Assume temp T = T0 (75 0) at Noon (00), To find out T after 4 hours at 4 p.m. or at the 8th period, each period being 0.5 hour. If the relationship is linear, i.e if temperature is dropping linearly each half an hour by 3 0, the T is a staight line T=m*t+ T0, in which the slope m=∇T/∇t = (-30/0.5 hr) = -60/hour (negative slope decreasing)
So at 4 p.m. for t=4, temp T = 750 -60/hour*4 hour= 750-240= 510
