Yes, 1000 is correct. Good job!
Hey guys, need some help with this.
So, first I'm assuming I have to solve for A so:
505 = A / 1+10^k(2000-2000)
505 = A/2
1010 = A, then I substitute 1010 in for A getting:
N(t) = 1010 / 1+(10^(-1(2002-2000)))
N(t) = 1010 / 1.01
= 1000?? Can anyone verify this for me please?