Plastic parts manufactured using an injection molding process may exhibit one or more defects, including sinks, scratches, black spots, and son on. Let x represent the number of defects on a single part, and suppose the distribution of x is as follows: x: 0 1 2 3 4 p(x): 0.79 0.09 0.05 0.04 0.03 What is the expected value, E(X), for the number of defects on a randomly selected plastic part

The expected value for the number of defects on a randomly selected plastic part is 0.43.

Step-by-step explanation:

We have the following probability distribution:

P(X = 0) = 0.79

P(X = 1) = 0.09

P(X = 2) = 0.05

P(X = 3) = 0.04

P(X = 4) = 0.03

What is the expected value, E(X), for the number of defects on a randomly selected plastic part?

Each value multiplied by its probability. So

The expected value for the number of defects on a randomly selected plastic part is 0.43.

part 1 answer = 8

part 2 answer = 72

step-by-step explanation:

part 1 explanation = add all of the numbers, then divide by how many numbers you have. so, 200/25 = 8.

part 2 explanation = not gonna lie, this is hard to explain, so just google how to do it.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3