<span>n = 5
The formula for the confidence interval (CI) is
CI = m ± z*d/sqrt(n)
where
CI = confidence interval
m = mean
z = z value in standard normal table for desired confidence
n = number of samples
Since we want a 95% confidence interval, we need to divide that in half to get
95/2 = 47.5
Looking up 0.475 in a standard normal table gives us a z value of 1.96
Since we want the margin of error to be ± 0.0001, we want the expression ± z*d/sqrt(n) to also be ± 0.0001. And to simplify things, we can omit the ± and use the formula
0.0001 = z*d/sqrt(n)
Substitute the value z that we looked up, and get
0.0001 = 1.96*d/sqrt(n)
Substitute the standard deviation that we were given and
0.0001 = 1.96*0.001/sqrt(n)
0.0001 = 0.00196/sqrt(n)
Solve for n
0.0001*sqrt(n) = 0.00196
sqrt(n) = 19.6
n = 4.427188724
Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>
Answer:
54.5454%
Step-by-step explanation:
6/11=0.545454545454 and that as a percentage is 54.5454%
I think the answer is 14 hope this helps
Answer:
326
Step-by-step explanation:
Given: Microscopes. Each one costs $200 and resells online for $526.
To find: What is the mark-up percentage?
Solution: Simply take the sales price minus the unit cost, and divide that number by the unit cost. Then, multiply by 100 to determine the markup percentage.
Net Profit Margin = 61.98%
Net Profit = $326.00
Profit Percentage = 163.00%
Find more here: brainly.com/question/23345732
