + 987742 = 1003150
__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\
||
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From James Dow Allen@21:1/5 to Ilan Mayer on Sun Jun 28 12:38:11 2020
On Monday, June 29, 2020 at 12:39:51 AM UTC+7, Ilan Mayer wrote:
On Sunday, 28 June 2020 05:17:23 UTC-4, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles: >> testit.
[et cetera]
( CARL * R ) + GINNOW = COINER -> ( 7891 * 9 ) + 654420 = 725439
( CARL * R ) + GINNOW = ATTRACT -> ( 5136 * 3 ) + 987742 = 1003150
ILAN x MAYER = EXEMPLAR
BRAINY + ILAN + MAYER = MARVEL
James
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From James Dow Allen@21:1/5 to James Dow Allen on Mon Jun 29 07:53:48 2020
On Monday, June 29, 2020 at 2:38:13 AM UTC+7, James Dow Allen wrote:
On Monday, June 29, 2020 at 12:39:51 AM UTC+7, Ilan Mayer wrote:
On Sunday, 28 June 2020 05:17:23 UTC-4, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles:
testit.
[et cetera]
( CARL * R ) + GINNOW = COINER -> ( 7891 * 9 ) + 654420 = 725439
( CARL * R ) + GINNOW = ATTRACT -> ( 5136 * 3 ) + 987742 = 1003150
ILAN x MAYER = EXEMPLAR
BRAINY + ILAN + MAYER = MARVEL
I'm disappointed nobody but Edward Murphy and Ilan Mayer
have joined in the fun. But I won't give up yet.
Now is the season for campaign mottoes. Wouldn't it add to
the fun if such mottoes had to be valid base-10 word arithmetic
puzzles with unique solutions? Here's one for the incumbent:
MAKE x AMERIKA = INEPT x AGAIN
It will be easier to make a motto for the Blue-shirts when we
know who their V.P. nominee will be. Here are two possibilities:
PRAISE = BIDEN + HARRIS
BIDEN + BALDWIN = LOVABLE
James Dow Allen
--- SoupGate-Win32 v1.05
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• From Richard Heathfield@21:1/5 to James Dow Allen on Tue Jun 30 01:45:28 2020
On 29/06/2020 15:53, James Dow Allen wrote:
<snip>
Now is the season for campaign mottoes. Wouldn't it add to
the fun if such mottoes had to be valid base-10 word arithmetic
puzzles with unique solutions? Here's one for the incumbent:
MAKE x AMERIKA = INEPT x AGAIN
Either you screwed up or I did. Probably me. I can find no valid
solutions. (I even brute-forced it to check.)
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From James Dow Allen@21:1/5 to Richard Heathfield on Mon Jun 29 21:48:42 2020
On Tuesday, June 30, 2020 at 7:45:34 AM UTC+7, Richard Heathfield wrote:
On 29/06/2020 15:53, James Dow Allen wrote:
MAKE x AMERIKA = INEPT x AGAIN
Either you screwed up or I did. Probably me. I can find no valid
solutions. (I even brute-forced it to check.)
I screwed up, not you. :-( :-(
My face is so red that I'd better crawl into a corner and pout.
Back in the day, I had the humility and diligence to check
my results, as well as the acuity to not actually need the checking.
(Maybe Viagra would have helped?)
Mea culpa. Sorry for the time-wasting.
I don't blame any of you for never trusting me again, but this
useless one I did double-check:
MAKE x AMERIKA = MANTIS x AGAIN
In a vain effort to make amends, I offer this:
Richard -- what is your middle name? If it is 'Dan'
then this works in Base-12:
RICHARD * DAN = HEATHFIELD
Or would you like to change your middle name to 'Veil'?
I think this works in Base-11:
RICHARD * VEIL = HEATHFIELD
Yes, I am double-checking these now. :-(
I think there are some nifty middle-names for
you in Base-13 if you want to go that far! :-)
James
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Carl G.@21:1/5 to James Dow Allen on Thu Jul 2 13:59:03 2020
On 6/28/2020 2:17 AM, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in rec.puzzles:
testit.
Dammit, I'm a coder not a mind-reader! What are you testing?
Given that
(a) Carl frequently posted word problems,
(b) He is an excellent COINER of all sorts of puzzles, and
(c) His surname might be Ginnow
... I'm going to guess that he's testing whether rec.puzzles
is still capable of designing word arithmetic puzzles, with
unique solutions. I think we are:
(CARL * R) + GINNOW = COINER
Since Carl may be trying to ATTRACT interest in rec.p,
I'll submit another, which may be rather easy.
(CARL * R) + GINNOW = ATTRACT
I hope others also post word arithmetic puzzles -- I'd like
to show off my own! :-)
But again, I'm not a mind-reader. Perhaps Carl is annoyed
that nobody has responded to some of his puzzles,
e.g. "Resistance is Futile":
...
My excuse is that I was AWOL from rec,puzzles in 2000.
I'll to look into this puzzle tomorrow.
James Dow Allen
I was installing Mozilla Thunderbird on an old laptop computer and
accidentally sent a test message when in my news account, rather than my default account.
OBPuzzle: Anagram/alphametic below:
CARL+GINNOW = A*CLOWN+GRIN
--
Carl G.
--
Carl G.
--
This email has been checked for viruses by AVG.
https://www.avg.com
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Richard Heathfield@21:1/5 to Carl G. on Fri Jul 3 00:46:29 2020
On 02/07/2020 21:59, Carl G. wrote:
On 6/28/2020 2:17 AM, James Dow Allen wrote:
Recently, champion puzzler Carl G. posted a cryptic message in
rec.puzzles:
testit.
Dammit, I'm a coder not a mind-reader! What are you testing?
Given that
(a) Carl frequently posted word problems,
(b) He is an excellent COINER of all sorts of puzzles, and
(c) His surname might be Ginnow
... I'm going to guess that he's testing whether rec.puzzles
is still capable of designing word arithmetic puzzles, with
unique solutions. I think we are:
(CARL * R) + GINNOW = COINER
Since Carl may be trying to ATTRACT interest in rec.p,
I'll submit another, which may be rather easy.
(CARL * R) + GINNOW = ATTRACT
I hope others also post word arithmetic puzzles -- I'd like
to show off my own! :-)
But again, I'm not a mind-reader. Perhaps Carl is annoyed
that nobody has responded to some of his puzzles,
e.g. "Resistance is Futile":
...
My excuse is that I was AWOL from rec,puzzles in 2000.
I'll to look into this puzzle tomorrow.
James Dow Allen
I was installing Mozilla Thunderbird on an old laptop computer and accidentally sent a test message when in my news account, rather than my default account.
OBPuzzle: Anagram/alphametic below:
CARL+GINNOW = A*CLOWN+GRIN
s
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0
CARL+GINNOW = A*CLOWN+GRIN
8274 + 163395 = 2 * 84953 + 1763
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5) | 4,071 | 13,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-30 | latest | en | 0.940107 |
http://web2.0calc.com/questions/help_31362 | 1,516,226,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886979.9/warc/CC-MAIN-20180117212700-20180117232700-00235.warc.gz | 372,170,268 | 5,463 | +0
# help
0
38
1
Problem:
The quadratic \( x^2+7x+20\) has the same roots as the quadratic \( Ax^2+Bx+1\). What is \(A+B\)?
Guest Dec 28, 2017
Sort:
#1
+80905
+1
x^2 + 7x + 20 = 0
The sum of the roots = -7/1 = -7
The product of the roots is 20/1 = 20
So.....this implies that in Ax^2 + Bx + 1 = 0
-B/A = -7 ⇒ B/A = 7 ⇒ B = 7A
And
1/A = 20 ⇒ A = 1/20
So..... B = 7(1/20) = 7/20
So..... A + B = 1/20 + 7/20 = 8/20 = 2/5
CPhill Dec 28, 2017
### 22 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 320 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-05 | latest | en | 0.475678 |
https://discourse.pymc.io/t/how-to-get-on-line-prediction-estimates-for-data-over-time/5070 | 1,656,608,512,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00681.warc.gz | 274,970,483 | 7,737 | # How to get "on-line" prediction estimates for data over time?
Hi, I’m new to pymc3 so forgive me if this is somehow a stupid question:
I’m trying to model a production process, where I’m using three parameters A, B, and C to predict an output measurement Y. Each product has two relevant dates: the production date and the measurement date. At the time of production, I want to make my best guess at what the measured value will be, and then at the measurement date when I get the true measured value, I will update my model.
So what I’m trying to do is: for each product, on the production date P give the best estimate for Y based off observed data from every product whose measurement date is before P.
What is the best way to solve this kind of problem?
Since time does not seem to be a relevant variable in your model (A, B, C don’t change over time), you can ignore it. Just make your model predict observed Ys given their A, B, and C. Then use the posterior to predict new unobserved Ys given their A, B, and C. As you get more observed Ys you can keep updating your model. This gives you an idea of how you might do it in PyMC:
https://docs.pymc.io/notebooks/updating_priors.html
Hi Ricardo, thanks for the response. I should have specified, the impact of certain parameters may indeed change over time (e.g. in a chemical process the precursor may age and its properties will change, so the batch of chemical C may be a time-dependent parameter,) while other variables are time-independent (e.g. the solvent A being used in the reaction.)
Do you know how I would write the code to take into account a time-varying parameter?
I am afraid that is too general of a question for me to offer specific usable advice. You will need to decide how to model your time effects. Two examples include: linear function based on time, or random walk prior…
Let me study your example. I wasn’t sure if a for loop was the best way to do this, but from your example it seems like a for loop is the way to go. I also didn’t know how to use the old posterior as the new prior, but I think by studying your example I should be able to figure that out (it looks like it will involve using the Interpolated function?)
If I still can’t figure it out, I will come back with more context and paste in my code.
One more question: if I’m running pm.sample thousands of times in my for loop (all on similar types of data,) do I have to tune each time? Or is there a way to bypass the tuning step by explicitly setting the parameters it’s optimizing?
My current difficulty: in the linked example they are estimating three variables: alpha, beta0, and beta1. In my case different data points may use different recipes (e.g. depending on the solvent used for one data point, I would want to use B[1] or B[2]) so there will be a vector instead of just one parameter.
Here’s an example of what my code looks like: on the second iteration (when I’m trying to retrieve the posteriors from the first run to use as priors) it gives me:
“MissingInputError: Input 0 of the graph (indices start from 0), used to compute Elemwise{exp,no_inplace}(eps_log__), was not provided and not given a value.”
(This is my first project with pymc3, so if I’m doing anything else stupid in the code please let me know.)
``````import numpy as np
import pymc3 as pm
import pandas as pd
from scipy import stats
def from_posterior(param, samples):
smin, smax = np.min(samples), np.max(samples)
width = smax - smin
x = np.linspace(smin, smax, 100)
y = stats.gaussian_kde(samples)(x)
# what was never sampled should have a small probability but not 0,
# so we'll extend the domain and use linear approximation of density on it
x = np.concatenate([[x[0] - 3 * width], x, [x[-1] + 3 * width]])
y = np.concatenate([[0], y, [0]])
return pm.distributions.Interpolated(param, x, y)
if __name__ == '__main__':
Data = pd.DataFrame([["A1", "B1", -1], ["A1", "B2", 1]], columns=["A", "B", "Y"])
Distributions = {}
InitialSigma = 2
InitialNoise = 2
for i in Data.index:
with pm.Model() as my_model:
try:
A = Distributions[Data["A"].loc[i]]
except:
A = pm.Normal('A', 0, InitialSigma)
try:
B = Distributions[Data["B"].loc[i]]
except:
B = pm.Normal('B', 0, InitialSigma)
try:
eps = Distributions["Noise"]
except:
eps = pm.HalfCauchy('eps', InitialNoise)
Y = A + B
y = pm.Normal('y', Y, sigma=eps, observed=Data.loc[i,"Y"])
with my_model:
trace = pm.sample(draws=1000, tune=200)
Distributions[Data["A"].loc[i]] = A
Distributions[Data["B"].loc[i]] = B
Distributions["Noise"] = eps``````
I modified the example you provided to handle a situation in which there are multiple independent parameter settings with different alpha, beta0, and beta1. The initial sampling works well, however the code is erroring out when it gets to the from_posterior() function because my trace[‘alpha’] is shape (4000,2) and the function is expecting trace[‘alpha’] to be shape (4000,).
Would you be able to help me modify the from_posterior() function to handle multidimensional sets of distributions? I feel like it would be simple enough to make vectors out of the variables smin, smax, width, x, and y but I don’t know how to use the pymc3 Interpolated function to combine those into a set of distributions.
``````import matplotlib.pyplot as plt
import matplotlib as mpl
import pymc3 as pm
from pymc3 import Model, Normal, Slice
from pymc3 import sample
from pymc3 import traceplot
from pymc3.distributions import Interpolated
from theano import as_op
import theano.tensor as tt
import numpy as np
from scipy import stats
from random import choice
import pandas as pd
if __name__ == '__main__':
plt.style.use('seaborn-darkgrid')
print('Running on PyMC3 v{}'.format(pm.__version__))
# Initialize random number generator
np.random.seed(93457)
# True parameter values
alpha_true = [5, -5]
beta0_true = [7, 17]
beta1_true = [13, 23]
# Size of dataset
size = 100
# Predictor variable
X1 = np.random.randn(size)
X2 = np.random.randn(size) * 0.2
# for different data points, different known parameter settings are applied, which means there are two values for alpha, two values for beta0, and two values for beta1
Data = pd.DataFrame(columns=["Y", "alpha", "beta0", "beta1"], index=range(0,size))
for point in range(0,size):
# randomly choose which alphas and betas are applied as parameters for this data point
alpha_choice = choice([0,1])
beta0_choice = choice([0,1])
beta1_choice = choice([0,1])
this_alpha = alpha_true[alpha_choice]
this_beta0 = beta0_true[beta0_choice]
this_beta1 = beta1_true[beta1_choice]
# Simulate outcome variable
this_Y = alpha_true[alpha_choice] + beta0_true[beta0_choice] * X1[point] + beta1_true[beta1_choice] * X2[point] + np.random.randn()
Data.loc[point] = [this_Y, alpha_choice, beta0_choice, beta1_choice]
basic_model = Model()
with basic_model:
# Priors for unknown model parameters
alpha = Normal('alpha', mu=0, sigma=1, shape=2)
beta0 = Normal('beta0', mu=12, sigma=1, shape=2)
beta1 = Normal('beta1', mu=18, sigma=1, shape=2)
# Expected value of outcome
mu = alpha[Data["alpha"].values.astype(int)] + beta0[Data["beta0"].values.astype(int)] + beta1[Data["beta1"].values.astype(int)]
# Likelihood (sampling distribution) of observations
Y_obs = Normal('Y_obs', mu=mu, sigma=1, observed=Data["Y"])
# draw 1000 posterior samples
trace = sample(1000)
traceplot(trace);
def from_posterior(param, samples):
smin, smax = np.min(samples), np.max(samples)
width = smax - smin
x = np.linspace(smin, smax, 100)
y = stats.gaussian_kde(samples)(x)
# what was never sampled should have a small probability but not 0,
# so we'll extend the domain and use linear approximation of density on it
x = np.concatenate([[x[0] - 3 * width], x, [x[-1] + 3 * width]])
y = np.concatenate([[0], y, [0]])
return Interpolated(param, x, y)
traces = [trace]
for _ in range(10):
# generate more data
X1 = np.random.randn(size)
X2 = np.random.randn(size) * 0.2
for point in range(0,size):
alpha_choice = choice([0,1])
beta0_choice = choice([0,1])
beta1_choice = choice([0,1])
this_alpha = alpha_true[alpha_choice]
this_beta0 = beta0_true[beta0_choice]
this_beta1 = beta1_true[beta1_choice]
this_Y = alpha_true[alpha_choice] + beta0_true[beta0_choice] * X1[point] + beta1_true[beta1_choice] * X2[point] + np.random.randn()
Data.loc[point] = [this_Y, alpha_choice, beta0_choice, beta1_choice]
model = Model()
with model:
# Priors are posteriors from previous iteration
alpha = from_posterior('alpha', trace['alpha'])
beta0 = from_posterior('beta0', trace['beta0'])
beta1 = from_posterior('beta1', trace['beta1'])
# Expected value of outcome
mu = alpha[Data["alpha"].values.astype(int)] + beta0[Data["beta0"].values.astype(int)] + beta1[Data["beta1"].values.astype(int)]
# Likelihood (sampling distribution) of observations
Y_obs = Normal('Y_obs', mu=mu, sigma=1, observed=Data["Y"])
# draw 1000 posterior samples
trace = sample(1000)
traces.append(trace)
print('Posterior distributions after ' + str(len(traces)) + ' iterations.')
cmap = mpl.cm.autumn
for param in ['alpha', 'beta0', 'beta1']:
plt.figure(figsize=(8, 2))
for update_i, trace in enumerate(traces):
samples = trace[param]
smin, smax = np.min(samples), np.max(samples)
x = np.linspace(smin, smax, 100)
y = stats.gaussian_kde(samples)(x)
plt.plot(x, y, color=cmap(1 - update_i / len(traces)))
plt.axvline({'alpha': alpha_true, 'beta0': beta0_true, 'beta1': beta1_true}[param], c='k')
plt.ylabel('Frequency')
plt.title(param)
plt.tight_layout();`````` | 2,551 | 9,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.917165 |
http://www.chegg.com/homework-help/definitions/third-law-of-motion-2 | 1,441,191,752,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645261055.52/warc/CC-MAIN-20150827031421-00184-ip-10-171-96-226.ec2.internal.warc.gz | 363,838,389 | 14,529 | # Definition of Third Law of Motion
Isaac Newton's third law of motion states that forces always occur in pairs. If an object A exerts force on an object B, then object B exerts an equal and opposite force on object A. The law is also commonly stated as "To every action there is an equal and opposite action." However, although the forces are exactly equal and opposite in both magnitude and direction, the acceleration due to the forces is not (by Newton's second law of motion). For example, when people jump, they apply a force to the earth below them, and the earth applies an equal and opposite force on them. However, although they move, the earth doesn't (at least not very much) because its mass is so much greater.
### Get Definitions of Key Science Concepts from Chegg
In science there are many key concepts and terms that are crucial for students to know and understand. Often it can be hard to determine what the most important science concepts and terms are, and even once you’ve identified them you still need to understand what they mean. To help you learn and understand key science terms and concepts, we’ve identified some of the most important ones and provided detailed definitions for them, written and compiled by Chegg experts. | 256 | 1,254 | {"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-2015-35 | longest | en | 0.96207 |
https://examandmore.com/result/ap-statistics-exam-answers-2009-semester-1st | 1,611,459,330,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00028.warc.gz | 336,263,916 | 4,333 | # Preparing for Exams
### Other results for Ap Statistics Exam Answers 2009 Semester 1St:
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### AP statistics 1st semester exam Flashcards | Quizlet
Start studying AP statistics 1st semester exam. Learn vocabulary, terms, and more with flashcards, games, and other study tools.
Date: 15-3-2020
Tags: test
### AP Statistics: The Exam | AP Central – The College Board
Visit Taking Online AP Exams for the latest exam information. New Secure Practice Exam The international 2019 AP Statistics Exam is now available on the AP Course Audit site and in the AP Classroom question bank, along with scoring guidelines and a scoring worksheet. You continue to have access to the international 2018 and 2017 exams.
Date: 27-1-2020
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### AP Statistics, First Semester Exam 2010-11
AP Statistics, First Semester Exam 2010-11 ... Use the following to answer question 1. During the early part of the 1994 baseball season, many sports fans and baseball players noticed that the number of home runs being hit seemed to be unusually large. Below are the team-by-team statistics on home runs hit through Friday, June 3, 1994 (from the ...
Date: 14-3-2020
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### AP stat - First Semester Review (Part A)
AP stat - First Semester Review (Part A) AP stat - First Semester Review (Part A) ... Multiple Choice Strategies for the AP Statistics Exam - Duration: 7:07. Mrmathblog 24,672 views.
Date: 11-2-2020
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### AP Statistics First Semester Final Exam Review
AP Statistics First Semester Final Exam Review Author: Someone Else Created Date: 4/12/2017 9:06:52 PM ...
Date: 17-3-2020
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### AP Statistics First Semester Review
I will be discussing some details of my first semester exam for AP Statistics, answering questions from my students, and going over a few example FRQs.
Date: 16-1-2020
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### Every AP Statistics Practice Test Available: Free and Official
2012 AP Statistics Released Exam. 1997 AP Statistics Released Exam. Both links include the complete exam, an answer key, and scoring information. Both of these are very useful study resources, even the 1997 exam since the AP Stats exam hasn't changed much since then. This is the current exam is three hours long with two sections.
Date: 27-2-2020
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### AP Statistics 2018 Free-Response Questions
(I) Descriptive Statistics-3-xi x n Ç 1 2 sxx x n 1Çi 22 1212 12 11 11 p ns n s s nn 01 y ˆ bbx 1 2 ii i x xy y b xx Ç Ç 01 bybx 1 1 ii xy x xy y r ns s ÈØ ÈØ ÇÉÙÉÙ ÊÚÊÚ 1 y x s br s 2 1 2 ˆ 2 ii b i yy s n x x Ç Ç 2018 AP ® STATISTICS FREE-RESPONSE QUESTIONS
Date: 23-1-2020
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### AP Statistics - Wikipedia
Advanced Placement Statistics (AP Statistics, AP Stat or AP Stats) is a college-level high school statistics course offered in the United States through the College Board's Advanced Placement program. This course is equivalent to a one semester, non-calculus-based introductory college statistics course and is normally offered to sophomores, juniors and seniors in high school.
Date: 19-1-2020
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### Nelson, Howard / AP Stats Review
AP Statistics Summer Assignment; AP Statistics Chapter 0; AP Statistics HW Assignment; AP Statistics Chapter 1; AP Statistics Chapter 2; AP Statistics Chapter 3; AP Statistics Chapter 4; AP Statistics Chapter 5; Second Semester Project; AP Statistics Chapter 6; AP Statistics Chapter 8; AP Statistics Chapter 7; Inference; AP Stats Review ... | 923 | 3,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-04 | latest | en | 0.880424 |
https://forum.alphasoftware.com/showthread.php?63337-Getting-a-Quick-Sum-of-a-Column&s=054010b19347d32491b4a9ced9007217 | 1,563,903,085,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529481.73/warc/CC-MAIN-20190723172209-20190723194209-00019.warc.gz | 396,102,362 | 16,642 | # Thread: Getting a Quick Sum of a Column
1. ## Getting a Quick Sum of a Column
Greetings,
Is there a way, in A5V7, to get a sum of a column, without creating a report or writing Xbasic to read every record in the file?
I have a transaction file that has several hundred rows and I have a value of those transactions based on a monetary filed in the data.
I promise to wash my mouth afterwards, in MS-Access you can do a Quick Query and sum the column and have a total in seconds with designing a report per-sae or writing code... can you do the equivalent in A5?
Regards,
Keith
2. ## Re: Getting a Quick Sum of a Column
Keith,
Not sure if this will suffice but on the toolbar next to the excel button is a yellow button for field statistics. Place cursor in the field you are interested in and press the button. The item second from the bottom is total.
3. ## Re: Getting a Quick Sum of a Column
Originally Posted by Tim Kiebert
Keith,
Not sure if this will suffice but on the toolbar next to the excel button is a yellow button for field statistics. Place cursor in the field you are interested in and press the button. The item second from the bottom is total.
Hi Tim,
THANK-YOU !!
How in the world that I missed that is unbelievable, the Total and Average are just what I needed! Now, I need to look into what the Expression Tab will allow, this WILL make my life easier !!
Regards,
Keith
4. ## Re: Getting a Quick Sum of a Column
Originally Posted by KeithW
Hi Tim,
THANK-YOU !!
How in the world that I missed that is unbelievable, the Total and Average are just what I needed! Now, I need to look into what the Expression Tab will allow, this WILL make my life easier !!
Regards,
Keith
Field statistics.
5. ## Re: Getting a Quick Sum of a Column
Keith, Happy to help. After a bit of a play I found that the result is based on the active record set. So you can filter the records your favourite way and the resulting records will be used in the statistics. This applies to embeded browses as well. Check out this page in the help on the related function
6. ## Re: Getting a Quick Sum of a Column
Wonder what that related function might be... Would like to be able to recreate the statistics button on an xdialog browse. But just copying the code from the toolbar doesn't do it - returns an argument incorrect data type error on line 43 using either addin.run("field_statistics") or a5_field_statistics()
7. ## Re: Getting a Quick Sum of a Column
Here's a few ideas I've played with.
Add this table to a workspace and look at the code in the browse.
testgrptotal.zip
8. ## Re: Getting a Quick Sum of a Column
That's pretty cool, but that is a button event that is part of the browse. I just want a stat button to work like it does on the browse toolbar but as part of the xdialog. Alpha auto adds in the navigation buttons, save and refresh in the dlg code when it gives you the xml definition. So it seems a stat button ought be possible too.
9. ## Re: Getting a Quick Sum of a Column
Robin,
I just want a stat button to work like it does on the browse toolbar but as part of the xdialog.
10. ## Re: Getting a Quick Sum of a Column
Hi John,
As I said in post in #6, I tried that but the code errs. I tried the code from both of the system form and browse toolbars.
11. ## Re: Getting a Quick Sum of a Column
Originally Posted by MoGrace
That's pretty cool, but that is a button event that is part of the browse. I just want a stat button to work like it does on the browse toolbar but as part of the xdialog. Alpha auto adds in the navigation buttons, save and refresh in the dlg code when it gives you the xml definition. So it seems a stat button ought be possible too.
This wasn't meant to be the final answer, but some options to get to it.
Can you post what you are starting with and info on how you want to improve it?
12. ## Re: Getting a Quick Sum of a Column
I found another function that might do it COMPUTE_FIELD_STATISTICS - I will try that then let you know how it went.
Edit: In using the script recorder on the xd to see what the sort buttons returned, I get this instead of the pointer to the browse object in the script:
generatedbrowse.quick_sort()
Well the Wiki is wrong. This function above returns a dot variable and is thus a pointer. How do I get the pointer results from the IW into a list I can display in a msgbox()? I am thinking I need an array? Where is Stan when you need him? But I am now off this thread since I haven't found a way to get a total SIMPLY which is what the OP was about.
...More Info from the IW using a numeric field - and it works with the current query:
Code:
```brw1 = topparent.this
v = brw1.GetActiveObjParent().active_drilldown()
?typeof(v)
= "C"
?v
= ":ApptList:HH_CNT"
stats = compute_field_statistics(v)
?typeof(stats)
= "P"
?stats
= Average = 0.363802559414991
count = 547
First = 5
Last = 1
Maximum = 6
Minimum = 0
Standard = 0.921610634766173
Total = 199
Variance = 0.849366162114108```
now with a regular browse open I can use
a5_field_statistics()
and it brings up the statistics dialog - from the IW. Notice there is not any arguments for this function or a pointer needed, so when the toolbar uses this function it gets the info it needs on its own just like the other toolbar buttons do. But of course not if the browse is embedded in an xdialog. I noticed the script recorder only showed results from some of the buttons Alpha added.
13. ## Re: Getting a Quick Sum of a Column
I am getting a bit closer to creating a function:
Code:
```'Note: brw1 is the pointer I used in the xdialog for the xml browse definition so
'defining it here is just for testing purposes
brw1 = topparent.this
objname = brw1.GetActiveObjParent().active_drilldown()
t = brw1.Table_Get()
tname = t.name_get()
fld = word(objname,-1,":")
fptr = a5_field_info(tname,fld)
stats = compute_field_statistics(fld)
x=properties_enum(stats,"R,O=p: v")
msgbox("Field Stats",ut(fld)+" Statistics:"+crlf(2)+x,0)```
If you open a standalone browse and set a query, you can copy this to the IW and run it. If a numeric field is selected then
stats.total will return the value.
14. ## Re: Getting a Quick Sum of a Column
I have since added another embedded xml browse in my XD so I changed the statistics button to read the field stats from either one:
Code:
```brw = topparent.this
objname = brw.GetActiveObjParent().active_drilldown()
t = brw.Table_Get()
tname = t.name_get()
fld = word(objname,-1,":")
fptr = a5_field_info(tname,fld)
stats = compute_field_statistics(fld)
x=properties_enum(stats,"R,O=p: v")
msgbox("Field Stats",ut(fld)+" Statistics:"+crlf(2)+x,0)```
Note that now 'topparent.this' returns whichever browse is active.
15. ## Re: Getting a Quick Sum of a Column
Originally Posted by MoGrace
Wonder what that related function might be... Would like to be able to recreate the statistics button on an xdialog browse. But just copying the code from the toolbar doesn't do it - returns an argument incorrect data type error on line 43 using either addin.run("field_statistics") or a5_field_statistics()
Finally got an error message that reveals something - from the IW without setting the session I got:
a5_field_statistics()
ERROR: Script:field_statistics line:43
topparent.child_enum("r,e=this.field_get().fullname_get()"): Argument is incorrect data type
It seems that the valid pointer for 'topparent' is what that function expects to find.
Edit: And that valid pointer for an xml browse in an Xdialog is 'generatedbrowse.this'
so here is my event code for the Statistics button on my XD that actually works:
Code:
```If a_dlg_button = "b_Stats" then
if brw1.active() <> "" then
fld = brw1.active()
brw = generatedbrowse.this
else
end
end if
'get info for the msgbox
t = brw.Table_Get()
tname = t.name_get()
'get the fully qualified object name
objname = brw.GetActiveObjParent().active_drilldown()
stats = compute_field_statistics(objname)
x=properties_enum(stats,"R,O=p: v")
delete stats
msgbox("Field Stats",ut(tname)-"->"-ut(fld)+" Statistics:"+crlf(2)+x,0)
end if```
I couldn't however, get a valid pointer to the 2nd browse in the XD, but didn't really need it anyway. It seems since both browses have an active object, but only the 1st one defined gets used.
Note that a numeric field gets a Total entry:
Stat_Btn.jpg
#### Posting Permissions
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• | 2,112 | 8,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-30 | latest | en | 0.912409 |
https://leetcode.ca/2024-07-05-3196-Maximize-Total-Cost-of-Alternating-Subarrays/ | 1,721,889,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00608.warc.gz | 311,565,416 | 8,495 | # 3196. Maximize Total Cost of Alternating Subarrays
## Description
You are given an integer array nums with length n.
The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as:
cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (−1)r − l
Your task is to split nums into subarrays such that the total cost of the subarrays is maximized, ensuring each element belongs to exactly one subarray.
Formally, if nums is split into k subarrays, where k > 1, at indices i1, i2, ..., ik − 1, where 0 <= i1 < i2 < ... < ik - 1 < n - 1, then the total cost will be:
cost(0, i1) + cost(i1 + 1, i2) + ... + cost(ik − 1 + 1, n − 1)
Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally.
Note: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).
Example 1:
Input: nums = [1,-2,3,4]
Output: 10
Explanation:
One way to maximize the total cost is by splitting [1, -2, 3, 4] into subarrays [1, -2, 3] and [4]. The total cost will be (1 + 2 + 3) + 4 = 10.
Example 2:
Input: nums = [1,-1,1,-1]
Output: 4
Explanation:
One way to maximize the total cost is by splitting [1, -1, 1, -1] into subarrays [1, -1] and [1, -1]. The total cost will be (1 + 1) + (1 + 1) = 4.
Example 3:
Input: nums = [0]
Output: 0
Explanation:
We cannot split the array further, so the answer is 0.
Example 4:
Input: nums = [1,-1]
Output: 2
Explanation:
Selecting the whole array gives a total cost of 1 + 1 = 2, which is the maximum.
Constraints:
• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109
## Solutions
### Solution 1: Memoization
Based on the problem description, if the current number has not been flipped, then the next one can either be flipped or not flipped; if the current number has been flipped, then the next one can only remain unflipped.
Therefore, we define a function $\text{dfs}(i, j)$, which represents starting from the $i$-th number, whether the $i$-th number can be flipped, where $j$ indicates whether the $i$-th number is flipped. If $j = 0$, it means the $i$-th number cannot be flipped, otherwise, it can be flipped. The answer is $\text{dfs}(0, 0)$.
The execution process of the function $dfs(i, j)$ is as follows:
• If $i \geq \text{len}(nums)$, it means the array has been fully traversed, return $0$;
• Otherwise, the $i$-th number can remain unflipped, in which case the answer is $nums[i] + \text{dfs}(i + 1, 1)$; if $j = 1$, it means the $i$-th number can be flipped, in which case the answer is $\max(\text{dfs}(i + 1, 0) - nums[i])$. We take the maximum of the two.
To avoid repeated calculations, we can use memoization to save the results that have already been computed.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.
• class Solution {
private Long[][] f;
private int[] nums;
private int n;
public long maximumTotalCost(int[] nums) {
n = nums.length;
this.nums = nums;
f = new Long[n][2];
return dfs(0, 0);
}
private long dfs(int i, int j) {
if (i >= n) {
return 0;
}
if (f[i][j] != null) {
return f[i][j];
}
f[i][j] = nums[i] + dfs(i + 1, 1);
if (j == 1) {
f[i][j] = Math.max(f[i][j], -nums[i] + dfs(i + 1, 0));
}
return f[i][j];
}
}
• class Solution {
public:
long long maximumTotalCost(vector<int>& nums) {
int n = nums.size();
long long f[n][2];
fill(f[0], f[n], LLONG_MIN);
auto dfs = [&](auto&& dfs, int i, int j) -> long long {
if (i >= n) {
return 0;
}
if (f[i][j] != LLONG_MIN) {
return f[i][j];
}
f[i][j] = nums[i] + dfs(dfs, i + 1, 1);
if (j) {
f[i][j] = max(f[i][j], -nums[i] + dfs(dfs, i + 1, 0));
}
return f[i][j];
};
return dfs(dfs, 0, 0);
}
};
• class Solution:
def maximumTotalCost(self, nums: List[int]) -> int:
@cache
def dfs(i: int, j: int) -> int:
if i >= len(nums):
return 0
ans = nums[i] + dfs(i + 1, 1)
if j == 1:
ans = max(ans, -nums[i] + dfs(i + 1, 0))
return ans
return dfs(0, 0)
• func maximumTotalCost(nums []int) int64 {
n := len(nums)
f := make([][2]int64, n)
for i := range f {
f[i] = [2]int64{-1e18, -1e18}
}
var dfs func(int, int) int64
dfs = func(i, j int) int64 {
if i >= n {
return 0
}
if f[i][j] != -1e18 {
return f[i][j]
}
f[i][j] = int64(nums[i]) + dfs(i+1, 1)
if j > 0 {
f[i][j] = max(f[i][j], int64(-nums[i])+dfs(i+1, 0))
}
return f[i][j]
}
return dfs(0, 0)
}
• function maximumTotalCost(nums: number[]): number {
const n = nums.length;
const f: number[][] = Array.from({ length: n }, () => Array(2).fill(-Infinity));
const dfs = (i: number, j: number): number => {
if (i >= n) {
return 0;
}
if (f[i][j] !== -Infinity) {
return f[i][j];
}
f[i][j] = nums[i] + dfs(i + 1, 1);
if (j) {
f[i][j] = Math.max(f[i][j], -nums[i] + dfs(i + 1, 0));
}
return f[i][j];
};
return dfs(0, 0);
} | 1,630 | 4,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-30 | latest | en | 0.725977 |
http://tonetastic.info/read/how-to-wire-a-light-switch-to-volt-battery-sportissimo-html-rh-thisoldtractor-basic-12-volt-wiring-diagrams-wiring-volt-batteries-for-86 | 1,679,486,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00663.warc.gz | 52,960,594 | 6,388 | # How To Wire A Light Switch To, Volt Battery New Sportissimo Html Rh Thisoldtractor, Basic 12 Volt Wiring Diagrams Wiring, Volt Batteries For Photos
## Other recommended diagram ideas:
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Now we come to led riding unit. Right here, we use equal logic of ahead biased diode to lessen voltage. The battery voltage of 6volt smf battery is around 6.Four-6.6volts. Right here we connect five ahead biased in4007 diode to lessen voltage with the aid of appx 3volt, so the led is fed with round 3.5volts. If we require more brightness, we can quick any person of the diodes to growth the voltage by using zero.6volt. The motive force transistor here used is d880, we've got used it in many circuits, as it's far a without difficulty available npn transistor with high present day ratings. We bias the base via a 10k resistor from the 6v rail, and reverse bias it thru a ldr to gnd. As a consequence, when there may be mild on ldr, the transistor may be unsaturated, and vice versa. The entire parts cost in this circuit is 30-35 rupees. We can also make a pcb in preference to making it on vero.
12volt smf batteries comes around 650-seven hundred rupees, even as 6volt smf batteries comes in 150-2 hundred rupees, usually referred to as charger battery. These have 4ah rating at 6volt. That’s of the source and storage, now we consider the output. Yes, we are able to be the usage of a led of excessive leumens of route. 1watt or 3watt led calls for 3.Five-4volt forward voltage to glow at full depth. We additionally want to offer a manner that the circuit senses light and turns on the led on every occasion there's darkness(night time).
Within the diagram, we will see 12volt deliver from solar panel(5watt) is regulated by way of lm7806 i/c to charge 6volt battery. Considering that 6volt batteries charge at 7.2-7.5volt, we added two ahead biased in4007 diode to reference degree, so that the pair makes a zener breakdown at 1.2volts, making the output at 7.2volts. That voltage charges battery at most 5/12=400ma(appx) which is a good modern-day level, as it's miles precisely c/10 of that of battery. So, this will never overcharge the battery. | 623 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.903327 |
http://www.jsoftware.com/help/dictionary/dictg.htm | 1,386,491,732,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163057372/warc/CC-MAIN-20131204131737-00034-ip-10-33-133-15.ec2.internal.warc.gz | 397,604,358 | 3,109 | >> << Ndx Usr Pri JfC LJ Phr Dic Rel Voc !: wd Help Dictionary
G. Extended and Rational Arithmetic
Extended precision integer constants can be entered as a sequence of decimal digits terminated by an x. The monad x: applies to integers and produces extended integers. For example, the 2-element vector 1234x 56x (or 1234 56x) is equivalent to x: 1234 56. Various primitives produce extended results if the argument(s) are extended. Some verbs f produce floating point (inexact) results on some extended arguments because the result is not integral; however, <.@f and >.@f produce extended integer results when applied to extended integer arguments. Comparisons involving extended integers are exact. For example:
``` !40
8.15915e47
!40x
815915283247897734345611269596115894272000000000
*/ x: >: i.40
815915283247897734345611269596115894272000000000
0j_25 ": ! 2000x * i. 5 1 NB. Exponent format, 25 digits
1.0000000000000000000000000e0
3.3162750924506332411753934e5735
1.8288019515140650133147432e12673
2.6839997657267395961163166e20065
5.1841810604808769398058198e27752
] r=: <.@%: 2 * 10 ^ 56x
14142135623730950488016887242
,. *: r + _1 0 1
199999999999999999999999999968972697904100132394908592081
199999999999999999999999999997256969151562033370942366564
200000000000000000000000000025541240399023934346976141049
```
Rational constants can be entered as the decimal digits of the numerator and denominator, separated by an r and preceded by an optional sign. Thus 3r4 is the rational number three-quarters and _12r5 is negative 12 divided by 5. Rational numbers are stored and displayed in a standard form, with the numerator and denominator relatively prime and the denominator positive. Thus:
``` 1r2 _1r2 2r4 2r_4 _2r_4 0r9 5 _5
1r2 _1r2 1r2 _1r2 1r2 0 5 _5
```
Various primitive verbs produce (exact) rational results if the argument(s) are rational; non-rational verbs produce (inexact) floating point or complex results when applied to rationals, if the verb only has a limited number of rational arguments that produce rational results. (For example, %:y is rational if the atoms of y are perfect squares; ^0r1 is floating point.) The quotient of two extended integers is an extended integer (if evenly divisible) or rational (if not). Comparisons involving two rationals are exact. Dyadic verbs (e.g. + - * % , = <) that require argument type conversions do so according to the following table:
``` | B I X Q D Z
---+------------------
B | B I X Q D Z B - boolean
I | I I X Q D Z I - integer
X | X X X Q D Z X - extended integer
Q | Q Q Q Q D Z Q - rational
D | D D D D D Z D - floating point
Z | Z Z Z Z Z Z Z - complex
```
For example, in the expression 2.5+1r2, the 1r2 is converted to 0.5 before being added to 2.5, resulting in a floating point 3. And in the expression 2+1r2, the 2 is converted to 2r1 before being added to 1r2, resulting in 5r2.
In particular, a comparison involving a rational and a floating point number is tolerant, because the rational argument is first converted into a floating point number.
The verb x: produces rational approximations to non-rational arguments.
``` 2%3
0.666667
2%3x
2r3
(+%)/\10\$1 NB. Floating point convergents to golden mean
1 2 1.5 1.66667 1.6 1.625 1.61538 1.61905 1.61765 1.61818
(+%)/\x: 10\$1 NB. Rational versions of same
1 2 3r2 5r3 8r5 13r8 21r13 34r21 55r34 89r55
|: 2 x: (+%)/\x: 10\$1
1 2 3 5 8 13 21 34 55 89
1 1 2 3 5 8 13 21 34 55
(+%)/ 100\$1r1
573147844013817084101r354224848179261915075
0j30 ": (+%)/100\$1r1 NB. Display 30 decimal places
1.618033988749894848204586834366
H=: % @: >: @: (+/~) @: i. @ x: NB. Hilbert matrix of order n
] h=: H 6
1 1r2 1r3 1r4 1r5 1r6
1r2 1r3 1r4 1r5 1r6 1r7
1r3 1r4 1r5 1r6 1r7 1r8
1r4 1r5 1r6 1r7 1r8 1r9
1r5 1r6 1r7 1r8 1r9 1r10
1r6 1r7 1r8 1r9 1r10 1r11
-/ .* h NB. Determinant of h
1r186313420339200000
~. q: % -/ .* h NB. Unique prime factors of reciprocal of det
2 3 5 7 11
i.&.(p:^:_1) 2*#h NB. Primes less than 2*n
2 3 5 7 11
^ 2r1 NB. ^y is floating point or complex
7.38906
%: 49r25 NB. %: on a rational perfect square is rational
7r5
%: 49r25 10r9
1.4 1.05409
%: _2r1
0j1.41421
1 = 1+10r1^_15 NB. Exact (rational) comparison
0
(1.5-0.5) = 1+10r1^_15 NB. Tolerant (floating point) comparison
1
0.5 = 1r2
1
```
>> << Ndx Usr Pri JfC LJ Phr Dic Rel Voc !: wd Help Dictionary | 1,687 | 4,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-48 | longest | en | 0.637815 |
https://www.reddit.com/user/Goonder | 1,521,485,507,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647044.86/warc/CC-MAIN-20180319175337-20180319195337-00541.warc.gz | 884,486,138 | 27,911 | ×
[–] 1 point2 points (0 children)
Nice thanks, also the car looks dope man!
[–] 0 points1 point (0 children)
how much did it take to do the wheels? was looking to do mine soon also a SEL
[–] 0 points1 point (0 children)
Must be a super old facility because we only have a area like that for very small items in ours.
[–] 1 point2 points (0 children)
How old is the FC you work at that the pickers don't just drive OP's around in? We have 40 ft high bins we drive around and pick orders out of so if by running you mean standing on a machine and driving around for 10 hours then sure but its not that bad I do it lol
[–] 1 point2 points (0 children)
Most of the time people who watch and understand the sport can spot the moves that are commonly used to pin someone so when those start happening it gets exciting but at the same time there is a point system. So if time runs out without each person getting pinned the person with more points wins so you also have to factor that into watching since stuff that could seem very little and simple could lead to the whole match changing. It's mostly based on who has control of who throughout the match so gaining and losing control will earn you or your opponent points.
[–] 1 point2 points (0 children)
Winning a match in wrestling is actually like the WWE. If you pin the other person on their back for a 3 count they lose, but more specifically in HS/college wrestling not only do you have to be on your back but both of your shoulders have to be against the mat. So most of the time the ref during matches will get down like this to see the angle of both shoulders but while this happens they can start moving while trying to get both shoulders down. Instead of getting up and moving to a new position this ref just decided to slide across the mat to keep his angle. Also since there are 2 refs not just a single one this match was probably a pretty high valued one so that could factor into him doing this.
[–][S] 0 points1 point (0 children)
haha no its a joke since these shoes are rare to find
[–] -1 points0 points (0 children)
Then go learn net code programming and solve the problem that has been a bug in the gaming industry since MM existed or stop complaining about the most known issue that if you did any research probably has some explanation as to why this can rarely happen.
[–] 1 point2 points (0 children)
Have you played any other game that uses a matchmaking system like this? All of them can do this, its common fucking sense to just re queue if you every played another MM game. This shit goes back to Halo 2 MM this is not a new problem or specifically just a problem for siege's MM system.
[–] 3 points4 points (0 children)
They make the DLCs so more 12 year olds use their parents cards to buy fake money for all the over priced shit they come out with that adds nothing to the game. You could pay the same amount for a shitty sharkcard to a modder and get 10x the money. The discord I have is selling 100 million for \$20.
[–] 19 points20 points (0 children)
1k hours building doors
[–][S] 1 point2 points (0 children)
Well for air shows they open the base up to the public when you normally need a military ID to get on the base so it makes sense to have some show of force when dealing with civilians on government property.
[–][S] 0 points1 point (0 children)
I should upload the photo I got of them doing a formation break that actual does look terrifying when coming straight at you.
[–][S] 0 points1 point (0 children)
Also they had 3 on display so about 3 billion dollars worth of government property.
[–][S] 0 points1 point (0 children)
Seeing how the event was on the Air Force base which would be government property I'm sure they could use lethal force. I asked my dad, who is 20+ years retired from USAF, as a joke what would happen if someone were to go under the ropes and he said they wouldn't even ask you to go back but just straight up tackle you and probably arrest you.
[–][S] 1 point2 points (0 children)
The chairs were under the far left one which had 3 other guys sitting in them lol
[–] 7 points8 points (0 children)
Also it was just a hearing, there wasn't anything to really watch I just listened to the live stream while working.
[–] 8 points9 points (0 children)
Should tell your team to up their actually hockey skills so you make it past the first round
[–] 0 points1 point (0 children)
Also besides the points everyone else has already made if you know what the old CODs used to be like and how some of them were much more suitable for a competitive fps, makes you wish they would have never added all the extra bullshit into the game and kept it simple. That's the main reason I can't watch it because I wish they would have supported the pro mod version on PC with COD 4 where the skill gap and curve was much higher like CS. The most skillful part of COD nowadays is just knowing spawn points and having decent teammates. | 1,193 | 4,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-13 | latest | en | 0.973596 |
https://doc.qt.io/qt-6/qmediatimerange-interval.html | 1,675,757,129,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00197.warc.gz | 234,591,395 | 7,204 | Interval Struct
struct QMediaTimeRange::Interval
The QMediaTimeRange::Interval class represents a time interval with integer precision. More...
Public Functions
Interval(qint64 start, qint64 end) bool contains(qint64 time) const qint64 end() const bool isNormal() const Interval normalized() const qint64 start() const Interval translated(qint64 offset) const
bool operator!=(Interval lhs, Interval rhs) bool operator==(Interval lhs, Interval rhs)
Detailed Description
An interval is specified by an inclusive start() and end() time. These must be set in the constructor, as this is an immutable class. The specific units of time represented by the class have not been defined - it is suitable for any times which can be represented by a signed 64 bit integer.
The isNormal() method determines if a time interval is normal (a normal time interval has start() <= end()). A normal interval can be received from an abnormal interval by calling the normalized() method.
The contains() method determines if a specified time lies within the time interval.
The translated() method returns a time interval which has been translated forwards or backwards through time by a specified offset.
Member Function Documentation
Interval::Interval(qint64start, qint64end)
Constructs an interval with the specified start and end times.
bool Interval::contains(qint64time) const
Returns true if the time interval contains the specified time. That is, start() <= time <= end().
qint64 Interval::end() const
Returns the end time of the interval.
bool Interval::isNormal() const
Returns true if this time interval is normal. A normal time interval has start() <= end().
Interval Interval::normalized() const
Returns a normalized version of this interval.
If the start() time of the interval is greater than the end() time, then the returned interval has the start and end times swapped.
qint64 Interval::start() const
Returns the start time of the interval.
Interval Interval::translated(qint64offset) const
Returns a copy of this time interval, translated by a value of offset. An interval can be moved forward through time with a positive offset, or backward through time with a negative offset.
Related Non-Members
booloperator!=(Intervallhs, Intervalrhs)
Returns true if lhs is not exactly equal to rhs.
booloperator==(Intervallhs, Intervalrhs)
Returns true if lhs is exactly equal to rhs. | 500 | 2,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | longest | en | 0.721273 |
http://www.msnucleus.org/membership/ngss/sixth_ngss/06motion.html | 1,516,129,660,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00129.warc.gz | 504,784,720 | 6,501 | Weather, Climate and Change
Density of different layers of ocean can cause movement.
SIXTH GRADE - OCEANS IN MOTION
OBJECTIVES:
· Exploring and identifying water masses and currents.
·
Discovering how different water masses cause circulation throughout the oceans.
VOCABULARY:
·
currents
·
Ekman Spiral.
·
Water mass
·
salinity
MATERIALS:
·
density timers (with 3 colors)
·
big test tubes (two per pair of kids)
· food coloring
· salt
· spoons
·
worksheet (optional)
· powerpoint
https://msnucleus.org/membership/html/jh/earth/oceanography/lesson5/oceanography5a.html
BACKGROUND:
The oceans are always moving. The movement is graceful and subject to the principles of fluid motion. Fluid motion refers to the response of liquid to forces of wind, density, and rotation of the Earth in space. Locally you must include factors such as topography and tidal forces.
The waters of the oceans are not homogeneous. There are ocean masses that move as units, called currents. The ocean’s movement is predictable based on mathematical modeling. The model takes into account many factors that control currents not only on the surface but with depth
Coriolis does not alone explain movements, but other factors not related to movement such as density of water, wind and local submarine topography need to be included.
Wind also moves water around, especially the upper portions of the water column. However, wind is generally moving in the same direction caused by the rotation of the Earth, so they "add" forces. In areas where land and water meet, the direction of the wind is controlled by temperature differences of the land and water. Strong winds can influence the movement of coastal waters.
Wind is actually a bizarre phenomenon. It can change quickly depending on the balance of heat. A mountain can stop wind and add to the chaos of wind motion. Wind produces energy throughout the water column in a predictable pattern called the Ekman spiral. The transfer of wind energy from the atmosphere to the waters can cause motion far from where the energy first entered the system
Temperature and density can define a water mass. Warmer water can hold more salt, and colder water holds less salt. Salt water is more dense than fresh water of the same temperature. The salt water will layer itself below the fresh water. Warm water is less dense than cold water. So cold water will layer under warm water. Add the combination of different amounts of salt and different temperature and you have a layering effect in the water column. The world’s oceans are a three-dimensional nightmare of layers of different water masses that can move in different directions.
Bottom topography can act as a barrier to water masses already in motion. If a cold saline water mass is moving along the bottom and "hits" a mountain it would be forced upwards. This is one form of upwelling. This would displace the water masses above it, causing lots of movement.
All these reasons cause the movement of the oceans to spin into smaller eddies. Please keep in mind that although this motion looks chaotic, it really is responding to natural forces, which we can mathematically model.
Seawater density varies from place to place because it is affected by salinity and temperature. High salinity makes water denser. This is because there is more salt packed into the water.
.
PROCEDURE:
Go over powerpoint presentation on Ocean Movements
1. First slide review Coriolis effect (earth rotation) covered last time (should take about 2 minutes max)
2. Second slide. Remind the students about the barriers to ping pong balls from last time. What stopped them. Did the pinp pong balls just stop or move to the side or bounce back? Underwater has barriers too that influence ocean currents. Ask them to name some. (example, under water ridges, the continents, ice sheets, etc.)
3. Slides 3 to 6 Differences in the density of water produces water currents. cold, high saline water is very dense. When icebergs form, that means fresh water is taken out of the system, concentrating the salt in the water. This water is so dense it drops to the bottom and remains there for tens of years, as it travels on the bottom. The diagram shows "new" water starts in the North Atlantic and moves along the bottom. The cold water warms up slowly and emerges in the Pacific Ocean only to start circulating through different routes.
4. Dense water sinks below less dense water. This is the principle that drives the deep ocean currents that circulate around the world. A combination of high salinity and low temperature near the surface makes seawater dense enough to sink into the deep ocean and flow along the bottom of the basins
FIRST ACTIVITY—density timers
5. Give students a density timer with a warning not to shake the time violently. That ruins the density difference if small bubbles arise. Ask them to look at the 3 different liquids and see if they can figure out which is denser than the other. (The slower color to fall is the least dense.)
6. Put the density timer on its side and see if you can get the bottom layer to come upward. If they rock it back and forth the bottom current, will “upwell.” The sides of the timer are analogous to bottom denser water hitting to an underwater mountains and the water is forced upward. The rocking is basically a steady wind blowing in one direction which can bring up bottom water. Density in the oceans, wind, and bathymetry (underwater mountains) also help in making the motion in the oceans change.
SECOND ACTIVITY how different densities of water cause ocean currents.
PREPARATION: Fill all the test tubes up with fresh water to the line second from the top (the second one down) before the class. Rinse out the test tubes and re-apply fresh water between classes.
1. They will now work in teams of two. Each team gets two large test tubes one that says fresh on top and one that says salt. Ask them to swirl the test tubes to see if they can produce a “tornado” shape.
2. Take the salt test tube and ask them to add ½ teaspoon (it doesn’t have to be exact if you don’t have measuring spoons). They should reapply the cap tightly and then swirl the test tube around until all the salt is dissolved. When it is all dissolved, ask them to swirl the test tube and look at the “tornado”. Is it easier to form? Ask them to compare it to the freshwater one.
3. Ask them to add another ½ teaspoon of salt and repeat all steps.
4. Ask them to add another ½ teaspoon of salt and repeat all steps
5. Ask students which is denser, the water with salt or without (salt is denser)
6. Docent should then go around and add food coloring to the fresh water (about 4 to 8 drops (enough so it is a dark color).
7. We are now going to combine the two. Ask students to predict what will happen. The students should tilt the fresh water to the side and SLOWLY add the salt water to the fresh by pouring it onto the inside of the tilted test tube. (that is don’t pour it in the middle or it will get all mixed up.) What happens? The clear salty water should sink to the bottom.
8. Tell them you are going to give them an ice cube to add to the test tube. Ask them to predict what will happen. Which is more dense, the ice cube or the water (the water). As the water melts it is colder. Which is more dense, the cold water or the colored water? (the cold water) Add an ice cube on top (tell them to let it slip in slowly. Don’t drop it in to not disturb the layers.
9. Put the test tube back in the stand and watch. What do they think will happen as the ice cube melts?
10. Discuss: what happened with the melted ice? What happened with the layers. Why. How does this help them understand the ocean currents better? | 1,752 | 7,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-05 | latest | en | 0.902523 |
https://hp41programs.yolasite.com/batrachions.php | 1,726,675,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00822.warc.gz | 269,896,400 | 5,696 | Batrachions
# 3 Batrachions for the HP-41
Overview
2°) Conway's Batrachion
3°) Mallows's Batrachion
-These programs compute the successive terms of 3 recursive sequences
-The sequence is defined by H(1) = H(2) = 1 ; H(n) = H(n-H(n-1)) + H(n-H(n-2))
Data Registers: R00 = 0 ; R01 = H(1) ; R02 = H(2) ; .......... ; Rnn = H(n)
Flags: /
Subroutines: /
01 LBL "HOF" 02 .1 03 % 04 1 05 STO 01 06 ST+ Y 07 0 08 STO 00 09 STO 02 10 LBL 01 11 RCL Z 12 X<>Y 13 - 14 RDN 15 RCL IND T 16 X<>Y 17 ST- T 18 X<> L 19 X<>Y 20 RCL IND T 21 + 22 STO IND Z 23 ISG Z 24 GTO 01 25 END
( 43 bytes / SIZE nnn+1 )
STACK INPUTS OUTPUTS Z / 1+n.nnn Y / H(n-1) X n H(n)
Example: 100 XEQ "HOF" >>>> H(100) = 56 ( in 39 seconds ) and H(n) is in register Rnn for n < 101
-The firs few terms are:
n 1 2 3 4 5 6 7 8 9 10 H(n) 1 1 2 3 3 4 5 5 6 6
2°) Conway's Batrachion
-The sequence is defined by C(1) = C(2) = 1 ; C(n) = C(C(n-1)) + C(n-C(n-1)) ( n > 2 )
Data Registers: R00 unused ; R01 = C(1) ; R02 = C(2) ; .......... ; Rnn = C(n)
Flags: /
Subroutines: /
01 LBL "CNW" 02 3 03 X<=Y? 04 GTO 00 05 SIGN 06 RTN 07 LBL 00 08 X<>Y 09 E3 10 / 11 + 12 1 13 STO 01 14 STO 02 15 LBL 01 16 RCL Y 17 X<>Y 18 - 19 X<>Y 20 RCL IND Y 21 RCL IND L 22 + 23 STO IND Y 24 ISG Y 25 GTO 01 26 END
( 42 bytes / SIZE nnn+1 )
STACK INPUTS OUTPUTS Y / 1+n.nnn* X n C(n)
* if n > 2
Example: 100 XEQ "CNW" >>>> C(100) = 57 ( in 30 seconds ) and C(n) is in register Rnn for n < 101
-The first few terms are
n 1 2 3 4 5 6 7 8 9 10 C(n) 1 1 2 2 3 4 4 4 5 6
-We have C(2k) = 2k-1
-The graph of the ratio C(n)/n looks like a hopping frog and C(n)/n tends to 1/2 as n tends to infinity.
C(n)/n
| *
| * * *
| * * * * *
| * * * * * *
| * ** * * *
--|*-------*----------*------------*------------------------- n
3°) Mallows's Batrachion
-The sequence is defined by M(1) = M(2) = 1 ; M(n) = M(M(n-2)) + M(n-M(n-2))
Data Registers: R00 = 0 ; R01 = M(1) ; R02 = M(2) ; .......... ; Rnn = M(n)
Flags: /
Subroutines: /
01 LBL "MLW" 02 .1 03 % 04 1 05 STO 01 06 STO 02 07 ST+ Y 08 0 09 STO 00 10 LBL 01 11 X<>Y 12 RCL Z 13 X<>Y 14 - 15 RDN 16 RCL IND T 17 RCL IND L 18 + 19 STO IND Z 20 ISG Z 21 GTO 01 22 END
( 38 bytes / SIZE nnn+1 )
STACK INPUTS OUTPUTS Z / 1+n.nnn Y / M(n-1) X n M(n)
Example: 260 XEQ "MLW" >>>> M(260) = 180 ( in 85 seconds ) and M(n) is in register Rnn for n < 261
-The first few terms are
n 1 2 3 4 5 6 7 8 9 10 M(n) 1 1 2 3 3 4 5 6 6 7
Reference:
[1] Clifford A. Pickover - "Keys to Infinity" - John Wiley & Sons - ISBN 0-471-11857-5 | 1,344 | 2,897 | {"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-2024-38 | latest | en | 0.119524 |
https://socratic.org/questions/naphthalene-moth-balls-contains-93-71-carbon-and-6-29-hydrogen-its-molecular-mas | 1,571,632,868,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987756350.80/warc/CC-MAIN-20191021043233-20191021070733-00437.warc.gz | 707,510,771 | 6,583 | # Naphthalene (moth balls) contains 93.71% carbon and 6.29% hydrogen. Its molecular mass is 128 g mol-1, calculate its molecular formula?
Aug 17, 2017
${C}_{10} {H}_{8}$
#### Explanation:
Assuming that is a mass amount, divide each proportion by the atomic weight of the component. Normalize to obtain the empirical formula and then use the molecular mass to determine the actual molecular formula.
$\frac{93.91}{12} = 7.83$ relative moles of carbon
$\frac{6.29}{1} = 6.29$ relative moles of hydrogen
Ratio of C:H is 7.83:6.29 or 1.2. Nearest whole multiple is 5, resulting in ratio of 6:5. 6-C and 5-H would have a molecular weight of $\left(6 \times 12\right) + \left(5 \times 1\right) = 77$ The actual molecular weight is 128, so the formula must be $\frac{128}{77} = 1.67 \times {C}_{6} {H}_{5}$.
${C}_{10} {H}_{8}$
Subsequent note on Whole Multiple:
A chemical compound MUST have whole numbers of atoms - we can't have a compound with a "partial" atom! The observed/calculated math just gives us ratios of the mass. We need to insure that our final mass contains WHOLE atoms. Thus, when we find our molar ratio of C:H is 7.83:6.29 or 1.2:1 we need to convert that to one with whole numbers in the numerator as well as the denominator.
In this case, with the denominator already "normalized" at 1, we only need to find a factor to make the 1.2 a whole number. The key that 0.2 part - we need to make it $1.0$.
0.2 xx ? = 1.0
? = 1.0/0.2 = 5 Applying this to our original $\frac{1.2}{1}$ ratio we obtain:
$5 \times \frac{1.2}{1} = \frac{6}{5}$ So, our atomic molar ratio is 6-C and 5-H, as used in the rest of the original answer derivation. | 514 | 1,652 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-43 | longest | en | 0.881996 |
http://www.lmfdb.org/ArtinRepresentation/9.2e12_3e12_7e3.12t165.1c1 | 1,596,605,366,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735909.19/warc/CC-MAIN-20200805035535-20200805065535-00325.warc.gz | 144,225,990 | 9,264 | # Properties
Label 9.746636341248.12t165.a.a Dimension 9 Group $(A_4\wr C_2):C_2$ Conductor $2^{12} \cdot 3^{12} \cdot 7^{3}$ Root number 1 Frobenius-Schur indicator 1
# Related objects
## Basic invariants
Dimension: $9$ Group: $(A_4\wr C_2):C_2$ Conductor: $746636341248= 2^{12} \cdot 3^{12} \cdot 7^{3}$ Artin number field: Splitting field of 8.4.3186376704.2 defined by $f= x^{8} - 2 x^{7} - 4 x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 4 x^{2} - 8 x - 6$ over $\Q$ Size of Galois orbit: 1 Smallest containing permutation representation: 12T165 Parity: Even Determinant: 1.28.2t1.a.a Projective image: $(A_4\wr C_2):C_2$ Projective field: Galois closure of 8.4.3186376704.2
## Galois action
### Roots of defining polynomial
The roots of $f$ are computed in an extension of $\Q_{ 29 }$ to precision 75.
Minimal polynomial of a generator $a$ of $K$ over $\mathbb{Q}_{ 29 }$: $x^{3} + 2 x + 27$
Roots:
$r_{ 1 }$ $=$ $26 a^{2} + 19 a + 22 + \left(8 a^{2} + 11 a + 21\right)\cdot 29 + \left(27 a^{2} + 14 a + 22\right)\cdot 29^{2} + \left(2 a^{2} + 4 a + 19\right)\cdot 29^{3} + \left(18 a^{2} + a + 10\right)\cdot 29^{4} + \left(9 a^{2} + 17 a + 6\right)\cdot 29^{5} + \left(23 a^{2} + 21 a + 4\right)\cdot 29^{6} + \left(24 a^{2} + 27 a + 23\right)\cdot 29^{7} + \left(4 a^{2} + 22 a + 27\right)\cdot 29^{8} + \left(21 a^{2} + 28 a + 21\right)\cdot 29^{9} + \left(23 a^{2} + 2 a + 20\right)\cdot 29^{10} + \left(18 a^{2} + 19 a + 5\right)\cdot 29^{11} + \left(2 a + 19\right)\cdot 29^{12} + \left(12 a^{2} + 26 a + 24\right)\cdot 29^{13} + \left(13 a^{2} + 20 a\right)\cdot 29^{14} + \left(6 a^{2} + 22 a + 14\right)\cdot 29^{15} + \left(20 a^{2} + 13 a + 19\right)\cdot 29^{16} + \left(24 a^{2} + 16 a + 12\right)\cdot 29^{17} + \left(13 a^{2} + 24 a + 22\right)\cdot 29^{18} + \left(28 a^{2} + 5 a + 1\right)\cdot 29^{19} + \left(23 a^{2} + 28 a\right)\cdot 29^{20} + \left(2 a^{2} + 26 a + 27\right)\cdot 29^{21} + \left(16 a^{2} + 11 a + 8\right)\cdot 29^{22} + \left(25 a^{2} + 9 a + 27\right)\cdot 29^{23} + \left(25 a^{2} + 13 a\right)\cdot 29^{24} + \left(24 a^{2} + 28 a + 6\right)\cdot 29^{25} + \left(24 a^{2} + 26 a + 19\right)\cdot 29^{26} + \left(23 a^{2} + 18 a + 14\right)\cdot 29^{27} + \left(25 a^{2} + 8 a + 2\right)\cdot 29^{28} + \left(9 a^{2} + 16\right)\cdot 29^{29} + \left(3 a^{2} + 6 a + 8\right)\cdot 29^{30} + \left(23 a^{2} + 13 a + 4\right)\cdot 29^{31} + \left(2 a^{2} + 13 a + 4\right)\cdot 29^{32} + \left(26 a^{2} + 5 a + 25\right)\cdot 29^{33} + \left(18 a^{2} + 17 a\right)\cdot 29^{34} + \left(3 a^{2} + a + 10\right)\cdot 29^{35} + \left(10 a^{2} + 3 a + 21\right)\cdot 29^{36} + \left(22 a^{2} + 21 a + 1\right)\cdot 29^{37} + \left(3 a^{2} + 24\right)\cdot 29^{38} + \left(20 a^{2} + 4 a + 3\right)\cdot 29^{39} + \left(19 a^{2} + 19 a + 5\right)\cdot 29^{40} + \left(2 a^{2} + 28 a + 16\right)\cdot 29^{41} + \left(13 a^{2} + 6 a + 11\right)\cdot 29^{42} + \left(10 a^{2} + 11 a + 13\right)\cdot 29^{43} + \left(12 a^{2} + 6 a + 25\right)\cdot 29^{44} + \left(24 a^{2} + 22 a + 24\right)\cdot 29^{45} + \left(17 a^{2} + 3 a + 13\right)\cdot 29^{46} + \left(25 a^{2} + 8 a + 17\right)\cdot 29^{47} + \left(23 a^{2} + a + 14\right)\cdot 29^{48} + \left(23 a^{2} + a + 18\right)\cdot 29^{49} + \left(21 a^{2} + 27 a + 25\right)\cdot 29^{50} + \left(5 a^{2} + 23 a + 13\right)\cdot 29^{51} + \left(27 a^{2} + 22 a + 26\right)\cdot 29^{52} + \left(26 a^{2} + 26 a + 20\right)\cdot 29^{53} + \left(25 a^{2} + 10 a + 5\right)\cdot 29^{54} + \left(23 a^{2} + 5 a + 3\right)\cdot 29^{55} + \left(20 a^{2} + 8 a + 23\right)\cdot 29^{56} + \left(15 a + 22\right)\cdot 29^{57} + \left(8 a^{2} + 26 a + 22\right)\cdot 29^{58} + \left(24 a + 18\right)\cdot 29^{59} + \left(27 a^{2} + 21 a + 27\right)\cdot 29^{60} + \left(4 a^{2} + 10 a + 16\right)\cdot 29^{61} + \left(3 a^{2} + 21 a\right)\cdot 29^{62} + \left(20 a^{2} + 4 a + 12\right)\cdot 29^{63} + \left(14 a^{2} + 7 a + 9\right)\cdot 29^{64} + \left(12 a^{2} + \frac{175}{16} a + 14\right)\cdot 29^{65} + \left(27 a^{2} + a + 5\right)\cdot 29^{66} + \left(9 a^{2} + 21 a + 23\right)\cdot 29^{67} + \left(13 a^{2} + 25 a + 15\right)\cdot 29^{68} + \left(4 a^{2} + 14 a + 17\right)\cdot 29^{69} + \left(15 a^{2} + 24 a + 18\right)\cdot 29^{70} + \left(7 a^{2} + 10 a + 6\right)\cdot 29^{71} + \left(25 a^{2} + 21 a + 9\right)\cdot 29^{72} + \left(25 a^{2} + 2 a + 10\right)\cdot 29^{73} + \left(a^{2} + 28 a + 14\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 2 }$ $=$ $16 + 29 + 14\cdot 29^{2} + 17\cdot 29^{3} + 20\cdot 29^{4} + 22\cdot 29^{5} + 20\cdot 29^{6} + 22\cdot 29^{7} + 26\cdot 29^{8} + 8\cdot 29^{9} + 18\cdot 29^{10} + 22\cdot 29^{11} + 23\cdot 29^{12} + 19\cdot 29^{13} + 24\cdot 29^{14} + 10\cdot 29^{15} + 22\cdot 29^{16} + 10\cdot 29^{17} + 13\cdot 29^{18} + 9\cdot 29^{19} + 26\cdot 29^{20} + 2\cdot 29^{21} + 20\cdot 29^{22} + 5\cdot 29^{23} + 14\cdot 29^{24} + 15\cdot 29^{25} + 25\cdot 29^{26} + 26\cdot 29^{27} + 21\cdot 29^{28} + 22\cdot 29^{29} + 5\cdot 29^{30} + 21\cdot 29^{31} + 28\cdot 29^{32} + 22\cdot 29^{33} + 20\cdot 29^{34} + 22\cdot 29^{35} + 3\cdot 29^{36} + 6\cdot 29^{37} + 29^{38} + 10\cdot 29^{39} + 9\cdot 29^{40} + 6\cdot 29^{41} + 16\cdot 29^{42} + 29^{43} + 10\cdot 29^{44} + 3\cdot 29^{45} + 9\cdot 29^{46} + 22\cdot 29^{47} + 17\cdot 29^{48} + 3\cdot 29^{49} + 10\cdot 29^{50} + 22\cdot 29^{51} + 17\cdot 29^{52} + 8\cdot 29^{53} + 23\cdot 29^{54} + 4\cdot 29^{55} + 28\cdot 29^{56} + 17\cdot 29^{57} + 28\cdot 29^{58} + 28\cdot 29^{59} + 23\cdot 29^{60} + 13\cdot 29^{61} + 18\cdot 29^{62} + 4\cdot 29^{63} + 14\cdot 29^{64} + 18\cdot 29^{65} + 14\cdot 29^{66} + 8\cdot 29^{67} + 27\cdot 29^{68} + 17\cdot 29^{69} + 21\cdot 29^{70} + 22\cdot 29^{71} + 12\cdot 29^{72} + 4\cdot 29^{73} + 19\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 3 }$ $=$ $25 a^{2} + 5 + \left(10 a^{2} + 8 a + 4\right)\cdot 29 + \left(17 a^{2} + a + 12\right)\cdot 29^{2} + \left(2 a^{2} + 25 a + 10\right)\cdot 29^{3} + \left(a^{2} + 13 a + 20\right)\cdot 29^{4} + \left(11 a^{2} + 25 a + 24\right)\cdot 29^{5} + \left(3 a^{2} + 18 a + 21\right)\cdot 29^{6} + \left(19 a^{2} + 17 a + 27\right)\cdot 29^{7} + \left(24 a^{2} + 20 a + 10\right)\cdot 29^{8} + \left(17 a^{2} + 9 a + 11\right)\cdot 29^{9} + 4 a\cdot 29^{10} + \left(11 a^{2} + a + 9\right)\cdot 29^{11} + \left(7 a^{2} + 10 a + 3\right)\cdot 29^{12} + \left(12 a^{2} + 17 a + 7\right)\cdot 29^{13} + \left(4 a^{2} + 26 a + 14\right)\cdot 29^{14} + \left(16 a^{2} + 12 a + 14\right)\cdot 29^{15} + \left(22 a^{2} + 15 a + 5\right)\cdot 29^{16} + \left(23 a^{2} + 5 a + 12\right)\cdot 29^{17} + \left(5 a^{2} + 25 a + 17\right)\cdot 29^{18} + \left(13 a^{2} + 2 a + 14\right)\cdot 29^{19} + \left(20 a^{2} + 23 a + 25\right)\cdot 29^{20} + \left(13 a^{2} + 21 a + 11\right)\cdot 29^{21} + \left(22 a^{2} + 24 a + 18\right)\cdot 29^{22} + \left(9 a^{2} + 8 a + 9\right)\cdot 29^{23} + \left(23 a + 26\right)\cdot 29^{24} + \left(3 a^{2} + 3 a + 23\right)\cdot 29^{25} + \left(7 a^{2} + 12 a + 22\right)\cdot 29^{26} + \left(5 a^{2} + 5 a + 1\right)\cdot 29^{27} + \left(8 a^{2} + 9 a + 12\right)\cdot 29^{28} + \left(12 a^{2} + 7 a + 9\right)\cdot 29^{29} + \left(8 a^{2} + 28 a + 6\right)\cdot 29^{30} + \left(22 a^{2} + 23 a + 20\right)\cdot 29^{31} + \left(9 a^{2} + 26 a\right)\cdot 29^{32} + \left(4 a^{2} + 16 a + 24\right)\cdot 29^{33} + \left(12 a^{2} + 5 a + 19\right)\cdot 29^{34} + \left(19 a^{2} + 16\right)\cdot 29^{35} + \left(15 a^{2} + 9 a + 23\right)\cdot 29^{36} + \left(22 a^{2} + 10 a + 28\right)\cdot 29^{37} + \left(21 a^{2} + 8 a + 2\right)\cdot 29^{38} + \left(17 a^{2} + 13 a + 2\right)\cdot 29^{39} + \left(2 a^{2} + 23 a + 9\right)\cdot 29^{40} + \left(17 a^{2} + 8 a + 9\right)\cdot 29^{41} + \left(9 a^{2} + 25\right)\cdot 29^{42} + \left(20 a^{2} + 23 a + 27\right)\cdot 29^{43} + \left(12 a^{2} + 24 a + 17\right)\cdot 29^{44} + \left(5 a^{2} + 6 a + 20\right)\cdot 29^{45} + \left(27 a^{2} + 16 a + 3\right)\cdot 29^{46} + \left(28 a^{2} + 25 a + 5\right)\cdot 29^{47} + \left(12 a^{2} + 8 a + 3\right)\cdot 29^{48} + \left(5 a^{2} + 15 a + 26\right)\cdot 29^{49} + \left(3 a^{2} + 6 a + 12\right)\cdot 29^{50} + \left(9 a^{2} + 22 a + 7\right)\cdot 29^{51} + \left(a^{2} + 10 a + 11\right)\cdot 29^{52} + \left(5 a^{2} + 28 a + 21\right)\cdot 29^{53} + \left(7 a^{2} + 7 a + 14\right)\cdot 29^{54} + \left(5 a^{2} + a + 6\right)\cdot 29^{55} + \left(11 a^{2} + 22 a + 28\right)\cdot 29^{56} + \left(19 a^{2} + 9 a + 11\right)\cdot 29^{57} + \left(4 a^{2} + 28 a + 16\right)\cdot 29^{58} + \left(26 a^{2} + 27 a + 26\right)\cdot 29^{59} + \left(26 a^{2} + 2 a + 25\right)\cdot 29^{60} + \left(22 a^{2} + 15 a + 2\right)\cdot 29^{61} + \left(12 a^{2} + 20 a + 6\right)\cdot 29^{62} + \left(14 a^{2} + 23 a + 9\right)\cdot 29^{63} + \left(25 a^{2} + \frac{95}{16} a + 17\right)\cdot 29^{64} + \left(24 a^{2} + 20 a + 9\right)\cdot 29^{65} + \left(21 a^{2} + 17 a + 27\right)\cdot 29^{66} + \left(24 a^{2} + 26 a + 10\right)\cdot 29^{67} + \left(6 a^{2} + 25 a + 10\right)\cdot 29^{68} + \left(3 a^{2} + 13 a + 7\right)\cdot 29^{69} + \left(14 a^{2} + 15 a + 11\right)\cdot 29^{70} + \left(8 a^{2} + 21 a + 4\right)\cdot 29^{71} + \left(25 a^{2} + 22 a + 14\right)\cdot 29^{72} + \left(26 a^{2} + 28 a + 27\right)\cdot 29^{73} + \left(6 a^{2} + 7\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 4 }$ $=$ $19 a^{2} + 9 a + 3 + \left(3 a^{2} + 28 a + 6\right)\cdot 29 + \left(5 a^{2} + 3 a + 26\right)\cdot 29^{2} + \left(20 a^{2} + 10 a + 12\right)\cdot 29^{3} + \left(7 a^{2} + 13 a + 10\right)\cdot 29^{4} + \left(28 a^{2} + 22 a + 9\right)\cdot 29^{5} + \left(5 a^{2} + 13 a + 27\right)\cdot 29^{6} + \left(2 a^{2} + 23 a + 28\right)\cdot 29^{7} + \left(20 a^{2} + 21 a + 13\right)\cdot 29^{8} + \left(27 a^{2} + a + 21\right)\cdot 29^{9} + \left(3 a^{2} + 21 a + 14\right)\cdot 29^{10} + \left(5 a^{2} + 4 a + 19\right)\cdot 29^{11} + \left(a^{2} + 22 a + 11\right)\cdot 29^{12} + \left(27 a^{2} + 21 a + 3\right)\cdot 29^{13} + \left(4 a^{2} + 15 a + 5\right)\cdot 29^{14} + \left(5 a^{2} + 6 a + 18\right)\cdot 29^{15} + \left(2 a^{2} + 17 a + 21\right)\cdot 29^{16} + \left(21 a^{2} + 4 a + 27\right)\cdot 29^{17} + \left(21 a^{2} + 26 a + 26\right)\cdot 29^{18} + \left(24 a + 23\right)\cdot 29^{19} + \left(27 a^{2} + 17 a + 1\right)\cdot 29^{20} + \left(12 a^{2} + 6 a + 10\right)\cdot 29^{21} + \left(24 a^{2} + 10 a + 18\right)\cdot 29^{22} + \left(11 a^{2} + 16 a + 26\right)\cdot 29^{23} + \left(14 a^{2} + 23 a + 22\right)\cdot 29^{24} + \left(9 a^{2} + 15 a + 6\right)\cdot 29^{25} + \left(a^{2} + 27 a + 12\right)\cdot 29^{26} + \left(6 a^{2} + 2 a + 2\right)\cdot 29^{27} + \left(4 a^{2} + 23 a + 6\right)\cdot 29^{28} + \left(8 a^{2} + 24 a + 24\right)\cdot 29^{29} + \left(13 a^{2} + 10 a + 21\right)\cdot 29^{30} + \left(24 a^{2} + a + 1\right)\cdot 29^{31} + \left(24 a^{2} + 10 a + 24\right)\cdot 29^{32} + \left(26 a^{2} + 15 a\right)\cdot 29^{33} + \left(16 a^{2} + 7 a + 13\right)\cdot 29^{34} + \left(18 a^{2} + 16 a + 24\right)\cdot 29^{35} + \left(9 a^{2} + 2 a + 27\right)\cdot 29^{36} + \left(24 a^{2} + 25 a\right)\cdot 29^{37} + \left(27 a^{2} + 9 a + 3\right)\cdot 29^{38} + \left(12 a^{2} + 12 a + 14\right)\cdot 29^{39} + \left(27 a^{2} + 28 a + 10\right)\cdot 29^{40} + \left(2 a^{2} + 18 a + 25\right)\cdot 29^{41} + \left(23 a^{2} + 3 a + 21\right)\cdot 29^{42} + \left(4 a^{2} + 19 a + 7\right)\cdot 29^{43} + \left(23 a^{2} + 23 a + 22\right)\cdot 29^{44} + \left(22 a^{2} + 26 a + 19\right)\cdot 29^{45} + \left(12 a^{2} + 23 a + 1\right)\cdot 29^{46} + \left(21 a^{2} + 2 a + 16\right)\cdot 29^{47} + \left(14 a^{2} + 4 a + 26\right)\cdot 29^{48} + \left(5 a^{2} + 2\right)\cdot 29^{49} + \left(25 a^{2} + a + 4\right)\cdot 29^{50} + \left(12 a^{2} + 20 a + 5\right)\cdot 29^{51} + \left(4 a^{2} + 18 a + 19\right)\cdot 29^{52} + \left(4 a^{2} + 25 a + 26\right)\cdot 29^{53} + \left(13 a^{2} + 6 a + 17\right)\cdot 29^{54} + \left(27 a^{2} + 23 a + 11\right)\cdot 29^{55} + \left(18 a^{2} + 2 a + 3\right)\cdot 29^{56} + \left(15 a + 24\right)\cdot 29^{57} + \left(9 a^{2} + 22 a + 10\right)\cdot 29^{58} + \left(16 a + 12\right)\cdot 29^{59} + \left(21 a^{2} + 3 a + 16\right)\cdot 29^{60} + \left(25 a^{2} + 6 a + 1\right)\cdot 29^{61} + \left(11 a^{2} + 28 a + 26\right)\cdot 29^{62} + \left(12 a^{2} + 11 a + 9\right)\cdot 29^{63} + \left(5 a^{2} + 15 a + 8\right)\cdot 29^{64} + \left(18 a^{2} + 13 a + 2\right)\cdot 29^{65} + \left(8 a^{2} + 19\right)\cdot 29^{66} + \left(16 a^{2} + a + 2\right)\cdot 29^{67} + \left(16 a^{2} + 9 a + 20\right)\cdot 29^{68} + \left(5 a^{2} + 16 a + 28\right)\cdot 29^{69} + \left(4 a^{2} + 23 a + 3\right)\cdot 29^{70} + \left(a^{2} + 2 a + 27\right)\cdot 29^{71} + \left(12 a^{2} + 16 a + 10\right)\cdot 29^{72} + \left(6 a^{2} + 20 a + 13\right)\cdot 29^{73} + \left(17 a^{2} + 6 a + 15\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 5 }$ $=$ $17 a + 20 + \left(20 a^{2} + 16 a + 6\right)\cdot 29 + \left(16 a^{2} + 6 a + 3\right)\cdot 29^{2} + \left(a^{2} + 24 a + 5\right)\cdot 29^{3} + \left(14 a^{2} + 4 a + 3\right)\cdot 29^{4} + \left(28 a^{2} + 5 a + 12\right)\cdot 29^{5} + \left(2 a^{2} + 11 a + 23\right)\cdot 29^{6} + \left(26 a^{2} + 24 a + 10\right)\cdot 29^{7} + \left(20 a^{2} + 13 a + 23\right)\cdot 29^{8} + \left(23 a^{2} + 18 a + 21\right)\cdot 29^{9} + \left(26 a^{2} + a + 7\right)\cdot 29^{10} + \left(28 a^{2} + 2 a + 16\right)\cdot 29^{11} + \left(13 a^{2} + 14\right)\cdot 29^{12} + \left(26 a^{2} + 2 a + 25\right)\cdot 29^{13} + \left(27 a^{2} + 4 a + 18\right)\cdot 29^{14} + \left(20 a^{2} + 13 a + 17\right)\cdot 29^{15} + \left(7 a^{2} + 15 a + 17\right)\cdot 29^{16} + \left(6 a^{2} + 26 a + 10\right)\cdot 29^{17} + \left(20 a^{2} + 11 a + 24\right)\cdot 29^{18} + \left(16 a^{2} + 16 a + 28\right)\cdot 29^{19} + \left(16 a^{2} + 19 a + 11\right)\cdot 29^{20} + \left(13 a^{2} + 20 a + 5\right)\cdot 29^{21} + \left(21 a^{2} + 16 a + 21\right)\cdot 29^{22} + \left(24 a^{2} + 8 a + 4\right)\cdot 29^{23} + \left(20 a^{2} + 17 a + 18\right)\cdot 29^{24} + \left(20 a^{2} + 24 a + 11\right)\cdot 29^{25} + \left(6 a^{2} + 14 a + 28\right)\cdot 29^{26} + \left(7 a^{2} + 19\right)\cdot 29^{27} + \left(6 a^{2} + 21 a + 23\right)\cdot 29^{28} + \left(18 a^{2} + 11 a + 9\right)\cdot 29^{29} + \left(6 a^{2} + 12 a + 20\right)\cdot 29^{30} + \left(5 a^{2} + 5 a + 28\right)\cdot 29^{31} + \left(5 a^{2} + 3 a + 21\right)\cdot 29^{32} + \left(3 a^{2} + 2 a + 6\right)\cdot 29^{33} + \left(3 a^{2} + 5 a + 28\right)\cdot 29^{34} + \left(10 a^{2} + 26 a + 11\right)\cdot 29^{35} + \left(7 a^{2} + 14 a + 18\right)\cdot 29^{36} + \left(5 a^{2} + 12 a + 5\right)\cdot 29^{37} + \left(21 a^{2} + 5 a + 10\right)\cdot 29^{38} + \left(18 a^{2} + 7 a + 2\right)\cdot 29^{39} + \left(19 a^{2} + 3 a + 8\right)\cdot 29^{40} + \left(6 a^{2} + 25 a + 12\right)\cdot 29^{41} + \left(5 a^{2} + 8 a + 16\right)\cdot 29^{42} + \left(2 a^{2} + 23 a + 26\right)\cdot 29^{43} + \left(14 a^{2} + 5 a + 27\right)\cdot 29^{44} + \left(22 a^{2} + 23 a + 4\right)\cdot 29^{45} + \left(17 a^{2} + 4 a + 20\right)\cdot 29^{46} + \left(14 a + 12\right)\cdot 29^{47} + \left(21 a^{2} + 12 a + 6\right)\cdot 29^{48} + \left(19 a^{2} + 21 a + 11\right)\cdot 29^{49} + \left(9 a^{2} + a + 22\right)\cdot 29^{50} + \left(10 a^{2} + 23 a + 28\right)\cdot 29^{51} + \left(10 a^{2} + 5 a + 22\right)\cdot 29^{52} + \left(7 a^{2} + 16 a + 4\right)\cdot 29^{53} + \left(13 a^{2} + 26 a\right)\cdot 29^{54} + \left(8 a^{2} + 15 a + 9\right)\cdot 29^{55} + \left(21 a^{2} + 13 a + 9\right)\cdot 29^{56} + \left(7 a^{2} + 2 a + 19\right)\cdot 29^{57} + \left(5 a^{2} + 13 a + 8\right)\cdot 29^{58} + \left(27 a^{2} + 11 a + 25\right)\cdot 29^{59} + \left(22 a^{2} + 24 a + 13\right)\cdot 29^{60} + \left(18 a^{2} + 15 a + 24\right)\cdot 29^{61} + \left(19 a^{2} + 15 a + 2\right)\cdot 29^{62} + \left(2 a^{2} + 25 a + 5\right)\cdot 29^{63} + \left(27 a^{2} + 5 a + 19\right)\cdot 29^{64} + \left(22 a^{2} + 21 a + 16\right)\cdot 29^{65} + \left(18 a^{2} + 13\right)\cdot 29^{66} + \left(11 a^{2} + 26 a + 22\right)\cdot 29^{67} + \left(16 a^{2} + 5 a + 3\right)\cdot 29^{68} + \left(a^{2} + 2 a + 5\right)\cdot 29^{69} + \left(19 a^{2} + 23 a + 8\right)\cdot 29^{70} + \left(8 a^{2} + 17 a + 14\right)\cdot 29^{71} + \left(5 a^{2} + 17 a + 16\right)\cdot 29^{72} + \left(a^{2} + 3 a + 12\right)\cdot 29^{73} + \left(18 a^{2} + 16 a + 3\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 6 }$ $=$ $4 a^{2} + 12 a + 6 + \left(15 a^{2} + 3 a + 25\right)\cdot 29 + \left(25 a^{2} + 2 a + 20\right)\cdot 29^{2} + \left(6 a^{2} + 17 a + 21\right)\cdot 29^{3} + \left(23 a^{2} + 16 a + 15\right)\cdot 29^{4} + \left(26 a^{2} + 22 a + 25\right)\cdot 29^{5} + \left(22 a^{2} + 6 a + 8\right)\cdot 29^{6} + \left(24 a^{2} + 11 a + 18\right)\cdot 29^{7} + \left(15 a^{2} + 9 a + 24\right)\cdot 29^{8} + \left(5 a^{2} + 13 a + 25\right)\cdot 29^{9} + \left(13 a^{2} + 17 a + 10\right)\cdot 29^{10} + \left(27 a^{2} + 27 a + 15\right)\cdot 29^{11} + \left(25 a^{2} + 20 a + 8\right)\cdot 29^{12} + \left(9 a^{2} + 23 a + 17\right)\cdot 29^{13} + \left(6 a^{2} + 6 a + 15\right)\cdot 29^{14} + \left(2 a^{2} + 10 a + 6\right)\cdot 29^{15} + \left(8 a^{2} + 12 a + 19\right)\cdot 29^{16} + \left(15 a^{2} + 11 a + 15\right)\cdot 29^{17} + \left(16 a^{2} + 28\right)\cdot 29^{18} + \left(25 a + 23\right)\cdot 29^{19} + \left(13 a^{2} + 26 a + 12\right)\cdot 29^{20} + \left(28 a^{2} + 8 a + 16\right)\cdot 29^{21} + \left(13 a^{2} + 8 a + 18\right)\cdot 29^{22} + \left(10 a^{2} + 3 a + 25\right)\cdot 29^{23} + \left(27 a^{2} + 6 a + 1\right)\cdot 29^{24} + \left(8 a^{2} + 27 a + 14\right)\cdot 29^{25} + \left(13 a^{2} + 20 a + 25\right)\cdot 29^{26} + \left(20 a^{2} + 24 a + 17\right)\cdot 29^{27} + \left(17 a^{2} + 8 a + 23\right)\cdot 29^{28} + \left(18 a^{2} + 26 a + 3\right)\cdot 29^{29} + \left(3 a^{2} + 9 a + 24\right)\cdot 29^{30} + \left(9 a^{2} + 19 a + 21\right)\cdot 29^{31} + \left(27 a^{2} + 17 a + 16\right)\cdot 29^{32} + \left(21 a^{2} + 12 a + 17\right)\cdot 29^{33} + \left(18 a + 23\right)\cdot 29^{34} + \left(17 a^{2} + a + 13\right)\cdot 29^{35} + \left(12 a^{2} + 20 a + 18\right)\cdot 29^{36} + \left(11 a^{2} + 18 a + 16\right)\cdot 29^{37} + \left(24 a^{2} + 14 a + 18\right)\cdot 29^{38} + \left(6 a^{2} + 3 a + 1\right)\cdot 29^{39} + \left(17 a^{2} + 4 a + 18\right)\cdot 29^{40} + \left(18 a^{2} + 20 a + 12\right)\cdot 29^{41} + \left(14 a^{2} + 7 a + 20\right)\cdot 29^{42} + \left(15 a^{2} + 27 a + 24\right)\cdot 29^{43} + \left(13 a^{2} + 7 a + 26\right)\cdot 29^{44} + \left(11 a^{2} + 22 a + 20\right)\cdot 29^{45} + \left(11 a^{2} + 5 a + 26\right)\cdot 29^{46} + \left(16 a^{2} + 24 a + 10\right)\cdot 29^{47} + \left(2 a^{2} + 21 a + 24\right)\cdot 29^{48} + \left(25 a^{2} + 21 a + 7\right)\cdot 29^{49} + \left(7 a^{2} + 25 a + 6\right)\cdot 29^{50} + \left(24 a^{2} + 26 a + 10\right)\cdot 29^{51} + \left(19 a^{2} + 8 a + 22\right)\cdot 29^{52} + \left(14 a^{2} + 17 a + 6\right)\cdot 29^{53} + \left(15 a^{2} + 9 a + 16\right)\cdot 29^{54} + \left(4 a^{2} + 28 a\right)\cdot 29^{55} + \left(13 a^{2} + 22 a + 1\right)\cdot 29^{56} + \left(16 a^{2} + 20 a + 2\right)\cdot 29^{57} + \left(24 a^{2} + 21 a + 9\right)\cdot 29^{58} + \left(22 a^{2} + 17 a + 21\right)\cdot 29^{59} + \left(15 a^{2} + 9 a + 28\right)\cdot 29^{60} + \left(15 a^{2} + 24 a + 7\right)\cdot 29^{61} + \left(16 a^{2} + 3 a\right)\cdot 29^{62} + \left(9 a^{2} + 12 a + 13\right)\cdot 29^{63} + \left(4 a^{2} + 13 a + 26\right)\cdot 29^{64} + \left(10 a^{2} + 16 a + 18\right)\cdot 29^{65} + \left(17 a^{2} + 10 a + 11\right)\cdot 29^{66} + \left(21 a^{2} + 5 a + 16\right)\cdot 29^{67} + \left(5 a^{2} + 26 a + 18\right)\cdot 29^{68} + \left(24 a^{2} + 12 a + 25\right)\cdot 29^{69} + \left(24 a^{2} + 19 a + 15\right)\cdot 29^{70} + \left(11 a^{2} + 18 a + 18\right)\cdot 29^{71} + \left(27 a^{2} + 17 a + 26\right)\cdot 29^{72} + \left(25 a + 21\right)\cdot 29^{73} + \left(4 a^{2} + 11 a + 13\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 7 }$ $=$ $13 a^{2} + a + 24 + \left(15 a^{2} + 11 a + 4\right)\cdot 29 + \left(26 a^{2} + 24 a + 15\right)\cdot 29^{2} + \left(18 a^{2} + 6 a + 24\right)\cdot 29^{3} + \left(15 a^{2} + 8 a + 9\right)\cdot 29^{4} + \left(22 a^{2} + 27 a + 4\right)\cdot 29^{5} + \left(10 a^{2} + 12 a + 24\right)\cdot 29^{6} + \left(7 a^{2} + 17 a + 5\right)\cdot 29^{7} + \left(7 a^{2} + 9 a + 17\right)\cdot 29^{8} + \left(8 a^{2} + 14 a + 4\right)\cdot 29^{9} + \left(26 a^{2} + 12 a + 24\right)\cdot 29^{10} + \left(26 a^{2} + 27 a + 17\right)\cdot 29^{11} + \left(17 a^{2} + 28 a + 4\right)\cdot 29^{12} + \left(14 a^{2} + 15 a + 9\right)\cdot 29^{13} + \left(18 a^{2} + 25 a\right)\cdot 29^{14} + \left(3 a^{2} + 24 a + 20\right)\cdot 29^{15} + \left(a^{2} + 7 a + 2\right)\cdot 29^{16} + \left(9 a^{2} + 9 a + 13\right)\cdot 29^{17} + \left(a^{2} + 8 a + 27\right)\cdot 29^{18} + \left(25 a^{2} + 16 a + 18\right)\cdot 29^{19} + \left(26 a^{2} + 22 a + 21\right)\cdot 29^{20} + \left(20 a + 9\right)\cdot 29^{21} + \left(17 a^{2} + 12 a + 24\right)\cdot 29^{22} + \left(15 a^{2} + 21 a + 16\right)\cdot 29^{23} + \left(2 a^{2} + 20 a + 22\right)\cdot 29^{24} + \left(23 a^{2} + 10 a + 22\right)\cdot 29^{25} + \left(22 a^{2} + 25 a + 10\right)\cdot 29^{26} + \left(27 a^{2} + 23 a + 18\right)\cdot 29^{27} + \left(25 a^{2} + 6 a + 13\right)\cdot 29^{28} + \left(a^{2} + 17 a + 18\right)\cdot 29^{29} + \left(20 a^{2} + 26 a + 19\right)\cdot 29^{30} + \left(19 a^{2} + 16 a + 23\right)\cdot 29^{31} + \left(16 a + 14\right)\cdot 29^{32} + \left(17 a^{2} + 23 a + 27\right)\cdot 29^{33} + \left(14 a^{2} + 2 a + 3\right)\cdot 29^{34} + \left(21 a^{2} + 4 a + 8\right)\cdot 29^{35} + \left(16 a^{2} + 28 a + 11\right)\cdot 29^{36} + \left(14 a^{2} + 8 a + 16\right)\cdot 29^{37} + \left(22 a^{2} + 18 a + 24\right)\cdot 29^{38} + \left(28 a^{2} + 2 a + 6\right)\cdot 29^{39} + \left(24 a^{2} + 6 a + 15\right)\cdot 29^{40} + \left(24 a^{2} + a + 24\right)\cdot 29^{41} + \left(13 a^{2} + 20 a + 20\right)\cdot 29^{42} + \left(5 a^{2} + 9 a + 14\right)\cdot 29^{43} + \left(12 a^{2} + 18 a + 11\right)\cdot 29^{44} + \left(14 a^{2} + 28 a + 4\right)\cdot 29^{45} + \left(17 a^{2} + 14 a + 25\right)\cdot 29^{46} + \left(27 a^{2} + 23 a + 14\right)\cdot 29^{47} + \left(4 a^{2} + 20 a + 13\right)\cdot 29^{48} + \left(18 a^{2} + 11 a + 27\right)\cdot 29^{49} + \left(13 a + 22\right)\cdot 29^{50} + \left(2 a^{2} + 14 a + 26\right)\cdot 29^{51} + \left(21 a^{2} + 8 a + 24\right)\cdot 29^{52} + \left(26 a^{2} + 6 a + 19\right)\cdot 29^{53} + \left(24 a^{2} + 3 a + 26\right)\cdot 29^{54} + \left(a^{2} + 19 a + 17\right)\cdot 29^{55} + \left(3 a^{2} + 21\right)\cdot 29^{56} + \left(10 a^{2} + 28 a + 4\right)\cdot 29^{57} + \left(15 a^{2} + 12 a + 15\right)\cdot 29^{58} + \left(20 a^{2} + a + 9\right)\cdot 29^{59} + \left(a^{2} + 27 a + 15\right)\cdot 29^{60} + \left(2 a^{2} + 10\right)\cdot 29^{61} + \left(20 a^{2} + 15 a + 14\right)\cdot 29^{62} + \left(24 a^{2} + 6 a + 25\right)\cdot 29^{63} + \left(\frac{25}{2} a^{2} + 17 a + 2\right)\cdot 29^{64} + \left(27 a^{2} + \frac{47}{16} a + 5\right)\cdot 29^{65} + \left(21 a^{2} + 27 a + 27\right)\cdot 29^{66} + \left(2 a^{2} + 6 a + 3\right)\cdot 29^{67} + \left(28 a^{2} + 23 a + 16\right)\cdot 29^{68} + \left(18 a^{2} + 26 a + 17\right)\cdot 29^{69} + \left(9 a^{2} + 9 a + 1\right)\cdot 29^{70} + \left(20 a^{2} + 15 a + 14\right)\cdot 29^{71} + \left(20 a^{2} + 20 a + 22\right)\cdot 29^{72} + \left(25 a^{2} + 5 a + 19\right)\cdot 29^{73} + \left(9 a^{2} + 23 a + 5\right)\cdot 29^{74} +O\left(29^{ 75 }\right)$ $r_{ 8 }$ $=$ $22 + 10\cdot 29 + 7\cdot 29^{2} + 7\cdot 29^{3} + 20\cdot 29^{4} + 6\cdot 29^{5} + 15\cdot 29^{6} + 24\cdot 29^{7} + 20\cdot 29^{8} + 17\cdot 29^{9} + 4\cdot 29^{10} + 22\cdot 29^{11} + 7\cdot 29^{12} + 2\cdot 29^{13} + 28\cdot 29^{14} + 11\cdot 29^{15} + 27\cdot 29^{16} + 26\cdot 29^{17} + 20\cdot 29^{18} + 24\cdot 29^{19} + 5\cdot 29^{20} + 28\cdot 29^{21} + 23\cdot 29^{22} + 9\cdot 29^{23} + 8\cdot 29^{24} + 3\cdot 29^{25} + 21\cdot 29^{26} + 6\cdot 29^{27} + 28\cdot 29^{28} + 12\cdot 29^{29} + 24\cdot 29^{30} + 7\cdot 29^{31} + 10\cdot 29^{32} + 21\cdot 29^{33} + 7\cdot 29^{34} + 2\cdot 29^{35} + 27\cdot 29^{36} + 2\cdot 29^{37} + 25\cdot 29^{38} + 13\cdot 29^{39} + 11\cdot 29^{40} + 11\cdot 29^{41} + 27\cdot 29^{42} + 22\cdot 29^{43} + 28\cdot 29^{44} + 3\cdot 29^{45} + 20\cdot 29^{46} + 26\cdot 29^{47} + 15\cdot 29^{48} + 3\cdot 29^{49} + 21\cdot 29^{50} + 19\cdot 29^{51} + 11\cdot 29^{52} + 25\cdot 29^{53} + 11\cdot 29^{54} + 21\cdot 29^{55} + 10\cdot 29^{56} + 17\cdot 29^{57} + 17\cdot 29^{58} + 25\cdot 29^{59} + 11\cdot 29^{60} + 15\cdot 29^{61} + 16\cdot 29^{62} + 14\cdot 29^{63} + 15\cdot 29^{64} + 29^{65} + 26\cdot 29^{66} + 27\cdot 29^{67} + 3\cdot 29^{68} + 25\cdot 29^{69} + 5\cdot 29^{70} + 8\cdot 29^{71} + 3\cdot 29^{72} + 6\cdot 29^{73} + 7\cdot 29^{74} +O\left(29^{ 75 }\right)$
### Generators of the action on the roots $r_1, \ldots, r_{ 8 }$
Cycle notation $(3,6)(5,8)$ $(1,3)(2,5)(4,6)(7,8)$ $(1,5)(2,6)(3,7)(4,8)$ $(3,6,8)$
### Character values on conjugacy classes
Size Order Action on $r_1, \ldots, r_{ 8 }$ Character value $1$ $1$ $()$ $9$ $6$ $2$ $(3,6)(5,8)$ $-3$ $9$ $2$ $(1,4)(2,7)(3,6)(5,8)$ $1$ $12$ $2$ $(1,3)(2,5)(4,6)(7,8)$ $3$ $12$ $2$ $(1,5)(2,6)(3,7)(4,8)$ $3$ $36$ $2$ $(1,7)(5,8)$ $1$ $16$ $3$ $(3,8,6)$ $0$ $32$ $3$ $(1,2,4)(3,5,6)$ $0$ $32$ $3$ $(1,2,4)(5,6,8)$ $0$ $36$ $4$ $(1,3,4,6)(2,5,7,8)$ $-1$ $36$ $4$ $(1,5,4,8)(2,6,7,3)$ $-1$ $36$ $4$ $(1,7,4,2)(3,5,6,8)$ $1$ $72$ $4$ $(3,5,6,8)(4,7)$ $-1$ $48$ $6$ $(1,4)(2,7)(3,8,6)$ $0$ $96$ $6$ $(1,5,2,6,4,3)(7,8)$ $0$ $96$ $6$ $(1,6,2,8,4,5)(3,7)$ $0$
The blue line marks the conjugacy class containing complex conjugation. | 13,380 | 25,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-34 | latest | en | 0.268212 |
https://www.solvedlib.com/n/in-each-pnt-below-g1e-mxn-matitx-in-neluaed-tow-cchelon,15207316 | 1,696,102,256,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00134.warc.gz | 1,045,673,689 | 28,279 | # In each Pnt below , g1e MXn matitx In neluaed Tow cchelon formn satisfying the given conditions briefly explain why
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##### Please solve using given picture The following information applies to the next eight (8) questions worth...
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##### Beaker represents the equivalence point At the moles of weak acid will be the same a5 the moles of OH : In thls experiment 180.0 mL ol the acetlc ucid has &n Initlal colicentratlon of 0.5 M: As YOu cun sce trom the plot on the previous page, the equivalence point Is met at [00.0 mL ot NaOH, so what [s the concentration of NaOH in this experiment? Show workAt the equivalence point; all ol acetic acid and hydraxlde molecules have neutralized. This lcaves only the acetate Ion and water: This cu
Beaker represents the equivalence point At the moles of weak acid will be the same a5 the moles of OH : In thls experiment 180.0 mL ol the acetlc ucid has &n Initlal colicentratlon of 0.5 M: As YOu cun sce trom the plot on the previous page, the equivalence point Is met at [00.0 mL ot NaOH, so w...
##### Please help with the discussion question as follows: You are the risk manager of a hospital....
Please help with the discussion question as follows: You are the risk manager of a hospital. A nurse from the operating room reports that, during a surgery, the head surgeon did not conduct a “time out” to confirm the side and site of the surgery. You question the surgeon, and he denies ... | 1,485 | 4,812 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.710181 |
https://steel-machinery-spareparts.com/es/productos/prensa-tipo-garaje-266529 | 1,674,992,526,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00435.warc.gz | 541,476,390 | 4,128 | # College algebra problem solving
College algebra problem solving can help students to understand the material and improve their grades. We can solving math problem.
## The Best College algebra problem solving
In this blog post, we will show you how to work with College algebra problem solving. We all know that exponents are a quick way to multiply numbers by themselves, but how do we solve for them? The answer lies in logs. Logs are basically just exponents in reverse, so solving for an exponent is the same as solving for a log. For example, if we want to find out what 2^5 is, we can take the log of both sides of the equation to get: 5 = log2(2^5). Then, we can just solve for 5 to get: 5 = log2(32). Therefore, 2^5 = 32. Logs may seem like a complicated concept, but they can be very useful in solving problems with exponents.
How to solve by elimination is a method of problem solving where you systematically remove possible answers or solutions until only the correct answer is left. This can be useful when you are trying to narrow down a list of possibilities, such as when you are trying to find the culprit in a whodunit novel. To solve by elimination, you need to first identify all of the possible answers or solutions. Once you have a list, you can start to eliminate the ones that are not viable options. For example, if you were trying to figure out who stole a cookie from the cookie jar, and you had a list of suspects that included a cat, a dog, and a baby, you could eliminate the cat and the dog because they would not be able to reach thecookie jar. This would leave you with the baby as your only suspect. How to solve by elimination is a simple yet effective way to narrow down your options and find the right answer.
Math home work can be a tricky thing for some students. Math is a difficult subject for some, so doing homework on it can be frustrating. Some tips to help with math homework are to get a tutor, practice at home, and try to understand the concepts. A tutor can help go over the material and help with any confusion. Also, practicing math problems at home can be helpful. Doing a few problems each night can help solidify the material. Lastly, trying to understand the concepts can be very helpful. If a student understands why they are doing a certain math problem, it can make the problem much easier. Math homework can be tough, but these tips can make it a little bit easier.
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Another method is to use exponential equations. Exponential equations are equivalent to log equations, so they can be manipulated in the same way. By using these methods, you can solve natural log equations with relative ease.
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Ivanka Watson | 862 | 4,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-06 | latest | en | 0.96617 |
https://www.physicsforums.com/threads/pumping-a-cylindrical-tank.202742/ | 1,539,978,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00408.warc.gz | 1,024,949,827 | 13,742 | # Homework Help: Pumping a cylindrical tank
1. Dec 5, 2007
### kuahji
A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.
My work
I set the bottom of the tank at the point (0,0) & then drew a circle with a radius of 4.
dV=2x(20)dy
then solved the the circle's equation for x, x=$$\sqrt{(16-y^2)}$$
dV=40$$\sqrt{(16-y^2)}$$
F(y)=57(40)$$\sqrt{(16-y^2)}$$
10-y should be the distance the work must do
W=2280 $$\int$$(10-y)$$\sqrt{(16-y^2)}$$
Then I distributed the (10-y)
W=22800$$\int$$$$\sqrt{(16-y^2)}$$- 2280$$\int$$y$$\sqrt{(16-y^2)}$$
For part two, I set u=16-y^2 & got
W=22800$$\int$$$$\sqrt{(16-y^2)}$$+ 1140$$\int$$y$$\sqrt{(16-y^2)}$$
This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4$$\pi$$ as follows
22800(4$$\pi$$)+ 1140$$\int$$y$$\sqrt{(16-y^2)}$$ (evaluated from 0 to -4), I tried from (0 to 4) in my solution.
I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?
2. Dec 5, 2007
### Shooting Star
The problem can be solved easily in a different way if you calculate position of the centre of mass of the oil by integration. After that, work done = Mgh, where M is the total mass and h the height the CM has to rise. This would be the Physicist's approach.
3. Dec 5, 2007
### kuahji
Wouldn't you have to calculate it for a 3d object though? I don't have those skills yet. So far all I've dealt with was thin plats, 2d objects.
4. Dec 6, 2007
### Shooting Star
Yes, for 3d objects, but you have to do that anyway. And because of symmetry, the integration would be only for 2d.
I don't notice any value of g in your calcs? Also, I'm not very sure what you are trying to do.
Let me know if you need more help, but after explaining what is the method you are following.
5. Dec 6, 2007
### kuahji
The value of g should already be in the 57 lbs, as weight. Guess I'm kind of lost myself, but up unto there, I everything the solution manual has.
6. Jul 9, 2008
### dudicuff
Hi everyone.
I have a very similar problem to the one kuahji posted, but my problem is a storage tank completely full of oil. Would I then calculate using (6+8 -y) or (14-y) or would I use the same (10-y)? Also, wouldn't my integral be from -4 to 4 ( or 2* [0 to 4])? Thanks for any help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 846 | 2,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-43 | latest | en | 0.910802 |
https://corsetacademy.net/tutorial-7-positioning-calculating-and-cutting-the-first-ruffle/ | 1,618,211,684,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00228.warc.gz | 298,182,002 | 9,196 | # Tutorial 7. Positioning, Calculating, and Cutting the First Ruffle.
## Tutorial 7. Positioning, Calculating, and Cutting the First Ruffle.
I have pressed all Rigilene bones on the petticoat.
The next step is to sew a hard mesh ruffle along the bottom edge.
This ruffle is only supposed to prevent the hem from curling in and not to increase the fullness.
The bottom ruffle will be sewn on along the line which marks a tunnel for hoop 2 on the face on the petticoat. The edge of the ruffle will be therefore overlaid by a bias tape. The only function of this tunnel is to reinforce the mesh fabric. I will sew another one on the inside and insert a steel hoop in it.
First of all, let us calculate the width of the ruffle. It is very important to get it right.
The second hoop is positioned 5cm above the bottom edge at the front and 25cm above the bottom edge at the back. The bottom edge of the ruffle is supposed to stay about 3cm above floor level at the front. The petticoat itself must not drag along the floor. If a stiff construction (a stiff cage in our case) touches the floor, the skirt will bump along in a very unpretty way when the bride walks around. It will be very obvious that there is a stiff construction hiding underneath the skirt. On the other hand, you cannot do without this stiff construction because otherwise the hem of the main skirt will keep curling up.
If the second hoop is positioned 25cm above floor level and the bottom edge of the ruffle must be positioned horizontally and stay 3cm above the floor along the full perimeter, then the width of the main bottom ruffle should be 22cm.
But in this case, nearly the whole back of the petticoat will be left without ruffles and the train will not be reinforced.
To solve this problem, I will sew an additional 22cm wide ruffle between the side seams onto the back. It will be placed below the level of the second hoop. This additional ruffle will rest upon the floor and give support to the train and the lining of the dress.
I will most likely add one more ruffle and sew it on at an angle: it will be positioned 5cm below the second hoop at the back and 3cm below it at the side. In this case, this additional ruffle will basically reach down to the floor at the back.
Now that I know the widths of the main and the additional ruffle, I need to decide on their lengths and figure out fabric consumption with regard to the implied gathering ratio.
I know all necessary measurements: the second hoop (and the main ruffle) has a circumference of 240cm, and the additional ruffle should be 110cm long.
This makes a total length of 350cm.
With a 3 to 1 gathering ratio (which is more than enough), I will need a strip of mesh fabric with a length of 105cm and a width of 22cm.
Since my mesh fabric has a width of 180cm, I need to cut six 22cm strips. This will make a total length of 108cm.
Here is how much mesh fabric I need considering its width:
22cm × 6 = 132cm ≈ 140cm
I cut six 22cm wide strips of mesh fabric.
Next, I need to sew them together in a single strip. I sew the strips together with a presser foot width seam allowance.
Remember to fold all seam allowances to the same side when doing this.
Let me explain it in more detail for those who are not sure how to do it right. Put two strips of fabric face to face and join them. The seam allowance stays on the wrong side of the fabric. Take the free end of the top strip and open the joined strips so as to have them lying face side up. Take the third strip and put it face down over the strip that was at the top. Sew them together. Again, the seam allowance stays on the wrong side of the fabric. Continue until you reach the last strip.
All seam allowances are on the inside.
The result is a ruffle with a total length of 110cm and with all seam allowances on the same side.
I switch to a ruffler presser foot. The gathering ratio is about 3 to 1.
I test it on a small area and start gathering the ruffle. You will be able to gather more thickly if you slow down the feeding by placing your finger behind the presser foot and pushing the ruffle in with the other hand.
The ruffle is ready! I am happy with the result. If I decide to make it thicker, I will simply gather it one more time.
And now, what concerns finishing the bottom edge of the ruffle. The mesh is rather itchy and may feel uncomfortable in wear. You can finish the bottom edge with a bias tape. But my dress has a tulle face and the bias tape edge of the ruffle will show if the hem curls up. I don't like it. I wouldn't want to make it too obvious that there is a petticoat under the skirt. I will leave the edge of the ruffle unfinished for now. It is so light and delicate. When my client comes to pick up the dress and tries it on, she will tell me whether the raw edge disturbs her. It won't take long to finish it with a bias tape if it causes her legs to itch.
I will measure the length of the ruffle to see whether I need to make it longer and continue working. | 1,177 | 5,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-17 | latest | en | 0.89595 |
https://www.coursehero.com/tutors-problems/Math/7647164-DO-PROBLEM-V/ | 1,553,096,268,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202433.77/warc/CC-MAIN-20190320150106-20190320172028-00061.warc.gz | 737,498,282 | 77,950 | View the step-by-step solution to:
# MATH 140A (Lecture B; course code 44775) Fall 2011 Instructor: Professor R. Reilly Assignment #7 COMPLETE Reading Assignments (Note: All section or...
DO PROBLEM V
MATH 140A (Lecture B; course code 44775) Fall 2011 Instructor: Professor R. C. Reilly Assignment #7 COMPLETE Reading Assignments (Note: All section or page numbers refer to the course text, “Elementary Analysis: The Theory of Calculus”, by K. Ross.) Monday 11/07/2011: Section 11 Wednesday 11/09/2011: Section 12 Friday 11/11/2011: HOLIDAY Monday 11/14/2011: Section 14 Some Extra Material Here is some more stuff that was mentioned in the lecture but does not appear to be stated explicitly in the text; or if it is, it is somewhat hidden. (1) Alternate Form of the Definition of ‘Cauchy Sequence’ The following condition, on a se- quence σ = ( s 1 ,s 2 , ...s n , ... ) of real numbers, is equivalent to the ‘Cauchy Condition’ described in Definition 10.8: For every ε > 0 there exists N such that if n > N then | s n + k - s n | < ε for all k in IN. Outline of the Proof Given in the Lecture In the text’s definition, one considers the inequality | s n - s m | < ε , where n,m N . However in this inequality there is no loss of generality in assuming that m n . Likewise, one can further reduce to the case in which m > n , since in the case m = n the desired inequality is trivially true. But if m > n , then m = n + k for some k in IN, namely k = m - n ; and, conversely, for each such k one has n + k > n . The desired result follows easily. (2) Theorem : Let A be an infinite subset of IN. Then A can be described, in exactly one way, in the form A = { n 1 ,n 2 , ...n k , ... } , where 1 n 1 < n 2 < ...n k < ... . Moreover, one has n k k for all k . Outline of the Proof Given in the Lecture The number n k is simply the k -th smallest element of the set A . The inequality n k k follows by induction after noting that n 1 1, since 1 is the smallest natural number, together with the fact that consecutive n k ’s differ by at least 1. (3) If A = { n 1 < n 2 < ...n k < ... } is an infinite subset of IN, with the terms written in strictly increasing order, and σ = ( s 1 ,s 2 , ...s n , ... ) is a sequence, then the subsequence of σ associated with A , denoted σ A , is the sequence ( s n 1 ,s n 2 , ...s n k , ... ). Remark The notation σ A used above is NOT standard. Indeed, about the only place you will find it with this meaning is in my Math 140A course. (4) Strengthened Version of Theorem 11.2 Let σ = ( s 1 ,s 2 , ...s n , ... ) be a sequence of real numbers. If L is either a real number or one of the quantities + or -∞ , and if lim n →∞ s n = L , then for every subsequence τ = ( t 1 ,t 2 , ...t k , ... ) of σ , one also has lim k →∞ t k = L . Note The text proves this only when L is finite. The proof of this stronger version which I gave in the lecture used the result of Hand-in Exercise (I) in Assignment (4). (5) Alternate Proof of Theorem 11.5, the Bolzano-Weierstrass Theorem for Sequences The statement of the theorem in question is as in the text: Every bounded sequence of real numbers has a convergent subsequence. Alternate Proof Let σ = ( s 1 ,s 2 , ...s n , ... ) be the sequence in question. Then, by the ‘bound- edness’ hypothesis, there exist numbers m and M such that m < s n < M for all n . Let a 1 = m and b 1 = M , so that a 1 < b 1 , and s n is in the closed interval I 1 = [ a 1 ,b 1 ] for all n . Let c 1 be
the midpoint of the interval I 1 , so c 1 = ( a 1 + b 1 ) / 2. Then for at least one of the half-intervals [ a 1 ,c 1 ] or [ c 1 ,b 1 ], the following condition holds: there are infinitely many indices n such that s n is an element of that half-interval. If this condition holds for the left half-interval [ a 1 ,c 1 ], let a 2 = a 1 and b 2 = c 1 . If, instead, this condition does not hold for the left-half interval [ a 1 ,c 1 ], so that it must hold for the right-half interval [ c 1 ,b 1 ], let a 1 = c 1 and b 2 = b 1 . Either way, one gets a new subinterval, [ a 2 ,b 2 ], half the length of the original interval [ m,M ], such that s n [ a 2 ,b 2 ] for infinitely many n . Continuing this process indefinitely, one obtains an infinite sequence of intervals [ a k ,b k ], with k IN, such that [ a 1 ,b 1 ] = [ m,M ], and for each k IN the interval [ a k +1 ,b k +1 ] is one of the two halves of [ a k ,b k ], and for each k IN there are infinitely many n IN such that s n [ a k ,b k ]. Clearly b k - a k = ( M - m ) / 2 k - 1 for each k , so that lim k →∞ ( b k - a k ) = 0. Thus it follows from Discussion Exercise (4) from Tuesday, 10/27/2011, that the intersection of these intervals consists of a single point; call it d . Note that d = lim k →∞ a k = lim k →∞ b k . Now define a subsequence ( s n 1 ,s n 2 , ...s n k , ... ) of σ by the following rule: First, set n 1 = 1; note that n 1 is the first index such that s n 1 [ a 1 ,b 1 ]. Likewise, define n 2 to be the first index larger that n 1 such that s n 2 [ a 2 ,b 2 ]. Suppose now that indices n 1 , n 2 , . . . n k have already been constructed so that n 1 < n 2 < ... < n k , and s n j [ a j ,b j ] for each j = 1 , 2 , ...k . Then define the index n k +1 to be the smallest index such that n k < n k +1 and s n k +1 [ a k +1 ,b k +1 ]. (The existence of such an index which is strictly larger than n k follows from the fact that we have constructed these intervals so that for each of them, s n in the interval for infinitely many indices n ; in particular, n k cannot be the largest such index.) It follows that n 1 < n 2 < ... < n k < ... for all k , so that τ = ( s n 1 ,s n 2 , ...s n k ... ) is a subsequence of σ . It is also clear from the construction that s n k [ a k ,b k ] for each k ; that is, a k s n k leq b k for each k . It now follows from the Squeeze Theorem, i.e., the result of Exercise 8.5(a), that lim k →∞ s k = d as well, as required. Remarks (a) The proof outlined above, of the Bolzano-Weierstrass Theorem for Sequences, is pretty much the standard one, found in most analysis texts. (b) I’ve appended the phrase ‘for Sequences’ to the text’s name of this theorem because there are very similar theorems, found in other texts, which are also called ‘the Bolzano-Weierstrass Theorem’, but which apply in other situations. Be careful, when reading other texts or analysis papers, to know to which version their use of the name ‘Bolzano-Weierstrass Theorem’ refers. Things to Prepare for the Discussions (These are not to be handed in.) Note: These are problems/examples/topics which the Teaching Assistant (TA) will cover in the discussion sections. You are expected to work on them before you go to the discussions. (A) (Items for the discussion on Tuesday 11/08/2011. Do not hand in.) (1) Exercise 11.2 (2) Exercise 12.3 (a, b, d, e) Note: You do not need the results of Section 12 to do this exercise. It could have been placed it in Section 10 just as well. (3) Go over Example 4 in Section 11 (B) (Items for the discussion on Thursday 11/10/2011. Do not hand in.) REVISED (1) Let σ = ( s 1 ,s 2 , ... ) be a sequence of real numbers. Let τ = ( t 1 ,t 2 , ... ) be the subsequence of σ given by the odd-order terms of σ , and let ξ = ( x 1 ,x 2 , ... ) be the subsequence of σ given by the even-order terms of σ . That is, τ = ( s 1 ,s 3 ,s 5 , ...s 2 k - 1 , ... ) and ξ = ( s 2 ,s 4 ,s 6 , ...s 2 k , ... ) ,
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Browse Documents | 2,373 | 7,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-13 | latest | en | 0.880247 |
https://math.stackexchange.com/questions/3140891/find-the-points-on-two-line-segments-that-are-exactly-r-units-apart | 1,725,839,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00118.warc.gz | 353,232,678 | 35,551 | # Find the points on two line segments that are exactly r units apart
I am trying to use this technique in order to find the collision point of a fast moving 2D circle and a line segment. Because the circle is fast moving I only know it’s start and end point but the collision may have happened somewhere between the two. The thought is that at the point where the fast moving circle intersects with the line the distance between the circles center (and thus a point on the particles movement line) will be r.
This is sort of a derivative of the formula where you are finding the closest points between two lines. However in this case you are trying to find the points that are r units apart.
I can’t figure out how to even begin solving this.
Keep in mind these are line segments and both lines do have endpoints.
I do realize there could be a number of outcomes from this formula including cases where there are 2 sets of points that are r units apart. | 198 | 959 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.958896 |
https://www.missmamosworld.com/how-many-tortilla-chips-per-person.html | 1,660,893,681,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00611.warc.gz | 748,559,198 | 22,594 | Tortilla
# How many tortilla chips per person?
A serving of tortilla chips is equal to 1 ounce, which works out to between 10 and 15 chips, depending on size. Clearly, restaurants don’t abide by this suggested serving size, which means you might end up eating two, three or more servings of tortilla chips at one time.
## How many bags of chips do I need per person?
A regular-sized bag has eight ounces and a single serving is one ounce, meaning, a regular bag of chips can serve eight people in a group – but this is just not reality – drink some beer or wine and chip consumption goes up dramatically.
## How many chips equal a serving?
You can expect about 7-10 chips per 1 oz. serving.
## How many chips in a Lays bag?
Lay’s Oven Baked Original (1.125 oz bag) = 20 chips.
## How do you estimate food for a crowd?
Take the number of your guest list and multiply it by three to calculate how many servings you will need for the appetizers. A rule of thumb is to always calculate a big group meal as though it is a buffet, even if it is a sit down affair.
## How much food do I need for a buffet?
Rules to Plan By Each adult will consume 1 pound of food total; children, about 1/2 pound. The more options you have, the less you need of each; decrease the main course portion sizes by 1 to 2 ounces if served on a buffet. Guests will always eat — and drink — more at night than during the day.
## How do you plan a taco bar for 100?
1. 18 Pounds 80/20 ground beef.
2. 20 cups Mexican Cheese.
3. 100 Taco shells – soft & crunchy.
4. 6-7 8oz bags Shredded Lettuce.
5. 6 Large Tomatoes.
6. 62 oz Salsa with an 8 oz bottle of taco sauce.
7. 64 oz sour cream.
## How big is a regular bag of chips?
Regular Lay’s are sold in 10-ounce bags; flavored Lay’s are sold in 9.5-ounce bags; and both are sold for the same \$4.29 price. That might not sound like a lot, but it will with a bit of simple math. Americans buy some \$1.6 billion worth of Lay’s potato chips every year.
## How many tortilla chips is 28 grams?
on the Border- Tortilla Chips – Chips 1 oz (28g) 7 Chips.
## How many Pringles is 1oz chips?
Potato Chips Nutrition Facts The standard serving size for potato chips on nutrition labels is 1 ounce — that is about 28 chips.
## How many appetizers do I need for 30 guests?
A general rule here is about 3 different appetizer recipes for 10 people, 5 different appetizer recipes for 30 people and 7 different appetizer recipes for 50 people.
## How much does meat weight per person?
When Meat Is the Main: When cooking something like steak, roast, chicken, or pork, where meat is the main feature of the meal and paired with a few side dishes, we recommend about 1/2 pound (eight ounces) per person, up to 3/4 (12 ounces) pound for bigger appetites and those who love leftovers.
## How much does a bag of Lays chips cost?
Lay’s potato chips bring in over \$1 billion annually in retail sales, equivalent to over 200 million bags, if the average price per bag is somewhere around \$4.
## What are the healthiest chips?
1. Barnana pink salt plantain chips. Price: \$
2. Jackson’s Honest sweet potato chips. Price: \$
3. Safe + Fair olive oil and sea salt popcorn quinoa chips. Price: \$
4. Lesser Evil Paleo Puffs. Price: \$
5. Made in Nature Veggie Pops.
6. Siete tortilla chips.
8. Forager Project grain-free greens chips.
## Why are chips bags so empty?
The Reason You’re Looking at a Bag Half-Full In the manufacturing industry, “slack fill” is empty space that’s intentionally placed around a product. The idea is that the extra room can as a buffer to protect your Lays, Ruffles or Tostitos from damage.
## What is a good meal to feed a large crowd?
1. White Pesto Spinach Lasagna (10 Servings)
2. Slow-Cooker Crispy Carnitas (10 Servings)
3. Baked Sausage and Cheese Rigatoni (8 Servings)
4. Caramelized Shiitake Mushroom Risotto (8 Servings)
5. Korean Beef Tacos (12 Servings)
6. Healthy Instant-Pot Turkey Chili (8 Servings) | 1,022 | 3,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.948858 |
https://articlebiz.com/article/1051654706-understanding-betting-odds-for-successful-online-gambling | 1,726,069,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00736.warc.gz | 87,225,903 | 8,984 | ### Understanding Betting Odds For Successful Online Gambling
• Author John White
• Published March 27, 2020
• Word count 719
For those of you interested in online betting, understanding betting odds can be a daunting task. However, this needn't be the case, as a basic understanding of how odds are calculated should allow even the beginner to understand and implement them in their betting activities.
What are Odds?
In essence, odds are a reflection of the likelihood of a certain outcome taking place in a specific event.
"What on earth does that mean?", you might ask.
In every event where there is betting involved, all outcomes have a certain chance of taking place. Odds are simply an interpretation of those chances, and the odds presented by bookmakers merely reflect such chances to the best of the bookmaker's abilities, minus the bookmaker's edge. Most online bookmakers offer up to three different choices on how you want to view your betting odds: Decimal, Fractional or American.
Decimal Odds
Decimal odds are commonly used in Europe and are therefore sometimes referred to as European odds.
To convert a chance into decimal odds just put the probability as a percentage and divide it into 100.
100/%Chance = decimal odds
So, if you believe that something has a 50% chance of winning, then:
100/50 = 2 or odds of 2.0.
Say then that you want to place a bet on a selection that has decimal odds of 2.0. If you win, for every dollar that you stake you will receive 2 dollars back. Stake \$100 and you will receive back \$200. This amount received back includes your original stake in the odds.
Fractional Odds
The more traditional fractional odds are often still used in the UK, and can therefore also be referred to as British odds, UK odds or traditional odds. These odds quote the net total that will be paid out to the bettor should he win, relative to his initial stake. Using the 50% chance example listed above again, the fractional odds equivalent of 2.0 are 1/1 which is also knows as evens or even money.
For example, you want to place a bet that has odds of 1/1. If you win, for every dollar you stake you will win \$1 and you will receive your initial \$1 stake back, giving you a total return of \$2. If you place a bet of \$100 at 1/1, then you will win \$100 and have your \$100 stake returned, giving a total return of \$200.
American Odds
Also known as moneyline odds, these odds are favoured by US bookmakers as their name suggests. These odds show either a negative or a positive figure when quoting the odds of a certain outcome taking place.
If US Odds are indicated with a + sign then they show the amount you would win for a \$100 stake. If there is a - sign then they show how much you need to stake to win \$100.
So if you are betting on an event which has decimal odds of 2, or fractional odds of 1/1, the US Odds would be +100 (i.e. you would win \$100 if you bet \$100). If you are betting at decimal odds of 1.5, or fractional odds of 1/2, then the US odds would be -200 (you need to bet \$200 in order to win \$100 more)
Conclusion
Betting is often known as a sucker's game and with good cause. The sheer chance of any outcome taking place in an event and the difficulty in predicting which outcome will occur, as well as the bookmaker's inherent edge in all betting events, often make it extremely difficult for individuals to come out with a long-term profit from their betting activities.
However, if you are interested in online sports betting and do think that you have what it takes to profit, then it's important that you educate yourself first before starting to bet. This will give you the best possible chance of coming out ahead in the long run.
Find bookmakers that offer odds with the least edge and therefore present the best value to you. Also, take advantage of bookmaker's sometimes generous free bets on offer, as these can increase your profit if used wisely.
Finally, do in-depth study on the event on which you wish to bet. Read statistics, past performances and anything at all that could have an impact on the result. Being informed in this way will allow you to make the best possible choice, which will in turn give you the best possible chance of winning on the day.
Article Source: http://EzineArticles.com/1806391
Oddstify is the best odds comparison website. Its main purpose is to help people who are betting to always get the highest odds on whatever bet they make and never lose out on the best opportunities from the leading bookmakers.
https://oddstify.com/
Article source: https://articlebiz.com | 1,016 | 4,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.96066 |
http://www.compadre.org/portal/items/detail.cfm?ID=11157 | 1,540,231,752,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515375.86/warc/CC-MAIN-20181022180558-20181022202058-00221.warc.gz | 435,565,261 | 8,680 | ## Detail Page
written by Tom Henderson
This unique activity presents eleven interactive challenges designed to help students understand the language of motion. Each challenge requires the student to match the motion of an animated car to the correct verbal description of that motion. After all matches have been completed, students check their answers and try again in case of a mismatch.
Editor's Note: This activity is ideal for promoting constructivism in the secondary science classroom. It offers learners a low-risk environment for building their own understanding of velocity as a vector quantity, positive and negative acceleration, constant speed, and constant acceleration. It can be adapted for use in both middle and high school. The activity sheet allows teachers to insert the animation quickly into existing lessons.
This resource is part of The Physics Classroom web site, a growing collection of resources for teachers and learners of introductory physics.
Please note that this resource requires Shockwave.
Subjects Levels Resource Types
Classical Mechanics
- Motion in One Dimension
= Acceleration
= Position & Displacement
= Velocity
- High School
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- Lower Undergraduate
- Instructional Material
= Activity
= Interactive Simulation
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© 2011 Tom Henderson
Keywords:
acceleration, constant acceleration, constant velocity, interactive animations, kinematics animations, language of kinematics, language of motion, motion animations, velocity
Record Cloner:
Metadata instance created April 18, 2011 by Caroline Hall
Record Updated:
May 3, 2011 by Lyle Barbato
Last Update
when Cataloged:
February 28, 2011
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### AAAS Benchmark Alignments (2008 Version)
#### 4. The Physical Setting
4F. Motion
• 6-8: 4F/M3a. An unbalanced force acting on an object changes its speed or direction of motion, or both.
• 9-12: 4F/H8. Any object maintains a constant speed and direction of motion unless an unbalanced outside force acts on it.
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When i went to a interview the interviewer ask me a
question for that i cant able to answer him if anyone knows
How will you design a capacitor bank for a plant? if my p.f
is 0.6 i have to improve it to 0.9 or 1 what kind of steps i
have to do? and is there any formula for calculating this?
Answers were Sorted based on User's Feedback
When i went to a interview the interviewer ask me a question for that i cant able to answer him if..
yes there is an answer fir that, the answer is
KVAR=KW(inv(tan(PF2)-(inv(tan(PF1)))
where tana is desired power factor and tanb is your current
pf.
now to calaulate your a and b parameter we use this formula
power factor =cosa and power factor=cosb
i.e., inv(cos)=power factor
put this in above equation uyou will get your KVAR RATING
but you should the KW of the load.
Is This Answer Correct ? 6 Yes 2 No
When i went to a interview the interviewer ask me a question for that i cant able to answer him if..
this is also calculated from power triangle. i.e.,
KVAR=KW/COS(@).
Is This Answer Correct ? 1 Yes 1 No
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what is b checks and m checks in diesel generator maintenance?
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what is working of Master trip relay ?
What is power factor? How to improve Power factor? draw a circuit diagram showing the connection of power factor improvement equipment to the electrical instillation.
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Why is the electric shock from the DC supply is more dangerous than the shock from AC supply.
for opening electrical shop, tips needed.
What is the role of ESP(Electro Static Precipitator)in steam power plants? Where it is located? How it works?
what is isochronous mod? what is voltage drop mode and p.f mode?
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# Étude de cas - amortissement
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• Notes d'étude
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Tashika B. 0 0 The case study is very interesting.
BabaJide Martins F. 0 0 The accumulated depreciation represents the amount of the cost of an asset allocated as an expense added up over a number of accounting periods.
Alice B. 0 0 Please can some one tell me if there are two ways of calculating depreciation under the straight line method because on previous topic an example was given and after using this method I got the result of the diminishing balance.
Morne V. 0 0 information about diminishing and straight line depreciation method through illustration is clearly given.
Sunday O. 0 0 Case study - Depreciation
Roselyn S. 0 0 Thank you it helps to my further understanding..
Odongo M. 0 0 Accounting -> Case study - Depreciation Case study - Depreciation Example A Pizza Oven is bought on 30 September 2001 for \$10 000. It costs a further \$5000 to install. The residual value is estimated at \$3000 and the estimated life is 6 years. Required: calculate depreciation for the year ended 30.6.2002 and 30.6.2003 using both the straight line method and the diminishing balance (25% per annum) present these calculations in the general journal show the affect on the Profit and Loss statement (extract provided) show how an extract may appear in the balance sheet Straight line Profit and Loss for year ended 30.6.2002 \$ \$ Sales 8 000 less Expenses 4 000 + depreciation of pizza oven Total expenses Net profit Straight line Profit and Loss for year ended 30.6.2003 \$ \$ Sales 8 000 less Expenses 4 000 + depreciation of pizza oven Total expenses Net profit Diminishing balance Profit and Loss for year ended 30.6.2002 \$ \$ Sales 8 000 less Expenses 4 000 + depreciation of pizza oven Total expenses Net profit Diminishing balance Profit and Loss for year ended 30.6.2003 \$ \$ Sales 8 000 less Expenses 4 000 + depreciation of pizza oven Total expenses Net profit A related theory question would ask students to explain the relative effects on profit of the selection of the depreciation method. View the word document for solution to Case study - Depreciation
Saul K. 0 0 great
Manish K. 0 0 information about diminishing and straight line depreciation method through illustration is clearly given.
Davidson Ade O. 0 0 understood | 705 | 2,948 | {"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} | 2.921875 | 3 | CC-MAIN-2020-34 | latest | en | 0.764683 |
https://byjus.com/maths/metric-system/ | 1,722,944,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00642.warc.gz | 128,703,809 | 115,038 | # Metric System
Metric System is basically a system used for measuring distance, length, volume, weight and temperature. It is based on three basic units with which we can measure almost everything in the world
1. M- meter, used to measure the length
2. Kg- kilogram, used to measure the mass
3. S- second, used to measure time
The metric system was originated in the year 1799. Over the years, this system has been extended to incorporate many units. Although a number of variants of this system were discovered in the late 19th and early 20th centuries, the term – metric system is used as another word for SI or the International System of Units.
## SI Units Metric System
The SI Units (standard International System of Units) give proper definitions for meter, kilogram and the second. It also specifies and defines four different additional units :
Kelvin- temperature
Ampere – Electric Current
Candela- luminous Intensity
Mole – Material Quantity
## Metric Number Prefixes
In order to remember the proper movement of units, arrange the prefixes from the largest to the smallest.
Kilo hecto deca Unit deci centi milli 103 102 101 100 10-1 10-2 10–3
Thus, the conversion from one unit to the other unit is done by multiplying or dividing the powers of 10.
Now, let us discuss some of the units for length, weight, volume, and time.
Length:
The most common units used to measure the length are as follows:
Kilometer (km) Hectometer (hm) Decameter (dam) Meter (m) Decimeter (dm) Centimeter (cm) Millimeter (mm) 1000 100 10 1 1/10 1/100 1/1000
Also, check: Centimeter
Volume or Capacity
The most common units used to measure the capacity or volume of any object are as follows:
Kiloliter (kl) Hectoliter (hl) Decaliter(dal) Liter (l) Deciliter (dl) Centiliter (cl) Milliliter(ml) 1000 100 10 1 1/10 1/100 1/1000
Weight
The most common units used to measure the weight of any object are as follows:
Kilogram (kg) Hectogram (hg) Decagram (dag) Gram (g) Decigram (dg) Centigram (cg) Milligram (mg) 1000 100 10 1 1/10 1/100 1/1000
Time
To calculate the time, the base unit is second. Now, let us discuss some other units of time.
1 minute = 60 seconds
1 hour = 60 minutes
1 day = 24 hours
1 week = 7 days
1 month = 30 or 31 days (Note: February has 28 days, but in leap year February has 29 days)
1 year = 12 months (or)
1 year = 365 days (But, in leap year, 1 year = 366 days)
Whether you want to measure very small or very big things, You can go for metric number prefixes—for example- kilo and milli.
There are some common units that are based on the kilogram, meter and second.
1. Area = Square Meter (Area= length by length) Therefore the basic unit of area is equal to meters by meters = m2 (square meters).
1. Volume = Cubic Meter. Therefore, the basic unit of volume = m3 (cubic meters).
1. Liter = one-thousandth of a cubic meter(1 m3 = 1,000 Liters) Therefore, 1 Liter = 1/1000 m3
1. Time = Hour (1 hour=60 minutes, 1 minutes=60 seconds) Therefore,1 hour= 60×60= 3600 seconds
1. Day = (1 day= 24 hours) Therefore, 1 day= 24 x 60 x 60 = 86400 seconds
Important Things to Remember
### Metric System Conversion
We can follow different methods for metric conversions. The below metric conversion chart shows the conversion of Metric Units to Metric Units for Length.
## Conversion Ratio
In the measurement of units, the metric conversion ratio is a ratio, which is equal to 1. It is also known as unit factors. It is used to convert the units between the same system, as well as the conversion between the metric and the English systems. It is based on the concept of equivalent values. Similarly, a conversion factor is defined as a number that is used to convert one unit to another unit by dividing or multiplying to get the equivalent values.
For example, to convert 5 meters to centimetres, a conversion factor is required.
We know that 1 meter = 100 centimetres
Thus, the conversion factor is “multiplying by 100”.
(i.e.,) 5 m = 5 x 100 cm
5 m = 500 cm
Therefore, the conversion from 5 m to cm is 500 cm.
In terms of conversion ratio, it can be written as:
5m / 500 cm = 1
### Metric System Examples
The following examples explain the conversion from one unit to the other unit.
Example 1:
Convert 356 centimetres to meters.
Solution:
We know that, 1 cm = 0.01 m
Thus, 356 cm = 356 x 0.01
356 cm = 3.56 m.
Therefore, 356 cm is equivalent to 3.56 meters.
Example 2:
Convert 4000 grams to kilograms.
Solution:
We know that, 1 gram = 0.001 kilogram
Thus, 4000 g = 4000 x 0.001
4000 g = 4 kilogram
Therefore, 4000 grams is equivalent to 4 kilogram
Example 3:
Convert 50 kilolitres to liters.
Solution:
We know that 1 kilolitre = 1000 liters
Thus, 50 kilolitres = 50 x 1000 litres
50 kilolitres = 50000 litres
Therefore, 50 kilolitres are equivalent to 50000 liters.
## Frequently Asked Questions on Metric Systems
Q1
### What is meant by the metric system?
The metric system is defined as the system of measurements to calculate the mass, distance, and volume of any object. We generally use the metric system to measure smaller or larger quantities.
Q2
### What are the base units for length, weight, and volume in a metric system?
In the metric system, the base units for length, weight, and volume are meters, grams, and liters respectively.
Q3
### Mention the US standard units for length, weight, and volume.
In US customary systems, the units used are:
Distance or length in miles, yards, feet, inches
Mass or weight in pounds, tons, ounces
Capacity or volume in cups, gallons or quarts, pints, fluid ounces.
Q4
### What is meant by metric number prefixes?
The metric number prefix is defined as a unit prefix, which precedes a base unit to indicate the multiple or the submultiple of a unit. Some of the metric prefixes are:
Kilo (1000 or 103)
Hecto (100 or 102)
Deca (10 or 101)
Base unit (1)
Deci (1/10 or 10-1)
Centi (1/100 or 10-2)
Milli (1/1000 or 10-3)
Q5
### Convert 80 grams to milligrams.
We know that 1 gram = 1000 milligrams
Therefore, 80 grams = 80 x 1000 milligrams
80 grams = 80000 milligrams
Thus, 800 grams is equivalent to 80000 milligrams.
Quiz on Metric system | 1,679 | 6,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-33 | latest | en | 0.848388 |
jordivila.cat | 1,490,256,540,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186841.66/warc/CC-MAIN-20170322212946-00000-ip-10-233-31-227.ec2.internal.warc.gz | 188,647,932 | 2,830 | ### Fun Trig
Fun Trig is a simple native app for the iPhone that uses animation to help K-12 Pre-Calculus students to explore The Unit Circle and understand Circle Trigonometry better in order to visualize the meaning of the 'sine wave' which is so important in describing many different natural phenomena. The sine and cosine of an angle are defined as the lengths of the y (orange) and x (green) coordinates of a point rotating counterclockwise around a circle of radius one (Unit Circle) centered at the origin. When you click on the Graphs button you can watch the sine (orange) and cosine (green) graphs in a separate screen, where the mapping between the coordinates of a point on the unit circle and the distance travelled around it (yellow) can be watched through the animation.
By using the app and its features (like visualizing the angle of reference for any given angle), students might just 'watch' the answers to some usual questions they might be required to answer while learning Trigonometry such as:
• What is the largest value of the sine function? What is the smallest value?
• Does the sine of an angle increase or decrease as the angle increases from 180° to 270°?
• Find all values of an angle between 0° and 360° such that its sine is equal to -0.8192
• TWhat is the reference angle of 238°
For simplicity only sine and cosine are drawn. Tap on the screen to pause/resume the animation.
### Why Fun Trig?
Prior to move to Software Engineering, for a number of years back in the 80s, I worked as a Calculus and Pascal professor for the well-known Massachusetts Prep School Phillips Andover Academy (andover.edu). My first apps for the Macintosh in the mid-eighties were in fact Math educational software and also applications for the Science Museum of Barcelona. This little free app is a way to remember those very happy days of my professional life. To my surprise, the number of downloads of the app so far seems to have justified the effort! | 430 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-13 | longest | en | 0.920033 |
http://sites.millersville.edu/bikenaga/number-theory/linear-congruences/linear-congruences.html | 1,550,758,181,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00453.warc.gz | 237,882,389 | 3,845 | # Linear Congruences
Theorem. Let , and consider the equation
(a) If , there are no solutions.
(b) If , there are exactly d distinct solutions mod m.
Proof. Observe that
Hence, (a) follows immediately from the corresponding result on linear Diophantine equations. The result on linear Diophantine equations which corresponds to (b) says that if is a particular solution, then there are infinitely many integer solutions
I need to show that of these infinitely many solutions, there are exactly d distinct solutions mod m.
Suppose two solutions of this form are congruent mod m:
Then
Now divides both sides, and , so I can divide this congruence through by to obtain
Going the other way, suppose . This means that and differ by a multiple of d:
So
This implies that
So
Let me summarize what I've just shown. I've proven that two solutions of the above form are equal mod m if and only if their parameter values are equal mod d. That is, if I let t range over a complete system of residues mod d, then ranges over all possible solutions mod m. To be very specific, all the solutions mod m are given by
Example. Solve .
Since , there are no solutions.
Example. Solve .
Since , there will be 1 solutions mod 4. I'll find it in three different ways.
Using linear Diophantine equations.
By inspection , is a particular solution. , so the general solution is
The y equation is irrelevant. The x equation says
Using the Euclidean algorithm. Since , some linear combination of 3 and 4 is equal to 1. In fact,
This tells me how to juggle the coefficient of x to get :
(I used the fact that .
Using inverses mod 4. Here is a multiplication table mod 4:
I see that , so I multiply the equation by 3:
Theorem. Let , and consider the equation
(a) If , there are no solutions.
(b) If , there are exactly distinct solutions mod m.
I won't give the proof; it follows from the corresponding result on linear Diophantine equations.
Example. Solve
, so there are solutions mod 10. I'll solve the equation using a reduction trick similar to the one I used to solve two variable linear Diophantine equations.
The given equation is equivalent to
Set
Then
, , is a particular solution. The general solution is
Substitute for w:
, , is a particular solution. The general solution is
will produce distinct values of y mod 10. Note, however, that s and produce and , which are congruent mod 10. That is, adding a multiple of 2 to a given value of s makes the term in x repeat itself mod 10. So I can get all possibilities for x mod 10 by letting .
All together, the distinct solutions mod 10 are
Remarks: I saw the particular solution , by inspection. In general, you can get one using the Extended Euclidean algorithm. For example, in this case
Multiply by (to match ) to get
So a particular solution is and .
In general, it can be tricky to determine the parameter ranges which give the correct number of solutions; it may require some trial-and-error, or careful analysis of the general solution.
Contact information | 688 | 3,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-09 | latest | en | 0.901475 |
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Volume 10, Issue 45; November 10, 2010: How to Make Good Guesses: Tactics
# How to Make Good Guesses: Tactics
Making good guesses probably does take talent to be among the first rank of those who make guesses. But being in the second rank is pretty good, too, and we can learn how to do that. Here are some tactics for guessing.
In the first part of this series, we examined strategies for making good guesses — overall approaches that lead to excellent conjectures. Let's now turn to tactics for making good guesses based on what you see — and what you don't.
Look for what's not there
Many guesses involve recognizing the absence of something important. Some missing factors are obvious, such as gaps in a sequence, or something missing that's usually implied by something that's present. See "On Noticing," Point Lookout for May 2, 2012, for more.
Other missing items are more difficult to notice. For instance, consider two factors present in the situation before you. Then ask, if these two are connected in some way, what would that connecting feature imply? That implied attribute of the situation might be missing. If it is, what does that tell you? See "On Noticing," Point Lookout for May 2, 2012, for more.
Examine temporal sequences
A temporal sequence is a sequence in which time of occurrence determines position in the sequence. Since time of occurrence is often confused with time of discovery or time of recognition, the first thing to sort out is temporal order.
Once you know the order, you can reverse it, and consider whether the reversed sequence is actually possible. If the reversed sequence or any subset of it could have happened in that order, it's possible that the order you believe you have is actually incorrect. What if it is? What does that tell you?
For people, focus on situation, not character
When most of us conjecture what others will do in a given situation, we tend to put too much weight on their character or motivation, and too little weight on how they experience that situation. This error is so common that it has a name: the Fundamental Attribution Error.
Since disregarding Many guesses involve
recognizing the absence
of something important
character or motivation is also an error, keep it in the mix. But think much more about how the situation looks to the people in question. What will they know? What will they not know? How will their past experiences influence what they notice or don't notice? What are others hiding? What disinformation is present? Focus on trying to see things from their vantage point, and then project the decisions they're likely to make based on the information they have.
Most important, watch others. You probably know someone who makes consistently good guesses. Actually, you probably know more such people than you imagine you do. Many great guessers conceal from others — and sometimes themselves — that they're guessing. They present a demeanor of knowledge and confidence designed to conceal their guessing.
When someone appears to "know" something you think they might not actually know, make a note of it. Later, imagine how you would have made that guess. This exercise, repeated over time, gives you a chance to build your guessing skills.
Are your projects always (or almost always) late and over budget? Are your project teams plagued by turnover, burnout, and high defect rates? Turn your culture around. Read 52 Tips for Leaders of Project-Oriented Organizations, filled with tips and techniques for organizational leaders. Order Now!
Thank you for reading this article. I hope you enjoyed it and found it useful, and that you'll consider recommending it to a friend.
This article in its entirety was written by a human being. No machine intelligence was involved in any way.
Point Lookout is a free weekly email newsletter. Browse the archive of past issues. Subscribe for free.
Support Point Lookout by joining the Friends of Point Lookout, as an individual or as an organization.
Do you face a complex interpersonal situation? Send it in, anonymously if you like, and I'll give you my two cents.
## Related articles
More articles on Personal, Team, and Organizational Effectiveness:
Time Management in a Hurry
Many of us own books on time management. Here are five tips on time management for those of us who don't have time to read the time management books we've already bought.
In the Groove
Under stress, we sometimes make choices that we later regret. And we wonder, "Will I ever learn?" Fortunately, the problem usually isn't a failure to learn. Changing just takes practice.
Contextual Causes of Conflict: I
When destructive conflict erupts, we usually hold responsible only the people directly involved. But the choices of others, and general circumstances, can be the real causes of destructive conflict.
Still More Things I've Learned Along the Way
When I have an important insight, or when I'm taught a lesson, I write it down. Here's another batch from my personal collection.
Brain Clutter
The capacity of the human mind is astonishing. Our ability to accomplish great things while simultaneously fretting about mountains of trivia is perhaps among the best evidence of that capacity. Just imagine what we could accomplish if we could control the fretting…
See also Personal, Team, and Organizational Effectiveness and Personal, Team, and Organizational Effectiveness for more related articles.
## Forthcoming issues of Point Lookout
Coming September 4: Beating the Layoffs: I
If you work in an organization likely to conduct layoffs soon, keep in mind that exiting voluntarily before the layoffs can carry significant advantages. Here are some that relate to self-esteem, financial anxiety, and future employment. Available here and by RSS on September 4.
And on September 11: Beating the Layoffs: II
If you work in an organization likely to conduct layoffs soon, keep in mind that exiting voluntarily can carry advantages. Here are some advantages that relate to collegial relationships, future interviews, health, and severance packages. Available here and by RSS on September 11.
## Coaching services
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## Get the ebook!
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# How many different four-letter words can be formed (the words don't
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How many different four-letter words can be formed (the words don't [#permalink]
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12 Dec 2007, 11:15
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How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(3!*2!*2!*2!)
[Reveal] Spoiler: OA
Last edited by Bunuel on 05 Apr 2015, 04:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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12 Dec 2007, 12:21
Difficult one
M E D I T R A N
E R A N
E
E _ _ R
M 7
E 8
D 7
I 7
T 7
R 7
A 8
N 8
So 5x7 + 3x8=35+24=59
Cannot figure out anything other than brute force.
Tried 11C2 which made logic to me, but got 11C2=55
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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12 Dec 2007, 20:19
My way:
Available letters:
M E D I T R A N (8 letters)
E _ _ R
We have 1 combination for E and 1 combination for R, and also we have 8 combinations for the 2nd letter and 7 combinations for the last letter, so:
E 8 7 R = 1 * 8 * 7 * 1 = 56
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 01:28
can somebody explain how to solve this one? would appreciate
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 02:41
young_gun wrote:
How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/2!*2!*2!
C. 56
D. 23
E. 11!/3!*2!*2!*2!
Please, could you explain that to me so that I can easily understand?? I am very bad at perms!
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 03:23
marcodonzelli wrote:
Please, could you explain that to me so that I can easily understand?? I am very bad at perms!
we should complete word E _ _ R using set {M-1, E-2 (one E we use as the first letter), D-1, I-1,T-1,R-1 (one R we use as the last letter) ,A-2,N-2}
So, the set consist of 5 single letters and 3 pairs of letters.
1. for second position we have 8 cases (or 5+3)
2. for third position we have either 8 cases (second letter is from a pair) or 7 cases (second letter is single letter).
Therefore,
N=(3*8+5*7)=59
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 04:17
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Please, could you explain that to me so that I can easily understand?? I am very bad at perms![/quote]
we should complete word E _ _ R using set {M-1, E-2 (one E we use as the first letter), D-1, I-1,T-1,R-1 (one R we use as the last letter) ,A-2,N-2}
So, the set consist of 5 single letters and 3 pairs of letters.
1. for second position we have 8 cases (or 5+3)
2. for third position we have either 8 cases (second letter is from a pair) or 7 cases (second letter is single letter).
Therefore,
N=(3*8+5*7)=59
I understand point 1 and point 2 as well...but why N=(3*8+5*7)?thanks
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 04:37
marcodonzelli wrote:
I understand point 1 and point 2 as well...but why N=(3*8+5*7)?thanks
for letters of E,A,N at second position we have 8 cases for third one. So, 3*8
for letters of M,D,I,T,R at second position we have 7 cases for third one (we cannot use, for example, M twice). So, 5*7
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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13 Dec 2007, 13:32
5
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young_gun wrote:
How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/2!*2!*2!
C. 56
D. 23
E. 11!/3!*2!*2!*2!
We have 11 letters after E and R occupied their places. But E, A and N show up twice each. So we have 8 distinct letters for 2 places.
For the second place - 8 letters
for the third - 7 letters
Number of variants - 8*7=56, but we have to take into account additional 3 variants with double letters EAAR, ENNR, EEER.
So the ultimate calculation is 56+3=59
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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04 Apr 2015, 16:54
anybody plz explain why position 2nd and 3rd not like this : 11*10 = 110 ways !
many thanks !
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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06 Apr 2015, 13:38
Hi Bunuel, could you please explain the solution in your words ?
Thanks.
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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07 Apr 2015, 04:33
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Swaroopdev wrote:
Hi Bunuel, could you please explain the solution in your words ?
Thanks.
How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(3!*2!*2!*2!)
E - - R
We are left with the following 11 letters: {M, D, I, T, R, EE, AA, NN} out of which 8 are distinct: {M, D, I, T, R, E, A, N}.
We should consider two cases:
1. If the two middle letters are the same, we'd have 3 words: EEER, EAAR and ENNR.
2. If the two middle letters are distinct, then we are basically choosing 2 letters out of 8 when the order of the selection matters, so it's 8P2 = 56.
Total = 56 + 3 = 59.
Hope it's clear.
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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08 Apr 2015, 01:50
In the above problem, if the letters of the word MEDITERRANEAN are allowed to be used multiple times irrespective of their count in the parent word (commonly referred as ‘repetition’ in the P&C parlance), the answer would change. Let me explain the solution for such a case.
We need to fill the 2nd and the 3rd place with letters present in the word MEDITERRANEAN. Since, there are 8 different letters (M, E, D, I, T, R, A, N) in the word MEDITERRANEAN, the 2nd place can be filled with 8 possible letters and the 3rd place can also be filled with 8 possible letters (because, in the case we are discussing here, the letters can be used multiple times, even if they are present only once in the word MEDITERRANEAN).
So, we will have a total of 8*8 = 8^2= 64 possible set of words
Similarly, if the above case is extended to the first and the last letter as well (i.e. we don’t have the constraint of having ‘E’ as the first letter and ‘R’ as the last letter), we will have 8^4 possible sets of words which we can form from the word MEDITERRANEAN.
The key here is to be careful on two points:
Whether letters can be used more than their count in the parent word, in this case MEDITERRANEAN.
If yes, then we need to focus only on different letters present in the parent word, in this case the 8 different letters in the parent word MEDITERRANEAN.
Hope it helps!
Regards
Harsh
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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11 Apr 2015, 03:04
I used a diffrent method:
we can also solve this question with combinatorics fairly easy:
after E and R are set as the first and the last letters we are left with the two middle ones.
since both E and R show up more then once we can still use all the original letters for the two remeaining blanks.
actually our bank of letters will now look as so:
M=1
E=2
D=1
I=1
T=1
R=1
A=2
N=2
if all remaining letters would have shown up just once the answer would have been:
#=8P2=8!/(8-2)!=56
but since we are left with 3 letters that show up more then once (E,A,N) we need to add the possibilty of using the same letter twice, meaning:
#=8P2+3=59
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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Re: How many different four-letter words can be formed (the words don't [#permalink]
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04 Jun 2017, 18:14
Hello all,
First, sorry for posting on this post again, idk if it is allowed
I understood the correct answer, but I cannot figure out why my approach is incorrect.
We know the total of letters as well as the total of repeated letters.
M - 1
E - 3
D - 1
I - 1
T - 1
R - 2
A - 2
N - 2
And we know that E and R have been already used once, so we now have 2 E's and 1 R.
If the question was: how many different 13-letter words can be formed, I would calculate like this:
E x x x x x x x x x x x R - > 11! / (2! * 2! * 2!) (and there is an answer for that)
As the question if for 4-letter word, and we have just to spaces left, I would just do this:
11 * 10 / (2! * 2! * 2!)
I know that this is incorrect (also because it isn't an integer number ). But I do not know why!
Thanks!
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Re: How many different four-letter words can be formed (the words don't [#permalink]
### Show Tags
05 Jun 2017, 06:53
young_gun wrote:
How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(3!*2!*2!*2!)
We have to foind total ways of filling E _ _ R
After taking one E and one R out we get 8 unique letters M E D I T R A N and 3 double letters E A N
Taking 2 letters from set 1 and arranging them = $$_{8}C_{2} * 2!$$ = 56
Taking 2 letters from set 2 and arranging them(arrangement in this case does not matter) = 1+1+1=3
Anwer 56+3 = 59
A correct
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Re: How many different four-letter words can be formed (the words don't [#permalink] 05 Jun 2017, 06:53
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# Sebastian
### Gesellschaft für Akustikforschung Dresden mbH (AFD)
38 total contributions since 2011
Professional Interests: Acoustics
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más de 4 años ago | 1,562 | 5,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-05 | latest | en | 0.597801 |
https://discourse.mc-stan.org/t/ito-process-as-numerical-solution-of-stochastic-differential-equation/9192 | 1,652,836,391,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00799.warc.gz | 278,612,783 | 12,241 | # Ito process as numerical solution of stochastic differential equation
Just pushed the first cut of an Euler scheme for SDE
Unit test is based on simple model dX = \mu X dt + \sigma X dW.
sde_euler_1d.pdf (8.7 KB)
The proposed signature of the function ito_process_euler is
matrix ito_process_euler(F1 drift, F2 diffusion, vector y0,
matrix wiener_normal,
real[] theta1, real[] theta2, real t)
where F1 and F2 are functions for drift and diffusion with theta1 and theta2 being their parameters, respectively. y0 is initial condition at t=0, wiener_normal is a matrix of iid standard normals for Weiner process steps. With t being the final time, the step size equals to t divided by the number of columns of wiener_normal.
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The 2nd example/test is based on a stochastic SIR model, with parameter vector \theta=(a,b,c,d), the SDE is characterized by
f_{\text{drift}}(x) = (-ax_1x_2 + d, ax_1x_2 -(c+d)x_2, cx_2-dx_3)^t
f_{\text{diffusion}}(x) = (-bx_1x_2 , bx_1x_2, 0)^t
numerical results from ito_process_euler below show Ito processes that range from stable to unstable.
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I don’t know anything about SDEs, so I’m not going to be much help here.
Is there an issue defining what you want the behavior to be? Any chance we can get @charlesm93 or @bbbales2 or @wds15 to help review the code? I’m also looping in @syclik because he’s in charge of the math lib and also has a strong enough math background for this kind of work. @betanalpha would be another possibility for someone with enough C++ and math to handle this.
I’m about to submit an issue but since I’m getting zero feedback from discourse Performance of stochastic volatility model I’m hesitating to submit a PR. Currently my main concern is
• This implementation could be completely written in Stan lang, is it still worth to add it as a function.
• The Wiener process generation asks for a potentially large matrix/vector of iid normals, how’s this going to affect the performance in MCMC?
Everyone’s overwhelmed by volume and providing only spotty attention. You’ll need to tap some specific people to review this, and I think we only have a few core devs who are qualified.
The place to start is with an issue, not with a PR. Then when there’s agreement the issue is something we want to implement, then create the PR. @syclik is in the charge of the math lib and the one to contact if the process gets stuck.
That’s true of almost all the functions we write. By adding to Stan, we get something much easier to use and hopefully more reliable and efficient. But it’s a judgement call, which is why I’m suggesting starting with an issue and with pinging @syclik to review the intent.
You mean part of the implementation requires generating iid normals? We’d have to work out how to pass in an RNG in that case. They’re only in scope in transformed data and generated quantities as things are now.
I’ve considered this option but decided to let user pass in a real array that is supposed to be of iid normal distribution.
Can that be reused? One way to generate iid normals if they’re fixed is to use transformed data—that allows an RNG.
If one’s fitting a SDE strong solution with given trajectory this iid normal array is going to be parameter.
The 3rd example solves a stochastic volatility model, in particular, the Heston model with truncated volatility
dy_1 = \mu y_1dt + \sqrt{\max(y_2,0)}y_1dW_1 \\ dy_2 = \kappa(\theta - \max(y_2,0)) dt + \xi\sqrt{\max(y_2,0)}dW_2
This model can be seen as an extension of the volatility model in Stan’s user guide. Unlike previous two examples, here we have two correlated Wiener processes dW_1 & dW_2 driving the evolution, therefore the return of f_2 functor should be a 2-by-2 matrix(in this model it’s diagonal).
With asset spot price 100.0 and volatility 0.02, figures below show the solution processes in 1-year period, with 1-day time step. The two driving Wiener processes have correlation -0.7 for leverage effect.
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@Charles_Driver, this looks like something that you’d know about. If you have an interest in chiming in, please do :D.
I’m not familiar with SDE solvers, so I can’t review a PR on the subject, at least not in a short time span. I suggest bringing these results up during the Stan meeting and maybe adding “reviewer for SDE solver” as an item on the agenda.
What are some motivating problems for this feature? Anything in pharmacometrics I might be familiar with?
@bbbales2 Ok I can try :) it seems nice to have to me, though two points confuse me a little:
Since the output per sample seems to be stochastic, how does this work with an mcmc type sampling procedure wrapped around it? I’m more familiar with the situation where we integrate over the stochastic nature of the SDE to obtain expectations and covariances, so in a sense we understand the complete picture of the SDE implied by the parameters given the sampled values of the parameters.
What’s the point of passing in the Wiener random noise elements, does this add some flexibility? It seems like an unnecessary specification for the user.
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You can do a similar implementation of an integrated ou process. In a pure c++ version the iid normals are a detail much like the internal parameterization of the simplex type. When implementing in Stan code you have to pass in the iid normals just like you do when you implement the non centered parameterization in Stan
The output X per sample is the same as the other parameters: the sample of (an array of) random variables. When the iid normal for the wiener process is parameter, per sample we get one path of the wiener process, then use it to solve the SDE.
This is along the line of weak approximation, as one seek \tilde{X} to approximate X in the sense of reaching small error ||E(f(\tilde{X}))-E(f({X}))||. For this purpose \tilde{X} doesn’t have to solve SDE with the given wiener process but just some process. This is different from what’s proposed here: we seek a strong approximation for a given wiener process.
As @sakrejda points out this is just to present a finite points(of size n) of a wiener process W_n as a reparameterization as an array of iid normals N_n, as the W_n=LN_n, with L being a lower triangular matrix filled with square root time steps. As @Bob_Carpenter mentioned that when the wiener process is data we can do this as RNG. However, we need wiener process to be parameters when we want to fit a given Ito process path. Also it does add flexibility, for example, user can specify the relationship of multiple wiener processes: in the 3rd example above, W_1 and W_2 are correlated to reflect that asset price often falls when volatility rises. In Stan lang this can be achieved by passing in a matrix N_n\in R^{2\times n} created by multiplying the cholesky factor to an R^{2\times n} matrix of iid normals.
Yes ok all makes more sense if you’re talking about the case where the Wiener process is made up of parameters. Some discussion above re random number generation had me thinking that wasn’t the general intention…
There has definitely been non-zero interest in SDE models over the past few years, with applications spanning physics and biology and epidemiology amongst others, so having a self-contained implementation isn’t a bad idea.
The Euler-Mayorama discretization is very simple but for a stochastic process without much additional structure (like the structure one would get in a Langevin process) it’s the typical tool for which people seem to reach in practice.
One potential issue is with the interface of the function – as most of us don’t use SDEs we’re not particularly well suited to comment on the interface design. In particular should it return a single realization, an ensemble of realizations, moments of the realizations, etc, and then in what format should those outputs be given. I think it would help to have some documentation that more carefully defines the output of the function along with the proposed implementation and a few hypothetical Stan programs using that implementation to help us get a handle on the appropriateness of the interface. Comments from users would also be super helpful.
Finally the biggest concern is the differentiability of the function, which is somewhat related to the interface question above. In general a single realization of a stochastic process won’t be differentiable with respect to the input parameters, hence an ensemble of realizations won’t be either. Conditioning on the stochasticity (i.e. passing in precomputed PRNGs) offers one workaround, as the conditional paths might then be smoothly related to the input parameters. As would returning moments when the discontinuous behavior averages out.
Ultimately because of lack of expertise on these topics within the typical dev community I think we would all benefit from a bit more doc in order to better understand the approach instead of having to deconvolve the method from the code itself. It would certainly help me verify the proposal.
Sounds like you are referring the pathwise sensitivies when calculating greeks in finance, and the rationale can be found, for example, http://www.columbia.edu/~mh2078/MonteCarlo/MCS_Greeks.pdf section 2.
Good point, I’ll write more doc on this.
NONMEN had SDE + Kalman filter for a while, though I haven’t used it.
There are numerous approaches and seemingly even more numerous notations/terminologies. At the end of the day whatever solution the solver returns just has to be locally smooth with respect to the parameters.
Can you point me to some other notations/terminologies? Calculating greeks is the only context/application that I’m familiar with. | 2,300 | 9,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.88813 |
https://www.codeproject.com/Articles/38514/The-beauty-of-fractals-A-simple-fractal-rendering?fid=1544381&df=90&mpp=10&noise=1&prof=True&sort=Position&view=Expanded&spc=None&fr=11 | 1,529,839,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00503.warc.gz | 769,159,824 | 29,391 | 13,597,197 members
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Posted 26 Jul 2009
Licenced CPOL
# The beauty of fractals - A simple fractal rendering program done in C#
, 27 Jul 2009
A fractal rendering application demonstrating many .NET programming techniques.
## Introduction
I have written many fractal applications in my life. Starting on the C64, on the Amiga, on the PC, using different languages (even on a TI calculator in BASIC). Rendering fractals is a good practice when learning programming. Last week, I spent some time writing a simple fractal engine in C#. The purpose was to pack many of the programming techniques I was using during the last years into one application. In this application, you will find many interesting topics covered like:
• General C# programming
• MDI applications done in .NET
• Usage of interface classes in C#
• Use of delegate functions
• Asynchronous calculations
• Making use of multiple CPU cores
• Bilinear filtering of bitmaps
• Complex number class done in C#
• Mathematic background of Mandelbrot, Julia, and Newton fractals
• and many more ...
## Background
I have been fascinated by fractals ever since I was a kid. I can remember the first Mandelbrot engines on the commodore 64. Again and again, I am fascinated by the beauty of the outcome of such simple algorithms:
A Newton fractal
The Mandelbrot fractal from a bird's eye view
The Julia fractal
A Mandelbrot fractal without and with 'exact' rendering
Exact rendering means that the number of iteration steps is not interpreted as `int` but as `double`, and the color used to represent the result is interpolated between the previous and the next color in the palette. Using exact rendering does not need much more CPU power, but greatly improves the quality of the outcome.
Somewhere inside the Mandelbrot fractal
## The code
I don't want to put any code snippets here. Just open the solution and have a look into it. I think the code is very simple and clear, everything should be self-explanatory. I think it will be a good tutorial for beginners going through the code. You will find many things here that took me a long time to learn ;-)
There is one interface class `IFractal` from which a base fractal class will be derived. From this base fractal class, the Mandelbrot, Julia, and Newton fractals are derived.
You have an MDI container and a general MDI child form that can handle displaying dragging and zooming of fractals. You can drag a rectangle with the left mouse button pressed to zoom into a fractal. You can move the mouse with the right button pressed to scroll the fractal around. You can move the mouse with the middle parameter pressed, to change the control point used for the Julia fractal.
Fractals can be stored to and loaded from XML files. Fractal renderings can be stored as PNG files.
The engine renders at decent speed. It detects the number of available CPU cores and creates one rendering thread for each. The time needed to render a fractal nearly halves when going from a single core to a dual core machine. On my dual core Notebook, rendering a bird's eye Mandelbrot fractal with 2000x2000 pixels and 256 iterations per pixel takes approximately 650 ms, which is an acceptable speed.
This program is far from being finished, and probably will never be, due to lack of time. But, it is already capable of doing very beautiful fractal renderings. And, it shows how simple things can be done in C#. I remember doing the same in C++ some years ago, and it was a lot more work.
## Next steps ...
There are some things that I want to implement in the near future, for the next release (I hope I will find the time):
• A screensaver generating random fractals.
• A fractal movie renderer.
• Handle dynamic palette sizes up to an unlimited amount of colors. Right now, the palette is fixed to 256 colors, though the palette form can display an arbitrary amount of colors.
If you have any good suggestions, just post them here.
## History
• 27.07.2009: Initial version.
## Share
Software Developer (Senior) Austria
I have started programming at the age of 13 on the commodore 64.
Ever since then I have been programming on many systems in many languages.
During the last 12 years I have been working as professional programmer in different companies and different areas.
Now I am working as freelancer programmer / consultant
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Re: Modern art? zimstep28-Jul-09 7:32 zimstep 28-Jul-09 7:32 | 1,597 | 5,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-26 | latest | en | 0.91565 |
https://dictionary.iucr.org/index.php?title=Bravais_arithmetic_class&oldid=4807 | 1,580,173,883,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00050.warc.gz | 412,700,667 | 6,163 | # Bravais arithmetic class
### From Online Dictionary of Crystallography
Classe de Bravais (Fr). Bravais-Klasse (Ge). Classe di Bravais (It). ブラベー類 (Ja). Clase de Bravais (Sp).
## Definition
An arithmetic crystal class with matrix group of lattices is called a Bravais arithmetic crystal class, or Bravais class for short.
Each lattice is associated with a Bravais class, and each matrix group of a Bravais class represents the point group of a lattice referred to an appropriate primitive basis.
There exist 5 Bravais classes in two dimensions:
• 2p
• 2mmp
• 2mmc
• 4mmp
• 6mmh
There exist 14 Bravais classes in three dimensions:
• ${\bar 1}$P
• 2/mP
• 2/mS
• mmmP
• mmmS
• mmmI
• mmmF
• 4/mmmP
• 4/mmmI
• ${\bar 3}$mR
• 6/mmmP
• m${\bar 3}$mP
• m${\bar 3}$mI
• m${\bar 3}$mF | 284 | 785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | longest | en | 0.457037 |
http://www.101computing.net/optical-illusions/ | 1,532,213,215,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00296.warc.gz | 391,146,017 | 11,013 | # Optical Illusions
Which of the above two purple circles is bigger?
#### Learning Objectives
In this challenge you will use Python Turtle to draw geometric shapes and optical illusions.
• drawLine() to draw a straight lines between two points,
• drawSquare() to draw a square of a given size and colour,
• drawCircle() to draw a circle of a given size and colour.
To position these shapes on the screen you will need to use x,y coordinates and pass these coordinates when using the three functions given above.
#### (X,Y) Coordinates
The canvas we are drawing on (using Python Turtle) is 400 pixels wide by 400 pixels high.
Look at the canvas below to understand how (x,y) coordinates work:
#### Check the code
Complete the code given to draw the following optical illusions:
Which of the above two purple circles is bigger?
Which of the above two lines is longer?
Which of the above two lines is longer?
Are both purple lines perfectly straight?
Are all the white dots really white?
Tagged with: , | 215 | 1,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.883406 |
https://exercism.org/tracks/unison/exercises/roman-numerals | 1,709,157,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00248.warc.gz | 237,046,899 | 10,476 | Tracks
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Unison
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Exercises
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Roman Numerals
# Roman Numerals
Easy
## Introduction
Today, most people in the world use Arabic numerals (0–9). But if you travelled back two thousand years, you'd find that most Europeans were using Roman numerals instead.
To write a Roman numeral we use the following Latin letters, each of which has a value:
M D C L X V I
1000 500 100 50 10 5 1
A Roman numeral is a sequence of these letters, and its value is the sum of the letters' values. For example, `XVIII` has the value 18 (`10 + 5 + 1 + 1 + 1 = 18`).
There's one rule that makes things trickier though, and that's that the same letter cannot be used more than three times in succession. That means that we can't express numbers such as 4 with the seemingly natural `IIII`. Instead, for those numbers, we use a subtraction method between two letters. So we think of `4` not as `1 + 1 + 1 + 1` but instead as `5 - 1`. And slightly confusingly to our modern thinking, we write the smaller number first. This applies only in the following cases: 4 (`IV`), 9 (`IX`), 40 (`XL`), 90 (`XC`), 400 (`CD`) and 900 (`CM`).
Order matters in Roman numerals! Letters (and the special compounds above) must be ordered by decreasing value from left to right.
Here are some examples:
`````` 105 => CV
---- => --
100 => C
+ 5 => V
``````
`````` 106 => CVI
---- => --
100 => C
+ 5 => V
+ 1 => I
``````
`````` 104 => CIV
---- => ---
100 => C
+ 4 => IV
``````
And a final more complex example:
`````` 1996 => MCMXCVI
----- => -------
1000 => M
+ 900 => CM
+ 90 => XC
+ 5 => V
+ 1 => I
``````
## Instructions
Your task is to convert a number from Arabic numerals to Roman numerals.
For this exercise, we are only concerned about traditional Roman numerals, in which the largest number is MMMCMXCIX (or 3,999).
Note
There are lots of different ways to convert between Arabic and Roman numerals. We recommend taking a naive approach first to familiarise yourself with the concept of Roman numerals and then search for more efficient methods.
Make sure to check out our Deep Dive video at the end to explore the different approaches you can take!
### Source
The Roman Numeral Kata
Edit via GitHub
### Ready to start Roman Numerals?
Sign up to Exercism to learn and master Unison with 40 exercises, and real human mentoring, all for free.
Deep Dive into Roman Numerals!
Explore a variety of different solutions to this tricky exercise, ranging from recursion to table-based lookups, one solution that makes use of mixed radix numbers, and some simplistic cheating by Common Lisp. | 703 | 2,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.898583 |
https://antitheology.wordpress.com/2011/12/07/concept-language-multiple-variables/ | 1,642,542,569,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00644.warc.gz | 156,844,413 | 21,283 | Concept Language: Multiple Variables
Phrases have many variables, and the values of these variables need to be able to have mutually interacting constraints. As a stepping stone towards this, let us define concepts over not only single natural numbers, but pairs, triples, etc., of natural numbers. These concepts will define sub-sets of N, N x N, N x N x N, etc.
An appropriate visualization is an n-dimensional figure demarcated into blocks, with some of these blocks being white (satisfies the concept) and others being black (does not satisfy the concept).
So let us consider a concept over variables x, y, z, …
As before, the two primitive concepts are Y and N.
Then for each of the variables we have a sequencing operator: x[C1, C2, …], y[C1, C2, …], z[C1, C2, …], … To make more sense of this, let us consider a two-dimensional concept with variables x and y. Then x[C1, C2, …] means “horizontal sequence,” and y[C1, C2, …] means “vertical sequence.” By combining horizontal and vertical sequences, we can make arbitrary two-dimensional patterns.
Then for each of the variables we have a repetition operator: x(C * n), y(C * n), z(C * n), … Only now n does not have to be a constant; it can be a complex expression referring to any variables other than the variable which the repetition operator is over. We will say that this complex expression is a function which can use the following syntax:
* Constants and variables.
* Operators: +, -, *, /, %.
* Conditionals: if (C1) { E1 } else if (C2) { E2 } … else { En }
The conditionals are composed of comparisons (=, <, <=) joined by boolean operators (and, or, xor, not).
If it’s less confusing, we can also factor out the determinants of the three variables. Then the concept has separate expressions for each variable. Those expressions consist of Y/N, sequences, and repetitions whose coefficients are functions over the variables.
Performing analysis on these concepts is a straightforward extension of the method given for one-dimensional concepts.
Performing synthesis to determine a value for one variable is straightforward if we already have fixed values for all of the other variables. If we don’t have such fixed values, then things are trickier.
One solution is to restrict the variables so that they are computed in a specific order. So if we have variables a, b, c, d, in computing a we can refer to b, c, d, in computing b to c, d, in computing c d to d, in computing d to nothing. Then performing synthesis is straightforward. This restriction is probably acceptable.
1. Concept Language Redux « antitheology | 591 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-05 | longest | en | 0.878951 |
https://studysoup.com/tsg/993165/modern-physics-for-scientists-and-engineers-4-edition-chapter-4-problem-30 | 1,675,571,310,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00004.warc.gz | 571,407,455 | 9,101 | Modern Physics For Scientists And Engineers - 4 Edition - Chapter 4 - Problem 30
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9781133103721
Light from a Nd: Yag laser with a wavelength of 397nm is incident upon a hydrogen atom
Modern Physics for Scientists and Engineers | 4th Edition
Problem 30
Light from a Nd: Yag laser with a wavelength of 397nm is incident upon a hydrogen atom in the n 2state at rest. What is the highest state to which hydrogencan be excited?
Accepted Solution
Step-by-Step Solution:
Step 1 of 3
Physics 222 2/05/19 Electric Current Currents: consider a conducting wire of crosssectional area A having n free charge carrying particles per unit volume with each particle having a charge q with particle moving at drift speed v d o Current is the motion of any charge, positive or negative, from one point to another. o Current direction is along (opposite to) the motion of the positive (negative) charges. o Current is defined to be the amount of charge that passes a given area in a given amount of time o Current has units of Ampere = 1 Coulomb/1 sec o When electrons move from smalldiameter to large
Chapter 4, Problem 30 is Solved
Step 2 of 3
Step 3 of 3
Unlock Textbook Solution | 322 | 1,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.871042 |
https://math.okstate.edu/people/lebl/osu4143-f15/ | 1,590,527,976,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00137.warc.gz | 452,745,352 | 3,360 | # Math 4143/5043 - Advanced Calculus I
http://math.okstate.edu/people/lebl/osu4143-f15/
Lecture: MWF 10:30-11:20AM, MSCS 422.
## Lecturer:
Jiří Lebl
Web: http://math.okstate.edu/people/lebl/
Office: MSCS 505
Office hours: Monday 4-5pm, Wednesday 11:30-12:30pm, Thursday 4-5pm, and by appointment at other times.
Office phone: (405) 744-7750
Email: lebl at okstate dot edu
## Text:
We will be using my book, Basic Analysis: Introduction to Real Analysis.
or you can get a printed copy on amazon.
The grading scheme is given below:
$\text{Grade} = 0.2 \times \text{(Homework)} + 0.2 \times \text{(Exam 1)} + 0.2 \times \text{(Exam 2)} + 0.4 \times \text{(Final Exam)}$
To account for (bad exam day, etc..) an alternative grade will be computed as follows
$\text{Grade} = 0.2 \times \text{(Homework)} + 0.1 \times \text{(Exam 1)} + 0.1 \times \text{(Exam 2)} + 0.58 \times \text{(Final Exam)}$
The higher of the two will be used for your grade. Notice that in the second scheme, the score does not sum to 100 percent. That is on purpose! You should count on the first scheme, the second scheme is only to account for things going terribly terribly wrong on one of your exams.
## Exams:
Exam 1: Friday, September 18th, (same time/room as class), 20% of your grade.
Exam 2: Friday, October 30th, (same time/room as class), 20% of your grade.
Final Exam: Friday, December 11th, 10:00am - 11:50am (same room as the class), 40% of your grade. (Comprehensive, think of the final exam as half exam 3 and half comprehensive final)
Exam Policies: No books, calculators or computers allowed on the exams or the final. One page (one sided) of handwritten notes allowed on the exams.
## Homework:
Assigned weekly (some weeks may be skipped).
Homework page (hw.html)
Worth 20%, spot checked (spot checked means: some spot(s) of each homework checked, and all will be collected). Part of the grade will be simply for turning the homework in. Lowest 2 homework grades dropped (so no late homeworks).
## Missed Work:
No makeup or late homework (two lowest are dropped anyhow), but feel free to turn homework in early if you you cannot for whatever reason turn it in on time. For exams, there will be reasonable accommodation if you have a valid and documented reason, and the documentation is provided in advance unless absolutely impossible. If you have a university approved (see the syllabus attachment) final conflict exam, you must tell me at least two weeks befre the final exam week, so so that we can figure out what to do. | 698 | 2,534 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-24 | latest | en | 0.866268 |
https://owlcation.com/stem/math/?page=2 | 1,582,043,117,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00524.warc.gz | 506,492,203 | 45,652 | # Math
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### What Is Calculus? A Beginner's Guide to Limits and Differentiation
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### What Is It About the Tower of Hanoi?
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### How to Analyze a Statistical Survey: Standard Deviation, Outliers, and More
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### Univariate and Multivariate Linear Regression
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### 8 ÷ 2(2 + 2) = 1 and Only 1. The Sad Legacy of Calculator Dependence.
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### Math Help: How to Calculate the Area of Circle and Get an Answer in Terms of Pi
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### 1ppm – Have You Ever Wondered What 1ppm Means?
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### How to create magic squares
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### Parabola Equations and Graphs, Directrix and Focus and How to Find Roots of Quadratic Equations
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### How to Do Division on the Abacus in Easy Steps
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4
working | 7,338 | 33,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-10 | latest | en | 0.864901 |
https://www.sacredgeometry.blog/legend-33/ | 1,721,748,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00579.warc.gz | 800,099,745 | 15,903 | # Legend 33
The point (0D) implies the pure potentiality of life itself:
The line (1D) implies the 33 vertebrae of the human spine:
A centred dodecahedral number (2D) is a centred figurate number that represents trust. The figure below has 33 balls and reminds us of the alchemical symbol for gold:
In geometry, the (3D) rhombic dodecahedron is a convex polyhedron with 12 congruent rhombic faces. The rhombic dodecahedron can be used to tessellate 3D space—it can be stacked to fill a space much like hexagons fill a plane. A rhombic dodecahedron’s centre-face-pyramids can be inverted to form a cube. It can also be seen as the 33 balls of faith:
In geometry, a hyperdodecahedron is composed of 120 dodecahedral cells with four meeting at each vertex. It is the 4D analogue of the regular dodecahedron, since just as a dodecahedron has 12 pentagonal facets, with three around each vertex, the dodecaplex has 120 dodecahedral facets, with three around each edge. The hyperdodecahedron numbers are 4D figurate numbers based on the 120-cell regular polytope. Starting at a(0) = 33 we find truth:
This hyperdodecahedron incorporates the geometries of every convex regular polytope in the first four dimensions. Conversely, it can also be understood by first working with each of its predecessors and the sequence of increasingly complex symmetries they exhibit. So the 120-cell is like a 4D Swiss Army Knife—it contains examples of every relationship among all the convex polytopes found in the first four dimensions.
The unit-radius 120-cell has edge length 1/φ2√2 ≈ 0.270 which can be seen in the animated ratios of our solar system. When working with a number that embodies the universe itself, 33 is a good place to start because it represents life across all dimensions.
You, yourself, as much as anybody in the entire universe, deserve your love and affection.Buddha | 479 | 1,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.931647 |
https://fenq.com/one-skill-a-day-binary-bias-to-the-left-is-binary-search-also-useful-in-distributed-systems/ | 1,656,771,387,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00314.warc.gz | 297,669,919 | 24,541 | # One skill a day: binary bias to the left, is binary search also useful in distributed systems?
I believe everyone knows binary search. In an ordered list, using binary search can quickly determine whether the target is in the list with O(logN) time complexity.
The code for binary search is very simple, and it only takes a few lines of code to use recursion:
` 1 2 3 4 5 6 7 8 9 10 11 12 13` ` def binary_search (sorted_list, target) : """ sorted_list is a monotonically increasing list """ if not sorted_list: return False mid = len(sorted_list) // 2 if target > sorted_list[mid]: return binary_search(sorted_list[mid + 1 :], target) elif target < sorted_list[mid]: return binary_search(sorted_list[:mid], target) else : return True`
The operation effect is shown in the following figure:
Python comes with a binary search module called `bisect` , which can also implement binary search, but its execution result is a bit different from the effect of our code above:
` 1 2 3 4 5 6 7 8` ` import bisect a = [ 41 , 46 , 67 , 74 , 75 , 76 , 80 , 86 , 92 , 100 ] index = bisect.bisect(a, 75 ) print(index) index = bisect.bisect(a, 82 ) print(index)`
The operation effect is shown in the following figure:
As you can see, `bisect.bisect()` returns an index. If the number to be searched is already in the list, then it returns the index of`最右边`target number in the list + 1. Taking the list `[41, 46, 67, 74, 75, 76, 80, 86, 92, 100]` for example, to search for `75` . Since the index of `75` in the original list is `4` . So return the`索引+1` which is 5. If `75` appears many times in the original list, such as `[41, 46, 67, 74, 75, 75, 76, 80, 86, 92, 100]` then`最右边`is returned That `75` corresponds to the index `+1` , which is `6` .
If the number you are looking for is not in the original list, then `bisect.bisect()` will return an index. When we insert the target number into the list at the corresponding index position, the list is still in order. For example `[41, 46, 67, 74, 75, 76, 80, 86, 92, 100]` , we look for `82` . It returns `7` . The position of index 7 in the original list is the number `86` We insert 82 into this position, and the original data is shifted one place backward in turn. At this time, the list is still in order.
The `bisect` module also has a function called `bisect.bisect_left()` . If the target number is in the original list, then the index corresponding to the leftmost number is returned. If it is not in the list, the returned index is still in order after being inserted into the target number, as shown in the following figure:
This function looks very simple, but you may not know it, it also has important uses in distributed systems.
Suppose now you have 10 single nodes of Redis for distributed caching. For some reason, you can’t do clustering. When you want to search for a piece of data, you must first make sure that the data is in Redis. If it is, read the data directly from Redis; if not, read it in the database first, and then cache it in Redis.
Because of the large amount of data, you cannot store the same data in 10 Redis nodes at the same time, so you need to design an algorithm to store different data in different Redis nodes.
When you want to query data, you can query which Redis the data (if in the cache) should be stored in according to this algorithm.
Students with a little experience in distributed system design will definitely think that this is simple, 10 Redis nodes number 0-9. Calculate the hash value for the `key` , this hash value is a 32-bit hexadecimal number, which can be converted into After decimal, find the remainder of 10, and what the remainder is, put it in the corresponding node.
In this way, as long as a new data comes, you only need to go to the Redis corresponding to the remainder to determine whether it has a cache or not.
But here comes the problem. If you start to use this method and there is already data in Redis, the number of your Redis nodes cannot be changed. Once you add or subtract 1 node, all the remainders are changed, and the Redis node found by the new data must be wrong. For example, the hash value of the key was originally divided by 10, and the remainder was 2. Now it is divided by 9, and the remainder is 1. You should have gone to Redis No. 2 to find the cache, but now you go to Redis No. 1 to find the cache, you must not find it.
How to solve this problem? Let’s demonstrate with a simple example. Suppose I now have a list: `[200, 250, 300, 400, 500, 530, 600]` . Each number represents a house at that price point. The unit is ten thousand. You want to buy a house, but the cheap house is too shabby and the good house is too expensive. So you only look for houses with a price equal to your expectations, or slightly higher than your expectations but with the smallest gap.
Suppose now your expectation is 2.5 million, and a house happens to sell for 2.5 million, so you can buy it.
Suppose now your expectation is 4.7 million, then your only option is a 5 million house.
It should be pretty well understood by now, so let’s add or subtract candidates:
1. The 5 million house was bought by someone else. The list becomes `[200, 250, 300, 400, 530, 600]` , so the only thing that works for you is a house of 5.3 million.
2. If the 2.5 million house is now bought, the list becomes `[200, 300, 400, 500, 530, 600]` . At this time, it has no effect on you, and the house suitable for you is still a 5 million house.
3. If now a 4.8 million house is added, the list becomes `[200, 250, 300, 400, 480, 500, 530, 600]` . Then the right house for you now becomes 4.8 million.
4. If a 2.4 million house is now added, the list becomes `[200, 240, 250, 300, 400, 500, 530, 600]` . At this time, it has no effect on you, and the house that is suitable for you is still 5 million.
In this scenario, we can just use `bisect.bisect_left()` ! Results as shown below:
When an alternate option is changed, only the houses near your target option are affected. Changes in houses smaller than your candidate and more expensive houses will have no effect on you.
Did you notice that this scenario is very similar to our distributed cache adding and removing Redis nodes. We used to have 10 Redis, and now we have added one and it has become 11. Then only part of the cache of one Redis will be migrated to this newly added Redis. The other 9 Redis caches do not need to be changed.
Similarly, when we delete a Redis node, the data in the deleted node only needs to be synchronized to another Redis node next to it. The other 8 Redis nodes do not need any modifications! (It can also be out of sync. Only a small number of keys will not find the data due to deleting this node, and the database will be re-read. 80% of the cache will not be affected in any way.)
This is the consistent Hash algorithm.
Let me briefly describe the implementation of this algorithm. First, we create 10 connection objects using `redis.Redis(不同redis的连接参数)` . Then create a map with each connection object and a Hash value, as shown in the following figure:
Then, we sort these 10 hash values and put them into a list. As shown below:
Now, there is a new cache query request, we calculate the hash value corresponding to the key, and then use `bisect.bisect_left()` to find the index of the hash value of the corresponding Redis node in the list. If the returned index is equal to the length of the list, then let the index equal to 0. After finding the index, get the Hash of the corresponding Redis node, and finally use this Hash to find the corresponding Redis node. The simplified code is as follows:
`
If you add or delete Redis nodes, you only need to update `node_map` and `cycle` . Only a small data migration occurs, and it does not affect the vast majority of caches. For example, I now delete the Hash corresponding to the`第1个Redis链接对象`: `fbef6b15be1abe9edc8f6aaac6a86357` from `node_map` and `cycle` . If you query again, you will find that the Redis node numbered 6 is still found.
Consistent Hash is widely used in distributed systems. But you may not think that its core principle is `bisect_left` in binary search.
Of course, the above is just a simplified algorithm. The complete algorithm of consistent hash also involves virtual nodes and algorithms to avoid data skew. If you are interested, I can also write an article to fully explain its algorithm implementation. | 2,153 | 8,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-27 | latest | en | 0.748416 |
http://web2.0calc.com/questions/angle-of-altitude-of-an-object | 1,513,122,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00016.warc.gz | 308,703,555 | 5,919 | +0
Angle of Altitude of an object.
0
134
5
There are people who think the earth is flat and that the sun orbits the earth at 3000 miles. If the sun is over the equator and I'm 2000 miles, at what angle would I see the sun?
The adjacent would be 2000, the opposite 3000. so it would be, o/a, then use the inverse of tan. Is this correct?
If so, I can use this procedure for any distance value for the adjacent?
Guest Jun 13, 2017
edited by Guest Jun 13, 2017
edited by Guest Jun 13, 2017
edited by Guest Jun 13, 2017
edited by Guest Jun 13, 2017
edited by Guest Jun 15, 2017
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#1
+79784
+1
Actually, because of the Earth's curvature - and other factors - the procedure is a little more complicated.
http://www.pveducation.org/pvcdrom/properties-of-sunlight/elevation-angle
CPhill Jun 13, 2017
#2
+2
But the premise is that the earth is flat. Could u respond from that angle, pun intented...:)
Guest Jun 13, 2017
#3
+79784
+2
HAHAHA!!.....yeah...if you're amongst the "Flat Earthers"....then your procedure is "valid"....
CPhill Jun 13, 2017
#4
+1
No, flat earth theory makes no sense...:)...and thank you! I think one of the things that makes no sense is one person in particular thinks that a sun orbiting over a flat earth at 3000 miles, fits the current path of the sun with a sunrise and sunset and all.
Guest Jun 13, 2017
edited by Guest Jun 13, 2017
#5
+1
To finish my equation:
3000/2000 = 1.5, using the inverse of tan = 56.3°
Guest Jun 13, 2017
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We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 528 | 1,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-51 | latest | en | 0.889138 |
https://gateoverflow.in/15991/set-all-bit-strings-with-even-number-regular-expression-will?show=216868 | 1,547,651,134,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00031.warc.gz | 506,702,954 | 20,871 | +1 vote
1.5k views
Options are
a) (0*10*1)*
b) 0*(10*10*)
c) 0*(10*1)*0*
d) 0*(10*1)*10*
I have doubt in b and c.which one is correct and why.?
IT would be option c
Answer is C, Check for string 110 neither a) nor b) can express it, and since d) is clearly expressing strings containing odd number of 0's so option c is correct. Moreover, option b) is expressing strings containing only two 1's, in case you have not done a typo. ;)
0
in c check for 110011
0
It can be represented.
0
how?
0
Use the RE once to generate 110 and then use it to generate 011, & concatenate them.
0
you are using RE 2 times .
after generating 110 it will over but if the RE will like this (0*(10*1)*0*)* then you can do it .
0
Yes, you are right, I am sorry,I did not cared that we can only use one time.
0
so none of the options are matchin
0
Yeah! None of the options are matching for sure. Possibly there is some mistake in the question.
ans is None of the options are matching .
For A check with 110
For B and C check with 110011
for D this is pretty much obvious that it can support odd length also .
0
You are right..and if b) were 0*(10*10*)*..then it would be answer.
clearly none of the answers actually match .try 110011 as already mentioned. .. I dont think ambiguous questions will be given in GATE.
Option b and d is eliminated because it doesn't produce epsilon . Epsilon has also even no of 1's.
Option a can't produce 110. so a is also eliminated.
So Ans is option C
Option A will accept either epsilon or a string that is ending with 1 only,It may happen that string have even no. of 1's but it is ending with 0,So it won't be right answer.
Option B will generate string having only 2 1's,so it won't be the right answer because we are looking for that solution that may generate even strings having even no. of 1's.
Option C will generate string having even no. of 1's so it will be the correct answer.
Option D will generate string having odd no. of 1's so definitely it is not going to be the right answer.
So,according to me C will be the right Regex for generating string having even no. of 1's.
Correct me if I'm wrong.
1
2
+1 vote | 597 | 2,158 | {"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} | 2.921875 | 3 | CC-MAIN-2019-04 | latest | en | 0.940486 |
https://www.semanticscholar.org/paper/On-searching-for-solutions-of-the-Diophantine-x3-%2B-Koyama/de1272a60d25b10be260581e4dbe5f9026114d9a | 1,652,990,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00179.warc.gz | 1,165,430,007 | 65,537 | # On searching for solutions of the Diophantine equation x3 + y3 +2z3 = n
```@article{Koyama2000OnSF,
title={On searching for solutions of the Diophantine equation x3 + y3 +2z3 = n},
author={Kenji Koyama},
journal={Math. Comput.},
year={2000},
volume={69},
pages={1735-1742}
}```
• K. Koyama
• Published 2000
• Mathematics
• Math. Comput.
We propose an efficient search algorithm to solve the equation x3 + y3 + 2z3 = n for a fixed value of n > 0. By parametrizing |z|, this algorithm obtains |x| and |y| (if they exist) by solving a quadratic equation derived from divisors of 2|z|3±n. Thanks to the use of several efficient numbertheoretic sieves, the new algorithm is much faster on average than previous straightforward algorithms. We performed a computer search for six values of n below 1000 for which no solution had previously…
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On searching for solutions of the Diophantine equation x3 + y3 + z3 = n
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A new search algorithm to solve the equation x 3 + y 3 + z 3 = n for a fixed value of n > 0.5 is proposed, using several efficient number-theoretic sieves and much faster on average than previous straightforward algorithms.
Nonexistence conditions of a solution for the congruence x1k + ... + xsk = N (mod pn)
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We obtain nonexistence conditions of a solution for of the congruence x k 1 +... + x k s ≡ N (mod p n ), where k ≥ 2, s ≥ 2 and N are integers, and p n is a prime power. We give nonexistence
A Note on the Diophantine Equation: x n + y n + z n = 3
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In this note solutions for the Diophantine equation 3 v3 + 9 = 3 are sought along planes x + v + z = 3m, m E Z. This was done for Iml < 50000, and no new solutions were found.
Computational number theory at CWI in 1970--1994
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A concise survey of the research in Computational Number Theory, carried out at CWI in the period 1970 to 1994, with updates to the present state-of-the-art of the various subjects, if necessary.
Solutions of the Diophantine equation
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Unsolved problems in number theory
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• 1999 | 1,604 | 5,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.851378 |
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