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2 years ago

Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

1 answer:

6 0

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$

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A) the man is 40 and the daughter is 8
b)in 8 years when the man is 48 and the daughter is 16

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3 years ago

The market and Stock J have the following probability distributions:

denis-greek [22]

Answer:

1) $E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

2) $E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

3) $E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

4) $E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

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3 years ago

What is an equation of the line that passes through the point (1,-3) and is perpendicular to the line x+3y=21

docker41 [41]

The linear equation that is perpendicular to the line x+3y=21 is:

y = 3*x - 6

<h3>How to find the equation of the line?</h3>

A general line in the slope-intercept form is written as:

y = m*x + b

Where m is the slope and b is the y-intercept.

Two linear equations are perpendicular if the product between the two slopes is equal to -1.

Rewriting the given line we can get:

x +3y = 21

3y = 21 - x

y = 21/3 - x/3

y = (-1/3)*x + 21/3

Then the slope is (-1/3), if our line is perpendicular to this one, then:

m*(-1/3) = -1

m = 3

our line is:

y = 3*x + b

To find the value of b, we use the fact that our line passes through (1, - 3)

-3 = 3*1 + b

-3 - 3 = b

-6 = b

The line is y = 3*x - 6

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