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riadik2000 [5.3K]
3 years ago
7

Define ????: ℝ → {0,1} as the characteristic function of ℤ+, the set of positive integers. That is, ????(x) = ????A (x), for the

set A of positive integers = { n | n ∈ ℤ+ }.

Mathematics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

Step-by-step explanation: steps explained in attached file

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C(4+3c) -0.75 (c+3) =<br><br> plzz help
Andreyy89

Answer:

if I am not mistaken the answer should be

11.25C?

Step-by-step explanation:

I did like terms to add 3c to 3c plus 4c minus .75 which would equal 11.25 c if it were to ask for expression then 12c-.75

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3 years ago
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If anyone knows how to obtain the answer, please notify me.
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Answer:

Area =  

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Step-by-step explanation:

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The new e-reader measures 8 inches long, 5 inches wide, and 2 inches deep. which is the correct unit of measurement to use when
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Inches cubed. Or inches^3
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Show that if X is a geometric random variable with parameter p, then
Lubov Fominskaja [6]

Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

\sum_{k=1}^{\infty} b^{k-1}

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

And with this we have the requiered proof.

And since b=1-p we have that:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

4 0
3 years ago
What could this answer be??
bekas [8.4K]

The number in the parentheses is the rate of change. Because this number is less than 1 it is a decrease, so it is a decay.

The percent decrease is 1 - 0.63 = 0.37 = 37% decrease

5 0
3 years ago
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