A personnel manager has found that, historically, the scores on aptitude tests given to applicants for entry-level positions follow a normal distribution with standard deviation 21.73 points. A random sample of 14 test scores from the current group of applicants had a mean score of 74.57 points. Based on these sample results, a statistician found for the population mean a confidence interval extending from 65.7425 to 83.3975 points. Find the confidence level of this interval. Round off your answer to two decimal places.

The value is

Step-by-step explanation:

From the question we are told that

  The standard deviation is  

  The sample size is  n  =  14

  The mean score is  

  The confidence interval is  65.7425 to 83.3975

Generally the degree of freedom is mathematically represented as

=>  

=>  

Generally the margin of error is mathematically evaluated as

=>

Generally this margin of error is mathematically represented as

=>  

=>  

From the t distribution table the probability of    at a degree of freedom of 13  is  

=>  

=>  

So the confidence level is evaluated as  

=>  

=>  

Converting to percentage

=>  

=>  

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation: