Answer:
Mean track length for this rock specimen is between 10.463 and 13.537
Step-by-step explanation:
99% confidence interval for the mean track length for rock specimen can be calculated using the formula:
M±
where
- M is the average track length (12 μm) in the report
- t is the two tailed t-score in 99% confidence interval (2.977)
- s is the standard deviation of track lengths in the report (2 μm)
- N is the total number of tracks (15)
putting these numbers in the formula, we get confidence interval in 99% confidence as:
12±
=12±1.537
Therefore, mean track length for this rock specimen is between 10.463 and 13.537
Answer:
11 13/15
Step-by-step explanation:
-3 2/3+b = 8 1/5
Add 3 2/3 to each side
-3 2/3 + 3 2/3+b = 8 1/5+ 3 2/3
b = 8 1/5 + 3 2/3
We need to get a common denominator of 15
8 1/5 = 8 3/15
3 2/3 = 3 10/15
----------------
11 13/15
Answer:
x = 9 m
y = 1 m
Step-by-step explanation:
6/1 = x/1.5
x = 1.5 × 6
x = 9
6/1 = 6/y
6y = 6
y = 1
I believe the answer is 3 3/4.
No because there has to be a remaining opposite persent