Answer:
The answer to your question is the second option 
Step-by-step explanation:
Expression
![[\frac{(x^{2}y^{3})^{-2}}{(x^{6}y^{3}z)^{2}}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%28x%5E%7B2%7Dy%5E%7B3%7D%29%5E%7B-2%7D%7D%7B%28x%5E%7B6%7Dy%5E%7B3%7Dz%29%5E%7B2%7D%7D%5D%5E%7B3%7D)
Process
1.- Divide the fraction in numerator and denominator
a) Numerator
[(x²y³)⁻²]³ = (x⁻⁴y⁻⁶)³ = x⁻¹²y⁻¹⁸
b) Denominator
[(x⁶y³z)²]²= (x¹²y⁶z²)³ = x³⁶y¹⁸z⁶
2.- Simplify like terms
a) x⁻¹²x⁻³⁶ = x⁻⁴⁸
b) y⁻¹⁸y⁻¹⁸= y⁻³⁶
c) z⁻⁶
3.- Write the fraction

B and D are both rational.
A rational number is a number that can be represented as a ratio fo two integers.
-7/3 is rational because both -7 and 3 are integers. 2.777... equals 25/9. 25 and 9 are both rational numbers.
We can start with:
x + y = N
where
x is the first 3 digit number
y is the second 3 digit number
In order to get the maximum value of N, the digits used must 9,8,7,6,5,4
To maximize, 9 and 8 must be added. This can only happen if 9 is the first digit of the first number and 8 is the first digit of the second number. The same idea is applied to the other digits. So
x must be 975 and
y must be 864
N = 975 + 864 = 1839
Answer:
The upper 20% of the weighs are weights of at least X, which is
, in which
is the standard deviation of all weights and
is the mean.
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.
Upper 20% of weights:
The upper 20% of the weighs are weighs of at least X, which is found when Z has a p-value of 0.8. So X when Z = 0.84. Then



The upper 20% of the weighs are weights of at least X, which is
, in which
is the standard deviation of all weights and
is the mean.