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:
8 * (7 + 4)
See process below
Step-by-step explanation:
We start by writing each number in PRIME factor form:
56 = 2 * 2 * 2 * 7
32 = 2 * 2 * 2 * 2 * 2
Notice that the factors that are common to BOTH numbers are 2 * 2 * 2 (the product of three factors of 2).Therefore we see that the greatest common factor for the given numbers is : 2 * 2 * 2 = 8
Using this, we can write the two numbers as the product of this common factor (8) times the factors that are left on each:
56 = 8 * 7
32 = 8 * 2 * 2 = 8 * 4
We can then use distributive property to "extract" that common factor (8) from the given addition as shown below:
56 + 32
8 * 7 + 8 * 4
8 * (7 + 4)
8 * (11)
88
The denominator is the one on the bottom so 9 is the denominator the top is the numerator