Answer:
The standard deviation is used in conjunction with the mean to numerically describe distributions that are bell shaped. The mean measures the center of the distribution, while the standard deviation measures the spread of the distribution.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by
The mean is the average value of the measures while the standard deviation measures how spread the measures are from the mean. So
The standard deviation is used in conjunction with the mean to numerically describe distributions that are bell shaped. The mean measures the center of the distribution, while the standard deviation measures the spread of the distribution.
This is an incomplete problem
Problem 1
<h3>Answer: 7.3</h3>
Explanation: Apply the square root to the area to get the side length. This only applies to areas that are squares (hence the name).
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Problem 2
<h3>Answer: C) 1.3</h3>
Explanation: Use your calculator to find that choices A,B,D plugged into the square root function yield terminating decimal values. "Terminating" means "stop". This implies that they are perfect squares (though not perfect squares in the sense of whole number perfect squares which you may be used to). Choice C is the only value that has a square root that leads to a non-terminating decimal. The digits of this decimal go on forever without any pattern. The value is irrational.
- sqrt(5.29) = 2.3 terminating decimal
- sqrt(13.69) = 3.7 terminating decimal
- sqrt(1.3) = 1.140175425 keeps going forever without any pattern
- sqrt(0.09) = 0.3 terminating decimal
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Problem 3
<h3>Answer: 23.6 feet approximately</h3>
Explanation: Apply the square root to 15.5 to get roughly 3.937; this is the approximate side length of one square. Six of these tiles placed together will lead to a total length of roughly 6*3.937 = 23.622 which rounds to 23.6 feet. Like with problem 1, the square root being used like this only works for square areas.
|sin(x)−sin(y)|=|2sin(x−y2)cos(x+y2)|≤|2sin(x−y2)|
