Ask question

Ask question

All categories

3 years ago

15

A Student select a marble from a bag, keeps it and select another. The bag contains 5 Green marbles 4 black marbles and 2 blue m

arbles. Find the probability of selecting a green marble on the first trial and a black marble on the second trial.

1 answer:

Serga [27]3 years ago

7 0

Answer:

Yes because yes.

Step-by-step explanation:

Y + e + s = Yes

You might be interested in

the tiger is can grow up to 10 4/5 feet long. express this amount as a mixed number in simplest form. please explain how you did

andreev551 [17]

I believe that the number itself is already in it's simplest form, 10 4/5. This is because you can not divide 10 4/5 by any number for it to be more reduced. Otherwise, the only way I could think of it the number written as a improper fraction which can be written as this: 54/5.

Also, may I be marked brainliest please? I need them to rank up :( Thank you for understanding

4 0

3 years ago

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

ankoles [38]

Answer:

Jebb-44/82

11/2 x 41/4=44/82

6 0

3 years ago

PLEASE HELP!!!!!!!!!! ON TIME LIMIT!!!!!!!!!!

kvv77 [185]

<span> 18 + 6n ≥ 42, so n ≥ 4 is the answer I believe.
I hope Im not too late.

I hope this helps :)</span>

7 0

4 years ago

Read 2 more answers

Find the value of X in the proportion check your answer 28/x=7/8

kondaur [170]

X=32 that’s the answer for this equation

6 0

3 years ago

Read 2 more answers

HELP ASAP 20 POINTS CAUSE IM DESPERATE ToT

Elenna [48]

1. By de Moivre's theorem,

$\left(4\left(\cos\left(\dfrac{7\pi}9\right) + i \sin\left(\dfrac{7\pi}9\right)\right)\right)^3 = 4^3 \left(\cos\left(\dfrac{21\pi}9\right) + i \sin\left(\dfrac{21\pi}9\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac{7\pi}3\right) + i \sin\left(\dfrac{7\pi}3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac\pi3\right) + i \sin\left(\dfrac\pi3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \boxed{32 + 32\sqrt3\,i}$

2. First write the given number in exponential/trigonometric form.

$z = 5-5\sqrt3\,i$

has modulus

$|z| = \sqrt{5^2 + \left(-5\sqrt3\right)^2} = \sqrt{100} = 10$

and since it lies in the second quadrant of the complex plane, its argument is

$\arg(z) = \pi + \tan^{-1}\left(-\dfrac{5\sqrt3}5\right) = \pi + \tan^{-1}\left(-\sqrt3\right) = \pi - \dfrac\pi3 = \dfrac{2\pi}3$

So, we have

$z = 5 - 5\sqrt3\,i = 10 e^{i2\pi/3} = 10 \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right)$

Now we apply de Moivre's theorem again, and make sure to account for the multivalued-ness of the exponential function. For $k\in\{0,1,2,3,4\}$ , the fifth roots of $z$ are

$z^{1/5} = 10^{1/5} e^{i(2\pi/3 + 2\pi k)/5}$

$k=0 \implies z^{1/5} = 10^{1/5} e^{i2\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{2\pi}{15}\right) + i \sin\left(\dfrac{2\pi}{15}\right)\right)}$

$k=1 \implies z^{1/5} = 10^{1/5} e^{i8\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{8\pi}{15}\right) + i \sin\left(\dfrac{8\pi}{15}\right)\right)}$

$k=2 \implies z^{1/5} = 10^{1/5} e^{i14\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{14\pi}{15}\right) + i \sin\left(\dfrac{14\pi}{15}\right)\right)}$

$k=3 \implies z^{1/5} = 10^{1/5} e^{i20\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{4\pi}3\right) + i \sin\left(\dfrac{4\pi}3\right)\right)}$ $k=4 \implies z^{1/5} = 10^{1/5} e^{i26\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{26\pi}{15}\right) + i \sin\left(\dfrac{26\pi}{15}\right)\right)}$

6 0

2 years ago