Asoccer field for a youth league has an area of 4000 square yards and is 60% longer than it is wide what is the length of the field

Let

width,w

length, l = 1.6w ...eqn 1

area, a = l × w ...eqn 2

subst for l in eqn 2

a = 1.6 w × w = 4000

4000 = 1.6 w^2

w^2 = 2500

w = 50

subst for w in eqn 1

l = 50 x 1.6 = 80

length = 80 yds

width = 50 yds

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,