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# WASSCE CHEMISTRY MAY/JUNE 2015 (SSCE WAEC MAY/JUNE 2015 CHEMISTRY)

WASSCE CHEMISTRY MAY/JUNE  2015 (SSCE WAEC MAY/JUNE 2015 CHEMISTRY)

1.The negatively charged particle in an atom is the

A. Electron

B. Neutron

C. Positron

D. Proton

2. How many orbits are contained in an atom with atomic number 13?

A. 7

B. 6

C. 5

D. 3

3. The Van der waals' forces are dominant intermolecular forces in_______?

A. Ammonium chloride

B. Chlorine

C. Sodium chloride

D. Water

4. The compound formed between 14X and 16Y is _______?

A. XY

B. XY2

C. X2Y

D. X4Y6

5

5. The shape of a molecule of water is _______?

A. Non-linear

B. Octahedral

C. Pyramidal

D. Tetrahedral

6. A metallic ion Z2+ with an inert gas structure contains 18 electrons. How many protons are contained in this ion?

A. 20

B. 18

C. 16

D. 2

7. Which of the following pairs of compounds belongs to the same homologous series?

A. C3H8 and C3H6

B. C4H10 and C5H10

C. C2H4 and C4H10

D. C2H6 and C4H10

8. Which of the following pairs of elements forms amphoteric oxides?

A. Be and Mg

B. Na and K

C. B and Al

D. Si and Pb

9. The following transition metal ions would be coloured in aqueous solution except________?

A. Cr3+

B. Fe3+

C. Mn3+

D. Sc3+

10. The gas given off when ethanol reacts with sodium is

A. Carbon(IV)oxide

B. Hydrogen

C. Methane

D. Oxygen

000000000

11. Which of the following halogens is solid at room temperature?

A. Bromine

B. Chlorine

C. Flourine

D. Iodine

12. The alkaline earth metals have similar chemical properties because________?

A. They are in the same period

B. Their salts are colourless

C. They have the same number of valence electrons

D. They are very reactive

13. The number of unpaired electrons in an atom of an element 8Q is _______?

A. 2

B. 4

C. 6

D. 8

14

14. The bond formed when ammonia reacts with hydrogen ion to form ammonium ion is

A. Covalent

B. Dative

C. Hydrogen bond

D. Ionic

15. To which group and period respectively does an element with 15 electrons belong?

A. 3 and 3

B. 3 and 5

C. 5 and 3

D. 5 and 5

16. The shape of a graphite crystal is

A. Tetrahedral

B. Pyramidal

C. Octahedral

D. Hexagonal

17. Which of the following oxides is ionic?

A. P4O10

B. MgO

C. Al2O3

D. SO2

18. Which of the following substances when boiled with aqueous solution of sodium hydroxide would be hydrolyzed?
i. Protein

ii. Fat

iii. Polythene

A. i

B. ii

C. i and ii

D. ii and iii

19. Which of the following trioxonitrate(V) salts would decompose on heating to form a metal?

A. Cu(NO3)2

B. AgNO3

C. Pb(NO3)2

D. KNO3

20. Consider the following reaction equation:
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)
The volume of CO2(g) produced at s.t.p when 0.05 moles of C2H4(g) was burnt in O2(g) is
[Molar Volume of gas = 22.4dm3]

A. 1.12dm3

B. 2.24dm3

C. 3.72dm3

D. 4.48dm3

1

21. The number of oxygen molecules present in 16.0g of the gas is ______?
[Avogadro's number = 6.02 × 1023]

A. 6.02 × 1022

B. 6.02 × 1023

C. 3.01 × 1023

D. 1.51 × 1023

22. Consider the following reaction equation:
SO42-(aq) + 2H+(aq) + ye- →SO42-(aq) + H2O(1)
The value of y in the equation is

A. 2

B. 3

C. 4

D. 6

23. The general gas equation was derived from______?

A. Boyle's and Gay Lussac's laws

B. Boyle's and Graham's laws

C. Boyle's and Charles' laws

D. Dalton's atomic theory

24. The vapour pressure of a liquid depends on:
I. Temperature

II. rate of condensation

III. Cohesive forces holding the particles together.

A. I

B. I and II

C. I and III

D. II and III

25. Which of the following gases will diffuse most rapidly?
[H = 1.00, C = 12.0, O = 16.0, S = 32.0, Cl = 35.5]

A. Cl2

B. SO2

C. CH4

D. C2H6

26. When a reaction is endothermic

A. Enthalpy change, ΔΔH is negative

B. Heat content of a product is less than the heat content of a reactant

C. Heat content of reactants is less than the heat content of product

D. The reaction is non-spontaneous

27. Which of the following statements about intermolecular distances and cohesive forces between gas is correct? They are

A. Both large

B. Both negligible

C. Constant and negligible

D. Large and negligible

28

28.

The following diagram illustrates the rate curve that was obtained when Mg reacted with excess dilute HCl.
The diagram became horizontal at X because

A. The reaction was slowed down

B. All the dilute HCl has reacted

C. All the Mg has reacted

D. Hydrogen gas is produced at a steady rate

29. An example of an acid salt is

A. CH3COOONa

B. Mg(OH)Cl

C. NaHSO4

D. (NH4)2SO4

30. Which of the following oxides can be reduced by hydrogen?

A. Aluminium oxide

B. Magnesium oxide

C. Sodium oxide

D. Silver oxide

31. Solubility is practically applied in

A. Fractional distillation

B. The determination of pH

C. The determination of saturation in hydrocarbons

D. Solvent extraction

32. Which of the following compounds is the least soluble in water?

A. CaCl2

B. CaSO4

C. NaCl

D. Na2SO4

33

33. A substance which dissolves readily in organic solvent would

A. Be a covalent compound

B. Have strong electrostatic forces of attraction

C. have a high melting point

D. Conduct electricity in molten state

34. Consider the following equilibrium system:
2SO2(g) + O2(g) ⇌2SO3(g)
The addition of more O2(g) to the system will shift the equilibrium position to the

A. Right leading to the production of more SO3(g)

B. Right leading to the production of more SO2(g)

C. Left leading to the production of more SO2(g)

D. Left leading to the production of more SO3(g)

35. A change in the temperature of a saturated solution disturbs the equilibrium between the

A. Undissolved solute and the solvent

B. Dissolved solute and the solvent

C. Dissolved solute and the undissolved solute

D. Dissolved solute and the solution

36. Which of the following statements about an electrochemical cell is correct? Oxidation occurs

A. At the anode

B. At the cathode

C. Through the salt bridge

D. In the aqueous solution

37. Consider the following reaction equation:
CuO(s) + H2(g) $\rightleftharpoons$ Cu(s) + H2O(1)
Which substance is oxidized?

A. Cu

B. CuO

C. H2

D. H2O

38. Which of the following metals is the strongest reducing agent?

A. Sodium

B. Silver

C. Potassium

D. Copper

38

39. The complete hydrogenation of benzene gives

A. Cyclohexene

B. Cyclohexane

C. Hexene

D. Hexane

40. A compound has an empirical formular CH2O and molecular mass of 90.
[H = 1.00, C = 12.0, O = 16.0]

A. C4H10O2

B. C3H10O2

C. C3H6O3

D. C2H2O4

40

41. Which of the following reactions would take place when concentrated sodium hydroxide solution is added to palm oil?

A. Esterification

B. Neutralization

C. Polymerization

D. Saponification

42. Starch could be converted to glucose by the process of

A. Condensation

B. Dehydration

C. Fermentation

D. Hydrolysis

43. Which of the following compounds is a secondary alkanol?

A. Ethanol

B. 2-methylbutan-2-ol

C. 3-methylpentan-2-ol

D. Propan-1-ol

44. Which of the following substances is a heavy chemical?

A. Ammonia

B. Barium Hydroxide

C. Hydrochloric acid

D. Tetraoxosulphate(VI)acid

45. Which of the following process does not take place in domestic water treatment?

A. Chlorination

B. Flocculation

C. Neutralization

D. Sedimentation

46. A substance responsible for the sour taste of unripe orange is

A. Alkene

B. Alkanol

C. Alkanoic acid

D. Alkanoate

47. Which of the following products of biotechnology can be used as a fuel in place of petrol?

A. Butane

B. Ethanol

C. Ethene

D. Propanol

48. Which of the following polymers in thermosetting?

A. Bakelite

B. Nylon

C. Polypropene

D. Polystyrene

49. The correct balanced equation for the reaction between aluminum metal and hot concentrated tetraoxosulphate(VI)acid is ?

A. 2AI(s) + 6H2SO4(1) → AI2(SO4)3(aq) + 6H2O(1) + 3SO2(g)

B. 2AI(s) + 3H2SO4(1) → AI2(SO4)3(aq) + 6H2O(1) + 3SO2(g)

C. 2AI(s) + 4H2SO4(1) → AI2(SO4)3(aq) + 8H2O(1) + 3SO2(g)

D. 2AI(s) + 5H2SO4(1) → AI2(SO4)3(aq) + 8H2O(1) + 3SO2(g)

50. Which of the following gases is monoatomic?

A. Argon

B. Chlorine

C. Nitrogen

D. Oxygen

WAEC SUBJECTS:
Year of WASSCE EXAM:


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# circle trig definitions

Circle Trig Definitions

A set of definitions of the six trig functions: sine, cosine, tangent, cosecant, secant, and cotangent.

format: plain


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# Synchrotron Polarization

## Cyclotron Polarization

Recall for cyclotron radiation, the emitted power pattern as a function of angle looks like a torus centered around the acceleration vector (right), that is if the center hole’s size is infinitely small. This behavior is due to the ${\displaystyle \sin(\theta )^{2}}$ term in the Larmor Formula. The polarization of Larmor (or cyclotron) radiation is linear and directed along the “kink” in the electric field created by the acceleration of the charge. For Larmor radiation, the polarization is therefore parallel to the acceleration vector (i.e. along the red line in the figure to the right). One way to remember this is that polarization of Larmor radiation is linear and follows a “donut cut” along the emitted torus-shaped power profile (red line), as opposed to a bagel cut (blue line).

Now imagine one non-relativistic electron spiralling around a magnetic field. If we were to look at this electron along the magnetic field direction we would see the electron carve out a circular path around the magnetic field lines (i.e. face-on). Now try to picture the behavior of the donut-shaped power pattern as the electron spirals around the magnetic field. If the electron’s orbit radius is small enough, it’s almost as if we took the donut and spun it on its side, like a nickel on a table top. If the red-line in the figure above represents the linear polarization of Larmor radiation, then we can see that Cyclotron radiation viewed face-on is circularly polarized. Remember that observing face-on refers to being not face-on with respect to the donut, but being face-on with respect to the circular path the electron traces out in its trajectory (which are perpendicular to each other!). This is similar to how a spinning nickel viewed from above has an observed edge that rotates over time.

Now image we view the same non-relativistic electron spiralling around a magnetic field but view it from the side, such that it traces out a trajectory that resembles a dipole (edge-on). You can picture the Larmor radiation pattern and see that the observed polarization is always the same throughout one cycle of the electron: it has linear polarization that is constant over time. Another way to state this is that the polarization of Larmor radiation is linear along the direction of acceleration, and because the projected acceleration vector always lies along a single line for cyclotron emission viewed edge-on, the polarization is also linear and constant over time.

If we don’t view the electron directly edge-on we get elliptical polarization that depends on the inclination angle of our observation.

## Synchrotron Polarization

Synchrotron emission due to highly relativistic electrons focuses the donut-shaped emission pattern into highly collimated beams. While we typically work with diagrams that show the beaming to occur in the plane of the orbit, in practice highly relativistic electrons also have large velocities along the magnetic field lines, which means their beamed radiation is viewed neither directly face-on nor edge-on (see Figure 6.5 in Rybicki and Lightman). This means that synchrotron polarization, for a single electron, is generally elliptically polarized. The handed-ness of the polarization, however, depends on whether or not us as observers lie directly above or below the center-line of the focused beam: the point of "maximal radiation" (again Figure 6.5). Because synchrotron polarization is highly collimated, for any reasonable distribution of particles (that also have a distribution of pitch angles) we can assume that we will view synchrotron radiation above for the center-line for some electrons and below it for other electrons, meaning we will be seeing both left-handed and right-handed polarized emission at the same time. These work to cancel themselves out to form mostly linearly polarized emission. For an ensemble of particles following a power-law distribution in energy to the ${\displaystyle -p}$ degree, the intensity of radiation that is linearly polarized can be shown to be about 70 percent of the total intensity: ${\displaystyle \Pi =(p+1)/(p+7/3)}$. Knowing that in practice ${\displaystyle p}$ ranges from 2 to 3, we get back that the degree of polarization is roughly 70 percent (Rybicki and Lightman pg 181). To derive this relationship yourself, see problem 6.5 in Rybicki and Lightman.


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Issue No. 01 - January-March (2009 vol. 6)
ISSN: 1545-5963
pp: 46-61
Mercè Llabrés , University of the Balearic Islands, Palma de Mallorca
Francesc Rosselló , University of the Balearic Islands, Palma de Mallorca
Gabriel Valiente , Technical University of Catalonia, Barcelona
Gabriel Cardona , University of the Balearic Islands, Palma de Mallorca
ABSTRACT
The assessment of phylogenetic network reconstruction methods requires the ability to compare phylogenetic networks. This is the first in a series of papers devoted to the analysis and comparison of metrics for tree-child time consistent phylogenetic networks on the same set of taxa. In this paper, we study three metrics that have already been introduced in the literature: the Robinson-Foulds distance, the tripartitions distance and the $\mu$-distance. They generalize to networks the classical Robinson-Foulds or partition distance for phylogenetic trees. We analyze the behavior of these metrics by studying their least and largest values and when they achieve them. As a by-product of this study, we obtain tight bounds on the size of a tree-child time consistent phylogenetic network.
INDEX TERMS
Biology and genetics, Graph algorithms, Trees
CITATION
Mercè Llabrés, Francesc Rosselló, Gabriel Valiente, Gabriel Cardona, "Metrics for Phylogenetic Networks I: Generalizations of the Robinson-Foulds Metric", IEEE/ACM Transactions on Computational Biology and Bioinformatics, vol. 6, no. , pp. 46-61, January-March 2009, doi:10.1109/TCBB.2008.70


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# How to include paragraph break in command argument?

Here is my MWE:

\documentclass{article}

\setlength{\parindent}{0pt}
\setlength{\parskip}{3pt}

\begin{document}

Here is the first paragraph.

Here is the second.

\newcommand*{\testing}[1]{#1}

\testing{\par Here is the third.\par Here is the fourth.}

\end{document}

The output looks correct in this case:

However, I have a compiler error:

Runaway argument?

! Paragraph ended before \testing was complete

I've tried using \\\\ instead of \par, but it creates incorrect space between the paragraphs.

How can I define multiple paragraphs in a command argument and have it come out formatted just as though I had typed those paragraphs in directly?

• Have you tried with \newcommand{\testing}[1]{#1} (no star)? – Mensch Jun 14 at 2:57
• @Kurt Oh, damn - I forgot to try that. And indeed it works. Thanks! – me-- Jun 14 at 3:01
• I have to wonder, what did you want the * to do? – David Carlisle Jun 14 at 6:59
• If a macro is not defined in terms of \long, the token \par is not allowed inside that macro's arguments—this is regardless \par's meaning/regardless \par's definition. \newcommand*, i.e., the starred variant, tells LaTeX explicitly to define the command not in terms of \long. As the \long-mechanism is only about the token \par itself and not about meanings of whatsoever tokens you can do: \testing{\csname par\endcsname Here is the third.\csname par\endcsname Here is the fourth.} or \let\parcopy=\par \testing{\parcopy Here is the third.\parcopy Here is the fourth.} – Ulrich Diez Jun 14 at 10:25

Command \newcommand*{\testing}[1]{#1} (the starred version!) does simply not allow an \par in code #1.

If you need \par use command \newcommand{\testing}[1]{#1} (no star) instead, it is defined to allow \pars ...

See the corrected code

\documentclass{article}

\setlength{\parindent}{0pt}
\setlength{\parskip}{3pt}

\begin{document}

Here is the first paragraph.

Here is the second.

\newcommand{\testing}[1]{#1} % <======================================== no star

\testing{\par Here is the third.\par Here is the fourth.}

\end{document}

and its with no errors compiled result:


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# Choosing the “Correct” Seed for Reproducible Research/Results

I have created this post to spark a discussion around an idea I had regarding the choice of the seed number used to replicate the results of a statistical model. Here is some background on how I came up with this idea and how it may be applied.

I have a data set, called data, which has 506 observations, 13 predictor variables, and 1 continuous response variable. I split the data into a training and test set by randomly sampling 380 indexes without replacement, storing the observations in the rows equal to the 380 indexes from the data dataframe into a dataframe called train, and storing the remaining 126 observations into a dataframe called test.

Once I created the training and test sets, I built a linear model using the glm function, predicted the outcomes of the test set using the predict function, and computed the MSE between the test set's predicted values and its actual values. In this instance, I obtained an MSE of 16.65786. I left the code here for the night as it was time to go home.

The R code for this is:

library(MASS)
data <- Boston

index <- sample(x = 1:nrow(data), size = round(nrow(data)*0.75), replace = FALSE)
train <- data[index, ]
test <- data[-index, ]

lm1.fit <- glm(medv ~ ., data = train)

pr.lm1 <- predict(lm1.fit, test)
(MSE.lm1 <- sum((pr.lm1 - test$medv)^2) / nrow(test))  The following day, I realized that I should have set a seed so that the random sample of indexes (rows) used to create the training and test sets are the same each time the code is run. So I chose a seed of 1 and reran the code above. This time I received an MSE of 28.72933, a huge difference from the MSE I obtained the night before. This sparked the question of what if the seed I chose was one that lead to an outlier case or that the unseeded result that I obtained was an outlier case. Maybe both results were outlier cases! So I did some investigating. I used a for loop 10,000 times, each time getting a new random sample of indices, created a new training and test set, built a linear model, predicted outcomes for the test set, and calculated the MSE. I stored each of the 10,000 MSEs in a dataframe and computed the mean and median MSE. I also looked at the distribution of the MSEs which appeared to be approximately normal. I reran the forloop again just to make sure that the mean and median of the MSEs did not change much, they didn't. The first 10,000 MSEs had a mean of 24.13 and a median of 23.51. The second 10,000 MSEs had a mean of 24.01 and a median of 23.48. The R code for this is: error <- data.frame(err = rep(0, 10000)) for(i in 1:10000) { index <- sample(x = 1:nrow(data), size = round(nrow(data)*0.75), replace = FALSE) train <- data[index, ] test <- data[-index, ] #Linear Model 1 - No Data Cleaning lm1.fit <- glm(medv ~ ., data = train) summary(lm1.fit) pr.lm1 <- predict(lm1.fit, test) error[i, 1] <- sum((pr.lm1 - test$medv)^2) / nrow(test)
}
summary(error[, 1])

The last thing I did was loop through this same model building process until I found a seed that lead to an MSE in between the mean and the median of the 20,000 MSEs.

The R code for this is:

for(i in 1:10000)
{
set.seed(i)
index <- sample(x = 1:nrow(data), size = round(nrow(data)*0.75),
replace = FALSE)
train <- data[index, ]
test <- data[-index, ]

#Linear Model 1 - No Data Cleaning
lm1.fit <- glm(medv ~ ., data = train)
summary(lm1.fit)

pr.lm1 <- predict(lm1.fit, test)
error <- sum((pr.lm1 - test$medv)^2) / nrow(test) if(error > 23 & error < 24){print(i)} }  I chose to use a seed of 47 which lead to an MSE of 23.63085. I believe that this method for obtaining a seed more accurately depicts a model's ability to accurately predict versus just randomly picking a seed or finding a seed which gives the best results (which I view as "cheating"). My question to all you statisticians out there is does this seem like a reasonable ideology / approach? Is there a more appropriate method out there? Is there a downside to picking a seed this way? Thank you everyone in advance for your input! [Scortchi] Thanks - I was trying to be pithy & conflated two points: (1) Specifically the OP's re-inventing Monte-Carlo cross-validation; & (2) the sensitivity of the out-of-sample MSE estimate to the seed used shows that they should be using some form of re-sampling validation rather than a train/test split. – Scortchi [Brandon] @Scortchi So what you're saying is I should simply use the following code instead: library(MASS) data <- Boston lm1.fit <- glm(medv ~ ., data = data) library(boot) set.seed(1) cvResult <- cv.glm(data = data, glmfit = lm1.fit, K = 10) cvResult$delta

Which leads to a cv prediction error of 23.38074, very close to the prediction error obtained using my long method above. Am I correct?

• Two objections come immediately to mind. First, when you select the seed based on the results, the simulation no longer is random. The approach appears to be self-defeating. Second, why keep only the 1/10000 part of the simulation results? What a waste of effort! Use them all. For another take on setting seeds see stats.stackexchange.com/questions/80407. – whuber Mar 21 '18 at 19:50
• @MichaelChernick: Good pseudo-random number generator with "convenient seeds" give "convenient results". The OP here is for example using the Mersenne Twister which is an excellent RNG for general-purpose Statistics. The quality of the RNG is not the issue here. – usεr11852 says Reinstate Monic Mar 21 '18 at 20:47
• I've known people to choose random seeds by rolling a 10-sided die six (or whatever) times, recording the digits sequentially, and using that at the seed. One might argue that because this procedure is uniform over all such die outcomes, it's preferable. – Reinstate Monica Mar 21 '18 at 21:12
• The moral ought to be not to use sample splitting on such small data-sets: use cross-validation - which you're close to re-inventing! – Scortchi - Reinstate Monica Mar 21 '18 at 22:04
• @Scortchi: cross-validation is great, but there's plenty of cases in which the expected difference between CV error is on the same scale as the standard deviation of CV error. In those cases "seed optimization" is still very viable, although giving your script with the seed declared at the top kinda kills that. – Cliff AB Mar 22 '18 at 0:12

The seed choice should not affect the ultimate result, otherwise you have an issue. Then why do we need a seed at all? The reason is mainly for debugging and trouble shooting.

What do I call an ultimate result? Suppose that you're analyzing a drug efficacy, and using Monte Carlo simulation to come up with some kind of a critical value. The seed shouldn't matter to your conclusion about whether the drug is efficient or not.

Consider this: what if you're using R and I'm suing Excel, how would I use your seed? I should be able to reproduce the ultimate result regardless of the differences in our software packages and platforms.

Of course, the seed will impact the exact value obtained for the critical value, e.g. 3.1278765876 instead of 3.128765987 with another seed, but this is not the ultimate result of your study. Suppose the test statistic is 20.5, then the difference in critical values between two seeds is not material. Actually, you should not even report the critical value beyond the third digit, just show 3.13, unless it's for debugging purposes.

However, if somehow the difference in fourth significant digit makes it or breaks it for this drug, then we have an issue. It means that we can't use this critical value, OR that we need to increase the number of simulations to nail the fourth digit and reduce the variation between seeds to a much smaller number.

Therefore, you need to set other parameters of simulation in such a way that seed wouldn't matter for the conclusion of your study. Do not "optimize" the seed in any way.

In my practice I had funny projects, where the client would insist on reporting all the calculated digits, e.g. \$31,456,890.01. The simulations would yield maybe 1% precision, hence the number would be changing in second or third digit between different seeds. Moreover, the model itself had accuracy around 10% at best, so to a physicist I would have reported 3e7. However, accountants would complain that the number looks "rounded" which drives auditors nuts: it looks to them as if someone's "cooking the books." We ended up reporting the numbers to the cents, and storing the seeds so that the numbers would be reproducible upon audit requests.

• "The seed choice should not affect the ultimate result, otherwise you have an issue." For something like MCMC, this should be a reasonable assumption. For looking at something like cross-validated MSE for competing state-of-art prediction algorithms, probably not. – Cliff AB Mar 21 '18 at 23:39
• @CliffAB, so will you choose different algorithms at different seeds? – Aksakal Mar 21 '18 at 23:42
• @CliffAB, if it's true, it's horrible. – Aksakal Mar 21 '18 at 23:59
• @user1993951: I think there's some major wires getting crossed at this point. I'm not saying you should use seed optimization or anything like that. What I'm saying is that occasionally, decisions will be made with something like cross-validation, choosing between several estimators with very similar predictive power. Then it's very possible that your final outcome will be different based on what seed you used. If this completely changes your predictions, that's should be of concern... – Cliff AB Mar 23 '18 at 21:38
• ...but if the predictions themselves are very similar, then it's really not a big deal that some part of the process was completely different even though a different seed was used. – Cliff AB Mar 23 '18 at 21:39

You certainly can use "seed optimization" to be deceptive about performance. For example, suppose I'm comparing two estimators that just give back pure noise. Fifty percent of my seeds will say estimator A is better than B when using cross validation, so all I need to do is make sure I pick a good seed and then go publish my paper!

Does that mean the use of saved seeds is bad? Not really. The whole point of the seeds is that you've got a script that someone can use to completely reproduce the exact results you got. If you claim the cross validated MSE was 1.2, and a researcher says "hmm, I don't believe that...", just rerun the script right in front of them and they will see it's what you got.

Now, could you have seed optimized to get that 1.2 MSE? Absolutely. But in the exact same manner, they can take your same script, change the seed a few times, and make sure the results almost replicate! So you can use a seed to replicate exactly the results you saw...and then test how reliable that is by randomly using other seeds on the exact same script.


text:
# Which of the following is a product that has resulted or improved because of recombinant DNA...

###### Question:

Which of the following is a product that has resulted or improved because of recombinant DNA technology?
a.   genetically engineered crops
b.   specialized microorganisms
c.   pharmaceuticals such as insulin and human growth hormone
d.   genetically engineered animals
e.   all of the above

The disease Xeroderma pigmentosum results from DNA-repair defects. People with this disease are sensitive to UV light and have atypical skin pigmentation. Thus these people have higher incidences of what other disease?
a.   albinism
b.   skin cancer
d.   none of the above

Gene therapy involves transferring desired genes into human cells to treat disease. Which of the following is NOT true about this technology?
a.   It involves transfer of genes into reproductive (germline) cells so that the correction is heritable.
b.   Many clinical trials for treatment of a variety of human diseases are underway.
c.   The gene transfer procedures involved are technically difficult.
d.   Safety concerns include the possibility of immune reactions and development of leukemia.
e.   Gene therapy was first successfully implemented in 1990.

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text:
Delayed printf for real-time logging

You often debug by adding a few printfs and looking at the logs. In some real-time/low-level contexts though, you don't have time for text formatting.

You don't want prints to affect timing too much, because then timing-related bugs you're chasing might disappear. And you certainly don't want the system to stop functioning altogether because prints cause it to miss real-time deadlines.

A common alternative to prints is more "raw" logging - an event buffer, where event is a union keeping the event type, and a few variables depending on the type. This is fast, but not nearly as readable (or writable) as free-form printf messages.

"Delayed printf" is a way to get it all - printf's convenience together with event buffers' speed. What we do is we take printf's arguments and save them into a buffer like this:

format_string_pointer
num_args
arg0
arg1
...
format_string_pointer
num_args
...

Then, outside of the program logging these messages, we can read the actual format strings from the executable file, and call printf to get formatted text. As long as all the format strings and %s arguments are constant, this works fine - and is very fast.

I wrote a working example at GitHub - reading and writing log files takes just 100+ LOC. In what follows, we'll walk through the implementation.

How do we read strings from the executable file? We could parse the file - it's a bunch of data sections, that is, base addresses and the bytes to be copied to those addresses by the program loader. For instance, on Linux, #include <elf.h> has the structs describing program sections in ELF binaries.

But there's a simpler way to lay our hands on the strings - just ask the debugger. In gdb, "print (char*)0x8045008" will print the string starting at that address (or garbage if a string doesn't start there.) We'll use something similar, except we'll script gdb in Python:

charp = gdb.lookup_type('char').pointer()
return gdb.Value(ptrbits).cast(charp).string()

Ugly, ain't it? Well, we could use the shorter

gdb.parse_and_eval('(char*)0x%x'%ptrbits).string()

...but parse_and_eval is dog-slow - slow enough to make the longer incantations well worth it.

Armed with read_string, we can take a raw log buffer - a list of words (32-bit in my examples) - and convert it to text:

def print_messages(words):
i = 0
while i<len(words):
n = words[i+1]
args = words[i+2:i+2+n]
print fmt % convert_args(fmt, args),
i += n+2

This simply reads a format string pointer, a number of arguments, and then that number of arguments until it runs out of words.

Why do we need convert_args? Because args is a bunch of integers, which is fine with %d or %x but not with %s or %f.

With %s, the integer is really a string pointer, and so we want to read_string. And with %f, we want to reinterpret the word bits as a floating point number - what C does with *(float*)&int_var.

def convert_args(fmt, args):
types = [s[0] for s in fmt.split('%')[1:]]
unpack = {'s':read_string, 'd':int, 'x':int, 'f':to_float}
return tuple([unpack[t](a) for t,a in zip(types,args)])

def to_float(floatbits):
return struct.unpack('f',struct.pack('I',floatbits))[0]

This is a tad naive - it doesn't support things like %.2f - but you get the idea.

And now we can add a user-defined gdb command called dprintf like so:

class dprintf(gdb.Command):
def __init__(self):
gdb.Command.__init__(self, 'dprintf', gdb.COMMAND_DATA)

def invoke(self, arg, from_tty):
def word(b): return struct.unpack('i',b)
words = [word(bytes[i:i+4]) for i in xrange(0, len(bytes), 4)]
print_messages(words)

dprintf()

We derive a class from gdb.Command so that __init__ registers a command named dprintf. When the command "dprintf mylog.raw" is entered, we open mylog.raw, read its words into a Python list, and call print_messages. We don't need to be in an interactive gdb session to decode log files - we can issue the command directly from the shell:

gdb -q prog -ex 'py execfile("dprintf.py")' -ex 'dprintf mylog.raw' -ex q

This can be made shorter by importing dprintf.py (the file with all the Python code above) somewhere in gdb's initialization scripts - then we won't need the execfile part.

Writing log files

To save printf arguments, we need a way to iterate over them. I figured C++11 variadic templates are faster than fiddling with C's va_list, which wants you to know the argument types to access them, which means you must read the format string, which I'd rather not. So here's the C++11 version:

typedef int32_t dcell;
struct dbuf { dcell *start, *curr, *finish; };
extern dbuf g_dbuf;

template<class... Rest>
void dprintf(const char* fmt, Rest... args) {
int n = sizeof...(args);
if(p + sz < g_dbuf.finish) {
*p++ = (dcell)fmt;
*p++ = n;
dwrite(p, args...);
}
}

I'm using a global buffer, g_dbuf - of course you could prefer passing a buffer as a dprintf's argument. But actually, a global buffer isn't that bad because dprintf updates the buffer atomically, with GNU's __sync_fetch_and_add. So it's OK to call dprintf from multiple threads and even from interrupt handlers (the latter you wouldn't do with a lock-protected buffer, because an interrupt meeting a buffer locked by a suspended thread is game over. But it's fine with a lock-free buffer - the point of lock-freedom being that someone busy with the buffer can never block someone else by getting suspended).

As to sizeof...(args) - in case you've never seen that, it's the number of a variadic template's arguments, and has nothing to do with their sizes in bytes.

dprintf uses dwrite to write the arguments into a buffer:

inline void dwrite(dcell* p) {}

template<class T, class... Rest>
inline void dwrite(dcell* p, T first, Rest... rest) {
*p++ = tocell(first);
dwrite(p, rest...);
}

This is actually a perfectly prosaic recursively self-instantiating variadic template - "the" way to iterate over C++ variadic template function arguments (is there any other way?)

In turn, tocell takes an argument of an arbitrary type and extracts the 32 bits. In my example, I overloaded it for ints, floats and char*, along the lines of:

inline dcell tocell(int x) { return x; }
inline dcell tocell(const char* p) { return (dcell)p; }
inline dcell tocell(float x) { union { int n; float f; } nf; nf.f = x; return nf.n; }

That's it; the only other thing we need is a dflush function writing the buffer out to a file, and we can use dprintf:

int i;
for(i=0; i<10; ++i) {
dprintf("i=%d i/2=%f %s\n", i, i/2., (i&1)?"odd":"even");
}
dflush();

dprintf("going to crash...\n");
volatile int* p=0;
dprintf("i=%d p=0x%x\n", i, (dcell)p);
p[i]=0;

What's with the "going to crash" part? Well, I think a really important part of logging is being able to look at not-yet-flushed messages in a debugger - either in a core dump or in a live debugging session. (Not being able to do this is an annoying thing about buffered printfs; of course you can work around this by sprintfing into a buffer of your own.)

So let's extend our dprintf gdb command to print messages directly from g_dbuf instead of reading from a file:

intpp = gdb.lookup_type('int').pointer().pointer()
def invoke(self, arg, from_tty):
if arg:
else: # no file passed
buf = gdb.parse_and_eval('&g_dbuf').cast(intpp)      start = buf[0]      curr = buf[1]
n = curr - start
words = [int(start[i]) for i in xrange(n)]
print_messages(words)

Here we do an ugly thing - we treat &g_dbuf as an int** instead of a dbuf*. Why do that - why not avoid any casts and simply access start and curr with buf['start'] and buf['curr']? Because that only works if we have DWARF debug info - that is, if we compiled with -g. But the uglier way above works even without -g. The debugger only needs to know where g_dbuf is, but nothing about its type. And where g_dbuf is debuggers know from the ELF symbol table - which is always there unless someone actively stripped the executable.

And now we can see the last messages in a core dump like so:

gdb -q progname core -ex dprintf -ex q

...and in a live debugging session, we'd simply type "dprintf". With our example, this should print:

going to crash...i=10 p=0x0

Conclusions

• You can debug with free-form text prints in the lowest-level contexts - interrupt handlers, systems with tiny memories, etc. etc.
• If you use C++, it's good to upgrade to C++11
• If you use gdb, it's good to have a repository of user-defined Python commands
• A debugger can be very useful even with release builds

If your strings are logged from a shared object, then the logged string pointers will not match the addresses stored in the .so file - there'll be an offset. It's OK if you have a debugged process or a core dump, because gdb knows where the .so was loaded. The problem is when you parse files logged by a dead process. One workaround is to log the runtime address of some global variable with a known name, and use that to compute the offset relative to the address of that name in the .so.

Previous post by Yossi Kreinin:
How FPGAs work, and why you'll buy one

[ - ]
Comment by October 25, 2013
How would you handle delayed printf of C strings that happen to be on the stack?
The memory will be thrashed by the time you print it out later.
[ - ]
Comment by October 25, 2013
I wouldn't :) As the article says, "As long as all the format strings and %s arguments are constant, this works fine" - I could have said "literals" but that's not strictly true, and "constants" isn't strictly true, either; well, we both know what I mean... In my production C code, I'm trying to make as many strings (and other pointers as well) point to static locations precisely because these addresses are the only ones you can always make sense of when things go awry.
[ - ]
Comment by August 5, 2015
Really interesting article, thanks!

To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.


text:
: com net

# :

1. ## :

$Q61)9^{\left | x+3 \right |}=\sqrt{9}^{-2x+1}$

$Q62)2^{2-4x} \sqrt{9}^{6x-3}=1$

2. ## The Following User Says Thank You to For This Useful Post:

(12-02-2017)

4. ## The Following 3 Users Say Thank You to For This Useful Post:

(12-02-2017),  (11-30-2017),  (12-01-2017)

6. ## The Following 2 Users Say Thank You to Nuha Ahmed For This Useful Post:

(12-02-2017),  (12-01-2017)

7. ## :

[
URL="http://alnasiry.net/forums"][/URL]

8. ## The Following User Says Thank You to Nuha Ahmed For This Useful Post:

(12-01-2017)

9. ## :

$Q63)solve .in. [0,2\pi ]:log_{sinx}cosx=1$

10. ## The Following 2 Users Say Thank You to For This Useful Post:

(12-25-2017),  (01-05-2018)

12. ## The Following User Says Thank You to . For This Useful Post:

(01-05-2018)

14. ## The Following 2 Users Say Thank You to For This Useful Post:

(02-14-2018),  (12-26-2017)

15. ## :

ҡ
: -

$\begin{array}{l} 1.\,\,\,0 < \cos x < 1\,\,\,\, \Rightarrow {D_1}: - \frac{\pi }{2} < x < \frac{\pi }{2}\,\,\,\,\,\,\, \\ 2.\,\,\,0 < \sin x < 1\,\,\,\, \Rightarrow {D_2}:0 < x < \frac{\pi }{2}\,\,\,\,\,\,\,or\,\,\,\,\,\frac{\pi }{2} < x < \pi \\ {D_1} \cap \,{D_2} = \left\{ {x:0 < x < \frac{\pi }{2}} \right\} \\ \end{array}$

16. ## The Following 2 Users Say Thank You to For This Useful Post:

(02-14-2018),  (02-10-2018)

17. ## :

$Q64)(26+15\sqrt{3})^{x}-5(7+4\sqrt{3})^{x}+6(2+\sqrt{3})^{x}+(2-\sqrt{3})^{x}=5$

$Q65)x^{2}-\frac{1}{2}=\sqrt{x+\frac{1}{2}}$

18. ## The Following 2 Users Say Thank You to For This Useful Post:

(01-06-2018),  (01-05-2018)

20. ## The Following User Says Thank You to For This Useful Post:

(01-06-2018)

21. ## :

$Q66)x^{log_{7}4}+5(2^{log_{7}x})-4=0$

22. ## :

$Q67)\left \lfloor x+y+4 \right \rfloor=18-y$
$\left \lfloor x+1 \right \rfloor+\left \lfloor y-1 \right \rfloor=18-x-y$

: 1 (0 1 )


text:
Rescaling \check symbol

I would like to rescale \check symbol but not for the whole document. For example,

\documentclass{article}

\usepackage{mathtools}
\usepackage{asana-math}

\begin{document}

this is a normal \check symbol $\check{u}$, but in $\check{U}$
I want the check symbol to be larger

\end{document}

• How much larger would you like the symbol to be?
– Mico
Nov 16, 2016 at 19:28
• the same as \widehat Nov 16, 2016 at 19:31

The mathabx package provides a \widecheck macro.

Since I don't have the asana-math package on my system, the following example uses LuaLaTeX and the unicode-math package to specify Asana Math as the OpenType math font.

\documentclass{article}
\usepackage{mathabx} % for '\widecheck' macro
\usepackage{unicode-math}
\setmathfont[Scale=MatchLowercase]{Asana Math}
\begin{document}
$\check{u}$ $\widecheck{U}$
\end{document}

• the problem is that i need to define my own theorems and so i need to load \mathtools in order to use \newtheorem command. However, i cannot use both mathabx and mathtools packages Nov 16, 2016 at 19:49
• @Yorgos - Have you tried loading mathtools before mathabx?
– Mico
Nov 16, 2016 at 19:53
• yes.. i get an error that \circ is already defined Nov 16, 2016 at 19:55
• @Yorgos - The loading sequence ordering should be mathtools, mathabx, unicode-math -- or asana-math, I suppose, but that's a package I don't have and hence can't test. Is that's what you're doing?
– Mico
Nov 16, 2016 at 20:04
• standard warning that mathabx changes all symbols. to avoid that: Importing a single symbol from a different font Nov 16, 2016 at 21:37

Strangely enough, unicode-math doesn't define \widecheck; you can do it yourself.

\documentclass{article}

\usepackage{mathtools}
\usepackage{unicode-math}
\setmainfont{TeX Gyre Pagella}
\setmathfont{Asana Math}

\def\removefixed#1fixed #2\removefixed#3{%
\protected\def#3{#1 #2}%
}

\AtBeginDocument{\expandafter\removefixed\check\removefixed\widecheck}

\begin{document}

this is a normal \verb|\check| symbol $\check{u}$, but in $\widecheck{U}$
I want the check symbol to be larger

\end{document}

• that was my main concern. i didn't know how possible is to define your own sympbols. i thought that you hane to 'stuck' with its package's symbols Nov 17, 2016 at 9:29
• @Yorgos I think that unicode-math should provide it. Nov 17, 2016 at 9:30
• i couldn't agree more Nov 17, 2016 at 9:31
• @Yorgos Filed issue report: github.com/wspr/unicode-math/issues/372 Nov 17, 2016 at 12:03


text:
#### Vol. 4, No. 6, 2010

 Recent Issues
 The Journal Cover Editorial Board Editors' Addresses Editors' Interests About the Journal Scientific Advantages Submission Guidelines Submission Form Subscriptions Editorial Login Contacts Author Index To Appear ISSN: 1944-7833 (e-only) ISSN: 1937-0652 (print)
Parabolic induction and Hecke modules in characteristic $p$ for $p$-adic GL$_n$

### Rachel Ollivier

Vol. 4 (2010), No. 6, 701–742
##### Abstract

We classify the simple supersingular modules for the pro-$p$-Iwahori Hecke algebra $\mathsc{ℋ}$ of $p$-adic ${GL}_{n}$ by proving a conjecture by Vignéras about a mod $p$ numerical Langlands correspondence on the side of the Hecke modules. We define a process of induction for $\mathsc{ℋ}$-modules in characteristic $p$ that reflects the parabolic induction for representations of the $p$-adic general linear group and explore the semisimplification of the standard nonsupersingular $\mathsc{ℋ}$-modules in light of this process.

##### Keywords
mod $p$ representations of Hecke algebras and $p$-adic groups, parabolic induction, integral Bernstein presentation, integral Satake transform
##### Mathematical Subject Classification 2000
Primary: 20C08
Secondary: 20G05, 22E50


text:
# floor sum of first $11$ terms of Harmonic Series

Value of $$\bigg\lfloor 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{11}\bigg\rfloor$$

Where $\lfloor x \rfloor$ Represent floor function of $x$.

$\bf{My\; Try::}$ Plotting $\displaystyle y = f(x) = \frac{1}{x}$ in Coordinate axis, We get

$$\int_1^{12}\frac{1}{x}dx<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots + \frac{1}{11} <1+\int_{1}^{11}\frac{1}{x}dx$$

So we get

\begin{align} \ln(12) & =2.303\log_{10}(12)<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{11} \\[10pt] & <1+\ln(11) =1+2.303\times \log_{10}(12) \end{align}

How can i solve after that, Help Required, Thanks

• You can just compute the quantity directly, since $1 + 1/2 + ... + 1/11 = 83711 / 27720 \approx 3.02$.... – user296602 Jun 6 '16 at 4:20
• I have added like $1+0.5+.33+.125+.......,$ But getting $<3$. – juantheron Jun 6 '16 at 4:22
• Then it seems your computation is incorrect somewhere. – user296602 Jun 6 '16 at 4:23
• Your approximations by integrals may give away too much. You can do better with $1+\frac{1}{2}+\text{approximations}$. Maybe overkill for $11$. – André Nicolas Jun 6 '16 at 4:26
• The integral approximation will not be good enough here as the error between $\int_{k}^{k+1}\frac{{\rm d}x}{x}$ and $\frac{1}{k}$ is $\sim \frac{1}{2k^2}$ which is quite large for all the terms. You need the total error to be smaller than $0.02$ to get the result. Just sum it! More fancy tecniques are going to take much more time anyway. – Winther Jun 6 '16 at 4:33

The exact answer is $$H_{11} = \frac{83711}{27720} > 3$$ so the answer is $3$


text:
# longtable and filecontents inside a newenvironment

I'm trying to create a new environment for longtable enclosed by a filecontents environment. But I'm stuck and don't know exactly how to solve my problem.

\documentclass[a4paper,parskip]{scrartcl}
\usepackage{polyglossia}

\usepackage{longtable, ltxtable, filecontents, colortbl}

\newenvironment{tbl}{%
\begin{filecontents}{\jobname-ltxtable.tex}%
\begin{longtable}{|p{2cm}|p{2cm}|p{2cm}|p{2cm}|}%
\hline\rowcolor[gray]{.8}%
Column 1 & Column 2 & Column 3 & Column 4\\\hline%
}{%
%\caption{#1} %
\end{longtable}%
\end{filecontents}%
\LTXtable{\textwidth}{\jobname-ltxtable.tex}%
}

\begin{document}

\begin{tbl}
1 & 2 & 3 & 4 \\\hline
1 & 2 & 3 & 4 \\\hline
1 & 2 & 3 & 4 \\\hline
\end{tbl}

\end{document}

I found out that using filecontents inside a macro leads to problems but I need something like filecontents to write out the longtable. The table shall later consist X columns. I reduced all the complexity to generate a minimal (non-working) example.

I hope someone can help me with this.

Thx + Bye!

• unrelated to longtable, filecontents is a "verbatim" environment and like all verbatim codes can not be nested inside another command – David Carlisle Jun 3 '16 at 8:19
• For a discussion regarding filecontents and newenvironment, see there: tex.stackexchange.com/q/130294/9075 – koppor Mar 7 '18 at 14:25

You can use the environ package to collect the body and then write it. Be aware that the body is written in one long line.

\documentclass[]{article}
\usepackage{polyglossia}

\usepackage{ltxtable,longtable,colortbl,environ,etoolbox}
\newwrite\mytabularwrite
\NewEnviron{tbl}
{%
\immediate\openout\mytabularwrite=\jobname-ltxtable.tex
\immediate\write\mytabularwrite{%
\unexpanded{%
\begin{longtable}{|p{2cm}|p{2cm}|p{2cm}|p{2cm}|}%
\hline\rowcolor[gray]{.8}%
Column 1 & Column 2 & Column 3 & Column 4\\\hline%
}
\immediate\write\mytabularwrite{\unexpanded\expandafter{\BODY}}%
\immediate\write\mytabularwrite{\unexpanded{\end{longtable}}}%
\immediate\closeout\mytabularwrite
}

\AfterEndEnvironment{tbl}{\LTXtable{\textwidth}{\jobname-ltxtable2.tex}}
\begin{document}

\begin{tbl}
1äöüß€ & \textbf{2} & 3 & 4 \\\hline
1 & 2 & 3 & 4 \\\hline
1 & 2 & 3 & 4 \\\hline
\end{tbl}

\end{document}

• sry for the late answer! Your code runs very well. Thanks a lot, Ulrike! – aronadaal Jun 29 '16 at 12:28


text:
My Math Forum measure of unbounded set

 October 13th, 2013, 01:09 PM #1 Newbie   Joined: Oct 2013 Posts: 25 Thanks: 0 measure of unbounded set Suppose A is not a bounded set and m(A?B)?(3/4)m(B) for every B. what is m(A)?? here, m is Lebesgue Outer Measure My attemption is : Let An=A?[-n,n], then m(A)=lim m(An)= lim m(An?[-n,n]) ? lim (3/4)m([-n,n]) = infinite. is my solution right? I am confusing m(A) < infinite , it doest make sence for me. Could someone help me???
October 14th, 2013, 02:30 AM   #2
Senior Member

Joined: Feb 2013

Posts: 281
Thanks: 0

Re: measure of unbounded set

Quote:
 m(A?B)?(3/4)m(B) for every B
choosing B = A, you get m(A) <=.75m(A), that is m(A)=0.

Quote:
 I am confusing m(A) < infinite , it doest make sence for me.
You didn't show that m(A)<inf, you did show that m(A)<=inf.

October 14th, 2013, 02:34 AM   #3
Newbie

Joined: Oct 2013

Posts: 25
Thanks: 0

Re: measure of unbounded set

Quote:
Originally Posted by csak
Quote:
 m(A?B)?(3/4)m(B) for every B
choosing B = A, you get m(A) <=.75m(A), that is m(A)=0.

[quote:1ulqxtmx] I am confusing m(A) < infinite , it doest make sence for me.
You didn't show that m(A)<inf, you did show that m(A)<=inf.[/quote:1ulqxtmx]

Thanks , i get the point. !!

 Tags measure, set, unbounded

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text:
## Algebra and Trigonometry 10th Edition

$3$ and $\theta =60^{\circ}$
$\overrightarrow{v} = \dfrac{3}{2 } \ \widehat{i} +\dfrac{3 \sqrt 3}{2} \ \widehat{j}$ Magnitude: $||\overrightarrow{v}|| = \sqrt {(\dfrac{3}{2 })^{2} + (\dfrac{3 \sqrt 3}{2} )^{2}} = 3$ The direction can now be found: $\tan \ \theta = \dfrac{\dfrac{3 \sqrt 3}{2} }{3/2} = \sqrt 3 \implies \theta =60^{\circ}$


text:
### Solution(By Apex Team)

Marked price is Rs. 500 Discount = Rs. 40% Selling Price = Marked Price − Discount = Rs. 500 – 300 = Rs. 200 Two successive discount 30% and 10% on 500 would be 500 ===30% discount ⇒ Rs. 350 350 ===10% discount = 315 Total discount in first case = Rs. 200 Total discount in second case = 500 – 315 = Rs. 185 So, difference = Rs. 15

## Related Questions on Profit and Loss

A. 45 : 56
B. 45 : 51
C. 47 : 56
D. 47 : 51

A. Rs. 2600
B. Rs. 2700
C. Rs. 2800
D. Rs. 3000

### A sells an article to B at a profit of 10% B sells the article back to A at a loss of 10%. In this transaction:

A. A neither losses nor gains
B. A makes a profit of 11%
C. A makes a profit of 20%
D. B loses 20%


text:
# How can I improve the resolution of a spectrogram of a low-frequency signal with high sampling frequency (Python)?

Imagine you have signals $$x_1(t)$$ and $$x_2(t)$$ that contain a limited number of periods of low frequency oscillations and are sampled at 44100Hz (coming from an external source) and you want to obtain their spectrograms.

I chose nperseg=8820 to capture one period of 5Hz (44100/5=8820) in an analysis window and noverlap=1 to move in steps of one sample.

• As shown in the figure below, after zooming the frequency axis, the frequency resolution in this low-frequency region is poor. Are there ways to improve this resolution? By setting special parameters of the spectrogram function or by downsampling for instance?

• And why does the spectrogram of $$x_2(t)$$ change color over time?

import numpy

import scipy.signal
import matplotlib.pyplot as pplot

fs = 44100
t = numpy.linspace(0, 100000, 100001) / fs
x1 = numpy.sin(2*numpy.pi*5*t)
x2 = numpy.sin(2*numpy.pi*5*t**2)

f1, t1, Sxx1 = scipy.signal.spectrogram(x1, fs, nperseg=8820, noverlap=1)
f2, t2, Sxx2 = scipy.signal.spectrogram(x2, fs, nperseg=8820, noverlap=1)

pplot.figure(1)
pplot.clf()
pplot.subplot(2,2,1)
pplot.plot(t, x1)
pplot.subplot(2,2,2)
pplot.plot(t, x2)
pplot.subplot(2,2,3)
pplot.pcolormesh(t1, f1, Sxx1)
pplot.ylim([0, 30])
pplot.subplot(2,2,4)
pplot.pcolormesh(t2, f2, Sxx2)
pplot.ylim([0, 30])
pplot.show()

• Probably yhere is a parameter defining the zero pedding. You can increase it. In some applications it is called nfft... – Gideon Genadi Kogan Mar 14 at 20:42


text:
## The pressure gradient system.(English)Zbl 1227.35208

The authors of this very interesting paper consider the two-dimensional compressible Euler system
\begin{aligned} \rho_t&+\nabla\cdot (\rho {\mathbf u})=0,\\ (\rho{\mathbf u})_t&+\nabla\cdot (\rho{\mathbf u}\times{\mathbf u}+pI)=0,\\ (\rho E)_t&+\nabla\cdot (\rho E{\mathbf u}+ p{\mathbf u})=0, \end{aligned}
where $$\rho$$ is the density, $${\mathbf u}$$ is the velocity vector, $$p$$ is the pressure, $$E=|{\mathbf u}|^2/2+p\rho^{-1}/(\gamma -1)$$ is the total energy density per unit mass, and $$\gamma >1$$ is the gas constant. The so-called pressure gradient system is a subsystem of the compressible Euler system $u_t+p_x=0, \quad v_t+p_y=0, \quad E_t+(up)_x+(vp)_y=0,$
where $$E=p+(u^2+v^2)/2$$. The authors provide numerical simulations, basic characteristic analysis and physical considerations for the Riemann problems to the considered model to find out appropriate internal conditions at the origin. This study reveals subtle structures of the velocity. Both components exhibit discontinuities at the origin. Comparing to the roll-up of shear waves or vortex-sheets of the Euler system, it turns out that the singularities are mild and occur only along rays from the origin. Numerical examples illustrate the theory, and sets of central contour lines from the pressure graph are shown in a very precise form.

### MSC:

 35L65 Hyperbolic conservation laws 35Q31 Euler equations 35L45 Initial value problems for first-order hyperbolic systems 35L60 First-order nonlinear hyperbolic equations 76N15 Gas dynamics (general theory)
Full Text:


text: is it 1???? … F22-d. O22+e. Verify … Ask Study Doubts; Sample Papers; Past Year Papers; Textbook Solutions; Sign Up. You can specify conditions of storing and accessing cookies in your browser. Q. Join Yahoo Answers and get 100 points today. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Do you expect these molecules to exist in the. For example, in diatomic nitrogen N≡N the bond number is 3, in ethyne H−C≡C−H the bond number between the two carbon atoms is also 3, and the C−H bond order is 1. What Is The Bond Order Of Li2−? Enter Your Answer As A Decimal. ONE PLZ..............plz solve this question in ATTACHMENT, public ans fast.......my questions r collecting....in line..,, Calculate the percentage of monochloro isomers producs during the chlorination of tertiary butane, ohh u like to eat skull's???? what is the bond order of Li2+ ion? Just take electrons that are in each MO, and . A bond order greater than zero means that more electrons occupy bonding MOs (stable) than antibonding MOs(unstable). Since both have a bond order of $+\frac{1}{2},$ both $\mathrm{Be}_{2}^{+}$ and $\mathrm{Be}_{2}^{-}$ will exist in gas phase. Q:-Density of a gas is found to be 5.46 g/dm 3 at 27 °C at 2 bar pressure. What is the relationship between bond order and the dissociation energy of a molecule? All chemical compounds are made up of bonds. Kindly Sign up for a personalised experience. Draw an MO energy diagram and predict the bond order of Be2 + and Be2 -. 10 0. similar to Be 2 2+ just remove 2 electrons from 2 s sigma anti bonding and gives bond order of 1 diamagnetic. If so the bond order is 1. Is n2 stable or unstable? ... We're trying to predict whether we expect brilliant two plus and brilliant to minus to be stable and if we … ?#Real Dpif yes ur welcomed....., any technique to study types of unit cell. Bond order= [No of bonding electrons]- [No of anti bonding electrons]/ 2 B.O= 2–2/2 B.O= 0/2 = 0 The bond order for He2 is 0. 14 valence electrons. Favourite answer. Bond Order. What is the bond order of Be2 −? Cloudflare Ray ID: 609f8c896d1f051f The bond order can be calculated in a simple manner. The bond order is 22−2 = 0. This site is using cookies under cookie policy. 10 valence electrons.b. Ne22+c. Join. Answer. Topics. See drawing. If you mean Li2 with a +1 charge, YES. From the electronic configuration it is clear that there is no singly filled atomic orbital present in beryllium. Calculation of bond order. Get answers by asking now. 1 … dhyeysonani is waiting for your help. Q. • Which of the following species has the shortest bond length? Which of the following gases will have the lowest rate of diffusion? Lv 4. Bonds order means the number of bonds that exist in two bonded atoms in a molecule. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Do you expect these molecules to exist in the gas phase? See Answer. 6 years ago. Robert. Is Li2− Paramagnetic Or Diamagnetic? Ask question + 100. Draw an MO energy diagram and predict the bond order of Be2 + and Be2 -. Answer to Draw an MO energy diagram and predict the bond order of Be2+ and Be2−. Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen. What is the relationship between bond order and the dissociation energy of a molecule? What is the bond order of Be 2 +? for each electron in a bonding MO, it adds #0.5# to the bond order, because more bonding character strengthens the bond...; for each electron in an antibonding MO, it subtracts #0.5# from the bond order, because more … Is Li2− paramagnetic or diamagnetic? We’re being asked to determine the bond order of He 2 +. Step 3: Calculate the bond order of the molecule/ion. For this, we need to do the following steps: Step 1: Calculate the total number of valence electrons present. Expert Answer . Update: no i mean a Li^2+ charge... Answer Save. Here we consider the molecular orbital diagram (MO) of #"Li"_2#:. F22+b. 12 valence electrons.c. the theory, molecular orbitals … Recall that the formula for bond order is: asked Jun 8, 2018 in Chemistry by Nisa (59.6k points) chemical bonding and molecular structure; class-11; 0 votes. Also asked, does be2 2+ exist? Assertion: H2, Li2 and B2 each has a bond order of 1 and hence are equally stable. What Is The Bond Order Of Be2+? Previous Question Next Question. Step 2: Draw the molecular orbital diagram. Want to see the step-by-step answer? A bond order of zero is obtained by placing the available electrons in the bonding and antibonding levels, indicating that dihelium does not exist according to valence bond and bond order theory. From the above MO diagram we can see that number of elctrons in the bonding and antibonding orbital is same and hence Be does not form Be2 molecule (for. Draw an MO energy diagram and predict the bond order of Be2+ and Be2-. Do you expect these molecules to exist in the gas phase? Add your answer and earn points. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability. The valence shell of each beryllium atom is 2s2 so there are a total of four valence shell electrons for which we Furthermore, does be2 2+ exist? Please enable Cookies and reload the page. As a result of the stability of the N 2 molecule, many nitrogen compounds are unstable, some explosively so. See the answer. Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. Kindly Sign up for a personalised experience. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Q:- Enter Your Answer As A Decimal. From the electronic configuration it is clear that there is no singly filled atomic orbital present in beryllium. Popular Questions of Class Chemistry. Do you expect these molecules to exist in the gas phase? Bond number gives an … Differentiate between bonding and antibonding molecular orbitals. Question: What Is The Bond Order Of Li2−? 1 shows that bond-order gradually increases to 1 in the range (0-2) electrons then falls to zero in the range (2-4) electrons then it further rises to 1 for (4-6) electrons and once again falls to zero for (6-8) electrons then again rises to 3 in the range (8-14) electrons and then finally falls to zero for (14-20) electrons. We we draw the molecular orbital diagram for neutral di-beryllium we find a bond order of zero since there are four bonding electrons and four anti-bonding electrons. Answer and Explanation: The bond order for Be2 is 0 (zero). 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text:
# [NTG-context] References title in Norwegian

Trond Lossius lossius at bek.no
Fri Feb 22 08:26:51 CET 2008

Thanks, this works.

I was just about to ask how to do the same concerning the list of
contents when I realized that it automagically translates itself when using

\mainlanguage[no]

Best,
Trond

> \mainlanguage[no]
> \setupheadtext[no][pubs=Litteratur]
>
>> I'm writing a publication in Norwegian containing citations and
>> references. I use
>>
>> \completepublications[criterium=all]
>>
>> at the end of the tex file to get the references. The only problem is
>> that the title of it when typeset show up as "References" in English. Is
>> it possible somehow to substitute that for the Norwgeian equivalent

More information about the ntg-context mailing list


text:
# (Random Walk) Probability of Returning to Origin

I want to find out the probability that a 1-dimensional asymmetric random walk, which steps to the right with probability $p > \frac{1}{2}$ and to the left with probability $1-p$, ever returns to the origin.

I tried to solve this by defining the probability of return to the origin after exactly $n$ steps, and taking the summation over these probabilities. This led me to

$$\mathbb{P}(\text{"return to origin"}) = \sum_{n=0}^{\infty} \binom{2n}{n} p^{n} (1-p)^{n}$$

However, for any value for $p$ greater than $\frac{1}{2}$, the series converges to a number greater than 1. (Tested with WolframAlpha.) What am I doing wrong? How can I solve this problem?

• You are confusing the probability to return at the origin with the mean number of times one visits the origin. To compute $p_0=P_0(T_0<\infty)$, use the Markov property after one step, thus $p_0=pp_1+(1-p)p_{-1}$ where $p_x=P_x(T_0<\infty)$. Now, $p_{-1}=1$ should be clear and to compute $p_1$, you might want to use the Markov property after one step once more. – Did May 9 '16 at 20:03
• Another option is to note that the RHS in your post is $1/(1-p_0)$ hence recognizing $\sum\limits_{n\geqslant0}{2n\choose n}x^n$ for $1/\sqrt{1-4x}$ yields $p_0=2(1-p)$ right away. – Did May 9 '16 at 20:06
• @Did, the recurrence should read $p_0= p\cdot p_1 + (1-p)p_{-1}$. – zhoraster May 9 '16 at 20:07
• @zhoraster Indeed, thanks. – Did May 9 '16 at 20:08
• That is a much easier method, thanks for bringing it up. – Tiamo P. May 9 '16 at 20:22

What you're doing wrong is that you're double-counting those walks that return to the origin twice or more. Instead of $\binom{2n}{n}$, you may want to use the Catalan numbers to construct the number of paths that return to the origin for the first time after exactly $n$ steps.
• Aha, stupid of me to miss that. The series, iff $p > \frac{1}{2}$, indeed converges to the right number now! – Tiamo P. May 9 '16 at 20:21


text:
## A particle is projected at angle 60º with speed 10 √3, from the point ‘ A ‘ as shown in the fig….

Q: A particle is projected at angle 60º with speed 10 √3, from the point ‘ A ‘ as shown in the fig. At the same time the wedge is made to move with speed 10 √3 towards right as shown in the figure. Then the time after which particle will strike with wedge is
(g = 10 m/sec2) :

(a) 2 sec

(b) 2 √3 sec

(c) 4/ √3 sec

(d) none of these

Click to See Solution :
Ans: (a)

Sol: Suppose particle strikes wedge at height ‘ S ‘ after time t.

$S = 15 t – \frac{1}{2}.10 . t^2$

$S = 15 t – 5 t^2$

During this time , distance travelled by particle in horizontal direction = 5 √3 t.

Also wedge has travelled extra distance

$x = \frac{S}{tan 30^o}$

$x = \frac{15 t – 5 t^2}{\frac{1}{\sqrt{3}}}$

Total distance travelled by wedge in time t = 10 √3 t

= 5 √3 t + √3 (15 t – 5 t2)

t = 2 sec

## A particle moves along the parabolic path y = ax2 in such a way that the y-component of the velocity remains constant ….

Q: A particle moves along the parabolic path y = ax2 in such a way that the y-component of the velocity remains constant, say c. The x and y coordinates are in meters. Then acceleration of the particle at x = 1 m is

(a) $\displaystyle ac \hat{k}$

(b) $\displaystyle 2 a c^2 \hat{j}$

(c) $\displaystyle – \frac{c^2}{4 a^2} \hat{i}$

(d) $\displaystyle – \frac{c}{2 a} \hat{i}$

Click to See Solution :
Ans: (c)
Sol: y = a x2

$\displaystyle \frac{dy}{dt} = c$ (given )

$\displaystyle \frac{d^2y}{dt^2} = 0$

$\displaystyle \frac{dy}{dt} = 2 a x\frac{dx}{dt}$

$\displaystyle 2 a (\frac{dx}{dt})^2 + 2ax \frac{d^2 x}{dt^2} = 0$

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{1}{x} (\frac{dx}{dt})^2$

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{1}{x} (\frac{c}{2ax})^2$

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{c^2}{4 a^2 x^3}$

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{c^2}{4 a^2 }$

## Starting at time t = 0 from the origin with speed 1 m/s a particle follows a two-dimensional trajectory in the x-y plane ….

Q: Starting at time t = 0 from the origin with speed 1 m/s a particle follows a two-dimensional trajectory in the x-y plane so that its co-ordinates are related by the equation $y = \frac{x^2}{2}$ .The x and y components of its acceleration are denoted by ax and ay respectively . Then

(a) ax = 1 m/s^2 implies that when the particle is at the origin , ay = 1 m/s^2

(b) ax = 0 implies ay = 1 m/s^2 at all times

(c) at t = 0 the particle’s velocity points in the x-direction

(d) ax = 0 implies that at t= 1 s , the angle between the particle’s velocity and the x axis is 45°

Click to See Solution :
Ans: (a,b,c,d)

Sol: The given equation is $y = \frac{x^2}{2}$

$\frac{dy}{dt} = x \frac{dx}{dt}$

vy = x vx

Differentiating w.r.t time

$\frac{dv_y}{dt} = x \frac{dv_x}{dt} + \frac{dx}{dt} v_x$

ay = x ax + vx2

## The distance x covered by a particle in one dimensional motion varies with time t as x^2 = at^2 + 2 bt + c ….

Q: The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2 bt + c . If the acceleration of particle depends on x as x-n , where n is an integer , the value of n is …….

Sol: x2 = at2 + 2 bt + c

$\large x = (at^2 + 2 bt + c)^{1/2}$

Differentiating with respect to time t ,

$\large \frac{dx}{dt} = \frac{1}{2}(at^2 + 2 bt + c)^{-1/2}(2at + 2b)$

Acceleration , $\displaystyle A = \frac{d^2x}{dt^2} = \frac{1}{2}[ (at^2 + 2 bt + c)^{-1/2}(2a) + (2at + 2b)(-\frac{1}{2})(at^2 + 2bt + c)^{-3/2}(2at + 2b)]$

$\large Ax = a – (\frac{at+b}{x})^2$

$\large Ax = \frac{ax^2 -(at+b)^2 }{x^2}$

$\large Ax = \frac{a(at^2 + 2bt + c)-(at+b)^2}{x^2}$

$\large A = \frac{ac-b^2}{x^3}$

$\large A \propto \frac{1}{x^3}$

$\large A \propto x^{-3}$

Hence , n = 3


text:
# Upward Mobility: Android for iOS Developers, Part 2

## Turning Hello World into something useful

When we last left our application, it was running on the emulator, but didn’t do much.  This week, we’ll add some more controls to our activity and wire up some functionality.

As a reminder, activities are roughly equivalent to view controllers in iOS. Right now, there’s not much in our activity, because all we have is a single label in our activity. As iOS developers, we’re used to never looking at our XIB files in the raw, because they’re pretty much human-unreadable. By contrast Android layout files (which end up in the res/layout folder in a standard Eclipse project) are both readable and intended to be edited. At the moment, here’s what ours looks like:

It’s a basic layout with a text view in the upper left hand corner. To start, we can center the text view and bump it up against the top border. If you drag it around in the graphical layout, you’ll see the same kinds of guides as you do in Interface Builder, and it’s pretty easy to just drag the control where you want it, so that it ends up looking like this:

Now we can switch back to the raw XML view, and see how things have changed:

There are now two new attributes, aligning to the parent top and horizontally centering the control. Now we can add a control that lets the user select just which world they are saying hello to. The right control for this is a spinner, which we can drop right under the text view. By default, the spinner will be called spinner1, but I’d like to give it a more informative name, so I go over to the Properties pane and change the id field to @+id/planetSpinner. When I do this, I immediately get a prompt:

What is happening is that Eclipse is going to globally change that id, including in a bunch of autogenerated code. As a result, it wants to make sure that we want to do this. If you say yes, you can see that new id has shown up in the XML version of the layout:

Now we’re ready to go into our MainActivity class and do something with this control, such as populate it with a list of planets. If we look at MainActivity.java, we can see the onCreate method:

The setContentView call is what relates the Java class to the layout file. Once we’ve set the content view, we can get a handle on objects inside the layout and play with them. The first thing we need is an array of string that we can feed into the Spinner. In Android, this is usually done through a string resource file (similar to a string properties file in iOS). You’ll find a strings.xml file hiding in the res/values folder. Once we add our new string array, it will look like this:

We now have four planets that our user can choose from. Before we go any further, make sure that you have saved both the string resource and layout files! When you do, a magic class file called R will be updated, creating ids automatically for these new entities we’ve added:

The last step is to give those planet names to our new spinner, in our onCreate method:

Walking through the code, we first get a handle on the spinner by using the automatically generated id in the R class file. We create an array adapter, which is a special resource that is going to return spinner items from our planet names. Finally, we tell the spinner to use that adapter, and when we rerun the code in the emulator, we get our new drop-down.

Personally, I think that this is a bit complicated for a simple drop-down, requiring me to create an adapter to provide a simple list of strings, and is representative of how Java-based platforms tend to over-architect things that should be easy. I’m sure the adapter idea is there so that you can put more complicated things such as images into your drop-down, but in iOS, you normally have a single way of doing something that covers the 95% use case, and a more involved way to do the complex stuff.  Why I can’t just hand a spinner an ArrayList of strings is beyond me. Let it handle the adapters and such for me, at least for the simple cases.

In any event, editorializing aside, next week we’ll find out how to listen for changes on that spinner and use it to drive actions in our activity.

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# Math Help - Set Counting Problem

1. ## Set Counting Problem

The problem is:

In carrying out a survey of the efficiency of the lights, brakes and steering of motor vehicles, 100 vehicles were found defective as follows:

35 defective lights
8 had defective lights and brakes
7 had defective lights and steering
6 had defective brakes and steering

a) how many had defective lights, brakes and steering?
b) how many vehicles had defective lights only?

Now I know the answers are 5 and 25 respectively but I cannot form the correct equation to arrive at them.

I think the equation has to be in the form of

x(A u B u C)= x(A) + x(B) + x(C) - x(A n B) - x(A n C) - x(B n C) + (A n B n C)

I would appeciate help with the formulae.

Many thanks.

2. Hello, barc0de!

In carrying out a survey of the efficiency of the lights, brakes and steering of motor vehicles,
100 vehicles were found defective as follows:

35 defective lights
8 had defective lights and brakes
7 had defective lights and steering
6 had defective brakes and steering

a) How many had defective lights, brakes and steering?

b) How many vehicles had defective lights only?

Now I know the answers are 5 and 25 respectively,
but I cannot form the correct equation to arrive at them.

I think the equation has to be in the form of:
. . $x(A\cup B \cup C) \:= \:x(A) + x(B) + x(C) - x(A\cap B) - x(A \cap C)$ $-\; x(B \cap C) + (A \cap B \cap C)$
. . Right!

We are given:. . $\begin{array}{ccc}
x(L\cup B\cup S) &=& 100 \\ x(L) &=& 35 \\ x(B) &=&40 \\x(S) &=&41 \\
x(L\cap B) &=&8 \\ x(L\cap S) &=& 7 \\ x(B \cap S) &=& 6

Our equation is:

. . $x(L \cup B \cup S) \;=\;x(L) + x(B) + x(S) - x(L \cap B)$ $-\; x(L \cup S) - x(B \cap S) + x(L \cap B \cap S)$

. . . . . $100 \qquad\;=\quad 35 \;+\; 40 \;\;+\;\; 41 \quad\;-\quad\;\; 8 \quad\;\; -\quad\;\;\; 7 \quad\;\;-\quad\;\;\; 6 \quad\;\;+$ $x(L \cap B \cap S)$

(a) And we have: . $100\:=\:95 + x(L \cap B \cap S) \quad\Rightarrow\quad \boxed{x(L\cap B\cap S) \:=\:5}$

The best way solve part (b) is with a Venn diagram.

Or we can invent a formula for this situation . . .

We want: . $x(L \cap B' \cap S')$

And: . $x(L \cap B' \cap S') \;=\;x(L) - x(L \cap B) - x(L \cap S) + x(L \cap B \cap S)$

. . . . . . . . . . . . . . $= \;\;\; 35 \quad\;-\;\quad 8 \quad\;-\qquad 7 \quad\; +\qquad\;\; 5\quad=\quad\boxed{25}$


text:
Question

# Solve differential equation xu'(x)= u^2-4

First order differential equations
Solve differential equation $$xu'(x)= u^2-4$$

2021-03-08

$$xu'(x)= u^2-4$$
$$xu'(x) \frac{1}{u^2-4}=1$$
$$\frac{1}{u^2-4} du=\frac{1}{x} dx$$
Integrating on both sides
$$\int \frac{1}{u^2-4}du= \int \frac{1}{x} dx$$
The partial fraction of the left side $$\frac{1}{(u+2)(u−2)}= −\frac{1}{4(x+2)}+\frac{1}{4(x−2)}$$
$$-\frac{1}{4} \int \frac{1}{u+2}du +\frac{1}{4} \int \frac{1}{u-2}du= \int \frac{1}{x} dx$$
$$-\frac{1}{4} \ln (u+2)du+\frac{1}{4} \ln(u-2)= \ln(x)+C_1$$
$$-\ln(u+2)du+ln(u-2)= 4 \ln(x)+4C_1$$
$$\ln(\frac{u-2}{u+2})= \ln(x^4)+C_2$$
$$(\frac{u-2}{u+2})= (x^4)+e^{C_2}$$
$$= x^4+C$$
$$u= - \frac{2(C+x^4)}{-C+x^4}$$


text:
## Can you craft magic items without specific abilities–and therefore, requirements/costs?

Ghostwalk’s glove of the master strategist is described as follows:

This black leather glove is decorated with purple silk thread. It acts as a glove of storing and allows the wearer to use a true strike effect once per day.

Caster Level: 6th; Prerequisites: Craft Wondrous Item, shrink item, true strike; Market Price: 3,600 gp; Weight: —.

Let’s say you don’t need the true strike ability. The glove doesn’t say that the true strike effect requires a command, so it seems safe to assume it’s use-activated. As a 1st level spell (CL 1st) with one use per day, magic item creation rules imply that this ability is worth 1 × 1 × 2000 / (5/1) = 400 gp.

As such, could you craft what’s essentially a glove of storing at CL 6th with only the Craft Wondrous Item feat, preparation of the shrink item spell, and 1,600 gp + 128 XP (half of the base price of 3,200 gp, plus XP costs)?

If so, what would happen if you gave the glove a requirement that the user have at least 1 rank in the Sleight of Hand skill? Is the 10% discount based on the “normal” cost of 1,800 gp (180 gp discount, seems unlikely) or the “intended” cost of 1,600 gp (160 gp discount, seems likely), or does doing so not apply a discount in this way (seems possible)?

## Would creating undead minions be considered an evil action in PFS, and therefore earn Infamy?

Would creating undead minions be considered an evil action in PFS, and therefore earn Infamy? I’m not keen on GMing for a necromancer or anti-paladin type character, but also don’t want to deny a player’s choices outright. It seems like a necromancer (sans undead minions) is distinctly different than someone with an undead horde. This line of questioning extends to fiendish summoning, etc.

## Does Dual Wielding count as two attacks, therefore two actions?

I’m not sure if two weapon fighting and duel wielding are the same thing.

I’m very new, like first game new. What I’m wondering is: if I have the Dual Wielding Feature, do I make two separate weapon attack rolls? Or do i make one attack roll with two instances of damage.

This came up because I attempted to like stab someone with both at once, and since my whole group is new we had no idea.

## When is optimization not premature and therefore not evil?

“Premature optimization is root of all evil” is something almost all of us have heard/read. What I am curious what kind of optimization not premature, i.e. at every stage of software development (high level design, detailed design, high level implementation, detailed implementation etc) what is extent of optimization we can consider without it crossing over to dark side.

## State true or false –> x=θ(\$n^4\$), y=θ(\$n^2\$) therefore x/y=θ(\$n^2\$)

State true or false:

• $$x=\Theta(n^4)$$, $$y=\Theta(n^2)$$ therefore $$x/y=\Theta(n^2)$$.

I think statement ans is true, but the book it says false. What is the explanation?

My reason for thinking it to be true is that say for $$x$$ we have a polynomial $$n^4+\dots$$ and for $$y$$ we have a polynomial like $$n^2+\dots$$ so if we divide $$x/y$$ the highest power in the result should be $$n^2$$ and the resultant polynomial will be like $$n^2+\dots$$, so it should be $$\Theta(n^2)$$.

## Updating from such a repository can’t be done securely, and is therefore disabled by default

Err:16 http://us.archive.ubuntu.com/ubuntu precise-backports InRelease The following signatures were invalid: 630239CC130E1A7FD81A27B140976EAF437D05B5

W: GPG error: http://us.archive.ubuntu.com/ubuntu precise-backports InRelease: The following signatures were invalid: 630239CC130E1A7FD81A27B140976EAF437D05B5

E: The repository ‘http://us.archive.ubuntu.com/ubuntu precise-backports InRelease’ is not signed.

N: Updating from such a repository can’t be done securely, and is therefore disabled by default.

N: See apt-secure(8) manpage for repository creation and user configuration details.

W: GPG error: http://us.archive.ubuntu.com/ubuntu precise Release: The following signatures were invalid: 630239CC130E1A7FD81A27B140976EAF437D05B5

E: The repository ‘http://us.archive.ubuntu.com/ubuntu precise Release’ is not signed.

N: Updating from such a repository can’t be done securely, and is therefore disabled by default.

N: See apt-secure(8) manpage for repository creation and user configuration details.

## Samsung A5 2016 phone doesnt pick up any network (Vodacom) and therefore cannot connect to my data

Baseband Version – A510FXXU7CRJ2 – phone description code – Security warning unauthorised actions

## PHP Fatal error: Class Mock_LoggerInterface contains 8 abstract >methods and must therefore be declared abstract or implement the remaining

Preconditions (*)

1. Magento 2.3.1 upgraded from Magento 2.3.0
2. Php 7.2.15
3. OS : Ubuntu 18.04 Bionic
4. Phpstorm 2019.1.1
5. Evnironemnt : Magento development
6. PHPUnit 6.5.0
7. magento/magento2-functional-testing-framework”: “~2.3.13”

Steps to reproduce (*)

• Follow the code to reproduce the error

Error

PHP Fatal error: Class Mock_LoggerInterface_a49cf619 contains 8 abstract >methods and must therefore be declared abstract or implement the remaining >methods (Psr\Log\LoggerInterface::emergency, Psr\Log>\LoggerInterface::alert, Psr\Log\LoggerInterface::critical, …) in /var/www/html/myproject/vendor/phpunit/phpunit-mock-objects>/src/Generator.php(264) : eval()’d code on line 1

My class code is given below

``use Psr\Log\LoggerInterface;  class Foo {       private \$  logger;      public function __construct(LoggerInterface \$  logger)     {         \$  this->logger = \$  logger;     }     public function Bar ()     {       // some code       \$  logger->debug ('debug message')      } } ``

My Test Class Code is

``use PHPUnit\Framework\TestCase;  class FooTest extends TestCase {      private \$  logger;     public function setUp()     {         \$  this->logger = \$  this->getMockBuilder('\Psr\Log\LoggerInterface')             ->getMock();          \$  this->logger->expects(\$  this->any())             ->method('debug')             ->willReturn('Message Logged');     }      \$  this->myclassObject = \$  this->getMockBuilder('MyVendor\MyModule\Model\Foo')     ->setMethods(['__construct'])             ->setConstructorArgs(['\$  logger'])     ->disableOriginalConstructor()     ->getMock();      public function testBar()     {          \$  this->assertEquals(\$  expected_result,\$  this->myclassObject->Bar());     } } ``

I’m expecting a successful unit test.

## Let n be the number of vertices of G. Recall that G denotes the complement of G. Show that a) n ≤ χ(G)χ(G) and therefore, b) χ(G) + χ(G) ≥ 2√n.

This question was on my graph theory exam and I didn’t know how to approach solving it. The question in Tex is as follows:

Let n be the number of vertices of G. Recall that G denotes the complement of G. Show that a) n ≤ χ(G)χ(G) and therefore, b) χ(G) + χ(G) ≥ 2√n.

Any tips on how to approach this would be greatly appreciated!

## Can’t get Hadoop Datanode but didn’t show any error .Therefore can’t know which is resource of error

I have a question. I’m currently taking a project to build data governance.

I have already connected master ip and slave1 ip to each other. Then, I have made ssh keys and copied.

I don’t think I have any errors about network.

Next, I tried to use hadoop as follow.

``     hdfs namenode -format ``

and I got a warning indicating

``   WARN util.NativeCodeLoader: Unable to load native-hadoop library for your platform... using builtin-java classes where applicable ``

Although I got a warning sign, i don’t think it is a problem. However, I can’t get datanode after running start-dfs.sh and start-yarn even without any issue

[problem]https://imgur.com/08V7fC8

[run start-dfs and yarn]https://imgur.com/A85KARw

Can someone tell me why I get this problem?

Hadoop configure I get is as below:

[yarn_site.xml] https://imgur.com/DEBEhSe

[mapred_site.xml] https://imgur.com/nxgdBdM

[core_site.xml] https://imgur.com/Y0l02rJ

[hdfs_site.xml] https://imgur.com/truB0zz