Can someone proove it to me thank you for your help

Mathematics

1 answer:

Dovator [93]4 years ago

3 0

$\sin(2x + x) = \sin(2x) \cos(x) + \\ \cos(2x) \sin(x) = 2 \sin(x) { \cos(x) }^{2} \\ + \sin(x) { \cos(x) }^{2} - { \sin(x) }^{3} = \\ 3 \sin(x) { \cos(x) }^{2} - { \sin(x) }^{3} = \\ 3 \sin(x) - 3 { \sin(x) }^{3} - { \sin(x) }^{3} = \\ 3 \sin(x) - 4 { \sin(x) }^{3}$

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A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n indepe

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Step-by-step explanation:

Binomial distribution formula :-

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.

Given : A binomial probability experiment is conducted with the given parameters.

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Now, $P(x\leq3)=P(3)+P(2)+P(1)+P(0)$

$=^9C_3(0.8)^3(1-0.8)^{9-3}+^9C_2(0.8)^2(1-0.8)^{9-2}+^9C_1(0.8)^1(1-0.8)^{9-1}+^9C_0(0.8)^0(1-0.8)^9\\\\=\dfrac{9!}{3!6!}(0.8)^3(0.2)^6+\dfrac{9!}{2!7!}(0.8)^2(0.2)^7+\dfrac{9!}{1!8!}(0.8)(0.2)^8+\dfrac{9!}{0!9!}(0.2)^9=0.003066368\approx0.0031$