Answer with explanation:
We are given that 
Sample size : 
The mean of the sampling distribution of the sample proportions is given by :-

The mean of the sampling distribution of the sample proportions is <u>0.91</u>
The standard deviation of the sampling distribution of the sample proportions :

Hence, the standard deviation of the sampling distribution of the sample proportions is <u>0.0238</u>
Answer:
Confidence interval is 128±2.326 days (between 125.674 days and 130.326 days) in 90% confidence level
Step-by-step explanation:
Confidence interval can be calculated as M±ME where
- M is the average pregnancy days of smoking women and
- ME is the margin of error in 90% confidence level.
Margin of Error in 90% confidence level can be calculated using the formula:
ME=
where
- z is the z-value in 90% confidence level (1.645)
- s is the standard deviation of pregnancy length (16)
- N is the sample size (128)
putting these numbers in the formula, we get:
ME=
=2.326
then confidence interval is 128±2.326 ( between 125.674 and 130.326 ) in 90% confidence level
<h2>The answer is D. 54d + 2</h2><h3>Hope it helps!! </h3>
Interval notation:
−
∞
,
−
2
)
∪
(
−
2
,
2
)
∪
(
2
,
∞
)
Hope this help I haven’t done this in awhile
Answer:
2(3 - 2d)
Step-by-step explanation:
All you need to do is factor out 2 from the original expression, so 6 - 4d.