A box contains five balls numbered 1, 2, 3, 4, 5. Three balls are drawn randomly at the same time from the box.

1 answer:

6 0

The possible outcomes could be:{123,124,125,134,135,145,234,235,245,345}
Their sums are as follows respectively:{6,7,8,8,9,10,9,10,11,12}=7
The odd sums are{7,9,11}=3
the probability of the sum being odd is 3/7
b.The number of outcomes are 10
the number of outcomes in which L is 4 are 3
So the probability of L being 4 is 3/10

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How to integrate 1/(1-y)^2

Akimi4 [234]

If you're using the app, try seeing this answer through your browser: brainly.com/question/866966

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!\dfrac{1}{(1-y)^2}\,dy}\\\\\\ =\mathsf{\displaystyle\int\!\dfrac{-1}{(1-y)^2}\cdot (-1)\,dy}\\\\\\ =\mathsf{\displaystyle- \int\!(1-y)^{-2}\cdot (-1)\,dy\qquad\quad(i)}$

Substitution:

$\begin{array}{lcl} \mathsf{1-y=u}&\quad\Rightarrow\quad&\mathsf{y=1-u}\\\\ &&\mathsf{dy=-1\,du} \end{array}$

So the integral $\mathsf{(i)}$ becomes

$=\mathsf{\displaystyle -\int\! u^{-2}\,du}\\\\\\ =\mathsf{-\,\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\ =\mathsf{-\,\dfrac{u^{-1}}{-1}+C}\\\\\\ =\mathsf{u^{-1}+C}\\\\\\ =\mathsf{\dfrac{1}{u}+C}$

$=\mathsf{\dfrac{1}{1-y}+C}\\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\!\dfrac{1}{(1-y)^2}\,dy=\dfrac{1}{1-y}+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>indefinite integral fraction rational substitution power differential calculus</em>

3 0

2 years ago

Determine la ecuación de la recta normal a la curva f(x)=ax(bx+a) en el punto de abscisa x=1. Dar como respuesta el intercepto a

AysviL [449]

Answer:

$f(x) = ax(bx + a) \\ a = 1 \: b = 10 \: x = 1 \\ f(x) = ax(bx + a) \\ f(1) = 7(1)(10(1) + 7) \\ f(1)= 7(10 + 7) \\ f(1)= 70 + 49 \\ f(1)= 149$