Of the 20 members of a kitchen crew, 17 can use the meat-cutting machine, 18 can use the bread-slicing machine, and 15 can use both machines. If one member of the crew is to be chosen at random, what is the probability that the member chosen will be someone who cannot use either machine?

Step-by-step explanation:

The total number of members in the crew  = 20

The number of people able to use meat machine  = 17    ⇒n (M)  = 17

The number of people able to use bread machine  = 18  ⇒n (B)  = 18

The number of people able to use both machine  = 15 ⇒n (M∩B)  = 15

Now, a person is chosen at random:

Let us assume the number of people who cannot use either machine = K

Now,as we know:

So, the number of people who can use either of the machines  

=n (M) + n(B)  -  n(M∩B)  = 17 + 18 - 15  = 20

So, there are 20 people who can use either of the two machines.

Also, as given the total number of people in the crew = 20.

Hence, from the above statement it is clear that there exists no person who can not use either of the machines.

So, such probability = 0

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

$15 each

step-by-step explanation:

$45/3 children= $15 each