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Thanks Dak for typing that up. I cannot access anything in safe mode. It is just a blank screen. I went into command prompt. I cannot access C:\windows\ServicePackFiles\i386. The command always starts with C:\DOCUMENTS AND SETTINGS . . . .etc The error message is exactely: WININET.dll How do I run the antivirus/ antispyware from command prompt. Should I reinstall the whole OS ???

I tried running system restore through safe mode with a command prompt %systemroot%\system32\restore\rstrui.exe The system restore window comes up but its another blank screen (this time a white blank screen) Just great.

In normal mode, explorer.exe (the browser) fails to start. An error message comes up when I click on my login saying 'the application has failed to start because WININET.exe dll was not found' Maybe the Dynamic library has been deleted ????

I tried safe mode. There appears to be a 4th user called 'Administrator' (Maybe a system hijack?) Safe mode is a black screen with windows service pack 2 (and a whole bunch of other stuff -description of system etc) written at the top and safe mode written at all 4 sides. No icons are displayed. The fan is working overtime and the system is frozen again. I'll try looking around the web for other help. Any more suggestions from here?

I need some help (I work on a Windows XP hp pavillion laptop (zd7000)) I kept getting all these spyware warnings over and over. Then the screen froze down. I tried running Ad aware. Everthing froze. I removed the battery which broke the circuit and shut the laptop down. (I'm not sure if this was the best move). I then reinserted the laptop battery and everthing was fine and it came to the Xp login screen. I then clicked on my login (limited account) The screen is just an enlarged version of the wallpaper with no icons showing. And its frozen in this state. The day before I had run ad aware 3x and quarantined all spyware. I don't know why this happened? This is the same for all the other accounts. This is a big problem because all my work is on the HD (only some of it in a portable storage system) Please could someone help. Much appreciated.

407 = 395 x 76 + 24x _____________ 100 30020 + 24x ___________ = 407 100 407 x 100 = 40700 40700 - 30020 = 10680 / 24 = 445 Mean weakly wage for woman = £445 Thanks a lot everyone - you really are a bunch of clever people:mad: I'm sure you probably figure that I was being sarcastic since you're all clever en all. I took the liberty of working it out myself. Thanks for nothing.

A factory employs 100 people. 76 of these people are men. 24 are women. The mean weekly wage of all the 100 people is £407 The mean weekly wage of the men is £395 work out the mean weekly wage of the 24 women.? __________________________________________________ I'm sure there are people here who could do this question in their heads with their eyes closed. Please reply with a full explanation. Thanks in advance. (You've got 5 hours) MAX.

This is very simple maths but I just can't get my head around it. 1) My first question (KEEP IN MIND THAT > SHOULD ACTUALLY INCLUDE EQUALS TO ALSO - I JUST COULDN'T PUT THE UNDERSCORE) I have a frequency table as follows: 1Q MIDPOINT FREQUENCY FX CUMULATIVE FREQUENCY 80 > 85 82.5 1 1 85 > 90 87.5 - - 90 > 95 92.5 7 8 95 > 100 97.5 4 12 THE QUESTION: The 85 > 90 CLASS INTERVAL HAS NO FREQUENCY. How would I deal with this when I draw the cumulative frequency curve (graph). Would I start at 1 and go down to 0 and then draw the curve up to 8 and then to 12. WOULD I HAVE TO GO DOWN TO 0. My final QUESTION concerns the midpoints. ARE THEY CORRECT? Since the cumulative frequency graph is plotted at the class interval boundary (e.g. 85.5, 90.5, 95.5 etc) - will the midpoint change to 82.75, 90.75. instead of82.5 etc) I took a long time writing this post so please answer both questions. Thanks in advance.

Is there any free/trial software that I can download that will enable me to draw up cumulative frequency curves.I tried doing this is Excel and it worked for values over 95 (on the X axis) but anything below that the whole graph included values all the way to 0 which made the data series very tall instead of being a curve. (all my values for the curve are between 95 and 120 in most cases) but some are lower and the graphs end up looking weird).

The market at the moment is all systems go for mobile phones. Mobile phones are THE thing to have. (It is projected that atleast 200 million people wil be playing games on their mobile on a regular basis by 2006).) When looking at it in terms off money: Each mobile game is sold at around a fiver each. If at least 5% of that 200 million gaming community buys my game(which is likely considering the game is reasonably good) - we are looking at 10 million in revenue. If you contrast that to shareware games - the other way I can develop. I would probably sell at fiver a game still. The probablility that someone will buy my game is much MUCH lower when they have all the newer games out (World of Warcraft, Worms 4 etc.). Currently there are about the same amount of PC gamers but most of them will be buying all the new games. I could project that if my game is good enough - a tight knit community of around 500 gamers would purchase my game. That's a tidy profit of 2500 but nearly not as good as 10 million. Which one will prove more better in the long term (Financially). Obviously mobile phones. Breaking into PC is much harder because of the technology learning curve and the stiff competition. Mobile phones do have stiff competition but as long as you know a programming language you're half way there already. And there are so many ways to get the mobile game out on the market (operators, website, publishers, affliated organisations). As opposed to my shareware PC title where the only way would be to sell via my own website. I doubt that any publisher will be willing to take on my 'crap' shareware game for the latest Fable game (e.g. Microsoft). If you were in my shoes what path would you take - considering the majority of you like PC games.

PC games are just as hard to develop - probably even harder (World of Warcraft). And making a game on your own - there is no way you can make a game as good as World of Warcraft or the orginal call of duty. Besides -you need a budget of atleast half a million just to get started. That's why I've opted to either do mobile games (budget starting off is less than 10,000) and also PC shareware (cost of development is minimal). But don't let this influence your decision. (PC shareware games are definitely not as good as full blown commercial games) - So when posting in this thread just think which PLATFORM you would prefer playing any game on - whether it is 2D/3D, Simple/amazingly complex. Mobile phones or PC ?

Don't know about the basketball thing - not into basketball (Shaq maybe? - just a wild guess and the fact that he's the only basketball player I know - does Micheal jordan still count). Tennis however - I think that its a tie between John McEnroe and Boris Becker. (Thats the old players). Although I like Olivier Rochus because he gives inspiration to a short tennis players like me. (I'm the same height as him).

Obviously console games are the best platform (most people play on consoles) But developing for consoles is much harder. Take Xbox = not only do you need to be proficient in C++ but also you need to know how to programme C++ with the Microsoft direct 3D API. (Its crazy - and absolutely difficult, takes years to learn and master).

There IS water. (I just have no proof.)

There must be an alternative - Medicinal marijuana also brings lots of unnecessary legal issues

T1000 (the only one I can think of off the top of my head). Another great villian - the "director" in the Truman show (if you count him as one). INCREASE THE WIND - FULL ON STORM - DO IT!!!

Just to follow up on the thread "programming languages" Could I just ask for opinions on one question: What platform would you rather play video games on (enjoy playing)?: PC or mobile phones Thanks for any replies. PS: Can I just say that there are twice as many mobile phone users in the world than PC users (in general - obviously a lot use PCs and mobiles). 0.60 billion PC users 1.28 billion mobile phones

I've been trying to write programs for 2 years now and am struggling to find the right language to start with. I began with C and learnt the basics. But this wasn't ideal for my purpose. I want to develop software for mobile phones. This has its own set of programming languages as you probably know Java (J2SE) becomes J2ME C++ becomes Symbian, Brew or Mophun (c++) Python becomes python Which language would be the easiest to learn for me(I'm pretty much a complete beginner - some very light knowledge of C). I heard that python is quite easy. I've looked at the code for J2ME and Symbian/Brew/mophun and its very hard. To learn J2ME I need to learn most of Java first. For the rest I need to know C/C++ first. For python - I think I can jump straight in. IF THERE IS AN ALTERNATIVE THAT YOU COULD SUGGEST THATS EASIER THAN ALL OF THESE - What language should I use? (I now know that javascript cannot be used for Java-enabled mobile phones).

Well? Upon the 10 degrees, the enzyme also helps speed up the rate of a reaction by lowering the activation energy. WHY DOES IT WORK . . .and HOW DO I CALCULATED THE AMOUNT BY WHICH THE ACTIVATION ENERGY IS LOWERED without using the arrehius equation etc

It does say at the top in CAPITALS It a QUANTITATIVE PREDICTION -(just after the prediction). (which means it in the planning section) or P. (I've completed O and E properly. Since most of P and A are the same (and have the most marks) - I need this theory right. (YOU say you're not good at biology.) - (well most of this quantitave work is chemisty anyway - (biochemistry to be exact). IS IT CORRECT from your knowledge of chemistry.

TO 5614, ONLY 26/27 !!! WHAT DO YOU MEAN ONLY! I'd kill for that kind of mark...(kind off). Thats about a high A grade (WHAT DO YOU MEAN "ONLY"!) You were just a mark from an A*. (I think someones a boffin). I worked my ass off and only managed to get 22/30. You are right in saying that its hard to score marks - (most of the theory wanders into the region of AS/A level Science - especially in BIOLOGY - which I posted a thread on and NOONE is replying - yeah thats right - I'm advertising my THREAD in this THREAD . . .so get yourselfs over and help me out of my low B grade).

This is A-level Physics . . .right? As you might have read I got 22/30 recently on my GCSE coursework so I'm pretty much crap at this. I don't even know why I'm posting a reply. Just for the sake of it perhaps. Anyway - could you have a look at my post about BIOLOGY COURSEWORK - if you help me out - I might be able to help you out . . .kind off. (I recently got my year 10 exam result in Maths back - and it was Goooood.) a very high A grade. (just a few marks from A*).

I've done this - I got 22/30 which is around a Grade C / B (a very low B). You will need to do some experiments on circuits and resistance. I did the affect of voltage on current. Do you know about the four sections - P O A E Planning - involves your preliminary work and PREDICTION - with heavy theory Obtaining evidence - just the results and an average graph Analysis - looking at the graph and writing as much as you can about what you see Evaluation - identifying anamolous results and saying what you could have done better The prediction and the conclusion (in the analysis section) are pretty much the same heavy theory. (BE SURE TO INCLUDE CALCULATIONS). This may not help -but it gives you some direction.

I need some help with the theory section of my biology coursework on catalase (an enzyme). I need to make a QUANTATATIVE PREDICITON. THIS IS WHAT I HAVE SO FAR: So far I have wrote about the 10 degree rule Quantitative prediction I predict, in quantitative terms, that at every ten degrees Celsius increment the reaction rate will double. This is based on the activation energy needed to start a reaction. The activation energy is the amount of energy needed in a collision for hydrogen peroxide and catalase molecules to start a successful reaction. If this energy is not reached by the molecules, a reaction will not take place. In most cases, the majority of molecules will not attain the activation energy needed as shown in this example graph which plots a line of a certain temperature against the number of molecules(y axis) and energy (x-axis). GRAPH Blue area = molecules which haven’t attained enough energy to activate Grey area = molecules which have attained activation and are reactable The graph shows that most of the molecules haven’t attained the activation energy required. The grey region shows the number of molecules that have reached the activation energy which is significantly less than the blue area. The activation energy is a certain amount and therefore has been plotted to show the barrier between successful reactions and unsuccessful molecules. So if, for example, the temperature was raised by 1 degree Celsius each molecule will have proportionally 1 more joule of kinetic energy (if 1 degree = 1 joule). Therefore each particle will have greater energy so more will pass the activation energy point: graph From the above graph, it is clear that more molecules have successfully reacted. The 10 degree rule for chemical reactions states: “for every ten degrees celsius the reaction rate doubles” This is understandable because if I was to present another graph showing what would be ten times that of the above graph (therefore being a 10 degrees celsius increment upon the first graph) it would show that the amount of molecules that have successfully reached the activation energy has doubled – since the amount of molecules have doubled so has the rate: GRAPH This should also apply to biochemical reactions such as the one which takes place between catalase and hydrogen peroxide. In this reaction catalase decreases the activation energy and this means that more molecules make the activation boundary. However, catalase does this by binding with the hydrogen peroxide molecule into an ES complex. This reduces the amount of energy needed for the activation to happen. So my quantitative prediction is: “At every ten degree increment up to the optimum the activation energy will be decreased by half [by the catalase], therefore doubling the amount of molecules attaining the activation energy and so doubling the overall rate of reaction.” Graph showing my quantitative prediction: GRAPH THE END IS THE ABOVE CORRECT. What I don't understand is why enzymes lower the activation energy. Is what I've written about that correct?. Please help me tie of all the lose ends. Corect all mistakes in the above etc. ITS DUE IN ON WEDNESDAY 15TH JUNE 2005 So HELP!. (all theory is based on going to the site http://www.s-cool.co.uk and going to A-level chemistry - reaction kinetics - affect of temperatue on rate. (Bear in mind - I'm a GCSE student doing higher double science).