Try our Free Online Math Solver!

Thank you for visiting our site! You landed on this page because you entered a search term similar to this: how to solve exponential probability in TI83 Plus calculator, here's the result:
Weight W in newtons  Length l in meters 
20  2042 
40  2086 
60  2132 
80  2176 
100  2219 
If we were to plot this data on a graph of weight versus length, we would see that our data points would lie on (or very close to) a straight line. We would therefore conclude that a linear relationship exists between W and l. Many relationships exist in nature that are linear, but many relationships in nature are nonlinear as well. Bacteria, for example, grow in an exponential fashion, as will be covered later in this section.
We will now use the data given above in an actual example involving Hooke’s law. One form of Hooke’s law can be written in the following form
l = m W + l
The problem is to find values for m and l based on the data given above. We will then obtain a model for which we can calculate the extension of the wire produced by a weight of 75 newtons. Since l = m W + l is a linear relationship between l and W, then the graph of this relation will be a straight line with slope m and yintercept l. If we plot the experimental data as a graph and draw the best fitting line connecting all of the data points, we see that the line crosses the l axis at l = 2000 meters. l is therefore given the value 2000 meters, which is to be expected since the unextended length of the wire was given as 2 meters. Using the points (0, 2000) and (100, 2219), we can compute the line’s slope as follows:
m =
We can therefore rewrite our mathematical model as l = 000219W + 20 00
If we evaluate our model by setting W = 75 newtons, l becomes 2164 meters. This implies that the wire’s length has been extended by 0164 meters (1.64 mm) since 2164  2000 = 0164. If we take the ration of the extended length to the unextended length, we see that the wire’s length has increased by much less than 1% of its original length, i.e., 2164/2000 = 1082.
Interpolation and extrapolation are two very critical concepts in data analysis and the modeling of physical phenomena based on experiment. We now examine these concepts in relation to the linear problem posed above. Plot the data given above on a graph of l versus W, and label each point P , …, P respectively from top to bottom in our data table.
Now suppose that we are interested in the value of l for an imposed load somewhere between 40 and 60 newtons. Intuitively, we conclude that this value of l should lie somewhere between 2086 and 2132 meters given the data above. Now assuming Hooke’s law is valid, we can connect the points P and P with a straight line. We can now read directly from the graph the value of l which corresponds to a given value of W. The method used to obtain information between collected data points in this fashion is called linear interpolation. The term linear is used here to imply that we are using the method of interpolation with a linear relationship.
Notice that the experiment was conducted using weight values of up to 100 newtons. Suppose that we are interested in a certain value of l given by a value of W greater than 100 newtons. We would use the method of linear extrapolation in this case. The method is as follows. By Hooke’s law, we can connect P and P with a straight line. We can then extend the line to any value W desired to obtain a corresponding value for l.
It is important to realize that the methods of analysis outlined above give approximations and not precise values. Part of this is due to the fact that the values of l measured in an experiment may contain errors of varying degrees. We may however want to use interpolation or extrapolation to actually calculate more precise values of the dependent variable in question. To do so, we use the interpolation/extrapolation formula.
To use the interpolation/extrapolation formula, we must first have an experimentally based data set. Let’s call our independent variable x and our dependent variable y. Let’s also assume that we have at least two values for x, which we will denote x and x, and two for y, denoted y and . Now suppose that we are interested in the value of y, denoted , that corresponds to some value of x between x and x, which we will call . Note that the distance  x is just a fraction ( x)/( x  x) of the distance x  x. We can therefore conclude that the corresponding distance  y is the same fraction of  y. Setting these two ratio’s equal to each other and rearranging gives
, which gives
This is the interpolation/extrapolation formula, and it works well for both methods. Now let’s use this formula to calculate the extended length of the same steel wire as before, but for an attached weight of 30 newtons. We therefore take x=20 and x=40 newtons. We also take y=2042 and = 2086 meters, with =30 newtons in this case. Thus
= 2042+ = 2064 meters.
Notice that our answer makes perfect sense since the calculated value lies in between y and . Note also that we used interpolation in this case. To solve a similar problem using extrapolation, we would have chosen to be some weight value greater than those used in the experiment, say 200 newtons for instance.
One final note regarding interpolation and extrapolation: not all relationships are entirely linear. It turns out that the relationship between weight and the corresponding values for extended length of a steel wire is linear for a certain range of weights. As the weight load gets heavier and the wire gets closer to breaking, the graph of this relationship becomes nonlinear and more exponential in nature. This therefore implies that linear extrapolation may involve an associated amount of error as a result of working outside the scope of the data. Linear interpolation is useful for approximating values of the dependent variable for intermediate values of the independent variable. In both cases we are dealing with approximations, hence care should be taken when using either method.
Lesson 2: Bacterial Growth and Models Involving Exponentials
Standards/Benchmarks Addressed: 15
Suppose a microbiologist is interested in the rate at which Escherichia coli (a type of bacteria) grows under optimal growing conditions. Now lets suppose that the microbiologist measures the number of bacteria cells every ten minutes, and creates the following table based on these observations:
Time Elapsed (min)  0  10  20  30  40  50  60  70  80  90  100 
Number of cells  7  10  14  20  27  39  54  76  108  151  213 