y = -3/7 x + 18 would be the equation of the line y intercept is where x = 0 at 0,18
x intercept occurs where y = 0
0 = -3/7x +18
18 = 3/7 x x = 18 (7/3) = 42 the x intercept is 42,0
Area = h x b and h = 3b-3
(3b-3) * b = 60
3b^2 - 3b - 60 = 0 solve for b the h = 3b-3 (factor or use quadratic fromula to find 'b')
x x+ 2 x+4
4x- ( x+4) = 2 (x+2)+6
3x - 4 = 2x+10
x = 14 14 16 18
x = original amount
2x-100 = after 1st temple
(2)(2x-100) - 100 after second temple = 4x-200-100 = 4x-300
2(4x-300) -100 = 8x-600-100 = 8x-700 after third temple
2 (8x-700) - 100 = 0 after last temple
16x-1500 = 0
16x = 1500
x = 93.75
More correctly:
Expected value
probability of a 3 is .8 .8 x3 = 2.4
9 is .2 .2 x9 = 1.8 expected value 2.4 + 1.8 = 4.2
add a nine prob of a 3 is now 8/11 8/11 x 3 = 2.1818
prob of a 9 is now 3/11 3/11 x 9 = 2.4545 expected value 2.1818+2.4545 = 4.63
I think expected would be (3 + 3+ 3+ 3+ 3+3 +3 +3 +9 +9 )/10= 4.2
add a nine : (3 + 3+ 3+ 3+ 3+3 +3 +3 +9 +9 +9)/11 = 4.63
add two nines = 5
etc .....
If you think about it, this makes sense.... John runs a meter faster than David does....that is why he wins the 100 meter in the first place.
adding ONE meter to each of their distances will result in John running THAT meter faster than David too....and John wins again !
\(\sum_{0}^{44}\) n =990
\(\sum_{0}^{45}\) n = 1035 HE MUST HAVE COUNTED '10' TWICE !
Let 'x' be the time john runs 100 'x' is also the time david runs 99
rate = 100/x rate = 99/x
for 101 meters for 100 m
time = 101m/ (100/x) = 101/100 = 1.01 x time = 100 m / (99/x) = 100/99 x = 1.010101x
So John will STILL win the race ! (by a hair !) because david's time is greater......
Hmmmm ... isnt it just 30 sec/120 sec = 1/4 ?
