A.) P(t) = P0e^(kt)
P(20/60) = 40 e^(20k/60)
80 = 40 e^(k/3)
e^(k/3) = 80/40 = 2
k/3 = ln(2)
k = 3ln(2)
b.) P(8) = 40(2)^24 = 40(16777216) = 671088640 cells
d.) Rate of change = e^(8k) = e^(8(3ln(2))) = e^(24ln(2)) = e^(16.6355) = 16777216 cells/hour
e.) P(t) = 40(2)^(3t); t in hours
1,000,000 = 40(8)^t
25,000 = 8^t
ln(25,000) = t ln(8)
t = ln(25,000)/ln(8) = 4.87 hours
Each LED bulb, along with installation labor, is priced at
.. $6.95 +$3 = $9.95
For 100 bulbs over a span of 10 years, that equals (100*10) = 1000 bulb·years. At $9.95 per bulb, 5 bulb·years are obtained, and thus the projected total cost for 1000 bulb·years is
.. (1000 b·y)*($9.95/(5 b·y)) = $1990
In summary, for a decade, the installation and changes of 200 bulbs in 100 lamps amount to $1990. Therefore, the yearly cost is...
.. $1990/(10 yr) = $199/yr
The response indicates that the test comprises 10 questions worth 3 points each and 14 questions worth 5 points. If there are any queries, feel free to ask!!! Thank you!
