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AnnZ [28]
3 years ago
11

Find the sum of the infinite geometric series, if it exists.

Mathematics
1 answer:
weeeeeb [17]3 years ago
4 0

So the series looks something like this:

\Sigma_{n=4}^{\infty}\frac{n-1}{16} \\\frac{1}{16}\Sigma_{n=4}^{\infty}n-1

If \lim_{n\rightarrow\infty}\neq than \Sigma{a_n} diverges. So we must apply limit infinity property:

\lim_{n\rightarrow\infty}(ax^n+\dots+bx+c)=\infty, a>0 and n is odd.

So...

\lim_{n\rightarrow\infty}n-1=\infty

The series diverges.

Hope this helps.

r3t40

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Cerrena [4.2K]
Suppose D is the event that a given patient has the disease, and P is the event of a positive test result.

We're given that

\mathbb P(D)=0.01
\mathbb P(P\mid D)=0.98
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where A^C denotes the complement of an event A.

a. We want to find \mathbb P(P^C). By the law of total probability, we have

\mathbb P(P^C)=\mathbb P(P^C\cap D)+\mathbb P(P^C\cap D^C)

That is, in order for P^C to occur, it must be the case that either D also occurs, or D^C does. Then from the definition of conditional probability we expand this as

\mathbb P(P^C)=\mathbb P(D)\mathbb P(P^C\mid D)+\mathbb P(D^C)\mathbb P(P^C\mid D^C)

so we get

\mathbb P(P^C)=0.01\cdot0.02+0.99\cdot0.95=0.9407

b. We want to find \mathbb P(D\mid P). Now, we can use Bayes' rule, but if you're like me and you find the formula a bit harder to remember, we can easily derive it.

By the definition of conditional probability,

\mathbb P(D\mid P)=\dfrac{\mathbb P(D\cap P)}{\mathbb P(P)}

We have the probabilities of P/P^C occurring given that D/D^C occurs, but not vice versa. However, we can expand the probability in the numerator to get a probability in terms of P being conditioned on D:

\mathbb P(D\cap P)=\mathbb P(D)\mathbb P(P\mid D)

Meanwhile, the law of total probability lets us rewrite the denominator as

\mathbb P(P)=\mathbb P(P\cap D)+\mathbb P(P\cap D^C)

or in terms of conditional probabilities,

\mathbb P(P)=\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)

so that

\mathbb P(D\mid P)=\dfrac{\mathbb P(D)\mathbb P(P\mid D)}{\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)}

which is exactly what Bayes' rule states. So we get

\mathbb P(D\mid P)=\dfrac{0.01\cdot0.98}{0.01\cdot0.98+0.99\cdot0.05}\approx0.1653
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Brrunno [24]

Answer:

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Step-by-step explanation:

We are asked to determine whether the given side lengths are sides of a right triangle or not.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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