In a sheet metal operation, three identical notches and four identical bends are required. if the operations can be done in any order, how many different possible sequences are there to complete the manufacturing?

Step-by-step explanation:

Three identical notches and four identical bends are required in the sheet metal operation.

In total 7 things are required in the metal operation.

We can think it as we need to put the 3 notches and 4 bends in 7 places.

First, lets put the 3 notches in 3 places.

In order to do so, we need to choose 3 places from the 7 places.

We can choose 3 places in ways.

The 3 notches can be arrange in 3! = 6 ways.

The 4 bends can arrange in 4! = 24 ways.

Thus, in total different possible sequences.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497