The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 59 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 59 and 62

About 34%

Step-by-step explanation:

Given that :

Mean, m = 59 ; Standard deviation, s = 3

Bulb replacement numbering between 59 and 62

For, x = 59

Zscore = (x - m) / s

Z = (59 - 59) / 3 = 0

For x = 62

Z = (62 - 59) / 3 = 1

P(Z < 1) - P(Z < 0)

0.84134 - 0.5

= 0.34134

About 34%

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation:

answer: c

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but i think the formula is f(x)=6x