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3 years ago

12

A coin is fixed so that the probability to get heads in two throws is double the probability. Count the probability that there w

ill be head after the second throw.

1 answer:

Setler [38]3 years ago

5 0

Answer:

no clue try looking it up on here see what you find

Step-by-step explanation:

i just wanna help but im not very smart

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Use the distributive property and combining like terms to solve the following equation for mm.

tangare [24]

Answer:

there are no solutions to m

Step-by-step explanation:

$- 2(m + 4) - 9 = \\ 5 - 2(m + 4) - 9 = 5$

$- 2m - 8 - 9 = 5 - 2m - 8 - 9 = 5$

$- 2m - 17 = - 2m + 5 - 8 - 9 = 5$

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$- 2m - 17 = - 2m - 12 = 5$

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$- 2(0 + 4) - 9 = 5 \\ - 2(0 + 4) - 9 = 5 \\ \\ 0 - 8 - 9 = 5 \\ 0 - 8 - 9 = 5 \\ - 17 = 5 \\ there \: are \: no \: solutions$

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Solve the simultaneous equations C^2+d^2=5 and 3c+4d=2

Naily [24]

If 3c+4d=2 then c=(2-4d)/3
And if you put that into the other equation
${\frac{2 - 4d}{3}}^{2} + {d}^{2} = 5$
Then you multiply everything by 3, getting

${(2 - 4d)}^{2} + 3 {d}^{2} = 15$
And then

$2 - 4d + \sqrt{3} d = \sqrt{15}$
Leaving
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I dont have a calc with me right now but you just pot this into the calculator:
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And when you get the number you insert it into the nest eqation to get c

$c = \frac{2 - 4d}{3}$
So basically

$c = \frac{2 - 4 \frac{ \sqrt{15} - 2}{ \sqrt{3} - 4} }{3}$

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