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Answer with explanation:</h2>
Formula to find the confidence interval for population proportion (p) is given by :-

, where z* = Critical value.
= Sample proportion.
SE= Standard error.
Let p be the true population proportionof U.S. adults who live with one or more chronic conditions.
As per given , we have

SE=0.012
By z-table , the critical value for 95% confidence interval : z* = 1.96
Now , a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.:



Hence, a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.
Interpretation : Pew Research Foundation can be 95% confident that the true population proportion (p) of U.S. adults who live with one or more chronic conditions lies between 0.42648 and 0.47352 .
Answer:
Undifined
Step-by-step explanation:
Answer:
65% profit
-25% discount
Step-by-step explanation:
cost price = 40+25 = 65% higher than marked price
discount = -25%
Answer:
3(2c+d)−d
Distribute:
=(3)(2c)+(3)(d)+−d
=6c+3d+−d
Combine Like Terms:
=6c+3d+−d
=(6c)+(3d+−d)
=6c+2d
=6c+2d
Step-by-step explanation: