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:
a
n = 4
n + 33
Step-by-step explanation:
n = number of jumps after the 0 term of '33'
aka: add 4 to the previous number
Answer:
The answer is cosx cot²x ⇒ the first answer
Step-by-step explanation:
∵ cot²x = cos²x/sin²x
∵ secx = 1/cosx
∴ cot²x secx - cosx = (cos²x/sin²x)(1/cosx) - cosx
= (cosx/sin²x) - cosx
Take cosx as a common factor
∴ cosx[(1/sin²x) - 1] ⇒ use L.C.M
∴ cosx[1-sin²x/sin²x]
∵ 1 - sin²x = cos²x
∴ cosx(cos²x/sin²x) = cosx cot²x