In January of 2012, the second stop for a Republican to get votes to.pdf

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In January of 2012, the second stop for a Republican to get votes toward the presidential nomination was at the New Hampshire Primary. The following exhibit shows the votes several candidates received from registered Republicans and Independents. Party Affiliation Mitt Romney Ron Paul Jon Huntsman Others Republicans 62645 16792 12733 27610 Not Republican 34887 40056 29212 24550 Source: ABC News What is the probability that a randomly selected voter voted for Ron Paul? Given that a randomly selected voter is not a Republican, what is the probability that he/she voted for Ron Paul? If a randomly selected voter voted for Jon Huntsman, what is the probability he is a Republican? Party Affiliation Mitt Romney Ron Paul Jon Huntsman Others Republicans 62645 16792 12733 27610 Not Republican 34887 40056 29212 24550 Solution.

In January of 2012, the second stop for a Republican to get votes toward the presidential
nomination was at the New Hampshire Primary. The following exhibit shows the votes several
candidates received from registered Republicans and Independents.
Party Affiliation
Mitt Romney
Ron Paul
Jon Huntsman
Others
Republicans
62645
16792
12733
27610
Not Republican
34887
40056
29212
24550
Source: ABC News
What is the probability that a randomly selected voter voted for Ron Paul?
Given that a randomly selected voter is not a Republican, what is the probability that he/she
voted for Ron Paul?
If a randomly selected voter voted for Jon Huntsman, what is the probability he is a Republican?
Party Affiliation
Mitt Romney
Ron Paul
Jon Huntsman
Others
Republicans
62645
16792
12733
27610
Not Republican
34887

40056
29212
24550
Solution

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All answers are in the form of True/False with a explantion as to why: 1) An ideal band-pass filter centered (at ± 0) can be created by just multiplying the sinc function by cos(0t) in the the time domain. 2) Nyquist theorem deals with windowing of a signal in the time domain, which is translated into leakage in the frequency domain. 3) The global frequency response of an LTI system composed of two LTI systems in parallel with each other is simply the addition of the individual frequency responses of the systems. 4) We can find the H(j) of an unknown LTI system by stimulating it with a impulse and taking the Fourier Transform of its response. Another more realistic way is to stimulate the LTI system with a unit step function (signal), take the Fourier Transform of its response, multiply it by (j) for every 0 and divide it by pi at =0. ^I believe that one is true, but I am needing help forming an explanation. 5) A first step towards sampling an unknown analog signal without aliasing is to low-pass filter the signal. Solution 1) False Because fourier transform of sinc function is rectanctangular function in frequency domain, and fourier transform of cos function is impuse function. So when when multiply in time domain that means convoution in frequency domain. And if ve convolude rectangular function with impulse function then it would be rectangular pulse train. Which is not ideal band paas filter at w=0. 2) False It would be either window or leakage in time domain or frequency domain. 3) TRUE It is simply understand by block reduction method in control system. When there is another forward block we add both maid forward block with secondry forward block. OR LTI is linear in nature so add in parallel and multiply in series. 4) TRUE Its because the response or tansfer function of any system is called impulse response. It means when the input is Impuse function what is the output of that system. Same as the relation between unit imulse response and impuse respone in just fouries tranfer of unit step function. Because the fourier transform of impulse is unity. so if input is unit step function multiply it by (j) for every 0 and divide it by pi at =0.(1/(i)+(),) 5)FALSE Low paas filter use after sampling to recover the original signal..

All answers are in the form of TrueFalse with a explantion as to wh.pdf

All answers are in the form of TrueFalse with a explantion as to wh.pdf

All answers are in the form of TrueFalse with a explantion as to wh.pdf

Consider miRNA target sites, How do these compare to transcription factor binding sites? Imagine that both miRNA target sites and transcription factor binding sites were dispersed randomly in the genome ? Which would be likely to appear more frequent? Solution mRNA localization: mRAN location is critical for regulation of gene expression by allowing spatially regulated protein production. Through mRAN localization protein are transcribed in their intended target site of the cell. This is especially important during early developement when rapid cell cleavages give different cells various combinations of mRNA which can then lead to drastically different cell fates.RBPs are critical in the location of this mRNA that insures protein are only transcribed in ttheir intended regions. One of these proteins is ZBp1 binds to beta- actin mRNA at the site of transcription and moves with mRNA into the cytoplasm. It then localizes this mRNA to the lamella region of several asymmetric cell types where it can then be translated. FMRP is another RBP involved in RNA localization. It was shown thatt in addition to other functions for FMRP in RNA metabolism, FMRP is involved in the stimulus-induced localization of several dendritic mRNAs in neuronal dendrites. Transcription factors binding site: Transcrption factors bind to either enhancer or promoter regions of DNA adjacent gene is either up- regulated. Transcription factors use a varity of mechanisms for the regulation of gene expression. These mechanisms include: Histone acetyltransferase (HAT) activity - acetylates histone proteins, which weakens the association of DNA with histones, which make the DNA more accessible to transcription, thereby up- regulating transcription. Hiestone deacetylase (HDAC) activity - deacetylates histone proteins, which strengthens the asociation of DNA with histones, which make the DNA less accessible to transcription, thereby down- regulating transcription. Activation: trancsription factors may be activated (or deactivated) through their signal- sensing domain by a number of mechanism including:.

Consider miRNA target sites, How do these compare to transcription fa.pdf

Consider miRNA target sites, How do these compare to transcription fa.pdf

Consider miRNA target sites, How do these compare to transcription fa.pdf

Anaerobic Respiration & Fermentation (Microbiology) What are the bubbles in fermented medium made of? Describe the process that they come from. Solution Answer: The bubbles in the fermented medium is made of carbon dioxide and hydrogen. Microbes ferment glucose/dextrose to produce gases such as carbon dioxide and hydrogen. The medium most commonly used is phenol red glucose broth and is composed of 0.5-1.0% glucose. As the broth gets fermented, acidic end products are produced. This reduces the pH of the media which can be detected as the phenol red changes its colour from red to yellow. Inverted tubes called Durham tubes can be attched to the tubes of the broth which trap any carbon dioxide produced as a result of fermentation..

Anaerobic Respiration & Fermentation (Microbiology)What are the bu.pdf

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4. Pick out any text-mining tool that impressed you the most and discuss an overview of its text mining process, sentimental & speech analytics. Solution LanguageWare is a natural language processing (NLP) technology developed by ibm, which allows applications to process natural language text. It comprises a set of Java libraries which provide a range of nlp functions: language identification, text segmentation/tokenization, normalization, entity and relationship extraction, and semantic analysis and disambiguation. The analysis engine uses Finite state machine approach at multiple levels, which aids its performance characteristics, while maintaining a reasonably small footprint. Speech analytics: Speech analytics allows users to analyse and extract information from both live and recorded conversations. It is being used effectively to gather intelligence for security purposes, to enhance the presentation and utility of rich media applications, and perhaps most significantly, to deliver quantitative business intelligence through the analysis of the millions of recorded calls . Sentiment analysis: Sentiment analysis is a branch of speech analytics that focuses specifically on assessing the emotional states displayed in a conversation. One common use of sentiment analysis within contact centres is to provide insight into a customer’s feelings about an organisation, its products, services, customer service processes, as well as its individual agent behaviours..

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