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:
2 hundreths
Step-by-step explanation:
Answer:
84%
Step-by-step explanation:
What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.
We have that z is equal to:
z = (x - m) / (sd)
x is the value to evaluate, m the mean, sd the standard deviation
So for 190000 we have:
z = (190000 - 200000) / (10000)
z = -1
and this value represents 0.1587
for 230000 we have:
z = (230000 - 200000) / (10000)
z = 3
and this value represents 0.9987
we subtract:
0.9987 - 0.1587 = 0.84
Which means that it represents 84% of the houses
Answer:
8y = -6
Step-by-step explanation:
x + 6y -x +2y = 9 +(-15) ➡ 8y = -6
Answer:
Step-by-step explanation:
Assuming we want to find the equation (74,2) and (4,9)
This is an equation of a straight.
Let this equation be y=mx+b.
We substitute the slope and (4,9) to get:
We rewrite in standard form to get:
