<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:
6
Step-by-step explanation:
I am not completely sure.
3x
if x=4 then it would be 12
if x=6 then it would be 18
if x=9 then it would be 27
if x=10 then it would be 30
the side is shorter than 19 but definitely longer than 6. To me, it looks like a little more than double, more of triple and 6 times 3 is 18.
Answer:
20 cents
Step-by-step explanation:
Divide 3 dollars by 15 caterpillars.
3/15 = 0.2
20 cents
Answer:
m = 8
Step-by-step explanation:
-3m - 4 = -28
+4 +4
-3m = -24
÷-3 ÷-3
m = 8
Answer:
B. -6
Step-by-step explanation:
-6 + 10 =4
