1.928•10^7 in standard notation

Show that if X is a geometric random variable with parameter p, then

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

Marco states that 7.696696669…… is a rational number because it is a repeating decimal. Is he correct? Justify your answer.

Allisa [31]

Answer:

no because at the end it is 669 not 696 so it is not repeating decimal

3 0

3 years ago

Read 2 more answers

Anthony is an executive assistant. He makes $15.85 per hour. If he worked 47.50 hours this week, what is his (a) OT pay? What ab

alexdok [17]

Anthony's OT pay will be $118.875 and his overall pay will be $752.875.

Number of hours Anthony has to work in a week = 5 × 8 = 40 hours.

He worked for = 47.50 hours.

Overtime = 47.50 hours - 40 hours = 7.50 hours

Pay per hour = $15.85 / hour

Overtime pay (OT pay) = $15.85 × 7.50 = $118.875

Overall pay = $15.85 × 47.50 = $752.875

Therefore, Anthony's OT pay is $118.875 and his overall pay is $752.875.

Learn more about overtime pay here -

brainly.com/question/19022439

#SPJ9

5 0

1 year ago

Solve this for Brainliest! See the attachment [Level - GCSE]

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

Using Pythagoras

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

Using quadratic formula

$x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$

$x = 3 + 4\sqrt{3} = 9.928(4s.f.)$

OR

$x = 3 - 4\sqrt{3} = -3.928(4s.f.)$ (Length can't be negative)

∴ $x = 3 + 4\sqrt{3}$

3 0

2 years ago

Read 2 more answers

Solve for h and c to find what h and c equals; 34.50=4h+7c and 30.50=8h+3c

Irina18 [472]

34.5=4h+7c
(34.5-7c)/4=h
Put h= (34.5-7c)/4 into another equation
30.5=8h+3c
30.5=2(34.5-7c)+3c
30.5=69-14c+3c
11c=38.5
c=38.5/11
Put c=38.5/11 into one of the equations
to calculate the h.

4 0

3 years ago