<span>The number of dollars collected can be modelled by both a linear model and an exponential model.
To calculate the number of dollars to be calculated on the 6th day based on a linear model, we recall that the formula for the equation of a line is given by (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), where (x1, y1) = (1, 2) and (x2, y2) = (3, 8)
The equation of the line representing the model = (y - 2) / (x - 1) = (8 - 2) / (3 - 1) = 6 / 2 = 3
y - 2 = 3(x - 1) = 3x - 3
y = 3x - 3 + 2 = 3x - 1
Therefore, the amount of dollars to be collected on the 6th day based on the linear model is given by y = 3(6) - 1 = 18 - 1 = $17
To calculate the number of dollars to be calculated on the 6th day based on an exponential model, we recall that the formula for exponential growth is given by y = ar^(x-1), where y is the number of dollars collected and x represent each collection day and a is the amount collected on the first day = $2.
8 = 2r^(3 - 1) = 2r^2
r^2 = 8/2 = 4
r = sqrt(4) = 2
Therefore, the amount of dollars to be collected on the 6th day based on the exponential model is given by y = 2(2)^(5 - 1) = 2(2)^4 = 2(16) = $32</span>
Answer:
The maximum possible error of in measurement of the angle is 
Step-by-step explanation:
From the question we are told that
The angle of elevation is 
The height of the tree is h
The distance from the base is D
h is mathematically represented as
Note : this evaluated using SOHCAHTOA i,e

Generally for small angles the series approximation of 

So given that 


=> 
Now from the question the relative error of height should be at most
%
=> 
=> 
=> 
So for 

substituting values
![d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p](https://tex.z-dn.net/?f=d%20%5B%5Cfrac%7B%5Cpi%7D%7B12%7D%20%5D%20%20%3D%20%20%5Cpm%20%20%5Cfrac%7B%5B%5Cfrac%7B%5Cpi%7D%7B12%7D%20%5D%20%2B%20%20%5Cfrac%7B%5B%5Cfrac%7B%5Cpi%7D%7B12%7D%20%5D%5E3%20%7D%7B3%7D%20%7D%7B1%2B%20%5B%5Cfrac%7B%5Cpi%7D%7B12%7D%20%5D%20%5E2%7D%20%2A%20%20%20%20%5C%20p)
=> 
Converting to degree


If you are talking about a graph:
The input would be the number of adult tickets.
The output would be the amount of money.
If you talking about it normally:
The input would be the amount of money.
The output would be the number of adult tickets.