Over a 12 hour period from 8 pm to 8 am the tempeture fell a steady rate 8°f to 16°f if the temperature fell at the same rate every hour what was the temperature at 4 am

What we know:
12 hour period from 8pm to 8am
temperature dropped from 8°F to 16°F from 8pm to 8am

We need to find temperature at 4 am.

We can start by setting up points:
8pm is are starting point with 8°F, we can express it as (0,8), 0 represents initial time from 0 to 12 hour span.
8am is the ending point with 16°F, we can express it as (12,16), 12 represents the end time of 0 to 12 hours span.
We will use these points to find slope.
slope=m=(16-8)/(12-0)=8/12=2/3

Now, we can set up an expression to find any temperature at a specific time.  Aslo, x represents the hours not the the specific time of 4am. We will use 8 since 4am is the 8th hour of the 12 hour span. Using slope of 2/3 and the y intercept of (0,8) since we were already at 8°F at the initial time of 0 we have the function:

f(x)=2/3x+8 

f(8)=2/3(8)+8= 40/3≈13.3°

Just a second zzz