A 33 gram sample has k value of .1124
(Actually I think the problem SHOULD be worded: A 33 gram sample has k value of -.1124/days)
See attached graphic:
Half-Life = ln (.5) / k
Half-Life = -.693147 / -.1124/days
Half-Life =
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6.1667882562
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days
Source:
http://www.1728.org/halflif2.htm
Answer:
There can be 14,040,000 different passwords
Step-by-step explanation:
Number of permutations to order 3 letters and 2 numbers (total 5)
(AAANN, AANNA,AANAN,...)
= 5! / (3! 2!)
= 120 / (6*2)
= 10
For each permutation, the three distinct (English) letters can be arranged in
26!/(26-3)! = 26!/23! = 26*25*24 = 15600 ways
For each permutation, the two distinct digits can be arranged in
10!/(10-2)! = 10!/8! = 10*9 = 90 ways.
So the total number of distinct passwords is the product of all three permutations,
N = 10 * 15600 * 90 = 14,040,000
Ratio=2 ft/s
ratio=2 ft/s * (1 mile / 5280 ft)*(3600 s / 1 h)=1.36363636.. miles/h≈1.36 miles/h.
Answer: 1.36 miles/h
Answer:
no not that wuch
Step-by-step explanation: