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4 years ago

A bag contains 5 white, 3 black, and 2 green balls. Balls are picked at random.

Mathematics

1 answer:

Soloha48 [4]4 years ago

6 0

Answer:

look down there

Step-by-step explanation:

First ball:

Probability of drawing a white ball is 5/8

Probability of drawing a black ball is 3/8

Second ball:

This depends on the first ball drawn, lets say you drew a white ball initially, 4 white balls are left out of 7 balls in total. The probability of a white ball in the second pick is 4/7.

Total probability of drawing two white balls is 5/8*4/7 (since they are independent events).

If you picked a black ball initially, picking another black ball would have a probability of 2/7, on similar grounds , total prob for 2 blacks would be 3/8*2/7.

The probability that you pick 2 balls of same color is (5/14 + 3/28) = 13/28. (Since they are mutually exclusive events)

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maksim [4K]

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Polygon X is similar to polygon.

And Polygon X is three times the size of polygon Y.

So, it becomes,

$\dfrac{\text{Perimeter of polygon X}}{\text{Perimeter of polygon Y}}=\dfrac{3}{1}\\\\\dfrac{81}{x}=\dfrac{3}{1}\\\\x=\dfrac{81}{3}\\\\x=27$

Hence, perimeter of polygon Y is 27.

Therefore, Option 'C' is correct.

5 0

3 years ago

Read 2 more answers

John Burroughs $6.50 from his friends to pay for snacks and $11. 75 from sister to go to the movies. How much money does he need

Korvikt [17]

Answer:

18.25

Step-by-step explanation:

......ur welcome.............

4 0

3 years ago

Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

4 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago

What number should be added to both sides of the equation to complete the square?<br> x2 + 12x = 11

makkiz [27]

given:

x² + 12x = 11

perfect square:

a² + 2ab + b²

a² = x² ⇒ x * x

2ab = 12x ⇒ 2(6)x

b² = 6² ⇒ 36

x² + 12x + 36 = 11 + 36

(x+6)(x+6) = 47

Both sides must be added with 36.

6 0

3 years ago

Read 2 more answers