Asked by Layne
Water boils at 212 degrees F at sea level (altitude 0 ft), 204.6 degrees F at 40000 ft, 197.2 deg. F at 8000 ft, and 189.8 deg. F at 12,000 ft. Find a good model relating the boiling temperature t of water to the altitude a.Is the model linear or quadratic?
Answers
Answered by
RickP
0 ft altitude to 4,000 ft (I assume you made a typo) is an increase of 4,000 ft and a decrease of 7.4 degrees F.
4,000 ft to 8,000 ft is an increase of 4,000 ft and a decrease of 7.4 degrees.
8,000 ft to 12,000 ft is an increase of 4,000 ft and a decrease of 7.4 degrees.
So, from the data given, the trend is that each increase in 4,000 ft altitude results in a drop of 7.4 degrees F.
Let y = degrees F and x = altitude in ft. Then y = 212 - 7.4(x / 4000)
Which can be reexpressed as:
y = 212 - 7.4x / 4000
or
y = 212 - (7.4/4000)x
or
y = (-7.4/4000)x + 212
or
y = (-74/40000)x + 212
or finally
y = (-37/20000)x + 212
There's you functional model
Now, does that look like it fits the equation of a line, y = mx + b???
4,000 ft to 8,000 ft is an increase of 4,000 ft and a decrease of 7.4 degrees.
8,000 ft to 12,000 ft is an increase of 4,000 ft and a decrease of 7.4 degrees.
So, from the data given, the trend is that each increase in 4,000 ft altitude results in a drop of 7.4 degrees F.
Let y = degrees F and x = altitude in ft. Then y = 212 - 7.4(x / 4000)
Which can be reexpressed as:
y = 212 - 7.4x / 4000
or
y = 212 - (7.4/4000)x
or
y = (-7.4/4000)x + 212
or
y = (-74/40000)x + 212
or finally
y = (-37/20000)x + 212
There's you functional model
Now, does that look like it fits the equation of a line, y = mx + b???
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