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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 = <span> <span> <span> 6.1667882562 </span> </span> </span> days
Source:
http://www.1728.org/halflif2.htm
(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 = <span> <span> <span> 6.1667882562 </span> </span> </span> days
Source:
http://www.1728.org/halflif2.htm
Answer:
95% Confidence interval = (23.4,26.2)
Step-by-step explanation:
In this problem we have to develop a 95% CI for the mean.
The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.
We know that for a 95% CI, the z-value is 1.96.
The CI is