Two airplanes are flying in the same direction in adjacent parallel corridors. at time t = 0, the first airplane is 10 km ahead of the second one. suppose the speed of the first plane (km/hr) is normally distributed with mean 550 and standard deviation 9 and the second plane's speed is also normally distributed with mean and standard deviation 535 and 9, respectively. a) what is the probability that after 2 hr of flying, the second plane has not caught up to the first plane? (round your answer to four decimal places.)b) determine the probability that the planes are separated by at most 10 km after 2 hr. (round your answer to four decimal places.)

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet