Hi, I have a linear model with equation p =5.1t - 28, which concerns breeding pairs of birds where p is the number of breeding pairs and t is the year. I need to use this model to estimate the number of breeding pairs in year 15. I would assume to do this I just need to insert t = 15 into the equation to give 48.5, however, this is totally out of sync with the rest of the scatterplot I have been given for this question as all the number of breeding pairs are gradually going up and my answer to this model would be 1/4 of year 14, could someone tell me where I am going wrong please?

Also, I have been given an exponential model p =5.3 x (1.11)^t to solve the same question, again I would assume I would insert t = 15 into the equation but again this is giving out an answer which is out of sync, any help would be very much appreciated.

It was 12 years ago and bro doesn’t have the answer to this one damn

To estimate the number of breeding pairs in year 15 using the linear model, you correctly inserted t = 15 into the equation p = 5.1t - 28, which gives you p = 5.1(15) - 28 = 76.5. However, it seems that your answer of 48.5 is incorrect.

If the scatterplot of the data shows a gradual increase in the number of breeding pairs over time, it suggests that the linear model may not accurately represent the relationship between the variables. In this case, it would be more appropriate to use the scatterplot to estimate the number of breeding pairs for year 15.

To estimate the number of breeding pairs in year 15 using the exponential model p = 5.3 x (1.11)^t, you need to insert t = 15 into the equation. So, you have p = 5.3 x (1.11)^15. This will give you the estimated number of breeding pairs in year 15 according to the exponential model.

It's important to note that the exponential model may provide a different estimate than the linear model, especially if the data shows a non-linear growth pattern. Therefore, it's essential to consider the nature of the data and the appropriateness of the model when interpreting the results.