Y = - x + 6
m = - 1
0 = - 1 ( 0 ) + c
c = 0
y = - x
x + y = 0
Do the same to the other side
<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:
x = 1/4 ± 1/4√233
Step-by-step explanation:
(2x + 5)(x - 3) = 14
~Use FOIL on the left side
2x² - 6x + 5x - 15 = 14
~Combine like terms
2x² - x - 15 = 14
~Subtract 14 to both sides
2x² - x - 29 = 0
~Use the quadratic formula and simplify
x = 1/4 ± 1/4√233
Best of Luck!
Answer:
5
Step-by-step explanation:
z = -3 + 8 = 5
