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Verizon [17]
2 years ago
14

Anser rigt know plesew

Mathematics
1 answer:
Debora [2.8K]2 years ago
8 0

Answer:

honarol

Step-by-step explanation:

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How many variables in are the following expression? 8 + b -12c + m
Alex

Answer:

3 variables.

Step-by-step explanation:

b, c, and m are all variables.

6 0
2 years ago
PLEASE HELP ME WITH THIS!!!
ELEN [110]
You do 48 divided by 2 so that you can put in your answer. any other questions ask me
3 0
3 years ago
Translate this sentence into an equation.
gregori [183]
63 = 10 + m
hope this helps :)
4 0
3 years ago
At noon, the temperature in Ebonsville, Texas was 29℉. By midnight, the temperature had fallen to −10℉. What was the change in t
UkoKoshka [18]

Answer:

i believe it's 39

Step-by-step explanation:

29 -39 = -10

8 0
3 years ago
Somebody please assist me here
Anettt [7]

The base case of n=1 is trivially true, since

\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)

but I think the case of n=2 may be a bit more convincing in this role. We have by the inclusion/exclusion principle

\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)

with equality if E_1\cap E_2=\emptyset.

Now assume the case of n=k is true, that

\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)

We want to use this to prove the claim for n=k+1, that

\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

The I/EP tells us

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)

and by the same argument as in the n=2 case, this leads to

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})

By the induction hypothesis, we have an upper bound for the probability of the union of the E_1 through E_k. The result follows.

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

5 0
2 years ago
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