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D = Juk0.38 0291%0 < IDi @Time left 0.31.09...

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D = Juk0.38 0291%0 < IDi @Time left 0.31.09

d = Juk 0.38 0291%0 < IDi @ Time left 0.31.09



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Use the formula $I=P r t$. Solve for the time, $t$ for(a) $I=\$ 624, P=\$ 6,000$ $r=5.2 \%$ (b) in general

Okay, well, again, start with determining the decay constant from the half life minions of one over minutes. We can then use the first order, integrated rate mine in order to calculate the number of radioactive new Clyde's that are left after a particular time passes. Knowing we have 4.2 times 10 to 12 magnesium, 27 new Clyde's at time. Zero. That will be for the negative K, which we just determined multiplied by the time that elapsed, which was 30 minutes. And we can solve for the number of radioactive magnesium 27 new Clyde's that are left after 30 minutes doing a little bit of algebra. 4.71 times 10 to the 11 radioactive new Clyde's. Now to determine the rate at time zero and after 30 minutes rate is equal to the decay constant multiplied by the number of radioactive new clients that we have. So what time? Zero. We have 4.2 times 10 to the 12 so that gives us 3.7 tended the 11 to case per minute, Then the next rate will be for the number of radioactive new Clyde's that we have after 30 minutes, which we just calculated up here 4.71 times, 10 to the 12 and we get 3.44 times 10 to the 10 to kase per minute. Now what we're going to do to determine the probability is we could determine how maney radioactive new Clyde's decay at time zero after a second passes based upon how many radioactive new clients we have. Or we could calculate the number of radioactive new clients that decay at 30 minutes based upon its rate and the number of radioactive new clients that we have, if I'm going to do it at time zero that a case that would occur in one second The assumption that we're going to make is the rate is not gonna change. The rate is going to stay constant. So the which it doesn't. But that's what we're assuming. So we could assume then that in one second with a rate of 3.7 times 10 to the 11 decays per minute, which is what we calculated at time, zero going a second passes, 1/60 of a minute has passed, and so the rate are the number of radioactive nuclear sites that would have decayed would be the rate multiplied by the time or 5.11 times 10 to the ninth decays. Now, if that many decayed how many decayed based on how many we had? So we calculate a ratio. This money decayed when we started with this many, and we yet 1.22 times 10 the negative three. And if we take the reciprocal of that, that will tell us the ratio between the number that decayed and the number that we had, so essentially one decays out of 822.

Hello and welcome to another difference equation problem. And this one we have Y prime equals T. Y four minus Y. All over one plus teeth. Our initial condition is why of zero is equal to why not? But this why not? Is specifically positive? All right. So the first question is, what is the behavior of why? As T approaches infinity? Well, the way that we look at this is we look at the value of why here and here. So if why is positive, then we should look at what will happen to why over time. So if y starts out at a positive number, this will be positive. And this right here, the four months late term will be negative. So no matter what positive term we use, the slope will always be decreasing why prime will always be decreasing until it hits either uh Y equals zero or wife was four. So we can write this. Is that um why will approach four? If uh if it's a really large initial condition, like why not is equal to a million? Because this ah this differential equation will have a wide prime approach four. There's a positive slope here and a negative slope here. So will decrease until it's four. If it's a number in between zero and four, for example, it will Like one. it will increase until it hits for uh let's show this uh the sign of a of this being uh of why being one is positive, 4 -1 is also positive. It will increase until it hits this four minus four mark here. So rent why approaches for as T approaches infinity for part B. We have to find uh the initial condition exactly. So we're going to have to solve this difference equation. I'm going to speed through this because the point right now is not to know the integration techniques but to really analyze this problem. So we will have um do I over why Times four -Y. Is equal to uh T t times t over two plus one? All right. And then we will integrate both sides. This left side has to be uh sold with partial fractions. So you'll be the integral of do I quickly to these parts fractions? Um If you're good at partial fractions you can just do on the fly. Um Otherwise just right at the uh the A. And B do it the long life for this. This will be for Y. It's a big plus. Uh Do I On four times 4 -Y. Yeah. All right. Let's look at this other integral. Um For this there's a little trick that you can do. You can actually add one and then subtract one. So this will look like um T plus one Over 2-plus 1 -1 of her. two Plus 1. Really we just added one plus 21 over the quantity two plus one. And we subtracted that. We really didn't alter the value. You can still integrate this. This will be one. This is just a simple log rhythm. This will be T minus natural log of T. Plus one plus C. This over here will be um 1/4. I'm just going to do this quickly will be times the natural log of. Why Over four -Y. And that's equal to this. All right so now we can substitute this and for um Why not equals two. And why is equal to 3.99. Um So but first we actually have to find this value of C. I apologize for this confusing order. So we'll evaluate um T. As zero and why is why not? So we have um 1/4 L. N. Why not? All over four minus? Why not? That's equal to well T0 here will be zero minus Ln of zero. Or Ellen of one Ellen of 10.0. We have C. So C. Is this right here? Okay. And now we can do part B. The exact substitution Part B is if why not equals two Fund the time at which the solution 1st reaches 3.99. All right. So let's do that now. Um I don't think we have the full Equation right now but it's 1/4. Why not over four minus? Why apologizes not? Why not there? This is going to be equal to this t minus Ln of t plus one term plus this C term that we just found which is 1/4 Ellen of Why not? Or 4- Why not? All right. Now we have that we can um We can evaluate the why not? At two we can find why at 3.99. So of 1 4th times Ellen of 3.99 over 0.001. This is going to be a lot of 399 as you call to. T well we don't know what T is. That's what we're trying to find. Two minus Ln of two plus one plus 1/4 times its value of. Well well It will be a line of two or 2. So this will just be Ellen of 10. So we can cut out that term. Great. And now we have to um we have to figure out what T. S and the best way to do this would be um to raise everything to an exponent. All right. So we're going to have um eat the to this and E. To this. All right. And I'm going to actually skip the rest of this derivation because it is a it's quite difficult to get an exact value. I recommend plotting this on a graphing calculator in finding the results, but in the end t will Uh resolved to around 2.84. Alright. Now under part C. To do part C. We have to plug in um 3.99 for Y. And T. S. Two. Doing that with our equation up here. Um Well results in just a lot of uh annoying calculations but I'm going to write out the form uh for you We'll have 1 4th Ln of Uh well if we're gonna YB 3.99 We'll get 3.99. All over four 3.99. Which is your .01. And this is going to be equal to ah 2 -6 of three. Okay. Plus 1 4th. This uh strange term at the end 1 4th Ln of why not over 4-? Why not? All right. We're going to do the same process for For y equals 4.01 And solve for this. Why not term? It's going to be a lot of really ugly calculation. And right now, I can tell you if we're doing this, we're going to subject the two at the Ellen three multiplied by four. Take the uh exponent of all this and we're gonna multiply this four minutes. Why not turn over? And then I factor that along with the why not term here. Move it all back over and sulfur. Why not? It's really gross. Um And that's not really the point of this problem. But to realize that To get these value end points, we have to substitute 4.01 in and 3.99 in here to get the range of T. Values Um of 3.66, approximately To 4.44. Yeah. Um And that concludes this problem.

Hello students in this question we have given that if only one atom of atoms of radioactive element radio active element is left left then this item will disintegrate this integrated after this integrate after. So now we can see here that only one atom of radioactive element is lead. Okay. The rest of the atoms has been uh has been decayed also or disintegrate also. So we know that the element or the element has a smallest thing that is atom. Okay. And this atom cannot be break in sub atoms by radioactive theory. Okay. So hence we can say that this Acton will disintegrate in a time which cannot be specified. Okay, cannot be specified. Or we can say that this will not disintegrate at all. Or in the other time we can say that it will take in finite time. Okay so we can say from the given options that is options. See is the answer of this problem. Okay so this one atom is left so this will not be this will not disintegrate after that. So this time cannot be specified. Okay thank you.

Here we are told that the duration for some process follows a three parameter Waibel distribution with parameter gamma or the shift equal to 3.5. The shape parameter alfa is equal to two, and the scale parameter beta is equal to 1.5 for part. A were asked to find the cumulative distribution function for X X being the duration of the process. Now this is similar to the CDF for the two parameter Waibel distribution, except instead of X. We will substitute in X minus gamma, so it comes out as follows and this holds for X is greater than or equal to zero. It's zero for X is less than zero. Now, if we substitute in our parameter values, we have. So that is our a cumulative distribution function now for Part B were asked to find the expected value of X and the variance of X. Now, this three parameter Waibel distribution is just the regular what two parameter Waibel distribution, except shifted to the right by a 3.5. So let's first find the expected Valley for the regular two parameter Waibel distribution. With these parameters using the formula, it looks like this That comes out to 3/4 times the square root of pi that's using the effect that gamma of three over to is equal to half of gamma of half it was half times the square root of pi. And now to find the expectation on X, which is thes e three parameter Waibel distribution given to us in the problem, we have X is equal to why plus 3.5 That's due to this shift in the distribution to the right due to the linearity of expectation is equal to the expected value of why plus 3.5 that comes out to 4.8 to 9. Now, to find the variance of excellence first, find the variance of why which is the two parameter Waibel random variable. Remember that are variants. Formula is as follows So substituting in our parameter values here I've used the fact that this is equal to tu minus one factorial or one factorial and this is equal to this here. Now this comes out to 0.483 now to solve for the variance of X now following the scaling properties of variance, this is equal to simply the variance of why again, that's 0.483 So that answer is part A and part B. Now, see, we're asked for the probability that X is greater than five. Now we've found the CDF for X in part A. So we just have to substitute in the number five for the variable X in that in that function and we get the following this comes out to zero point 368 and last for part D were. As for the probability that X is between five and eight. So first solving for the cumulative probability at eight and now we subtract the cumulative probability at five, which we have already solved for right here came out 2.632 and we get a probability of zero point 368 which is basically the same answer is part C, which tells us that the probability of X being greater than eight is very close to zero and therefore these probabilities are approximately the same


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