A survey of 1000 air travelers1 found that prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is . Use a normal distribution to find a 95% confidence interval for the proportion of air travelers who prefer a window seat.

Round your answers to three decimal places.

The 95% confidence interval is

s

step-by-step explanation:

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}

step-by-step explanation:

for exempl this   system of equations   y = x - 3

                                                                y = 2x - 5

look   the line : y = x -3     ( red)

                            the line : y = 2x -5 ( bleus)

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.