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3 years ago

Steve has 20 marbles in a bag, all the same size and shape. There are 8 red,

1 answer:

3 0

50/50 chance of getting a yellow marble.

Explanation:

There are 10 yellow marbles in the bag, and the other half are other colors. This would create a 50/50 ratio.

I am honestly super rusty with probability but oh well...

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Please help! Its a test that’s due today.

Eduardwww [97]

Answer:

it's b

Step-by-step explanation:

the denimonater is the root of the radical and the power

Hope this helps :)

3 0

3 years ago

Read 2 more answers

What is the answer for this question because it is very hard

Vladimir79 [104]

Answer:

Daniella: 0.2

Dwayne: 0.25

Angel: 0.2

Daniella and Angel have the same amount of scores. And Dwayne score is not. Dwayne has a higher score than Daniella and Angel.

Hope this helps

-Amelia

7 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

What is the missing value?

madam [21]

the answer is 40

X 1 (+1) 2 (+1) 3 (+1) 4 (+1) 5

Y 4 (+9) 13 (+9) 22 (+9) 31 (+9) 40

3 0

2 years ago

Read 2 more answers

6(x-2=4(x+3) what does x equal

Delicious77 [7]

Hey mate here is your answer.

➛ 6(x-2) = 4(x+3)

➛ 6x-12 = 4x + 12

➛ 6x - 4x = 12 + 12

➛ 2x = 24

➛ x = $\frac{24}{2}$

⛬ x = 12

4 0

4 years ago