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3 years ago

Explain when it would be more useful to use the Counting Principle than a tree diagram to count possible outcomes of an event.

Mathematics

1 answer:

Kipish [7]3 years ago

8 0

Because a tree diagram is mostly for finding how many possibilities or choices one has.

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The product of victors saving and 9 is 435

denpristay [2]

=3,915

product is use for multiplying

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2 years ago

List all the factors of 16

Ronch [10]

Answer:

1,2,4,8,16

Step-by-step explanation:

prime factorization: 2x2x2x2 =16

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2 years ago

PLEASEEEE help Jack, Jill, and Bill each carried a 48-ounce bucket full of water down the hill. By the time they reached the bot

g100num [7]

Hold on I think I know this

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2 years ago

If the acceleration of an object is given by dv/dt = −2v, Find the position function s(t) if v(0) = 7 and s(0) = 0.

kirill [66]

Answer:

Step-by-step explanation:

Given that acceleration of an object is

$\frac{dv}{dt} =-2v\\\frac{dv}{v} =-2dt\\ln v = -2t+C\\$

is the solution to the differential equation

Since v(0) =7

we get ln 7 = C

Hence $lnv = -2t+ln 7\\v=7e^{-2t}$

since velocity is rate of change of distance s we have

$v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[$

substitute t=0 and s=0

$C=7/2$

So solution for distance is

$s(t) =\frac{-7}{2} (e^{-2t}-1)$

4 0

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).

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