A number that is a factor of 16 and is prime would be 2.
2 x 8 = 16 (So it's a factor of 16)
Hope this helps! :D
~PutarPotato
9514 1404 393
Answer:
x - 4
Step-by-step explanation:
The expression simplifies to ...
![\sqrt{x^2-8x+16}=\sqrt{(x-4)^2}=|x-4|](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-8x%2B16%7D%3D%5Csqrt%7B%28x-4%29%5E2%7D%3D%7Cx-4%7C)
For x ≥ 4, the argument of the absolute value function is non-negative, so it remains unchanged. The simplified expression is ...
x - 4 . . . . for x ≥ 4
Answer:
y
=
−
2
x
+
9
Step-by-step explanation:
Write in slope-intercept form, y
=
m
x
+
b
.
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 ![\mu = 100, \sigma = 15](https://tex.z-dn.net/?f=%5Cmu%20%3D%20100%2C%20%5Csigma%20%3D%2015)
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.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![2.054 = \frac{X - 100}{15}](https://tex.z-dn.net/?f=2.054%20%3D%20%5Cfrac%7BX%20-%20100%7D%7B15%7D)
![X - 100 = 15*2.054](https://tex.z-dn.net/?f=X%20-%20100%20%3D%2015%2A2.054)
![X = 130.81](https://tex.z-dn.net/?f=X%20%3D%20130.81)
IQ scores of at least 130.81 are identified with the upper 2%.
The answer is either c of d but if I had to pick one I would say C.