A manufacturer makes three types of screws. Type A comes in a bulk pack of 1000 screws. Type B comes in a bulk pack of 500 screws. Type C comes in a bulk pack of 800 screws. Type A screws each have a probability of .001 of having a flaw. The probability is .003 for Type B and .005 for Type C.
(a) If a contractor buys one pack of each type of screws, find the expected total number of defective screws among the three packs.
(b) Find the probability that there are no more than 5 defective screws total among the three packs.
(c) Find the expected number of packs that have no more than 5 defective screws. (Your answer will be between 0 and 3.)

Step-by-step explanation:

Given that a manufacturer makes three types of screws

Type             A        B           CTotal

Pack         1000500  800  

P for flaw0.0010.030.005  

Pack*p                 115                    4     20

a) Expected total number of defective screws = 20

b) For one pack each no of defective screws =20

Hence for 5 expected number number of packs form each should be 1/3

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

it is true

step-by-step explanation:

all the combinations add up i used   calculator