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
The formula for an area of a circle is A= pi(r)^2
so if you plug in 5 to r (because it’s the radius) you should get A= 78.5
P(sum = 6) = 5/36 ; P(sum = 12) = 1/36
expected payoff = (5/36)(5) + (1/36)(9) = $0.94
Answer:
Step-by-step explanation:
x²-8x-20
=x²-10x+2x-20
=x(x-10)+2(x-10)
=(x-10)(x+2)
second factor is x+2
Answer:
$8.95
Step-by-step explanation:
you add 7+6+9=22 196.90 divided by 22 is 8.95 hope this helps
