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3 years ago

What is the probability of pulling out two yellow marbles?

Mathematics

2 answers:

IgorC [24]3 years ago

5 0

Answer:

Step-by-step explanation:

n(Y) = 8

Total marble = 8+4+5 = 17

Prob(1st Y) = 8/17

2nd been yellow means number of yellow left is 7 ant total number of marble is 16

Prob(2nd Y) = 7/16

Prob(1st Y and 2nd Y)= 8/17 × 7/16

= 7/34

GenaCL600 [577]3 years ago

3 0

2/ 17 because it’s 17 marbles and I hav a 2/17 chance of pulling 2 yellows

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Express the complex number in trigonometric form. -6 + 6 square root of 3 i ?

bagirrra123 [75]

$\bf \begin{array}{clclll}
-6&+&6\sqrt{3}\ i\\
\uparrow &&\uparrow \\
a&&b
\end{array}\qquad 
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\
-------------------------------\\\\$

$\bf r=\sqrt{(-6)^2+(6\sqrt{3})^2}\implies r=\sqrt{36+(6^2\cdot 3)}\implies r=\sqrt{144}
\\\\\\
r=12\leftarrow 
\\\\\\
\theta =tan^{-1}\left( \frac{6\sqrt{3}}{-6} \right)\implies \theta =tan^{-1}\left( -\sqrt{3}\right)\implies \theta =
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\frac{2\pi }{3}\leftarrow \\
\frac{5\pi }{3}
\end{cases}$

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.

thus $\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]$

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4 years ago

An experiment consist of rolling two fair number cubes. The diagram shows the sample space of all equally likely outcomes. Find

Tema [17]

The answer Is D ....

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4 years ago

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pshichka [43]

To find the answer of this problem, we need to distibute and multiply all of the numbers with each other.

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3 years ago

If x-y=√3, x²-y²=√5 & a²-a√3+1=0, then find out the value of x+y

Lemur [1.5K]

Answer:

√15/3

Step-by-step explanation:

x² - y² =√5

(x +y)*(x -y) = √5

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x+y = √5 / √3

x +y = √5*√3 / √3*√3

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3 years ago

Roots of Quadratics 50 PTS!!!

noname [10]

The values of k for the different quadratic equation solutions are as follows

a the equation 2x² - x + 3k = 0 has two distinct real roots

k < 1/24

b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots

k = 3/5

ci the equation -x² + 3x + (k + 1) = 0 has real roots

k > -3.25

d the equation 3kx² - 3x + 2 = 0 has no real solutions

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<h3>How to solve quadratic equations to get different answers</h3>

Quadratic equations of the form ax² + bx + c = 0 is solved using the formula

$-b+\frac{\sqrt{b^{2}-4ac } }{2a}$ OR $-b-\frac{\sqrt{b^{2}-4ac } }{2a}$

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b² - 4ac > 0

substituting the values gives

(-1)² - 4 * 2 * 3k > 0

1 - 24k > 0

1 > 24k

divide through by coefficient of k

k < 1/24

To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0

b² - 4ac = 0

substituting the values gives

(-2)² - 4 * 5 * (2k - 1) = 0

4 - 40k + 20 = 0

-40k = -24

divide through by coefficient of k

k = 3/5

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b² - 4ac > 0

substituting the values gives

(3)² - 4 * -1 * (k+1) > 0

9 + 4k + 4> 0

4k > -13

divide through by coefficient of k

k > -3.25

To solve the equation and get no real solutions 3kx² - 3x + 2 = 0

b² - 4ac < 0

substituting the values gives

(-3)² - 4 * 3k * 2 < 0

9 - 24k² > 0

9 > 24k²

divide through by coefficient of k²

k² < 24/9

k < ± 1.633

Learn more about roots of quadratic equations: brainly.com/question/26926523

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1 year ago

Read 2 more answers