Answer:
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
p+/-M.E
Given that;
M.E = margin of error
Proportion p = 52% = 0.52
Number of samples n = 700
Confidence interval = 95%
z value (at 95% confidence) = 1.96
Substituting the values we have;
0.52 +/- 1.96√(0.52(1-0.52)/700)
0.52 +/- 1.96(0.0189)
0.52 +/- 0.0370
( 0.4830, 0.5570)
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
Answer:
The answer is "(27,3]"
Step-by-step explanation:
Please find the complete question in the attached file.
![\to f(-5) = (-5)^2 + 2 = 25 +2 =27\\\\\to f(7)= 7-4 = 3\\\\answer \ is = ( 27, 3]\](https://tex.z-dn.net/?f=%5Cto%20f%28-5%29%20%3D%20%28-5%29%5E2%20%2B%202%20%3D%2025%20%2B2%20%3D27%5C%5C%5C%5C%5Cto%20f%287%29%3D%207-4%20%3D%203%5C%5C%5C%5Canswer%20%5C%20is%20%3D%20%28%2027%2C%203%5D%5C)
The last one should be it
Answer:
x = -18
Step-by-step explanation: