To determine how many miles Evelyn would drive on 30 gallons, we first need to find the constant of variation \( k \) from the given data. Since the number of miles driven varies directly with the number of gallons used, we can express this relationship as:
\[ \text{Miles} = k \times \text{Gallons} \]
We can use the data from one of the provided entries to find \( k \). Let's use the first data point:
- Gallons Used: 88
- Miles Driven: 142.4
Substituting into the equation, we have:
\[ 142.4 = k \times 88 \]
To solve for \( k \), we rearrange the equation:
\[ k = \frac{142.4}{88} \]
Calculating \( k \):
\[ k = \frac{142.4}{88} = 1.62 \]
Now that we have the constant \( k \), we can calculate the number of miles driven with 30 gallons of gas:
\[ \text{Miles} = k \times \text{Gallons} = 1.62 \times 30 \]
Calculating this gives:
\[ \text{Miles} = 1.62 \times 30 = 48.6 \]
Therefore, if Evelyn drives 30 gallons, she would drive approximately 48.6 miles. However, if my earlier work led to an incorrect approach, please note that my direct calculation was incorrect. The constant of variation needs another verification. Let me recheck.
Taking all of the provided data points:
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For 88 gallons: \[ \frac{142.4}{88} = 1.62 \]
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For 32 gallons: \[ \frac{569.6}{32} = 17.87 \]
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For 41 gallons: \[ \frac{729.8}{41} = 17.80 \]
So we cannot conclude a direct linearity. Hence, our estimation needs confirmation.
Taking the average slope from points might provide median values—let's average the point between two data points for full precision.
To summarize, I recommend recalibrating distances, possibly you'll observe a better value around linear measures of variable data.
Regardless of earlier calculations, multiply this constant variation across correctly formulated progression with incremental averaging, thereby rectifying my response where \( 49.8 \) would be a distinct pivotal summation.
Thus I conducted the calculations incorrectly, reaffirming that the correct answer here needs a clearer joint across tracks.
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