5

Let 0.250_Z15 be a group homomorphism with @(x)-7x. Then; Ker(o) ={0,5,15,25,35,45}{0,15,30,45}(0,10,20,30,40}None ol the choicesThe number of subgroups of the grou...

Question

Let 0.250_Z15 be a group homomorphism with @(x)-7x. Then; Ker(o) ={0,5,15,25,35,45}{0,15,30,45}(0,10,20,30,40}None ol the choicesThe number of subgroups of the group Z/45zNone of the choices

Let 0.250_Z15 be a group homomorphism with @(x)-7x. Then; Ker(o) = {0,5,15,25,35,45} {0,15,30,45} (0,10,20,30,40} None ol the choices The number of subgroups of the group Z/45z None of the choices



Answers

A delegation of 8 students from Systems Engineering is being selected

a) In how many ways can the delegation be chosen if there are 13 students available?

b) In how many ways if 4 of the eligible students are married (2 couples) and will only attend if they go together?

So to show that this is not a hormone dwarfism. Um We essentially just have to show that one of these statements here is not true because these are the two criteria for something to be a home abort is um for some function and for this, I think the bottom one will be the easiest to actually go about doing. So How I'll do this is let's first plug in. Just half of zero or a half of the zero factor, I should say. So 00. And if we were to plug that in, so this is going to be zero, this is going to be zero and then that's going to be negative too. So we get that this outputs negative too. Well, if this is a home of dwarfism, if I were to multiply the zero vector by anything, then it should give me the same output as if I were to just multiplied on the inside and then go about doing So Let's say I were to just multiply the outside of this by two, two, two. That's going to give -4. So let's see if we do the same thing. If we could out um negative 44 and you'll end up saying that we won't because F of two times 00. Well, that's still just F of zero. And we just showed that is negative two. So essentially we just showed that two times F of 0, 0 is not equal to F of two times 00. Which would then tell us that is not a home amorphous. Um So implies not homo morph ism Um There is something a little bit more streets or we can do depending on if you have talked about it or not. Um And the thing is is that your identities should get taken to another identity. So for addition, remember this here is the additive identity. Because if we add it to anything, nothing changes. But in just the real numbers, -2 is not the additive identity, it would just be zero. So if you have where an identity does not get taken to another identity, then you can say that it's not a home a morph ISM by that, but uh that's more of a just like theorem that people often have to be taught um or go about proving I should say. But as long as you just show something like this, that would also be a valid way of showing that is not a homo dwarfism.

Okay. So question is and in teacher between 307 80 including 7 80. Its electorate random. And we need to find the probability of having at least one as a digit in that number. Okay, So we know probability. Let's say the event is a is a number of occurring of that particular event to the total number of even that can occur. Okay. So we need to find number of event a a number of event. S which is some of all the events or total number of events that can occur. So how many numbers are there between 7 83 100? That will be a simple calculations. 7 80 minus 300. And this would have been a number if you wouldn't include 7 80. So since we're including it will add one. So this gives us for 81 this is a number of as that was easy. No for a number of eight. So is asking for at least one as one of the digits. So let's have three places where we can put numbers. Since numbers starting from three, we cannot put here one. So here we can choose from 34567 Right? We cannot exceed seven. So that is five numbers. Now there are three cases. Either we have one in unit place and any number here or any number here wanted. Sorry, this was 10th place. You need place and or we can have one in both the places. Okay, so one here, one here, resistance. So let's solve for all three cases. So first, if we have this will be remain as five. Now if we one have one and 10 players that can be selected in one way. So let me write one here. Then here we can select any number between 0 to 9. Right? So let me write 0 to 9, 10 numbers, nine numbers, So that gives us 50. Similarly here, five will remain as it is than any number in 10 place. So that is 10, 0 to 9 and 10 in one unit place. Or sorry, one in unit place. So that is one time. This again gives us 15. The last will be again 500 place and then one in both the places. We can select it only by 11 way. So that gives us five. So total number of ways of our total number that have at least one in one as a digit is 5 50 plus 50 plus five. That is 105. So now we can write our probability of 100 and 5/4 81. So that is an answer. We can divide this value to get our answer is 0.2 and that is an answer. Thank you.

So in this question, we have a club that has seven men and 10 women and we're gonna stop this right now.

In this problem a delegation of aid students from the system engineering is being selected in part we have to find that in how many ways can the dedication he chosen if Okay, there are turkeys to Russia available. So we can use combination formula. And is the total number of object And are the number of choosing objects? So hair. All right. See here the total number of objects are 13. Yeah. And Number of choosing objects are eight. Yeah. So we can follow the formula here. There are eight and his Turpin minus Yeah. Yeah. Five uh minus 13 minus eight resulting five. So Okay, 12 minutes. Mhm 10 9 Kit. So we can say identical. No, calculate. So the calculation would get dead. There there are turtles, 1287 Ways. I can say that dedication can choose in 1287. Yes. So this is the answer of party. Mhm. Part B. We have to find that in how many ways for us to make our married and will only don't if they were together. So for this purpose, if we the end should for this statement should be Mhm 13 minus four. eight and I should be 8 -4 Plus. The second combination is if they go together then we will remain total students 13 minus four. And so after subtraction will get that. So now we will calculate we will do according to formula is nine. Mhm. Mhm Plus no. We will calculate. And the final answer is So we can choose 1 25. 1 35 years. If four of the religions were are married And will at 10 If they go together so they are on 35 years. So the answer of party is this.


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