What is the question if it 98-49 than it would be 49
Well, to solve, you would subtract 8 from both sides, which means x would end up equaling 32.
X=32
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: -4/15
Step-by-step explanation: To add these two fractions together,
we start by finding their <em>Least Common Denominator</em> (LCD).
Since our denominators of 5 and 3 have no factors in common,
our least common denominator is 5 · 3 or 15.
In order to get a denominator of 15 in each fraction,
we multiply top and bottom of the first fraction by 3
and top and bottom of the second fraction by 5.
That gives us -9/15 + 5/15 which simplifies to -4/15.
As your last step, make sure that the fraction
that you end up with doesn't reduce.
In this case, -4/15 does not reduce but watch out for this last step.