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:
In parallelogram ABCD
FD is perpendicular to BC
BE is perpendicular to CD
Consider triangle BEC and triangle DFC
FC = EC (Given)
Angle BEC = Angle DFC (=90°)
Angle BCE = Angle DCF (common)
Therefore triangle BEC is congruent to triangle DFC (AAS congruency)
DF = BE (CPCT)
Since the altitudes are equal their bases will also be equal
Therefore BC = DC
Therefore BC = DC = AD = AB
Therefore ABCD is a rhombus
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D and B
Explanation:
when you look at that point two lines pass through it and those two lines that pass through it are b and d.
Answer:
4 hours.
Step-by-step explanation:
The difference between 6 and 2 is 4.
There were...
11cards left.
heres why.
they started with 24 cards....
so...
24+8=32
32-12=20
20-9=11
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