Radiative Zone Lab Report

The design relied on two Schmitt triggers to generate the two different tones while using the transistors to act as a switch. This causes it to trigger continuously between two unstable states, allowing automatic switching between two frequencies producing two different tones. The RC values between the two Schmitt triggers will differ. Capacitors charge and discharge faster when it’s resistance is smaller.

For most sequences at position 4 and 5 we observe only the nucleotides G and T, respectively. There may be rare cases where other nucleotides may also be found. To consider such observations, we need to do a process called additive smoothing or Laplace smoothing to smooth the categorical data. [9] In this case, we add 4 sequences: AAAAAAAAA, CCCCCCCCC, GGGGGGGG, TTTTTTTTT.

I need to find the area of rectangle ABCD. I know that ABCD is a rectangle with diagonals intersecting at point E. Segment DE equals 4x-5, segment BC equals 2x+6, and segment AC equals 6x. I predict that To find the area of rectangle ABCD I need to find out the base and height of the rectangle. The first step is to find what x equals. Since I know the intersecting line segments AC and DB are congruent that means when I times the equation 4x-5 for segment DE by two it will equal the equation 6x for segment AC.

Experiment 7 In this experiment we configured several DC circuits consisting of an emf and a network of resistors. The circuits were composed of a power supply, two DMMs, a circuit board, an SPST switch, and an assortment of known resistors along with one unknown resistor. We measured the current and voltage of the entire circuit as well as the potential drops across each resistor to determine the parameters of the circuit including the resistance, voltage, and current for each component.

%% Init % clear all; close all; Fs = 4e3; Time = 40; NumSamp = Time * Fs; load Hd; x1 = 3.5*ecg(2700). ' ; % gen synth ECG signal y1 = sgolayfilt(kron(ones(1,ceil(NumSamp/2700)+1),x1),0,21); % repeat for NumSamp length and smooth n = 1:Time*Fs '; del = round(2700*rand(1)); % pick a random offset mhb = y1(n + del) '; %construct the ecg signal from some offset t = 1/

(a) 3Mbps / 150Kbpa =3 X 1024 / 150 = 3072 / 150 =20.48 20 Users can be supported 150Kbps dedicated. (b)

Figure shows the intersection of line joining the camera center and image points ${\bf x}$ and ${\bf x'}$ which will be the 3D point ${\bf X}$.\\ \end{figure} The ‘gold standard’ reconstruction algorithm minimizes the sum of squared errors between the measured and predicted image positions of the 3D point in all views in which it is visible, i.e.\\ \begin{equation} {\bf X=\textrm{arg min} \sum_{i} ||x_i-\hat{x_i}(P_i,X)||^2} \end{equation} Where ${\bf x_i}$ and ${\bf \hat{x_i}(P_i,X)}$ are the measured and predicted image positions in view $i$ under the assumption that image coordinate measurement noise is Gaussian-distributed, this approach gives the maximum likelihood solution for ${\bf X}$. Hartley and Sturm [3] describe a non-iterative

Using the data provided in each one of these tests it can be assumed that one has done the steps to be able to determine the magnitude and orientation of the charges of the tape in each test, thus, allowing them to apply the same principle to any object they so desired. Their results would line up with the following; that if the two pieces of tape are torn from the same 40 centimeter strip then the tops of both pieces of tape would be positive and the bottoms of both pieces of tape would be negative and that if they would double the tape the attraction or repulsion in general would lower due to the increased density. Their data would also show that two pieces of tape ripped from each other would result in one piece being entirely positive and the other being entirely negative, they would also be able to state that the orientation of how the tape is paired up doesn’t matter.

1. The test subjects will prepare for sleep by acquiring everything needed for the subjects’ sleep preferences. 2. The test subjects will all set alarms on their smartphones for approximately 6, 8, and 10 hours after the subjects’ enter the resting period (Subjects may wake during the resting period for the bathroom, but they must not stay awake for more than ten minutes at a time to prevent as much deviation as possible.). 3.

determine each pixel belongs to background or foreground. Wis the weights between the pattern and summationneurons, which are used to point out with which a pattern belongs to the background or foreground. They areupdated when each new value of a pixel at a certain position received by implementing the following function:Wt+1ib =fc(1−βNpn)Wib+MAtβ!(37)Wt+1i f=(1−Wt+1ib)(38)whereWtibis the weight between theith pattern neuron and the background summation neuron at timet,βisthe learning rate,Npnis the number of the pattern neurons of BNN,fcis the following function:fc(x)1,x>1x,x≤1(39)MAtindicates the neuron with the maximum response (activation potential) at frame t, according to:MAt1,f or neuron with maximum response0,otherwise(40)Function

Suppose you need to find the fractional European call and the fractional European put options. Let the Hurst parameter be $H=0.85$, the $\sigma=0,25$, $r=0.10$, $S_{fbm} = 100$, $K = 95$, we have \begin{eqnarray*} d_1^{fBm} & = & \frac{\ln{\frac{S}{K}} + \frac{1}{2}(r( T - t) + \frac{(1)\sigma^2{( T^{2H} - t^{2H})}}{2})}{\sigma{\sqrt{T^{2H} - t^{2H}}}}\\ & = & \frac{\ln(\frac{105}{100}) + (0.10(0.25 -0) + \frac{(1){0.25^2}{0.25^{2(0.85)} - (1)0.25^{2(0.85)}}}{2}}{(0.25){\sqrt{0.25^{2(0.85)} - 0}})} \end{eqnarray*} we obtain $d^{fBm}_1= 1.0558$. We find in the normal distribution that $N(1.0558)= 0.8544$ and $N(-1.0558) = 0.1456.$

1. There are two ways of maximizing points in this experiment. The first one is that I should connect myself to a vertex that is in the biggest component and purchases immunization. Since the probability of being infected is based off of expected value, I would have less than 1% chance of getting infected. The second way is that I try to make myself stay in the second-largest connected component.

1. Identify the range of senses involved in communication • Sight (visual communication), Touch (tactile communication), Taste, Hearing (auditory communication), Smell (olfactory communication) 2. Identify the limited range of wavelengths and named parts of the electromagnetic spectrum detected by humans and compare this range with those of THREE other named vertebrates and TWO named invertebrates. Figure 1: the electromagnetic spectrum source: www.ces.fau.edu Vertebrates Human Japanese Dace Fish Rattlesnake Zebra Finch Part of electromagnetic spectrum detected ROYGBV (visible light) detected by light sensitive cells in the eye called rods and cones.

The disk of the sun was dull silver as though it were nearer and not so hot, yet the air stifled.”

Lab 1 helps create a better understand of the changes in crystal structures when the annealing and quenching process is applied to 1020 and 1080 steel. The numbered steel refers to the ASTM grain-size number. Formula 1 is used to solve for the grain size. n=2^(G-1) Equation (1) at 100x magnification Crystal structures change shapes which changes the strength of the material and its properties. The metal might become soft, brittle, hard, or ductile.