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Of 29 (26 complete)HW Score: 74.698, 31,33 0f 42 p6.3.15-BE 'Queston Hep Abasketball team sels tickets that cost $10, 520, Chapte sale; are 566.240.How MArIN l...

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Of 29 (26 complete)HW Score: 74.698, 31,33 0f 42 p6.3.15-BE 'Queston Hep Abasketball team sels tickets that cost $10, 520, Chapte sale; are 566.240.How MArIN lickets for VIP seats, $30 , Tha leam has sold 3402 Iickets Lindtave cuerol De#n scld? nae4zMore 520 [kets [nsn S10 lukcts 03/04/1 How many 510 lickets wcra sold? 03/03/1Zore: 76,9894UnllmitHow many 20 5 TicKets Were Sold ?How 30f Tic Kets were Sold ? many

of 29 (26 complete) HW Score: 74.698, 31,33 0f 42 p 6.3.15-BE 'Queston Hep Abasketball team sels tickets that cost $10, 520, Chapte sale; are 566.240.How MArIN lickets for VIP seats, $30 , Tha leam has sold 3402 Iickets Lindtave cuerol De#n scld? nae4zMore 520 [kets [nsn S10 lukcts 03/04/1 How many 510 lickets wcra sold? 03/03/1 Zore: 76,9894 Unllmit How many 20 5 TicKets Were Sold ? How 30f Tic Kets were Sold ? many



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A local band sells out for their concert. They sell all $1,175$ tickets for a total purse of $\$ 28,112.50$ . The tickets were priced at $\$ 20$ for student tickets, $\$ 22.50$ for children, and $\$ 29$ for adult tickets. If the band sold twice as many adult as children tickets, how many of each type was sold?

Okay, let axe equal the number of student tickets. What? Why? Equals the number of children tickets and then left Zi, pull the number of adult tickets. Okay, we know that all the tickets together is experts y policy. And this is 1175 because they saw 1025 tickets total now substituting and the prices they've given. As you can see, $20 for student tickets substitute. In that information, you get 20 acts plus 22.5. Why? Plus 29 equals Giving Z equals two. Why? Okay, it's gonna be important for later. No, we know that we have ex post wide plus two y equals +1175 betweens expose three. Why equals 11 75. This means that acts equals 11 75 minus three. What now? Substitute in two. The original equation 20. 20 acts and you should end up with if we know the axe is 1025 Minuses. Three. Why would substitute end to the original equation? As the as that is, the value acts this should give us why equals 225. Okay, now we have one variable. You have two more variables to do. You know that Z equals to why? I underline that earlier. The two times 1 25 is going to be 4 50 So for 50 equals e. Okay, now solving for X X equals 1175 minus three. Why? But now we have the value of y so he can simply plug in and they get X equals 500. So the answer is 500 student, 225 children and then 450 adult tickets.

Alright here we are going to solve a word problem by setting up assistance of equation. Okay so we know that there are two that there are a couple of variables here. Okay so we have students student tickets and we also have adult tickets. Then we also have um student ticket prices and then we also have adult ticket prices. So it's important to differentiate these different factors here. So let's assign our student ticket as X. And our adult ticket as why. The question tells us that the student ticket price um is $2 so that's going to be two X. And our adult ticket price is $5. So that's going to be five. Why? All right now let's set let's create some equations based on what the question is telling us. It says that the number of students was three less than 10 times the number of adults. The number of students um is exit could be X equals three less than 10 times the number of adults. So 10 y minus three. That's our first equation. Yeah. And then the total the total money that we made from ticket sales of 619. So maybe two X plus five Y equals 619. All right. So that we have our equations just blow this up a little bit so we can solve this is a system of equations because we have two different variables X and Y. We have two equations. Okay, so um to solve for our systems of equations, we first want to put it in form A X plus B Y equals C. This is in that form already but we want to move this around a bit so that we can put that in that form. So we're gonna circle this here and bring it down so we can do that. So to do that, I'm going to move our X to this side and our negative three to this side. So once we do that, we're going to have three equals 10 Y minus x. I'm going to flip that so that it becomes 10 y minus X equals three. And then we're just gonna re copy the other equation down below it so that we have everything nice and close together. Okay, so now that we have our two equations, we're going to solve the systems of equations by elimination. It's exactly what it sounds like. We want to eliminate one of the variables that we only have one variable and then we can solve so we can see here that we have and negative X. And a two X. So if we multiply this entire first equation by two, this is going to become 20 y minus two. X. Equal six. Remember if you do one thing to every side of an equation, you're changing how the equation looks. But you're not changing any of the values. All right. Let's rewrite the second equation under And then let's add this equation together. We're gonna add everything together. Remember we add like terms. So we're gonna first add the 20 Y. And the five Y. Teal 25 Y. Then we're gonna add our negative two X. And R. Two X. Which is going to give us zero X. Okay? So we've eliminated one of our variables. Now we're gonna add six and 619 which will give us 6 25. Alright? Now that we've eliminated the variable we can solve for the remaining variables. So to solve for what we're gonna do 6 25 divided by 25. We're going to get the Y. Equals 25. Okay? Now that we know that why equals 25 we can go back and plug it in for something. So let's let's go back and look so we can plug it into either of these equations to solve for X. Let's plug it into the second one, it might be a little bit easier. So remember that why equals 25. So let's put that in here. X equals 10 time, is 25 minus three, right, So X. Is going to be 247. So now we want to know how many tickets were sold for each one. So now we know that 25 adult tickets were sold, and 247 student tickets were sold, so 247 tickets. That's right. 2047 Yeah, student tickets in 25 adult ticket. Mhm. And that is our final answer.

In this question, we're gonna be taking a look at the distribution of scores on achievement test using a relative frequency table and a percentile graph or a relative frequency graph. So we're going to start with our data and we're going to construct our team a lot of frequencies and then calculate a relative frequencies to make our graph. So relative frequency is just a fancy way of talking about percents. Cumulative frequency means you're just gonna add these numbers up. So we're going to bring the five over, and then we're going to keep adding them as we got so five plus 17 it's 20 to 22 plus 22 is 44 out on another 48 we get 92 add on another 22 or upto 1 14 and at on six, and that gives us a grand total of 20. So that means we are represented representing 120 pieces of data. So that is our end in this case, to find a relative frequency, we're gonna take each one of these cumulative frequencies and divide by 1 20 So, for our first interval five divided by 1 20 is approximately 200.4 So just to show you where that's coming from, I'm rounding these because we're going to be graphing on this graph where everything is estimated, so we don't have to be that precise. Eso 22 divided by 1 20 is 18% or 180.18 44 divided by 20th. About 440.37 or 37% 92 divided by 1 20 is about 77%. 1 14 divided by 20 is 95% and 1 20 divided by 1 20 is one. Now we're gonna graph this over here and we're starting with 1 96 But there are no achievement test scores below 1 96.5 So that's gonna get a zero. There's nothing before. And then as we go to the end of each interval, we're going to continue to build up our graphics. We move forward. So at the end of the next interval work, four percents about here at the end of the next interval were at 18% so just below 20% about here at the end of the next interval were at 37%. So about here, the end of the next interval were at 77%. So about here, the end of the next interval were 95% almost to the tall. And at the end of the last interval, we are at 100% now. One of the things that's important to know is that each of these intervals represents even other kind of knot standard ways with numb label, a scale they do represent consistently spaced out intervals. So I want to try to connect these with a straight line as much as I possibly can free so and this is what we make our cumulative frequency graph. Now, the reason we connect these with a straight line is because we are assuming that all of the achievement tests in the interval in between are evenly distributed. That's our best way of making a guess here. That may not be actually the case, but that's what we're going to use to make our estimation. Okay, so now we have our frequency graph curved, and we can use our frequency curve draft on. We can use that to make our estimates. So first thing that we want to estimate is what is the approximate percentile for somebody who scored 2 20 So to 17 to 2 38 to 59. What? Um, what we need to kind of know is how big that interval is. So we did say that they were evenly spaced on. That means that in between each of those intervals, we're looking at about 21 points. So each one of those lines is representing seven. So to 20 plus seven is to 24. So when I'm looking for to 20 I'm gonna be looking at the score of 2 20 It's gonna be about here, so I'm gonna follow to 20 up to the curve, and then I'm gonna follow that line over 2% all trying to draw this line a straight is that possibly can, and it looks like it's between 5% and 10% a little bit closer to 5%. So let's estimate that at about the sixth percentile, question B is asking for the percentile associated with a score of 2 45 So to 45 is going to be about here, Right? So about 7 to 38.7 about 2 45 about there. So again, we're gonna follow from a test score up to the graph and then follow from the curve over to the percent axis. So that looks to be at approximately 25. So about the 25th percentile. Her question. See, we're looking for the percentile associated with score of 2 76 So to 76 is going to be somewhere over here. So a little bit below to 80. I'm so I'm gonna estimate about here. You should not quite so straight there. Tried to even it out some of the end on. Then we'll move that over a little bit below 70%. So I'm gonna estimate that to be about the 68th percentile. These are just estimates. So if you're not exactly spot on, um, that's okay, that's that's a reasonable, uh, thing to happen with this kind of graph. So now what is the percentile associated with to 80? So, I mean, that's just like here, right? So we can kind of see that there's kind of a big gap that's happening there. So we're all the way up here now. All right, so we should really be drawing this line all the way down. This isn't much better done with a ruler on paper, then digitally, unless you were doing something like on Power Point or something. But we're above 75%. So maybe, like the 76 there, 77th percentile, and then the final score we're going to try to estimate the percentile for is 300. So three hundreds gonna be a little bit below here, and we're gonna follow that up a bit wonky. But it's a little bit before this line hears about their and then we're gonna follow that over, and we know that we're below 95% but not a lot below 95%. So maybe 94 93%. Ah, percent. All right, now we've got all those lines on there that we're gonna have to try to ignore a little bit, but we're gonna work the other direction, and we're going to try to, um, estimate the test score now given a percent. So if we have the 15th percentile, what is the approximate test for that goes with that? So we're gonna work in the opposite direction. 15th percentile. Follow that line over to the graph, and then we're gonna come down here and estimate. So we remember we said that each one of these lines is about seven. So this would be like to 24. This would be like to 31 2 38 So somewhere between 2 31 2 38 little closer, 2 to 31. So maybe, like 232 G. Um, we're gonna estimate that 29th percent up, so you can see, like, all this estimation, all we're going to be able to do is kind of guess right, the 29th percentile. Follow that line right below the 30th percentile over Teoh about here, maybe, and I will follow that down. And so this is to 38 to 45. The next one would be, what to 52. So we're a little bit before 2 50 like maybe 2 49 The important thing to kind of remember is that when you have the percentile, you work. You started to have relative frequency and work to the data. When you have the data, you started the data and you work to the relative frequency. Okay, so, H, we are looking at the 43rd percentile so the 43rd percent tells somewhere between 40 and 45 and we're going to come over to the graph over here and then follow that line down. And if this is to 59 this would be to 66. And so we're a little bit I don't know in the low in the low to sixties. So let's say like to 64 I were looking for the 65th percentile. So 65th percentile is here. Follow that over to the curve and then from the curve down. So it's gonna be a little bit above this line right here. So that was 2 59 to 66 to 70 three. So a little bit above to 73 I don't know. Let's say to 74 and then finally, the last one the 80th percentile. So we'll follow from the 80th percentile to the curve, but here, then we'll follow that down and like to 87 a little bit below 2 87 So let's say to 85. So, as you can see is the percentile was increasing. So where the test scores and that makes sense because we're accumulating more and more test scores as we go. Um, and the same was true here. The lower the test score, the lower the percentile that it was associated with.

So this problem is about tickets being sold to three movies. And so I'm using X, Y and Z to stand for the number of tickets sold for movies 12 and three. And we have a bunch of information about the numbers of tickets sold that will use to create equations. So first of all, we are told that they sell 100 tickets to the first movie, so that's pretty simple. Equation X equals 100. We're also told that the total number of tickets is 642 so we add the number of tickets to movie one, plus the number of tickets to movie two, plus the number of tickets to movie three X plus y plus C. We get 6 42 and finally, we're told the amount of money they charge for each kind of ticket and the total revenue. So tickets to the first movie we're $5 so that revenue is five times acts. Tickets to the second movie were $11 so that revenue is 11. Why, on tickets to the third movie were $12 so that revenue is 12 z on the total revenue is 6774. So here we have our three equations, and the first thing we want to do is find out whether or not there's a unique solution. And we can do that by finding the determinant of the coefficient matrix. So from the first equation, the coefficients on X, Y and Z are 10 and zero. From the second equation. The coefficients on X Y and Z are 11 and one. And from the third equation we have 5 11 and 12. And to find that determine it by hand, we would rewrite column one and call him to on the outside. And then we would find all of the sums Ah, the sum of all of the products going down into the right and then subtract all the products of the diagonals going down into the left. And when you do that, you end up with 12 plus zero plus zero minus zero, minus 11 minus zero, and that is one. And because that turned out to be a non zero number, we know that there is a unique solution. So now that we know that there is a unique solution, we're going to go ahead and find it using Cramer's rule. For all of the values of A of XB, X, Y and Z, well, we don't really need to do it for X because we were already told in the problem that X is 100. So that's good. Now let's find the values of why Enzi, and to get the value of why we have a three by three determine in our numerator, which will be made up of first in the first column, the coefficients on X 11 and five. And then in the second column, we don't use the coefficients on why we use the constants 100 6 42 and 6774 that were on the other side of the equations. And then for the Z column We use the Z coefficients, which were 01 and 12 and that's the numerator And the denominator is the determinant of the coefficient matrix, which we already computed. And we found the answer was one. So let's find the determined of this three by three matrix. We would rewrite the first column and the second column, as you saw me do earlier, and we would find all the products and we end up with 230. So that's over one. So that tells us that why is to 30? Which means they sold 230 tickets to the second movie and above. When I said X was 100 I should probably interpret that and say that they sold 100 tickets to the first movie. And then, lastly, we want to find the value of Z, even though we could just subtract what we have here already from 6 42 to get the value of Z. We're supposed to use Cramer's rule, so the numerator for Z will have the X coefficients in their column, and we'll have the Y coefficients in their column, and that would be 01 11. And then instead of the Z coefficients, we have the constants 106 42 6774. And then, like you saw above, we would rewrite Columns one and two and calculate the determinant of that. And that will be our numerator in its 3 12 And our denominator is one on this one as well from the coefficient matrix. And so that tells us that there were 312 tickets sold to the third movie, so we have all three and they should add up to 642.


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