Team Cape

do whatever you want lol. fly had like 15 feedback and people trusted him for some reason. now he banned af

For feedback to be valid, it has to involve a trade of 1m+.

@@Saiyan if you dont do that i will fuck you up

Congrats bro I'm probably late on this because i couldn't be on much in the past few days, but yooooooo congrats

I will respond to all messages as soon as I can

gp is only worth anything because bots get banned.

autist

Use the GE API in the snippets section

law of av·er·ag·es noun the principle that supposes most future events are likely to balance any past deviation from a presumed average.

post a bug report

ran into this awhile ago. I realized that if an object has a name, then it has the examine option, so you might want to try if(object.getName() != null) {}, because for whatever reason, it doesnt appear in the list of interact options (i checked this many times). if you want to interact and examine it, there is also a method for that - object.examine(). Hope this helped

try it without the pathpreferenceprofile and see if it works. also can you copy and paste the error you get in the logger

both

theres a correlation

If you just had a 2-day ban, if they catch you botting again, they will always give you a permanent ban. If you want to bot, and you care about the account, it would be best to wait until the 2-day ban expires. They will be monitoring your account more than they were previously because you've already been caught botting as well.

Code looks nonmalicious good release bro

:P

Cool idea

Ahhh here we go. For #3: u is an unbiased estimator so we can assume the mean is the same. Since the parent distribution is x also normal, we also know that it's normal. This should follow the 10% rule as well. So the distribution we use is ~ Nor(300, 3/sqrt(6)) which is ~ Nor(300, 1.225). So now we can plug this into the calculator. normalcdf(-infinity, 299, 300, 1.225) = 20.711% chance. I'm fairly sure this is correct.

1. I assumed the first one was binomial, so i used binomcdf(100, 0.6, 60) on my caluclator and got a 53.79% chance. Not totally sure on this one, but this is what I got. 2. It's given that it's normal, so use normalcdf(-infinity, 299, 300, 3) (or you could use a table but thats very tedious). Comes out to 36.944% - This means that 36.944% of bottles will contain less than 299mL. This one is definitely right. 3. I multiplied the distribution by 6 (so it would be ~ Nor(1800, 18)), and I decided to look for 1794. I believe this should be the same probability (someone can correct me if I'm wrong). I plugged in normalcdf(-infinity, 1794, 1800, 18), and I got 36.944% Edit: I did all this in AP Stats like a couple months ago, so my memory is hazy. This is what I got though. I don't think #3 is right.

You'll have an issue with that if you use the 'Make all' option. Here's what to do: Every time the inventory decreases by 1 in amount of whatever resource you're using, record the time using System.currentTimeMillis(). Then, when you go back in the loop, check if your inventory resource count has decreased by 1 again. If it has, change the time to System.currentTimeMillis() again. Then create a check to see if it has not changed in the past ~ 5 seconds. If it hasn't, then you know you've stopped making it for whatever reason. You could do this in a conditionalsleep or in the regular loop cycle if you want - it's your choice. That's your first option. Your second option (less stable but easier to make) would be to check if the interaction with both item returned 'true' and if it did, then conditional sleep until your inventory doesn't contain one of them anymore OR dialogues/random events, and if either condition is met, stop sleeping.

"The Russian government just tweeted an image of a white supremacist frog" http://www.vox.com/world/2017/1/9/14212496/russia-embassy-pepe

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