Find the point, M, that divides segment AB into a ratio of 2:3 if A is at (0,15) and B is at (20,0)
1 answer:
The ratio has a total of 2+3=5 units, of which 2 correspond to the length of segment AM. The difference of x-coordinates Bx-Ax = 20-0 = 20. 2/5 of that amount is (2/5)*20 = 8, so the x-coordinate of M is
Mx = 8 + Ax = 8 + 0 = 8
The difference of y-coordinates is 0 - 15 = -15. 2/5 of that amount is (2/5)*(-15) = -6, so the y-coordinate of M is
... My = -6 + Ay = -6 + 15 = 9
The point M is (Mx, My) = (8, 9).
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<h3><u>Answer:</u></h3>
- The chance of getting a 5 or 6 in the next turn is about 40%.
- => (5 + 3) ÷ (3 + 3 + 0 + 6 + 3 + 5)
- => 8 ÷ 20
- => 8/20
- => 4/10
- => 0.40
- => 0.40 x 100
- => 40%
Therefore, the chance of getting a 5 or 6 in the next turn is about 40%.
Hoped this helped
there's a quick graph below of the bounds and the tangent at "c"
not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant
