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# Tutor profile: Nick K.

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Nick K.
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## Questions

### Subject:Calculus

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Question:

Take the derivative of y(x) = x * ln(x).

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Nick K.

Remembering the product rule for derivatives, for some function f(x) = g(x) * h(x) we have f'(x) = [g'(x) * h(x)] + [g(x) * h'(x)]. Now we can solve for y'(x). y'(x) = [1 * ln(x)] + [x * 1/x] = ln(x) + 1.

### Subject:Statistics

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Question:

A bag contains 4 blue marbles, 6 red marbles, and 10 green marbles. Mark selects 2 marbles from the bag without replacement. What is the probability that the two marbles are the same color.

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Nick K.

We'll break this problem into three parts. We can find out the probabilities of selecting 2 blue marbles, 2 red marbles, and 2 green marbles. Then we can add the probabilities together because each of the results is independent from the others (Mark cannot select 2 blue marbles and 2 red marbles at the same time because he is only choosing 2 marbles). Let's start with the blue marbles. There are 20 total marbles, so there is a (4/20) chance that Mark selects a blue marble on the first pick. Now there are 3 blue marbles left and 19 total marbles, so there is a (3/19) chance that Mark selects a blue marble on the second pick. So the probability of Mark selecting a blue marble on the first pick and second pick is: (4/20) * (3/19) = 12/380 = 3/95 (or about 0.0316). Following the same process, the probability of selecting 2 red marbles is: (6/20) * (5/19) = 30/380 = 3/38 (or about 0.0789), and the probability of selecting 2 green marbles is: (10/20) * (9/19) = 90/380 = 9/38 (or about 0.2368). So the probability that Mark selects 2 marbles of the same color is: (12/380) + (30/380) + (90/380) = 132/380 = 33/95 or 0.347.

### Subject:Differential Equations

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Question:

Solve the following initial value problem: ty' + y = 0, y(1) = 1.

Inactive
Nick K.

Our goal is to get all of the 'y' terms on one side of the equation, and all the 't' terms on the other side. We can do this with some algebra: ty' + y = 0 ==> ty' = -y ==> y' = -y/t ==> y'/y = -1/t We now integrate both sides, and using the proper antiderivates we get: ln(y) = -ln(t) + c ==> ln(y) = ln(1/t) + c Now we put both sides with base e to get rid of the natural logs. e^(ln(y)) = e^(ln(1/t) + c) ==> y = e^(ln(1/t)) * e^c. Let's let C = e^c: ==> y = C * 1/t, y(1) = 1. Substituting in the initial conditions, (1) = C * 1/(1) ==> C = 1. So our solution is y = 1/t.

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