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3) Find thl area Ok the shudud_region_Whwe (=-31 6n d bzLs7mnt: normal cdrpro babilitU thata_studt uSts smwthinring Uhr nttonng on reg busis 1S 0.3Y. In gwup Uk 20...

Question

3) Find thl area Ok the shudud_region_Whwe (=-31 6n d bzLs7mnt: normal cdrpro babilitU thata_studt uSts smwthinring Uhr nttonng on reg busis 1S 0.3Y. In gwup Uk 20 studuts Whot The_prv thl F exudTTy OK them MSUsmanthinkng OK |oe MuwMng Uh rey: blSis ?

3) Find thl area Ok the shudud_region_Whwe (=-31 6n d bzLs7 mnt: normal cdr pro babilitU thata_studt uSts smwthinring Uhr nttonng on reg busis 1S 0.3Y. In gwup Uk 20 studuts Whot The_prv thl F exudTTy OK them MSUsmanthinkng OK |oe MuwMng Uh rey: blSis ?



Answers

$\triangle \mathrm{OBD}$ is encased in rectangle OACE as shown. a Find the areas of regions I, II, and III. b Find the area of $\triangle O B D$. (FIGURE CANT COPY)

All right. What is a standard normal curve? Well, it looks something like this, right where the meaner zero, they're the mean is zero. Okay, Now we have to find the area under the standard normal curve that lies to the right of We have been given for values. So the first one is minus 3.1 So minus 3.1 will lie somewhere around here. So I want the area of the right, which means this overall area. How do I find this? What I can do is I can use the P value method to get the area and the left tail that is area to the left. This area, this will be the P value. Right? And I can subtract that value from one. So this would be one minus p value. Okay, So what is the P value for this? I can either use a set table or statistical package or an online tool. I'm using an online tool over here. So this is minus 3.1 I hit Enter and my P value is 0.0 1306 So I want one minus 10.1306 or this gives me 0.99 87 But he is the answer. This 0.9987 Okay, moving on to part B, that is equal to minus 1.59 minus 1.59 will be here. This is minus 1.59 Again, I'll find the P value, which will be the area and the left tail and one minus the p value. So when I do that, my answer turns out to be my p value turns out to be 0.5 0.56 something, Uh, sorry. 0.56 My P value turns out to be 0.5 So this will be one minus 0.56 Something close to something like this and all right. And my answer becomes 0.944 My answer becomes zero. Oh, just told me. Yeah. Okay, So this is 0.944 Okay, Now, moving on to part C. Part C is 1.78 variable 1.78 line. It will lie somewhere around here. This is going to be 1.78 So p value will be area of the right. So that's all we need. We put in the value 1.78 and we get 0.37 We get the okay. 0.37 0.37 And how about part D? Now? We have 3.113 point 11 means somewhere around here, right, 3.11 So this is going to be very small value. And again, the P value turns out to be 0.93 0.93 And these are my answers.

Hi, I'm tim And today we're gonna be looking at problem number 10. So we're looking for the area Between the curves. The two equations given our one plus square root X and three plus x. All over three. So we can go ahead and start graphing them on this little graph here. So when X equals zero on the first one, why simply equals one? And then acts like a regular square root graft. It's just shifted up one. So it looks approximately like that. Remember with square root grafts? Because of the fact that In a regular function you can't have two Outputs for one input, it's simply just the positive half of it, the part that goes up. So you know that it looks roughly like that. And then for the second one, when X equals zero, Y equals one. So we can continue with that point. And then it looks like X in all the slope is one of her three. So for every three x's increases by one Y. So it's going to look something approximately like that. So we can roughly tell intersects around they're obviously not exact point that we're gonna be looking for the area in here. So normal with them would look like that. Uh for Riemann sum, you can determine whatever value you want to use Riemann sums with a problem like this aren't exactly the most funds. So we're just gonna go with regular old integration. And so in order to find those two points already know zero comma one but we don't know the second point. So we're gonna set the two equations equal to each other one plus square X equals three plus X over three. So three plus three. Red x -3 -1 Equal zero. If we have to set that equal to zero and find out the solution obviously three's cancel out. So it becomes three screw x equals x. If we add that back to the other side we're gonna make it easier so we can find that. Oh there are several ways we can go about it. I'm just going to go ahead and divide by screw X on their side. So we're left with three equals X over square root X. Which just equals square X. So we know if we square both sides That we end up getting nine equals x. So that's the second point. That's the second bound we're going to be using So we know zero and 9 are the bounds the upper equation, the one that's higher is one plus square X. And then the lower is three plus X. All over three the X. Since it's just always easier to do it within the expound unless it's not possible. There are some occasions where it is easier to go with respect to Y. But we simply haven't run into that yet. So we're gonna go ahead and integrate each of these square root X just equals X. To the 1/2. So when you integrate each of these, one becomes X plus screw X. Since it's X to the 1/2 and you're adding one That's X to the 3/2. Then you carry down that coefficient to over three. And so that you can move on to the next part minus 3/3. You can just simplify this to 3/3 X. And then X over three It becomes x squared over six. Alright we're gonna go ahead And put on the bound 09 for evaluating this entire equation at 09. Before we do that however we gotta go and simplify just a little bit so we know that X and then 3/3 X the same value. So those to actually cancel out. So we're left with two X to the 3/2 oh three minus X squared over six. We go ahead and evaluate it at those bounds nine and zero Some throw in nine Squared of nine is 3 and cubes so 27. So we get two times 27/3. When we said student, the first one Then nine squared is 81/6 -81/6. And then if we would evaluate at zero then we would get zero in both terms. We can ignore the second part. Just go straight into simplifying this. So two times 3, cute essentially Is over three gives us two times 9 -81/6 and we could take out three from 81 and a three from 6. That gives us or we can make the 1st 1 18 And then 81 is essentially 9th square. That's what we had earlier With that. We can actually go down to 27 times three equals 81 And then over three times 2. So that can give us 27/2. And with that we just have to simplify a little smidge more So we can make 1836 over two. So I'm gonna continue to work right over here since we're the finishing stretch -27, all of it over to Now, we get nine over to. So that is the final answer for the area between those bounds. It's going to go ahead and fill in the area that we're looking at In that area is equivalent to nine over to. All right, well, I hope that helped and help you have a nice day.

For this problem, we are asked to find the area of the region bounded by f of X equals three times X cubed minus X and G of x equals zero. So we are also given this sort of little, I have a little sketch of the region that we're looking at. So we can see that there are two sort of separate areas where F of X and G of X will swamp Which one is greater. So the way that we can address this is by splitting our overall area into two areas that will calculate separately A one and A two. We can also See that because of the symmetry of three x cubed -X. A one is actually going to equal a two. So in fact we just need to calculate one of the areas and multiply it by two. So we'll have two times the integral from negative one up to zero. For instance, of just ffx, it would be F of x minus G of X, but G of X zero. So it'll just be the integral of three X cubed minus six or three X cubed minus X. Excuse me, D. X. So we'll have a six out front from that three and then it becomes the integral of negative or integral from negative one up to zero of X cubed minus X. D. X. So that integral we can apply our power rule for integration to get extra power for over four minus X squared over two, Evaluated from -1 up to zero. So that when we plug in the ax equals zero point. That will just evaluate to zero. We'll have zero minus uh negative one to the power of four. So that's going to still end up being just 1/4 then minus 1/2. So that one quarter minus one half is going to turn into negative one half, one second here. Excuse me, that should be a negative one quarter. So we have six times negative negative one quarter, so we'd have just 6/4 or 3/2 as our final result.

Welcome to New Merida. In the current problem we have to find the areas that are under the normal girl for specific points and mostly we will use the rights right till the distribution. So we have it ready over here. If in case we need the left one, we will go for this. So the first question asks is media uh lift wolf maybe is equal to minus two point and it was going to nine ft or oh right off. That is the calls to 2.94 It's like it's minus a nice. So minus 2.94 will be here plus 2.94 will be here now left off 2.194 means this area right of two point informants this area. Now they are seen. So basically both uh huh seen. Therefore the probability that greater than right right segments, right means 2.94 and afterwards 2.9 food will be over here. We get to point line the 01234 This is our thank you. Point usually 16 So 0.16 The next problem asks. Uh huh minus that is equal to minus 1.68 What night? Oh z is equal to 3.5 Okay. So if you see for this lift off minus 1.6 side, will they? Right of minus 1.680 The because 1.68 that is probability is greater than 1.68 So 1.6 is over here. 0123 45678 This is the 10 point 046 France. Okay, so right of 3.5 No. Right off. 3.0. Is this Okay, So almost we can right probability said later than 3.5 is equal to probability the head of approximately goes through a three point zero, correct? Which is equal to 0.13 Hey, the powder problems in the set? Yes, lift off. That is. He goes to minus. You don't want it quick. What? Mhm. Off that is equal to 1.23 That is this. We can think to be there right off the the call through 0.88 That is probability the greater than 0.88 So 0.8 is over here and 0.88 will be this 0.1894 And for this we have that greater than 1.23 1.2 is over here. So 1.3 would be this. So it is 0.1093 So I hope you could understand how we are calculating each of these values. Let me know if you have any questions. These favor dances.


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