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
<span><span><span>2r - 9 > -6
</span><span>2r - 9 = -6
</span>2r = 3</span><span>
r = 3/2 = 1.5</span></span><span><span>
r > 1.5</span></span>
<span><span /></span><span><span>9x-5 < -41
</span><span>9x-5 = -41
9x = -36
x = -36/9 = -4
x < -4</span></span>
<span><span>3x + 13 > 7
3x + 13 = 7
3x = -6
x = -6/3 = -2
x > -2</span></span>
<span><span>4x + 3 > -17
4x + 3 = -17
4x = -20
x = -20/4 = -5
x > -5</span></span>
<span><span>7x - 4 < 10
7x - 4 = 10
7x = 14
x = 14/7 = 2
x < 2</span></span><span>
</span>
The answer of this question is 284
Answer:
9 ÷ 1504 = 0 R 9
0.00598404255
Step-by-step explanation:
Shown above is remainder form and decimal form...
Answer:
you cannot see the graph. So I cannot answer this question
Step-by-step explanation:
if you add the graph maybe I can help :D
