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%.
1/10 of a pound or 1.6 ounces
Answer:
im not sure
Step-by-step explanation:
Answer:
-21X + 9
Step-by-step explanation:
Answer:
X=the smaller
y=the larger
The bigger of two numbers is four more than the smaller, then:
y=x+4
one more than twice the smaller is the same as the larger, then
1+2x=y
We have the following system of equations:
y=x+4
y=2x+1
We solve this system by equal values method:
x+4=2x+1
x-2x=1-4
-x=-3
x=3
We find the value of "y"now:
y=x+4
y=3+4
y=7
the values of x and y are:
x=3
y=7
Step-by-step explanation: