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3 years ago

AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?

Mathematics

1 answer:

alekssr [168]3 years ago

4 0

Given a slope $m$ , the slope $m'$ of a line perpendicular to $m$ is the opposite of the inverse of $m$

In other words:

$mm' = -1 \iff m' = -\dfrac{1}{m}$

So, the slope of CD is -5/2.

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One invests 100 shares of IBM stocks today. He expects that there could be five possible opening prices with the respective prob

GuDViN [60]

Answer:

$E(x) = 177.130$

$Var(x) = 484.551$

$\sigma = 22.013$

Step-by-step explanation:

Given

The attached table

Solving (a): The mean

This is calculated as:

$E(x) = \sum x * p(x)$

So, we have:

$E(x) = 182.11 * 13\% + 163.88 * 19\% + 180.30 * 33\% + 216.08 * 17\% + 144.92 * 18\%$

Using a calculator, we have:

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$E(x) = 177.130$ --- approximated

<em>The average opening price is $177.130</em>

Solving (b): The Variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x^2) = \sum x^2 * p(x)$

$E(x^2) = 182.11^2 * 13\% + 163.88^2 * 19\% + 180.30^2 * 33\% + 216.08^2 * 17\% + 144.92^2 * 18\%$

$E(x^2) = 31859.482249$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 31859.482249 - 177.1297^2$

$Var(x) = 31859.482249 - 31374.9306221$

$Var(x) = 484.5516269$

$Var(x) = 484.551$ --- approximated

Solving (c): standard deviation

The standard deviation is:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{484.5516269}$

$\sigma = 22.0125418796$

Approximate

$\sigma = 22.013$

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AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?​

AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?