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3 years ago

What is the lcm of each group of numbers 3,9,15

Mathematics

2 answers:

Nina [5.8K]3 years ago

7 0

Answer: 45

Step-by-step explanation:

By factor method,

3 = 1, 3

9= 3, 3

15= 3, 5

LCM= 3*3*5

=45

mamaluj [8]3 years ago

5 0

Answer:

Step-by-step explanation:

Multiples of 3: 3, 6, 9, 12, 15, 18, 27, 30, 33, 36, 39 , 41, 45, ...

Multiples of 9: 9, 18, 27, 36, 45...

Multiples of 15: 15, 30, 45, 60,...

The lowest multiple they have in common is 45

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A used-boat dealership buys a boat for $2600 and then sells it for $3800. What is the percent increase?

Vladimir79 [104]

46% Find the difference of 2600 and 3800= 1200. Now divide 1200 into 2600. If you round the number it will be 0.46= 46%

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3 years ago

The grassy area of a playground has a side length of x and a side of width of x + 3 what is the area of the playground?

Kipish [7]

Answer:

Step-by-step explanation:

We assume this playground is a rectangle. The area of a rectangle is of the form: A = lw, where l is the length and w is the width.

Here, the length is l = x and the width is w = x + 3. So, plug these in:

A = lw

A = x(x + 3)

Expand this out:

A = x² + 3x

The answer is thus A.

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3 years ago

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qaws [65]

Next step would be 5x10x5

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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).

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

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2 years ago

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KonstantinChe [14]

Answer:

is B))))

Step-by-step explanation:

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2 years ago

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