 This problem, we need to calculate composite r-value. The wall consists of four layers, 1 half inch wood siding. And its r-value is given straight away 4 half inch as 0.81. And we have 3 quarters inch plywood. And this plywood's r-value is also given as 0.94. This is 4 3 quarters inch. Whereas the fiberglass, each inch has an r-value of 3.7. We are using 3 and 1 half inches. So 3 and 1 half times 3.7 would give us about 13 r-value. And the last layer would be 1 half inch plasterboard, which is also drywall. And its r-value is given as 0.45. So when you add these up, you get 15.2, which means the composite r-value of this wall is 15.2 degrees Fahrenheit foot square hour over BTUs.