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seraphim [82]
3 years ago
9

Please show your work and explain it.

Mathematics
1 answer:
Maru [420]3 years ago
6 0

Answer:

f(x)=\dfrac{x+2}{2(x-2)}

Step-by-step explanation:

Remember when you divide fractions, you need to get the reciprocal of the divisor and multiply. So your first simplification would be:

\dfrac{x^2+4x+4}{x^2-6x+8}\div\dfrac{6x+12}{3x-12}\\\\=\dfrac{x^2+4x+4}{x^2-6x+8}\times\dfrac{3x-12}{6x+12}\\\\=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}

Next we factor what we can so we can further simplify the rest of the equation:

=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}\\\\=\dfrac{(x+2)(x+2)(3x-12)}{(x^2-6x+8)(6(x+2))}\\\\

We can now cancel out (x+2)

=\dfrac{(x+2)(3x-12)}{(x^2-6x+8)(6)}

Next we factor out even more:

=\dfrac{(x+2)(3)(x-4)}{(x-2)(x-4)(6)}

We cancel out x-4 and reduce the 3 and 6 into simpler terms:

=\dfrac{(x+2)(1)}{(x-2)(2)}

And we can now simplify it to:

=\dfrac{x+2}{2(x-2)}

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Step-by-step explanation:

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Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

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To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

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If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

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Then the area between 1250 and 1300 is:

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Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

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To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

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Amount in tip jar = Amount in tip jar at noon + Additional amount made after noon = $20 + $0.50n

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