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denpristay [2]
3 years ago
13

Prior to the Academy Awards ceremony in 2009, the United Kingdom bookmaker Ladbrokes reported the following odds for winning an

Oscar in the category of best actress (The Wall Street Journal, February 20, 2009).
Best Actress Movie Odds
Anne Hathaway Rachel Getting Married 2:11
Angelina Jolie Changeling 1:20
Melissa Leo Frozen River 1:33
Meryl Streep Doubt 3:10
Kate Winslet The Reader 5:2
a. Express the odds for each actress winning as a probability. (Round your answers to 3 decimal places.)
Best Actress Probability
Anne Hathaway
Angelina Jolie
Melissa Leo
Meryl Streep
Kate Winslet
Mathematics
1 answer:
hodyreva [135]3 years ago
3 0

Answer:

Anne Hathaway: 0.154 = 15.4%

Angelina Jolie: 0.048 = 4.8%

Melissa Leo: 0.029 = 2.9%

Meryl Streep: 0.231 = 23.1%

Kate Winslet: 0.714 = 71.4%

Step-by-step explanation:

Odds to probability:

Odds of a:b

The probability is given by:

p = \frac{a}{a+b}

Anne Hathaway 2:11

So p = \frac{2}{2+11} = \frac{2}{13} = 0.154

Angelina Jolie 1:20

So p = \frac{1}{1+20} = \frac{1}{21} = 0.048

Melissa Leo 1:33

So p = \frac{1}{1+33} = \frac{1}{34} = 0.029

Meryl Streep 3:10

So p = \frac{3}{3+10} = \frac{3}{13} = 0.231

Kate Winslet 5:2

So p = \frac{5}{5+2} = \frac{5}{7} = 0.714

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\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)

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\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}

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\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0

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