<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>
The inner circle circumference is 18 and the outer is 4
Answer:
For part 1- y=50x+75 For part 2- y=325
Step-by-step explanation:
the cost is increasing 50 for every hour
there is a flat rate of 75 to answer a house call
y=50x+75
y=50(5)+75
y=250+75
y=325
<h3><u>D</u></h3>
If ΔXYZ ≅ ΔDEF, then X = D, Y = E and Z = F.
