Write a radical expression as an exponential expression

A number cube with faces labeled from 1 to 6 will be rolled once.

tresset_1 [31]

Question:

A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling a number greater than 2.

If there is more than one element in the set, separate them with commas.

Answer:

$S = \{1,2,3,4,5,6\}$

$Greater2 = \{3,4,5,6\}$

Step-by-step explanation:

Given

A roll of a 6 sided number cube

Solving (a): The sample space

This implies that we list out all number on the number cube.

So:

$S = \{1,2,3,4,5,6\}$

Solving (b): Outcomes greater than 2

This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.

So:

$Greater2 = \{3,4,5,6\}$

3 0

2 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:

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

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

2 years ago

You’re walking down the hall and you overhear someone talking about you behind your back. What would you do and why?

Alina [70]

Answer: Go up and confront them because i hate when people do that!

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The diagram shows the measurements of a frame into which concrete is poured to make steps. The steps are shaped like rectangular

Ivahew [28]

(6 x 16 x 6) + (14 x 16 x 7)
Answer = B

4 0

2 years ago

The equation for the cost in dollars of producing computer chips is y =.000015x^2-.03x+35. Where x is the number of chips produc

Anna35 [415]

In order to find the number of chips that would result in the minimum cost, we take the first derivative of the given equation. Note that the derivative refers to the slope of the graph at a given point. We can utilize this concept knowing that at the minimum or maximum point of a graph, the slope is zero.

Taking the derivative of the given equation and equating it to zero, we have:

y' = (0.000015)(2)x - (0.03)x° + 0
0 = (0.00003)x - 0.03

Solving for x or the number of chips produced, we have x = 1000. We then substitute this value in the given equation, such that,

y = (0.000015)(1000)² - (0.03)(1000) + 35

The minimized cost, y, to produce 1000 chips is then calculated to be $20.

6 0

2 years ago