Answer:
2.9
Step-by-step explanation:
3.6 - 0.7 = 2.9
Using the relation between standard deviation and variance it is concluded that the standard deviation for the population for the given variance is 7.1
<h3>What is the relation between standard deviation and sample variance? </h3>
In statistics, the two most crucial metrics are variance and standard deviation. While the variance is a measurement of how data points vary from the mean, standard deviation is a measure of the distribution of statistical data.
The square root of the variance yields the standard deviation, i.e.
Standard deviation = 
Given that sample, the variance is 49.7 and we have to calculate the standard deviation.
Standard deviation (σ) =
= 
= 7.0498
=7.1
Hence, the standard deviation for the population for the given variance is 7.1
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Answer:
Step-by-step explanation:
<u>Using slope formula to find the value of k:</u>
- (-9 - k)/(-2 - (-4)) = - 9
- (-9 - k)/2 = - 9
- - 9 - k = - 18
- k = 18 - 9
- k = 9
Answer:
C
Step-by-step explanation:
To get rid of the absolute value bars you have to get two different equations which would be 2x+4>-8 AND 2x+4>8. Pay attention to the signs, if there is greater than or equal to then it would have a line underneath the sign, which neither of the equations should have since you haven't moved a negative. solve normally and you'll get -6 blank circle and 2 blank circle.
Answer:
90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]
Step-by-step explanation:
The data given is
Mean Std Dev
Before 177.25 29.325
After 169.50 22.431
Difference 7.75 8.598
Hence d`= 7.75 and sd= 8.598
The 90% confidence interval for the difference in means for the paired observation is given by
d` ± t∝/2(n-1) *sd/√n
Here t∝/2(n-1)=1.895 where n-1= 8-1= 7 d.f
and ∝/2= 0.1/2=0.05
Putting the values
d` ± t∝/2(n-1) *sd/√n
7.75 ±1.895 * 8.598 /√8
7.75 ± 5.7605
1.989 ,13.5105
90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]
