9514 1404 393
Answer:
5.6 miles
Explanation:
Let the following distances between exits be defined:
a = exit 5 to 6
b = exit 6 to 7
c = exit 7 to 8
Then the given relations are ...
a = 3.2
b = 1.1 +a = 4.3
(b +c) = 2.4 +(a +b) ⇒ c = a +2.4 = 5.6
The distance between exits 7 and 8 is 5.6 miles.