Question

1. use the graph paper on the back of this page to plot the data in the table at right. make the graph fill the page as much as possible; label the axes and choose a uniform scale for each axis. use a ruler to draw a best - fit line, then answer the following. (draw dotted lines on the graph showing how you find each answer.)

a. what is the density of gasoline with 9.2% additive?
b. what is the percent additive in gasoline with density 1.04 g/ml? (handwritten “15%” near this part)
c. what is the density of pure gasoline?
table:
% fuel additive: 2, 4, 6, 8, 10, 12, 14, 16
density (g/ml): 0.82, 0.85, 0.89, 0.92, 0.95, 0.99, 1.02, 1.06

Explanation:

Part a:

Step1: Identify the data point

We have a table with % Fuel Additive and Density. For 9.2% additive, we look at the table. Wait, the table has % Fuel Additive as 2,4,6,8,10,12,14,16 and corresponding densities. Wait, maybe there's a typo or maybe we need to use linear interpolation. Wait, the % Fuel Additive values are 2,4,6,8,10,12,14,16. So 8% additive has density 0.92 g/mL, 10% has 0.95 g/mL. Wait, 9.2% is between 8% and 10%. Let's find the slope between 8% (d=0.92) and 10% (d=0.95). The change in % is 10 - 8 = 2, change in density is 0.95 - 0.92 = 0.03. So per 1% change, density changes by 0.03/2 = 0.015. Now, 9.2% - 8% = 1.2%. So the density increase is 1.2 * 0.015 = 0.018. So density at 9.2% is 0.92 + 0.018 = 0.938 g/mL. Wait, but maybe the table is misread. Wait, the % Fuel Additive: 2 (0.82), 4 (0.85), 6 (0.89), 8 (0.92), 10 (0.95), 12 (0.99), 14 (1.02), 16 (1.06). So the pattern: from 2 to 4 (2% increase), density increases by 0.03; 4 to 6 (2% increase), density increases by 0.04; 6 to 8 (2% increase), density increases by 0.03; 8 to 10 (2% increase), density increases by 0.03; 10 to 12 (2% increase), density increases by 0.04; 12 to 14 (2% increase), density increases by 0.03; 14 to 16 (2% increase), density increases by 0.04. Hmm, maybe linear regression? Let's calculate the linear equation. Let x be % additive, y be density.

Points: (2, 0.82), (4, 0.85), (6, 0.89), (8, 0.92), (10, 0.95), (12, 0.99), (14, 1.02), (16, 1.06)

Calculate slope (m) and intercept (b) of y = mx + b.

Sum of x: 2+4+6+8+10+12+14+16 = 72

Sum of y: 0.82+0.85+0.89+0.92+0.95+0.99+1.02+1.06 = let's calculate:

0.82+0.85=1.67; +0.89=2.56; +0.92=3.48; +0.95=4.43; +0.99=5.42; +1.02=6.44; +1.06=7.5

Sum of x^2: 4 + 16 + 36 + 64 + 100 + 144 + 196 + 256 = 4+16=20; +36=56; +64=120; +100=220; +144=364; +196=560; +256=816

Sum of xy: (20.82)+(40.85)+(60.89)+(80.92)+(100.95)+(120.99)+(141.02)+(161.06)

Calculate each term:

20.82=1.64; 40.85=3.4; 60.89=5.34; 80.92=7.36; 100.95=9.5; 120.99=11.88; 141.02=14.28; 161.06=16.96

Sum of xy: 1.64+3.4=5.04; +5.34=10.38; +7.36=17.74; +9.5=27.24; +11.88=39.12; +14.28=53.4; +16.96=70.36

n=8

Slope m = (nSum(xy) - Sum(x)Sum(y)) / (n*Sum(x^2) - (Sum(x))^2)

m = (870.36 - 727.5) / (8*816 - 72^2)

Calculate numerator: 870.36=562.88; 727.5=540; 562.88 - 540=22.88

Denominator: 8*816=6528; 72^2=5184; 6528 - 5184=1344

m=22.88 / 1344 ≈ 0.01702

Intercept b = (Sum(y) - mSum(x))/n = (7.5 - 0.0170272)/8

0.01702*72≈1.22544; 7.5 - 1.22544≈6.27456; 6.27456/8≈0.7843

So equation: y = 0.01702x + 0.7843

Now, for x=9.2, y=0.01702*9.2 + 0.7843 ≈ 0.1566 + 0.7843 ≈ 0.9409 g/mL. Approximately 0.94 g/mL.

But maybe the table is such that at 8% it's 0.92, 10% is 0.95. So the difference between 8 and 10 is 2% additive, 0.03 density. So per 1% additive, 0.015 density. So 9.2% is 8% + 1.2%, so 0.92 + 1.2*0.015 = 0.92 + 0.018 = 0.938 ≈ 0.94 g/mL.

Step2: Conclusion

Using linear interpolation or regression, the density at 9.2% additive is approximately 0.94 g/mL.

Step1: Identify the density

We need to find % additive when density is 1.04 g/mL. Using the linear equation y = 0.01702x + 0.7843, where y=1.04.

So 1.04 = 0.01702x + 0.7843

Subtract 0.7843: 1.04 - 0.7843 = 0.2557 = 0.01702x

x = 0.2557 / 0.01702 ≈ 15.02% ≈ 15%

Alternatively, looking at the table: 14% additive has density 1.02, 16% has 1.06. So 1.04 is halfway between 1.02 and 1.06? Wait, 1.02 (14%) and 1.06 (16%). The difference in density is 0.04 over 2% additive. So per 0.01 density, 0.5% additive. 1.04 - 1.02 = 0.02, so 0.02 / 0.04 * 2% = 1%. So 14% + 1% = 15%. So that matches the earlier calculation.

Step2: Conclusion

The % additive when density is 1.04 g/mL is approximately 15%.

Step1: Pure gasoline means 0% additive

Using the linear equation y = 0.01702x + 0.7843, when x=0 (0% additive), y=0.7843 g/mL.

Alternatively, looking at the trend: at 2% additive, density is 0.82. The density at 0% would be 0.82 - (0.85 - 0.82) [since from 0 to 2% is 2%, and from 2 to 4% is 0.03 increase]. Wait, from 2% (0.82) to 0% (2% decrease), density would decrease by the same as from 0 to 2%. The increase from 0 to 2%: let's see the first two points: (2, 0.82) and (4, 0.85). The slope between 2 and 4 is (0.85 - 0.82)/(4 - 2) = 0.03/2 = 0.015 per % additive. So from 0 to 2% (x=0 to x=2), density increases by 0.0152 = 0.03. So density at x=0 is 0.82 - 0.03 = 0.79 g/mL? Wait, but the linear regression gave 0.7843, which is close to 0.79. Wait, maybe the table's first point is 2% (0.82), so pure gasoline (0%) would be 0.82 - (0.85 - 0.82) = 0.82 - 0.03 = 0.79? Wait, no, the slope from 2 to 4 is 0.015 per %, so from 0 to 2, it's also 0.015 per %, so 0% would be 0.82 - 20.015 = 0.82 - 0.03 = 0.79 g/mL. But the linear regression gave ~0.784, which is ~0.78 - 0.79.

Wait, let's check the linear equation again. When x=2, y=0.017022 + 0.7843 ≈ 0.034 + 0.7843 ≈ 0.8183, which is close to 0.82. When x=4, 0.017024 + 0.7843 ≈ 0.068 + 0.7843 ≈ 0.8523, close to 0.85. So the equation is good. So at x=0, y=0.7843 ≈ 0.78 g/mL.

Alternatively, looking at the table, the density increases with % additive. So pure gasoline (0% additive) should be less than 0.82 g/mL. The linear regression gives ~0.78 g/mL, which is reasonable (gasoline density is typically around 0.7 - 0.8 g/mL).

Step2: Conclusion

The density of pure gasoline (0% additive) is approximately 0.78 - 0.79 g/mL, more precisely ~0.784 g/mL.

Answer:

Approximately 0.94 g/mL (or more precise 0.938 - 0.94 g/mL)

Part b: