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3 years ago

Can u please help thank you

Mathematics

1 answer:

torisob [31]3 years ago

8 0

Answer : 6 is opposite of 1

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The amazon rainforest covered 6.42 million square kilometers in 1994. In 2014, it covered 50/59 as much. Which is closest to the

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The Powerball lottery is played twice each week in 28 states, the Virgin Islands, and the District of Columbia. To play Powerbal

Alecsey [184]

Answer:

There are 3478761 ways to select the first 5 numbers

Step-by-step explanation:

As understood from the statement of this problem we assume that it does not matter the order in which the first 5 white balls are selected.

In this case it is a combination.

So, what we want to know is how many ways you can choose 5 white balls out of 55.

Then we use the formula of combinations:

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

Where you have n elements and choose x from them.

Then we look for:

$C(55, 5) = \frac{55!}{5!(55-5)!}\\\\C(55, 5) = \frac{55!}{5!(50)!}\\\\C(55, 5) = 3478761$

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2 years ago

What is 6 divided by 756

Alexxx [7]

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3 years ago

Hi need help for this maths question

Sauron [17]

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

$\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}$

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

$\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}$

c) The mean is given by the integral,

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy$

Integrate by parts, with

$u = y \implies du = dy$

$dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}$

Then

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)$

$\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy$

$\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}$

8 0

1 year ago