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Bogdan [553]
3 years ago
5

Six possibilities are equally likely and have payoffs of $2, $4, $6, $8, $10, and $12. the expected value is:

Mathematics
1 answer:
igomit [66]3 years ago
8 0

The expected value is even

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The weight is 3000 grams.
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Calculate the following limit:
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\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\
=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1


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3 years ago
A small swimming pool has a circumference of 3π feet. Select the true statement about the pool.
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