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:
it's probably b
Step-by-step explanation:
I don't know how to do it so don't do b
The answer is 21
By comparing the 4 vertices you can determine the rectangle x-length and y-length. The coordinate for the vertice is:
1,+4i
-2, +4i
-2, -3i
1, -3i
From the coordinate, you can see there are 2 x coordinate(1 and -2) and 2 y coordinate(4i and -3i) that used.
The x-length would be: 1 - (-2)= 3
The y-length would be: 4i- (-3i)= 7i
Then the area of the rectangle would be:
area= x-length * y-length= 3*7=21
Answer:
35x+72y
Step-by-step explanation:

Brainliest if it helped you!