Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
Hello!
We have the following data:
Paula weight (before training) = 65.5 kg
Weight lost by Paula (in training) = 3500 grams = 3.5 kg
Paula's weight (after training) =?
65.5 kg - 3.5 kg = 62 kg (Current weight)
Yes
I’m assuming 16 is the amount he earned and (-16) is the amount he spent
16 + (-16) = 16-16 = 0
Juan spent the same amount of money as he earned that day and broke even
