1. ## Another..AlgebraII-Help

Carl was running an experiment in his physics class. He was observing the amount of water that remained in a cylinder after a small hole was made in the bottom. Carl collected the following data.

Time (in sec.) Volume (in Liters)
20 15.15
25 14.05
40 11.20
50 9.63
60 8.28
75 6.60
80 6.11
100 4.52

a.) find the model for the data

b.)estimate the amonf of water remaining after 3 minutes

Thanks for the Help :]

2. Originally Posted by ep78
Carl was running an experiment in his physics class. He was observing the amount of water that remained in a cylinder after a small hole was made in the bottom. Carl collected the following data.

Time (in sec.) Volume (in Liters)
20 15.15
25 14.05
40 11.20
50 9.63
60 8.28
75 6.60
80 6.11
100 4.52

a.) find the model for the data

b.)estimate the amonf of water remaining after 3 minutes

Thanks for the Help :]
Here again, I uses a TI-84+ calculator to plot the bivariate data into a scatter plot. The plot seemed linear, so I performed a linear regession analysis and came up with this equation:

$\displaystyle y=-.1347632184x+17.02293103$

with a correlation coefficient of .9854

When x=180, y= -7.2 (already empty)

As near as I can tell, the cylinder went empty somewhere between 126 seconds and 127 seconds.

Ignore this post. Read the next one!

3. Originally Posted by ep78
Carl was running an experiment in his physics class. He was observing the amount of water that remained in a cylinder after a small hole was made in the bottom. Carl collected the following data.

Time (in sec.) Volume (in Liters)
20 15.15
25 14.05
40 11.20
50 9.63
60 8.28
75 6.60
80 6.11
100 4.52

a.) find the model for the data

b.)estimate the amonf of water remaining after 3 minutes

Thanks for the Help :]
Ok, hold everything. After looking closer at the plot, it seems that an exponential regession model works better with this equation:

$\displaystyle y=20.50690223(.984990842)^x$

with a correlation coefficient of .999999

Prediction of capacity after 180 seconds (3 minutes) = 1.348015 liters

4. oh okay..i don't understand what the correlation coefficient of .999999 is..or why you use it..?

Originally Posted by masters
Ok, hold everything. After looking closer at the plot, it seems that an exponential regession model works better with this equation:

$\displaystyle y=20.50690223(.984990842)^x$

with a correlation coefficient of .999999

Prediction of capacity after 180 seconds (3 minutes) = 1.348015 liters

5. sorry forget that question..i have a different one..i understand how to get the first equation that you gave me, but the second one..not so sure..cuz on the first one i picked points to get the slope..but i don't think you do that for the other one you gave me..right?

Originally Posted by ep78
oh okay..i don't understand what the correlation coefficient of .999999 is..or why you use it..?

6. Originally Posted by ep78
sorry forget that question..i have a different one..i understand how to get the first equation that you gave me, but the second one..not so sure..cuz on the first one i picked points to get the slope..but i don't think you do that for the other one you gave me..right?
The correlation coefficient, r, is .9999993314 which places the correlation into the "strong" category. (0.8 or greater is a "strong" correlation)
The coefficient of determination, r 2, is .9999986628 which means that 99.9% of the total variation in y can be explained by the relationship between x and y. Yes, it is a very "good fit".

You can use the slopes to determine a relationship and a line of best fit if the data is linear. The data in the second exercise was not linear. It followed an exponential curve. You would need some tool (statistics/graphing calculator) to complete this analysis. The algebra involved would be messy by hand.