Answer:
95% Confidence interval = (23.4,26.2)
Step-by-step explanation:
In this problem we have to develop a 95% CI for the mean.
The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.
We know that for a 95% CI, the z-value is 1.96.
The CI is

Complete Question
Table of Annual CPI values
2003-184.00
2004-188.90
2005-195.3
2006-201.6
2007-207.342
2008-215.303
2009-214.537
2010-218.056
2011-224.939
2012-229.594
2013-232.957
2014-236.736
QRINC offered new employees a starting salary of $34,862 in 2013. What would a comparable starting salary have been in 2003?
Answer:
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Step-by-step explanation:
From the question we are told that
CPI for 2003(index)=2003-184.00
CPI for 2013(index)=2013-232.957
Starting salary in 2013 at $34,862
Generally comparable starting salary C is given as


Therefore C the comparable starting salary is givrn to be
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
Answer:
3.14
Step-by-step explanation:
The answer is going to be 3.14
Hope it helps!
Answer:
3 gallons of orange juice
Given are:
250 adults and 180 parents
so we can solve the number of not parents
not parents = 250 - 180
not parents = 70
then we can number of people of votes for no
no not parents = 70 - 40 = 30
no parents = 180 - 150 = 30
yes no
parents 150 30
not parents 40 30