It would be 0.2222 so it's your answer
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.
If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?
Given Information:
standard deviation = σ = 30 hours
confidence level = 99%
Margin of error = 6 hours
Required Information:
sample size = n = ?
Answer:
sample size = n ≈ 165
Step-by-step explanation:
We know that margin of error is given by
Margin of error = z*(σ/√n)
Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size
√n = z*σ/Margin of error
squaring both sides
n = (z*σ/Margin of error)²
For 99% confidence level the z-score is 2.576
n = (2.576*30/6)²
n = 164.73
since number of bulbs cannot be in fraction so rounding off yields
n ≈ 165
Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.
Answer:
Hello! answer: y = 76
Because there vertical angles they Will have the same measure therefore y = 76 Hope that helps!
Answer:
The answer is c
Step-by-step explanation:
cuzzzzzzz I toooooookkkkk aaaa test
The sum of a geometric sequence is:
s(n)=a(1-r^n)/(1-r) in our case:
s(n)=3(1-4^n)/(1-4)
s(n)=-(1-4^n)
s(n)=(4^n)-1 and s=1023
(4^n)-1=1023
4^n=1024
n ln4=ln1024
n=(ln1024)/(ln4)
n=5
