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:
16
Step-by-step explanation:
First, simplify by finding the absolute values:
2|-9| - |-2|
2(9) - (2)
Then, simplify this by multiplying 2 and 9:
18 - 2
= 16
So, the answer is 16
That looks like a translation; let's check. We have
A(-5,1), B(-3,7), A'(3,-1), B'(5,5)
If it's a translation by T(x,y) we'd have
A' = A + T
B' = B + T
so
T = A' - A = (3,-1) - (-5,1) = (8,-2)
and also
T = B' - B = (5, 5) - (-3, 7) = (8,-2)
They're the same so we've verified this transformation is a translation by (8,-2), eight right, two down.
Option e: associative property of addition
Answer:
350 yards
Step-by-step explanation: