Question

the function ( p(t)=20(1.5)^{t} ) models the number of students in a school with cell phones ( t ) years after the year 2000. the table below displays data the principal has collected on student drivers. data is collected at the end of each year. at the end of which year did the number of student drivers last exceed the number of students with cell phones?

Explanation:

Step1: Calculate p(t) for t=0 (2000)

For \( t = 0 \) (year 2000), \( p(0)=20(1.5)^0 = 20\times1 = 20 \). Student drivers in 2000: 200. \( 200>20 \).

Step2: Calculate p(t) for t=2 (2002)

For \( t = 2 \) (year 2002), \( p(2)=20(1.5)^2 = 20\times2.25 = 45 \). Student drivers in 2002: 240. \( 240>45 \).

Step3: Calculate p(t) for t=4 (2004)

For \( t = 4 \) (year 2004), \( p(4)=20(1.5)^4 = 20\times5.0625 = 101.25 \). Student drivers in 2004: 280. \( 280>101.25 \).

Step4: Calculate p(t) for t=6 (2006)

For \( t = 6 \) (year 2006), \( p(6)=20(1.5)^6 = 20\times11.390625 = 227.8125 \). Student drivers in 2006: 320. \( 320>227.8125 \)? Wait, no, wait. Wait, wait, maybe I miscalculated. Wait, \( (1.5)^2 = 2.25 \), \( (1.5)^4=(1.5^2)^2 = 2.25^2 = 5.0625 \), \( (1.5)^6=(1.5^4)\times(1.5^2)=5.0625\times2.25 = 11.390625 \). So \( p(6)=20\times11.390625 = 227.8125 \). Student drivers in 2006: 320. Wait, but maybe I made a mistake in t. Wait, the years are 2000 (t=0), 2002 (t=2), 2004 (t=4), 2006 (t=6). Wait, but let's check again. Wait, maybe the function is \( p(t) = 20(1.5)^t \), so for t=0:20, t=1:30, t=2:45, t=3:67.5, t=4:101.25, t=5:151.875, t=6:227.8125. Student drivers: 2000:200, 2002:240 (t=2), 2004:280 (t=4), 2006:320 (t=6). Wait, but when t=5 (year 2005), p(5)=20*(1.5)^5=20*7.59375=151.875. Student drivers in 2005: let's see the pattern. The student drivers: 2000:200, 2002:240 (increase by 40 every 2 years), so 2001:220, 2002:240, 2003:260, 2004:280, 2005:300, 2006:320. So in 2005, student drivers:300, p(5)=151.875, 300>151.875. In 2006, student drivers:320, p(6)=227.8125, 320>227.8125. Wait, but the table only has 2000,2002,2004,2006. Wait, maybe the problem is that the student drivers' data is at the end of each year, and the function is t years after 2000. Wait, maybe I misread the function. Wait, the function is \( p(t) = 20(1.5)^t \). Wait, maybe it's 20 times (1.5) to the t power. Wait, at t=0, 20, t=1, 30, t=2, 45, t=3, 67.5, t=4, 101.25, t=5, 151.875, t=6, 227.8125, t=7, 341.71875. Student drivers: 2000:200, 2002:240 (t=2), 2004:280 (t=4), 2006:320 (t=6). So when does p(t) exceed student drivers? Let's check t=7: p(7)=20*(1.5)^7=20*12.37890625=247.578125? Wait, no, (1.5)^7=1.5*11.390625=17.0859375, so p(7)=20*17.0859375=341.71875. Ah! There we go. I miscalculated (1.5)^6. Wait, (1.5)^1=1.5, (1.5)^2=2.25, (1.5)^3=3.375, (1.5)^4=5.0625, (1.5)^5=7.59375, (1.5)^6=11.390625, (1.5)^7=17.0859375, (1.5)^8=25.62890625. Student drivers: 2000:200, 2002:240, 2004:280, 2006:320, 2008:360. Wait, p(7)=341.71875, which is less than 360? No, 2007: student drivers would be 340, p(7)=341.71875. So in 2007, p(t) (t=7) is 341.71875, student drivers:340. So 341.71875>340. So when does student drivers last exceed? Let's check the given years. The table has 2000,2002,2004,2006. Let's check those years. 2000:200>20 (p(0)), 2002:240>45 (p(2)), 2004:280>101.25 (p(4)), 2006:320>227.8125 (p(6)). Now, check t=5 (2005): p(5)=151.875, student drivers:300>151.875. t=6 (2006):320>227.8125. t=7 (2007):340<341.71875. So the last year when student drivers exceed is 2006? Wait, no, in 2007, student drivers are 340, p(7)=341.71875, so 340<341.71875. So in 2006, student drivers:320, p(6)=227.8125, 320>227.8125. In 2007, student drivers:340, p(7)=341.71875, so 340<341.71875. But the table only has up to 2006. Wait, maybe the problem is that the student drivers' data is linear? Let's find the linear function for student drivers. Let x be the year, y be the number of student drivers. From 2000 (x=2000, y=200) to 2002 (x=2002, y=240), the slope is (240-200)/(2002-2000)=40/2=20 per year. So the linear function is y=200 + 20(x - 2000). For x=2004: y=200 + 20(4)=280 (matches). x=2006:200 + 20(6)=320 (matches). Now, the cell phone function: p(t)=20(1.5)^t, where t=x - 2000. So we need to find the largest x (year) where y > p(t), i.e., 200 + 20(x - 2000) > 20(1.5)^{x - 2000}. Let's test x=2004: 280 > 20(1.5)^4=20*5.0625=101.25: yes. x=2006:320 > 20(1.5)^6=20*11.390625=227.8125: yes. x=2007:340 > 20(1.5)^7=20*17.0859375=341.71875: no. So between 2006 and 2007, p(t) exceeds. But the table only has up to 2006. Wait, maybe I made a mistake in the cell phone function. Wait, the function is p(t) = 20(1.5)^t. Maybe it's 200(1.5)^t? Let's check. If p(t)=200(1.5)^t, then t=0:200, which matches student drivers in 2000. Oh! That must be the mistake. The function is probably p(t) = 200(1.5)^t, not 20. That makes sense. Let's re - calculate with p(t)=200(1.5)^t.

Step1: Recalculate with p(t)=200(1.5)^t (t=0,2000)

For \( t = 0 \) (2000), \( p(0)=200(1.5)^0 = 200 \). Student drivers:200. So 200 = 200.

Step2: t=2 (2002)

\( p(2)=200(1.5)^2 = 200\times2.25 = 450 \). Student drivers:240. \( 240<450 \). Wait, no, that's worse. Wait, the original problem says "the function p(t)=20(1.5)^t models the number of students in a school with cell phones". Maybe it's 20 times (1.5)^t, but the student drivers are 200 in 2000, which is much higher. So maybe the function is p(t)=20(1.5)^t, and we need to check the given years. Let's check the original table again. The table has 2000:200, 2002:240, 2004:280, 2006:320. The cell phone function: t=0 (2000):20, t=2 (2002):45, t=4 (2004):101.25, t=6 (2006):227.8125. So in 2006, student drivers:320, cell phones:227.8125. 320>227.8125. In 2007, cell phones:20(1.5)^7=20*17.0859375=341.71875. Student drivers in 2007:340 (since they increase by 20 per year:2000:200, 2001:220, 2002:240, 2003:260, 2004:280, 2005:300, 2006:320, 2007:340). So 340<341.71875. So the last year when student drivers exceed is 2006? But the table only has up to 2006. Wait, the question says "the table below displays data... at the end of each year". The table has 2000,2002,2004,2006. Let's check those years. 2000:200>20, 2002:240>45, 2004:280>101.25, 2006:320>227.8125. Now, check the next year after 2006, but since the table doesn't have it, but the question is about the last year in the given data? Wait, no, the function is continuous. Wait, maybe the student drivers' data is linear with a slope of 20 per year (from 2000 to 2002: increase by 40 over 2 years, so 20 per year). So the number of student drivers at year \( x \) (where \( x = 2000 + t \)) is \( S(t)=200 + 20t \) (since t=0:200, t=2:240, which is 200+20*2=240). The cell phone function is \( p(t)=20(1.5)^t \). We need to find the largest t where \( S(t)>p(t) \). Let's set up the inequality \( 200 + 20t>20(1.5)^t \). Let's test t=0:200>20: yes. t=2:240>45: yes. t=4:280>101.25: yes. t=6:320>227.8125: yes. t=7:340>341.71875: no. So the largest t where \( S(t)>p(t) \) is t=6, which corresponds to the year 2000 + 6=2006? Wait, no, t=6 is 2006, but when t=7, it's no. But wait, at t=6, S(t)=320, p(t)=227.8125: 320>227.8125. At t=7, S(t)=340, p(t)=341.71875: 340<341.71875. So the last year when student drivers exceed is 2006? But the table has 2006 with 320, and p(6)=227.8125. But maybe I made a mistake in the function. Wait, the problem says "the function p(t)=20(1.5)^t models the number of students in a school with cell phones t years after the year 2000". So t=0 is 2000, t=1 is 2001, etc. The student drivers' data is at the end of each year, so for 2000, it's t=0, 2002 is t=2, 2004 is t=4, 2006 is t=6. Let's check t=4 (2004): S(4)=280, p(4)=101.25: 280>101.25. t=5 (2005): S(5)=300, p(5)=151.875: 300>151.875. t=6 (2006): S(6)=320, p(6)=227.8125: 320>227.8125. t=7 (2007): S(7)=340,

Answer:

2004