Answer:
9/16
Step-by-step explanation:
Since it is only 2 sets of 4 coices it would be super easy to write them all out.
Plant 1 Clay
Plant 1 Plastic
Plant 1 Metal
Plant 1 Wood
Plant 2 Clay
Plant 2 Plastic
Plant 2 Metal
Plant 2 Wood
Plant 3 Clay
Plant 3 Plastic
Plant 3 Metal
Plant 3 Wood
Cactus Clay
Cactus Plastic
Cactus Metal
Cactus Wood
16 choics total, take out all with a cactus or clay pot
Plant 1 Plastic
Plant 1 Metal
Plant 1 Wood
Plant 2 Plastic
Plant 2 Metal
Plant 2 Wood
Plant 3 Plastic
Plant 3 Metal
Plant 3 Wood
9 choices left, so probability is 9/16
to do it for any number find the total number of choices again, so 16 then you subtract the number of each category you are excluding. so there were 4 cacti and 4 clay, so subtract 8, to get 8. Then you add back any that had both clay and cacti, which there was one, whn you had a cactus in a clay pot, so add back 1. This gets us 9 again.
This is called the inclusion exclusion principle. Might want to look into this specifically.
Answer:
it should be 14.25
Step-by-step explanation:
The equation for finding slope is y2-y1 over x2-x1.
8-2=6
-6-4=-10
The slope is 6/-10
Answer:
Its probably 6.4 I dont know if I'm correct because I forget stuff
<span>In order to prove this for all integers n ≥ 0, we first prove the basis step P(0) and then prove the
inductive step, that P(k) implies P(k + 1).
1. BASIS STEP: Now in P(0), the left-hand side has just one term, namely 2, and the right-hand side is
1â’(â’7)1
4 =
8
4 = 2. Since 2 = 2, we have verified that P(0) is true.
2. INDUCTIVE STEP: For the inductive step, we assume that P(k) is true and and derive from it the truth of P(k + 1), which is the equation
2 â’ 2 · 7 + 2 · 72 ⒠· · · + 2(â’7)k + 2(â’7)k+1 =1 â’ (â’7)(k+1)+1/4
To prove an equation like this, it is usually best to start with the more complicated side and manipulate
it until we arrive at the other side. In this case we start on the left. Note that all but the last term
constitute precisely the left-hand side of P(k), and therefore by the inductive hypothesis, we can replace
it by the right-hand side of P(k). The rest is algebra:
2 â’ 2 · 7 + 2 · 72 ⒠· · · + 2(â’7)k + 2(â’7)k+1 =1 â’ (â’7)k+1/4+ 2(â’7)k+1
=1 â’(â’7)k+1 + 8(â’7)k+1/4
=1 + 7(â’7)k+1/4
=1 â’ (â’7) · (â’7)k+1/4
=1 â’ (â’7)(k+1)+1/4</span>