Question #5ae10

1 Answer
Sep 13, 2017

Answer:

I tried this:

Explanation:

The two scales are related but in a slight complicated way (because the difference in intervals between fixed points of the two). Anyway, we have:

#"Temp. in Celsius"=5/9xx("Temp. in Farhenheit"-32)#

you need to insert #98.6# into the formula and evaluate the Temperature in Celsius.

[You should get #37^@C#]