What is 17,242 rounded to the nearest 10,000

Jon performs 3 transformations on a pentagon. First, he reflects it over the x-axis. Then, he translates it 3 units to the left

Liono4ka [1.6K]

All the transformation performed on a pentagon does not affect on its size or its length
It is not stretched or shrunk
So,
The original and transformed images are congruent

8 0

3 years ago

The bounds for a discrete brownian walk is [-3, 4]. what is the probability that the escape happens at 4?

Akimi4 [234]

Using it's concept, there is a 0.125 = 12.5% probability that the escape happens at 4.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

Considering that the bounds for a discrete brownian walk is [-3, 4], the possible outcomes are:

-3, -2, -1, 0, 1, 2, 3, 4.

A escape at 4 is one outcome, hence the probability is given by:

p = 1/8 = 0.125.

0.125 = 12.5% probability that the escape happens at 4.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ4

3 0

2 years ago

6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul

drek231 [11]

Answer:

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

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

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that $n = 1012, \pi = \frac{374}{1012} = 0.3696$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

$E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}$

$E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}$

$\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

$n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

4 0

3 years ago

Louise, Tammy, Delores, and Sheryl score 78 points total in the tennis matches they play. They score a consecutive number of poi

irga5000 [103]

$\huge\boxed{18\ \text{points}}$

We'll represent Louise's, Tammy's, Delores's, and Sheryl's point values with $x$ , $x+1$ , $x+2$ , and $x+3$ respectively since each one is 1 point more than the last.