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2 years ago

How many times was the number cube rolled

Mathematics

1 answer:

snow_lady [41]2 years ago

5 0

The ball was rolled 50 times.

6+4+8+12+10+10 = 50

Hope this helps; have a great day!

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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

2 years ago

How many ways can four people (A, B, C, D) sit in a row at a move theater if C and D must sit next to each other?

Novay_Z [31]

Answer:

24 ways in four people.

6 0

2 years ago

What is the area of the triangle

ehidna [41]

Answer:

45.1 km^2

hope this helps

have a good day :)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A random sample of 864 births in a state included 426 boys. Construct a 95% confidence interval estimate of the proportion of bo

oksian1 [2.3K]

Using the z-distribution, it is found that the 95% confidence interval is (0.46, 0.526), and it does not provide strong evidence against that belief.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We have that a random sample of 864 births in a state included 426 boys, hence the parameters are given by:

$n = 864, \pi = \frac{426}{864} = 0.493$

Then, the bounds of the interval are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 - 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.46$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 + 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.526$

The 95% confidence interval estimate of the proportion of boys in all births is (0.46, 0.526). Since the interval contains 0.506, it does not provide strong evidence against that belief.

More can be learned about the z-distribution at brainly.com/question/25890103

4 0

2 years ago

Math problem I need help please it will be great if you help btw don't take this question down its is a school question

Maksim231197 [3]

The first question’s answer is the last option. And the range is 104

7 0

2 years ago