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Arisa [49]
3 years ago
11

Which of the following shows the division problem below in synthetic division form? 3x^2-10x+7/x+4

Mathematics
1 answer:
saul85 [17]3 years ago
7 0

Answer:

Step-by-step explanation:

The answer is at the bottom :)

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Gino bought 1/2 lb of​ candy, 3/4 lb of​ nuts, and 7/8 lb of granola to make a trail mix. How many pounds of trail mix did Gino​
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3 years ago
Select each quadrilateral that has exactly one pair of parallel sides.
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4 0
3 years ago
Write an equation of the line that passes through the given point and is perpendicular to the graph of the given equation. Pleas
Anna71 [15]

Answer:

y = \frac{1}{3}x + 4

Step-by-step explanation:

First, to figure out the equation of a line you must find the slope then plug in the coordinates.

In order to find the slope of the equation, you know that it is perpendicular to the equation y = -3x + 7.

Perpendicular slopes are basically reciprocals with opposite signs; you flip the number and give it the opposite symbol.

For example, the perpendicular slope of 2 would -1/2, because flipping the fraction 2/1 is 1/2 and the opposite sign of + would be -.

Therefore, the perpendicular slope of -3 would be 1/3, be 3/1 flipped upside down is 1/3 and the opposite sign of - is +.

Then you plug in the slope into the equation of a line along with the coordinate.

A line is y = mx + b

Since you found the slope, m = 1/3

They also give you the coordinate (3,5), meaning when x = 3, the y =5.

You plug everything into the equation to solve for b:

5 = (1/3)(3) + b

b = 5 - (1/3 * 3) = 5 - 1 = 4

Since you have values for m and b, the equation can now be constructed as

y = 1/3x + 4

6 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
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