Distance (x), Temperature(y): predict temp at 75 mi going E

[therese478]

If my distance (x) starts at 0 miles and ends at 25 miles. And my temperature (y) Starts at 85.2 and
ends at 83.6 degrees Fahrenheit. and I want to perdict the temperature at 75 miles going east. Would this mean my temperature is going up the y axis. Because the distance is going east. Would I go about doing this as. Y=-0.062*(75)+85.26 or do I use the start which is 85.2 or do I use temp at 83.6 which is my last known temperature.

 

[Deleted member 4993]

Re: Distance (x), Temperature(y)

If my distance (x) starts at 0 miles and ends at 25 miles. And my temperature (y) Starts at 85.2 and
ends at 83.6 degrees Fahrenheit. and I want to perdict the temperature at 75 miles going east. Would this mean my temperature is going up the y axis. Because the distance is going east. Would I go about doing this as. Y=-0.062*(75)+85.26 or do I use the start which is 85.2 or do I use temp at 83.6 which is my last known temperature.

Please post the exact problem - as posted it is not complete.

Does your statement:

distance (x) starts at 0 miles and ends at 25 miles. ...temperature (y) Starts at 85.2 and
ends at 83.6 degrees Fahrenheit

mean that at x = 0, y = 85.2 and at x = 25, y = 83.6 (providing a relationship - function - between x & y)

or is it merely your graphing window.

 

[galactus]

Re: Distance (x), Temperature(y)

Assuming you mean at 0 the temp is 85.2 and at 25 the temp is 83.16, then you have a negative slope because the temp is going down as x increases.

The slope would be \(\displaystyle m=\frac{y-y_{1}}{x-x_{1}}=\frac{83.16-85.2}{25-0}=-.0816=\frac{-51}{625}\)

Then we can use either data point to find the equation.

\(\displaystyle y=mx+b\)

We know b=85.2 because that is where the line crosses the y-axis.

In fractional form, which looks nicer, we have \(\displaystyle y=\frac{-51}{625}+\frac{426}{5}\)

See how it works now?.

Now, just plug in x=75 to find the temp there.