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3 years ago

You and a friend are playing a game by tossing two coins. If both coins land

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

3 0

c. No. You have a probability of winning, while your friend has a
probability of winning.
Or it’s A

klio [65]3 years ago

3 0

Answer:

Step-by-step explanation:

Let the random variable X be the number of heads that come up when a fair coin is tossed 4 times. Then X∼B(4,12). We want there to be exactly two heads (forcing the other two tosses to be tails), so P(X=2)=(42)(12)2(12)2=38.

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Anton [14]

Answer:

4.5

Step-by-step explanation:

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3 years ago

Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2

nlexa [21]

$\bold{[ \ Answer \ ]}$

$\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}$

$\bold{-------------------}$

$\bold{Rewrite}$

$\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Rewrite}$

$\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}$

$\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}$

$\bold{Rewrite}$

$\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}$

$\bold{Simplify}$

$\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}$

$\boxed{\bold{[] \ Eclipsed \ []}}$

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3 years ago

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3 years ago

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likoan [24]

Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

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