A random sample of 700 home owners in a particular city found 112 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. Express your results to the nearest hundredth of a percent.
1 answer:
Answer:
Step-by-step explanation:
Given : A random sample of 700 home owners in a particular city found 112 home owners who had a swimming pool in their backyard.
i.e. n= 700 and x= 112
Sample proportion :
z-value for 95% confidence interval :
Now, the 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard will be :-
Hence, 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard :
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Answer:
x=24/7
Step-by-step explanation:
3(x+2)+4(x-5)=10
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7x=10+14
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Answer:
Were is the graph
Step-by-step explanation:
show me the graph
x = 0.17(repeating 7) (1st)
10x = 1.77 (2nd)
subtract (1st) from (2n)
9x = 1.77 - 0.17
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x = 16/90
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answer
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