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
Step-by-step explanation:
answer is B
use unitary method to solve it
The probability that all marbles are yellow is
(7/20) (7/20) (7/20) = 343/8000
We are given
x = 2 and 3
u = 2.5
s = 0.5
Getting the z score
z = (2 - 2.5) / 0.5 and (3 - 2.5)/ 0.5
z = -1 and 1
The percentage is
1 - 0.1557 = 81.5%
Answer:
40(1.25-t)
Step-by-step explanation:
There are 3 components to consider; time, speed and distance
Time and Speed are given.
The distance has to be calculated.
- Speed to work = 55 miles per hour
- Time to work = 1.25-T
- Speed to home = 40 miles per hour
- Time to home = 1.25-t
- Total Time = T + t = 1.25
<u>Distance for trip to home</u>
Speed = Distance/Time
40 = Total Distance/1.25-t
Total Distance = 40(1.25-t)
<em>Therefore, 40(1.25-t) is the correct answer.</em>
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