The answer would be D.
you would start by choosing one of the variables to eliminate, i chose to eliminate X first. in order to eliminate X, i had to multiply the second equation by -2. X then cancelled out, and i solved for the Y the same way you would a 1 step equation. Y=-4
I then took the first original equations, plugged Y in, and solved for X. X=3.
Hope this helps!
Answer:
A, C are true . B is not true.
Step-by-step explanation:
Mean of a discrete random variable can be interpreted as the average outcome if the experiment is repeated many times. Expected value or average of the distribution is analogous to mean of the distribution.
The mean can be found using summation from nothing to nothing x times Upper P (x) , i.e ∑x•P(x).
Example : If two outcomes 100 & 50 occur with probabilities 0.5 each. Expected value (Average) (Mean) : ∑x•P(x) = (0.5)(100) + (0.5)(50) = 50 + 25 = 75
The mean may not be a possible value of the random variable.
Example : Mean of possible no.s on a die = ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 21/6 = 3.5, which is not a possible value of the random variable 'no. on a die'
X + y = 24
x - y = 2
Adding both equations
2x = 26
x = 13
y = 11
The delivery driver has to make deliveries at 5 locations <span>among the 6 locations. </span>This means the order of the probability is important because the route he will take from A to B is different with A to C.
So, you need to use permutation for this problem. The calculation would be:
6P5= 6!/ (6-5)!= 720 different routes
