Nt1310 Unit 4 Test Paper

Based on the data, the number of pill bugs on the dark side is higher at almost every minute compared the number of bugs on the light side. In the beginning, the pill bugs are evenly spread out on both sides. However, as time goes on, the number of pill bugs on the dark side increased rapidly, while the number on the light side decreases. Based on the graph, starting at about three minutes, the number of bugs on the dark side begins to increase significantly. The pill bugs quickly react and choose a best fit environment for themselves. Eventually, at nine minutes there are nine pill bugs on the dark side and only one on the light side. At the end of ten minutes, all ten of the bugs are on the dark side. If the experiment went on for a longer

This is then done for all trials. Then, once all five values of k are found, the average is taken by adding all five values of k and dividing by 5. The experimental k average is 0.105894M/s.

pH was recorded every time 1.00 mL of NaOH was added to beaker. When the amount of NaOH added to the beaker was about 5.00 mL away from the expected end point, NaOH was added very slowly. Approximately 0.20 mL of NaOH was added until the pH made a jump. The pH was recorded until it reached ~12. This was repeated two more times. The pKa of each trial are determined using the graphs made on excel.

Submission: The report from part 4 including all relevant graphs and numerical analysis along with interpretations.

3) What do the rate of change values you just calculated represent? Why are some positive and some negative?

The equation is dz/dx=ky(1-z/A). dz/dx is increasing rate of zombie, z is the population and A is the capacity (50). The k value we use is the average of the k values of the second to the fourth data point, k=1.0603. The function represent z in term of t is z=A-A/(B*e^kt+1). B is another constant. The B value we use is the average of each B value of each data point, B=0.0211. So z=50-50/(0.000256*e^(2.7477t)+1).

The state sequence of MKB is change as N0 ⊢ N1 ⊢ ··· ⊢ Nc, where the initial N0 is

10. Now, calculate the percentage of the cell cycle spent in each phase and record it in Data

You will need to spend some serious time with Figure 12.6. Use it to help you label this figure. Label each phase by name; then label the smaller structures. Finally, make 2 or 3 summary statements that indicate important features to note about the phase.

During the lag phase there is no increase in cell numbers, although the bacteria are synthesizing enzymes present in their environment in preparation for the exponential phase. During the exponential or logarithmic phase, the bacterial population grows at a rate that doubles the population during the generation time. The stationary phase incurs neither an increase nor a decrease in the cell population. The population growth cannot continue at the exponential rate since the nutrient supplies have been depleted and waste products have accumulated. The final phase of the bacterial population growth curve is the death phase, during which more cells die than are replaced by new cells.

Use a stopwatch to time 5 cycles and record the time in the data table.

Logistic equation is an example of one dimensional attractor. In order for us to study complicated phenomena we have to understand higher order attractor functions. Henon attractor is an example of two-dimensional attractor. It was introduced by a theoretical astronomer Michel Henon while he was studying stars movement in space [2]. On the other hand, Lorenz system is three-dimensional system published in the Journal of

The refined equation fits the data much more closely at the beginning, yet around the Age = 13 mark it starts to deviate again. Looking at this graph, one can come to the conclusion that it will take more than one mathematical function to graph the entirety of the data.

data. They suggested that a suitable value of k is approximately the square root of

max κ Using a permeability field κ whose initial value is shown in Figure 2, and the contrast minκ