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zubka84 [21]
3 years ago
9

The y-intercept of the line whose equation is 3x - 2y = 6 is -3 2 6

Mathematics
2 answers:
Ludmilka [50]3 years ago
7 0
The best way to answer this question is to ask yourself what the Y intercept is. The Y intercept is the point on the graph when the line crosses the Y axis. This point will always be wherever X is equal to 0.

So, to find the answer, we need to say that X = 0

3(0) - 2Y = 6

Zero times any number equals zero, so we are left with:

-2Y = 6

Now, we solve for Y by dividing both sides by -2.

Y = 6/(-2)

Y = -3

So, when X = 0, Y = -3.

Thus, the Y intercept is at point (0,-3).
yulyashka [42]3 years ago
4 0

Answer:

The y-intercept is -3

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Let S be the set of outcomes when two distinguishable dice are rolled, let E be the subset of outcomes in which at least one die
allochka39001 [22]

Answer:

E'∪F'=[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5], [2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6], [6, 2], [6, 4], [6, 6]

Step-by-step explanation:

Sample Space of the Two Dice is:

[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]

[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]

[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]

[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]

[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]

[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]

E=At least one die shows an even number.

Sample Space of E

[1, 2],  [1, 4], [1, 6]

[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]

[3, 2] [3, 4],  [3, 6]

[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]

[5, 2],  [5, 4],  [5, 6]

[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]

Complement of E

[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5]

F=At least one die shows an odd number

Sample Space of F

[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]

[2, 1], [2, 3], [2, 5],

[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]

[4, 1], [4, 3], [4, 5]

[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]

[6, 1], [6, 3], [6, 5],

Complement of F

[2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6]

[6, 2], [6, 4], [6, 6]

Therefore:

E'∪F'=[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5], [2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6], [6, 2], [6, 4], [6, 6]

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Step-by-step explanation:

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