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3 years ago

Consider three events A, B and C with following properties.

1 answer:

5 0

By the inclusion/exclusion principle,

$\mathbb P(D)=\mathbb P(A\cup B\cup C)$
$\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)$
$\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}$
$\mathbb P(D)=\dfrac{16}{39}\neq1$

So the union of A, B, and C does not constitute the entire sample space.

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jonny [76]

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2 years ago

The ratio of the circumference of two circles is 3:2.What is the ratio of their area

Nezavi [6.7K]

The ratio of their area is 9/4

<h2>Explanation:</h2>

We use ratios to compares values. In this exercise, we are comparing the circumference of two circles, which is:

$3:2$

and we want to know what is the ratio of their area. Recall that the circumference of a circle is given by:

$C=2\pi r \\ \\ \\ Where: \\ \\ C:Circumference \\ \\ r:Radius \ of \ the \ circle$

If we define:

$C_{1}: \ Circumference \ of \ circle \ 1 \\ \\ C_{2}: \ Circumference \ of \ circle \ 2 \\ \\ r_{1}: \ Radius \ of \ circle \ 1 \\ \\ r_{2}: \ Radius \ of \ circle \ 2$

Then, the ratio of the circumference of two circles is 3:2 is:

$\frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{3}{2} \\ \\ \therefore \frac{r_{1}}{r_{2}}=\frac{3}{2}$

The area of a circle is given by:

$A=\pi r^2 \\ \\ A:Area \\ \\ r:Radius$

So the ratio of their area can be found as:

$\frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^2}{\pi r_{2}^2} \\ \\ \\ A_{1}:Area \ of \ circle \ 1 \\ \\ A_{2}:Area \ of \ circle \ 2$

So:

$\frac{A_{1}}{A_{2}}=\frac{r_{1}^2}{r_{2}^2}=\left( \frac{r_{1}}{r_{2}} \right)^2 \\ \\ \frac{A_{1}}{A_{2}}=\left(\frac{3}{2}\right)^2 \\ \\ \boxed{\frac{A_{1}}{A_{2}}=\frac{9}{4}}$

<h2>Learn more:</h2>

Unit rate: brainly.com/question/13771948#

#LearnWithBrainly

5 0

3 years ago

How do you do this? Thanks for your time and effort!

joja [24]

Since log(0) is undefined, we can assume that there is a vertical asymptote there. So, to make the function log(0), x needs to be -3. There is a vertical asymptote at x=-3. Since the function has not been shifted anywhere, or flipped vertically/horizontally, the answer would most likely be A.

3 0

3 years ago

Read 2 more answers

See attached file. show all work.

andrew11 [14]

The point-slope form of the equation for a line can be written as

... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)

Your function gives

... f'(h) = m

... f(h) = k

a) The tangent line is then

... y = 5(x -2) +3

b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.

... y = (-1/5)(x -2) +3

_____

You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.

In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.

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3 years ago

Please help if you can ty!!

Slav-nsk [51]

To travel 13 mph it would take roughly 8.2 minutes !

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3 years ago