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:
h=60/7
w=45/7
Step-by-step explanation:
just try it.
Answer: c=19 and n=4
please mark as brainliest if it helped
3x+6=3(x+2)
3x+6=3x+6
minus 6 both sides
3x=3x
divide by 3
x=x
true
therefor all numbers work for x
there are infinite soltuions
Answer:
=7/6
Step-by-step explanation:
Join 4/3=1:7/3
7/3/2
Apply the fraction rule:=
7/3x2
Multiply the numbers=
7/6