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jekas [21]
3 years ago
11

Rewrite the expression as an equivalent power with a negative exponent. a. 1/49 b. 27/125

Mathematics
1 answer:
Hunter-Best [27]3 years ago
3 0
\dfrac{1}{49}=\left(\dfrac{1}{7}\right)^2=7^{-2}\\
\dfrac{27}{125}=\left(\dfrac{3}{5}\right)^3=\left(\dfrac{5}{3}\right)^{-3}
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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)

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kkurt [141]

Answer:

2 (4x+3y)

Step-by-step explanation:

Hope this helps :)

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