You're welcome.

You're welcome!

You're welcome. I'm glad to hear you found it helpful.

Thanks for pointing that out! The values in the column are correct for ~p so the final conclusion is the right one.

Sometimes when these videos are used in classrooms, they're listed out of order based on what that class is covering. You can find the complete Discrete Math playlist, in order, at kzbin.info/aero/PLjRL3OOnLvS8-FkldKhmTiyMjc4pmGrMe.

Thanks for pointing that out! The leading 1 on the third line, should have been a 2.

I'm glad to hear you found it helpful! Thanks for the feedback.

I use Microsoft PowerPoint and then output the results in full HD. Up until recently I was using just regular HD to keep the size down but I switched when I realized there was a small but noticeable difference in the quality. In this video, I just typed all the equations as text but, for more complicated ones, I use MathType's Office plug in. It's very similar to Word's equation editor but has more options both in symbols and formatting. It's been a while since I bought it but I remember the price being very reasonable, especially for something that I use this often.

Thanks, Cole!

That's awesome!

You're welcome!

This was a pretty straightforward application of the formula. If you've got another approach, we'd love to hear it.

@@whitecraneeducation2228 yea...just take a sample element dv , which has a volume of 2*pi*x*f(x)*dx. integrate using the same limits and we get the same result, rather than substracting the larger volume from smaller one and simplifying it, i think its more easy to visualize a small volume element and finding its volume using basic geometry.

@@bharathharish9905 Were you referring to this video or the one where I derived the formula? There wasn't any mention of inner versus outer radii here - I just went straight to the formula. In the other video, I can see how someone could jump to that formula based on what they've seen previously but relying on your intuition, especially for students who are relatively new to the material, is a risky choice. It's better to start with the definitions or theorems you've already derived and go forward from there. That's what I was trying to do in that lecture.

Thanks for the feedback!

"Why does a^2/c - x imply the distance from point (x,y) to the directrix" Don't think of it as the directrix just yet. I think it's helpful to connect this back to the definition of a parabola first. Remember that the points on a parabola were defined such that: (distance from point to focus) = (distance from point to a line called the directrix) Keep in mind that a^2 / c - x is the distance from the point (x, y) to the line x = a^2 / c. Now, at around the 3:20 mark, we get a similar equation to the parabola one but with one addition: (distance from point to focus) = (a constant) * (distance from point to the line x = a^2/c) Since the relationship is so similar, we're going to use the same terminology and call that line, x = a^2 / c, the directrix of the ellipse which makes the equation (distance from point to focus) = (a constant) * (distance from point to the directrix) Now we can see the key geometric connection between the two definitions of a directrix: This equation says that, in an ellipse, rather than being equal like they were in a parabola, the distance from every point to the focus *is proportional* to the distance from the point to the directrix. Does that answer your question?

@@whitecraneeducation2228 Thank you! This is extremely helpful to me. I think I struggle the most with the section 1:42-1:52 -- and maybe this question is redundant, but..how do we know that the equation inside the parentheses is the distance from point x,y to the directrix? How does setting it equal to zero help determine the location of the line? Thanks for bearing with me, I'm in my late 30s and love working through the "why" questions in math. Perhaps there is some algebra I'm just not recalling. Cheers, John

@@JFRCCR2015 Suppose you have two points on the x-axis, say (3, 0) and (5, 0). What's the distance between them? Since the two points are on a horizontal line, it's just 5 - 3 = 0. Now, shift that horizontal line up so that it goes through the point (x, y). That line is going to cross the line x = a^2 / c at the point (a^2 / c, y). By the same argument that gave us 5 - 3 = 2 as the distance between my first two points, the distance between those two points has to be a^2 / c - x. At the risk of being self-referential, you might find this video, kzbin.info/www/bejne/gYCydIifaN2cb6s, helpful. In it I talk about finding the distance between two horizontal points (and two vertical points) as the setup for deriving the more general distance formula. Hopefully that helps!

@@whitecraneeducation2228 I think I'm picking up what your're putting down. I was able to see it more clearly on a drawn graph as well. I was making the mistake of thinking of a^2/c as the x value in the (x,y) coordinate pair if that makes sense. Thank you for the help, appreciate your content.

Thanks for the feedback, AJ. We're glad to hear you found it helpful.

Thanks for the feedback! I'm glad to hear you found it helpful.

Thanks! I'm glad to hear you liked it.

@@whitecraneeducation2228 you illustrated the point that binary numbers are really not much different from decimal numbers much better than i could. Line binary numbers just made sense at a certain point but i find it sad that most professors never take the time to really explain what „binary number“ even means. You are giveb a set of rules for how they behave but i find that once you understand the clre idea the rest becomes intuitive and obvious(to me at least)

That equation comes from writing the definition of an ellipse, using the coordinates of the foci in this ellipse. Given an arbitrary point (x,y) on the ellipse the sum of its distances to the 2 foci will be the fixed quantity 2a. Once you have that down a little bit of algebra, such as squaring of the 2 sides, will give you the equation he has. I hope this helps.

Thanks! I'm glad to hear you found it useful.

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I'm glad to hear you found it helpful. Coming up with your own practice questions is a great studying technique.

lol, increased quality, fixed

@@davidmoerman8700 Awesome. Thanks for the feedback!

I'm glad you found it helpful! Was there something about the p = q step that I could clarify for you?

​@@whitecraneeducation2228I think explaining why having two fixed points would cause a contradiction that is false and how the derivative of the function being less than one proves uniqueness would really help me understand that 😅

I'm glad to hear you liked it, Chatoy.

I assume you're asking about right around the 05:20 mark where I calculated the value of p. In this question, we're asked about patients who "experienced failure" with the treatment so that's going to be what the p probability represents, i.e. p = "probability of a patient experiencing a failed result" The statement of the question gives the probability of a *successful* result (0.982) so p = "probability of a failed result" = 1 - "probability of a successful result" = 1 - 0.982 = 0.018

Thanks!

You can find the video on the corresponding Poisson parameters at kzbin.info/www/bejne/onirhYaDodeSmZo.

Statistics classes are a little weird like that in the beginning. I usually take a minute at the beginning of the second week to reassure the students that there will be numbers and formulas coming along shortly!

We're glad to hear it helped!

Thanks!

Thanks, Arian!

Thanks for the kind words, Aadeeshwar! We're glad you find the content helpful.

@@whitecraneeducation2228 You're welcome! I've gone through multiple Combinatorics tutorials with millions of views and I genuinely found this to be more helpful. Keep it up

You're welcome, Ivan. We're glad you found it.

We're glad to hear you found it useful!

We're glad to hear you found it helpful!

You're welcome.

I'm glad you found it useful, Jonathan.