algebra precalculus - Which is greater: $1000^ {1000}$ or $1001
Which is greater: $1000^ {1000}$ or $1001^ {999}$ Ask Question Asked 11 years, 5 months ago Modified 11 years, 5 months ago
probability - 1/1000 chance of a reaction. If you do the action …
A hypothetical example: You have a 1/1000 chance of being hit by a bus when crossing the street. However, if you perform the action of crossing the street 1000 times, then your chance …
Repeating something with (1/n)th chance of success n times
By definition, if given 1000 unique random trials, a particular event (result) occurs once, then the observed probability of that event occurring is 1/1000. The inverse is basically the application …
What does "X% faster" mean? - Mathematics Stack Exchange
That seems reasonable: 100% faster should mean twice the speed, so half the time; 1000% faster should mean eleven times the speed so 1/11 of the time, though I would always bear in mind …
If you toss $1000$ fair coins $10$ times each, what is the …
Essentially, $1000/1024$ is the average number (or "expected" number) of coins that will have come up all heads, but that includes the cases where more than one coin comes up heads all …
What does it mean when something says (in thousands)
It means "26 million thousands". Essentially just take all those values and multiply them by $1000$. So roughly $\$26$ billion in sales.
The final state of 1000 light bulbs switched on/off by 1000 people ...
The final state of 1000 light bulbs switched on/off by 1000 people passing by Ask Question Asked 14 years, 11 months ago Modified 8 years, 8 months ago
What is mathematical basis for the percent symbol (%)?
Percent means 1 part of 100 or 1/100 and is indicated with %. Per mille means 1 part of 1000 or 1/1000 and is indicated with ‰, so it seems that these symbols indicate the mathematical …
Which is bigger, $ \log_ {1000} 1001$ or $\log_ {999} 1000
It doesnt let me make it look correct, it's supposed to be log 1000 of 1001 and log 999 of 1000
Last two digits of $2^ {1000}$ via Chinese Remainder Theorem?
For the congruence modulo $4$ you don't even need to invoke Euler's Theorem; you can just note that since $2^2\equiv 0\pmod {4}$, then $2^ {1000}\equiv 0 \pmod {4}$.