A drag racer has two parachutes, a main and a backup, that are designed to bring the vehicle to a stop (combined with the brakes) after the end of a run. Suppose that the main chute deploys with probability 0 . 95, and that if the main fails to deploy, the backup deploys with probability 0.999. Required:
a. What is the probability that one of the two parachutes deploys?
b. What is the probability that the backup parachutes deploys?

a. 0.99995 = 99.995% probability that one of the two parachutes deploys

b.0.04995 = 4.995% probability that the backup parachutes deploys

Step-by-step explanation:

a. What is the probability that one of the two parachutes deploys?

Either the first deploys(with probability 95%), or the first does not deploy(probability 5%) and the second does(probability 99.9%). So

0.99995 = 99.995% probability that one of the two parachutes deploys

b. What is the probability that the backup parachutes deploys?

The first does not deploy(probability 5%) and the second does(probability 99.9%). So

0.04995 = 4.995% probability that the backup parachutes deploys

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

i think c. but don't trust it.

step-by-step explanation: