1/3 and 1/2 common denominators are the number 6 ,1/3x2=2/6 and 1/2 x3=3/6 so 2/6 divide by 2 =1/3 and 3/6 divide by 3 =1/2 I hope this helps
It’s 2 and 3 bc 6a+12 and then 3(2a+4) and you need to distribute for that so 3 times 2a equals 6a and 3 times 4 equals 12 so you get 6a+12
Answer:
A'B'C'D' was dilated by 4
Step-by-step explanation:
The distance from B' to C' is 1.
The distance from B to C is 4.
4/1=4
B. ASA
FGH = IHG
Please correct me if I'm wrong!! :)
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.
