Low Results Lab Report
Tn 4351 was originally isolated from bacteroides fragilis [30] . The transposon was successfully introduced into Cytophaga succinicans, Flavobacterium meningosepticum, Flexibacter canadiansis, Flexibacter strain SFI and Sporocytophaga myxococcoides by conjugation [25]. Tn 4351carries two antibiotic resistance gene. One of the codes for resistance to erythromycin and clindamycin which is expressed in bactroides but not in E.Coli. The other gene codes for resistance in tetracycline and is expressed in aerobically grpwn E. coli, but not in anaerobically grpwn E. coli or in bacteroides.
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.
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
Typical sample dimensions 9.51 × 4.83 mm2in surface area and1.58 mm in thickness were coated with conductive silver paint formetallic contacts. The dielectric constant of the sample was mea-sured for the applied frequency that varies from 100 Hz to 1 MHz atdifferent temperatures (40◦C, 60◦C, 80◦C). The observations weremade while cooling the sample. The dielectric constant εrwas cal-culated using the relation, εr =
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.
Discussion 1. Zn0 (s)+ Cu2+S6+O42-(aq) →Cu0(s) + Zn2+S6+O42-(aq) Zn0(s) → Zn2+(aq) + 2e- Cu2+(aq) + 2e- → Cu0(s) Zn0(s) + Cu2+(aq) → Zn2+(aq) + Cu0(s) Oxidant (oxidizing agent) is the element which reduces in experiment.
19.386526 -67.45 -44.1 20.53525 -68.39 -44.1 21.75204 -68.56 -44.1 23.04093 -67.97 -44.1 24.406191 -67.25 -44.1 25.852348 -66.75 -44.1 27.384196 -66.66 -44.1 29.006812 -66.79 -44.1 11.54782 -67.25 -44.1 12.232071 -66.3 -44.1 12.956867 -65.38 -44.1 13.72461 -64.56 -44.1 14.537844 -64.01 Adrian Bersiks_bersik_Acoustic Analysis_Excel.xlsx-44.1 15.399265 -63.86 -44.1 From the figure above there are no interpolation points above the reference line, which means the frequencies were bounded nicely under the maximum amplitude, and the greatest amplitude was captured on the sampling interval exactly, with a closer distribution in amplitudes. Again the 130Hz drop is consistent. Looking at the Excel spreadsheet, the resposnse almost mimics the
1. What area/aspect of this setting is the most challenging? 2. In the setting, you work in, is there a certain population of patients you see more? How does this affect you?
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.
Suppose we have a single-hop RCS where there is one AF relay that amplifies the signal received from a transmitter and forwards it to a receiver. Assume that the transmitter sends over the transmitter-to-relay channel a data symbol ${s_k}$, from a set of finite modulation alphabet, $S={S_1, S_2,ldots,S_{cal A}}$, where ${cal A}$ denotes the size of the modulation alphabet. The discrete-time baseband equivalent signal received by the relay, $z_k$, at time $k$ is given by egin{equation} z_k = h_{1,k}s_k + n_{1,k},~~~~for~~k=1,2,ldots,M label{relaySignal} end{equation} where $n_{1,k}sim {cal N}_c(0,sigma_{n1}^2)$ is a circularly-symmetric complex Gaussian noise added by the transmitter-to-relay channel, $h_{1,k}$ denotes the transmitter-to-relay channel, and
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 $N(-1.0558) = 0.1456.$
On April 6, 2016 at approximately 11:45am, a local police station got a call about a hostage situation at a local pharmacy. When police and medical examiners got to each crime scene, they learned that all of the hostages were given drugs and had overdosed on them. Some of the pills, in powder form, were found near the victims. One of the victims was stable enough to tell the investigators that the power on the floor were the drugs they were forced to take. The medical examiner found out each hostage was given either unknown A or unknown B.
4.0 Methods The experiment examined the efficiency of 10 test subject in memorising and reproducing a pattern of numbers, firstly in a grid of 6x6 boxes with the numbers 1-10 randomly ordered throughout. There were two trials in the experiment. Two different sequences were used for each trial which was no music, rock music and Mozart for the trial 1 but for trial 2 it was rock music, no music and then symphony. During the experiment, each subject was seated and a blank identical test paper was given to them.
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