If the probability of trowing a score of seven with 2 dices is ⅛ ,what is the probability of not throwing a score of seven
1 answer:
You might be interested in
The complete question in the attached figure
we know that
area each pasture=15000 ft²
area each pasture=x*y
so
15000=x*y-----> equation 1
perimeter two rectangular pastures=1050 ft
perimeter two rectangular pastures=2*[2x+y]+y----> 4x+2y+y---> 4x+3y
so
1050=4x+3y----> divide by 3----> 1050/3=(4/3)x+y
clear variable y
y=350-(4/3)x-----> equation 2
the answer part a)
y=350-(4/3)x------> <span>it was proved
</span>
Part b)Find the possible lengths and widths of each pasture.
substitute equation 2 in equation 1
15000=x*y-----> 15000=x*[350-(4/3)x]
15000=350x-(4/3)x²----> multiply by 3----> 45000=1050x-4x²
4x²-1050x+45000=0
using a graph tool----> to resolve the second order equation
see the attached figure
the solution is
x=53.942 ft
x=208.558 ft
15000=x*y-----> y=15000/x
for x=53.94 ft
y=278.09 ft
for x=208.56 ft
y=71.92 ft
the possible lengths and widths of each pasture are
case 1
x=53.94 ft
y=278.09 ft
case 2
x=208.56 ft
y=71.92 ft
we know that
area each pasture=15000 ft²
area each pasture=x*y
so
15000=x*y-----> equation 1
perimeter two rectangular pastures=1050 ft
perimeter two rectangular pastures=2*[2x+y]+y----> 4x+2y+y---> 4x+3y
so
1050=4x+3y----> divide by 3----> 1050/3=(4/3)x+y
clear variable y
y=350-(4/3)x-----> equation 2
the answer part a)
y=350-(4/3)x------> <span>it was proved
</span>
Part b)Find the possible lengths and widths of each pasture.
substitute equation 2 in equation 1
15000=x*y-----> 15000=x*[350-(4/3)x]
15000=350x-(4/3)x²----> multiply by 3----> 45000=1050x-4x²
4x²-1050x+45000=0
using a graph tool----> to resolve the second order equation
see the attached figure
the solution is
x=53.942 ft
x=208.558 ft
15000=x*y-----> y=15000/x
for x=53.94 ft
y=278.09 ft
for x=208.56 ft
y=71.92 ft
the possible lengths and widths of each pasture are
case 1
x=53.94 ft
y=278.09 ft
case 2
x=208.56 ft
y=71.92 ft