Answer: 49.85%
Step-by-step explanation:
Given : The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped ( normal distribution ) and has a mean of 61 and a standard deviation of 9.
i.e.
and 
To find : The approximate percentage of lightbulb replacement requests numbering between 34 and 61.
i.e. The approximate percentage of lightbulb replacement requests numbering between 34 and
.
i.e. i.e. The approximate percentage of lightbulb replacement requests numbering between
and
. (1)
According to the 68-95-99.7 rule, about 99.7% of the population lies within 3 standard deviations from the mean.
i.e. about 49.85% of the population lies below 3 standard deviations from mean and 49.85% of the population lies above 3 standard deviations from mean.
i.e.,The approximate percentage of lightbulb replacement requests numbering between
and
= 49.85%
⇒ The approximate percentage of lightbulb replacement requests numbering between 34 and 61.= 49.85%
Answer:
5 classes.
Step-by-step explanation:
You can use the
rule to determine the number of classes for a frequency distribution.
The
rule says that
where
is the number of classes
is the number of the data points
We know that the number of data points is
= 20.
Next, we start searching for
so that we can get a number 2 to the
that is larger that the number of data points.

This suggests that you should use 5 classes.
1. is B
2. is A
3. is C
4. is B
5. ? what graphs
Answer:
$15
Step-by-step explanation:
First you move the decimal in the percent 2 times to the left and get .25. And then you multiply that by 60 and get 15.
The area would be 25π inches.
Since the cross-section goes through the center of the sphere it will be a circle. The area of a circle is given by A = πr²; for our circle, since the diameter is 10 inches, the radius is 5;
A = π(5²) = 25π