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%
The square roots of 64 is A B and C.
Mean - add up all of the scores, and divide it by the number of members:
68 + 62 + 60 + 64 + 70 + 66 + 72 = 462
462 / 7 = 66
ANSWER: The mean is 66
Median - write out all the numbers in order, and select the middle value:
60, 62, 64, 66, 68, 70, 72
ANSWER: As you can see, 66 is the middle value.
Midrange - find the mean (average) of the smallest and largest number:
Largest number: 72
Smallest number: 60
Midrange: 72 + 60 = 132
132 / 2 = 66
ANSWER: So the midrange is 66
Answer:
? = 65°
Step-by-step explanation:
180 - 115 = 65° (115° and 65° are supplementary angles)
180 - 65 - 50 = 65° (all triangle interior angles total 180°)
Set f(x)=0
-4x+4=0
solve for x
-4x=-4
x=1