Amino acids, and proteins. Hope this helps! :)
Answer:
Mean of sampling distribution=3.5
Standard error of sampling distribution=0.219
Explanation:
We know that
The average of catching fish per fishing trip=μ=3.5
and
The standard deviation of catching fish per fishing trip= σ=1.2.
We have to find the mean and standard error of sampling distribution of 30 fishing trips i.e. μxbar=? and σxbar=?
Mean of sampling distribution=μxbar=μ=3.5
Standard error of sampling distribution=σxbar=σ/√n
Standard error of sampling distribution=σxbar=1.2/√30
Standard error of sampling distribution=σxbar=1.2/5.4772
Standard error of sampling distribution=σxbar=0.219
Thus, the mean and standard error for a sampling distribution of 30 fishing trips are 3.5 and 0.219.
I would probably go with B or A
Greeted by enthusiastic crowds, that’s what I read. I hope this helps
