All categories

3 years ago

12) A bag contains 3 red marbles, 4 black

2 answers:

Rudiy273 years ago

5 0

3+ 4 + 5 = 12

So the chance of picking a marble that is not black is minus it

3+5 = 8 marbles are not black

Take 8 divided to the whole number which is 12

8/12= 66%

Hope this helps!

Cloud [144]3 years ago

3 0

$|\Omega|=3+4+5=12\\|A|=3+5=8\\\\P(A)=\dfrac{8}{12}=\dfrac{2}{3}\approx66.7\%$

You might be interested in

Steve has received $200 in the form of a dividend. What type of income is this.

blagie [28]

Answer:

passive income

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

4^2 -6 (4 to the second power minus 6) *

uysha [10]

Answer: 10

Step-by-step explanation: you have to simplify for it to equal 10

5 0

3 years ago

If the slope of a line is 2 and its y-intercept is 5, what is the equation of the line?

CaHeK987 [17]

Hello,

y=2x+5
==========
(y-5)=2(x-0)

7 0

3 years ago

Read 2 more answers

MARKING BRAINLIEST HELP

Alenkinab [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago