Answer:
-13
Step-by-step explanation:
MrBillDoesMath!
Answer to #4: 81/256 * s^8 * t^ 12
Comments:
(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.
---------------------
(1) (27s^7t^11)^ (4/3)
= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)
As 27 = 3^3, 27 ^(4/3) = 3^4 = 81
(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)
As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256
So (1)/(2) =
81 * s^(28/3)* t^(44/3)
------------------------------- =
256 s^(4/3) * t^((8/3)
81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =
81/256 * s^(24/3) * t (36/3) =
81/256 * s^8 * t^ 12
MrB
Answer:
YG you and I don't know what to do with it when I get home from work and cheese for dinner and cheese for dinner yet lol I don't know what to do with it yyyyyyy lol I don't know
Answer:
Example:
A bag contains 3 black balls and 5 white balls. Paul picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.
a) Construct a probability tree of the problem.
b) Calculate the probability that Paul picks:
i) two black balls
ii) a black ball in his second draw
Solution:
tree diagram
a) Check that the probabilities in the last column add up to 1.
b) i) To find the probability of getting two black balls, first locate the B branch and then follow the second B branch. Since these are independent events we can multiply the probability of each branch.
ii) There are two outcomes where the second ball can be black.
Either (B, B) or (W, B)
From the probability tree diagram, we get:
P(second ball black)
= P(B, B) or P(W, B)
= P(B, B) + P(W, B)
Answer:
0
Step-by-step explanation:
−2−(−4)−2
=2−2
=0
