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2 years ago

Suppose that events a and b are mutually exclusive with p(a) = ½ and p(b) = ⅓.

1 answer:

4 0

No, a & b are not independent if they're mutually exclusive.

Mutually exclusive means the two events cannot occur simultaneously. So if "a" happens "b" cannot. In contrast, two events are independent if the occurrence of one has no effect on the occurrence of the other.

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The function f(x) = x2 − 12x + 5 written in vertex form is f(x) = (x − 6)2 − 31. What are the coordinates of the vertex?

Artist 52 [7]

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
f(x)=1(x-\stackrel{h}{6})^2\stackrel{k}{-31}\qquad \qquad vertex~(6,-31)$

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2 years ago

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The length of a rectangle is 4 cm more than the width and the perimeter is at least 34 cm

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It is 28 i think been 3 year since i did this

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2 years ago

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How can i differentiate this equation?

Dmitry_Shevchenko [17]

$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}$

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2 years ago

What is the inverse of x-2

SCORPION-xisa [38]

Answer:

y = x + 2

Step-by-step explanation:

The expression x-2 is from the equation y = x - 2.

To find the inverse of the equation, <u>switch the "x" and "y" variables</u>, then solve or <u>isolate for "y"</u>, meaning to separate it from the other numbers.

y = x - 2 Original equation

x = y - 2 Switched "x" and "y" variables

x + 2 = y - 2 + 2 Add 2 to both sides to cancel out '-2' on the side with "y"

x + 2 = y

y = x + 2 Variable is on left side in standard formatting

To isolate, do the reverse operation for the number you want to get rid of.

On the side with "y", I wanted to get rid of '-2' so I could separate "y" from everything else. '+2' is the reverse operation for '-2'.

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2 years ago

A box of highlighters originally cost

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