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:
With my predictions it is 2/11 hope it helps
Solve for m:3 m + 7/2 = 5/2 - 2 m
Put each term in 3 m + 7/2 over the common denominator 2: 3 m + 7/2 = (6 m)/2 + 7/2:(6 m)/2 + 7/2 = 5/2 - 2 m
(6 m)/2 + 7/2 = (6 m + 7)/2:(6 m + 7)/2 = 5/2 - 2 m
Put each term in 5/2 - 2 m over the common denominator 2: 5/2 - 2 m = 5/2 - (4 m)/2:(6 m + 7)/2 = 5/2 - (4 m)/2
5/2 - (4 m)/2 = (5 - 4 m)/2:(6 m + 7)/2 = (5 - 4 m)/2
Multiply both sides by 2:6 m + 7 = 5 - 4 m
Add 4 m to both sides:6 m + 4 m + 7 = (4 m - 4 m) + 5
4 m - 4 m = 0:6 m + 4 m + 7 = 5
6 m + 4 m = 10 m:10 m + 7 = 5
Subtract 7 from both sides:10 m + (7 - 7) = 5 - 7
7 - 7 = 0:10 m = 5 - 7
5 - 7 = -2:10 m = -2
Divide both sides of 10 m = -2 by 10:(10 m)/10 = (-2)/10
10/10 = 1:m = (-2)/10
The gcd of -2 and 10 is 2, so (-2)/10 = (2 (-1))/(2×5) = 2/2×(-1)/5 = (-1)/5:Answer: m = (-1)/5
Answer: The answer to this question is 15.53lt/sec.
Step-by-step explanation: 1 liter is equal to 0.264gallons
1 minutes is equal to 60 seconds.
To convert we have to multiply 246gal/min by 1min/60sec then by 1lt/0.264gal to get the answer in lt/s
Where the answer is 15.53lt/s
The question is asking to determine the dimension of a rectangular solid with a maximum volume if its surface area is 337.5 square centimeters and base on my further computation, I would say that the answer would be x= sqrt 56.25 and y = 7.5 and having a maximum volume of 421.875. I hope this would help
