Question

the table below shows the number of prom tickets sold over a ten-day period. use this data to answer questions #11 - 16.
prom ticket sales

day (x)	1	2	5	7	10

1. what is the line of best fit for this data?

a) $y = -\frac{9}{2}x + 27$
b) $y = \frac{9}{2}x + 27$
c) $y = -\frac{2}{9}x + 27$
d) $y = \frac{2}{9}x + 27$

1. estimate how many tickets will be sold after 16 days.

a) 108 tickets
b) 81 tickets
c) 99 tickets
d) 90 tickets

1. can you assume that the longer tickets are sold, the more tickets will be sold? explain your answer.

Explanation:

Question 11

Step1: Analyze the trend

As \( x \) (day) increases, \( y \) (tickets sold) increases, so the slope should be positive. Eliminate options a and c (negative slopes).

Step2: Test a point

Take \( x = 1 \), \( y = 30 \). Test option b: \( y=\frac{9}{2}(1)+27 = 4.5 + 27 = 31.5
eq30 \). Test option d: \( y=\frac{2}{9}(1)+27\approx27.22
eq30 \)? Wait, maybe better to check another point. Take \( x = 2 \), \( y = 35 \). Option b: \( \frac{9}{2}(2)+27 = 9 + 27 = 36\approx35 \). Option d: \( \frac{2}{9}(2)+27\approx27.44
eq35 \). Wait, maybe my calculation is wrong. Wait, let's recalculate the slope. The data points: (1,30), (2,35), (5,55), (7,60), (10,70). The slope between (1,30) and (2,35) is \( \frac{35 - 30}{2 - 1}=5 \). Between (2,35) and (5,55): \( \frac{55 - 35}{5 - 2}=\frac{20}{3}\approx6.67 \). Between (5,55) and (7,60): \( \frac{60 - 55}{7 - 5}=2.5 \). Between (7,60) and (10,70): \( \frac{70 - 60}{10 - 7}=\frac{10}{3}\approx3.33 \). The average slope or the line of best fit. Wait, maybe the options have a typo? Wait, maybe I misread. Wait, option b: \( y=\frac{9}{2}x + 27 \), slope 4.5; option d: \( \frac{2}{9}x + 27 \), slope ~0.22. Wait, the actual slope between (1,30) and (10,70) is \( \frac{70 - 30}{10 - 1}=\frac{40}{9}\approx4.44 \), which is close to \( \frac{9}{2}=4.5 \). Wait, maybe the intended answer is b? Wait, no, when x=1, y=30: option b gives 31.5, option d gives ~27.22. But the data has (1,30), (2,35) (difference 5), (5,55) (from x=2 to 5, 3 days, 20 tickets, ~6.67 per day), (7,60) (2 days, 5 tickets, 2.5 per day), (10,70) (3 days, 10 tickets, ~3.33 per day). The line of best fit should have a positive slope, and \( \frac{9}{2}=4.5 \) is closer to the average rate. Alternatively, maybe the question has a different approach. Wait, maybe the correct answer is b? Wait, no, let's check x=5: option b: \( \frac{9}{2}(5)+27 = 22.5 + 27 = 49.5
eq55 \). Option d: \( \frac{2}{9}(5)+27\approx27 + 1.11 = 28.11
eq55 \). Wait, this is confusing. Wait, maybe I made a mistake. Wait, the table has days 1,2,5,7,10. Let's calculate the mean of x: \( \bar{x}=\frac{1 + 2 + 5 + 7 + 10}{5}=\frac{25}{5}=5 \). Mean of y: \( \bar{y}=\frac{30 + 35 + 55 + 60 + 70}{5}=\frac{250}{5}=50 \). The slope \( m=\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \). Calculate \( (x_i - \bar{x})(y_i - \bar{y}) \):

• (1-5)(30-50)=(-4)(-20)=80
• (2-5)(35-50)=(-3)(-15)=45
• (5-5)(55-50)=0*5=0
• (7-5)(60-50)=2*10=20
• (10-5)(70-50)=5*20=100

Sum: 80 + 45 + 0 + 20 + 100 = 245

\( \sum (x_i - \bar{x})^2 \):

• (1-5)^2=16
• (2-5)^2=9
• (5-5)^2=0
• (7-5)^2=4
• (10-5)^2=25

Sum: 16 + 9 + 0 + 4 + 25 = 54

Slope \( m=\frac{245}{54}\approx4.537 \), which is approximately \( \frac{9}{2}=4.5 \). So the slope is \( \frac{9}{2} \), and the equation is \( y - \bar{y}=m(x - \bar{x}) \), so \( y - 50=\frac{245}{54}(x - 5) \). When x=5, y=50, which matches. Now, let's write it in slope-intercept form: \( y=\frac{245}{54}x - \frac{245}{54}*5 + 50 \). Calculate \( \frac{245}{54}*5=\frac{1225}{54}\approx22.69 \), so \( y=\frac{245}{54}x + 50 - 22.69\approx\frac{245}{54}x + 27.31 \). \( \frac{245}{54}\approx4.537 \), which is close to \( \frac{9}{2}=4.5 \). So the equation is approximately \( y=\frac{9}{2}x + 27 \) (since 4.5*5 +27=22.5+27=49.5, close to 50). So the answer is b.

Step1: Use the line of best fit from Q11

We found the line of best fit is \( y=\frac{9}{2}x + 27 \).

Step2: Substitute x=16

Calculate \( y=\frac{9}{2}(16) + 27 = 72 + 27 = 99 \).

To determine if longer sales mean more tickets, we analyze the line of best fit (positive slope, \( y=\frac{9}{2}x + 27 \)) and the data trend. As days (\( x \)) increase, tickets sold (\( y \)) increase in the data (1→30, 2→35, 5→55, 7→60, 10→70) and the line of best fit has a positive slope, indicating a positive correlation. However, we must consider if the trend continues. The data is over 10 days, and extrapolating assumes the linear trend holds, but in reality, ticket sales might saturate (e.g., limited attendees), but based on the given data and line of best fit, the trend is positive, so we can assume (with caution) that longer sales (within the model's scope) lead to more tickets, as the slope is positive and data shows increasing \( y \) with \( x \).

Answer:

b) \( y = \frac{9}{2}x + 27 \)

Question 12