Answer:
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Answer:I believe it is at least that is what I would
Step-by-step explanation:
We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.
First of all, we will find z-score corresponding to 38 and 56.


Now we will find z-score corresponding to 56.

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is
.
We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.
We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
Answer:
c
Step-by-step explanation:
1 + tan²theta = sec²theta
tan²theta = 3² - 1
tan²theta = 8
tan theta = sqrt(8)
Positive because Quadrant 1
sqrt(8) = sqrt(4×2) = sqrt(4)×sqrt(2)
= 2×sqrt(2)