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3 years ago

Find two numbers whose difference is 154 and whose product is a minimum.

1 answer:

7 0

Y - x = 154
let z = xy = x(154 + x) = x^2 + 154x

we need to find the minimum value of z

z' = 2x + 154 which = zero for minm or maxm value

x = -77 and y = 154 -77 = 77

so the 2 numbers are -77 and 77

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Disregard my earlier question.

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$\mathbb E((X-\mu)^2)$

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Now, recall that

$\mathbb V(X)=\mathbb E((X-\mu)^2)=\mathbb E(X^2-2\mu X+\mu^2)$
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$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

(a) $Z=2+10X$

$\mathbb V(Z)=\mathbb E((2+10X)^2)-\mathbb E(2+10X)^2$
$\mathbb V(Z)=\mathbb E(4+40X+100X^2)-(\mathbb E(2)+\mathbb E(10X))^2$
$\mathbb V(Z)=\mathbb E(4)+40\mathbb E(X)+100\mathbb E(X^2)-(\mathbb E(2)+10\mathbb E(X))^2$
$\mathbb V(Z)=100\mathbb E(X^2)-40000$
$\mathbb V(Z)=100\left(\mathbb E(X^2)-\mathbb E(X)^2+\mathbb E(X)^2\right)-40000$
$\mathbb V(Z)=100\left(\mathbb V(X)+\mathbb E(X)^2\right)-40000$
$\mathbb V(Z)=100\left(5^2+400\right)-40000$
$\mathbb V(Z)=2500$

The standard deviation is the square root of the variance, so for (a) you have

$\sqrt{\mathbb V(Z)}=\sqrt{2500}=50$

That should give you an idea as to how to figure out the rest.

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3 years ago

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Did that based on the info you gave sorry if its wrong

Step-by-step explanation:

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Considering the concepts of the mean and the median of a distribution, the distribution of dates would be skewed to the left because the <u>mean would be less than the median</u>.

<h3>What are the mean and the median of a distribution?</h3>

The mean of a data-set is given by the <u>sum of all observations divided by the number of observations</u>.

The median of a data-set is the middle value of the data-set, the value which 50% of the measures are less and 50% are greater.

A distribution is classified as left skewed if the <u>mean is less than the median</u>.

In this problem, we have that most coins would be from the current year, hence the median should be the current year, while there are no coins from a year ahead of the current year and there should be a few coins of older years, hence the mean would assume a value less than the median and the distribution would be left skewed.

More can be learned about the mean and the median of a distribution at brainly.com/question/24732674

#SPJ1

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Hope this helps! :)

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