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2 years ago

The graph of a line has a slope of 3 and passes through the point (-4,-5).

Mathematics

1 answer:

Goshia [24]2 years ago

5 0

Answer:

3x - y + 7 = 0

Step-by-step explanation:

The given point is , ( -4 , -5) and the slope is 3. The point slope form is ,

=> y - y₁ = m( x - x₁)

=> y - (-5) = 3 { x - (-4) }

=> y +5 = 3( x + 4 )

=> y + 5 = 3x + 12

=> 3x - y + 12 - 5 = 0

=> 3x - y +7 = 0

<h3>Hence the equation of the line is 3x - y + 7 = 0 .</h3>

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3 years ago

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:

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

$E(x) = p$

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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

Read more about probability distribution at:

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2 years ago

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