The Best Party Collection

UNIT 10 PROJECT

Target

As in the target for the project, I will determine which possible options would be cheaper for the school party. There would be 2 options provided for each accompaniment, the bounce house and food. But direction states that there would be optional options for the students, such as ice cream or a dessert for some kind. As our target is stated, we will determine the cheapest cost for the school party and funds.
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Statement of Opinion for "Grabbing the Grub"

I think that having Pizza Palace is a necessary and efficient company for pizza for the school party. For that fact that there would be a lot of students and teachers that would be in and for that party coming up, this company would more likely be cheaper because of the expense of the pizza per person.

I would conduct to this company because even though there is a setup fee that is more expensive than the Pizza Bonanza, Pizza Bonanza has a more expensive cost for every person attending ($10>$9.25).

For example, if the party were to have at least 56 people, which is most likely, the Pizza Palace would be cheaper. But if there would be less than 54, the other company would be better

Critical Thinking Q/A

1. Describe how you created each equation.

When creating my equations, I always find out which variable is the dependent, or the y, and which one is the independent or the x, I then plug the factors into the slope intercept formula (y=mx+b). I then find out the y-intercept, or the b, I first look at the fliers presented and read the part about cost. I knew that cost beforehand is the y-intercept because both of the fliers for food stated about having a set-up fee or a catering fee which is a beforehand demand of money.

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, there will be a number of students that will have a balanced cost for both of the companies. The number of students would be 55, I solved and got the number 55 because I first made the 2 equations into 1 equation without the ‘y’ so it looked like (10x+20=9.25x+61.25). I then solved the problem by isolating the equation by having the ‘x’ value on the left and the y-intercept on the right. I solved it to get 55,

3. If you only have a budget of $1000 for food, how many people, maximum, would you be able to have at the 7th Grade Blowout for each company? Can you have a decimal or fraction as part of your answer? Why or why not?

If you would have had $1000, the Pizza Bonanza Company would only apply to 98 people, but to the Pizza Palace company, would apply to 101 people or if you were to round 101.48 would round to 102 people. But since the math problem for the Pizza Palace ends up into the decimal, we have to round it.

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Statement Of Opinion for "Bounce Houses"

I think that using the Hoppin' Around Company because the party would only be an hour or two. And since the Jumpin' Jack has a set-up and removal fee. And for only 1 hour, that would be ridiculous to pay $138 while you can pay for $75 in the "Hoppin' Around" Company.

But if there would be more than 3 hours, the other company would be better

Critical Thinking Q/A

1. Describe how you created each equation.

I created each equation by finding the cost per hour on the fliers for the Bounce Houses. I found that information because the fliers says “per”, and the word ‘per’ means the constant rate of change or slope. So that is plugged into the slope formula, y=mx+b and the ROC I plugged into the m. In the Jumpin’ Jack’s Company, there is a y-intercept because of the set-up and removal fee. While the other company doesn’t have a beforehand fee, or y-intercept (b). So I plugged the fee into the b slot. That is the equation for the companies.

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, because you would solve to find the point of intersection. The point of intersection is a symbol for the number of hours that would make both of the companies has an equivalent fee. The point of intersection in this problem would be 3, if you would double-check; you would plug the 3 into the x and have the same total cost for both of the companies.

3. If you only have a budget of $750 for the bounce house, how many hours, maximum, would you be able to have at the 7th Grade Blowout for each company? Can you have a decimal or fraction as part of your answer? Why or why not?

In the company Hoppin’ around, if you were to have $750, you would b able to afford 10 hours. While on the other hand, in Jumpin’ Jacks Bounce Houses Company, you would be able to afford 12 1/3 hours. Since when solving for the hours for Jumpin’ Jacks, I got 12.3333(repeating), you can have a decimal since it is talking about time so in this problem, the hours for Jumpin’ Jacks Company would be 12 hours and 20 minutes.

Mathematics Work

Grabbing the Grub Work

Pizza Bonanza

y=10x+20


1) [50] -> y=10(50)+20

y=500+20

y=520

2) [100] -> y=10(100)+20

y=1000+20

y=1020

3) [150] -> y=10(150)+20

y=1500+20

y=1520

4) [200] -> y=10(200)+20

y=2000+20

y=2020

5) [250] -> y=10(250)+20

y=2500+20

y=2520


Pizza Palace


y=9.25x+61.25


1) [50] -> y=9.25(50)+61.25

y=462.5+61.25

y=523.75

2) [100] -> y=9.25(100)+61.25

y=925+61.25

y=986.25

3) [150] -> y=9.25(150)+61.25

y=1387.5+61.25

y=1448.75

4) [200] -> y=9.25(200)+61.25

y=1850+61.25

y=1911.25

5) [250] -> y=9.25(250)+61.25

y=2312.5+61.25

y=2737.75

Grabbing the Grub- Point of Intersection

(Read 1 line at a time, moving down [numbers not underlined is action, look above it to see what it is subtracting/dividing from] dots are to separate the numbers)


10x+20=9.25x+61.25

-9.25......-9.25

.75x+20=61.25

........-20..-20

.75x=41.25

/.75...../.75

x=55

the point of intersection

Bounce Houses Work

Jumpin' Jacks Bounce House


y=54x+84


1) [2] -> y=54(2)+84

y=108+84

y=192

2) [4] -> y=54(4)+84

y=216+84

y=300

3) [6] -> y=54(6)+84

y=324+84

y=408

4) [8] -> y=54(8)+84

y=432+84

y=516

5) [10] -> y=54(10)+84

y=540+84

y=624


Hoppin' Around


y=75x


1) [2] -> y=75(2)

y=150

2) [4] -> y=75(4)

y=300

3) [6] -> y=75(6)

y=450

4) [8] -> y=75(8)

y=600

5) [10] -> y=75(10)

y=750

Bounce Houses- Point of Intersection

(Read 1 line at a time, moving down [numbers not underlined is action, look above it to see what it is subtracting/dividing from] dots are to separate the numbers)


75x=54x+84

-54x. -54x

21x=84

/21.../21

x=4

the point of intersection