Solve the equation on the interval [0, 2π). cos x + 2 cos x sin x = 0

1 answer:

6 0

Take a common factor of cos(x)
$\cos(x) (1 + 2 \sin(x) ) = 0$
then by zero multplication rules
$\cos(x) = 0$
$x = \frac{\pi}{2} \: x = \frac{3\pi}{2}$
and when
$1 + 2 \sin(x) = 0$
so
$2 \sin(x ) = - 1$
$\sin(x) = \frac{ - 1}{2}$
so
$x = \frac{3\pi}{2} - \frac{\pi}{3}$
and
$x = \frac{3\pi}{2} + \frac{\pi}{3}$

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a) Binomial distribution B(n=12,p=0.01)

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Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

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Then,

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c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$