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2,56 MF12.0 HF24.04.25 HF8,35 MF7.22 UF...

Question

2,56 MF12.0 HF24.04.25 HF8,35 MF7.22 UF

2,56 MF 12.0 HF 24.0 4.25 HF 8,35 MF 7.22 UF



Answers

(1) 212 (2) 532 (3) 244 (4) 733

Okay to use a chi square procedure, we need to have systematic random sample, which we do, and now we need to check that the distribution is approximately normal. When we can do this is by plotting it with a normal probability plot calculator, you can find one online, just copy paste your data in there, and we get this kind of interesting line 1.0, and then it kind of goes up here and then it kind of goes up here. So it does have a little bit of a weird slope that does this, but it's not too dramatic that you could still draw a line right about here, and it would still get every point within reason. So we're going to say that it's fairly reasonable to use the chi square procedure.

In this problem. It is a multi step problem. You want to first graph the data set and you'll notice that in the data sets in the data table, there are a couple blank spots. So here we haven't input of 10.3, an output of unknown value down here. We haven't output of five with an input of unknown value. In order to determine this information, you need to first, we'll find the line of regression, which is exactly what the question is asking us to do. So in this question, you want to graph the data set estimates a line of regression. What is your y equals MX plus B for this, and then using that information, go back to the numbers and then calculate the approximate value of those data points. So, first of all, here is my data sets and a graph. When making the graph, you want to make sure that your window is appropriate. So starting here, I'm gonna make my data set go from ah X value of 9.5 up to 11.5, and then for my y axis, I'm gonna go from 4.5 up to 6.5 using this data. Now I can see that there's very clearly a slope that's negative. So this is going to start off as why equals negative something. What is that? Negative, Will It almost looks to be about a negative one slope. I see. Here it starts way up here at the 10.10 6 And as I go down, it almost looks like I could draw a straight line down to this point. And this point is 11 5 So that would be a negative one slope. So negative one x plus something. Now, I need to make sure that I shift this graph way high. So was going. Why? Why intercept? Will. Might be 15 and 15 is a little low. 15. A little low. Someone would raise this up to 16 and look at that. 16 seems to go perfectly through my data set. So I'm going to use this equation Negative one x plus 16 to calculated these unknown values. So I need to calculates where I haven't in points of 10.3, and I need to calculate a second time for when I havent output of five points. Oh, so that's what I have over here, I have my equation y equals negative X plus 16. And I've changed my endpoint value to 10.3. Calculating this out, I'm gonna get a value that goes, Why equals approximately five 0.7 over here. I'm gonna have to do some algebra work to get this X by itself. So again, when you're due out abroad, you're trying to get your variable by itself. So I need to move this 16 and get rid of this negative coefficient. So the first thing I'm going to do is I'm going to subtract 16 from both sides of the equal sign from the left side and the right side of the equal sign and I will get as a result, 11 negative 11 equals negative X. And I also need to get rid of this negative variable this negative coefficients. So I divide by negative one to both sides of the equal sign again, and I end up with a conclusion of X equals positive 11. So now I'm going to go back to my graphing calculator and make sure that these will fall on the line and it should show me that it's approximately correct. So for the input value of 10.3, I haven't outputs of 5.7. Let's check that out 5.7. And if you notice that's exactly what I got right here. 10.35 point seven. It's perfectly on that blue line. And then for my 2nd 1 I have X equals 11 when my input, when my output value is five. So I have 11 here and again, that point and this point fall exactly on the line, and they would satisfy my line of regression.

On this question were being asked to determine the relationship between the weight and thousands of pounds compared to the stopping distance and feet. So here is our weight and dozens of pounds of this would be £2000 in this one right here would represent 96 feet. So we're looking to determine what is the relationship and what is the lean yer value that we can use to determine this information. So part of this question is not only graphing but also to determine the line of best fit and then last not least finding these missing values in the table. So in order to do this, we need to do a little bit of guessing check work. I have scaled this out so that we're dealing with very large numbers were going from zero through five on my X axis, but 50 through 250 on my Y axis, which means that we're gonna be dealing with some very large slopes here to get from here up to here within two units on the X values, we have to have a very large slope. So I'm going try. Why equals 50 X and see what that gives me an and look at that. It's it's pretty close. Umm, I might want us bring this down a little bit. So I'm gonna go ahead and do Ah, minus three. And that looks pretty good. But as I look at this line, I'm thinking to myself, Okay, this doesn't quite look exactly like I wanted to. It's looking a little steeper. Someone dropped this to 49 I think that bad line looks more accurate according to the data points. So this is gonna be the line that I use. Now, the question says that you will have different results. Your answer should probably be somewhere around these two numbers 49 negative three for your y equals MX plus B. Now, I'm gonna use these to determine my input value for this 160 foot stopping distance, and I'm going to use these £4000 to determine how long we need to stop. So coming over here, I'm gonna start by inserting and substituting the 160 into this y that you saw 100 60 feet is how long it takes to stop what type of car though, or what type of vehicle is when I'm assuming this problem is asking us. And then I'm just gonna go through my algebra work of adding, subtracting and dividing to get my answer. So just quickly do that gonna add three to both sides, plus three plus three That gives me a grand total of 163 on the left hand side, On the right hand side, I'm left with 49 x and a nice thing about padding the positive and negative of the same number they add to zero. Now I want to divide both sides by 49 so that I get a coefficient of one. The nice thing about this, the reason we want to divide it so that this becomes one and so I'm left with X is approximately equal to, and I'm going go ahead and around to one decimal place because this is a significant digits that we were given. So 3.3. So coming over here, I'm gonna put 3.3 into this and look at that. It's pretty close. Not perfect, but it's pretty close to the line. And that has to do with rounding differences now for the 2nd 1 The second blink, uh, high. Remember, replace the X with a four. Because it's selling as we have £4000 something and we want to determine how long it takes to stop this. I'm guessing vehicle. So we need to most by this out and then combine our like terms. So 49 times four is 196 minus three, 196 minus three Will then give me 193. And this should fall exactly on the line. 193. And look at that. It does so. There you go. Ladies and gentlemen, thank you very much.

This is a multiplication problem here. So to make this a little easier, I am going to stack them when you are multiplying. It does not matter what order they're in, so I think it will be a little easier. Tough 9 43 at the top and then 601 at the bottom. Um, just because one is a lot easier to multiply, and then you have just a zero as a placeholder, so you can skip the tens place and then move on to your hundreds place. Either way, you're gonna get the same answer. Doesn't matter what Whether it's gonna be in so one times 31 times for one times nine. Right? So I'm done with my tens Place cycm because you're there. I have nothing in my tens plate or my ones ways. I'm nothing in my tons place, so I don't need to do anything with that. And I can move on to my hundreds place. So six times three is 18 carry the one that's 24 plus one is 25 period to 54 plus two is 56 34 17 through the 1665 Make sure my decimal is Are the comma is there? You should have 566,743.


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