Answer:
A.Rate of Mrs. Allan's car=28 mile/gallon
Rate of Mrs. Owen's car=35 mile/gallon
B.Mrs. Owen's car
C.Mrs.Allen's car used gas for 560 miles=20 gallons
Mrs.Owen's car used gas for 560 miles=16 gallons
Explanation:
We are given that
Mrs. Allan's car uses 8 gallons of gas for 224 mile tripe.
Mrs. Owens car uses 6 gallons of a gas for 210 mile tripe.
A.Rate=
![(Distance\;traveled)/(Gas;used)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wqu09rlpbscs9dudnf41jrq05mmwofdmx6.png)
Rate of Mrs. Allan's car=
![(224)/(8)=<strong>28 </strong>mile/gallon](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zsa7ivcmipxvsb3nvaanisih1n5ds05ob2.png)
Rate of Mrs. Owen's car=
![(210)/(6)=<strong>35</strong> mile/gallon](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ifjyyjk61qulyefkz6x9pidnv0rzx7dmpc.png)
B.Mrs.Allan's car travel in 1 gallon of gas =28 miles
Mrs. Owen' s car travel in one gallon of gas=35 miles
Therefore, Mrs. Owen's car travel large distance in one gallon of gas.Hence, it is better.
C.If distance traveled by the cars=560 miles
Mrs. Allen's car used gas for 28 mile=1 gallon
Mrs. Allen's car used gas for 1 mile=
![(1)/(28)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/46w83mxi3v2zeccgvkh10wxhvt1k1sfb78.png)
Mrs.Allen's car used gas for 560 mile=
![(560)/(28)=20 gallons](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cx7wr8pbkoxedgemg228rlox937m8r6pq5.png)
Mrs.Allen's car used gas for 560 miles=20 gallons
Mrs.Owen's car used gas for 35 miles=1 gallon
Mrs. Owen's car used gas for 1 mile=
![(1)/(35)](https://img.qammunity.org/2020/formulas/mathematics/college/dzb06j1jmi8p1qxw3u2g0l85e9shbalvvk.png)
Mrs. Owen's car used gas for 560 miles=
![(560)/(35)=16 gallons](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hrcfxae00bcpv075fpl403qacbh72vnngf.png)
Mrs.Owen's car used gas for 560 miles=16 gallons