Posted: 2/1/2005 6:00:15 PM EDT
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Conneticut pop: 31431000 area: 155973 sq. mi Wyoming pop: 476,000 area 97105 sq. mi How many people would have to move from CT to the free Gun loving state of WY to make the Population densities in the two states the same? Explain. Sheesh, someone help me so I can get this kid to bed. |
Total number of people in both states divided by total number of land area. Then multiply by WY's land area, yields the number of people that are supposed to be there. Subtract current WY population to get number who'd have to move. |
12 147 835 |
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Let C = Area of Conn let W = Area of Wyoming let A = the final population of Conn let B = the final population of Wyoming A/C = B/W if the poulation density is equal B = AW/C A+B = 31431000 + 476000, the total population of both states and B = 31431000 + 476000 - A So B=AW/C = 31431000 + 476000 - A => A = (31431000 + 476000) / (W/C +1) Subtract A from the original population of Connecticut to find the number of people that |
Aaah, the arrogance of youth. OK math whiz, here is a question for you . . . oh and no calculators, you have to do this in your head. You planned to fly an airplane at 210 knots, you are currently flying at 210 knots and find that you are 10 seconds late to where you are going. How much do you increase your speed by to get to your destination on time? Yes, you have all the information you need to do this problem. Yea, these sorts of problems are so much harder when you get over 30. 
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Don't be so hard on the kid... 1st) I'd ask the FMS cuz I don't have a clue and by the time I figured it out I'd be past where I was going anyway. 2nd) I would imagine (and I'm guessing) that his oldtimer comment comes from the perspective of someone who is still in college verses someone who is not and doesn't use math much anymore in the real world (as most of us don't). I was kick ass at calculus and all that jazz in HS and college... I don't use it at all anymore and it definately shows. I can, however, do a kickass xwind landing... that requires no math... 
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Speed= Distance divided by Time, or S=D/T... ahhh fuck it. |
I don't believe so, you don't include the distance you are going to travel, nor the original time it was going to take. Additionally, you don't mention whether or not you've started your trip. If you're ten seconds late and only have 100 miles to go, you would need to increase your speed more than if you were going to travel 600 miles. We recently did a lot of word problems in our class, some very similar to this. And if you think I'm arrogant, and maybe not so good as I think, I scored in the 99th percentile on my PSAT test taken earlier this year (versus other college-bound juniors)...and 99th percentile in my overall score. |
since you ARE 10 seconds late, then you have already arrived, so you can't speed up to arrive in time. these problems are so simple. ETA: but before you consider increasing your speed, I would suggest advising your passengers that you will be one hour late to your destination.....damn airlines. |
You Don't. You slow the engine(s) down. If you are 10 seconds late then you are either in the process of landing or have already landed. And anyway 10 seconds behind schedual is more then within acceptable time delays. |
Damn, did I skip that day??
I’m not saying that you are not smart, or good at math. You’re probably better than I am. Heck, I suck at math higher than jr. high level. Just don’t underestimate the over 30 crowd. I’m still close enough to 16 to remember the scorn I had for the older crowd . . . now I am one. On the surface you would think this is a simple time/distance problem, yet no distance was provided. You can still answer this problem, and the answer can be the same if you have 100 miles to go or 10,000. There are a couple of different answers to this problem. So what’s the answer? Here’s one of them: Simple answer: increase speed by 10 knots (to 220), and hold that speed increase for 3.5 minutes (roughly). Once you know that ratios and rules you can do this in your head in seconds. Complex answer: Figure out how fast you planned to go in knots per second, figure out how far behind you are in nautical miles by being 10 seconds late, then how much of a speed increase will gain that time back, and how long you will have to hold that speed increase. Put that in your 99th percentile pipe and smoke it.
Yes, your FMS can provide the answer, or you could say, “Hey nav . . .?” Navigator = FMS that breaths.
Not necessarily. I’ve been 10 min into a 55 min low level route and found myself 10 sec behind. Usually 10 sec is an acceptable delay. I have a 2 min window (+/- 1 min) to work with. However, I like to be a close to my TOT as possible as soon as possible. You never know when something else will pop up that will complicate your time control plan.
That’s one of the great things about my job – passenger comfort, and keeping them informed is not a consideration . . . unless they are DVs. |
Your answer could be wrong, if you were only a minute or two away, and by your logic the answer would be anything greater than 210. You said there would be an increase in speed - so you haven't answered anything. All you've given is an approximation of how long you would have to hold a representative speed, something which wasn't even asked for. Oh, and I think I will take my 99th percentile PSAT results, and smoke my 1490 SAT score out of it. |
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Let C = Area of Conn let W = Area of Wyoming let A = the population of Conn let B = the population of Wyoming let X = the quantity of people that must move to Wyoming A/C = B/W if the poulation density is equal A/C = (B+X)/W A*W = C*(B+X) A*W = C*B + C*X (A*W)-(C*B)= (C*X) ((A*W)-(C*B))/C = X The solution is that the population of WY must increase by 19,092,177 people to the same population density of CT. Not to bust your chops but this just goes to show why our country is losing technical jobs to China and India. Americans are not being taught enough math at an early enough age to be competitive. |
If you assume that he discovered his error early enough in the flight that his speed adjustment doesn't need to be greater than his aircraft is capable of... He only needs to fly the next leg, however long it is, 10 seconds faster than he would have at his planned airspeed. The over all flight time, or distance traveled isn't important. There could be several different answers. 220 knots for x number of minutes, or 221 knots for y. Basically, use the info he gave you to derive an equation, and then graph it. |