Planning A Party
By: Ameen Mahouch
The table goes with it as-well. You just multiply your x-value with your rate of change and add 20 to it to find your y-value Here is the table created
Using that equation, you can make a table with different x values. Here is the table:
Critical Thinking Questions: Grab Some Grub
I created the equation by using y=mx+b (or y=kx+b) format. The rate of change was found in the flyers of each restaurant. The Y-Intercept was the catering/set-up fee. Now that we have both our ROC and Y-Intercept, you can create the equations of y=10x+20 and y=9.25x+61.25 for the restaurants.
2) Will there ever be a number of students where both companies will cost the same? Describe the steps you would use and then solve for the number of students for which both companies will cost the same amount.
Yes. When there are 55 students who attend, the fundraiser's costs will be the same. I previously showed you how to solve that by putting both equations equal to each other. When you solve, you come up with x=55. If you want to find the actual cost, you must put the x value in one of the equations and solve for Y. You should get y=570 as the intersection. Here the steps to solve for x:
Basically, this question is asking what the x will be if y=1000. For Burger Bonanza, your equation would be 1000=10x+20. If you solve this, you should get 98 people attending. You can't have a decimal as your answer b the y=1000, you cannot round up. Same applies with Pizza Palace, the equation would be 1000=9.25x+61.25. Although you have a decimal for an answer, you cannot round up. So you'd be forced to round down and get x=101.
PICKING OUT THE FUN
Critical Thinking Questions: Bounce Houses
1) Describe how you created each equation.
I created both equations using the y=mx+b (or y=kx+b) format. I found both of the rate of changes from each company on their flyers. For Jumpin' Jack's, they had a set-up fee of 84, which means that is their Y-intercept. I came up with y=75x for Hoppin' Around and y=54x+84 for Jumpin' Jack's
2) Will there ever be a number of hours where both companies will cost the same? Describe the steps you would use and then solve for the number of hours for which both companies will cost the same amount.
Yes, as I previously answered this. To solve this, you would need to put both equations equal to each other and solve for x. After solving, you should have x=4. If you'd like to go further and know the cost of 4 hours, you just plug in your x value of 4 into one of your equations. You should get 300 for both. The coordinate for this would be (4,300). Here is how I solved for x:
Basically, this question is asking if y=750, then what would your x value be. For Hoppin' Around, you would have to create the equation 750=75x. To solve this, you would have to divide 75 from each side to get 10=x. This means to spend $750, you can rent it up to 10 hours without exceeding. For Jumpin' Jack's, you would have the equation 750=54x+84. After you subtract 84 from each side, you get 666=54x. Now you divide 54 on each side to get 12.333 (infinite)=x. You cannot round up, or it will affect the dependent value, (Total Cost) You must round down to 12=x to spend a little less than $750.
For the bounce houses: If you are renting it for 3 or less hours, you should DEFINITELY go to Hoppin' Around for a more cheap price. Although, if you are renting more than 4 hours, you should choose Jumpin' Jack's for the better price. Once again, this is all caused by their points of intersections.