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(e) As a result of a FMECA analysis; a fault tree analysis (FTA; Figure 1) was commissioned t0 examine the reliability of a design of a new back-up coolant system f...

Question

(e) As a result of a FMECA analysis; a fault tree analysis (FTA; Figure 1) was commissioned t0 examine the reliability of a design of a new back-up coolant system for the plant: The FTA top event failure (EVENT 1) was therefore "loss of cooling water supply t0 the plant" , in the presence of a mains electrical failure (EVENT 2) which occurs with an annual frequency (w) of 2.What are the probabilities of each GATE and the frequency of the Top Event using the probability data in Table 1?

(e) As a result of a FMECA analysis; a fault tree analysis (FTA; Figure 1) was commissioned t0 examine the reliability of a design of a new back-up coolant system for the plant: The FTA top event failure (EVENT 1) was therefore "loss of cooling water supply t0 the plant" , in the presence of a mains electrical failure (EVENT 2) which occurs with an annual frequency (w) of 2. What are the probabilities of each GATE and the frequency of the Top Event using the probability data in Table 1? Express your answers using scientific notation t0 3 decimal places and show your calculations (you do not need the FTA diagram in your answer/booklet). [30%] () Focussing on individual failures in the back-Up system; what criticality was revealed by Ihe FTA? Suggest mitigation actions t0 improve reliability citing any risk management principles you use t0 aid your decision making? H[159]



Answers

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50$\%$ of all
such batches contain no defective components, 30$\%$ contain one defective component, and 20$\%$ of contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with $0,1,$ and 2 defective components being in the batch under each of the following conditions?
(a) Neither tested component is defective.
(b) One of the two tested components is defective.
[Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

To solve this complicated word problem that we're faced with. We ought first to translate the given information of the world problem into mathematical language and then paused to translate the question itself into mathematical language. And we can go from there to figure out which mathematical formula or serums will be most useful for solving the problem. So you know, we were faced with here is a manufacturer whose producing widgets and shipping them in bundles of 10 half the time Those bundles of tender, flawless have zero broken widgets we're gonna do note that state is a zero and say to the probability of the state A zero is 50% half the time our bundles are flawless. We also know that 30% of the time each bundle of 10 will have a single broken budgets. So we'll call that a one for one Brooke and Ridge. It's probability of a one point free or 30% and then the remaining 20%. What they find is the two widgets in the bundle are broken. Problem also recommends that we set up a tree at least the first generation tree showing these probabilities that I've done for you right here. I've also expanded that to a second generation because we do have the information given in the question to define these conditional probabilities in the second generation of the tree. These, of course, the probabilities. Given that you've said selected from a particular bundle, one of the probability is you will find a certain number of broken widgets. The answer to these probabilities will depend on the final observation to make from the problem. Before we define the actual question were being asked. And that is that we have a quality control officer with sampling from each from a random bundle of widgets and determining how many are broken from that sample. Now this quality control officers selecting two regions at a time, which makes our lives just a little bit more complicated. Since there are two which is being selected, we have three possible states for B B for broken. We might find that zero of those widgets are broken. One of them is broken or two of them were broken. Second, consider, because of these nine conditional probabilities for any given state of the Ben itself, a quality control officer might find three different three different levels of broken rigidness. Now what are we actually being asked to find? Because our next bit here would be a relatively complicated bit of math. Let's make sure we need to discover that before we move on. But the question is actually asking us for is to suppose that our quality control officer doesn't know which been she's selecting from, and she selects her to widgets. She tests them and finds out that none of them are broken. Based on that information, we need to derive the expectation of which Benji most likely took things from. What are the chances that when she finds zero broken, she took them from a zero? Been in a one been or eight to bid, we can, intuitively saying when none of them were broken. More likely that she's picking from in a zero gun for several reasons. We'll come up with the actual percentage she's also found, uh, it's occasionally one of them is broken. So when she finds herself in a state of B one, we need to derive what the likelihood has shed. Selected for a zero a one or a two in mathematical language that looks like this over here. Given b zero, what are the chances of a zero? You can be zero. What a the chances of a one, etcetera. So we're looking for six different answers now. They haven't asked us to find, um, solutions for B two when she pulls out two objects and they're both broken. This may seem at first like they're cutting you some slack, but in fact, this is just eliminating the most trivial possibility s. So I would suggest that you take just a moment. Don't even do any math. Just think through in your head what would be the answer if b two were the descriptive state given be to what? The problem is, a zero a one a two that's Matthew could probably do in your head or indeed a solution you could find before taking a probability schools. All right, so we're looking for these probabilities, and that calls to mind our basic rule of probability of conditional probability, which is the situation that you can find in your textbook problem with. This, of course, is that we don't have a number for B of N and computing. This, uh, this intersection here actually relatively complicated s so what we need to do next is transformed this. Using the role of multiplication into base there based thorium is gonna be exactly what we're looking for. To get a solution of this, and the bass tells us is that the conditional probability of a given be is the conditional probably would be given a times in our ability of a divided by the sum of all conditional probabilities of be given the different states of a times, all the probabilities of different states of a. Now, the good thing about this is we actually now can, uh, come up with solutions for each of these component parts. This looks like a big, bewildering, unending, some at fraction. But if we think about it bit by bets, we can come up with a much more reasonable solution. And we can just grind it out step by step. So we start with the ah, top of attraction here. He of a we know all of our piece of a business there indicated here ah p of be given a That's what we're going to cinsault for here. So I need to come up with those numbers we don't need to find the ones for B two. We'll talk in a second about why we're going to do that anyway. And in the bottom part of our fraction here is just the son of all of the possible top parts. So if we can find the probabilities of the different sets of be given the different states of a, we already know that it probably is the difference. It's a week unplugs all into the equation on come up with a satisfying solution. So let's find these conditional probabilities and be given the different states. A. We're gonna start here with a zero because a zero is really the easiest of all of them to handle. If you take two widgets out of a box of 10 unbroken widgets, what are the chances that both are unbroken? Course? Common sense will tell us. That's 100% of the time. There aren't any broken bridges in here, so it's impossible to find any broken widgets. They don't exist. S o P B zero of a zero is one remaining. Probably lose for the difference. It would be we're all zero. All right, now, once we have actual broken widgets and Arbenz. This becomes a little more difficult. So we're selecting never a one. What are the chances the zero of my widgets have a broken? When I take two of here, we can use the rule of multiplication. If I take the first widgets, there's one broken widget in there. What are the chances that I don't pick that one? Well, there are nine unbroken, Which is that a pen. So the chances that a night dance now our next state, which is dependent on this first state we found no broken, which is the first time. What are the chances with Second, which it is also not broken? Well, there are only nine, which it's left. Eight of them are unbroken. So there are eight out of nine possibilities for that to happen and then weaken grain. This through reduce our fraction. We find that this is 8/10 or you're just destruction down here to ah, aches, tents or for or 0.8. If you look, either one is fine, of course, depends on what your instructor prefers. How you want express these actions now be one is actually a slightly tricky one to compute. Let's compute B two, given a one, even though they don't ask you this, this very easy thing to compute and can actually make it much simpler to solve for B one, given it one. The tents, of course, of B two. Given a 10 there's only one broken thing in the den, so there's no way I could ever pull up to Bergen Things. And because we know that B zero, B one and B to exhaust all possible outcomes, we know that the probably be one given a one has to add up one. When we added to the problem of the zero give name. What is this? Has a 2.2. You can skip all the fancy mathematics now the last one is the most complicated of them. The probability of a of B 01 or two, given a two, has a bit of a sidekick here. If you're really more comfortable with competent works, you can use combinatorics. Resolve this problem as long as you're careful. So let's solve first for B zero, I have two Broken ridge. It's in the bin, and I'm pulling to I'm too broken, which has been I'm pulling to a random out. What are the chances that the first of those two is unbroken there, eight unbroken ones in there. That's eight out of 10 and again conditional probabilities and multiply role here when I reach in for my second one, there are seven on Brooklyn's left and pulling out, uh, seven out of mine, Sebastian's of being what we look at their will solve for that fraction, a swell. We got 28 45th. So the 45 eyes that is a call back there to that combinatorics principle is all this comment works. The significance of 45 will be meaningful to you. Let's skip PB one again. Because again, that's the more complicated of the three states. What about probably of be too well when I reach in and I grabbed my first with its cancers of that one being broken. There are two broken Richards at a pen, said Stuart, him reach back into my second one. Now there are only nine wickets left, and only one of them is broken. So it's a 1/9 chance and by and that's to get one out of 45 is Ah, that's my That's my probably have grabbing two broken wickets, having one every 45 times on. And I could do a little bit of math here, 16 45th remaining of that middle at a great for beat one. What remains here is just do a whole bunch of math plug in. All of the different numbers that we need for this for the first three they were looking for appear for these first three answers. We have the same denominator for our Bayes theorem equation so we can calculate that real quick and what we do to do that is we take all of the second generation branches associated with B zero and multiply them by their first generation matches, and we add up those totals. This is going to be probability of B zero. So it actually is the denominator of this original equation right here on this is the trick of bays law. Basically, it's a lot of fancy mathematics and hand waving, but really, the finding the denominator is just about finding the overall probability of the state that you're looking for other the conditional state. What are the chances of P zero, which is the condition of our very problem here. So these air, we're gonna take the Beezer of state of the second generation That's going one multiplied by the first generation associated with it. He has a 0.5 second generation for B zero and a one is 0.8. And we're going to multiply that by the first generation X 0.3 and then finally third generation 28 45th. So I didn't want Teoh reduce that fraction because we will actually lose accuracy if we do that. But playing that by a point to if you're more comfortable, you could, of course, convert this all to either Approximate decimals 45 here is gonna make that approximate It's got nine involved there. Or you can convert everything fractions on my all straight. Both solutions. Here you grind to the mathematics. You find that the reduction distraction is 389 45th or a little over 86% of the time, and that that's our overall chances of finding a state of B zero on that makes up. Half of the batches have no broken things, and even the broken gadgets are mostly unbroken. Let's go and find ah p of B one as well, because that's gonna be our denominator. For these second three equations. The second few responses. The topic, of course, RB one state has a zero probability. So we don't need to worry about a zero in this case. Well illustrates today. One probability of B one given a one is to we multiply that by its first generation, which is 0.3. And then we find the same information for the state. A two, which was 16 year 16 45th. I am too. And again you can express. This is a fraction. Where you can express it is a repeating decimal, and what you should come up with is these two answers, which are essentially equivalent 59 450 deaths or a 0.1311 Now again, what this is these are the sums that we're getting at the bottom of days there. We're talking to them affront because we're going to need them for all three of these answers. And if we calculate them up front, we reduce the risk of forgetting what they are forgetting a calculated them and therefore reinventing the wheel over and over again to find out three individual responses. Then we just take this final answer here and we multiple. We divide, um, are given numerator by this answer here. Of course, the numerator we're going to need are the three entries in, uh, our discovery of the probably of easier and be one. So the probably of a zero given b zero is going to be one times 10.5, which is, of course, one half. I did buy this some which we found out to be. 389 450 deaths. Eso Why did thereby which was absolutely beautiful. Ah, really gnarly bit of fraction work to be to be got over there you should find is 225 389th, which is approximately equal 0.578 So a little less than 60% of the time. If you find no broken ridge, it's probably drawing from been a from one of the easier events I would go through into the same mathematics for the next ones. That's what you get is he's really just quite horrendous fractions, which I've taken the liberty here producing to approximate decibels. So a little over a quart of the time. If you have nothing broken, you've picked from a one bin, uh, little more than an eighth of the time you a picture of a tube in, uh so in this follows basically where we would have expected the proportion of a zero is a little bit ballooned because there are more a zero bins and there are fewer broken widgets. They're in on important final thing to adhere to check you work. These three numbers should add up to one. It was a little bit of a fraction error here because we're changing the decimals. As long as you're extremely close, you should feel confident If you're gonna take the time to compute these horrifying fractions into ah fraction answer, you should find an exact ah arrival at one problem. Because, of course, there are only three possibilities for a zero a 12 So these are all these partition the possible and then finally have grinded through the last couple of fractions here for you to find solutions for the problems of a zero Anyone, any to given the one that we've got one broken object? Of course, there should be zero for what are the chances of selecting from a zero when you've got one broken object? Well, there's nothing. There are new broken objects. In interesting here, though, is that even though their arm or a one bins because there are more broken objects in a two, it is actually slightly more likely when you've picked a single broken object that you're selecting from, uh, the A to been okay, s. So this is our final answer right over here on. I hope you enjoy this.

Okay, so um we have uh we can pick a batch Um of items of 10 items and 50%. So 0.5 Contain zero defective components and 0.3 contain one defective components. So I'm using the hint and drawing a tree diagram and point to contain two defective components. That's your percent chance. Now, once you've picked a batch, let's figure out how many If I pick two items from the batch. well uh there's uh here there's a 100% chance of Um neither being defective, 0% chance of one being defective and 0% chance of to being defective. Here. Let's see um there's one defective item. So how do I can't pick two defective items from here? Um There's a let's see, nine choose to Over 10, choose to chance that I picked no defective items from this batch with one defective item. Okay, so that's like nine times eight over Um 10 times nine because the two factorial would cancel. So that's .8 0.8 percent chance that I get no defective items, Which means there's a .2% chance that I get the one defective item. Okay, and over here, let's see there's an eight shoes too Over 10 choose to chance that I get no defective items. eight times 7/10 times nine is 28/45, 28/45. Chance of getting no defective items. Um What about getting both defective items? That's 1/10, choose to write There's only one pair. Okay, so then that's 1/45. Okay, so a total of 29/45, Which means the remaining 16/25. Over 45 is my chance of getting one defective items. Okay now I have all the probabilities. Now we're asking what is the probability uh if I know neither tested component is defective. So I'm looking here um let's see, my probability of ending up On that whole branch is .5 from the beginning, Probability of ending up on that whole branch is .24. And probability of ending up on that whole branch is uh let's see it 28/45 times five I guess. Um Is that to 25? I believe it is 28 Over 2 25. Okay so um that's my whole world in which I've tested nothing to be defective, neither of the two components to be effective. So I just need the part where there's zero components over this whole world. Okay so p of zero tested. Being defective Is 0.5 plus 0.24. I'm just going to raise this temporarily. Plus 28/2 25. And then what I need is .5 divided by that. So this is um over P. 00. This is the probability. Okay so that's 0.5784 Or 57 84%. This is the probability Of zero defective in batch. Given zero defective Of the two tested. Okay so I'm more than halfway. Sure there's no none defective in the batch And then this is um do the same thing with .24 over probability Of zero defective tested. Um and that's going to be approximately .2776 Or 27.76%. So this is the probability that is one defective in batch, Given zero defective tested. And then this last one over here. 2 28/2 25 divided by probability of zero or one minus the other two things, if you like, the age Over 2 25 over All. Right, so this is .1 for 40. I'm going around to Which is 14.40%. And then this is the probability of two defective in batch, so maybe we want to test more than two things um because even if we tested two things and neither were defective, there's still a decent chance that two were defective in the batch. Okay, let's do this all one more time. So part B1 of the two components is defective. So instead of these red ones, we're going to go now on the highlight of branches. Okay, So these probabilities are zero 0.0 six, is that right? Yes. And then um 16/2 25. Okay, so then the probability of one thing um of the two being tested defective is going to be point there was six plus 16/2 25 is um there And now um So okay probability of zero defective in batch given one defective tested is zero because Um because of that zero and that makes sense. If you tested two things, one was defective um You don't have a completely non defective batch. Okay, probability of one defective in batch given one defective tested is going to be a .06 over this probability, make sure I have the right parentheses in my calculator as you always should. Um Okay that is going to be .4576. Okay, so almost a half chance of just one thing being defective in the batch and then probability of two defective in batch given one defective tested. Hopefully it's about 54%. But we'll do it um 16 over to 25 over this probability of one. Let's see, yep, I'm getting approximately 542, 4 Or 54.24%. All right, so um you might want to test a little bit more for more information but this certainly gives us some

Okay, so that's what P. But a probability of a flaw being detected equal. Reports a were asked the probability of it being detected by the second fixation. That's better. It being detected on the first or not being detected on the first and being detected on the seconds that's is equal to P plus not being detected is one night. It's p times it being detected on a second. Okay, now we're asked report, Be it being detected on and fixation. It is equal to one, minus it not being detected. End times. No, I asked where the whole body of a flow item past persons inspection after Drea fixes that's equal to one witness. Be to part three. It's because it wasn't infected three times. And the property. If our random item, it's flawed probability of a random Oh, I don't applaud his 00.1. What is it? The probability of it passing after doing inspections that's the same as war minus P. There are three times this rate. Three times never are being problem ot of the flaw passing, given that P is equal support. Five. So that's one when it's fortified to Part three, which is able to 0.1 to fight

Okay, so, mhm. In this problem we need to decide whether or not by no mill model is reasonable. Okay. For a binomial model to make sense. We're looking for fixed number of independent for nearly trials, each with the same probability of success. And I'm going to use um when you use and capital and to refer to a fixed number and I'm going to use P to refer to that equal probability of success. Okay, let's look at the first scenario and see if we can matchup what's going on with the pieces we need for a binomial model. Right? So we've got a process that makes thousands of transducers and we want to use X to represent the number Uh huh. Non conforming Out of a sample of 30. Okay, because we're producing thousands of components that's going to let us use ah independence. So there was such a big number that the there is some dependence. So if we if we pick one that influences the chance that the others that we picked in the several 30 are not conforming, but because the pool that we're picking from, so big dependence is diluted, so it's really small. So it's reasonable to model These as independents. So, these 30 transducers that were picking a chance that any one of them is non conforming is basically the same. No matter what other transmitters we've already picked because we're picking on such a big pool. So the Bernoulli trial is non conforming or conforming, The fixed number is 30. We have 30. Bernoulli trials were checking 30 transducers whether they're nonconforming conforming and the equal probability of success as I was saying is legitimate because we're picking or such a big pool and that value p um we don't know what it is, but if we could check every single transducer out of this huge huge pool, um it would be the number non conforming out of the whole population of transducers. Okay, so by not only model makes sense for this one, we have fixed number of trials, 30, they're independent because we're picking them out of a really big pool and there's a constant probability of success. Again, we can model that probably success to constant because we're picking from really big pool and I'm emphasizing that so much because yeah, it's going to be the deciding factor in the next scenario, which is very similar again, X is going to represent the number of non conforming transducers Out of a sample of 30, but this time Instead of pulling from a huge pool of transducers were only pulling from a pool of 50 and we're um and we're sampling without replacement. So that means after we pick a transducer, We only have 49 left in the pool, we pick the next one, we don't put it back and shuffle them up, pick again. So maybe we could pick that same one again, that's not what we're doing, we're taking one, they're only 49 left, taking another only 48 left and so on Until we've picked 30, so that means that The Burnett We still have Bernoulli trials each time we pick one we're going to check is it conforming or nonconforming? So it's the same kind of Bernoulli trial, nonconforming or conforming but they're not independent because if we pick a non conforming transducer the first time we've just reduced the number of nonconforming transducers left. Yeah. So the probability that the next one we pick is non conforming is different than if the version we picked had not been nonconforming. So the probability that the next one is non conforming depends on whether the first thing we picked was not conforming. So these are not independent, they're not going to have equal probability of success. So by knowing model is not appropriate for the second scenario because we have a very small pool that we're picking from. It's no longer reasonable to model the 30 that we're grabbing as independent. Okay, let's go to the next one. We've got four components, they're all hooked up to the same controller and over the period of a month we want to know how many of those are going to fail. So when one fails, never fails out of four in a month. So when one fails, the controller switches to another component. Okay, so this one's a little bit ambiguous, we need a little bit more information. Um I'm going to assume that the controller is only using one component at a time. So it uses the first one. If the first one doesn't break for the whole month, Great, it only uses one if the first one breaks, it switches the second one and only then does it start using a second one at all? So each component, he's only at risk of failing? Well it's being used, so that means the second one can only fail if the first one failed, that means these aren't independent. Um and uh there's a non constant probability of failure across the four trials, because with each successive component there's less time left to fail. So binomial model is not appropriate for this data, Let's go to the next one for this fourth scenario, when you use X to represent number of crashes on some stretch of highway in a month. Okay, so this one is not a good fit for binomial model for a few reasons. First of all, what are our Bernoulli trials here? Um We could make something up, that would kind of work, we could say for each car that drives down the highway in this month, it either uh crashes or does not crash. Mhm Okay, so there are two outcomes, um but then we have to figure out what's the fixed number, so we need a fixed number of independent trials, we sort of made up something that could look like a burning the trial, but we don't know how many cars are going to drive down the highway in a month, so there's no no fixed number in. Yeah there's some other problems with this for a binomial model to we don't it's probably not reasonable to assume that the probability of crashing is the same for every car and it's probably not reasonable to treat the probability of crashing is independent for all cars because some cars could crash into each other. Um If you're driving by when there's been a big accident, maybe you're more likely to crash or less likely to crash when you will drive more carefully. So there is um probably not a fixed P. And there's no independence. The new model not good, not good for this scenario, I'm going to get rid of these four and right, go on to the next set. Okay, so in this scenario we're going to use our random variable too represent number of correct questions on a multiple choice test out of some fixed total number of questions, it doesn't tell us how many there are but that's okay. We know that usually the number of questions on a testis is fixed. Uh huh. Yeah so we'll just label it and even though we don't know what it is Now for some of these questions the student taking the test is able to rule out some of the incorrect answers but not all of them. So that means that they'll have more than one remaining answers. So for some questions student is picking from uh two or more options. Okay and for some other questions, the student rules out all the incorrect answers so they know the right answer. Okay, So every question they know the answer to, Well, they'll get that right, correct. So, the probability of success for those questions, The probability of success for those questions is one. Mhm. For these other questions where they're able to rule out some of the wrong answers, but not all of them, the probability of success if they're just guessing at random out of the remaining options is one over the number of options they have left. So if they have two options left, it would be one half. Okay, so for these questions, Probability of success is one. These questions probably a success could be one half or 1/3. It might be different for different questions. So, there's not a constant probability success here. Okay, So independence makes sense, potentially fixed number of trials makes sense, but equal probably is success because of the information we have. We know that's not reasonable. So, by Newman model, not great for this. Yeah. All right, next one. Okay, so, we have some chips that were making computer chips, Yeah, that have some of which have defects The effects occur randomly, But we can't find all the defects. By testing. We can only find 80%. Take a sample Yeah, 40 chips That each have one defect. And then we test them for defects. Okay, So, we know Okay, we're going to say x Is the number that uh test positive for a defect out of 40. Okay. And we know that in reality all of them have a defect. And we also know that probability of detecting a defect Is 0.8. Okay. So we don't have any reason to think that these aren't independent. Probability of missing a defect for one ship matters. Depends on no whether or not you miss a defect in another chip. So independence we think is okay here. So check for independence. We've got a fixed number of trials here. That's 40. Check and the probability of detecting a defect provocative success is the same for each of them. It's .8. So fixed P check. Yeah, I think by norman model. Fine for this scenario. Next one. Yeah. Okay. So we've got a process where we fill detergent packages to a fixed weight. Yeah. And we're gonna use a variable to represent the number that are under filled. Right? Well, we're not told that we're testing out of a fixed set of packages. So there's there's one problem, fixed number. No, they didn't tell us that. Fixed probability. Mm. Well, we don't know very much so. Probably in our first attempt at modeling we would we would try out a model where the probability of under filling was the same for each package. Are they independent? Well, again, because we don't have a lot of information would probably try a model where they're independent until we have a reason to think they aren't. So these to fit, but we've got no fixed then. So this is probably not a binomial model by no meal. Not appropriate for this scenario. Just a couple left here. Uh huh. Yeah. All right. In this scenario we've got random variable to represent the number of errors in 100 1000 bits. Okay, Okay. And we're also told that errors tend to come in bursts. So if there's one error, if we see an error, we expect several Mourners. So say I use that uh his green dot to represent something fine. Got all these bits are fine. And then here comes an error. We expect the next one to be more likely to be another error, given that we've just seen an error. So Harris cohen bursts, We expect them to be group. So if we see one that tells us something about the next couple coming down the line. All right. So let's check for pieces here. Fixed number. Yeah, we've got a fixed number. It's 100,000. Okay, what about independence? Uh no, we do not have independence because as I was saying, um if we know the value of the last bit that tells us something about what we expect the value for the next bit to be either error or non error. Okay, there is not independence. And if there's not independence. Yeah, we can't use by newer model. So finally model does not seem appropriate for this scenario. Last one. X is a number of surface flaws in some steel steel object. Okay. Tell you that a number of flaws. Um Mhm. What's the trial? Okay. We could set up a structure that created a traffic, we could divide this object up into a grid and then we could look any square the grid and say, is there a flaw there? Yes or no? That could be a trial. So we could impose some structure to create a Bernoulli trial but as stated, there's there's no Bernoulli trial here. Um So I'm going to say no. I mean you need you need to impose a lot to make this a binomial model. Um really the only thing that it has going for it is that um the value is going to be a positive integer, so not a binomial model. Yeah. All right. And that's all of them.


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You have 7(t) Find 3 (a) Find a (1 point) 280 altempted Preview My Answers or False Pdioina: = MAA, 2 Integral ameinc Iniedra 8 8 Flz,u) V integral 0f any equauon along equation I| Juaquqie vector narabola IolMC Next Problem 8 1 lield F Blodpied 1 WpmbaS 3 1 tromine Problem 1 Ineonicin [0 the oriqin lo (3,9) L0VOM origin Io Ine 8 1 point (3, point (3,9) using t as usino [ 45 a palumelei 2 3 amez 3 0
You have 7(t) Find 3 (a) Find a (1 point) 280 altempted Preview My Answers or False Pdioina: = MAA, 2 Integral ameinc Iniedra 8 8 Flz,u) V integral 0f any equauon along equation I| Juaquqie vector narabola IolMC Next Problem 8 1 lield F Blodpied 1 WpmbaS 3 1 tromine Problem 1 Ineonicin [0 the oriqin...
5 answers
4,) Determine if the following improper integrals diverge= converge; If an integral converges, detenine its value.{z5
4,) Determine if the following improper integrals diverge= converge; If an integral converges, detenine its value. {z5...
4 answers
Find a formula for the quadratic function with the following properties: zeros at -3 and 3; passes through the point (6,-16.2). Your equation must begin with "y='This question accepts equations E.g: y-2 = 5(x-4)+1, Help Switch to Equation Editor Preview
Find a formula for the quadratic function with the following properties: zeros at -3 and 3; passes through the point (6,-16.2). Your equation must begin with "y=' This question accepts equations E.g: y-2 = 5(x-4)+1, Help Switch to Equation Editor Preview...
5 answers
Find the rational number represented by the given repeating decimal.$0.25252525 ldots$
Find the rational number represented by the given repeating decimal. $0.25252525 ldots$...
4 answers
Suppose the graph of f() on the interval [L.S] is the semi-circle below:2 3 4Evaluale [3f(2x'+1) dx Here "f('+)" means / OF 4 +[ The parentheses do notmcan multiplication. Hints: Use u-sub. remembering to #change limits* of integration. This is & short problem: like three or four lines long:
Suppose the graph of f() on the interval [L.S] is the semi-circle below: 2 3 4 Evaluale [3f(2x'+1) dx Here "f('+)" means / OF 4 +[ The parentheses do not mcan multiplication. Hints: Use u-sub. remembering to #change limits* of integration. This is & short problem: like three...
5 answers
Find the difference between the upper and lower estimates of the distance traveled at velocity f(t) = 26 _ /2 on the interval <t <4,for n 500 subdivisions_Enter the exact answer:The difference between the upper and lower estimates is
Find the difference between the upper and lower estimates of the distance traveled at velocity f(t) = 26 _ /2 on the interval <t <4,for n 500 subdivisions_ Enter the exact answer: The difference between the upper and lower estimates is...
5 answers
(5) (15 points) Let Pz be the vector space of polynomials of degree less tan equal to 2 Let B {b1,b2, bs} be the basis for Pz with b[ 2 b2 F-+I b; I+2. Find the coordinate vector [vls o v = 1 +2r with respect to
(5) (15 points) Let Pz be the vector space of polynomials of degree less tan equal to 2 Let B {b1,b2, bs} be the basis for Pz with b[ 2 b2 F-+I b; I+2. Find the coordinate vector [vls o v = 1 +2r with respect to...
5 answers
Final exam study times are normally distributed with a mean of 30 hours and a standard deviation of 4.9 hours How many study hours is it such that 98% of students are studying less than this amount for a final? (or 98% of students study fewer hours than this amount)
Final exam study times are normally distributed with a mean of 30 hours and a standard deviation of 4.9 hours How many study hours is it such that 98% of students are studying less than this amount for a final? (or 98% of students study fewer hours than this amount)...
5 answers
Point)a. Find particular solution to the nonhomogeneous differential equation y6y + 9y = e_Yphelp (formulas)b. Find the most general solution to the associated homogeneous differential equation: Use C1 and C2 In your answer to denote arbitrary constants and enter them as c1 and c2.help (formulas)Find the most general solution to the original nonhomogeneous differential equation. Use C1 and C2 in your answer to denote arbitrary constantshelp (formulas)
point) a. Find particular solution to the nonhomogeneous differential equation y 6y + 9y = e_ Yp help (formulas) b. Find the most general solution to the associated homogeneous differential equation: Use C1 and C2 In your answer to denote arbitrary constants and enter them as c1 and c2. help (formul...
5 answers
Say the position, in cm, ofan object is changing at a rate of 4 e sin(e" ) for time tin seconds_Find the general formula that describes the position of the object:
Say the position, in cm, ofan object is changing at a rate of 4 e sin(e" ) for time tin seconds_ Find the general formula that describes the position of the object:...
5 answers
Let A and B be two matrices (bamoB: 4)square matrix with Tow cf ]smust pe invertiolBis an inverse for iff AB=BA=0If B nas zero column; then s0 dces AB ifit existsIf A and are invertible then so dces A-B ifit existsIf B nas zero column; then s0 dces BA itit exists
Let A and B be two matrices (bamoB: 4) square matrix with Tow cf ]smust pe invertiol Bis an inverse for iff AB=BA=0 If B nas zero column; then s0 dces AB ifit exists If A and are invertible then so dces A-B ifit exists If B nas zero column; then s0 dces BA itit exists...
5 answers
Of substances pick the one that has the In each of the following groups given property. Justify your answer: Rank from highest surface tension to lowest for: CaHjo CaHza or CaHiz b. Rank from highest vapor pressure to lowest: Nz CO, or COz Rank from Highest boiling point to lowest: HF HCI Or HBr
of substances pick the one that has the In each of the following groups given property. Justify your answer: Rank from highest surface tension to lowest for: CaHjo CaHza or CaHiz b. Rank from highest vapor pressure to lowest: Nz CO, or COz Rank from Highest boiling point to lowest: HF HCI Or HBr...
5 answers
2 7 Questlon 3 theze W 142 1 Give reasons the 3104 1 change3xy
2 7 Questlon 3 theze W 142 1 Give reasons the 3104 1 change 3xy...

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