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
x Is less than 4
Step-by-step explanation:
It will be an open circle and the arrow is going the left direction on the number 4
The answer for this problem is b
234 = a + s s = students a = adults
2a = s
after you get the equations you solve. plug 2a into the above equation.
so u get
234 = a + 2a
234=3a then divide 3 from each side
78 = a
so 78 adult tickets were sold
Answer:

Step-by-step explanation:
We can directly substitute y of the first equation to the second equation.
-2x - 1 = 3x - 16
5x = 15
x = 3
Substituting back to any of the two equations, we get y = -2(3)-1 = -7. If you check with the second equation, y = 3(3)-16 = -7 as well.
Therefore
.