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2 years ago

For x, y ∈ R, prove that ||x| − |y|| ≤ |x − y|, (Hint: consider x = y + (x − y)).

Mathematics

1 answer:

dlinn [17]2 years ago

7 0

Using the triangular inequality (and the given hint!), we have

$|x|=|(x-y)+y|\leq|x-y|+|y|\implies|x|-|y|\leq|x-y|$

Similarly,

$|y|=|(y-x)+x|\leq|x-y|+|x|\implies|y|-|x|\leq|x-y|\implies -|x-y| \leq |x|-|y|$

We managed to bound the quantities in this fashion:

$-b\leq a\leq b \implies |a|$

And thus we have

$||x|-|y||\leq |x-y|$

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2 years ago

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced

Dahasolnce [82]

Answer:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.86,0.13)$

Where $\mu=5.86$ and $\sigma=0.13$

We are interested on this probability

$P(5.5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

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2 years ago

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max2010maxim [7]

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2 years ago

For x, y ∈ R, prove that ||x| − |y|| ≤ |x − y|, (Hint: consider x = y + (x − y)).​

For x, y ∈ R, prove that ||x| − |y|| ≤ |x − y|, (Hint: consider x = y + (x − y)).