Isn't the half-life the halfway point between one and the other?
I'm not sure I quite undertand this question; between one and the other what? If you mean between one element (in this case Cs-137) and the one it decays into (in this case Ba-137) then no, that's a complete misundertandng of what a half-life is (more on what a half-life is below).
And a legitimate question: is the conversion rate constant?
No it isn't, the conversion rate (as measured by the activity of a sample) is proportional to the number of radioactive atoms that are present (because the more atoms you have the greter the chance that one of them will decay in a given time period), so as a sample decays away the rate at which it decays reduces. The net result of all this is that both the number of atoms of the radioactive isotope and the activity follow an exponential decay. This means that the time it takes for half of a sample to decay does not actually depend on the number of atoms present, so it will be contant over time. We call this the half-life of the isotope and it works roughly like this:
What's more, because activity is proportional to the number of undecayed atoms you have, your sample will have gotten steady less radioactive over time too, in fact for every half life you waited the number of click/s on your gaiger counter (the activity) will have halved (assuming the decay product doesn't itself decay, this will complicate things a bit if you can't tell the difference between "clicks from the decay of the original isotope" and "clicks from the decay of the decay product", if the thing that the decay product decays into itself decays that'll screw things up further and so on and so forth).
However, all of this is (somewhat) irrelevant because we weren't expecting the decay of the Cs-137 to have decreased the amount we see (although it would, but only about 3.5% of the original if my quick back of the envolope calculation is accurate) but because we would expect the caesium to have dispersed throughout the whole pacific ocean, which would drastically reduce the concentration down to levels that simply aren't worth worrying about. That it hasn't implies that either something is keeping it there (for example it's been taken into the sand on the ocean floor (which doesn't move much)) or (more worrying) it is being dispersed but something is still leaking and replacing the casesium as fast as it's lost into the wide ocean.
Each reviewer produces a binary recommendation within the same timestep: ’accept’ or ’reject’. If a paper gets 2 ’accept’ it is accepted, if it gets 2 ’reject’, it is rejected, if there is a tie (1 ’accept’ and 1 ’reject’) it gets accepted with a probability of 0.5.
If a single 'bad' reviewer (i.e. one that gives the 'wrong' answer as determined by the 'correct' method of reviewing mentioned as a control in the paper) can cause a paper to have a 50:50 chance of acceptance or rejection it doesn't seem too suprising to me that a relatively small number of them could cause the process to become '[not] much better than by accepting papers by throwing (an unbiased) coin' - because in their model, in the case of a reviewer disagreement, that's exactly what is happening!
We initially left the choice of using it up to you because there's a downside: https can make your mail slower since encrypted data doesn't travel across the web as quickly as unencrypted data. Over the last few months, we've been researching the security/latency tradeoff and decided that turning https on for everyone was the right thing to do.
I wonder if this has anything to do with the reports of Chinese users having their accounts hacked? ‘Only two Gmail accounts appear to have been accessed, and that activity was limited to account information (such as the date the account was created) and subject line, rather than the content of emails themselves,’ said David Drummond in that blog update. That does sound like it perhaps could be a result of insecure HTTP traffic being intercepted in transit between the users and Gmail’s servers.
Lead me not into temptation... I can find it myself.