Anature preserve has a population of fourteen black bears. they have been tagged #1 through #14, so they can be observed over time. two of them are randomly selected and captured for evaluation. one is tested for worms and one is tested for ticks. what is the probability that bear #3 is tested for worms and bear #5 is tested for ticks?

1/182 or 0.55%

Step-by-step explanation:

There are 2 ways to approach this problem (and both give the same answer!).

First method: probability for each test.

First test, worms: There's 1/14 chances for bear #3 to be selected, since there are 14 bears total.

Second test, ticks: There's 1/13 chances for bear #5 to be selected, since only 13 bears remain (#3 cannot be tested for both ticks and worms).

So, to obtain the total probability, we multiply 1/14 with 1/13 = 1/182

Second method: permutations

We can see the problem from a permutations perspective.  How many permutations can we have for 2 bears to be tested? It's a permutation since the order is important. (#3 ticks + #5 worms ≠ #3 worms + #5 ticks)

There are 182 permutations possible, ONE of which is bear #3 is tested for worms and bear #5 is tested for ticks.

So, once again, 1/182.

i really dont know

step-by-step explanation:

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2