By inspection, it's clear that the sequence must converge to
because
when
is arbitrarily large.
Now, for the limit as
to be equal to
is to say that for any
, there exists some
such that whenever
, it follows that
From this inequality, we get
As we're considering
, we can omit the first inequality.
We can then see that choosing
will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that
.
<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
</span>
You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
</span>
Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
</span><span>
To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
</span>
Answer:
aa is the repuestos yes ok
reduce the index of the radical and exponent with 3, rhen you get your answer, 6
