A study of the career paths of hotel general managers sent questionnaires to an SRS of 230 hotels belonging to major U. S. hotel chains. There were 119 responses. The average time these 119 general managers had spent with their current company was 11.63 years. (Take it as known that the standard deviation of time with the company for all general managers is 1.3 years.) Find the margin of error for a 90% confidence interval to estimate the mean time a general manager had spent with their current company.

The value is

Step-by-step explanation:

From the question we are told that

The sample size is n = 119

The sample mean is

The standard deviation is

Given that the confidence level is 90% then the level of significance is mathematically represented as

=>

The critical value of obtained from the normal distribution table is

Generally th margin of error is mathematically represented as

=>

=>

okay so 15.15 *()5: +5/6

step-by-step explanation:

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: