<span>There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).</span>
Answer:
Step-by-step explanation:
Ok so I think this is what you are looking for
y ≈ -6x + 30for the first one and
y ≈ -6x + -22
If this is not what you're looking for I'm sorry.
I try my best.
Answer:
i think its D
Step-by-step explanation:
19/4 = 4.75
√20 = 4.47213595
13/3 = 4.3 ( reapeating )
Answer:
0.9332
Step-by-step explanation:
2.5 is 10% of 25. 2.5 * 6 = 15 = 60%
25 - 15 = 10. You have 10 words left to memorize.
I hope this helps.
