Answer:
represent the random sample selected
represent the number of pots that were bare ground (no vegetation

And replacing we got:

So then the sample proportion of bare ground spots is 0.792 for this sample
Step-by-step explanation:
We have the following info given from the problem:
represent the random sample selected
represent the number of pots that were bare ground (no vegetation)
And for this case if we want to find the sample proportion of bare ground spots we can use this formula:

And replacing we got:

So then the sample proportion of bare ground spots is 0.792 for this sample
Answer:
0.04
Step-by-step explanation:
In this case we must use the following formula to be able to find the standard error, which would be the square root, of the probability for its complement divided by the population:
SE= 
Now, knowing that p = 0.552 and n = 160 we replace and we are left with:
SE =
SE = 
SE = 0.039
Which means that the standard error of the proportion is 0.04
Answer: (27- 4/3 pi) r^3
Step-by-step explanation: 1. Volume of a cube: V= a^3, V= 3^3 , V=27
2. Volume of a sphere: V=4/3 pi r^3 ......
Answer:
the new scale is the same length as the original scale
Answer: 289 units
Step-by-step explanation:
Given the following :
Inventory (I) = 180
Lead time (L) = 7 days
Review time (T) = 2 weeks = 14 days
Demand (D) = 20
Standard deviation (σ) = 5
Zscore for 95% probability = 1.645
Units to be ordered :
D(T + L) + z(σT+L)
(σT+L) = √(T + L)σ²
= √(14 + 7)5²
= √(21)25
= 22.9
D(T + L) + z(σT+L) - I
20(14 + 7) + 1.645(22.9 + 7) - I
= 420 + 49.1855 - 180
= 289.1855
= 289 quantities