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Tarun B.
Engineering Student with a strong background in math, Physical sciences, and Biomedical Engineering
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Statistics
TutorMe
Question:

Suppose a standardized college entry exam is approximately normally distributed with μ=125 (Population mean) and σ=17(Standard Deviation) for the Math Section. If a student wants to be competitive for his top school of choice, what score should he/she aim to receive in that particular section in order to be in the top 5%? The aforementioned question can be rephrased in the following mathematical terms: X~N(125,15)

Tarun B.
Answer:

In order to solve this particular question, having access to a Z-score table (table corresponding to the population distribution in regards to any particular characteristic (Math scores, in this case)) and a prior basic understanding of the usage of Z-score table is crucial. Additionally, we will also require the use of the Z-score formula which can be stated as follows: Z=(X-μ)/σ Wherein X= the particular math score that a student would need to achieve in order to be in the top 5% of the population μ=Population Mean σ=Standard Deviation Since, we're solving for X in this question, we need to obtain the Z-score value using the Z-score table and solve for X. The Z-score table indicates the lower tail probability. Since the question requires the student to be in the top 5% of the population Math scores, the student must do better than 95% of population. Therefore, we must pick a Z-score that corresponds to P=0.95 (Z-score corresponding to a Math score that is expected to be higher than other Math score 95% of the time). The correct corresponding Z-score is 1.645. Now that we have the correct Z-score, we can plug it into the Z-score formula and solve backwards to find the X-value. The resulting X-value that we obtain is 152.97. In order for a student to be competitive for his top school of choice, the student must obtain a score of at least 152.97 or higher.

Biology
TutorMe
Question:

Which of the following types of cell junctions would be primarily useful in relaying signals between cardiac muscles? Pick the appropriate answer and explain how it would do so. a. Gap Junctions b. Tight Junctions c. Anchoring Junctions

Tarun B.
Answer:

The correct answer for this question is Gap Junctions. Gap Junctions are structurally composed of proteins called connexons and allow for direct transportation of solutes along with direct communication between cells, thereby bypassing the extracellular matrix entirely. The gap junctions in heart muscles act as electrical synapses allowing the heart cells to transfer the electrical signals faster and directly. This allows the heart cells to work together in unison. Tight Junctions consist of protein complexes that are used to prevent any (or most) form of transportation between 2 cells or a cell and its extracellular environment. Anchoring Junctions (Desmosomes, Hemidesmosomes, etc.) consist of integral and peripheral proteins in plasma membrane such as Cadherins, Integrins etc. that help bind the intra-cellular cytoskeleton of a cell with that of another cell or with the extra-cellular matrix. These junctions primarily aid in cellular adhesion and provide cytoskeleton stability.

Chemistry
TutorMe
Question:

List the factors that affect the acidity of compounds and explain how each factor contributes to increasing (or decreasing) their acidity

Tarun B.
Answer:

Bond Strength is one of the factors that affects the acidity of compounds. Elements further down in a group in the periodic table tend to be bigger and, thus, have a longer bond length. A longer bond length corresponds to a weaker bond which results in increased acidity of a compound. Electronegativity (Inductive effects), though not as significant as bond strength, affects the acidity of a compound as well. A more electronegative element in a compound tends to attract the electron cloud towards itself (increased polarization) and away from the Hydrogen atom, thereby making it easier to be donated off as a proton. Resonance stabilization of the resulting conjugate base of the acid strongly influences the acidity as well. If the conjugate base of a particular acid has varying resonance structures as compared to another conjugate base of an acid that has none, the acid with conjugate base that has resonance structures will be more acidic. The varying resonance structures of the conjugate base provide electron delocalization within the compound, thereby spreading the negative charge within the compound evenly and stabilizing it. HA --> H+ + A- (Has resonance structures --> negative charge is spread evenly ->more stable) HB --> H+ + B- (No resonance structures --> conjugate base is not as stable) Since A- is more stable than B-, HA will be more likely to dissociate into H+ and A- than what HB will dissociate into H+ and B-.

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