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the table shows the estimated number of lines of code written by comput…

Question

the table shows the estimated number of lines of code written by computer programmers per hour when x people are working.
productivity
people working | lines of code written hourly
2 | 50
4 | 110
6 | 160
8 | 210
10 | 270
12 | 320
which model best represents the data?
○ $y = 47(1.191)^x$
○ $y = 34(1.204)^x$
○ $y = 26.9x - 1.3$
○ $y = 27x - 4$

Explanation:

Step1: Analyze linear models

First, check the linear models \( y = 26.9x - 1.3 \) and \( y = 27x - 4 \). For \( x = 2 \):

  • For \( y = 26.9(2)-1.3 = 53.8 - 1.3 = 52.5 \) (close to 50)
  • For \( y = 27(2)-4 = 54 - 4 = 50 \) (matches)

For \( x = 4 \):

  • \( y = 26.9(4)-1.3 = 107.6 - 1.3 = 106.3 \) (close to 110)
  • \( y = 27(4)-4 = 108 - 4 = 104 \) (close to 110, but 26.9x -1.3 is closer? Wait, no, wait 110: 26.94=107.6, 107.6-1.3=106.3; 274=108, 108-4=104. Wait, maybe I miscalculated. Wait 274=108, 108-4=104. 26.94=107.6, 107.6-1.3=106.3. The actual is 110. Now check exponential models.

Step2: Analyze exponential models

For \( y = 47(1.191)^x \), \( x=2 \): \( 47(1.191)^2 ≈ 471.418 ≈ 66.6 \) (not 50)
For \( y = 34(1.204)^x \), \( x=2 \): \( 34(1.204)^2 ≈ 341.449 ≈ 49.27 \) (close to 50)
\( x=4 \): \( 34(1.204)^4 ≈ 34(1.204^2)^2 ≈ 34(1.449)^2 ≈ 342.100 ≈ 71.4 \) (not 110)

Step3: Recheck linear models

For \( y = 26.9x -1.3 \), \( x=2 \): 52.5, \( x=4 \): 106.3, \( x=6 \): 26.96 -1.3=161.4 -1.3=160.1 (matches 160! Wow, that's close). \( x=8 \): 26.98 -1.3=215.2 -1.3=213.9 (close to 210). \( x=10 \): 26.910 -1.3=269 -1.3=267.7 (close to 270). \( x=12 \): 26.912 -1.3=322.8 -1.3=321.5 (close to 320).

For \( y = 27x -4 \), \( x=6 \): 276 -4=162 -4=158 (close to 160), \( x=8 \): 278 -4=216 -4=212 (close to 210), \( x=10 \): 2710 -4=266 (close to 270), \( x=12 \): 2712 -4=324 -4=320 (matches 320!). Wait, earlier miscalculation. Let's recalculate:

\( x=2 \): 27*2 -4=50 (matches)
\( x=4 \): 27*4 -4=108 -4=104 (actual 110, difference 6)
\( x=6 \): 27*6 -4=162 -4=158 (actual 160, difference 2)
\( x=8 \): 27*8 -4=216 -4=212 (actual 210, difference 2)
\( x=10 \):27*10 -4=266 (actual 270, difference 4)
\( x=12 \):27*12 -4=320 (matches 320)

For \( y = 26.9x -1.3 \):
\( x=2 \):52.5 (diff 2.5)
\( x=4 \):106.3 (diff 3.7)
\( x=6 \):160.1 (diff 0.1, perfect)
\( x=8 \):213.9 (diff 3.9)
\( x=10 \):267.7 (diff 2.3)
\( x=12 \):321.5 (diff 1.5)

The differences for \( y = 26.9x -1.3 \) are smaller overall? Wait \( x=6 \) is almost perfect. Let's check the pattern. The data points: 50,110,160,210,270,320. Let's see the differences between consecutive y-values: 110-50=60, 160-110=50, 210-160=50, 270-210=60, 320-270=50. Wait, not constant, but the linear model \( y=26.9x -1.3 \) fits very closely, especially at x=6 (160.1 vs 160), x=12 (321.5 vs 320), x=10 (267.7 vs 270), x=8 (213.9 vs 210), x=4 (106.3 vs 110), x=2 (52.5 vs 50). The exponential models are way off for x=4,6,8,10,12. The linear model \( y=26.9x -1.3 \) has very small residuals (differences between predicted and actual). Let's calculate residuals:

For \( y=26.9x -1.3 \):
x=2: 52.5 -50=2.5
x=4: 106.3 -110= -3.7
x=6: 160.1 -160=0.1
x=8: 213.9 -210=3.9
x=10:267.7 -270= -2.3
x=12:321.5 -320=1.5

For \( y=27x -4 \):
x=2:50 -50=0
x=4:104 -110= -6
x=6:158 -160= -2
x=8:212 -210=2
x=10:266 -270= -4
x=12:320 -320=0

The residuals for \( y=26.9x -1.3 \) are smaller in magnitude on average (2.5, 3.7, 0.1, 3.9, 2.3, 1.5) vs (0,6,2,2,4,0). Wait, but at x=6, \( y=26.9x -1.3 \) is almost perfect. The key is that the data looks roughly linear. Let's plot mentally: x=2 (50), x=4 (110), x=6 (160), x=8 (210), x=10 (270), x=12 (320). The slope between x=2 and x=4: (110-50)/(4-2)=30. Between x=4 and x=6: (160-110)/2=25. Between x=6 and x=8:25, x=8-10:30, x=10-12:25. So average slope around 27.5, which is close to 26.9 or 27. The model \( y=26.9x -1.3 \) is a better fit because at x=6 it's almost exact, and the residuals are smaller. The exponential models don't fit because the growth isn't…

Answer:

\( y = 26.9x - 1.3 \)