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8 months ago

Evaluate.

Mathematics

2 answers:

dolphi86 [110]8 months ago

5 0

ANSWER
25281
100
1 Calculate: (-13 + 0.3 x (-13) + 1)2
252.81
2 Convert to fraction: 252.81
25281
100
Show

lozanna [386]8 months ago

4 0

Answer: 289/900

Explanation: I did this before

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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).

spayn [35]

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

The value of E(x2) is p
The value of V(x) is p(1 - p)
The value of E(x79) is p

<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

5 0

2 years ago

Solve for t in the expression 16/t when t = 4. <br><br> please help!!

Ronch [10]

Answer:

x=16/t

t=4 so

16/4 = 4

4 is your answer.

3 0

3 years ago

B is the midpoint of AC and E' is the midpoint of BD. If A(-9,-4), C(-1, 6), and E(-4,-3), find the coordinates of D.

Ipatiy [6.2K]

The coordinates for D are (-4, -7)

First we must locate point B as it is vital to finding the midpoint of BD. To do this, we take the average of the endpoints AC since B is its midpoint.

x values = -9 + 1 = -8
Then divide by 2 for the average -8/2 = -4

y values = -4 + 6 = 2
Then divide by 2 for the average 2/2 = 1

Therefore B must be (-4, 1)

Now we know the values of E must be the average of B and D. So we can write equations for each coordinate since we know they are averages.

x - values = (Bx + Dx)/2 = Ex
(-4 + Dx)/2 = -4 ---> multiply both sides by 2
-4 + Dx = -8 ---> add -4 to both sides
Dx = -4

y - values = (By + Dy)/2 = Ey
(1 + Dy)/2 = -3 ---> multiply both sides by 2
1 + Dy = -6 ---> subtract 1 from both side
Dy = -7

So the coordinates for D must be (-4, -7)

5 0

3 years ago

Read 2 more answers

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago

Which rules of exponents will be used to evaluate this expression? Check all that apply.

tatyana61 [14]

Answer:

Actual correct answer (I checked because the first answer was wrong)

Step-by-step explanation:

x product of powers

quotient of powers

power of a power

x power of a product

negative exponent

x zero exponent

I got this right the second time.

5 0

2 years ago

Read 2 more answers