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2 years ago

10

Intergrate x×(x²-3)³

1 answer:

Kisachek [45]2 years ago

4 0

Answer: $\large \boldsymbol{\dfrac{1}{8 } (x^2-3)^4+c}$

Step-by-step explanation:

$\displaystyle \large \boldsymbol{} \int\limits {x(x^2-3)^2 } \, dx =\int\limits \frac{1}{2} {} \, (x^2-3)' (x^2-3)^3 dx =\frac{1}{8} (x^2-3)^4+c$

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Answer:

$P(X = 0) = 0.0263$

$P(X = 1) = 0.1407$

$P(X = 2) = 0.3012$

$P(X = 3) = 0.3224$

$P(X = 4) = 0.1725$

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Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 5, p = 0.517$

Distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263$

$P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407$

$P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012$

$P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224$

$P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725$

$P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369$

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3 years ago