7/3-<span>1.41
7/1.59
4.4
That's what i got...
</span>
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
Answer: The y-intercept is (0, -12)
Step-by-step explanation:
y = 40x - 12
y = 40(0) - 12
y = -12
90 ? Because it's rounding to the nearest tenth so .... it must be 90
