Enable contrast version

Tutor profile: Nick K.

Inactive
Nick K.
Ungraduate Mathematics Student
Tutor Satisfaction Guarantee

Questions

Subject: Calculus

TutorMe
Question:

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

Inactive
Nick K.
Answer:

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

TutorMe
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.

Inactive
Nick K.
Answer:

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

TutorMe
Question:

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

Inactive
Nick K.
Answer:

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.

Contact tutor

Send a message explaining your
needs and Nick will reply soon.
Contact Nick

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2020 TutorMe, LLC
High Contrast Mode
On
Off