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Anastaziya [24]
3 years ago
12

If 4 marbles are drawn at random all at once from a bag containing 8 white and 6 black marbles. in how many ways can we draw the

4 marbles so that 2 will be white and 2 will be black?
Mathematics
1 answer:
Andreas93 [3]3 years ago
3 0

You started out as if it was going to be a probability problem, but it turned out to be a simple binary counting problem.

It really doesn't matter how many of each color are in the bag, or how many  marbles there are in the bag all together.  I think you're just asking how many ways there are to pull out two white ones and two black ones.  That's exactly the same question as "How many 4-bit binary numbers can be written with two 1's and two 0's ?"

It's easy to draw up that list.  You start out like this:

0011,  0101,  1001,  0110,  1010,  1100 .

Gosh.  I just wanted to start out doing the list, but I think that's all of them.

There are six (6) ways.

WWBB,  WBWB,  BWWB,  WBBW,  BWBW, and  BBWW .

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Answer:

1200

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1/5 of a minute goes in to 1 minute 5 times

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Evaluate this expression for x=-5 and y=-5​
Talja [164]

Answer:

x - 2y=5

Step-by-step explanation:

Given

x = -5; y=-5

x - 2y --- missing from question

Required

Evaluate

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x - 2y

Substitute for x and y

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x - 2y=5

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3 years ago
Please someone help me...​
laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )}  -  \frac{1}{ \cot(6A)  -  \cot(2A) }

<u>Using trigonometric </u><u>identities</u>

That's

<h3>\cot(x)  =  \frac{1}{ \tan(x) }</h3>

<u>Rewrite the expression</u>

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )} -    \frac{1}{ \frac{1}{ \tan(6A) } }  -  \frac{1}{ \frac{1}{ \tan(2A) } }

We have

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{1}{ \frac{ \tan(2A) -  \tan(6A)  }{ \tan(6A) \tan(2A)  } }</h3>

<u>Rewrite the second fraction</u>

That's

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{ \tan(6A)  \tan(2A) }{ \tan(2A) -  \tan(6A)  }</h3>

Since they have the same denominator we can write the fraction as

-  \frac{1 +  \tan(6A) \tan(2A)  }{ \tan(2A) -  \tan(6A)  }

Using the identity

<h3>\frac{x}{y}  =  \frac{1}{ \frac{y}{x} }</h3>

<u>Rewrite the expression</u>

We have

<h3>-  \frac{1}{ \frac{ \tan(2A)  -  \tan(6A) }{1 +  \tan(6A) \tan(2A)  } }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{ \tan(x) -  \tan(y)  }{1 +  \tan(x)  \tan(y) }  =  \tan(x - y)</h3>

<u>Rewrite the expression</u>

That's

<h3>- \frac{1}{ \tan(2A -6A) }</h3>

Which is

<h3>-  \frac{1}{ \tan( - 4A) }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{1}{ \tan(x) }  =  \cot(x)</h3>

Rewrite the expression

That's

<h3>-  \cot( - 4A)</h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3>- ( -  \cot(4A) )</h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2>\cot(4A)</h2>

As proven

Hope this helps you

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Step-by-step explanation:

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