A survey of 1000 air travelers 1 found that 60% 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 SE = 0.015. Use a normal distribution to find a 99% confidence interval for the proportion of air travelers who prefer a window seat.

Round your answers to three decimal places. The 99% confidence interval is to .

What is x3+3x2−16x−48 divided by x−1?

What is 1/3x5 on anuber line what does it look like on a number line

Determine the common ratio and find the next three terms of the geometric sequence. 649-14-02-00-00_files/i0100000, 649-14-02-00-00_files/i0100001, 2, a. 3; 649-14-02-00-00_files/i0100002, 649-14-02-00-00_files/i0100003, 649-14-02-00-00_files/i0100004. c. 649-14-02-00-00_files/i0100005; 649-14-02-00-00_files/i0100006, 649-14-02-00-00_files/i0100007, 649-14-02-00-00_files/i0100008. b. 3; 6, 18, 54 d. 649-14-02-00-00_files/i0100009. 6, 18, 54 select the best answer from the choices provided a b c d