What do you invest your 401k in?

Maybe, Soren only posted it for the first time 13 hours ago tho…

Why not give credit to the artist?

Soren Iverson https://www.instagram.com/soren.iverson?igsh=eGd3d3N5dzc0ZDYy

I wasn’t sure if he admitted that as being the reason (even though it obviously is)

I never understood Musk’s reasoning for this decision. From my recollection it was basically “how do you decide who’s right when lidar and camera disagree?” And it felt so insane to say that the solution to conflicting data was not to figure out which is right but only to listen to one.

This is how Tile works as well. Not sure what their data harvesting policies are.

A tale as old as the industrial revolution

https://en.m.wikipedia.org/wiki/Luddite

I feel like there’s a great story behind each one of the five

Ok yeah you’re gonna die in your home

The difference between 3.5 and 4 is substantial. Here is what 4 says

To find the number of combinations in which four friends (A, B, C, and D) can stand in line such that A and C are not next to each other, we can use the following approach:

1. Total Combinations: First, calculate the total number of ways in which four people can be arranged in a line. This can be calculated by (4!) (4 factorial), since there are 4 slots to fill, each choice reducing the number of available choices by one for the next slot. [ 4! = 4 \times 3 \times 2 \times 1 = 24 ]

2. Unwanted Combinations (Where A and C are next to each other):

   • Consider A and C as a single unit. This effectively reduces the number of units to arrange from 4 to 3 (the AC unit, B, and D).
   • These three units can be arranged in (3!) ways: [ 3! = 3 \times 2 \times 1 = 6 ]
   • However, within the AC unit, A and C can switch places. So, there are 2 ways to arrange A and C within their unit.
   • Therefore, the total number of arrangements where A and C are next to each other is: [ 3! \times 2 = 6 \times 2 = 12 ]

3. Subtracting Unwanted Combinations: Subtract the number of unwanted combinations (where A and C are next to each other) from the total number of combinations: [ 24 - 12 = 12 ]

Thus, there are 12 combinations where A and C are not next to each other.