Robert earns $8.75 per hour. If he works 15.3 hours each week, how much money will Robert earn in one year? (1 year=52 weeks

You have been given data about gas prices in 15 states. Comparing the two data sets below, what does the larger mean and median

Oksanka [162]

Comparing the two data sets shown below

<h3>What is standard deviation and variance?</h3>

Variance is the average squared deviations from the mean, while standard deviation is the square root of this number.

Both the mean and the median can be used to describe where the “center” of a dataset is located.

So, the mean of data in 2022 is more than mean of data in 2018.

It’s best to use the mean when the distribution of the data values is symmetrical and there are no clear outliers.

As the variance of 2022 more than in variance in 2018.

A high variance indicates that the data points are very spread out from the mean, and from one another.

As the standard deviation of 2022 more than in standard deviation in 2018.

The high standard deviation indicates data are more spread out.

The bigger the range, the more spread out the data.

Learn more about standard deviation, variance here:

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6 0

2 years ago

Find the distance between ( -2, 5) (-2, 14).

WITCHER [35]

Answer:

9

Step-by-step explanation:

x didn't change in value, y changed by 9

the distance between the two points is 9

6 0

3 years ago

What is 48% of 60? This is for Algebra 1! :)

coldgirl [10]

Answer:

48% of 60 is 28.8.

4 0

3 years ago

Read 2 more answers

An insurance company has 25,000 automobile policy holders. If the yearly claim of a policy holder is a random variable with mean

telo118 [61]

Answer:

$P(T>8300000)=1-P(T$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Data given

n = 25000 represent the automobile policy holders

$\mu= 320$ represent the population mean

$\sigma =540$ represent the population standard deviation

Let T the variable that represent the total of interest on this case. We can assume that the random variable for an individual policy holder is given by:

$X\sim N(\mu = 540, \sigma=540)$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Solution to the problem

First we need to find the distribution for the random variable T like this:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

And the total T is given by:

$T=\sum_{i=1}^n X_i =n \bar X$

We can find the expected value, variance and deviation for this random variable like this:

$E(T)= n E(\bar X) = n \mu = 25000*320=8000000$

$Var(T)= Var(n\bar X)= n^2 Var(\bar X) = n^2 \frac{\sigma^2}{n}=n \sigma^2 =25000*(540^2)=7290000000$

$Sd(T)=\sqrt{7290000000}=85381.497$

And we are interested on this probability:

$P(T>8300000)$

And we can use the Z score formula given by:

$Z=\frac{T-E(T)}{\sigma_T}$

$P(T>8300000)=1-P(T$

6 0

3 years ago

For what value of the domain {–3, –1, 0, 2, 5} does f(x) = g(x) When f(x) = 3x – 4 and g(x) = x2 – 2

zavuch27 [327]

F(x) = g(x)
3(2) - 4 = (2)² - 2
6 - 4 = 4 - 2
2 = 2
I just substituted all the values in the domain to see which one would work. That's not really the quickest way but that's the only way I know

7 0

3 years ago