I need help, I really need help pleeease.

Question

The relationship between the number of tickets sold at a movie theater, t, and the total profit gained, P, in dollars is represented in four different ways below.

 

1) A movie theater has operating costs of $1,025 a day. Tickets cost $7.50 each. The movie theater’s profit each day depends on the number of tickets sold.	2) P = 7.5t - 1025
3)               Profit                                                               Tickets Sold	4)     t P 0 -$1025 50 -$650 100 -$275 150 $100 200 $475 250 $850

 

Question #1: Which of the four representations is most helpful for determining the following information:

1. The daily operating cost of the theater?
2. The number of tickets that must be sold for the theater to gain a profit of $500?
3. The daily break-even point for the movie theater?
4. The rate of change in the relationship?
5. The major family of functions (ie. linear, quadratic, exponential, etc.) to which this relationship belongs?

 

Question #2: For each of the three representations that your group did not choose as most helpful in Question #1, explain how you would find a-e. For example, if your group chose 1 as the most helpful for the rate of change, how can you find the rate of change in 2, 3, and 4?

 

Comments

Sorry. No can do. I can't solve this. Try asking my friend, Rod. He might help.

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It's a long question, so give me some time and I'll answer it some time today (UK time).

By the way, (3) does not show up on my device. There's no picture, so I can only use the information for (1), (2), (4) and guess that (3) probably shows a graph of the P function.

Answer 1

I assume that (3) is a graph, because it doesn't display on my tablet.

Question 1

a) 1 because the daily operating cost is stated specifically.

b) 3 because the profit of $500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204.

c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of $2.5, but 136 tickets would make a loss of $5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137.

d) The rate of change is 7.5 from (2). 

e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity.

Question 2

a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0.

b) For (1) you would need to find out how many tickets at $7.50 you would need to cover the operating costs of $1025 plus the profit of $500. That is, how many tickets make $1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=$500 somewhere between t=200 and 250 in the table.

c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150.

d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, $7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5.

e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add $375 to the profit.

Comments

The table below shows the elapsed time when two different cars pass a 10, 20, 30, 40 and 50-meter mark on a test track. 

a) For car 1, what is the average velocity (change in distance divided by change in time) between the 0 and 10-meter mark? Between the 0 and 50-meter mark? Between the 20 and 30-meter mark? Analyze the data to describe the motion of car 1. 

b) How does the velocity of car 1 compare to that of car 2?

	CAR 1	CAR 2
d	t	t
10	4.472	1.742
20	6.325	2.899
30	7.746	3.831
40	8.944	4.633
50	10.000	5.348

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a) CAR 1 av speed = 10/4.472=2.24 m/s approx (0-10m); 50/10=5 m/s (0-50m); (30-20)/(7.746-6.325)=10/1.421=7.04 m/s. The average speed is increasing so the car is accelerating. 

b) CAR 2 av speed=10/1.742=5.74 (0-10m); 50/5.348=9.35 m/s (0-50m); 10/(0.932)=10.73 m/s. CAR 2 has greater average speeds than CAR 1. Both cars are accelerating. 

The accelerations can be measured by calculating the change in speed over time. A graph or two would be useful in comparing the performance of the two cars.