Question

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Answer 1

Your answer is correct. As for the second part, the expected usage in ideal (non-windy) travel conditions for the round trip should be 2f. So it wouldn't be any different from the fuel economy on windy conditions because (f+w)+(f-w)=2f

Answer 2

" Compare how many gallons of fuel they used for the round trip to the expected usage in ideal (non-windy) travel conditions." sounds to me like \(\bf \cfrac{1287mi}{\frac{22mi}{gallon}}\implies \cfrac{1287}{22}gallons\quad \textit{one way, roundtrip}\implies \cfrac{1287}{22}gallons\times 2\)

Answer 3

hmmm \(\bf \cfrac{1287mi}{\frac{22mi}{gallon}}\implies \cfrac{1287}{22}gallons\quad \textit{one way, roundtrip}\\\quad\\\implies \cfrac{1287}{22}gallons\times 2\)

Answer 4

the 1st answer looks good though

Answer 5

so the 1st ROUNDTRIP uses \(\bf \cfrac{1287mi}{\frac{18mi}{gallon}}+\cfrac{1287mi}{\frac{26mi}{gallon}}\) gallons for the whole ROUNDTRIP

so see which one used more, the one with the windy days, or the one with no wind