<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:
I cant see whats the question
Step-by-step explanation:
Givens
Slower plane
r = r
t = 2.5 hours
d = d
Faster Plane
r_faster = 1.5 * r
t = 2.5 hours
d1 = d + 127.5
The time is the same for both
t = d/r
d1/r1 = d/r
(d+ 127.5)/1.5r = d/r Multiply each side by r
(d + 127.5)/1.5 = d Multiply both sides by 1.5
d + 127.5 = 1.5d Subtract d from both sides.
127.5 = 1.5d - d
127.5 = 0.5d Divide by 0.5
127.5 / 0.5 = d
255 = d
Now you can go back and figure out the rates.
First find d1
d1 = d + 127.5
d1 = 255 + 127.5
d1 = 382.5
<em><u>Rate of the slower plane</u></em>
d = 255
t = 2.5 hours
r = d/t
r = 255/2.5
r = 102 miles per hour.
<em><u>Faster plane</u></em>
d1 = 382.5 miles
t = 2.5 hours
r1 = d/t
r1 = 382.5/2.5 = 153 miles per hour.
The answer is 20 if that is an option
