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2 years ago

Use the distribution property to explain how 30 flowers in a bunch can be counted in groups of 2s and 3s.

Mathematics

1 answer:

Shtirlitz [24]2 years ago

8 0

The way 30 flowers in a bunch can be counted in groups of 2s and 3s is 2(3(3) + 3(2))

<h3>Prime factorization of numbers</h3>

From the question, the interpretation is that we are to write the prime factorization of 30

30 = 2 * 15

30 = 2{3 (3 + 2)}

Expand using the distribution property

30 = 2(3(3) + 3(2))

30 = 2(9 + 6)

Hence the way 30 flowers in a bunch can be counted in groups of 2s and 3s is 2(3(3) + 3(2))

Learn more on prime factorization here: brainly.com/question/92257

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Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

Calculates the total amount E, in dollars, that she earns for working h hours using the equation E = 10h

GrogVix [38]

So your equation is E=10h.

We know E is equal to 1 so let's plug that is.

1=10h

Now we just solve it.

1=10h Divide each side by 10

.1=h

Boom!

She has to work .1 hours or 1/10 of an hour to earn 1 dollar.

Hope I helped!

7 0

3 years ago

Help!<br> Algebraic Relationship <br> y= x/4-1

loris [4]

32, 7

0, - 1

20, 4

You just plug in the numbers and solve

3 0

3 years ago

Two-fifths of the students in your class are in the band. Of these, one-fourth play the saxophone.

vichka [17]

Answer:

a. 1/10 b. did not understand question

Step-by-step explanation:

It is as simple as two fifths of one fourth. (2/5) * (1/4)= 1 tenth.

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3 years ago

Emma, Steve, Maria, and George are comparing the solution of a math assignment problem below.

uranmaximum [27]

Is their a picture or an equation?

8 0

4 years ago

Read 2 more answers