F(x) = 1/(x+2) & g(x) = x/(x-3)

(f(x) + g(x) = 1/(x+2) + x/(x-3). Reduce to same denominator:

1/(x+2) + x/(x-3) =(x-3) + x(x-3)/(x+2).(x-3) ==> (x²+3x-3)/(x+2).(x-3)

7 0

2 years ago

(HELP!)

denis-greek [22]

Let $X$ be the random variable denoting the number of girls that are born. Then $X$ has a binomial distribution on $n=125$ babies and probability of getting a girl $p=0.5$ . The probability we want is

$\mathbb P(X\ge60)=\displaystyle\sum_{x=60}^{125}\binom{125}x0.5^{125-x}0.5^x\approx0.704=70.4\%$

7 0

2 years ago

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