Correct question is;
A bag contains 10 counters. 6 of them are white. A counter is taken at random and not replaced. A second counter is taken out of the bag at random. Calculate the probability that only one of the two counters are white
Answer:
probability that only one of the two counters is white = 8/15
Step-by-step explanation:
To solve this question, first of all, let's look at probability we would have to either draw two white counters or two non-white counters (4/10 * 3/9).
Probability(draw 2 white counters) = (6/10 × 5/9) = 30/90 = 1/3
Probability(draw 2 non-white counters) = (4/10 × 3/9) = 2/15
Now, In all other cases, we'll draw exactly one white and one non-white counter, so the odds of this would be;
P(one white counter and one non-white counter) = 1 - [1/3 + 2/15)
= 1 - 7/15 = 8/15
Not a 100% positive but I believe that your answer would be C.
4x-16+14=16-4x+6.
4x-16+14-16+4x-6=0(bring all values at one side)
4x+4x+14-16-16-6=0(arrange)
8x-24=0(solve)
8x=24(bring 8 on right hand side so x is alone)
X=24/8(solve)
X=3
Answer:
80
Step-by-step explanation:
-400 / -5 = 40
Hope that helps!
Do you have a picture I can’t answer just off of this sorry
