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Tutor profile: Mcsteve E.

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Mcsteve E.
Grad Student at Columbia University
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Questions

Subject: SAT

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Question:

For some teachers, they discover that the longer they teach, the more _____ it becomes. I have found the opposite to be my reality, as every new year I am faced with _____ challenges that reignite my passion. A. difficult; easy B. tiring; strenuous C. mundane; unfamiliar D. natural; irksome

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Mcsteve E.
Answer:

C. mundane; unfamiliar

Subject: Calculus

TutorMe
Question:

In light of the COVID-19 pandemic, Kodak decided to start the production of cylindrical vaccine vials, with covers. The pharmaceutical companies suggest that each vaccine contains 10mL. What dimensions should Kodak make their vials to minimize manufacturing costs? Assume the covers are part of the vial, and are not unscrewable.

Inactive
Mcsteve E.
Answer:

First we should note that 1 milliliter is 1 cubic centimeter, i.e. $1 mL = 1 cm^3$ So a vial that holds $10$ mL has a volume of $10$cm$^3$. The volume of a cylinder is $V={\pi} r^{2} h$, i.e. $10=\pi r^{2} h$ We want to minimize the dimensions of the cylinder. This means to use as little surface area as possible. Since the cylinder is covered on both ends, the surface area of a cylinder is $A = 2\pi r h + 2\pi r^2$ Going back to V, if we make $h$ the subject of the formula, we derive $h = \frac{10}{\pi \times \r^2} And substituting $h$ into the area formula, we derive $A = 2\pi r \frac{10}{\pi \times \r^2} + 2\pi r^2$ $A = \frac{20}{r} + 2\pi r^2$. This is our optimization function. To find the maximum and minimum points, we take the derivative of the optimization function (with respect to r) and set it to 0. Thus, $0 = A'(r) = -\frac{20}{r^2} + 4 \pi r$ $0 = \frac{-20+4\pi r^3}{r^2}$ $0 = -20+4\pi r^3$ $4\pi r^3 = 20$ $r = \sqrt[3]{\frac{20}{4}}$ $r=1.7099$cm Recall, $h = \frac{10}{\pi \times \r^2}$ Hence, $h = \frac{10}{\pi \times \1.71^2}$ $h = 1.0886$cm. Therefore, a radius of 1.71cm and height of 1.09 cm would provide Kodak with the least amount of glass to make a vial that holds 10 mL of the COVID-19 vaccine. This would not be a very good looking vial, Perhaps Kodak should stick to making lens.

Subject: Statistics

TutorMe
Question:

Only $40$ of students in a Stats class passed the final examination. The average score was $56$, and the standard deviation was $8$. Find the passing score.

Inactive
Mcsteve E.
Answer:

First, Let $x$ be our passing score, and $X$ be the number of students who scored less than the passing score, i.e. failed. Also, $\mu = 56$, and $\sigma = 8$ Then, $P(X <= x) = 1 - .4 = 0.6)$ Introduce normal distribution. $z(\theta) = z(\frac{x-\mu}{\sigma})$ Hence, $z(\theta) = .6$ and using the normal distribution table, we get $\theta = 0.2533$ So, $\theta = .2533 = \frac{x-56}{8}$ $x = (.2533*8)+56 = 58.03$

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