Mat 540 Final Paper

& { 2872(25$\%$)} & { 2499(22$\%$)} & { 5795(26$\%$)} & { 5100(23$\%$)}\\ $N_{\omega\to\pi^0\gamma}^{\circ}$ & { 4487(15$\%$)} & { 3590(12$\%$)} & { 1978(6$\%$)} & { 1721(5$\%$)} & { 5846(9$\%$)} & { 5145(8$\%$)} \\ \hline $BR^{measured}_{\omega\to\pi^0\gamma}$ & \textcolor{red}{ 1.07} & \textcolor{red}{ 0.78} & \textcolor{red}{ 0.52} & \textcolor{red}{ 0.43} & \textcolor{red}{ 0.73} & \textcolor{red}{ 0.61} \\ ($\%$) & \textcolor{red}{ (15$\%$)} & \textcolor{red}{ (11$\%$)} & \textcolor{red}{ (6$\%$)} & \textcolor{red}{ (5$\%$)} & \textcolor{red}{ (9$\%$)} & \textcolor{red}{ (8$\%$)} \\ \hline & \multicolumn{6}{c|} {\bf $\sigma_{dedp-sys}=\sigma^{av}_{rms}\times(1-\sigma_{fit-sys}^{rel})$ } \\ \hline \end{tabular} \caption[The standard deviation $\sigma^{av}_{rms}$ in ${N_{\omega\to\pi^0\gamma}}^{rec}$, ${N_{\omega\to\pi^0\gamma}}^{\circ}$ and $BR^{measured}_{\omega\to\pi^0\gamma}$ for the different energy-momentum conservation constraint are presented] { The standard deviation $\sigma^{av}_{rms}$ in ${N_{\omega\to\pi^0\gamma}}^{rec}$ and $BR^{measured}_{\omega\to\pi^0\gamma}$ for the different energy-momentum conservation

x 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 S A T C

To compute rho, the program GSC threshold.m denes 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

75 90 Doc 1 1.31556 1.80357 1.80357 Doc 2 1.75182 1.87012 1.87012 Doc 3 2.13338 2.27178 2.27178 Doc 4 1.56941 1.0743 1.0867

gen RHSB=4+(4/(50^.5)) gen out1=cond(meanaRHSA,1,0) gen out2=cond(meanbRHSB,1,0)

According to the Bohr model of hydrogen atom, the kinetic and potential energies of the electrons vary as .......... .. and... distance of electron from the nucleus. respectively. where r is the The square of the wavefunction, y}, gives us the Absorption of a given amount of energy gives rise to a larger increase in entropy the the initial temperature. Orbital energies in multi-electron atoms are mere difficult to calculate due to the additional Select the correct answers from the following multiple choice questions.

37 5. 18 6. 15 7.

Buggin’ Out (Buggy Lab) Purpose - To determine the motion of a battery powered buggy & use the data to determine when the two buggies would collide. Data - Position (m) Displacement (m) Time 1 (s) Time 2 (s) Time 3 (s) Time 4 (s) Average Time (s) Velocity (m/s) 1 0 0 0 0 0 0 -- .5 -0.5 1.74 1.38 1.44 1.66 1.53 -.333 0 -1 3.03 3.00 2.95 2.85 2.96 -.338 -.5 -1.5 4.44 4.28 4.44 4.41 4.39 -.342 -1 -2 5.81 5.25 5.65 5.83 5.64 -.354 3.

The following shows the marginal revenue (MR) and demand curves faced by a monopolist. [8 pnts] For the following questions, refer to the above figure and assume a constant marginal cost of $2. a. What is the optimal quantity that it should produce?

of CE(s) Eigenvalue Statistic Critical Value Prob . ** None * 0.650 32.562 27.584 0.011 At most 1 0.297 10.908 21.132 0.657 At most 2 0.171 5.824 14.265 0.636

(0.00, 6.10), (10.0, 6.00), (20.0, 5.75), (30.0, 5.65), (40.0, 5.30), (50.0, 4.85), (60.0, 4.40), (70.0, 4.10), (80.0, 3.55), (90.0, 2.80), (100, 0.05) base_new_customer_WOM = GRAPH(Awareness_of_E_Business) (0.00, 0.005), (10.0, 0.025), (20.0, 0.05), (30.0, 0.07), (40.0, 0.105), (50.0, 0.135), (60.0, 0.17), (70.0, 0.195), (80.0, 0.255), (90.0, 0.32), (100, 0.405) impact_of_lost_customers_on_WOM = GRAPH(Lost_online_Customers) (0.00, 1.00), (100000, 0.99), (200000, 0.982), (300000, 0.945), (400000, 0.907), (500000, 0.847), (600000, 0.795), (700000, 0.72), (800000, 0.675), (900000, 0.63), (1e+006, 0.585) instore_to_online__WOM = GRAPH(Awareness_of_E_Business*Instore_Customers) (0.00, 0.00), (1e+007, 0.09), (2e+007, 0.18), (3e+007, 0.255), (4e+007, 0.31), (5e+007, 0.345), (6e+007, 0.36), (7e+007, 0.39), (8e+007, 0.405), (9e+007, 0.41), (1e+008, 0.42) investment_spending = (100-marketing_%_of_budget)/100*total_budget marketing_%_of_budget = 75 marketing_spending = marketing_%_of_budget/100*total_budget online_customer_loss_fraction = GRAPH(E_Support_Quality) (0.00, 0.74), (10.0, 0.725), (20.0, 0.68), (30.0, 0.56), (40.0, 0.35), (50.0, 0.255), (60.0, 0.15), (70.0, 0.105), (80.0, 0.065), (90.0, 0.035), (100, 0.02) total_budget = 20 Total_Customers

-.334 Std. Error of Kurtosis .468 .468 The table shows that the in the data set, the GPA variable is a left skewed distribution of -.053 and a left kurtosis of -.811. The final variable indicates that the left skewed distribution stands at -.334 and a left kurtosis of -.334.

Use your results in Data Table 2 to support your answer.

The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different

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