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Jac asks: Which word is stressed in the sentence "are you okay?"
There is not a single answer to your question; which word is stressed depends on the context. It could be any of the three words, depending on the situation.
I'm hanging out with two friends - Joe and Moe. Joe looks really sad, so Moe says, "Are you okay?". Joe says, "Oh yes, I'm fine." But then, after Moe leaves, Joe tells me about all the horrible things that's been happening in his life. After listening to him for a few minutes, I say, "ARE you okay?" I emphasize the word 'ARE', because Joe has previously said that he is okay, and my emphasis on that word indicates that I doubt his statement about being okay.
Now Moe and I are in a car, and we get into a fender-bender. Moe bangs his head against the windshield, and then he looks at me and says, "Are you okay?" I reply that I'm fine, and then say, "Are YOU okay?" This time I emphasize 'YOU' because I'm thinking that I'm not the one we should be worrying about, it's Moe. It's a way of saying, "Never mind about me - you're the one we should be worrying about!"
This is probably the most common situation; the word 'okay' will have a natural upward inflection/stress because the sentence is a question. For a situation where the word 'okay' gets extra stress, imagine that Joe is telling me about all the terrible things that he's gone through, and I ask, "Are you OKAY?" (I'd be more likely to add the word 'but': "But are you OKAY?"). Emphasizing the word 'OKAY' in this situation might be a way of implying that I have doubts whether he's handling the situations in a healthy way.
There are a lot of sentences that can be stressed in different ways depending on the context. In most cases, we do it automatically without stopping to think about how we're stressing the words!
Thanks for the question, Jac.
"Are there any tricks that can help you easily factor three digit numbers (without using a calculator)?" ~Jay
Hi Jay, I assume you're talking about tricks besides the normal divisibility rules (for example, if the digits add to three, the number is a multiple of three, if it ends in 0 or 5 the number is divisble by 5, etc). If you're not familiar with those rules, you might want to take a look at this unit here: Divisibility Rules.
Beyond that, there are some tricks that sometimes help. Here's my favorite. Let's say you wanted to factor the number 483. Here's what I would do:
- Multiply the first and last digit: 4 x 3 = 12
- Find two numbers that multiply to 12 and add to the middle digit (8). The numbers are 6 and 2 (6 + 2 = 8 and 6 x 2 = 12).
- Now rewrite the number using those two numbers we just found: 483 = 460 + 23 (the tens place got split into two pieces using our numbers, and the entire number was rewritten as a sum of two numbers).
- Now factor the result: 460 + 23 = 23(20 + 1) = 23 x 21.
- Finish factoring: 23 x 7 x 3
Unfortunately, this doesn't always work. For example, it won't work for 648, because you can't find two numbers that add to 4 and multiply to 48. But maybe if we can find a way of regrouping this number, we might get around that. My first thought is to pull out one of the "hundreds" and put it into the tens place. So we're thinking of 648 as being rewritten 5(14)8. Now we do 5 x 8 = 40, and realize that our two numbers must be 4 and 10 (4 + 10 = 14 and 4 x 10 = 40). So we rewrite the number: (600 + 48 = 24(25 + 2) = 24 x 27. Then we just finish the prime factorization from there.
If the number is one of those special numbers (like 483) that can be factored without regrouping, it's a straightforward, foolproof process. But if the number has to be regrouped, it requires a bit of intuition to work it out. However, if you don't have a calculator, it might be worth doing!
Thabang from Lesotho writes, "how do we rationalize a denominator consisting of a cube root with another constant added to it or subtracted from it?"
Good morning, Thabang, and thank you for your question. This is actually something I don't remember ever seeing before, so I had to give it some thought before answering.
What you're looking for is, how do we rationalize the denominator, if the denominator is something like "The cube root of three, plus two" or "the cube root of three, minus two"?
In order to solve this, it's important to remember two factoring rules you may have learned in an Algebra class:
x3 + y3 = (x + y)(x2 - xy + y2)
x3 - y3 = (x - y)(x2 + xy + y2)
Let's say your denominator is the cube root of three, plus two. Then I'm going to do the following substitutions:
Let x = the cube root of three, let y = 2.
Now your denominator is x + y, and if you multiply the numerator and denominator of the fraction by (x2 - xy + y2), you will have turned the denominator into x3 + y3 = 3 + 8 = 11, which is rational.
That was using the first factoring rule shown above. If the denominator had a subtraction (the cube root of three, minus two), we'd just use the second factoring rule, and multiply by (x2 + xy + y2).
Thanks again for asking, Thabang.
Navya asks: "Why do we have names for numbers?"
There are two answers to this question. The first answer is: because it's impossible to talk about numbers verbally unless you have names for them. If we didn't have the names "one", "two", "three" and so forth, how would we ever say "I have five apples"?
That explanation is sufficient for why we have names for the numbers from zero to nine, but it's not sufficient for numbers like eleven, twelve, and so on. After all, if we didn't have the name "eleven", we could still say the number by saying "one one."
Thus, we would count like this (starting at ten): "one zero, one one, one two, one three..." and so forth.
And in some cases, that would be quicker. The name "eleven" has three syllables, while "one one" just has two. Even worse would be a number like "three thousand, four hundred, sixty three" which takes nine syllables instead of the four syllables required for "three four six three".
So, since the number names aren't always quicker to say than just reciting off the digits, why do we bother? The answer is that using number names allows us to get an immediate order-of-magnitude sense for how big the number is. Look at it this way - if I say "seven million, two hundred twenty three thousand, four hundred twelve," the moment I said "seven million" you had a very good sense for how big that number is. But if instead I had said "seven two two three four one two" you would not have any way of determining the number's magnitude until I was all done reciting the digits, and you knew how many digits there were. And if you lost track of how many digits there were, you still wouldn't have a good sense for how big the number is!
So number names are very helpful for order-of-magnitude sense of the size of a number.
Sarah asks, "On your site you asked the following question and i believe your answer is incorrect. "I have a drawer with 10 socks. 6 are blue and 4 are red. I draw a blue sock randomly and then I draw a red sock randomly. Are these independent or dependent events". You answered even though it is without replacement these are independent. I believe the answer should be dependent, since there is one less sock in the draw when you pick the red one. Am I correct?"
Hi Sarah, thanks for asking this question. I went back and looked at the page your question refers to (Independent and Dependent Events) and realized that the question you were asking about may be a bit ambiguous in how it's worded. Does it mean:
- I drew a sock specifically from the set of blue socks (in other words, I looked in the drawer to find the blue socks, and then randomly selected from that subset) or...
- I reached into the drawer without looking, and randomly pulled out a sock from the entire set, and that randomly selected sock happened to be blue.
I write competition math problems for various math leagues, and I always hate writing probability problems, because they can be so easily written in an ambiguous way (my proofreader hates proofreading them for the same reason). In this case, let's take a look at these two possible interpretations of the problem.
In the first case, the two events are clearly independent; it doesn't matter which of the blue socks was chosen; there are still 4 red socks, and the probability of choosing any particular red sock is 1/4. Thus, the second event is not affected by the first event.
In the second case, I randomly pulled a sock from the drawer, but now we're given the additional information that this sock happened to be blue. So this means that when I reach back into the drawer, there are now nine socks to choose from (not four, as in the previous case, because we assume I'm picking from the entire contents of the drawer.) Since we know that the sock I first chose is blue, there are still 4 red socks, so the probability of choosing a red is 4/9. We get a different answer if we read it this way, but we still have two events that don't affect each other. Since we know the first sock was blue, it doesn't matter which blue sock it was. The specifics of the draw don't affect the outcome of the second draw.
Here's how to make these two events dependent: don't specify that the first draw was blue. Now the result of the second draw is very much dependent on whether or not the first draw was blue. That's probably the situation you were thinking of.
Thank you for asking the question - as a result of your question, I'm going to do some tweaking in the wording of that problem. I don't want it to be ambiguous - especially since the second reading of the problem delves into conditional probabilities, which I don't address on that page!