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
Function would be f(x) = 5x+85
here, x= 15
5(15)+85 = 75+85 = $160
Answer:
73/80
Step-by-step explanation:
multiples of 11 between 1 and 80 are 11,22,33,44,55,66,77
number of terms=7
P(multiple of 11)=7/80
P(not a multiple of 11)=1-7/80=73/80