What is 45% of 748????
1 answer:
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Answer:
The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)
Answer:
see explanation
Step-by-step explanation:
Using the identity
cos2Θ = 1 - 2sin²Θ, then
1 - 2sin²( -
)
= cos [2( -
)]
= sos( - Θ )
= coscosΘ + sin
= 0 × cosΘ + 1 × sinΘ
= 0 + sinΘ
= sinΘ = right side