-2.54 is equal to -2 27/50
B)
2x6= 12
2x7= 14
2x8= 16
2x8= 18
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.
6(f+4) + 2(f+4)
Use the distributive property:
6f+24 + 2f + 8
Now combine like terms:
8f +32
The answer is 8f + 32
Answer:
If you mean by 1*2*3=6
Step-by-step explanation: