2x4x6x6x593838x52536733772737372737373838

Evaluate the mean of each data set mentally.<br> 61, 71, 81, 91, 101

goldfiish [28.3K]

Answer:

10

Step-by-step explanation:

they each go by ten

4 0

3 years ago

Find the area of the polygon with vertices (1, 2) (1, -5) (5, -5) (5,2)

Svetradugi [14.3K]

Answer:

28 square units

Step-by-step explanation:

These vertices are written in the form of their x and y coordinates. Let

A=(1,2) ; B =(1,-5) ; C =(5, -5) ; D=(5,2)

The x points are 1, 1, 5, 5

The y points are 2,-5, -5, 2

The x extremes range from 1 to 5. The length in-between is 5-1 = 4

The y extremes range from -5 to 2. The length in-between is 2-(-5) = 7

Area of the polygon enclosed by the coordinates

= 4 x 7 = 28 square units

5 0

3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

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

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

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".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

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

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$