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
Y coordinate on solving both equations comes out to be 0
-2x+3y=-6
3y = -6+2x
put the value of 3y in equation 2nd
5x-2(-6+2x) =15
5x+12-4x = 15
x=3
put value of X in 3y = -6+2x
3y = -6+2*3 = -6+6 = 0
thus y = 0
Answer:
-20° F
Step-by-step explanation:
the temperature would be :
=》-14 - 6 = -20° F
Answer:
Can Ou please explain.
Step-by-step explanation:
Answer:
d. 90 cm, 1.2 cm
Step-by-step explanation:
Given that:
The sample size n = 36
For the length of the link:
The mean 
= 2.5
The standard deviation s = 0.2 cm
In a chosen 36 links, The mean and standard deviation for the length chain is as follows:
Mean = n
Mean = 36×2.5
Mean = 90 cm
The Standard deviation = s×√n
The Standard deviation = 0.2×√36
The Standard deviation = 0.2×6
The Standard deviation = 1.2 cm
