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3 years ago

How would you write .54545454... (Repeating) as a fraction? Please explain the process!

Mathematics

1 answer:

Pani-rosa [81]3 years ago

5 0

You would write .54 with a line across the 54 because that's the number repeating

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David invests $7000 in two different accounts. The first account paid 4 %, the second account paid 8 % in

Irina-Kira [14]

What we know:

I = Invest (so we dont mess up on interest.)

So, I = $7,000

First Account is 4%

Second Account is 2%

End of the year David Earns $476 in interest

With percent's you can always turn them into a decimal.

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The Question: How much was in each account?

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Step by Step Answer:

I = amount invested at 4%

4% = 0.40

8% = 0.80

0.40 - $7,000 = 6999.6

0.80 - 6999.6 = 6998.8

6998.8 - 476

Amount in both accounts in total is 6522.80.

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HOPE THIS HELPS! =3

HAVE A FANTASTIC DAY!

6 0

3 years ago

A carnival game has 160 rubber ducks floating in a pool. The person playing the game takes out one duck and looks at it.

nata0808 [166]

Answer:

Here is the answer

Step-by-step explanation:

That will show you.

5 0

2 years ago

Read 2 more answers

Which of the following are solutions to the equation below?

Dovator [93]

Answer:

Apply quaratic equation

6 0

2 years ago

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

6x2(3x) simplify this expression

Papessa [141]

Answer: 3(2 • (2x))

Explanation:
6 • 2(3x)
6 • 6x
3(2 • (2x))

4 0

3 years ago