<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:
r ll s
Step-by-step explanation:
The easiest way to solve thi is to draw a picture .
P and Q are parallel, so draw those as two opposite seides of a square.
R and P are perpendictular so draw that as the top line of the square. Since it is perpendicular to P, R has to be perpendictular to Q.
Draw line S as the bottom line of the square and the same rules apply, so R is parallel to S
360-142 and you get ur answer unless it’s not a 360 degree
Answer: 42 cents
Step-by-step explanation: Take 10.08 divide it by 24 and there's your answer!
Answer:
It would be the transitive property! If you look up "If a = b, and b = c, then a = c" you'd see the transitive property. I hope this helps!
