Answer:
Grouping and then use the common factor. Therefor the answer is 6(2p+5)
Step-by-step explanation:
I hope this helps.
In 1, t<span>here are 6 outcomes for each die, so for three dice, the total combination is 6 x 6 x 6 = 216 outcomes. Hence, t</span><span>he probability of any individual outcome is 1/216 </span>
The outcomes that will add up to 6 are
<span>1+1+4 </span>
<span>1+4+1 </span>
<span>4+1+1 </span>
<span>1+2+3 </span>
<span>1+3+2 </span>
<span>2+1+3 </span>
<span>2+3+1 </span>
<span>3+1+2 </span>
<span>3+2+1 </span>
<span>2+2+2 </span>
<span>Hence the probability is </span><span>10/216 </span>
In 3, the minimum sum of the three dice is 3. so we start with this
<span>P(n = 3) </span>
<span>1+1+1 ; </span><span>1/216 </span>
<span>P(n = 4) </span>
<span>1+1+2 </span>
<span>1+2+1 </span>
<span>2+1+1 ; </span><span>3/216 </span>
<span>P(n = 5) </span>
<span>1+1+3 </span>
<span>1+3+1 </span>
<span>3+1+1 </span>
<span>1+2+2 </span>
<span>2+1+2 </span>
<span>2+2+1; </span><span>6/216
The sum in 3 is 10/216 or 5/108</span>
Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Answers:
x = 72
y = 83
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Explanation:
Angle VFG is 50 degrees. The angle adjacent to this is angle EFG which is 180-50 = 130 degrees.
Angle HDW is 77 degrees. The supplementary angle adjacent to this is 180-77 = 103 degrees which is angle EDH.
Pentagon EFGHD has the following five interior angles
- E = x
- F = 130
- G = 170
- H = 65
- D = 103
Note that angles F = 130 and D = 103 were angles EFG and EDH we calculated earlier.
For any pentagon, the interior angles always add to 180(n-2) = 180(5-2) = 180*3 = 540 degrees.
This means,
E+F+G+H+D = 540
x+130+170+65+103 = 540
x+468 = 540
x = 72
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Now focus your attention on triangle THS
We see that the interior angles are
The angle H is 65 degrees because it's paired with the other 65 degree angle shown. They are vertical angles.
For any triangle, the angles always add to 180
T+H+S = 180
y+65+32 = 180
y+97 = 180
y = 180-97
y = 83