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3 years ago

Help me please broskies

Mathematics

1 answer:

Luba_88 [7]3 years ago

6 0

From step 1 to step 2 they distributed the 2 (multiplied the inner numbers of the parentheses)
From step 2 to step 3 they added 6x to both sides,canceling out the -6x on the left side and adding 6x with the 5x
From step 3 to step 4 they subtracted 2 from both sides, canceling it out on the right side and subtracting 2 from 8 to get 6
From step 4 to step 5 they try and get x alone by dividing both sides by 11, canceling out the 11 on the right side and leaving a fraction the left
Leaving you with your answer of what x equals in the final step
Hope this helps!!

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Add:(3x2 - 5x + 6) + (9 - 8x - 4x2)

finlep [7]

You can see by the image how did I get the answer

3 0

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

What is the simplified answer to this problem 12 > -3x + 6

Verdich [7]

In this question, you're simplifying the inequality by solving for x.

Solve for x:

12 > -3x + 6

flip the equation:

-3x + 6 < 12

subtract 6 from both sides

-3x < 6

divide both sides by -3, while also flipping the inequality

x > -2

Answer:

x > -2

8 0

3 years ago

A sample space consists of 80 separate events that are equally likely. What is the probability of each? A sample space consists

yuradex [85]

Answer:

1/80

Step-by-step explanation:

7 0

2 years ago

Evaluate f(x) when x = -10

Sati [7]

So, when x = -10.

If you notice, -10 is NOT in the domain (the domain is (-3,13))

so, there is no solution.

3 0

3 years ago

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