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

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

96 39 cause 135 -96=39 im smart (i used a calculater)

replication and blocking

step-by-step explanation:

you should talk someone who owns the website about it. you did not deserve to have all that hard work and time taken away. id watch videos of little puppies to cool you down, and maybe some nacho cheese doritos.