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3 years ago

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The ages of 10 friends are consecutive integers. The sum of their ages is 710. they get together every Tuesday to play cards and

enjoy each other's company calculate their ages

1 answer:

Nady [450]3 years ago

7 0

Answer:

x = 66.5

x + 1 = 67.5

x + 2 = 68.5

x + 3 = 69.5

x + 4 = 70.5

x + 5 = 71.5

x + 6 = 72. 5

x + 7 = 73.5

x + 8 = 74.5

x + 9 = 84.5

Step-by-step explanation:

Given that :

Age of 10 friends are consecutive integers :

Let the ages be :

x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9

The sum of their ages = 710

Hence,

x + x+1 + x+2 + x+3 + x+4 + x+5 + x+6 + x+7 + x+8 + x+9 = 710

10x + 45 = 710

10x = 710 - 45

10x = 665

x = 665 / 10

x = 66.5

x + 1 = 67.5

x + 2 = 68.5

x + 3 = 69.5

x + 4 = 70.5

x + 5 = 71.5

x + 6 = 72. 5

x + 7 = 73.5

x + 8 = 74.5

x + 9 = 84.5

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Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

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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

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

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:

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

$E(x) = p$

Recall that:

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

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

Read more about probability distribution at:

brainly.com/question/15246027

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2 years ago

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