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
Line 1:
Expanding the vertex form, we have
x² + 2·1.5x + 1.5² - 0.25 = x² +3x +2
Expanding the factored form, we have
x² +(1+2)x +1·2 = x² +3x +2
Comparing these to x² +3x +2, we find ...
• the three expressions are equivalent on Line 1
Line 2:
Expanding the vertex form, we have
x² +2·2.5x +2.5² +6.25 = x² +5x +12.5
Expanding the factored form, we have
x² +(2+3)x +2·3 = x² +5x +6
Comparing these to x² +5x +6, we find ...
• the three expressions are NOT equivalent on Line 2
The appropriate choice is
Line 1 only
Answer:
He should expect to select 75 yellow cards.
Step-by-step explanation:
Since there is replacement, for each trial, we have:
1 yellow card
1 red card
1 blue card
1 green card.
If Greg does this 300 times, how many yellow cards should he expect to select?
Each trial, there are 4 cards, one of which is yellow. So 1/4 = 0.25 probability of selecting a yellow card.
Over 300 trials
0.25*300 = 75
He should expect to select 75 yellow cards.
Answer:
D
Step-by-step explanation:
