Comparing rates example

We're told that a conservationist has the hypothesis that when squirrels are more crowded together, they have higher rates of aggression. The table below shows the area of three parks and the number of squirrels in each; that's given right over here. Order the parks from least crowded to most crowded based on the crowdedness. In which park would the conservationist expect to see the highest rate of aggression? So pause this video and see if you can figure this out.

All right, now let's work through this together. So we want to order the parks from least crowded to most crowded. So how can we think about crowded? Well, we could think about it in terms of the number of squirrels per area, per hectare. Something with a lot of squirrels per hectare would be more crowded, and something with fewer squirrels per hectare would be less crowded.

You could also, if you wanted to, think about it in terms of hectares per squirrel. This would also be a legitimate way of tackling it. Of course, if you have more hectares per squirrel, that would be less crowded; it would be the other way around. However, if you had fewer hectares per squirrel, it would be more crowded. Looking at the numbers of squirrels, they are larger than the number of hectares in every scenario, so it might be a little bit easier to divide in this one, and this is also how my brain tends to think about it.

Let's calculate the number of squirrels per hectare for each of these parks. So first of all, let's think about Park A, and I will do that over here. For Park A, you have 54 squirrels (I'll write S, all right, squ for short) per every 8 hectares. This is going to be the same thing as 54 over 8 squirrels per hectare. We could try to estimate it, but it looks like they are all actually a little bit around seven if we divide the number of squirrels by the number of hectares. So we might need to get a little more precise.

Let’s see, 8 goes into 54. I will do it over here. Eight goes into 54; it goes 6 times; 6 times 8 is 48, and I subtract, I get a remainder of 6. Then let me put a little decimal here, and if I bring down that 0, 8 goes into 60 7 times; 7 times 8 is 56. I can keep going but let me see if this is enough precision for me to compare. So Park A is approximately (I'll make this little squiggle here for approximately) 6.7 squirrels per hectare.

Now, let me do Park B right over here. So for Park B, we have 20 squirrels (squ for short) for every 2.7 hectares. Now, one thing we can do to help simplify this so we don't have to deal with decimals is let's multiply both the numerator and the denominator by 10. Notice that's just equivalent to multiplying by one. So this is equivalent to saying you have 200 squirrels for every 27 hectares, or you could view it as 200 over 27 squirrels per hectare.

Let's take 27 into 200. If I were to estimate it, let’s see. 27 is close to 30; 30 would go into 200 six times. Let me try that out. It goes into 200 6 times; 6 times 7 is 42; 6 times 2 is 12 plus 4 is 16. If I subtract, I’m going to actually get, it looks like, 38, so maybe I can fit in one more 27 there. So let me do that, 7. So 7 times 7 is 49; 7 times 2 is 14 plus 4 is 18. Yup, that worked out nicely. If you subtract 189 from 200, you're going to get 11.

Now let me bring down a zero. So how many times does 27 go into 110? Well, it looks like it goes three times. I think three times 7 is twenty-one, and three times 2 is 6 plus 2 is 8. It looks like actually I could fit in one more, so let's see; it might go four times. So 110... so if I go four times, 4 times 7 is 28; 4 times 2 is 8 plus 2 is 10. Yeah, it went 4 times, so I get a remainder of 2. We could keep going, but this is approximately 7.4, so approximately 7.4 squirrels per hectare.

So we already see that Park B is more crowded than Park A, but now let's check out Park C. For Park C, we have 51 squirrels for every 6.8 hectares. We could do the same idea; let's multiply the numerator and the denominator by 10, which means that we have 510 squirrels for every 68 hectares.

So 68 will go into 510; I’m guessing I’m going to have to have some decimals here. It’s close to 70; 70 would go into 510 about seven times. Let me see how that works out. 7 times 8 is 56; 7 times 6 is 42 plus 5 is 47. I think I did well there. If I subtract here, I could do some regrouping or I could try to do it in my head. To go from 476 to 500, I would have to add 24 plus another 10, so I'm going to have 34 right over here.

Bring down a zero. If I'm thinking roughly, 70 goes into 340. Let's see... will it go? It might go five times, actually; let me try that out. If I say 7.5, 5 times 8 is 40, and then 5 times 6 is 30 plus 4. It went exactly 5 times. So that means in Park C we’re at 7.5 squirrels per hectare.

So what's the most crowded? If I wanted to order it, the most crowded is Park C, the second crowded is Park B, and the third most crowded is Park A. Based on crowdedness, in which park would I expect the highest rate of aggression? Well, in Park C, the squirrels are all much closer to each other; they might be fighting over things, who knows? But there we go, we answered the question.