Answer:
elk
Step-by-step explanation:
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:
5x + y = 8
Step-by-step explanation:
Given the equations:
x+3y=10 ____ (1)
2x−y=−1 ____ (2)
Multiply (2) by 2
2(2x - y = - 1) = 4x - 2y = - 2
4x - 2y = - 2 - - - (3)
Add (1) and (3)
4x - 2y = - 2
x + 3y = 10
___________
5x + y = 8
_________
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