Which equation below represents a line with a slope of - 2/3 that passes through the point (-4.5 , 10)?

Mathematics

1 answer:

Lostsunrise [7]2 years ago

8 0

Answer:

Option C

Step-by-step explanation:

If a line passes through a point, then that point is called a solution to the line's equation. Substituting the x and y values of that solution into the equation will give a true statement. So, to find out which option is correct, we can substitute the x and y values of (-4.5, 10) into each equation and see if the result is a true statement.

Let's try this with option C. To make things easier, convert -4.5 into decimal form: $-4\frac{1}{2}$ . Substitute $-4\frac{1}{2}$ for x and 10 for y in the equation, then solve:

$y = -\frac{2}{3} x+7\\\\10 = -\frac{2}{3} (-4\frac{1}{2} )+7\\10 = -\frac{2}{3} (-\frac{9}{2})+7\\10 = 3+7\\10 = 10$

10 does equal 10, so this is a true statement. Option C is the right answer.

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12^2<br><br> 28 questions left

Vesnalui [34]

Answer:

144

Step-by-step explanation:

12² is the same as 12 × 12

12 × 12 = 144

5 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

(5a + a2 - 14) (10a + a2 + 39)

yan [13]

15a+a4+25 is the answer

8 0

3 years ago

Do these three segments make a right triangle? 14cm, 23cm, and 25cm

Basile [38]

Answer:

This is not a right triangle(No)

Step-by-step explanation: