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.
i think the answer to this is 3.15
10/12, 15/18, 20/24, 25/30, 30/36, 35/42, 40/48, 45/54, 50/60, and so on ...<span>
Source: </span><span>Equivalent fractions for 5/6</span>
Answer:
$722444.49386776
Step-by-step explanation:
Use Compound Intrest Formula

- where p is the original amount.
- R is the amount of percentage compounded
- N is amount of times compounded per year.
- T is how long the interest last.
P is 400,00p
T is 12% or 0.12
N is 4 since it is compounded quarterly
T is 5.
Plug the values in

Ypu get
$722444.49386776