If a person weighs 185 lb on the surface of the earth (assume this to be 3960 mi from the center), how much will he weigh 1840 mi above the surface of the earth?

The problem can be solve using the equation of law of gravity:

F = G(m1 m2)/r^2

Whre G = gravitational constant = 6.674×10−11 N · (m/kg)^2)

F is the force between the masses

m1 is the first mass (person)

m2 is the second mass (earth) = 5.98 × 10^24 kg

r is the distance between the centers of the masses

convert the given value to si units

3960 + 1840 mi = 5800 mi = 9334195 m

185 lb = 83.9146 kg

F = (6.674×10−11 N · (m/kg)^2) *(83.9146 kg)*( 5.98 × 10^24 kg)/( 9334195 m)^2

F = 9.22x10^-21 N

M =  9.22x10^-21 N/9.81m/s2 =9.40x10^-22 kg

the fist one is not equalvalent but i think th second one isand the third isnt definetly or the last one is i think. hope this !

step-by-step explanation:

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