Answer:
The 95% confidence interval for mean compressive strength is:
Step-by-step explanation:
We have a sample of n=12 specimens, with sample mean=3259.9 and sample standard deviation = 35.57.
We know the population is normally distributed.
For a 95% confidence interval and df=n-1=11, the t-statistic is t=2.201.
Then, the CI can be expressed as:
Answer:
<em>DF = 10 units</em>
<em>EG = 5.04</em>
Step-by-step explanation:
<u>Properties of Rhombus es</u>
The image contains a rhombus with the following data (assume the center as point O):
DO = 5 units
GF = 5.6 units
4. Calculate DF
Applying property 4, diagonals bisect each other, thus the length of DF is double the length of DO, i.e. DF=2*5 = 10:
DF = 10 units
5. Calculate
Applying property 4 in triangle EFO, the center angle is 90°, thus angle EFO has a measure of 90°-62°=28°.
Applying property 5, this angle is half of the measure of angle EFG and angle DFG has the same measure of 28°.
6. FG is the hypotenuse of triangle OFG, thus:
EG is double OG: OG=2*2.52=5.04
EG = 5.04