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2 years ago

Ratio for 669 and 221

Mathematics

2 answers:

svetoff [14.1K]2 years ago

8 0

Answer:

Oi it's 3.03

Step-by-step explanation:

cruel reality

p.s sorry or lateness but people of the future may now know

alexandr402 [8]2 years ago

5 0

Hello from MrBillDoesMath!

Answer:

669/221

Discussion:

Ratio of 669 to 221 = 669/221.

Now, 669 = 3 * 223 and 221 = 13 * 17. Note that all numbers involved in the factorizations are distinct primes so nothing cancels out in their ratio. That is,

669/221 can not be reduced further.

Thank you,

MrB

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The Bernoulli distribution is a distribution whose random variable can only take 0 or 1

The value of E(x2) is p
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<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

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Add

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Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

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$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

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