A quadrilateral has angles that measure 44°, 89°, and 127°. What is the measure of the fourth angle?
2 answers:
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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.
A function 
is periodic if there is some constant 
such that 
for all 
in the domain of . Then is the "period" of .
Example:
If
, then we have 
, and so 
is periodic with period 
.
It gets a bit more complicated for a function like yours. We're looking for such that
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
, or more generally, when 
and 
(i.e. any multiple of 4).
The second two are satisfied whenever
, and more generally when 
with (any multiple of 10/7).
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.
Let's verify:
More generally, it can be shown that
is periodic with period
.
Example:
If
It gets a bit more complicated for a function like yours. We're looking for
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
The second two are satisfied whenever
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when
Let's verify:
More generally, it can be shown that
is periodic with period