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WHat is the uncertainty in a metre rule??

 Quote by Studiot Dickfore. If the result could be reported as 2 or 3 that is an uncertainty of 1.
http://en.wikipedia.org/wiki/Plus-mi...ion_indication

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 Quote by truesearch I am concerned when I read in post 6 (a PF Mentor) that measurements can be made to within +/- 0.1mm using a mm scale.
As a standard deviation, 0.2 is easy to achieve and 0.1 might be possible (all in mm).

As an example for 0.2 standard deviation, this means that 0.3 is usually (~70%) read somewhere between 0.1 and 0.5. It is sufficient to see that 0.3 is smaller than 0.5, but not close to 0, to do this.
0.5 should usually be read somewhere between 0.3 and 0.7 - which is everything not close to a mark on the scale.

If you want to give some "upper bound" for the error, you should use larger values, of course. But an upper bound is not always well-defined (apart from digital displays). And if you want to use the marks on the scale, you should format your numbers like $0.3^{+0.7}_{-0.3}mm$

 Yes indeed using ± notation. But if you are limited to scale divisions you cannot report half divisions, your statement has to be 2 or 3, not 2.5 ± 0.5. That implies you have to report 0,1,2,3,4 etc. So a report that the thickness is 2 implies that it is between 1 and 3 ie 2 ±1 Which is what I said. Incidentally you need to revise your statement on end standards.

 Quote by mfb As an example for 0.2 standard deviation, this means that 0.3 is usually (~70%) read somewhere between 0.1 and 0.5. It is sufficient to see that 0.3 is smaller than 0.5, but not close to 0, to do this.
I guess you meant 0.2 in the bolded number. Nevertheless, this is only true for a normal distribution. When we measure with a coarse scale such that we always get the length to be between the same two divisions, the error is not of statistical nature, and uncertainty has a different meaning from standard deviation. For a uniform distribution with ends a, and b, the standard deviation is:
$$\sigma_{U} = \frac{b - a}{2 \sqrt{3}}$$
Notice that [itex]2 \sqrt{3} \approx 3.5[/tex]

 Quote by Studiot Yes indeed using ± notation. But if you are limited to scale divisions you cannot report half divisions, your statement has to be 2 or 3, not 2.5 ± 0.5. That implies you have to report 0,1,2,3,4 etc. So a report that the thickness is 2 implies that it is between 1 and 3 ie 2 ±1 Which is what I said. Incidentally you need to revise your statement on end standards.
You can report half divisions, as it is customary in experimental physics.

As for end standards, we are not doing calibration of etalons. We are measuring the length of an object. Thus, we are free to slide the scale so that the left end coincides exactly with one of the ruler's divisions. Then, there is uncertainty in reading off only the right end.

There might be systematic errors due to the bad calibration of the rulers divisions, but that's another point.

 As for end standards, we are not doing calibration of etalons. We are measuring the length of an object. Thus, we are free to slide the scale so that the left end coincides exactly with one of the ruler's divisions. Then, there is uncertainty in reading off only the right end.
This is a fundamental error. The process of sliding still constitutes a 'reading' or alignment error. It is not as accurate as the engineering process of aligning an end standard ruler, even with a comparator microscope which engineers also use.

And BTW why mention etalons?

 There is no way anyone could say +/- 0.1mm
Actually there is but you require a draftsman's scale rule with diagonal scales. Have you heard of these?
However I have not seen one as long as 1 metre.

 Last few comments are missing the point and rely on unsubstantiated assumptions. Trying to see something between divisions which is not there! Assuming that there is some uniform scale within the division Assuming there is no distortion. The example of digital instruments should serve as a clue.... There is no way to 'eyeball' how close the last digit is to the one above or the one below. +/-1 max is a safe, sensible, objective bet.

 Quote by Studiot This is a fundamental error. The process of sliding still constitutes a 'reading' or alignment error.
Even if it does consists of an error, the error is of the order of the width of the mark, and not of the order of half the distance between two marks. I don't know about your rulers, but the marks on mine are pretty thin.

 I have used draughts mans scales ......I have also used verniers
 Recognitions: Science Advisor truesearch, error is not the absolute maximum or minimum you can be wrong by. It's the average error. So it behaves like a random walk. For a random walk of N steps of distance L each, the average distance you travel is L*Sqrt(N). So average error behaves like RMS. In practice, the actual value of how much you are off by will be normally distributed. The quoted error is the standard deviation of that distribution.

 Even if it does consists of an error, the error is of the order of the width of the mark, and not of the order of half the distance between two marks. I don't know about your rulers, but the marks on mine are pretty thin.
So why can't you 'read' the other end to the same precision?

As a matter of inerest how do you guarantee that the aligned 'zero' stays put while you read the other end?

I know how the navy does it for a traverse tape and how an engineering workshop does it for an engineering endstop rule and similarly how a drapers shop does it for a drapers endstop rule. Why do you think they do it this way with an end stop rather than your way?

 I have used draughts mans scales
So you know they are commonly calibrated in 1/100 inch or 0.1mm?

 Quote by Studiot So why can't you 'read' the other end to the same precision?
Because not all lengths in Nature are integer multiples of the divisions of our scale.

 Quote by Studiot As a matter of inerest how do you guarantee that the aligned 'zero' stays put while you read the other end?
You don't. But, you are describing sources of error that are order of magnitude smaller than the precision of the scale of the measuring instrument.

In fact, if the sources you allude to start giving comparable contributions, then it means your measuring scale is so precise that you do not get the same result as you repeat the measurement. In other words, you start getting statistical errors.

 Quote by Studiot I know how the navy does it for a traverse tape and how an engineering workshop does it for an engineering endstop rule and similarly how a drapers shop does it for a drapers endstop rule. Why do you think they do it this way with an end stop rather than your way?
Probably to account for the fact that in these cases the object being measured is violently moved during the measuring process. This, on the other hand, happens rarely in a Physics Lab.

 Dickfore all you are proving is that different people in different 'laboratories' use different techniques and thereby achieve (slightly) different results by going their own way. The whole object of calibration and standardisation is so that anyone anywhere can achieve the same result under the same conditions. This involves standardisation of measurement technique as well as tools in order to remove 'operator bias'. Measurement against a common stop end is one such standard. If a laboratory develops its own special techniques it needs to report these as part of the results. I once worked in such a laboratory measuring the lengths of bricks, accurate to less than a 10thou using the lab's specially developed technique. But we never pretended it was 'standard' or that the method should be widely adopted.

 Quote by Studiot The whole object of calibration and standardisation is so that anyone anywhere can achieve the same result under the same conditions.
Provided that the results are reported to same precision! What you are describing in the previous posts is comparing the result of a measurement done by a school ruler, to that of a micrometer screw gauge.

Let us say that (by the method I described), I get a length measurement of 3.5±0.5 mm.

Then, you come with your fancy equipment and get a result 3.329±0.007 mm (you had to repeat the experiment several times because you noticed that every time you get a different reading with your fine equipment. Then, you took the mean, and you found the standard deviation of the mean, and you took a 95% confidence interval for the mean.)

Does that make my measurement "wrong"?

Mentor
 Quote by truesearch I am concerned when I read in post 6 (a PF Mentor) that measurements can be made to within +/- 0.1mm using a mm scale.
Ultimately, the proper ± figure for scale-reading uncertainty depends on the person who is making the measurement and the instrument that he is using, and that person has to make a judgment about this.

I feel confident in assigning ±0.1 mm when using a metal scale with finely-engraved lines, in a way that eliminates or minimizes parallax error due to the thickness of the scale. It helps that I'm rather nearsighted so I can get my eye about 10 cm from the scale if I take off my eyeglasses. If I'm using a typical plastic ruler with relatively thick mm-lines, I might use ±0.2 mm. If I'm using a thick meter stick and can't lay it edgewise on the object being measured so that I have to sight across the thickness of the meter stick, I might use ±0.5 mm or even ±1.0 mm.

 Not at all to do with statistics. But everything to do with technique which contains inherent sources of error v technique which avoids these. When you place your test piece and ruler against the stop end you have a guaranteed square and reproducible 'zero'. When you estimate the alignment of two lines along a sight line that may or may not be square and hold the ruler and testpiece at some random (albeit small) angle to each other you have a recipe for variability of measurement. Notice I said 'sight line'. Two operators will align the pieces slightly differently by sight. They cannot do this with a stop end. Edit JTbell has just described the visual alignment issue to a T whilst I was posting.

 Quote by jtbell Ultimately, the proper ± figure for scale-reading uncertainty depends on the person who is making the measurement...
No. Wrong experimental procedure leads to systematic errors that are not a measure of the uncertainty.

 Quote by jtbell ...and the instrument that he is using,...
Yes. But:

 Quote by jtbell I feel confident in assigning ±0.1 mm when using a metal scale with finely-engraved lines, in a way that eliminates or minimizes parallax error due to the thickness of the scale.
This is definitely wrong if the "finely-engraved lines" are a distance 1 mm apart, as given in the OP.

 Tags metre rule, uncertainity


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## variational approximation to empirical likelihood ABC

Posted in Statistics with tags , , , , , , , , , , , , , , , , , , on October 1, 2021 by xi'an

Sanjay Chaudhuri and his colleagues from Singapore arXived last year a paper on a novel version of empirical likelihood ABC that I hadn’t yet found time to read. This proposal connects with our own, published with Kerrie Mengersen and Pierre Pudlo in 2013 in PNAS. It is presented as an attempt at approximating the posterior distribution based on a vector of (summary) statistics, the variational approximation (or information projection) appearing in the construction of the sampling distribution of the observed summary. (Along with a weird eyed-g symbol! I checked inside the original LaTeX file and it happens to be a mathbbmtt g, that is, the typewriter version of a blackboard computer modern g…) Which writes as an entropic correction of the true posterior distribution (in Theorem 1).

“First, the true log-joint density of the observed summary, the summaries of the i.i.d. replicates and the parameter have to be estimated. Second, we need to estimate the expectation of the above log-joint density with respect to the distribution of the data generating process. Finally, the differential entropy of the data generating density needs to be estimated from the m replicates…”

The density of the observed summary is estimated by empirical likelihood, but I do not understand the reasoning behind the moment condition used in this empirical likelihood. Indeed the moment made of the difference between the observed summaries and the observed ones is zero iff the true value of the parameter is used in the simulation. I also fail to understand the connection with our SAME procedure (Doucet, Godsill & X, 2002), in that the empirical likelihood is based on a sample made of pairs (observed,generated) where the observed part is repeated m times, indeed, but not with the intent of approximating a marginal likelihood estimator… The notion of using the actual data instead of the true expectation (i.e. as a unbiased estimator) at the true parameter value is appealing as it avoids specifying the exact (or analytical) value of this expectation (as in our approach), but I am missing the justification for the extension to any parameter value. Unless one uses an ancillary statistic, which does not sound pertinent… The differential entropy is estimated by a Kozachenko-Leonenko estimator implying k-nearest neighbours.

“The proposed empirical likelihood estimates weights by matching the moments of g(X¹), , g(X) with that of
g(X), without requiring a direct relationship with the parameter. (…) the constraints used in the construction of the empirical likelihood are based on the identity in (7), which can only be satisfied when θ = θ⁰. “

Although I am feeling like missing one argument, the later part of the paper seems to comfort my impression, as quoted above. Meaning that the approximation will fare well only in the vicinity of the true parameter. Which makes it untrustworthy for model choice purposes, I believe. (The paper uses the g-and-k benchmark without exploiting Pierre Jacob’s package that allows for exact MCMC implementation.)

## asymptotics of synthetic likelihood

Posted in pictures, Statistics, Travel with tags , , , , , , , , , , on March 11, 2019 by xi'an

David Nott, Chris Drovandi and Robert Kohn just arXived a paper on a comparison between ABC and synthetic likelihood, which is both interesting and timely given that synthetic likelihood seems to be lacking behind in terms of theoretical evaluation. I am however as puzzled by the results therein as I was by the earlier paper by Price et al. on the same topic. Maybe due to the Cambodia jetlag, which is where and when I read the paper.

My puzzlement, thus, comes from the difficulty in comparing both approaches on a strictly common ground. The paper first establishes convergence and asymptotic normality for synthetic likelihood, based on the 2003 MCMC paper of Chernozukov and Hong [which I never studied in details but that appears like the MCMC reference in the econometrics literature]. The results are similar to recent ABC convergence results, unsurprisingly when assuming a CLT on the summary statistic vector. One additional dimension of the paper is to consider convergence for a misspecified covariance matrix in the synthetic likelihood [and it will come back with a revenge]. And asymptotic normality of the synthetic score function. Which is obviously unavailable in intractable models.

The first point I have difficulty with is how the computing time required for approximating mean and variance in the synthetic likelihood, by Monte Carlo means, is not accounted for in the comparison between ABC and synthetic likelihood versions. Remember that ABC only requires one (or at most two) pseudo-samples per parameter simulation. The latter requires M, which is later constrained to increase to infinity with the sample size. Simulations that are usually the costliest in the algorithms. If ABC were to use M simulated samples as well, since it already relies on a kernel, it could as well construct [at least on principle] a similar estimator of the [summary statistic] density. Or else produce M times more pairs (parameter x pseudo-sample). The authors pointed out (once this post out) that they do account for the factor M when computing the effective sample size (before Lemma 4, page 12), but I still miss why the ESS converging to N=MN/M when M goes to infinity is such a positive feature.

Another point deals with the use of multiple approximate posteriors in the comparison. Since the approximations differ, it is unclear that convergence to a given approximation is all that should matter, if the approximation is less efficient [when compared with the original and out-of-reach posterior distribution]. Especially for a finite sample size n. This chasm in the targets becomes more evident when the authors discuss the use of a constrained synthetic likelihood covariance matrix towards requiring less pseudo-samples, i.e. lower values of M, because of a smaller number of parameters to estimate. This should be balanced against the loss in concentration of the synthetic approximation, as exemplified by the realistic examples in the paper. (It is also hard to see why M could be not of order √n for Monte Carlo reasons.)

The last section in the paper is revolving around diverse issues for misspecified models, from wrong covariance matrix to wrong generating model. As we just submitted a paper on ABC for misspecified models, I will not engage into a debate on this point but find the proposed strategy that goes through an approximation of the log-likelihood surface by a Gaussian process and a derivation of the covariance matrix of the score function apparently greedy in both calibration and computing. And not so clearly validated when the generating model is misspecified.

## Bayesian gan [gan style]

Posted in Books, pictures, Statistics, University life with tags , , , , , , , , , , , , , on June 26, 2018 by xi'an

In their paper Bayesian GANS, arXived a year ago, Saatchi and Wilson consider a Bayesian version of generative adversarial networks, putting priors on both the model and the discriminator parameters. While the prospect seems somewhat remote from genuine statistical inference, if the following statement is representative

“GANs transform white noise through a deep neural network to generate candidate samples from a data distribution. A discriminator learns, in a supervised manner, how to tune its parameters so as to correctly classify whether a given sample has come from the generator or the true data distribution. Meanwhile, the generator updates its parameters so as to fool the discriminator. As long as the generator has sufficient capacity, it can approximate the cdf inverse-cdf composition required to sample from a data distribution of interest.”

I figure the concept can also apply to a standard statistical model, where x=G(z,θ) rephrases the distributional assumption x~F(x;θ) via a white noise z. This makes resorting to a prior distribution on θ more relevant in the sense of using potential prior information on θ (although the successes of probabilistic numerics show formal priors can be used on purely numerical ground).

The “posterior distribution” that is central to the notion of Bayesian GANs is however unorthodox in that the distribution is associated with the following conditional posteriors

where D(x,θ) is the “discriminator”, that is, in GAN lingo, the probability to be allocated to the “true” data generating mechanism rather than to the one associated with G(·,θ). The generative conditional posterior (1) then aims at fooling the discriminator, i.e. favours generative parameter values that raise the probability of wrong allocation of the pseudo-data. The discriminative conditional posterior (2) is a standard Bayesian posterior based on the original sample and the generated sample. The authors then iteratively sample from these posteriors, effectively implementing a two-stage Gibbs sampler.

“By iteratively sampling from (1) and (2) at every step of an epoch one can, in the limit, obtain samples from the approximate posteriors over [both sets of parameters].”

What worries me about this approach is that  just cannot work, in the sense that (1) and (2) cannot be compatible conditional (posterior) distributions. There is no joint distribution for which (1) and (2) would be the conditionals, since the pseudo-data appears in D for (1) and (1-D) in (2). This means that the convergence of a Gibbs sampler is at best to a stationary σ-finite measure. And hence that the meaning of the chain is delicate to ascertain… Am I missing any fundamental point?! [I checked the reviews on NIPS webpage and could not spot this issue being raised.]

## ABCDE for approximate Bayesian conditional density estimation

Posted in Books, pictures, Statistics, Travel, University life with tags , , , , , , , , , , , , , on February 26, 2018 by xi'an

Another arXived paper I surprisingly (?) missed, by George Papamakarios and Iain Murray, on an ABCDE (my acronym!) substitute to ABC for generative models. The paper was reviewed [with reviews made available!] and accepted by NIPS 2016. (Most obviously, I was not one of the reviewers!)

“Conventional ABC algorithms such as the above suffer from three drawbacks. First, they only represent the parameter posterior as a set of (possibly weighted or correlated) samples [for which] it is not obvious how to perform some other computations using samples, such as combining posteriors from two separate analyses. Second, the parameter samples do not come from the correct Bayesian posterior (…) Third, as the ε-tolerance is reduced, it can become impractical to simulate the model enough times to match the observed data even once [when] simulations are expensive to perform”

The above criticisms are a wee bit overly harsh as, well…, Monte Carlo approximations remain a solution worth considering for all Bayesian purposes!, while the approximation [replacing the data with a ball] in ABC is replaced with an approximation of the true posterior as a mixture. Both requiring repeated [and likely expensive] simulations. The alternative is in iteratively simulating from pseudo-predictives towards learning better pseudo-posteriors, then used as new proposals at the next iteration modulo an importance sampling correction.  The approximation to the posterior chosen therein is a mixture density network, namely a mixture distribution with parameters obtained as neural networks based on the simulated pseudo-observations. Which the authors claim [p.4] requires no tuning. (Still, there are several aspects to tune, from the number of components to the hyper-parameter λ [p.11, eqn (35)], to the structure of the neural network [20 tanh? 50 tanh?], to the number of iterations, to the amount of X checking. As usual in NIPS papers, it is difficult to assess how arbitrary the choices made in the experiments are. Unless one starts experimenting with the codes provided.) All in all, I find the paper nonetheless exciting enough (!) to now start a summer student project on it in Dauphine and hope to check the performances of ABCDE on different models, as well as comparing this ABC implementation with a synthetic likelihood version.

As an addendum, let me point out the very pertinent analysis of this paper by Dennis Prangle, 18 months ago!

## 1500 nuances of gan [gan gan style]

Posted in Books, Statistics, University life with tags , , , , , , , , , , , on February 16, 2018 by xi'an

I recently realised that there is a currently very popular trend in machine learning called GAN [for generative adversarial networks] that strongly connects with ABC, at least in that it relies mostly on the availability of a generative model, i.e., a probability model that can be generated as in . For instance, there was a GANs tutorial at NIPS 2016 by Ian Goodfellow and many talks on the topic at recent NIPS, the 1500 in the title referring to the citations of the GAN paper by Goodfellow et al. (2014). (The name adversarial comes from opposing true model to generative model in the inference. )

If you remember Jeffreys‘s famous pique about classical tests as being based on improbable events that did not happen, GAN, like ABC,  is sort of the opposite in that it generates events until the one that was observed happens. More precisely, by generating pseudo-samples and switching parameters until these samples get as confused as possible between the data generating (“true”) distribution and the generative one. (In its original incarnation, GAN is indeed an optimisation scheme in .) A basic presentation of GAN is that it constructs a function D(x,ϕ) that represents the probability that x came from the true model p versus the generative model, ϕ being the parameter of a neural network trained to this effect, aimed at minimising in ϕ a two-term objective function

where the first expectation is taken under the true model and the second one under the generative model.

“The discriminator tries to best distinguish samples away from the generator. The generator tries to produce samples that are indistinguishable by the discriminator.” Edward

One ABC perception of this technique is that the confusion rate

is a form of distance between the data and the generative model. Which expectation can be approximated by repeated simulations from this generative model. Which suggests an extension from the optimisation approach to a ABCyesian version by selecting the smallest distances across a range of θ‘s simulated from the prior.

This notion relates to solution using classification tools as density ratio estimation, connecting for instance to Gutmann and Hyvärinen (2012). And ultimately with Geyer’s 1992 normalising constant estimator.

Another link between ABC and networks also came out during that trip. Proposed by Bishop (1994), mixture density networks (MDN) are mixture representations of the posterior [with component parameters functions of the data] trained on the prior predictive through a neural network. These MDNs can be trained on the ABC learning table [based on a specific if redundant choice of summary statistics] and used as substitutes to the posterior distribution, which brings an interesting alternative to Simon Wood’s synthetic likelihood. In a paper I missed Papamakarios and Murray suggest replacing regular ABC with this version…

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986 Views. The car is travelling at a speed of 66 km/hour. If each wheel rotates 400 times per minute, then the speed of the car (in km/hr) is (take p=22/7) Find. Radius (r) of the wheel of the car = 40 cm. Click hereðto get an answer to your question ï¸ The wheel of a car are of a diameter 80 cm each. areas related to circles; class-10; Share It On Facebook Twitter Email. View All. therefore distance covered by wheel in 10minutes (10 * 60) = 66*1000*100*10*60/3600 = 11* 10 5 cm. In the case of a size like 4.80-12, 4.80 is the section width of the tire in inches. Find the area of the design, https://www.zigya.com/share/TUFFTjEwMDUwOTE0. What is the number of complete revolution each wheel makes in 10 minutes?a)4275b)4350c)4375d)4450Correct answer is option 'C'. 89.0k VIEWS. The wheels of a car are of diameter 80 cm each. The wheels of a car are of diameter 80 cm each. Then, We have,r = Radius of the region representing Gold score = 10.5 cmâ´ r1Â = Radius of the region representing Gold and Red scoring areas = (10.5 + 10.5) cm = 21 cm = 2r cmr2Â = Radius of the region representing Gold, Red and Blue scoring areas = (21 + 10.5) cm = 31.5 cm = 3r cmr3Â = Radius of the region representing Gold, Red, Blue and Black scoring areas = (31.5 + 10.5) cm = 42 cm = 4r cmr4Â = Radius of the region representing Gold, Red, Blue, Black and white scoring areas = (42 + 10.5) cm = 52.5 cm = 5r cmNow, A, = Area of the region representing Gold scoring area, Const. Question bank for Defence. Circumference of the wheel of car = â´ Distance covered in one rotation = 220 cmâ´ Distance covered in 1 minute = (400 x 220) cmâ´ Distance covered in 1 hour = (400 x 220 x 60) cm â´ Speed of car = 52.8 kmph How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? Speed of car = 66 km/hour. A wheel has a diameter of 16 cm. Let OACBO be the given sector and AOB is the triangle. The wheel of a car are of diameter 80 cm each.how many complete revolution does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hours Report ; Posted by Priyanka Rani 6 months ago. Given diameter d = 80 cm Circumference of the circle c = pi*d = pi*80 = 80pi cm Speed of the car = 66 km//hr = 66 * 10^5/60 cm/min = 11*10^4 = 110000 cm/min speed = distance/time distance = speed * time = 110000*10 = 1100000 Number of complete revolutions made by wheel = distance/circumference = (1100000)/(80pi) =(1100000)/80*7/22 = 10000/8*7/2 =70000/16= â¦ 12.28). 2020 Zigya Technology Labs Pvt. speed of car = 66 km/hr = 66 * 1000 /3600 m/s = 66*1000*100/3600 cm/s. C= Answer by ikleyn(35177) (Show Source): You can put this solution on YOUR website!. is done on EduRev Study Group by Defence Students. You can study other questions, MCQs, videos and tests for Defence on EduRev and even discuss your questions like 7th. The diameter of a wheel of a car is 63 cm. Round the answer to the nearest integer. Then. agree to the. 502.4 cm B. The car is travelling at a speed of 66 km/hour. The wheels of a car are of diameter 80 cm each. A horse tied to a corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. DIAMETER OF WHEEL = 80 cm ; therefore radius = 40cm. community of Defence. 251.2 cm C. 100.48 cm D. 80 cm - 1109399 The wheels of a car is of diameter 80 cm each. Therefore, the increase in grazing area= (78.57 - 19.64) m2= 58.93 m2. Diameter of the wheel of the car = 80 cm. areas related to circles class-7 Each wheel of the car makes 770 revolutions per minute. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? Find the area of each of the five scoring regions. Can you explain this answer? Related Video. What is the number of complete revolution each wheel makes in 10 minutes?a)4275b)4350c)4375d)4450Correct answer is option 'C'. A car has wheels which are 80 cm in diameter. Can you explain this answer? Can you explain this answer? Correct answer is option 'C'. To the nearest tenth of an inch, what is the diameter of each car wheelâ¦ 6 600 000 / 6 pi. The diameter of each wheel of car is 70 cm, If each wheel rotates 400 times per minute, then the speed of the car (in km/hr) if$$\Large \left(Take\ \pi=\frac{22}{7}\right)$$ 0 votes . Switch; Flag; Bookmark; A horse tied to a corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 4. How far will the car move forward with one revolution of the wheel? Find the distance travelled by the car during the period, the wheel makes 1000 revolutions. 80 revolutions = 1 100 000 / 3.14159*80 revolutions = 4376.76 revolutions = 4376 full solutions (i) Â Â Â The area of that part of the field in which the horse can graze. The wheels of a car are of diameter 80 cm each. Ltd. Download books and chapters from book store. therefore distance covered by wheel in 1 sec = 66*1000*100/3600 cm. The final number indicates the wheel diameter, similar to the more common tire size notation. What is the number of complete revolution each wheel makes in 10 minutes? This discussion on The wheels of a car are of diameter 80 cm each. So, to travel 110cm, it's going to take 110/70pi revolutions, or roughly 1/2 revolution. soon. The wheels of a car are of diameter 80 cm each. Solution Each Wheel of a Car is of Diameter 80 Cm. © 12.28. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? the wheels of a car are of diameter 80cm each how many revolutions does each wheel will make in 10minutes when the car is at a speed of 66km per hour - Mathematics - TopperLearning.com | ortdsu99 The area of an equilateral triangle ABC is 17320.5 cm2 . Example: 6.5x15 Explanation: 6.5 is the wheel width x 15 is the wheel diameter Notes: The smaller size of the two is normally the width, with the larger being the diameter Wheel width is referred to in inches. Can you explain this answer? Perimeter and Area. Find the area of the shaded region (UseÂ Â = 3.14 andÂ 1.73205)Fig. How Many Completer Revolutions Does Each Wheel Make in 10 Minutes When the Car is Travelling at a Speed of 66 Km per Hour? If the first number of the size is 3 or 4 digits separated by a decimal point, â¦ Let the side of the equilateral triangle be 'a' cm, then. Fig. (ii) Â Â Â The increase in the grazing area if the rope were 10 m long instead of 5m. The car is travelling at a speed of 66 km/hour. Maths. The wheel of a car are of diameter 80 cm each.how many complete revolution does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? Can you explain this answer? Find the distance travelled by the car during the period, the wheel makes 1000 revolutions. The width of the wheel will dictate the tyre sizes that will fit on the wheel. Here, we haver = 12 cm and Ó© = 120Â°Let OACBO be the given sector and AOB is the triangle. Circumference of wheel = 2Ïr = 2Ï (40) = 80Ï cm. 12.3, depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. 12.11). 10 minutes if the car is travelling at speed of 66 km per hour. over here on EduRev! Approximately how far does it travel in 5 revolutions? Find the area of the corresponding segment of the circle. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. Question 1063766: A certain car tire is 80.4 cm in diameter. The wheels of a car are of diameter 80 cm each. The wheels of a car are of diameter 80 cm each. Wheel Width. 1 Answer. The car is travelling at a speed of 66 km/hour. How many complete revolutions does each wheel make in . The wheels of a car are of diameter 80 cm each. If the diameter is 70cm, the circumference is 70cm(pi) = 70pi cm, or roughly 217cm. Answers of The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 30 mins when the car is moving at the speed of - 15848516 Solution for The wheel of the car are of diameter 80 cm each .how many coplete revolutions does each wheel make in 10 minutes when the car is travelling at aâ¦ How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour ? 131.0k LIKES. AutoCAD 3D Sphere Command Tutorial Complete | Half Sphere; Diameter; Tan Tan Radius; 3 Points. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour? Vedant Agarwal moved 376) The wheels of a car are of diameter 80 cm each. What is the number of complete revolution each wheel makes in 10 minutes?a)4275b)4350c)4375d)4450Correct answer is option 'C'. are solved by group of students and teacher of Defence, which is also the largest student The diameter of a wheel of a car is 63 cm. Download the PDF Question Papers Free for off line practice and view the Solutions online. Sep 22,2020 - The wheels of a car are of diameter 80 cm each. A car is traveling at a speed of 50 miles per hour. The Questions and = 3.14*80.4 cm. Therefore, we have(i) Â Â Â â A = Ó© = 90Â°, r1Â = 5 cmNow,Area which can be grazed by the horse, (ii) We have, r2= 10 m, thenArea which can be grazed by the horse. How many complete revolutions does each wheel make in 10 minutes when a car is travelling at the speed of 66 km per hour? Click hereðto get an answer to your question ï¸ Each wheel of a car is of diameter 80 cm .How many complete revolutions does each wheel make in 10 minutes when the car â¦ The diameter of each wheel of a car is 70 cm. Concept: Circumference of a â¦ | EduRev Defence Question is disucussed on EduRev Study Group by 129 Defence Students. What is the number of complete revolution each wheel makes in 10 minutes?a)4275b)4350c)4375d)4450Correct answer is option 'C'. Apart from being the largest Defence community, EduRev has the largest solved Let the number of revolutions of the wheel of the car be n. distance covered in one revolution =circumference of wheel = 2 * 22/7 * 40 cm Distance travelled by the car in 10 minutes = 110000 × 10 = 1100000 cm. A 70-cm-diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4.0 s. Determine (a) its angular acceleration, and (b) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating. 12.11). Actions. Draw ODâ¥BC and join OB and OCIn âBOD and âCODOB = OC Â Â Â (radii)OD = OD Â Â Â (common)and Â Â Â â ODB = â ODC = 90Â° 802 Views, 232, Block C-3, Janakpuri, New Delhi, A. In this video, we have solved the question "The wheels of a car are of diameter 80 cm each. The car is travelling at a speed of 66 km/hour. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig 12.24. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? pls help. 89.0k SHARES. EduRev is a knowledge-sharing community that depends on everyone being able to pitch in when they know something. 376) The wheels of a car are of diameter 80 cm each. Concept: Circle - Circumference of a Circle. By continuing, I agree that I am at least 13 years old and have read and Perimeter and Area of a Circle. Report ; Posted by Raj Kumar 2 years, 9 months ago. how to do the following question the wheels of a car are of diameter 80 cm eachhow many complete revolutions does each wheel make in 10 minutes when t - Mathematics - TopperLearning.com | â¦ If the answer is not available please wait for a while and a community member will probably answer this Delhi - 110058. 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# University of Hertfordshire

## Constraints on the circumstellar dust around KIC 8462852

Research output: Contribution to journalLetterpeer-review

### Documents

• slw008

Final published version, 1.44 MB, PDF document

• M.A. Thompson
• Peter Scicluna
• F. Kemper
• James Geach
• M. M. Dunham
• Oscar Morata
• S. Ertel
• Paul T. P. Ho
• J. T. Dempsey
• I.M. Coulson
• G. Petitpas
• L. Kristensen
Original language English L39-L43 5 Monthly Notices of the Royal Astronomical Society 458 18 Feb 2016 https://doi.org/10.1093/mnrasl/slw008 Published - 1 May 2016

### Abstract

We present millimetre (SMA) and sub-millimetre (SCUBA-2) continuum observations of the peculiar star KIC 8462852 which displayed several deep and aperiodic dips in brightness during the Kepler mission. Our observations are approximately confusion-limited at 850 $\mu$m and are the deepest millimetre and sub-millimetre photometry of the star that has yet been carried out. No significant emission is detected towards KIC 8462852. We determine upper limits for dust between a few 10$^{-6}$ M$_{\oplus}$ and 10$^{-3}$ M$_{\oplus}$ for regions identified as the most likely to host occluding dust clumps and a total overall dust budget of $<$7.7 M$_{\oplus}$ within a radius of 200 AU. Such low limits for the inner system make the catastrophic planetary disruption hypothesis unlikely. Integrating over the Kepler lightcurve we determine that at least 10$^{-9}$ M$_{\oplus}$ of dust is required to cause the observed Q16 dip. This is consistent with the currently most favoured cometary breakup hypothesis, but nevertheless implies the complete breakup of $\sim$ 30 Comet 1/P Halley type objects. Finally, in the wide SCUBA-2 field-of-view we identify another candidate debris disc system that is potentially the largest yet discovered.

### Notes

© The Authors 2016. Published by Oxford University Press on behalf of The Royal Astronomical Society. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.

• ## Hertfordshire Astronomy 2015-18

Project: Research

• ## UH Centre for Astrophysics Research PATT-linked grant

Project: Research

ID: 9747906


text:
CAT 2020 Quant Question [Slot 3] with Solution 14

Question

The area, in sq. units, enclosed by the lines $x = 2,y = |x - 2| + 4$, the X -axis and the Y -axis is equal to

1. 8
2. 12
3. 10
4. 6
Option: 3
Solution:

The line $y = |x - 2| + 4$ intersects the $y$ -axis at (0,6) and intersects $x = 2$ at (2,4)

The other vertices are (0,0) and (2,0)

The figure formed is a trapezium of parallel sides 6 and 4 and the distance between the parallel sides is 2.

Required answer $= \frac{1}{2} \times 2 \times (6 + 4) = 10$

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