Activity for Tim Pederick‭

Type	On...	Excerpt	Status	Date
Answer	—	A: 25% probability that there was a chance of avoiding injury $\quad$ vs. $\quad$ 25% chance of avoiding injury One is a chance of doing something. The other is a chance of having a chance of doing something. Here are some analogies that may or may not help, depending on your familiarity with the kinds of situations described. Analogy 1: A tabletop roleplaying game (RPG). Your character is sneakin... (more)	—	almost 1 year ago
Answer	—	A: How can school children intuit why over 100, D is larger? But under 100, D% is larger? (I’m going to do this post in pounds so that I don’t have to escape dollar signs everywhere. But the currency doesn’t matter.) “Per cent” is, literally, “for each hundred”. Imagine making a literal pile of money from the cost of the item, and splitting it up into stacks of £100. So if the item cos... (more)	—	about 1 year ago
Answer	—	A: How to intuit P(win the same lottery twice) $= p^{2}$ vs. P(win the same lottery twice | you won the lottery once) $= p$? I may be misreading your post, but it seems like there is some confusion between these two things: 1. Winning the lottery twice 2. Winning the lottery a second time All of the remarks about the probability being the same are relevant to the second situation. Your calculations, however, address... (more)	—	about 1 year ago
Answer	—	A: Given two angles of a triangle, finding an angle formed by a median This is a synthetic geometry approach to the problem. It turned out more involved than I expected, and yet I haven’t found anything that could be omitted! I don’t say there isn’t a shorter way to do it, though, probably using some theorems that I haven’t thought of. (Maybe circle inversion? Seems lik... (more)	—	over 1 year ago
Question	—	Is posting multiple answers encouraged, and under what circumstances? Users are able to post multiple answers to a question. But is this encouraged, or discouraged? Does the answer change depending on the circumstances? For the sake of a concrete example: Say there’s a geometry question that could be approached in different ways: with synthetic geometry, with trigon... (more)	—	over 1 year ago
Answer	—	A: Isn't "any, some, or all" redundant? Why not write just "any"? Language is ambiguous. Mathematics should not be. We can either state our own definitions at the outset, or we can use wording that (hopefully) rules out any of the ambiguous cases. “Any” may mean “at least one” ($n\ge 1$) or “exactly one” ($n=1$). “Some” generally implies both “more than one” and... (more)	—	almost 2 years ago
Answer	—	A: Why 1. multiply the number of independent options? 2. add the number of exclusive options? First, a note: while the question is tagged “probability”, the quoted text talks about the “number of options”, which is equally applicable to counting and combinatorial problems. In what follows, I’ll talk about probability, but this too is equally applicable to counting. I don’t much like th... (more)	—	about 2 years ago
Answer	—	A: How $ijk=\sqrt{1}$? First, there’s a couple of algebraic errors in your question: $\sqrt{a} \sqrt{b} = \sqrt{ab}$ is only true for real numbers, not complex numbers or quaternions $x^2 = y$ does not necessarily imply that $x = \sqrt{y}$ Keep in mind that the radical sign $\surd$ means the principal root. Even whe... (more)	—	about 2 years ago
Answer	—	A: Missing a solution: Are A and B always equal if A-B=0 To directly answer the question in the title: yes, $a-b=0$ always implies $a=b$. (Well, okay, I won’t say there isn’t some abstract algebra, somewhere, where it doesn’t hold, and where $a-b+b=a$ isn’t necessarily true. But in our ordinary, everyday algebra, it’s true.) (more)	—	about 2 years ago
Answer	—	A: Out of 4 people, why does ways to choose a 2-person committee overcount by 2 the ways to divide the 4 into 2 teams of 2? The situations are different. In (a), two distinct groups of two are formed, “committee” and “not committee”. But in (b), we instead form two non-distinct “team” groups of two each. In fairness, the non-distinctness of the teams in (b) is not explicit in the question. You could certainly argue tha... (more)	—	over 2 years ago
Answer	—	A: How can you "easily see that such squares [of side length $\sqrt{13}$ and $\sqrt{18}$] will not fit into the [4 × 4] grid"? The key point here is what $m$ and $n$ represent. Why is the side length of each square $\sqrt{m^2+n^2}$? Why do they have to be less than 4? And why does the author say that swapping $m$ and $n$ gives an identical case? Because $m$ and $n$ are the legs of the right triangles that surround the squ... (more)	—	over 2 years ago