Mathematics, 25.12.2019 07:31 christopherarreola56
In 2013, the pew research foundation reported that "45% of u. s. adultsreport that they live with one or more chronic conditions", and the standard error for this estimate is 1.2%.identify each of the following statements as true or false. provide an explanation to justify each of youranswers.(a) we can say with certainty that the confidence interval from exercise 5.7 contains the true percentageof u. s. adults who suffer from a chronic illness.(b) if we repeated this study 1,000 times and constructed a 95% confidence interval for each study, thenapproximately 950 of those confidence intervals would contain the true fraction of u. s. adults who sufferfrom chronic illnesses.(c) the poll provides statistically significant evidence (at theα= 0.05 level) that the percentage of u. s.adults who suffer from chronic illnesses is below 50%.(d) since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty abouttheir answer.
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Mathematics, 21.06.2019 15:30
Will mark brainliest if you answer ! the taco shop sold 198 lunches friday. this number is 3 more than 3 times the number they sold monday. let n represent the number of lunches sold monday. which equation shows an equality between two different ways of expressing the number of lunches sold on friday? a. n – 3 = 198 b. 3n – 3 = 198 c. 3n + 3 = 198 d. n + 3 = 198
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In 2013, the pew research foundation reported that "45% of u. s. adultsreport that they live with on...
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