Answer:
I believe the pattern is linear and for each month that goes by she earns $50
Answer:
0.4 0.044 0.04 0.004
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
if it isnt D its B but im pretty sure its D
Answer:
i got the same problem
Step-by-step explanation:
Answer:
a) ![N= (5C1) *(5C1) *(5C1) = 5*5*5 = 125](https://tex.z-dn.net/?f=%20N%3D%20%285C1%29%20%2A%285C1%29%20%2A%285C1%29%20%3D%205%2A5%2A5%20%3D%20125)
b) ![N = (5C1)*(4C1) *(3C1) = 5*4*3 = 60](https://tex.z-dn.net/?f=%20N%20%3D%20%285C1%29%2A%284C1%29%20%2A%283C1%29%20%3D%205%2A4%2A3%20%3D%2060)
Step-by-step explanation:
For this case our sample space is the 5 letters given:
![S= [F,G, H, I, J]](https://tex.z-dn.net/?f=%20S%3D%20%5BF%2CG%2C%20H%2C%20I%2C%20J%5D)
And we want to find the number of three letter words can be made from the sample space with some conditions
Part a
For this case the repetition is allowed so then each time we will have 5 possibilites in order to select one letter so if we use combinatories we have:
![N= (5C1) *(5C1) *(5C1) = 5*5*5 = 125](https://tex.z-dn.net/?f=%20N%3D%20%285C1%29%20%2A%285C1%29%20%2A%285C1%29%20%3D%205%2A5%2A5%20%3D%20125)
So then we will have 125 possible combinations of 3 words letters with the 5 provided
We need to remember that ![nC x = \frac{n!}{(n-x)! x!}](https://tex.z-dn.net/?f=%20nC%20x%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-x%29%21%20x%21%7D)
Part b
For this case the repetition is not allowed so then the possible number of possibilities are:
![N = (5C1)*(4C1) *(3C1) = 5*4*3 = 60](https://tex.z-dn.net/?f=%20N%20%3D%20%285C1%29%2A%284C1%29%20%2A%283C1%29%20%3D%205%2A4%2A3%20%3D%2060)
So then we will have 60 possible combinations of 3 words letters with the 5 provided