Jake has decided to give a bonus to the employee who has worked the greatest total number of hours over the past six months. Fou
1 answer:
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A)
![\bf 1990-1910=80\leftarrow t \\\\\\ P(t)=8000(2)^{-\frac{t}{29}}\implies P(80)=8000(2)^{-\frac{80}{29}} \\\\\\ P(80)=8000\cdot \cfrac{1}{2^{\frac{80}{29}}}\implies P(80)=\cfrac{8000}{\sqrt[29]{2^{80}}}](https://tex.z-dn.net/?f=%5Cbf%201990-1910%3D80%5Cleftarrow%20t%0A%5C%5C%5C%5C%5C%5C%0AP%28t%29%3D8000%282%29%5E%7B-%5Cfrac%7Bt%7D%7B29%7D%7D%5Cimplies%20P%2880%29%3D8000%282%29%5E%7B-%5Cfrac%7B80%7D%7B29%7D%7D%0A%5C%5C%5C%5C%5C%5C%0AP%2880%29%3D8000%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5E%7B%5Cfrac%7B80%7D%7B29%7D%7D%7D%5Cimplies%20P%2880%29%3D%5Ccfrac%7B8000%7D%7B%5Csqrt%5B29%5D%7B2%5E%7B80%7D%7D%7D)
and surely you know how much that is.
b)
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.
and surely you know how much that is.
b)
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.