THIS IS THE COMPLETE QUESTION BELOW;
The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation of 425 dollars and an unknown population mean. A random sample of 22 sociologists is taken and gives a sample mean of 1520 dollars.
Find the margin of error for the confidence interval for the population mean with a 98% confidence level.
z0.10 z0.05 z0.025 z0.01 z0.005
1.282 1.645 1.960 2.326 2.576
You may use a calculator or the common z values above. Round the final answer to two decimal places.
Answer
Margin error =210.8
Given:
standard deviation of 425
sample mean x=1520 dollars.
random sample n= 22
From the question We need to to calculate the margin of error for the confidence interval for the population mean .
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
since the original price was ninety we have to multiply 5/6 by 90
90/1(5/6)
=450/6
=75
If you want the slope it's 1/-8
Answer:
Step-by-step explanation:
3x+2<11
3x<9
3x/3<9/3
X<3
X=3