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BlackZzzverrR [31]
3 years ago
9

I need first one and the second one please! I need it tomorrow!

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0
I have a total of 11 pencils = total outcome
a- since we have 6 Red, P(Red)=6/11
b- P(Black) = 3/11
c- P(Green) = 1/11
For the dice:
Total outcome 6 so :
P(3) =1/6
P(1) =1/6
P(less than5) means P( of 1 OR 2 OR 3 Or 4)
Since we have "OR" it means we can add the P(probabilities)====>
P( of 1 OR 2 OR 3 Or 4)=1/6+1/6+1/6+1/6=
= 4/6 = 2/3
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Solve the equation:
PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

4\sqrt{3-\frac{1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\4\sqrt{\frac{3x-1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\\\put ~\sqrt{\frac{3x-1}{x}} =a\\4a-\frac{1}{a} =3\\4a^2-1-3a=0\\4a^2-4a+1a-1=0\\4a(a-1)+1(a-1)=0\\(a-1)(4a+1)=0\\a=1,-\frac{1}{4}\\

\sqrt{\frac{3x-1}{x} } =1\\3x-1=x\\3x-x=1\\2x=1\\x=\frac{1}{2}

when a=-1/4

\sqrt{\frac{3x-1}{x} } =-\frac{1}{4} \\

\frac{3x-1}{x} =\frac{1}{16} \\48x-16=x\\47x=16\\x=\frac{16}{47}

\frac{(1-2x)^2 \times (x^2-9)}{-x^2-x+6} \geq 0,\\both ~numerator ~and~denominator ~are~of~same~sign.\\\frac{(1-2x)^2(x^2-9)}{-x^2-3x+2x+6} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{-x(x+3)+2(x+3)} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{(x+3)(-x+2)} \geq 0\\(1-2x)^2\geq 0~(always)\\x\neq -3\\case.~1.\\both~numerator~and~denominator >0\\\frac{x-3}{-x+2} \geq 0\\x-3\geq 0\\x\geq 3\\-x+2> 0\\-x>-2\\x

case 2.

both numerator and denominator are negative.

x-3\leq 0\\x\leq 3\\-x+2\leq 0\\-x\leq -2\\x\geq 2\\solution~is~(-\infty,-3)U(-3,3]U[2,\infty)

5 0
3 years ago
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Sever21 [200]

Answer:

6 4/5 hours

Step-by-step explanation:

If one pipe can fill the pool in 10 hours, then it can fill 1/5 of the pool in 2 hours.

If two pipe's can fill the pool in 6 hour's, then it can fill 1/5 of the pool 1 1/5 hours.

So:

6-1 1/5=4 4/5

4 4/5+2=6 4/5

8 0
3 years ago
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Answer:

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Step-by-step explanation:

Simply distribute the negative and add like terms together:

(f - g) = 3x - 1 - (x + 2)

(f - g) = 3x - 1 - x - 2

(f - g) = 2x -  3

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Answer:

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Step-by-step explanation:

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I am doing this right now, haha, Ill let you know when done!
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