Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that 
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.




IQ scores of at least 130.81 are identified with the upper 2%.
Answer:
2x+20+3x+10 = 180 (int. angles, parallel)
5x+30 = 180
5x=150
x=30
Answer:
Step-by-step explanation:
x - represents 1 point shots
(54 - x) - represents 2 point shots
x + 2(54 - x) = 89
x + 108 - 2x = 89
-x + 108 = 89
x = 19 (1 point shots)
54-x = 35 ( 2 point shots)
Test:
19 + 35 = 54 (total shots)
19 x 1 = 19
35 x 2 = 70
70 + 19 = 89 points