Answer:
20
Step-by-step explanation:
For the sake of the problem, let's make female workers "x" and male workers "y".
x+y<40 This equation shows that the total number of workers has a max of 40.
30x+20y<1,000 This equation shows that the total cost the manager pays ($30 to each woman, $20 to each man) has a max of $1,000.
Now you can solve for x and y.
X+y<40
-y -y
X<-y+40
Substitute -y+40 in for X in the second equation
30(-y+40)+20y<1,000
-30y+1200+20y<1,000 Distribute
-10y+1,200<1,000 Combine like terms
-10y<-200 Subtract 1,200
y>20 Divide by -10; flip the sign
Since y>20, and y=male workers, you now know that the minimum
number of male workers he should send is 20
Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.
The given sequence is:

a(2)=1
a(3)=3
a(4)=9
We are to find the average rate of change between n=3 and n=4 for the given function.
Average rate of change =

So the average rate of change for the given function from n = 3 to n = 4 is 6
I got 10! [_-_]-{-_-}-(_-_)