What is the answer for Expanding <br> 2t(6t-3)

On Julie's trip to the mountains she traveled 192 miles using 6 gallons of gas. On the return trip she traveled 235 miles on 8 g

Tamiku [17]

Julie travelled with 30.5 miler per gallon.

<u>SOLUTION: </u>

Given, On Julie's trip to the mountains she traveled 192 miles using 6 gallons of gas.

On the return trip she traveled 235 miles on 8 gallons of gas.

We have to find how many miles per gallon did Julie get on the way there?

We know that,

$\text { number of miles per gallon }=\frac{\text { total distance travelled }}{\text { total number of gallons }}$

Substituting the values we get,

$=\frac{192+235}{8+6}=\frac{427}{14}=30.5 \text { miles per gallon }$

Hence, Julie travelled with 30.5 miler per gallon on the way there on mountains.

5 0

3 years ago

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: