<img src="https://tex.z-dn.net/?f=4x%20-%205y%20%3D%20%20-%205%20%20%5C%5C%20y%20%3D%20x" id="TexFormula1" title="4x - 5y =

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tino4ka555 [31]

3 years ago

\\ y = x" alt="4x - 5y = - 5 \\ y = x" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

y=1 + $\frac{4x}{5}$

x=- $\frac{5}{4}$ + $\frac{5y}{4}$

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To prepare for a presentation, pete colored 1/2 of the poster and rebecca colored 1/3 of the poster, what fraction still needs t

WARRIOR [948]

Pete= 1/2

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3 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

Pls help i need to finish this !!

AlladinOne [14]

Answer:

Step-by-step explanation:

it's non-linear

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2 years ago

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How do you model the square root of 16

kotegsom [21]

$\sqrt{16} =\sqrt{4*4}=\sqrt{4}*\sqrt{4}=2*2=4$

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2 years ago

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Neil drives at an average speed of 60 miles/hour to reach his destination 480 miles away. On the way back, he decides to increas

daser333 [38]

Answer:

480/(x+60) ≤ 7

Step-by-step explanation:

We can use the relations ...

time = distance/speed

distance = speed×time

speed = distance/time

to write the required inequality any of several ways.

Since the problem is posed in terms of time (7 hours) and an increase in speed (x), we can write the time inequality as ...

480/(60+x) ≤ 7

Multiplying this by the denominator gives us a distance inequality:

7(60+x) ≥ 480 . . . . . . at his desired speed, Neil will go no less than 480 miles in 7 hours

Or, we can write an inequality for the increase in speed directly:

480/7 -60 ≤ x . . . . . . x is at least the difference between the speed of 480 miles in 7 hours and the speed of 60 miles per hour

___

Any of the above inequalities will give the desired value of x.

4 0

3 years ago