Question

emma recently purchased a new car. she decided to keep track of how many gallons of gas she used on five of her business trips. the results are shown in the table below.

miles driven	number of gallons used
200	10
400	19
600	29
1000	51

round your answer to the nearest hundredth. what is the slope of this equation?

Explanation:

Step1: Identify variables

Let \( x \) be miles driven (independent variable) and \( y \) be gallons used (dependent variable). We can use two points to calculate the slope, but since it's a linear relationship approximation (or using the concept of slope as \( \frac{\Delta y}{\Delta x} \), we can also use the formula for the slope of a line through multiple points, but here we can use the first and last points for simplicity (or any two, but let's check). Let's take the first point \((150, 7)\) and the last point \((1000, 51)\).

Step2: Apply slope formula

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \).

Substitute \( x_1 = 150 \), \( y_1 = 7 \), \( x_2 = 1000 \), \( y_2 = 51 \):

\( m=\frac{51 - 7}{1000 - 150}=\frac{44}{850}\approx0.05176\approx0.05 \) (Wait, no, wait, maybe I should check with another pair. Wait, maybe the relationship is linear, let's check the slope between (200,10) and (400,19): \( \frac{19 - 10}{400 - 200}=\frac{9}{200}=0.045 \). Between (400,19) and (600,29): \( \frac{29 - 19}{600 - 400}=\frac{10}{200}=0.05 \). Between (600,29) and (1000,51): \( \frac{51 - 29}{1000 - 600}=\frac{22}{400}=0.055 \). Wait, maybe we should use the least - squares method? Wait, the problem says "the slope of this equation", maybe assuming a linear regression? Wait, but maybe the problem expects using two points, or maybe it's a linear relationship. Wait, let's recalculate with (150,7) and (1000,51):

\( m=\frac{51 - 7}{1000 - 150}=\frac{44}{850}\approx0.0518 \approx 0.05\)? Wait, no, 44 divided by 850: 44÷850 = 0.0517647...≈0.05 (to nearest hundredth). Wait, but let's check with (200,10) and (1000,51): \( \frac{51 - 10}{1000 - 200}=\frac{41}{800}=0.05125\approx0.05 \). Wait, maybe the intended approach is to use the formula for slope as \( \frac{\text{change in } y}{\text{change in } x} \), and maybe the problem is assuming a linear model, so we can use the first and last points.

Wait, let's do the calculation properly:

\( x_1 = 150, y_1 = 7 \); \( x_2 = 1000, y_2 = 51 \)

\( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{51 - 7}{1000 - 150}=\frac{44}{850}\approx0.05 \) (rounded to nearest hundredth). Wait, but 44/850 = 0.05176... which is approximately 0.05 when rounded to the nearest hundredth? Wait, no, 0.05176 is closer to 0.05 or 0.06? The thousandth place is 1, so when rounding to the nearest hundredth (two decimal places), 0.05176≈0.05. Wait, but let's check with another pair. Let's take (600,29) and (1000,51):

\( m=\frac{51 - 29}{1000 - 600}=\frac{22}{400}=0.055\approx0.06 \). Hmm, there is a discrepancy. Wait, maybe the problem is expecting a linear regression? Let's calculate the slope using linear regression formula.

The formula for the slope \( m \) in linear regression is \( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}} \)

First, let's create a table:

| \( x \) (Miles Driven) | \( y \) (Gallons Used) | \( xy \) | \( x^{2} \) |
|---|---|---|---|
| 150 | 7 | \( 150\times7 = 1050 \) | \( 150^{2}=22500 \) |
| 200 | 10 | \( 200\times10 = 2000 \) | \( 200^{2}=40000 \) |
| 400 | 19 | \( 400\times19 = 7600 \) | \( 400^{2}=160000 \) |
| 600 | 29 | \( 600\times29 = 17400 \) | \( 600^{2}=360000 \) |
| 1000 | 51 | \( 1000\times51 = 51000 \) | \( 1000^{2}=1000000 \) |

Now, calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^{2} \)

\( \sum x=150 + 200+400 + 600+1000=2350 \)

\( \sum y=7 + 10+19 + 29+51=116 \)

\( \sum xy=1050+2000 + 7600+17400+51000=79050 \)

\( \sum x^{2}=22500+40000 + 160000+360000+1000000=1582500 \)

\( n = 5 \)

Now, substitute into the slope formula:

\( m=\frac{5\times79050-2350\times116}{5\times1582500-(2350)^{2}} \)

First, calculate numerator:

\( 5\times79050 = 395250 \)

\( 2350\times116=2350\times(100 + 16)=235000+37600 = 272600 \)

Numerator: \( 395250-272600 = 122650 \)

Denominator:

\( 5\times1582500=7912500 \)

\( (2350)^{2}=2350\times2350 = 5522500 \)

Denominator: \( 7912500 - 5522500=2390000 \)

Now, \( m=\frac{122650}{2390000}\approx0.0513 \approx 0.05 \) (rounded to nearest hundredth)

Wait, but when we calculated with (600,29) and (1000,51) we got 0.055, but the linear regression gives approximately 0.0513, which is approximately 0.05 when rounded to the nearest hundredth.

Alternatively, maybe the problem is assuming a linear relationship and we can use two points, say (150,7) and (1000,51):

\( m=\frac{51 - 7}{1000 - 150}=\frac{44}{850}\approx0.05 \) (rounded to nearest hundredth)

Answer:

\( 0.05 \) (or more accurately, if we consider the linear regression result, approximately \( 0.05 \) when rounded to the nearest hundredth. If we use (200,10) and (1000,51), \( \frac{41}{800}=0.05125\approx0.05 \) as well)