15. Survey Return Rate In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.

0.1352 ± 0.1516

Step-by-step explanation:

The best estimate of the population is calculated as:

P = x/n

P = 717 / 500

P = 0.1434

The best estimate of the population is 0.1434

The value of margin of error is calculated as:

Z-critical value = 90%, Confidence = 0.1434

E = Z(c) *

E = 1.645 *

E = 0.0082

A 90% confidence interval for the population proportion of returned survey is:

90%CI = P ± E

90%CI = 0.1434 ± 0.0082

90%CI = 0.1352 ± 0.1516

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: