Let 
be the number of students who eat cauliflower.
Therefore, 
.
Let 
be the total number of students surveyed.
Therefore, 
Thus, 
Now, for 90% confidence level, from the table we know that 
.
The formula for the interval range of proportion of students is :
Plugging in the values we get:
Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).
Answer:
Step-by-step explanation:
=3/16
Using Calculator
=0.1876
Answer:
30
Step-by-step explanation:
45+45 = 90
150-90 = 60
60/2 = 30
I think...
For C at 7pm they sold most and at 6 am they sold less
That would be 5/913. As a percent, that would be 0.5%.
I hope this answer helped you! If you have any further questions or concerns, feel free to ask! :)
