A factory received a shipment of 22 flanges, and the vendor who sold the items knows there are 6 flanges in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the flanges in the sample are defective, he will refuse the shipment. For each of the following, give your responses as reduced fractions. A. If a sample of 5 flanges is selected, find the probability that all in the sample are defective. B. If a sample of 5 flanges is selected, find the probability that none in the sample are defective.

A. 1/4,389

B. 728/4,389

Step-by-step explanation:

Total number of flanges = 22

Defective flanges = 6

Non-defective flanges = 16

A. The probability that all in the sample are defective is given by:

The probability is 1/4,389.

B. The probability that none in the sample are defective is given by:

The probability is 728/4,389.

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!