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 54 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 54, standard deviation = 3.

What is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?

63 = 54 + 3*3

So between the mean and 3 standard deviations above the mean.

The normal distribution is symmetric, which means that 50% of the values are below the mean and 50% are above.

Of those 50% above, 99.7% are below 63. So

0.5*0.997 = 0.4985

0.4985*100% = 49.85%

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

10 you would take 50/ .2

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~