5

Nhyt pnct- Hmen 44 lur Leir AJertnton cora? In tha Lutd lante MICI 4 patnadlq Tcqicna quyeh mple Isan of * {6.4H per_ Four. Auterielon Hecume Inat 0 | kown 31-A0 pe...

Question

Nhyt pnct- Hmen 44 lur Leir AJertnton cora? In tha Lutd lante MICI 4 patnadlq Tcqicna quyeh mple Isan of * {6.4H per_ Four. Auterielon Hecume Inat 0 | kown 31-A0 per 10J Founae And 4 Dool conidenccnen poulation mrun Eo thut Bunnela this rein Elelr Futermcon cron Lnan Wunst Atnt [14t4n Etor (In dolari (Fo tJc Inrnhl eriet nmithel Aountlenur Anerarnt Ra dcnnenties Dnt nUpoe Enut mamgin 0l corEnplc #In nictean Ju $ O0t" cuniene? lavd 7uiniber, Koed up t0 tho roarx ntol rumoer| [edanmalrnaEala

nhyt pnct- Hmen 44 lur Leir AJertnton cora? In tha Lutd lante MICI 4 patnadlq Tcqicna quyeh mple Isan of * {6.4H per_ Four. Auterielon Hecume Inat 0 | kown 31-A0 per 10J Founae And 4 Dool conidenccnen poulation mrun Eo thut Bunnela this rein Elelr Futermcon cron Lnan Wunst Atnt [14t4n Etor (In dolari (Fo tJc Inrnhl eriet nmithel Aountlenur Anerarnt Ra dcnnenties Dnt n Upoe Enut mamgin 0l cor Enplc #In nictean Ju $ O0t" cuniene? lavd 7uiniber, Koed up t0 tho roarx ntol rumoer| [edan malrna EalamA Man Nite T00 pcunds Hremeln Hentt (c) 4 {atmn brings 1S tuns c wateriekn t0 markel Fnd 90*8 Con"Juice Mterraltor {ne DCFU #un Iean Eash vaue 0l this Crol Oo-ansRing Tniau #ouner MFui tacn Mcaan numbic Roundcojranamento Iw) cacimal placer Untor umut Upet Micae Ire-e colln Anarin- mtom 0



Answers

Epidemics In $1840,$ Britain experienced a bovine (cattle and oxen ) epidemic called epizooty. The estimated number of new cases every 28 days is listed in the table. At the time, the London Daily made a dire prediction that the number of new cases would continue to increase indefinitely. William Farr correctly predicted when the number of new cases would peak. Of the two functions $$f(t)=653(1.028)^{t}$$$$\text { and } \quad g(t)=54,700 e^{-t t-200 j^{2} / 7500}$$ one models the newspaper's prediction and the other models Farr's prediction, where $t$ is in days with $t=0$ corresponding to August $12,1840$ $$\begin{array}{|c|r|}\hline \text { Date } & \text { New cases } \\\hline \text { Aug. } 12 & 506 \\\text { Sept. } 9 & 1289 \\\text { Oct. } 7 & 3487 \\\text { Nov. } 4 & 9597 \\\text { Dec. } 2 & 18,817 \\\text { Dec. } 30 & 33,835 \\\text { Jan. } 27 & 47,191\end{array}$$
(a) Graph each function, together with the data, in the viewing rectangle $[0,400,100]$ by $[0,60,000,10,000]$ (b) Determine which function better models Farr's prediction. (c) Determine the date on which the number of new cases peaked.

Ready. So problem 44 here seems a bit complicated at first rate, but there's a lot of words to read through it. What, you want to break it down? It's not not too bad. So I think the important thing to note here is that this f b s right. This bovine serum, this one disrupts okay, disrupts thes. So cycle clock. So this one disrupts the so cycle clock, whereas, uh, this deck PSA Metha sown here this in this other column this one disrupts the circadian clocks of this one disrupts circadian clock. All right, and so wants you. So we have that in mind that little bit easier to see what is happening. Okay, so we can go through this question asked us to go through for each a through G, sort of what that means. So in this first column first call him here, you can see that there's no disruption of the cell cycle, right? So for the no disruption of cell cycle here and there's no disruption of the circadian clock, right, So this one has no disruptions, and as a result, you can sort of see that the the period's there are red at 24 hours, Like with a little bit of an heir. See, that's what you'd expect. There's no disruptions, and they're both on this same the same 24 hour cycle, Okay. And so that if we move on to the next one, well, there's still no disruption to their circadian clock. But now we have a disruption here, uh, to the cell cycle. Okay, so So just the cell cycle is disrupted, but the results is that both the cell cycle and circadian period ended up being disrupted. Right? So they're different from that previous calm. So if we go down to the next column, we see again that there's no disruption to these circadian clock, and there is a disruption. Great. Her disruption to the cell cycle. Okay, so So there's, like, an increased cell cycle disruption. Okay. Ah, and the results here is an increased disruption to both Both periods. Both the cell cycle and circadian periods are even more disrupted from from the starting point. This column Mayweather's There was no disruptions. Okay. And then then we get down to D. C. That actually, both are being disrupted here. Right? So we have your description of the cell cycle, and, you know, they were adding the decks of methods. So we're getting this eruption of the circadian clock, and then we get some interesting results because there's almost no disruption here to the circadian, right? Very little. But there is. There is a pretty significant disruption here to the cell cycle. Period cases. One had had a cell cycle disruption. And then if we moved down to e again, we're getting disruptions in both. Right, So we're getting to disruptions here, and you can really see that e n f Both have the 20%? Yes. So they're kind of similar. They had two rounds of this same sort of outcome More same. Ah, same starting conditions. And and you'll come here in both Is is a disruption, uh, sort of across the board. So? So the cell cycle is a bit more disrupted, but both of these are quite a bit off of the 24 as well. Right, So So everything is disrupted here. Then there was this weird case, uh, here in G. Right. So we have a 10% and yes, here, which is actually the same starting conditions, is D, But we had this weird sort of confluence form which stopped the cell cycle. It's Wigan and A here in the stopping of the cell cycle. Uh, sort of left zero impact, then on on the circadian rhythm. Right, So we don't really know anything with cell cycle, but there was no disruption, so no disruption to the circadian. Okay, Sarah Cady and great. And so then that's sort of part B is to walk through very long Lee to walk through. Ah, all those pieces. But then part B wants us to based on all this data, Okay, How Queen justified claim that that wind cells are actively dividing. Okay, So when When there is a cell cycle period. Okay, then the circadian period is set by the cell cycle rather than the reverse. So the thing to note then is is how the circadian Pierre is reacting along with the cell cycle period. Right? So? So, if the circadian period was the influence there, then when there was no disruption to it here, we should have seen no disruption to it here. Right. But instead, instead, in both columns, B and C here, uh, we saw a disruption in the disruption correlated very precisely to the same disruption that the cell cycle was experiencing. Okay, And so? So that idea, you can sort of translate that as well, too, to the second half. Right? So here where it was being trapped in an A and I apologize because it is asking us specifically about win cell cycle will to divide. So that kind of eliminates this last column. Ah, so feel it here when it should be disrupted. Then we would expect everything in this column to be destructive, but we didn't see that here, Okay. And so that sort of gives us the impression that the circadian period is actually being set by the cell cycle period. Right? So the so say and be when our cell cycle period get disrupted, then are our circadian rhythm was subsequently disrupted. Despite despite that it was not given anything to disrupt it. Okay, so that's that's all the evidence that justifies that claim

So our probability of choosing I'm warm, that class is 61.22%. So that's gonna be 0.61 22. And we're assuming that this is going to be a binomial distribution since each each of the sources are assumed to be independent. And we also have 20 sources. Okay, So we can give our probability mass function. Yeah, by this probability that X is equal to any number of K. Or random variable is going to be equal to 20 shoes K. Then we're gonna have our probability to the power cake Times 1- or probability To 20-2. Then put down the probability but it's too long. So for per eight or asked what is the probability that we have at least one instance that we have class? So the probability that X is greater than or equal to one is simply going to be equal to the probability of one minus the probability that we have zero. Since that's the only thing behind that. So our probability that X0 is essentially gonna be zero as well. So we're just gonna have one to minus zero. And that's gonna be basically A 100% chance that we have at least one. Yeah. Now for the probability that we have at least three, That's going to be the probability of one. My sur probability of zero of course as to zero. So we can ignore it Most probably leave one and minus the probability of two, one. So we'll have a couple of different terms. And so only these two that are going to be important. So the probability of x equals one that is going to be 20 choose one. And then we'll have our probability To the power of one times 1- the probability to the power of 19. And then we'll have the probability that x equals two. Yeah, 20 choose to probability squared times one minus or probability to the 18th, mm hmm. And these things are going to equal, are these really small numbers? And this one is pretty much negligible, negligible. Sorry, we're just saying for this one, so we'll just count this one here. It's gonna be one minus, that's point oh three. That's going to be equal to 0.9999 seven. Yeah. All right. We got five nines. Yeah. Good. And then finally, the mean is going to be equal to our expectation, which is simply going to be and times p Which is 20 times our probability. And that's going to be equal to 12.244. And the deviation that's going to be equal to the square roots of our sample size owns the probability times one minus the probability. Yeah, yeah, we just copy that number down. Okay. And then we'll do that 1 -11-2. And this is going to get us a standard deviation for the population of the random variable Of 6: three. Yeah

Hi, everyone. This is the problem based on motion under gravity, without air resistance, When a particle is moving under gravity, it experienced the acceleration 9.8 m per second squared. So what's up? Center of the earth? And in this problem, we have to analyze the velocity and position off the particle. Let us see here. If there is no air resistance, then acceleration All that time off. Its flight is due to gravity. That is a skull toe minus tree are 9.8 m per second squared during flight. Golf went Yeah, but might That is 1609 m up on. Then one might don't back. You guys starting point? No, the speed just after lunch. No. Yeah. We finally square. It's called Toby. Initial square less to a final position. Minus initial position. Final velocity zero. Initially we have to calculate acceleration is 9.8 and this by final minus by initially 1609 m. So initial velocity, you will get 1 78 m per second. Just his time in the air may be fun. Mhm that time and air. Wow. Maybe phone by country motion from just after lunch, so just okay before impact. So we finally minus B initial building initial speed in two times plus half. It is square since launching time and impact time the same. So it's zero Initial velocity is 1 78 minus half 9.8 m per second square in tow. T square, so t you will get Yeah, 3 36.2 seconds. The rate off. Pay his rate off way, baby. Well, upon 36 0.2 seconds and we have toe. This is given an dollar per second 1027 six dollar per second. We can calculate per hour dollar per hour On its value will be 99 point three dollar per hour. That is the answer for this problem. Thanks for watching it.

Here is an example of a carnot heat engine cycle using an ideal gas. So a reminder that a carnot engine has to ice A thermal processes. One an expansion want a contraction during the expansion at high temperature heat is absorbed during the contraction heat is expelled to the cold reservoir at low temperature. Then there are two idiomatic processes where heat does not have a chance to exchange with the environment. One is again a contraction one. Again, it's an expansion. Usually when analyzing these cycles, a couple of tools are helpful. One is the ideal gas law provided an ideal gas is used. So I usually like to write that down and then find the pressure and volume and temperature At each of the four vortices shown in the cycle. Mm. Uh The other tool that often is useful is the first law of thermodynamics dealt to you is equal to Q. Heat minus work done by the engine. That could be helpful for processes or cycles that involve a idiomatic processes. It is also good to realize that pressure times volume raised to the idiomatic exponent is equal to a constant. It reminder that the automatic exponent is a ratio of cp two C. B. And depends on the nature of the gas. Here we are, given that this constant is 1.3. So this is usually how I like to start. I usually like to start with a table and fill out the three state variables pressure volume and temperature for each of the vertex is in the cycle Here. There are four that are labeled in the diagram for us. Now this problem is challenging because they don't give us outright numbers but they provide clues as to what's going on with this particular sample to put it mildly. Um And so the first thing they tell us is that there's the ambient Environment that the particle is interacting with at the .4. And they give us the ambient temperature as 123 Calvin. Uh now that's good because we now know the temperature at .1 because of the ice a thermal process. They also tell us that we are at one atmosphere at that point but they don't tell us what one atmosphere is other than yeah, it's ambient there. Mhm. Some other clues at .1 we are told there is something called the trigger volume and that's going to become important Because at .2 we get squeezed to half the trigger volume with the radius given As .08cm cubes. That's kind of no not cubed. Sorry. Yeah, that's kind of a weird way to give a clue. But at .2 they also give us the pressure. Yeah, As 20.3 kilo pascal's. So that's an outright calibrated measurement kilo pascal's at 0.3. They give us the clue that No, it's actually a .4. They give us the clue that the final radius expands from the trigger radius by three And since volume is R. Cubed, that tells us that we have 27 times the trigger volume. So we have some clues and I think these are the the most obvious clues. The rest. We're going to have to use some steps with the gas law and the idiomatic exponent to figure out. So looking at the table, the most information appears to be given. Well we'll start with two because there's a nice calibrated point right there. And so at point to what we know is the volume Is 4/3 pi r cubed. And that's going to unravel a lot. So we have 2.14 Times 10 to the -3 m3 figuring that out. I won't show all the math there and then what happens is we can now find the trigger volume By just multiplying it by two and that will give us the Entry for the volume at .1 and it will also give us then The volume at .4 Which is 27 times that. Yeah, 0.116 m3. Yeah. Yes. And we can fill in that entry into the table. Okay, so now we've got to do a little bit more work. Um we see that there's quite a bit of information given in steps one and 2 and they are connected through an idiomatic process. So what we can do between one and 2 because it is idiomatic, We can set the pressure one Times The Volume one raised to 1.3 is equal to the pressure 0.2 times the volume, it too Raised to the 1.3. And let's see that is going to give us of the unknowns. It will give us the pressure at 0.1. So we can solve for that. And we get uh huh 8.24 Killer Pascal's and bingo. Now we've got everything inside appoint one. Uh, we've got, yeah, definitely a bingo type situation. Everything is filled in so we can now use one that everything at that point to solve for the number of moles are, is just the gas constant 8.31 jules per bowl kelvin. And we could solve for that high temperature, which is a very important quantity. Um, once we find the number of moles. Okay, but actually we're using that point to saw for the number of moles. So let me be clear about that. Oh, okay. So we can solve for the number of moles And that's definitely important quantity 3.45 Times 10 to the -2 moles, which I'll just write off to the side because that is an important clue that we can then use in 0.2 to find the high temperature. It's like a mystery that unravels and doing that again, I won't show all the plugging and chugging, But we get a temperature of 151 0.5 kelvin. And that's nice because we now know both the low temperature and the high temperature and that is definitely important things for the operation of a carnot cycle. But while we're at it we have a few more things to fill out so we might as well go ahead and fill those out. There's a lot of information missing at .3. Um and as well as point for But what we see is .3 is connected 2.4 through a uh a dramatic so let's stop with one more step. What about .3? It's connected to four through idiomatic and it's connected to through ice a thermal. So what we can say is the following two relationships the unknown pressure at three Times the unknown volume at three raised to 1.3. That idiomatic exponent is equal to pressure it for volume it for Raised to the 1.3. Okay. Um well we don't know pressure it for, do we? But we can use the fact that four is connected to one through an ice a therm. And so we can say that pressure for volume for is equal to pressure one volume one. And we definitely know both of those at a .1 and we also know the volume at point for so we can solve for pressure for and again, I'll spare the details but we basically get mhm 301 pascal's Okay. Now what do we know? We don't know either pressure three or volume three. So we have to connect it to point to through the ISIS are okay. So we know both Pressure volume at .4. Same with two. We have two equations into unknowns and we can definitely then solve For both pressure three and volume three. And what are the steps to that? Well, I would probably take the bottom relationship And solve it for pressure three and put that back into the top equation. Yeah, 1.3 is the exponent And solving that we get volume three is equal to Okay. Uh huh. Uh huh. 5.84 Times 10 to the -2 cubic meters. And then we can solve for pressure three. Okay. So yeah, we've got the table full at this point. Yeah. So I'll put a little faced by that to show that that one is the most challenging and it's probably the least interesting too on top of it. So at this point, we have all the information about this cycle um we can do some other things with the cycle, basically knowing what the carnot efficiency is. I will use the expression without the percentage because certainly getting the fraction is good enough. But that efficiency is basically the difference between the high and the low temperature divided by the high temperature. So we know both of these temperatures and we can figure out that efficiency, It's about 18.8%. But .188 in terms of a fraction and a reminder that any engine has another definition of efficiency is the work that you get out divided by the heat absorbed at high temperature during the high temperature is a thermal expansion. Yeah. So yes, while the gas is hitting up, it is expanding and pushing against something um whether it's the wheels, the shaft of your car with a piston that's pushing or whether it's this little particle um doing something expanding. Anyway, what we're told is that the work out is 60 killer jewels per hour and that's really a rate at which work is being done, 60 killer jewels per hour Is the same as one killer jewel per minute. Just thinking about How hours translate two minutes. Um and then we can figure out the heat that is absorbed at least the rate. Just turning those into rates does not change the fraction there. And so qh dot the rate at which he dis absorbed is simply uh yeah 5.32 killer jewels per minute. And then we want to use the first law to figure out the difference between um the heat absorbed minus the heat expelled exhaust heat is equal to the output work. And really what you're doing with that is using the first law With Delta U. equals to zero. And solving for qc dot it's the heat absorbed minus the output work, energy must be conserved. And so this is four 32, kill the jewels permit it


Similar Solved Questions

5 answers
6. Prove that there are as many multiples of 5 as multiples of 20.
6. Prove that there are as many multiples of 5 as multiples of 20....
5 answers
Question 2:Natasha uncle has rectangular field that has an area &f square mile. The width of the field mile. How long is the field? Show your work:75+528Training Network 2015 @Copyright Nationa
Question 2: Natasha uncle has rectangular field that has an area &f square mile. The width of the field mile. How long is the field? Show your work: 75+528 Training Network 2015 @Copyright Nationa...
5 answers
Solve for x in the triangle Round your answer to the nearest tenth:679
Solve for x in the triangle Round your answer to the nearest tenth: 679...
5 answers
HwO8: Problem 14Previous ProblemProblemn ListNext Problempoint) Suppose the angle satisfies <A <2r. I cOs(A) ~0.83 and sin(A) > determine:The quadrant tor the angle A/ Quadrant?sin(A)sin(A/2)cos(A/2)tan(A/2)Be certain axpress dnswur least four docimal places: AJSo, be caraful regarding the 5ign5 for aach 0f your answers:Noto: You can earn parfial credit on this problom;Provlaw My AnsworsSubmlt AnswaraYou hava atiompted Ihls problemr - ImosThen,
HwO8: Problem 14 Previous Problem Problemn List Next Problem point) Suppose the angle satisfies <A <2r. I cOs(A) ~0.83 and sin(A) > determine: The quadrant tor the angle A/ Quadrant? sin(A) sin(A/2) cos(A/2) tan(A/2) Be certain axpress dnswur least four docimal places: AJSo, be caraful rega...
5 answers
1 (11 points) Consider the linesr(t) (2 + +,1 - 3t,-3 + 2t)anda(s) = (9 _ 2s,-14+4s,-10 + 3s) Does there exist plane containing these two lines? If s0, find such plane. If not _ explain why not .
1 (11 points) Consider the lines r(t) (2 + +,1 - 3t,-3 + 2t) and a(s) = (9 _ 2s,-14+4s,-10 + 3s) Does there exist plane containing these two lines? If s0, find such plane. If not _ explain why not ....
5 answers
La sigulente es una grafica de Arrhenlus de una roaccion de prlmor ordcn La consranie de rapk z % Ilto gn urdudog ds IaaDetermine el factor de (recuencla (A) para gara rcaccion: (R = 8,314 JiK 'mol)Seleccione una: 2.4* 1017 , ' 5.0 5-' 5.0 x 10" $-' 50 5-140 $7
La sigulente es una grafica de Arrhenlus de una roaccion de prlmor ordcn La consranie de rapk z % Ilto gn urdudog ds Iaa Determine el factor de (recuencla (A) para gara rcaccion: (R = 8,314 JiK 'mol) Seleccione una: 2.4* 1017 , ' 5.0 5-' 5.0 x 10" $-' 50 5-1 40 $7...
2 answers
4/6-25 DoinasTeeeeeeednaMNEltiVaentn nTinch Crracc adde (6,010,)rnmnxunhirucilalr Fucit4000smhoo {uo CaqubieJor [tnkmoktt tdaHerd 4te4p? Aalald
4/6-25 Doinas Teeeeeeedna MNElti Vaentn n Tinch Crracc adde (6,010,) rnmnxun hirucilalr Fucit 4000smhoo {uo Caqubie Jor [tnkmoktt tda Herd 4te4p? Aalald...
5 answers
3) Circle any shapes thal have rotational symmelry: Draw box around any shapes that have Iine symmelry- Il is possible to have both circle and square around ItlllThlnk of regular polygons and (helr possible lines of symmelry: Do you notice pattern In your answersm? you find onc, what is It?
3) Circle any shapes thal have rotational symmelry: Draw box around any shapes that have Iine symmelry- Il is possible to have both circle and square around Itlll Thlnk of regular polygons and (helr possible lines of symmelry: Do you notice pattern In your answersm? you find onc, what is It?...
5 answers
Deducc the thrce most stable oxidation states for the compounds of Arscnic (elemcnt 33). Identify cach cither noble gas, inert pair pseudo [ noble gas configuration_
Deducc the thrce most stable oxidation states for the compounds of Arscnic (elemcnt 33). Identify cach cither noble gas, inert pair pseudo [ noble gas configuration_...
2 answers
ACHC10 11 - 19Question 4Find Valuc of that atislic the conclusion the Menn Valuc Thewremnt for f(2) F On [3,5] _dcc nol LalYour answer:Clear answer
ACHC 10 11 - 19 Question 4 Find Valuc of that atislic the conclusion the Menn Valuc Thewremnt for f(2) F On [3,5] _ dcc nol Lal Your answer: Clear answer...
1 answers
How many cycles does each sine function have in the interval from 0 to 2$\pi ?$ Find the amplitude and period of each function. $$ y=\sin \theta $$
How many cycles does each sine function have in the interval from 0 to 2$\pi ?$ Find the amplitude and period of each function. $$ y=\sin \theta $$...
5 answers
Gn $ 1 2 1 Sin 3 0 2 3 Yn- T 0 2 8 (e9si 4 0 2 mea
Gn $ 1 2 1 Sin 3 0 2 3 Yn- T 0 2 8 (e9si 4 0 2 mea...
5 answers
For the following polynomial function, find a. f(-1), b. f(2), and c. f(0)f(x)= -5 x exponent 4 - 3x squared +6a. f(-1)b. f(2)c. f(0)
For the following polynomial function, find a. f(-1), b. f(2), and c. f(0) f(x)= -5 x exponent 4 - 3x squared +6 a. f(-1) b. f(2) c. f(0)...
5 answers
Zo (ue Iheqien TTof XhaWi tue luwwse. mafX Melld VGe8 4X1 ~ 6x7 =* 6 2X1 t Xa = Vak-Datc tuu peSult 0qivd te Ccus-Jordan 2lmihahion
zo (ue Iheqien TTof XhaWi tue luwwse. mafX Melld VGe8 4X1 ~ 6x7 =* 6 2X1 t Xa = Vak-Datc tuu peSult 0qivd te Ccus-Jordan 2lmihahion...
5 answers
Find the equation of the line tangent to Y=(X^3) (Y^3) + x at (0,0). Please show all steps
Find the equation of the line tangent to Y=(X^3) (Y^3) + x at (0,0). Please show all steps...
5 answers
0 Systematic Sampling 1 L Ji Sulpling1 Keybodrd Shortcult 1
0 Systematic Sampling 1 L Ji Sulpling 1 Keybodrd Shortcult 1...
5 answers
QuesIon 153p8What [s the charge of Cuin the complex [Cu(HzOl(COjICi]Question 169p1Wnat Is tne overal) charge of the Cobalt(Il) complex (CoBra(C_O421?
QuesIon 15 3p8 What [s the charge of Cuin the complex [Cu(HzOl(COjICi] Question 16 9p1 Wnat Is tne overal) charge of the Cobalt(Il) complex (CoBra(C_O421?...
5 answers
2 2 Ii 3 30 2 57 Jmtenya 1 Kene (Cnoose constnucted One chan these 3 ~data, interval 1 corscmtal 1 Jund Jmoueu than' 16 jv11
2 2 Ii 3 30 2 57 Jmtenya 1 Kene (Cnoose constnucted One chan these 3 ~data, interval 1 corscmtal 1 Jund Jmoueu than' 1 6 jv 1 1...
5 answers
5) Debts ofS1O,000 due 2 years ag0 and S20,.000 due today will be replaced by an unknown payment in 42 months and another Payment of Si5, Q0 in 5 years. If the cost of money (the loan) was 6% compounded quarterly; how large is the unknown payment? (Equivalent Value Question-use timeline) (best focal in 42 months)
5) Debts ofS1O,000 due 2 years ag0 and S20,.000 due today will be replaced by an unknown payment in 42 months and another Payment of Si5, Q0 in 5 years. If the cost of money (the loan) was 6% compounded quarterly; how large is the unknown payment? (Equivalent Value Question-use timeline) (best focal...

-- 0.019421--