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:
0.0869
Step-by-step explanation:
The arc electronic company had an income of 90 million dollars last year.
Mean(μ) = 75 million dollars
Standard deviation (σ) = 11 million dollars
Probability that the randomly selected will earn more than arc did last year = Pr(x>90)
Using normal distribution,
Z = (x - μ) / σ
Z = (90 - 75) / 11
Z = 15/11
Z = 1.36
From the normal distribution table, 1.36 = 0.4131
Φ(z) = 0.4131
Recall that when Z is positive, Pr(x>a) = 0.5 - Φ(z)
= 0.5 - 0.4131
= 0.0869
The answer is B.$7.47.
Let's first calculate how much in total 75 shirts cost:
- first 19 (less than 20) cost $10: 19 * $10 = $190
- the next 30 cost $8: 30 * $8 = $240
- the rest (26 = 75 - 19 - 30) cost $5: 26 * $5 = $130
So, in total 75 shirts cost $560:
19 * $10 + 30 * $8 + 21 * $5 = $190 + $240 + $130 = $560
Since 75 shirts are ordered, there must be 75 employees, so each employee must pay $7.47:
$565 ÷ 75 = $7.47
<h3>
Answer: 1/2</h3>
The midsegment is always exactly half as long compared to the side it's parallel to.
Put another way, the longer side (4x+20) is twice long as the midsegment (3x).
I don't fully understand the question but I hope this helps...
I believe this would mean he made $1405 that day.