A. 60(2/3) + 35(7/10) + 45(5/6) = $102.00
b. 60 + 35 + 45 = 140 = original price
140 - 102 = $30 saved
c. 102/140 = .7286 * 100 = 72.86 % (not sure how you need it rounded)
Answer: see proof below
<u>Step-by-step explanation:</u>

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.
In order to get the volume of a rectangular box, the formula would be V = LxWxH.
So given that Length L = 2w
H = 3-w
W = x
Let x be the width, so it will look like this.
V = (2x)(x-3)(x)
Multiply these and we will get 2x^3-6x^2.
Therefore, the answer would be option D. Hope this answer helps. Let me know if you need more help next time.
All you need to know is that domain is x and range is y, like a graph.