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# Tutor profile: Anastasios M.

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Anastasios M.
Statistics and Mathematics Tutor with over 14 years' experience
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## Questions

### Subject:Trigonometry

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Question:

Suppose we have a triangle ABC (where $$a$$ is the side opposite angle $$A$$, $$b$$ is opposite $$B$$, and $$c$$ opposite $$C$$). (1) If $$a=9$$, $$c=10$$, and $$B=25^\circ$$, find the length of side $$b$$. (2) Find the length of side $$b$$ if we knew instead that $$A=64.14^\circ$$, $$B=25^\circ$$, and $$a=9$$.

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Anastasios M.

1) Since we know the length of two sides and the measure of the angle between them, we can use the law of cosines: $( b^2=a^2+c^2-2ac \cos B.$) Here, we have \begin{align} b^2&= 9^2+10^2-2(9)(10)\cos(25^\circ)\\ &= 17.86 \mbox{ (correct to two decimal places)}. \end{align} Therefore, $$b=\sqrt{17.86}\approx 4.23.$$ 2) Since we know the measure of two angles and the length of one of the corresponding sides, we can use the law of sines to find the length of the other side: $( \frac{a}{\sin A}=\frac{b}{\sin B}.$) Solving for $$b$$, we have $$b= \frac{a\sin B}{\sin A}$$. Therefore, here, $( b = \frac{9\sin 25^\circ}{\sin 64.14^\circ} \approx 4.23.$)

### Subject:Calculus

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Question:

Find the derivative of $$\frac{x^2-1}{x^2+1}$$ with respect to $$x$$.

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Anastasios M.

Since this function is the quotient of two functions of $$x$$ (namely, $$f(x)=x^2-1$$ and $$g(x)=x^2+1$$), we can use the quotient rule for derivatives: $(\frac{d}{dx}(\frac{f}{g})= \frac{f' g - fg'}{g^2}.$) Thus, the derivative of $$\frac{x^2-1}{x^2+1}$$ with respect to $$x$$ is \begin{align} \frac{f' g - fg'}{g^2} &= \frac{(2x)(x^2+1)-(x^2-1)(2x)}{(x^2+1)^2}\\ &=\frac{2x^3+2x-2x^3+2x}{(x^2+1)^2}\\ &=\frac{4x}{(x^2+1)^2}. \end{align}

### Subject:Statistics

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Question:

Suppose we flip a fair coin 5 times. What is the probability we observe heads exactly twice?

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Anastasios M.

We recognize this as a binomial distribution problem because: 1) A procedure is repeated $$n$$ times (or "trials"). Here, the 5 coin flips are the trials; therefore, $$n=5$$. 2) There are 2 outcomes on each trial (a "success" or a "failure"). Here, we have "success"="heads", because it's the outcome we're interested in. 3) The probability of success, $$p$$, remains constant across the trials. This is clearly true here as we are flipping the same coin. 4) The trials are independent. For coin flips, we can generally assume this is true. -Since we are counting the number of heads, we let the random variable $$X$$ be "the number of heads (successes) in the $$n=5$$ trials". -We have $$p=0.5$$ because the coin is fair. Also, the probability of tails ("failure") is $$q=1-0.5$$ (because the two outcomes are complementary events). We want to find $$P(X=2)$$. Because of 1)-4), $$X$$ is a binomial random variable and we can use the binomial probability formula to solve the problem: $(P(X=a) = {n \choose a} p^a q^{n-a},$) where $$a$$ is the number of successes we want. Here, we have $(P(X=2) = {5 \choose 2} p^2 q^{5-2} = \frac{5!}{3!2!} (0.5)^2 (1-0.5)^3 = 0.3125.$) Therefore, the probability of observing heads exactly twice is 0.3125.

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