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Marysya12 [62]
3 years ago
12

Find the slope of the line.

Mathematics
1 answer:
pickupchik [31]3 years ago
6 0
There are several ways to do this.
I'll show you two methods.

1) Pick two points on the line and use the slope formula.
Look for two points that are easy to read. It is best if the points are on grid line intersections. For example, you can see points (-4, -1) and (0, -2) are easy to read.
Now we use the slope formula.

slope = m = (y2 - y1)/(x2 - x1)

Call one point (x1, y1), and call the other point (x2, y2).
Plug in the x1, x2, y1, y2 values in the formula and simplify the fraction.

Let's call point (-4, -1) point (x1, y1).
Then x1 = -4, and y1 = -1.
Let's call point (0, -2) point (x2, y2).
Then x2 = 0, and y2 = -2.

Plug in values into the formula:

m = (y2 - y1)/(x2 - x1) = (-2 - (-1))/(0 - (-4)) = (-2 + 1)/(0 + 4) = -1/4

The slope is -1/4

2) Pick two points on the graph and use rise over run.
The slope is equal to the rise divided by the run.
Run is how much you go up or down.
Rise is how much you go right or left.

Pick two easy to read points.
We can use the same points we used above, (-4, -1) and (-0, -2).
Start at point (0, -2).
How far up or down do you need to go to get to point (-4, -1)?
Answer: 1 unit up, or +1.
The rise is +1.
Now that we went up 1, how far do you go left or right top go to point (-4, -1)?
Answer: 4 units to the left. Going left is negative, so the run is -4.
Slope = rise/run = +1/-4 = -1/4

As you can see we got the same slope using both methods.
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If you roll a 6-sided die 54 times, what is the best prediction possible for the number of times you will roll an odd number?
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Which graph shows f(x) = 0.5|x + 3| – 2?
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This is the graph that the equation shows:

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A particle moves on a circle through points which been marked 0,1,2,3,4 (in a clockwise order). At each step it has a probabilit
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Answer:

Step-by-step explanation:

Given data:

SS={0,1,2,3,4}

Let probability of moving to the right be = P

Then probability of moving to the left is =1-P

The transition probability matrix is:

\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]

Calculating the limiting probabilities:

π0=π0+Pπ1                 eq(1)

π1=(1-P)π0+π1+Pπ2     eq(2)

π2=(1-P)π1+π2+Pπ3    eq(3)

π3=(1-P)π2+π3+Pπ4    eq(4)

π4=(1-P)π3+π4             eq(5)

π0+π1+π2+π3+π4=1

π0-π0-Pπ1=0

→π1 = 0

substituting value of π1  in eq(2)

(1-P)π0+Pπ2=0

from

π2=(1-P)π1+π2+Pπ3  

we get

(1-P)π1+Pπ3 = 0

from

π3=(1-P)π2+π3+Pπ4

we get

(1-P)π2+Pπ4 =0

from π4=(1-P)π3+π4  

→π3=0

substituting values of π1 and π3 in eq(3)

→π2=0

Now

π0+π1+π2+π3+π4=0

π0+π4=1

π0=0.5

π4=0.5

So limiting probabilities are {0.5,0,0,0,0.5}

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