Welcome to Windows 7 Forums. Our forum is dedicated to helping you find support and solutions for any problems regarding your Windows 7 PC be it Dell, HP, Acer, Asus or a custom build. We also provide an extensive Windows 7 tutorial section that covers a wide range of tips and tricks.

# Windows 7: Simple Test Problem

 07 Apr 2014 #11 Endzone Windows 7 32-bit 53 posts Quote: Originally Posted by MrWhoopee As a 30 year machinist and draftsman, we call that object a poiuyt or blivet. https://www.google.com/url?q=http://...M_VFkdKgGxyjlg Oh no, this isn't a trick problem. There are many solutions. My System Specs
 07 Apr 2014 #12 Endzone Windows 7 32-bit 53 posts Quote: Originally Posted by MrWhoopee Can't be done. .5 mi/30 mph=1 min. 1 mi/60 mph=1 min. To average 60mph over the whole mile, the car must get to the other end in 1 min., and that is exactly how much time it took to get to the top of the hill at 30 mph Alright, you're 1 for 3. The other 2 problems have very real solutions. I can't believe you can't get the orthographic projection problem if you were a draftsman for 30 years. My System Specs
 08 Apr 2014 #13 King Arthur Windows 7 Ultimate x64 SP1 603 posts Los Angeles, CA, USA Quote: Originally Posted by Endzone Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? Short answer: No. Long answer: Base chance of obtaining car: ~33.3333...% (100% / 3) Chance after door #3 is opened: 50% (100% / 2) Chance choosing to go through with door #1: 50% Chance choosing to switch to door #2: 50% Assuming I have not made any mathematical incorrections, the highest we can get the chance for the car is 50%. It is true the host knows which door has the car, but we can't scientifically assert that the host is actively trying to deny the car or not, thus we must discount the host's query as nothing more than words at face value. The problem also does not state that we must choose door #1 ("say No. 1", tone is that of an example), further devaluing the host's query as the host might as well have queried with door #1 instead of #2. ---    Quote: Originally Posted by Endzone In the diagram below, a car averages 30MPH uphill for the first 1/2 mile. What must be the car's average downhill speed in order to average 60MPH for the entire mile? Base equation: (30 + B) / 2 = 60 Solving for B (*2 to both sides): 30 + B = 120 Solving for B (-30 to both sides): B = 90 Plugging B into base equation: (30 + 90) / 2 = 60 Solving: (30 + 90) / 2 = 60 120 / 2 = 60 60 = 60 60 Average velocity needed in latter 1/2mi to achieve average velocity of 60mph over the entire given 1mi span: 90mph --- And if you're cleverly making us do your homework, may the deities have mercy. My System Specs
 .
 08 Apr 2014 #14 Endzone Windows 7 32-bit 53 posts King Arthur, thanks for playing, but you're wrong on both problems. If you don't switch doors, you have a 33.3% chance of winning the car. If you do switch doors, you have a 66.6% chance of winning the car. But, why is that? And the one with the car going up the hill and down the hill, it is impossible to average 60MPH for THAT particular mile. No, this isn't for homework. I'm 57, and I have been out of college for 30 years. My System Specs
 08 Apr 2014 #15 Stephanie Win 7 Pro x64, Win 10 Pro x64, Linux Light x86 5,961 posts England Quote: Originally Posted by Endzone King Arthur, thanks for playing, but you're wrong on both problems. If you don't switch doors, you have a 33.3% chance of winning the car. If you do switch doors, you have a 66.6% chance of winning the car. But, why is that? And the one with the car going up the hill and down the hill, it is impossible to average 60MPH for THAT particular mile. No, this isn't for homework. I'm 57, and I have been out of college for 30 years. Cool My System Specs
 08 Apr 2014 #16 Endzone Windows 7 32-bit 53 posts My System Specs
 08 Apr 2014 #17 King Arthur Windows 7 Ultimate x64 SP1 603 posts Los Angeles, CA, USA Wonderful explanation on the Monty Hall problem, thank you. I am however unconvinced that my answer to problem #3 is incorrect because it is (apparently?) mathematically provable. An "average" is the sum of a group of numbers divided by the number of numbers. In this case the 2 numbers are 30 (provided) and 90 (solved for) that when summed and then divided equals to an average 60. The problem as given in this thread did not specify any timeframe, so MrWhoopee's answer that incorporates minutes (a measure of time) is working with variables that were outright not provided. The "hill" itself is irrelevant as the focal point of the problem is the distance to be traveled, which can literally be in any shape possible so long as it's 1mi in length traversable from start to end. To put my answer differently, you travelled the first 1/2mi at 30mph, if you then hypothetically travel the following 1/2mi at 90mph you obtain 60mph average velocity over the 1mi span given because (30+90)/2=60. Yes, I realize it's physically impossible to instantly accelerate from 30mph to 90mph right at the 1/2mi mark, but this is the world of math-on-paper where some worldly limits can be exempted without skewing the results. I actually admit that I have an itching feeling that MrWhoopee is indeed correct and I am wrong, but I would like to hear an explanation for everyone's enlightenment. My System Specs
 08 Apr 2014 #18 Endzone Windows 7 32-bit 53 posts At 30MPH it takes 1 minute to get to the top of the hill. You can not take longer than 1 minute to go the entire mile because an average of 60MPH in 1 mile means you can't take longer than 60 seconds to do it. Time is up when the car gets to the top of the hill. The car would have to travel instantaneously down the back side to average 60MPH for the entire mile. My System Specs
 08 Apr 2014 #19 mjf Windows 7x64 Home Premium SP1 5,795 posts Australia Quote: Originally Posted by King Arthur I actually admit that I have an itching feeling that MrWhoopee is indeed correct and I am wrong, but I would like to hear an explanation for everyone's enlightenment. To drag it out: Let - t1 = time taken to travel up the hill t2 = time travel to travel down the hill 30mph = average speed up the hill v2 = average speed down the hill 60mph = average overall speed then the 3 equations are: 1) 30.t1 = 0.5 2) v2.t2 = 0.5 3) 60.(t1+t2) = 1 Solving: t1 = 1/60 t2 = 0 v2 -> infinity Conclusion: It cannot be done. Or the much simpler answer given by Mr Whoopee. Simpler is often smarter. ---------------------------- Perspective question. Need clarification. 1) should there be a mid line across all of the front view? If not 2) are you allowing a zero thickness folded shape like a folded tray with zero thickness sides? My System Specs
 08 Apr 2014 #20 King Arthur Windows 7 Ultimate x64 SP1 603 posts Los Angeles, CA, USA I understand the logic behind MrWhoopee's answer and can find no fault in it, but I also can't find fault in the way I came to the conclusion of 90mph as the latter of two variables needed to attain an average 60mph. I am going to assume that I severely misunderstood the problem somewhere because I am certain that the way I calculated averages at least is correct. Thank you for the explanations, was very enlightening. My System Specs