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Oduvanchick [21]
4 years ago
13

A box contains 25 marbles. there are 6 blue, 2 green, 8 red, 1 yellow, and 3 orange marbles. you randomly choose 3 marbles, one

after the other. each time, you replace the marble back in the box before choosing the next one. what is the probability that the first marble is green, the second marble is red, and the third marble is blue?
Mathematics
1 answer:
Elza [17]4 years ago
5 0
I don't understand your question. You state that a box contains 25 marbles but 6+2+8+1+3= 20!!!
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$2.68

Step-by-step explanation:

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3 years ago
Square ABCD and isosceles triangle BUC are drawn to create trapezoid AUCD. Square A B C D and triangle B U C are attached at sid
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Answer:

d) 135º

Step-by-step explanation:

Note that the angle DCU is the sum of the angles DCB and BCU. The angle DCB is 90º because A B C D is a square, then all its angles are equal to 90º.

After attaching B U C to A B C D, we obtain a trapezoid A U C D. Since A U C D has at least one pair of parallel sides, then AU should be parallel to CD, thus the angle CBU must be 90º.

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4 years ago
Read 2 more answers
How do I differentiate this?​
disa [49]

Answer:

\frac{8}{\left(-x+2\right)^2} & x = 1, x = 3

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)

Now remember the sum rule:

\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right)  = 0\\\frac{d}{dx}\left(3x+2\right) = 3

For this second bit we apply the chain rule:

\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}

If we substitute these values back into the expression...\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)

...we get the following:

3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)

The rest is just pure simplification:

3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}

Now let's equate this to equal 8 for the second bit and solve for x:

\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3

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Answer:

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Step-by-step explanation:

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