Question 963525
The outside dimension of the walk is (100'+6') x (150'+6'),
so the area of walk and lot is (106')(156')=16536ft^2
The area of the lot is (100')(150')=15000ft^2
So area of walk=(area of lot+walk)-(area of lot)=16536ft^2-15000ft^2=1536ft^2
There are 9ft^2 per yd^2:
(1536ft^2)(1yd^2/9ft^2)=170.67yd^2 
The area of the walk is 170.67 square yards.
The depth of the walk is (1/9)yard so the volume is:
(170.67yds^2)(1/9 yd)=18.96yd^3
Cost of concrete=(18.96yd^3)($24/yd^3)=$455.04
ANSWER: The cost of the concrete would be $455.04.