Question 136165
The formula for a half-life function is 



{{{A[r]=A[0](1/2)^(t/H)}}} where {{{A[r]}}} is the amount remaining, {{{A[0]}}} is the original amount of the substance, and H is the half-life



So in our case A=90 and H=17.5


{{{A[r]=90(1/2)^(t/17.5)}}} Plug in {{{A[0]=90}}} and H=17.5


{{{A[r]=90(1/2)^(6/17.5)}}}  Since we want to know how much is remaining after 6 days, simply plug in {{{t=6}}} 



{{{A[r]=90(1/2)^(0.34286)}}} Divide



{{{A[r]=90(0.78847)}}} Raise {{{1/2}}} to the 0.34286th power to get 0.78847



{{{A[r]=70.9623}}} Multiply 



So in 6 days, there will be about 70.9623 mg of arsenic left over