V3+4v2-5v+3v2+12v-15
v3+7v2+7v-15

and surely you know how much that is.
Answer:
Karina is correct
Step-by-step explanation:
<em>See attached</em>
<u>Area that Karina cleaned up:</u>
- 1/2*(6-0)(9-2) = 1/2*6*7 = 21
<u>Area that Petty cleaned up:</u>
As we see Karina cleaned up more area than Petty, so she is correct
Answer:
Step-by-step explanation:
Given that tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the event that the right leg tag is lost. Suppose those two events are independent and A2 the event that the right leg tag is lost. Suppose those two events are independent and P(A1) = P(A2) = 0.3
a) the probability that at least one leg tag is lost
= 
(since A1 and A2 are independent)
=0.3+0.3-0.09
=0.51
b) the probability that exactly one tag is lost, given that at least one tag is lost.
= P(one leg lost)/P(atleast one leg lost)
=
c) the probability that exactly one tag is lost, given that at most one tag is lost.
The answer to your question is c