Lab Report 1 Logarithmic Plotting Devin Edwards ENGR 3070L CRN: 27194 January 17, 2018 Dr. Margraves Objective The purpose of this experiment is to graph and look at the logarithmic plots and write corresponding exponential equations that match the “Best Fit” line of the data points. Theory The data in Table 1 can be represented by the exponential equation given in equation 1 below. Equation 1 is also used for Cartesian plots: Q=KH^n (1) On this type of plot a straight line is drawn representing the slope intercept form in equation 2: y=mx+b (2) The variables are switched out allowing it to represent the same relationship in logarithmic form, where y is log(Q), x is log(H), n is m, and b is log(K): log(Q)=log(K)+n[log(H)] (3) …show more content…
In the first method two random points are chosen along the best fit line of the data. The vertical distance and horizontal distance are measure in cm. The distance in the vertical direction is divided by the distance in the horizontal direction “rise over run”. Using this particular method I calculated that the value of n is .604. The second method was also used considering that two points along the best fit line were exactly one cycle away, meaning the value of H2 was ten times larger than the value of H1. Equation 3 was used in this method. It became easier since the value of log10 is equivalent to 1. After Q1 and Q2 were chosen at the H values, the next step is to solve for log Q2/Q1. This calculated value of n came out to be 0.5057. There were also two different methods for solving for the value of K using the best fit line given in Figure 1. By finding where the first K is equal to 1 and seeing what the Q value is at this given point, K can be solved. As said above when H equals 1 the Q value is the same as the K value. The K value was calculated to be 4.7. In the second method for solving for K, use the values calculated for the variables Q, K, and n and plug into Equation 1. By doing this the calculated value was 4.645. The value of K calculated in excel was calculated to be
Lets represent the ascii values from B to K in binary B = 66 = 01000001 C = 67 = 01000010 D = 68 = 01000011 E = 69 = 01000100 F = 70 = 01000101 G = 71 = 01000110 H = 72 = 01000111 I = 73 = 01001000 J = 74 = 01001001
I initialised the count value to ‘0’ and declared the k variable as “100000000” and Variable Nu is equal to “00000000” & Num because in this algorithm division function has low nominator compared to denominator so we end up with fraction values like 0.25, 0.33 etc. To do the binary division in this case is pretty challenging, one should have a better understanding and basic knowledge of how it’s done. Binary shift division is the solution to the problem here, whenever the nominator is lower than denominator value is shifted to the left by adding a ‘0’ to the last digit of the value until the nominator gets bigger than denominator and when denominator fits in nominator subtraction is done and a new value
In Table \ref{parameter_table} we present the values of the input parameters using the DAs of $N$. %In this section, we will only consider the central values of these parameters. \begin{table}[t] \addtolength{\tabcolsep}{10pt} \begin{tabular}{ccccccc} \hline\hline
To compute rho, the program GSC threshold.m denes two non-linear functions root2d and root2r as in Equation (15) and (16) of [1]. Each of these functions represents a system of non-linear equations in two variables. The program numerically solves these two by two systems of non-linear equations by using the inbuilt MatLab function f-solve. Since the probabilities, PNi are numerical solution computed by MatLab these values can be very very small numbers. To avoid these artifacts, the program replaces values of rho less than l_t by
gen RHSB=4+(4/(50^.5)) gen out1=cond(meanaRHSA,1,0) gen out2=cond(meanbRHSB,1,0)
This new formula would give
To graph population or disease, we needed to use exponents; in equation-form, the exponent was an X, but it could be substituted for any number, which would represent the year. You would also find the current population or number of cases and divide them by the amount the previous year (the starting number) and add that to one to find the rate, which would show you if it was growth or decay. Finally, you use the starting number as your constant or y-intercept. If you were trying to graph the decay of a population, the equation could be: y=150,000(1.5)x; if you were trying to graph decay, the equation could be: y=150,000(0.5)x. You can replace X with any number (number of years) to find the population in the future (positive number) or in the past (negative numbers).
Using FIT with theses adjusted patterns I feel confident that I will complete this assignment
The predicted and experimental responses are compared in order to validate the model and to calculate the prediction error. The prediction error was found to be below 7% indicating that the observed responses were very close to the predicted values. Percentage prediction error is useful in constituting the validity of generated equations and describes how close the predicted responses to that of actual values. The values of <15 are desirable to have closeness of the predicted values with the actual values
4. Calculate the difference amidst theoretical, simulated and practical values. 5. Consult Hishan to probe in details about problematic components. 6.
33.385 million =10.235 million * exp (0.12*10) is the value of the of the population when the rate changes by an increase of 2%. 38.950 million =10.235 million * exp (0.14*10) is valuation of the future population when the rate increases by 4% from its original rate of 10% for a total rate of 14%. Each input is accurate in comparison with its Excel counterpart, however, the Word calculations will have greater precision due to the estimation of the of the Excel counterparts.
Use your results in Data Table 2 to support your answer.
Predict/ roughly determine the Vmax and ½ Vmax values from the peak of the graph, where the slope of the graph levels off (the asymptotical line). Predict/ roughly determine the Km by reading off of the graph the corresponding substrate concentration on the x-axis for the ½ Vmax value. Plot a Lineweaver-Burke graph (the inverse of the velocity of the reaction vs. the inverse of the substrate concentration). Calculate accurate Vmax and Km values using the following equation for the Lineweaver-Burk
IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75
The graph shows the average volume of hydrogen that was produced from the 3 trials and the average volume of oxygen that was produced from the 3 trials across the voltage. I added the volumes of hydrogen in each trial and I divided them by 3 to get the average and I made the same thing for the volume of oxygen. The graph shows that the volume of hydrogen produced during the experiment is twice as much as the volume of oxygen. For example using the third data when I used 11 volts the average volume of hydrogen that was produced was 5.8 cm3 and the average volume of oxygen produced was 2.9 cm3