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Alekssandra [29.7K]
3 years ago
8

What is the probability that the arrow will land on 1 or 2?

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
4 0
Answer: 2/5
Explanation:
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Why won’t anyone answer my questions argh! how do you draw starlight line graphs
Oduvanchick [21]
The table tells us that the x coordinate. It also tells us that y is always x + 1.

For #1 you plot the coordinate (0, 1).
0 (the x coordinate) is given to us already.
1 (the y coordinate) is needed to be found by the equation.
You would then need to fill in the equation given with the x coorident. 
y = 0 + 1 
Then, solve for y.
0 + 1 = 1
The y coordinate is 1
Go to the horizontal line (x) and find 0.
Then go to the veridical line (y) and find 1.
Then match up the the x and y to plot the coordinate.

You would continue with this equation with the rest of the xs. 

This is a hard concept to explain in just words, so feel free to comment with any more questions. :D

4 0
3 years ago
Please simplify 6d²(-7d)
andrew-mc [135]
Simplifying-
 (6d^2)(-7d)=
= -(42d^2)(d)
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6 0
3 years ago
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Plz get correct for brainiest
Scorpion4ik [409]

Answer:

well all you need to do is try your best

Step-by-step explanation:

3 0
2 years ago
11. For each table Ariella waits on at a restaurant, she is paid $4.00 plus 18% of the
lisabon 2012 [21]

Answer:

m=0.18b+4

She'll earn $10.30

Step-by-step explanation:

m=0.18b+4

m=0.18(35)+4

m=6.3+4

m=10.3

5 0
3 years ago
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2−x2. What are the dimensions of su
Paul [167]

Answer:

Step-by-step explanation:

Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

y=2-x^2

the parabola is open down with vertex at (0,2)

We can find that the rectangle also will be symmetrical about y axis.

Let the vertices on x axis by (p,0) and (-p,0)

Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation

Now width = 2-p^2

Area = l*w = 2(2p-p^3)

Use derivative test

I derivative = 2(2-3p^2)

II derivative = -12p

Equate I derivative to 0 and consider positive value only since we want maximum

p = \sqrt{\frac{2}{3} }

Thus width= \sqrt{\frac{2}{3} }

Length =2\sqrt{\frac{2}{3} }

Width = 2-2/3 = 4/3

3 0
3 years ago
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