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I need help with my bio HW can you guys explain it n an easierway1. A population of Koalas on a small reserve west ofSydney Australia had a total count of 200 anim...

Question

I need help with my bio HW can you guys explain it n an easierway1. A population of Koalas on a small reserve west ofSydney Australia had a total count of 200 animals. Out of the200 animals, 128 were Gray and 72 were albinos (white) in the year2019. Gray is dominant=G and white (albino) is recessiveg. If this population were in Hardy Weinberg equilibrium,what would you expect the genotype frequencies (GG; Gg; gg) to beand what would the allele frequencies be (G and g)?2. There was a massive

I need help with my bio HW can you guys explain it n an easier way 1. A population of Koalas on a small reserve west of Sydney Australia had a total count of 200 animals. Out of the 200 animals, 128 were Gray and 72 were albinos (white) in the year 2019. Gray is dominant=G and white (albino) is recessive g. If this population were in Hardy Weinberg equilibrium, what would you expect the genotype frequencies (GG; Gg; gg) to be and what would the allele frequencies be (G and g)? 2. There was a massive wild fire out of control in unusually dry arid conditions in Australia this same year. After the fire, the remaining Koalas went through their breeding season. Their population numbers are only 100 total this next year 2020. You counted 84 Gray koalas and 16 albino (white) koalas. What would the genotype frequencies be of this new population (GG; Gg; gg) and the allele frequencies (G and g)? 3. Compare the two years population’s genotype and allele frequencies. Are these populations in Hardy Weinberg equilibrium (yes or no)? Explain your answer whether yes or no. If yes, what factors are keeping the populations in Hardy Weinberg equilibrium (name them all). If no, what major factor contributed to this change?



Answers

A scientist is studying two large populations of deer that are centralized in nearby forests. She takes blood samples from all of the deer in each population and records in how many individuals she finds allele A. She then computes the frequency of the alleles in each population. The frequencies observed over five years are shown in the tables below.
$\begin{array}{|c|c|}\hline & {\text { Population A }} \\ \hline \text { Year } & {\text { Allele $\mathbf{A}$} \\ { \text {Frequency} }} & {\text { Allele $\mathbf{B}$} \\ { \text {Frequency} }} \\ \hline 1 & {0.69} & {0.20} \\ \hline 2 & {0.71} & {0.29} \\ \hline 3 & {0.73} & {0.27} \\ \hline 4 & {0.75} & {0.25} \\ \hline 5 & {0.84} & {0.19} \\ \hline 6 & {0.84} & {0.16} \\ \hline\end{array}$
$\begin{array}{|c|c|}\hline & {\text { Population B }} \\ \hline \text { Year } & {\text { Allele $\mathbf{A}$} \\ { \text {Frequency} }} & {\text { Allele $\mathbf{B}$} \\ { \text {Frequency} }} \\ \hline 1 & {0.00} & {1.00} \\ \hline 2 & {0.00} & {1.00} \\ \hline 3 & {0.10} & {0.90} \\ \hline 4 & {0.16} & {0.84} \\ \hline 5 & {0.21} & {0.79} \\ \hline 6 & {0.25} & {0.75} \\ \hline\end{array}$
Which forms of evolution are most likely occurring in populations A and B? Explain your answer.
a. In population A, genetic drift is likely occurring, causing allele A to become more prevalent than allele B. In population B, mutation apparently occurred, introducing allele A to population B. Allele A also appears to be increasing due to genetic drift in population B.
b. In population A, natural selection is likely occurring, with allele A being favored over allele B. In population B, gene flow apparently occurred, allowing allele A to become established in population B. Allele A also appears to be favored by selection in population B.
c. In population A, gene flow apparently occurred, allowing allele B to become established in
population A. Allele A also appears to be favored by selection in population A. In population B genetic drift is likely occurring, causing allele A to become more prevalent than allele B.
d. In population A, mutation apparently occurred, introducing allele B to population A. Allele A also appears to be increasing due to genetic drift in population A. In population B natural selection is likely occurring, with allele A being favored over allele B.

So we can start this question by listing off the information given to us. Um, so first we know that there are two types of rabbits. We have white and brown rabbits. Um, second, we know that wolves are able to recognize rabbits that do not blend into their environment. Um, and third, we know that the current environment is changing in a way that reveals the brown rabbits to the wolves and sort of camouflages the white rabbits. So when does all this mean in terms of a little frequency? Well, we can look at this craft over here that relates on the X axis the amount of snow coverage to the proportion of the population of rabbits that the white and brown roberts make up. Um, so you can see that as the ground becomes covered in snow, the wolves are able to find the white rabbits less frequently, Um, and the brown rabbits more frequently, meaning that the bulls are eating of brown rabbits much more than the white rabbits in this snow covered environment. So if you look over here at this point in the graph, we see that the white rabbits are much more abundant. than the brown rabbits. Uh, and this means that the white only low frequency is increasing. So the answer to this question would be a

So if we take 42 Hey, if we breathe in a duty mouse with a non beauty mouse, that's one. We'll all be big, a little A and a beauty. And then the two will be free to one ratio, free of beauty, with one individual that has a big A big A in two individuals that are hitters, itis and one individual. It's non of beauty, and that one will have a Gina type of too little ace would be parts. It's similar. So we have a wild type mouse, which is black beauty course with a cinnamon Knauss and then in the F one. They will all be Harris, I guess, because the game it's a big being little B and they all have a wild typos, and then they have to. You'll have three that are wild type with one that 60 Bigbie to you. That hunters, I guess, and one that is cinnamon. And that individual will have the GM type two little beans for C heart. If you take a cinnamon mouse so big a big A to little bee's any time you have to. Little Bee's Cinnamon and too Little A's and two B's. So that's a non beauty mouse. In the F one, you will have a header. Psych it. That is a wild type black beauty and then in the F one star in the yet to When you breathe in, we'll look at what happens in deep part. So in deep part, what we're supposed to do is breed these two individuals together. So deep part, we're going to do a breeding between two headers against from C and what we get. It is this ratio we have. Nine. My set are Black Beauty, which is wild type. We have three mice that are black but Nana Beauty. We have three mice that are cinnamon. So remember cinnamon has these two little bees, and we have one individual that's a double process. It and that individual is chocolate Now. That's really important, because if you have a chocolate individual that shows up, that tells you a lot of information about your F one. So we're going to keep that information sort of handy so that when we move forward, we have it. Okay, so that's deep heart. So let's let it eat. Eat heart, you take. You have parents who are almost like this dominant, almost like a successive. So that individual is cinnamon, right? Because the two little babies read with a mouse that is homeless, I guess. Process it for one gene homicide. Istomin it for the other. And that individual is a black mouse. But he's a non a beauty. And then they asked one. They should all be lacking beauty because you have a header. Rows ICUs opens a library a little be there should have a headers, I guess. Which exhibits the wild type. Remember? Which is black beauty? No. And then in the F two, you should have a 9331 ratio Where nine. Exhibit the wild type color. So a something be something wild type and that distribution of Gina types looks like this. There should be one big a big a big, big me. There should be two headers, I guess a home is like a Solomon in beans to almost like a song to a header. Rows like this be and one header is up Food. I'm sorry. Four headers, I guess, for both traits. Now, if you're uncertain, just simply do it. Di hybrid cross between two headers like these individuals from the F one. Three of these individuals will be black Nana Beauty, and they will look like 12 little A's dominant, almost like this thing. And two homes, I guess. Process it A's headers, I guess. Fees. Three individuals will be cinnamon so big a something to little bee's and they will be like this. Want homicide installment A homeless like this process of B and two headers. I just a homeless, I guess processing creamy. And then the last individual will be chocolate. And that Gina type will be, too. It's a double recesses, so it's two homes, I guessed deans. Now we continue and we look It s it says You know what happens if you read a wild type mouse with the cinnamon Knauss and what happens is you get 1/4 homos, I guess. Dominant for a headers August for B 1/4 headers, I guess for both and both of those are wild types that there's 50% of your mice will be wild type color and then 50% will be cinnamon. And so they will have a big a big A into little bees and then headers. I guess for a but still to little bee's, because that's the only way you get cinnamon and then the other reading for yes, we take a wild type mouse, and we read it with a black Nana beauty mouse. And so here we have a little a 50 a little bit house with two little A's and two B's. And so remember, this is wild type. And this is Black Nana Beauty and in the offspring What it is. 1/4 big. A little ANC, too Big B's and that's wild type power. 1/4 big a little a big be a little bit. And so that's 50% wild type our and then 1/4. Too little a stick. Be big B and 142 Little Ace had ours. I kiss fee. And so these air 50% of your offspring and they're gonna be flat. Nana beauty. Okay. And then let's work on G part. Jeanne is what happens when you take a wild type males and you read it with a chocolate mass. And so what happens is you get 14 that are big, a little a big, beady little being, and that's gonna be wild type 14 will be big. A little a to little bee's that cinnamon 14 will be to little a speak, be a little baby. And because this tube a laser there, that's Black Nana Beauty and then 14 will be chocolate. What? And then H Part has four parts to it. So H. Ross one. The information we need here is that to be albino, they must have too little C's. Now I underline my recess of seeds because you can't tell the difference between my upper and lower case sees. So an underlying CD is excessive. So across one, the parents are albino somewhat underlined this recessive traits. We don't know what they are for A or B, so we'll leave them blank when they're crossed. With a parent as to its home is, I guess, dominant for all three James in the F one. We know that they were there. No albino, but they carry the albino gene. They have at least one big A and one big green in the F to you get 87 my slender wild type in power. 32 better cinnamon. He is 39. Better Alvina and so there's a couple things we could make a point of, and I'm going to just write them over here. Or maybe I'll just give it up in a minute. Difficulty where? Next to it. We know that cinnamon has to be too little beings. Oh, so in the F one, that means that the F one parents must have at least one piece that I have to be headers. I just will be. And so because the parent here is be Bigbie, this parent must be with the situation where they can give a little B so they haven't least one little being because that's where that one pain from. And so they're probably homeless, I guess, successive or be, because that's the only way you could get 100% headers like this for being in the F one. And so now we know that the F one parent is big. See little. See a question mark the big me a little bit. 1/4 of your progeny are expected to be albino, and 3/4 of them must be either a beauty or non 1/4 would be cinnamon a beauty or chocolate If Nana Beauty, if you notice we don't have any chocolate, so the F one parent must not carry the alil for non abuse so that F one parent must be with two big A's. Otherwise, you would have some Nana beauty offspring or chocolate offspring. So the original parent money off, I know Must Be Home is like it's recess it for seen homeless, I guess. Dominant for a homeless. I just process it for be read with this second parent who's homeless, I guess. Element for all three traits. Producing an offspring that looks like this in the F one. Okay, that's cross one. Let's look at a cost to across to This is still a church across two is an albino parent, and we don't know what A or B is. Cross the parent who is homeless Eyepiece. Dominant for all three traits. The F one we know it's fixed seat. Little seen a something be something. The F two. It's 60 to wild type. It's 18 albino, so we know the wild type is big. See something and albino is too little sees. This is a 3 to 1 ratio, so one genus, hetero Xigris and so the other has to be homeless, I guess. Dawn or processes And so well, smart legs. So the albino parent must be We'll see you, little c a little, a little, A little bit. A little being to make sure that these genes are headers, I guess because that's the only way you get a 3 to 1 ratio. So the albino parent must be homeless, like a successive for all of its rights and the other parent must be what we have it. Okay, across three, this is still a church. We haven't albino parents. We don't know what A and B are. Okay, And then that parent is red with another parents and the F one. Yes, A something and being something in the F to you have 96 wild type. You have 30 black and 41 albino. So we know the wild type has to have the capital cities, and albino has to have the two little A's. We also know that black is too little A's, and that's being wild type. You have to have one big A in one big B. And to be black, you have to have to recess. If the wheels for a and one dominant being, and we don't know for A and B for the albino. So for a black Nana beauty Bina, type two here in the act to the F one must have been headers, I guess. Who were the A gene. So here therefore it's genotype could be with this week. See little C make a little AP question Mark and the albino parent must be too little A's in order to get the big A little a combination have in love that one. So among the non albino offspring, which is wild type in the black, this is a 3 to 1 ratio of color. His 30 tons three is 90. And so that means that one of the two genes in the F one is header rows. Argus. So the F one must be big. See, little seen big A little a big be they be That means that the albino parent must be This header is a home is I guess process it to see homos I just processing for a home. So I guess dominant will be And homicide is dominant for all three teens in the other parent. All right, almost there. Let me erase this so that I have a different person out of whiteboard Space will do cross four. So across four, um, has a chocolate, so that shouldn't be too bad. Just one second, actually. No, this up since we'll be racing anyway. But I'm out of whiteboards face. That's how big and monstrous this question waas almost there. Oh, dear. Okay, Cross for last one. We made it across four. The parent is an albino and we don't know about A and B and the other parent is homeless, I guess. Solid it for all of the other trend of jeans. So we know in the F one. Have a letters I get. See something A and something be in the F two. We have 287 wild type. We're 86 black, we have 92 cinnamon, 29 chocolate and 164 albino. So let's write what we know. We know the wild type has to have a big C a big A in the big D. We know that Black has to have a big see homeless. I guess recessive a and home is I guess something or headers like this for being cinnamon has to be big C make a and too little bees. Always chocolate is big. See something and then a double home is like a successive. And then albino is too little. Sees something something we don't know. So they get chocolate offspring. The F one parent must be header rows I vis for all three traits. So we know this has to be head resigns. Otherwise, chocolate is not possible with that devil excessive. And so that means that excuse me, the albino parent must be homeless, I guess processed ID for all three genes because the other pair is homeless, I guess Dominant for all engines.

Question, and I value to look at the table as we go through this. I'm not gonna redraw the table. I'm just gonna go through this very quickly. Um, and, uh, make a few comments and a few notes as we go along the way here. But there's a lot to cover. So, um, when you look at these questions, you have to write whether they're in Hardy Weinberg or not. So the 1st 1 you could see everybody has a big A alil. So P is one. Hugh is zero. So there it's already where? Equilibrium? Because you wouldn't expect there to be any headers, I guess. Er, almost like it's recessive. Um, for the 2nd 1 the answer's no. It's non in Hardy Weinberg equilibrium. And the reason is that there headers, I guess turn, you'd expect when they, uh, mate and have offspring that some of those offspring would be. Homer Side gets for three answers. Yes, it is in hardy Wonder equilibrium and again, for the same reason is Number one is for four. It's not in Hardy Weinberg equilibrium, and you can simply do the basic math year to get P and Q. It's a little bit tedious, but not terrible for five. Again, not in Hardy Weinberg equilibrium. And if you look at those numbers, we would expect there to be more headers. I goods. And we're not getting that. Um, so again, these numbers here with people, he 375 and 0.6 to 5. All right, uh, for six. Um, this was also yes. As I lose track on my notes, this one was in Hardy Weinberg younger. And and you can win in their in equilibrium All you have to do to get the peace accused Take the square root of, ah, say the on the percentage proportion of him is, I guess, recesses. That'll give you Q And then you could just solve for P. So that's pretty straightforward. For seven. This is not in Hardy Weinberg equilibrium. You look at it. It's just such a strange set of numbers. I'm something is going on, um, and solving for it, you're gonna have half the Leo's will be, Uh ah. Big A and half will be little a eight is. Yes, it isn't Hardy Weinberg equilibrium. And again, it's easy when there are anywhere in your columbarium because you just have to take the square root. Let's say the number of from his, I guess in proportion of home a psychic processes to get the values, um, nine is yes, and it's simply the reverse of the number eight. So that's pretty straightforward. And then, finally, Number 10 has a tiny numbers involved with it, but it isn't Hardy Weinberg equilibrium, and that's plain 993 and playing 007 So that's a and B here. And in the questions, um, see is asking about unquestioned six. Um, remember the formulas here, and this is the mutation over selection formula of the balance. And then the selection coefficient equation, which is s equals one minus w and were given mu is five times 10 to the nine s six. So that's the mutation rate. And then we can solve for Q. Um, because where we have Q from the the questions were 4.9 times 10 to the minus fifth. Something like that, um, that we just have to play the sort of it equations in. That's you square. So, um so if we use this equation here, we have 4.9 times 10 to the minus fifth equals the mutation rate, which is five times 10 to the minus six. All right, it by s. And so if we saw for s, we get a value of 102 So there's some decker mint. There is selection acting against this Leo, and then we're just unplug s into this equation here. So, um, point 0.102 equals one minus w. And again, this is this equation here I'm using. Ah, And if we saw for W, we get 0.898 So that what we were shooting for in this question, and then finally d um, there's a little bit tedious, uh, and were given values or fitness of one. And then for the headers I goat. Well, you was 0.8 and then for the homeless, I guess. Recess, Ivo, it's 0.6. Um, and we're interested in trying to calculate p crime. That is the frequency of P and the following generation. Um, so they're different ways to do this, but I'm gonna use that to formula approach. First, I'm gonna calculate average fitness, and this is a formula that's in the textbook where you have P squared names. The Fitness of the Homicide, its dominant plus two p Q and the fitness of the Headers. I Go plus Q squared and the fitness of the home is that it's recessive and these air just frequencies cow times, a fitness values. And of course, that's going to give you your average fitness. And so if you plunge these numbers in and for time reasons, uh, I'm not going to, um, you end up with a value of 0.8, right? So this number goes here and then for peace squared, remember that will just be 0.5 and shoes squared would between five squared and those air those point fires come from the table about. So the average fitness is gonna re 0.8 and then we're going to use in that value, which we needed in our second equation. And again, this is from the textbook Ah, where you multiply the fitness of the homeless. I guess Tom meant going to pee and the fitness of the headers I go and divided by the average fitness that we just calculated. So keep prime equals 0.5, and again, that's from the table above, and we want to play all that by 0.5 times one. So that was the fitness of W. A. A. That's appear, plus Q. Just also 0.5, and to fitness of the um, header is I go. And that's 0.8. That's appear. And then if we divide that by the average fitness, which we just calculated his 0.8 a 0.8 and we do that math, then we get a P prime value or the P and the next generation of 25 6 I think the question asked us What the value for Q. Is that people cubicles one. We know that then the Q prime value will be zero point for for so ah, long question with a lot of work involved.

So here they give us an experiment, you can read through that. And we were given, um, results. And we're going to explain these results and indicate how we would test the hypothesis. Eso for a withy, nick to a little one, you have to know proto troughs at all. And this is a lack of Riverton's suggests either a, um delish in or inversion with energy. And then be these results are pretty long. Um, so we have three Protropin colonies and you cross feed from severally that wild strain, and then it gives us the results from those. So the understand these data were called that have the progeny should come from wall type parent. So for patrols A Because 100% of progeny are Perrotto trophic a reversion of the original immune site may have occurred for be have the progeny are parental proto drops and their mating credit troughs. 28% are the result of a new mutation. Notice that the 28% is approximately equal to 22% oxygen drops. The suggestion has at an unlinked, um, suppressor mutation occurred, yielding independent assortment with, uh, nick immune and lastly, a pro Trump c. So there are 496 river and proto troughs and four oxen drops. This suggested a suppressor mutation occurred in a site very close to the original mutation and was infrequently separated from the original mutation by recombination. So you could consider this a link suppressor mutation.


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