[(0.24*5)+ (0.66*5)] == 4.50
0.24x5 = 1.20
0.66x5 = 3.30
+ ____
4.50
27.00÷4.5 = 0.166666667
The highest common factor is found by multiplying all the factors which appear in both lists
a.(13-4b)(a+6)
13a-24b
13 and 24
so we have to factor this
13 =13×1
24=12×2=12 is 6×2=6 is 3×2=3 is 3×1=3×1
Therefore 1 and 1 get canceled
13×3=39
(a )39
like this we have to solve........
(m - 3)(m + 5)= 0
m= {3, -5}
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:
![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.
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
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.84 = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=0.84%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![X = 0.84\sigma + \mu](https://tex.z-dn.net/?f=X%20%3D%200.84%5Csigma%20%2B%20%5Cmu)
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.
There are n! ways to arrange n different elements
This means there are 3! or 6 different arrangements of any 3 elements.
In this case namely,
abc
acb
bac
bca
cab
cba