Membership in an elite organization requires a test score in the upper 30% range. if the mean is equal to 115 and the standard deviation is equal to 12, find the lowest acceptable score that would enable a candidate to apply for membership. assume the variable is normally distributed. (show work)

Lowest acceptable score = 121.3

Step-by-step explanation:

Mean test score (μ) = 115

Standard deviation (σ) = 12

The z-score for any given test score 'X' is defined as:  

In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation: