Answer:
A) The best way to picture this problem is with a probability tree, with two steps.
The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.
The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.
If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.
If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.
B) So a person can have a red marble and a blue marble in two ways:
1) Picking the red first and the blue last
2) Picking the blue first and the red last
C) P(R&B) = 3/5 = 60%
Step-by-step explanation:
C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5
Answer:
Step-by-step explanation:
![6k^3+10k^2-56k\\2k(3k^2+5k-28)\\=2k[3k^2+12k-7k-28]\\=2k[3k(k+4)-7(k+4)]\\=2k(k+4)(3k-7)](https://tex.z-dn.net/?f=6k%5E3%2B10k%5E2-56k%5C%5C2k%283k%5E2%2B5k-28%29%5C%5C%3D2k%5B3k%5E2%2B12k-7k-28%5D%5C%5C%3D2k%5B3k%28k%2B4%29-7%28k%2B4%29%5D%5C%5C%3D2k%28k%2B4%29%283k-7%29)
common factor is 3k-7
P = 4r + 3t
4r = p - 3t
r = (p - 3t)/4
what's up? the answer to this is 339+340+341
best of luck with your studies
Answer:
what is the quastion
Step-by-step explanation:
plz
