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3 years ago

8

Solve 7n 6 4n-2=2n 2(4n 6)

1 answer:

scoray [572]3 years ago

6 0

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Find the following limit. Limit of StartFraction StartRoot x + 2 EndRoot minus 3 Over x minus 7 EndFraction as x approaches 7 Wh

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3 years ago

Consider an urn containing 8 white balls, 7 red balls and 5 black balls.

weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

$=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053$

2) the probability of getting 2 white balls is equal to:

$=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465$

3) The probability of getting at least 72 white balls is:

$=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72} \times \left( \frac{7}{20} \right)^{78} +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73} \times \left( \frac{7}{20} \right)^{77} + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149} \times \left( \frac{7}{20} \right)^{1} +\left( \frac{8}{20} \right)^{150}$

$=\sum^{150}_{k=72} [C^{k}_{150}\times \left( \frac{8}{15} \right)^{k} \times \left( \frac{7}{15} \right)^{150-k}]$

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1 year ago

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